Heat And Mass Transfer R.k.rajput

Heat and Mass Transfer

350

Graetz number, G = Pe

r? -

Grashoff number, Gr = P ~ gP At L3 p2 3. The following similarities for testing of 'heat transfer equipment' must be ensured between the model and the prototype: (i) Geometric similarity (ii) Kinematic similarity (iii) Dynamic similarity (iv) Similarity of fluid entry conditions (v) Siniilarity of boundary temperature field. 1. 2. 3. 4. 5. 6.

7. 8. 9.

10. 11. 12.

'4

THEORETICAL QUESTIONS What is dimensional analysis? What are the uses of dimensional analysis? Explain the term dimensional homogeneity. Describe the Rayleigh's method for dimensional analysis. Describe Buckingham's method or x-theorem to formulate a dimensionally homogeneous equation between the various physical quantities effecting a certain phenomenon. What are dimensionless numbers? Discuss the physical significance of the following dimensionless number Re, Nu, Pr, St, Gr. Show by dimensional analysis for forced convection, Nu = (I (Re, Pr) Show by dimensional analysis for free convection, Nu = (I (Pr, Gr) What are the advantages and limitations of 'Dimensional analysis' ? What do you mean by 'Characteristic length or Equivalent diameter' ? For testing of 'heat transfer equipment' which of the similarities must be ensured between the model and the prototype ?

Forced Convection 7.1. Laminarflow over aflat plate: Introduction to boundary layer - Boundary layer definitions and characteristics - Momentum equation for hydrodynamic boundary layer over a flat plate - Blasius solution for laminar boundary layer flows - Von Karman integral momentum equation (Approximate hydrodynamic boundary layer analysis) - Thermal boundary layer - Energy equation of thermal boundary layer over a flat plate - Integral energy equation (Approximate solution of energy equation). 7.2. Laminar tube flow: Development of boundary layer - Velocity distribution - Temperature distribution, 7.3. Turbulent flow over aflat plate; Turbulent boundary layer - Total drag due to laminar and turbulent layers - Reynolds analogy. 7.4 Turbulent tubeflow. 7.5 Empirical correlations- Typical Examples- Highlights -Theoretical Questions - Unsolved Examples.

A. LAMINAR FLOW 7.1.

LAMINAR FLOW OVER A FLAT PLATE

7.1.1. Introduction to Boundary Layer The concept of boundary layer was first introduced by L. Prandtl in 1904 and since then it has been applied to several fluid flow problems. When a real fluid (viscous fluid) flows past a stationary solid boundary, a layer of fluid which comes in contact with the boundary surface, adheres to it (on account of viscosity) and condition of no slip occurs (The no-slip condition implies that the velocity of fluid at a solid boundary must be same as that of boundary itself). Thus the layer of fluid which cannot slip away from the boundary surface undergoes retardation; this retarded layer further causes retardation for the adjacetit layers of the fluid, thereby developing a small region in the immediate vicinity of the boundary surface in which the velocity of the flowing fluid increases rapidly from zero at the boundary surface and approaches the velocity of main stream. The layer adjacent to the boundary is known as boundary layer. Boundary layer is formed whenever there is relative motion , the fluid exerts a shear stress on the between the boundary and the fluid. Since zo = boundary and boundary exerts an equal and opposite force on fluid known as the shear resistance. According to boundary layer theory the extensive fluid medium around bodies moving in fluids can be divided into following two regions: ( i ) A thin layer adjoining the boundary called the boundary layer where the viscous shear takes place. (ii) A region outside the boundary layer where the flow behaviour is quite like that of an ideal fluid and the potentialflow theory is applicable.

7.1.1.1 Boundary Layer Definitions and Characteristics Consider the boundary layer formed on a flat plate kept parallel to flow of fluid of velocity U (Fig. 7.1) (Though the growth of a boundary layer depends upon the body shape, flow over a flat plate aligned in the direction of flow is considered, since most of the flow surface can be approximated to a Jlat plate and for simplicity).

'

Heat and Mass Transfer

352

- The edge facing the direction of flow is called leading edge. - The rear edge is called the trailing edge. - Near the leading edge of a flat plate, the boundary layer is wholly laminar. For a laminar - boundary layer the velocity distribution is parabolic. - The thickness of the boundary layer (6) increases with distance from the leading edge x, as more and mare fluid is slowed down by the viscous boundary, becomes unstable and breaks into turbulent boundary layer over a transition region. Laminar

--Tr?nsi4

boundary layer

tlon Y

Turbulentboundary layer 1-

I

- surface roughness, - pressure gradient,

- plate curvature, and - temperature difference between fluid and bounary.

(x) Though the velocity distribution would be a parabolic curve in the laminar sub-layer zone,

but in view of the very small thickness we can reasonably assume that velocity distribution is linear and so the velocity gradient can be considered constant. Boundary layer thickness (6) : The velocity within the boundary layer increases from zero at the boundary surface to the velocity of the main stream asymptotically. Therefore, the thickness of the boundary layer is arbitrarily defined as that distance from the boundary in which the velocity reaches 99 per cent of the velocity of the free stream (u = 0.99U). It is denoted by the symbol 6 This definition. however. nives an approximate value of the boundary layer thickness and hence 6 is generally " termed as nominal thickness of the boundary layer. The boundary layer thicknpss for greater accuracy is defined in terms of certain mathematical expressions which are the metsure of the boundary layer on the flow. The commonly adopted definitions of the boundary layer thickness are: 1. Displacement thickness (6*) 2. Momentum thickness (0) 3. Energy thickness (6,). ,

1

Fig. 7.1. Boundary layer on a flat plate.

For a turbulent boundary layer, if the boundary is smooth, the roughness projections are covered by a very thin layer which remains laminar, called laminar sublayer. The velocity distribution in the turbulent boundary layer is given by Log law of Prandtl's one-seventh power law. The characteristics of a boundary layer may be summarised as follows: (i) 6 (thickness of boundary layer) increases as distance from leading edge x increases. (ii) 6 decreases as U increases. (iii) 6 increases as kinematic viscosity ( v )increases.

1

I

[f)

(iv)\,q,= p hence rodecreases as x increases. However, when boundary layer becomes turbulent, it shows a sudden increase and then decreases with increasing x. ( v ) When U decreases in the downward direction, boundary layer growth is reduced. (vi) When U decreases in the downward direction. flow near the boundary is further retarded. boundary layer growth is faster and is susceptible to separation. (vii) The various characteristics of the boundary layer on flat plate (e.g.. variation of 6, $ or force F) are governed by inertial and viscous forces: hence they are functions of either

ux or -.UL -

I

353

Forced Convection

m

*

Displacement thickness (a*) : The displacement thickness can be defined as follws: "It is the distance, measured perpendicular to the boundby, by whi is displaced on account of formation of boundary layer." Or

"It is an additional "wall thickness" that would have to be added to compensate for the reduction in jlow rate on accounibf boundary layer formation. " The displacement thickness is denoted by 6*. Let fluid of density p flow past a stationary plate with velocity U as shown in Fig. 7.2. Consider an elementary strip of thickness dy at a distance y from the plate. Assuming unit width, the mass flow per second through the elementary strip ...( i ) = oudv

. .

Mass flow per second through the elementary strip (unit width) if the plate were not there . ...(ii) = p U.dy

v

v

(viii) 1f UI < 5 x lo5 ... boundary layer is laminar (velocity distribution is parabolic).

v

___,

Boundary layer

ux

If -> 5 x 10' ...boundary layer is turbulent on that portion (velocity distribution follows

v

Log law or a power law).

(ir) Critical value of

v at which boundary layer changes from laminar to turbulent depends

on : - turbulence in ambient flow,

___, Stationary plate

2

Fig. 7.2. Displacement thickness.

Heat and Mass Transfer

Forced Convection

Reduction of mass flow rate through the elementary strip = p(U-u)dy [The difference (U - u ) is called velocity of defect] Total reduction of mass flow rate due to introduction of plate

...(iii)

0

(if the fluid is incompressible) Let the plate is displaced by a distance S* and ,velocity of flow for the distance 6* is equal to the maidfree stream velocity (i.e., U).Then, loss of the mass of the fluidlsec. flowing through the distance S* = pU6* Equating Eqns. (iii) and (iv), we get

7

J

S

= J P ( uU-) d~

Energy thickness is defined as the distance, measured perpendicular to the boundary of the solid body, by which the boundary should be displaced to compensate for the reduction in K,E. o f the .,flowing- fluid on account of boundary layer formation. It is denoted by 6, Refer Fig. 7.2. Mass of flow per second through the elementaIy strip = pudy K.E. of this fluid inside the boundary layer 1 1 =2 rn u2 - 2 (pudy)u2

...(iv)

K.E. of the same mass of fluid before 1entering the boundary Iayer = 7 (pudy) tJ2

Loss of K.E. through elementary strip 1

pU6*

= j p(U-

1

=2 (pudy) u2- 5(pudy)u2 = ? p u (U -u 2) dy

6

u)dy

6

0

.;

A

I1

(u' - u2)dy ,

Total loss of K.E. of fluid = - -; pu A 0

Momentum thickness (0) : Momentum thickness is defined as the distance through which the total loss of momentum per second be equal to if it were passing a stationary plate. It is denoted by 8. It may also be defined as the distance, measured perpendicular to the boundary of the solid body, by which the bounhry should be displaced to compensate for reduction in momentum of the flowing fluid on account of boundary layer formation. Refer Fig. 7.2. Mass of flow per second through the elementary strip = p u dy Momentudsec of this fluid inside the boundary layer = p u dy x u = pu2 dy Momentum/sec of the same mass of fluid before entering boundary layer = pu Udy Loss of momentudsec = puUdy - pu2dy = pu (U - u ) dy :. Total loss of momentudsec.

Let 8, = distance by which the plate is displaced to c o r n p i a t e for the reduction in K.E. Then loss of K.E. through 6,of fluid flowing with velocity 0 1 ...(ii)

= I (pU6.) u2

Equating Eqns. (i) and (ii), we have s

IcpusA 2 u2= J

0

+pu(u2-u2)dy

S

= J ~ " (-0u ) d ~

.,.(i)

...(i)

0

Let 8 = distance by which plate is displaced when the fluid is flowing with a constant velocity U . Then loss of momentudsec of fluid flowing through distance 8 with a velocity U ...(ii) =p8U2 Equating Eqns. ( i ) and (ii), we have

u

Example 7.1. The velocity distribution in the boundoly layer is given by:

3 = i, where u

is the velocity at a distance y from the plate and u = U at y = 6, 6 being boundary layer thickness. Find: r*thickness, (ii) The momentum (i) The displacement thickness,

(iv) The value of 8 '

(iii) The energy thickness, and u Y Solution. Velocity distribution : 11 = 6

-

(i) The displacement thickness, 6*:

...(Given)

II

...I Eqn. (7.111 The momentum thickness is useful.in kinetics.

Heat and Mass Transfer

(ii) The momentum thickness, €I : S

€I =

j: 0

(l - :Idy

...[Eqn. (7.2)]

(iii)The energy thickness, 6, :

(iv) The value of

F . €I

'

'-

- - - '-I 2 €I

Example 7.2. The velocity distribution in the boundary layer is given by

-26

- -I y2 2 lj2'

U

6 - 120 Example 7.3. I f velocity distribution in laminar boundary layer over a flat plate is assumed to be given by second order polynomial u = a + by + c y , determine its form using the necessary boundary conditions. Solution. Velocity distribution: u = a + by + cyL The following boundary conditions must be satisfied: (i) At Y = 0, u=0

=

6 being boundary layer thickness.

Calculate the following:

..

(i) The ratio of displacement thickness to boundary layer thickness

(ii) ~t

(ii) The ratio of momentum thickness to boundary layer thickness

u = -3y Solution. Velocity distribution : U 26 (i) 6*/6:

-

Iy2 -262

6

6

( ):

6 . ~ 1l--dy=[ o

[

l - ? ~ + l d 26 2@)dy

--- - -

or

- 3.0 6/6

..

...(Given)

(iii) At

.. i

u = a + by + cy2 O=a+O+O ;.a=O y = 6, u=U U = b6 + c6' y = 6, -du dy= o =

d (a + by + cy2)= b + 2cy = b + 2c6 = 0 -

Substituting the value of b (= - 2 ~ 6from(ii) ) in (i), we get U = (- 2c6)6 + cs2 = - 2cs2 + c6' - c6' I

*

I

...(i)

...(dl)

Heat and Mass Transfer .

358

Forced Convection u

Then,

.

au +velocity ax . dx = velocity of fluid flow at the right hand face C D (since the flow au changes in the X-direction at the rate of change given by ax and the

,

Hence form of the velocity distribution is change in velocity during distance dr will be

G- )I

. dr .

Similarly, let, v = fluid velocity at the bottom face BC, then, v

7.1.2. Momentum Equation for Hydrodynamic Boundary Layer Over a Flat Plate Consider a fluid flowing over a stationary flat plate and the development of hydrodynamic boundary layer as shown in Fig. 7.3. (a). In order to derive a differential equation for the boundary layer, let us consider an elemental, two-dimensional control volume (dx x dy x unit depth) within the boundary layer region [enlarged view shown in Fig. 7.3 (b)]. Following assumptions are made: 1. The flow is steady and the fluid is incompressible. 2. The viscosity of the fluid is constant. 3. The pressure variations in the direction perpendicular to the plate are negligible. 4. Viscous-shear forces in the Y-direction are negligible. - 5. Fluid is continuous both in space and time. Refer Figure 7.3. (b). Let u = velocity of fluid flow at left hand face AB.

av dy = fluid velocity at the top-face AD. +y OY

_+

--b

...(7.6)

The mass flow rate along Y-direction, m, = pv (dr x 1 ) = pvdx The change of momentum of the mass my along -Y-direction is given by

= pv

Velocity distribution Stationary plate

...(7.4)

The mass flow rate along X-direction, m, = pu (dy ,X 1) = pu dy The change of momentum of the mass mx along X-direction is given by dM, = m, x change in vebcity in X-direction

-.au JY

.

h.dy

...(7.7)

Total viscous force along the X-direction is given by F, = [(z + 6t) z] x area

-

au a = [Ips& + 3-(.-$).dy)

- p$]

( h x 1)

a2~

...(7.8)

= p ---ia~ .dr.dy

Assuming the gravitational forces are balanced by buoyancy forces for equilibrium of the element, we have Inertia forces = viscous forces Elemental control volume

:.

au

B

au

or

u-+ ax

or

u-

au = p azu

v-

a~

-ST

P ay

au a 2 ~ au ax + v -ay =

'v

...(7.9)

'

(substituting v = P 1

U

4

azu

au

p u -ax . h . d y + p v - . h . da~y = p 5 . d x . d a~ y

r ' d x

.t b c (b)

'5

Fig. 7.3. Equation of motion for boundary layer.

Equation (7.9) is known as equation of motion or momentum equation for hydrodynamic boundary layer.

7.1.3. Blasius Exact Solution for Laminar Boundary Layer Flows The velocity distribution in the boundary layer can be obtained by solving the equation of

Heat and Mass Transfer

Forced Convection

motion for hydrodynamic boundary layer [Eqn. (7.9)]. The following boundary conditions should be satisfied. (ii) At y = 0, v = 0 (iii) At y = =, u = U (i) At y = 0, u = 0 The Blasius technique for an exact solution of the hydrodynamic boundary layer lies in the conversion of the following differential equations into a single differential equation.

au au a 2 ~ aX"5"p

I

The hydrodynamic equation for boundary layer: u Continuity equation:

...(Eqn. 7.9)

-du+ - =dvo ax

-

Here f is abbreviated as f (q)

ay

Prandtl suggested that the solution of Eqn. (7.9) can be obtained by reducing the number of variables with the help of magnitude analysis of the boundary layer thickness and transforming the partial differential equation into ordinary differentials. The inertia forces represented by the left terms, in the Eqn. (7.9), must be balanced by the viscous forces represented by the right terms. As u 2 v, therefore, we may write as

a ' --. - ( a &

Now,

ay

Similarly, @u

au u ax

a y m

fl

)

a &( ? ?) I .

a

h h a ~

u 8f ...(7.18)

d"f

Also as u = U and - = - along A plate length L, therefore, we have L' Again,

1

I I

i 1

1

From experiments it has been o b s e r v e W velocity profiles at different locations along the plate are geometrically similar, i.e., they differ only by a stretching factor in the Y-direction. u This implies that the dimensionless velocity can be expressed at any location x as a function U

-

of the dimensionless distance from the wall

6'

Substituting the value of 6 from Eqn. (7.1 1) in Eqn. (7.12), we obtain,

r

1

where, q = y g d e n o f e s the stretching factor. In order to account for the fact that the vertical camponent of velocity occurs in the boundary layer equation of motion (7.9)' it is essential to define a stream function \y such that,

-

-.

Inserting the values of u,

au au azu and v from Eqns. (7.15). (7.16), (7.17), (7.18) and a x 9 ay' a2

(7.19) in Eqn. (7.9), we get or J7.14 (a)] The continuous stream function is the mathematical postulation such that its partial differential with respect to x gives the velocity in the Y-direction (generally taken as negative) and, its partial differential with respect to y gives the velocity in the X-direction:

-

V vx

'

8f

4 3

Heat and Mass Transfer

362

Forced Convection

363

Table 7.1. Laminar boundary layer solution for a flat plate

or 2f"'+ ff" = 0 ...(7.20) which is an ordinary (but non-linear) differential equation for f. The number of primes on f denotes the number of successive derivatives off (q) with respect toy. The physical and transformed boundary conditions are: Physical boundary conditions Transformed boundary conditions

(ii) At y = 0,

v=O

At ? =l0.

f=O

(iii) At y = w

u=U

Atq =w,

g= f J =I 4

2. The graphlcurve I (i.e., the velocity distribution parallel to the surface) enables us to calculate the parameters : (i) Boundary layer thickness, 6 and (ii) skin friction coefficient,

c,.

(i) Boundary layer thickness, 6: The boundary layer thickness 6 is taken to. be the distance from the plate surface to a point U

at which the velocity is within 1% of the asymptotic limit, i.e., U = 0.99; it occurs at q = 5.0

-

(Fig. 7.4). Therefore, the value of q at the edge of boundary layer O, = 6) is given by Fig. 7.4. Velocity distribution in boundary layer on flat plate by Howarth

7

The numerical solution of Eqn. (7.20) with the corresponding values of u and v are plotted in Fig. 7.4, and the results are listed in table 7.1. The following results are of particular interest:

-

1. The single curve I1 shows the variation of normal velocity v It is to be noted that at the

u'

outer edge of the boundary layer where q + -, this does not go to zero but approaches the value

where

Ux

Re, = v is the local Reynolds number based on distance x from the leading

edge of the plate. (ii) Skin friction coefficient; Cf: The skin friction coefficient (Cf)is defined as the ratio of shear stress zo at the plate to the dyMmic head

5I p

Cfi at any value of x is

~ caused ' by free stream velocity. Thus the local skin friction coefficient

Heat and Mass Transfer

Cfx =

1 To

u 2

=

+

Forced Convection

365

Let ABCD be a small element of a boundary layer (the edge DC represents the outer edge of the boundary layer). 1 p2

...(7.23)

From Fig. 7.4, the gradient at 11 = 0 is

layer

flat plate

Fig. 7.5. Momentum equation for boundary layer by Von Karman.

Mass rate of fluid entering through AD

0

Mass rate of fluid leading through BC

ipudy2 l P u d y J h

=o

The average value of the skin friction coefficient Zf can be determined by integrating the local skin friction coefficient Cfx from x = 0 to x = L (where L is the plate length) and then dividing the integrated result by the plate length

:.

where ReL is the Reynolds number based upon total length L of the plate.

The fluid is entering through DC with a uniform velocity U . Momentum rate of fluid entering the control volume in x-direction through AD

Mass rate of fluid entering the control \ -1ume through the surface DC = mass rate of fluid through BC - mass rate of fluid through AD

6

7.1.4. Von Karman Integral Momentum equation (Approximate Hydrodynamic Boundary

Layer Analysis) Since it is difficult to obtain the exact solution of hydrodynamic boundary layer [Eqn. (7.9)] even for as simple geometry as flat plate (moreover the proper similarity variable is not known or does not exist. for many practical shapes), therefore, a substitute procedure entailing adequate accuracy has been developed which is known as "Approximate lntegral Method" and this is based upon a boundary-layer momentum equation derived by Von Karman. -von - Karman -suggested a method based on the momentum equation by the use of which the1 growth of a boundry layer along a flat plate, the wall shear stress and the drag force could be determined (when the velocity distribution in the boundary layer is known). Starting from the beginning of the plate, the method can be used for both laminar and turbulent boundnry layers Figure 7.5, shows a fluid flowing over a thin plate (placed at zero incidence) with a free stream velocity equal to U . Consider a small length dr of the plate at a distance x from the leading edge as shown in Fig. 7.5 (a); the enlarged view of the small length of the plate is shown in Fig. 7.5 (b). Consider unit width of plate perpendicular to the direction of flow.

=

I pu2 dy

Momentum rate of fluid leaving the control volume in x-direction through BC r6

8

1

Momentum rate of fluid entering the control volume through DC in x-direction r6

1

(':

?

:.

L" Rate of change of momentum of control volume

velocity = I / )

Heat and Mass Transfer

367

Forced Convection

= momentum rate of fluid through BC - momentum rate of fluid through AD - momentum rate of fluid through DC.

Euation (7.28) is known as Von Karman momentum equation for boundary layerflow, and is used to find out the frictional drag on smooth flat plate for both laminor and turbulent boundary u expresses the wall shear stress,% as a function of layers. Evidently, this integral equation (7.28) the non-dimensional velocity distribution U' u is a point velocity at the boundary layer and U is the velocity at the outer edge of boundary layer. The following boundary conditions must be satisfied for any assumed velocity distribution du u = 0, dY = finite value y = 0, (i) At the surface of the plate: -a

u=U

(ii) At the outer edge of boundary layer: y = 6, d

( p is constant for incompressible fluid) 0

As per momentum principle the rate of change of momentum on the control volume ABCD must be equal to the total force on the control volume in the same direction. The only external force acting on the control volume is the shear force acting on the side AB in the direction B to A (Fig. 7.56). The value of this force (drqg force) is given by

1

The shear stress zo for a given velocity profile in laminar, transition or turbulent zone is obtained from Eqn. [7.28 (a)]or [7.28 (b)].Then drag force on a small distance dr of a plate is given by A FD = shear stress x area = zo x (B x dx) = zo x B x dr (where B = width of the plate) :. Total drag on the plate ofL length L one side. F D = j ~o ~ D = ] b ~ ~ ~ &

AFD=z0xdr Thus the total external force in the direction of rate of change of momentum =-,r0xdx Equating the Eqns. (7.26) and (7.27), we have

..(7.29)

1

- The ratio of the shear stress zo to the quantity 2 pu2 is known as the "Local coefficient

...(7.27)

of drag" (or coeflcient of skin friction) and is denoted by Cfr

1

- The ratio of the total drag force to the quantity U called ~ 'Average coefficient of 2 ~ A is drag' - and is denote by CD.

FD ...(7.31) Cf = 7 -2 PAu2 p = mass density of fluid, where A = area of surfacelplate, and U = free stream velocity. It has been observed through experiments that for laminar boundary layer, the velocity along the plate are distribution is parabolic and the velocity profiles at different locations u 1.

...(7.28)

geometrically similar. This means that the dimensionless velocity

...[7.28 (a)]

! 6

But

(1

e.,

u can

be expressed at any

locatjon x as a function of the dimensionless distance from the wall, 6'

- $)dy = momentum thickness ( 8 )

\

...[7.28 (b)]

,

,

.

,

The constants can be evaluated by using the following boundary conditions:

Heat and Mass Transfer

(i) At y = 0 (wall surface),

Forced Convection

a2~

20

u = U and - = 0

a9

(ii) At y = 6 (outer edge of the boundary layer),

+ 4A4 = 3CLI/ 6

K ' on simplification, we get

p@

au u = U and - = 0

=o-

ay By applyiug boundary conditions the constants are evaluated which gives the velocity profile

2

0.646

7iq'

Therefore, the local skin friction coefficient.

as:

.,

,,

In order to determine the boundary layer thickness and average skin-friction coefficient for laminar flow over a flat plate, let us now use this above velocity profile in the Von Karman integral equation. Now putting the value of

U in Eqn. (7.28), we get

Average value of skin friction coefficient, 1 J 0.646 dx E f - L 0 cfx*=L o d $ z i f i

- jL

u

where

or

To

=

39 d6 pu2 du 280

($Izo 4% 3$)llv=o d

or

70

3

Y

p ( 6 )-

3CLLI

= 26

...(7.35)

From equations (7.34) and (7.35), we have

39

6

3pU % pu2dx - 26 = @.%fx 13 pu Since, 6 is a function of x only, integration yields

or

B -B+c 2

13pU By using the boundary condition 6 = 0 at x = 0, we obtain the integration constant C = 0. @ -B or 6 2 = 140x 2 x~ 2 13pU 13 PU This can be expressed in the non-dimensional form as

..

7

X

where Re, =

x U U r-

Mass pow through boundary layer The mass flow rate per unit width through the boundary layer is calculated assuming parabolic velocity distribution in the boundary layer as 3 1

3

=

ReL = k@! is Reynolds number based on total length L of the flate. u

J

...(7.34)

Newton's law of viscosity, at solid surface, gives =

--

...(7.36)

is the Reynolds number based on distance x from the leading edge of the plate.

Further, in order to estimate the value of zo, substituting the value of 6 Eqn. (7.36) in Eqn. (7.39, we get

-- 370

Heat and Mass Transfer

Example 7.4. The velocity profile for laminar boundary is in the form given below:

Find the thickness of boundary layer at the endof the plate 1.5 m long and 1 m wide when placed in water flowing with a velocity of 0.12 d s . Calculate the value of coefJicienr of drag also. Take p for water = 0.001 Ws/rn2 Solution. Velocity distribution

U u - 2 6 ) -

The length of the plate, L = 1.5 m The width of the plate, B = 1 m Free stream velocity, U = 0.12 m/s p for water = 0.001 Ns/m2

br

...( Given)

Thickness of the boundary layer, 6: Reynolds number at the end of the plate (i.e., at a distance of 1.5 m from the leading edge) is given by

Since ReL < 5 x lo5, therefore, this is case of laminar boundary layer. Thickness of boundary layer at a distance of 1.5 m is given by

Coefficient of drag

Cf :

,

Forced Convection

(i) The maximum distance from the leading edge upto which laminar boundary layer exists, X: 12 x x Re, = Ux or 2 x lo5 = v 0.15 x lo4

-

(ii) The maximum thickness of boundary layer, 6 : For the given velocity profile, the maximum thickness of boundary layer is given by 5.48 x =

Example 7.6. A plate of length 750 mm and width 250 mm has been placed longitudinally in a stream of crude oil which flows with a velocity of 5 m/s. I f the oil has a specific gravity of 0.8 and kinematic viscosity of 1 stoke, calculate: ( i ) Boundary layer thickness at the middle of plate, (ii) Shear stress at the middle of plate, and (iii) Friction drag on one side of the plate. Solution. Length of the plate, L = 750 mm = 0.75 m Width of the plate, B = 250 mm = 0.25 m U = 5 mls Velocity of oil, Specific gravity of oil = 0.8 Kinematic viscosity of oil, v = 1 stoke = 1 x lod4 m2/s ( i ) Boundary layer thickness at middle of the plate 6: Reynold number, (

Example 7.5. Air is flowing over a smooth flat plate with a velocity of 12 m/s. The velocity profile is in the form:

The length of the plate is 1.1 m and width 0.9mJf laminar boundary layer exists upto a value of Re = 2 x 16 and kinematic viscosity of air is 0.15 stokes, find: (i) The maximum distance from the leading edge upto which laminar boundary layer exists, and (ii) The maximum thickness of boundary layer. Solution. Velocity distribution:

U

Velocity of air, U = 12 m/s Length of plate, L = l.lm Width of plate, B = 0.9m Reynolds number upto which laminar boundary exists, Re = 2 x lo5 Kinematic viscosity of air, v = 0.15 stokes = 0.15 x lo4 m2/s

'.' At the middle of plate, x = 0.792 = 0.375 m)

Since Re, < 5 x lo5, therefore, boundary layer is of laminar character and Blasius solution gives

(ii) Shear stress at the middle of plate, T,: According to Blasius, the local coefficient of drag (C,J is given by

By definition,

3x2

Heat and Mass Transfer

(iii) Friction drag on one side of the plate, FD:

Solution. Length of the plate, L = 5 m B = 2.5 m Width of the plate, . U = 4 mls Velocity of air, p = 1.208 kg/m3 Density of air, Kinematic viscosity of air, v = 1.47 x 1w5 m2/s (i) Length of plate over which the boundary layer is laminar:

.

As the boundary layer is laminar even at the trailing edge, therefore, the average drag (friction) coefficient,

:.

Firction drag,

1 U2 x area of plate on one side FD = -' x 2P I

= 6.858 x

x

-21 x (0.8 x

Reynolds number

Re,

=uL V

Ux is laminar and on the rear, it is turbulent. Hence on the front portion, boundary layer

1000) x 52 x 0.75 x 0.25

Re, =

= 12.86 N (Note : If velocity profile is not given in the problem, but boundary layer is laminar then Blasius's solution is used.). Example 7.7. Atnzospheric air at 20°C is flowing parallel to a flat plate at a velocity of 2.8 4 s . Assuming cubic velocity profile and using exact Blasius solution, estimate the boundary layer thickness and the local coeficient of drag (or skin friction) at x = 1.2 m from the leading edge of the plate. Also find the deviation of the approximate solution from the exact solution. Take the kinenmtic viscosity of air at 20°C = 15.4 x lo4 m2/s Solution. Velocity of air, U = 2.8 m/s Distance from the leading edge of the plate, x = 1.2 m Reynolds number

Local coefficient of drag,

=5x v

1d

Hence the boundary layer is laminar on 1.837 m length of the plate. Thickness of the boundary layer (laminar), 6

(ii) Shear stress at the location where boundary layer ceases to be laminar, g : = O.CM0939 Local coefficient of drag, Cf, = 1 1 b = C f x X g p u ~ = o . w ~ ~ ~ ~~4~ x ~ x I . ~ o ~

K i F

Blasius solution: Boundary layer thickness,

1 . 44 7xx 150 - ~= 1 . 3 6 1 ~ 1 @

6=

-&

=

2 18 x 16

= 0.01285 m = 12.85 mm

0.664

= 0.00907 ~ / m ~ (iii) Total drag force on both sides of plate, FD 1 :

cfx= K = xlOI = 0.001422

Approximate solution (with assumption of cubic velocity profile):

-

F ~ = Z fX C2 X - ~ A U ~

1.328 where. Cf = average coefficient of drag (or skin friction) = 5 x 105 = 1.878 x l(r3 -

,

and A = area of the plate = 1.837 x 2.5 = 4.59 m2

..

The approximate solution deviates from the exact solution by 12.85 - 11.92 Derivation for 6: x 100 = 7.34% 12.85

Fo = 2 x 1.878 x

x

51 x 1.208 x 4.59 x 42 = 0.167 N

Example 7.9. Airflows over a plate 0.5 m long and 0.6 wide with a velocity of 4 d s . The u = sin -

Drviation for

u

Example 7.8. Air is flowing over a flat plate 5 m long and 2.5 m wide with a v&ocity of 4 m/s at 15°C. I f p = 1.208 kg/m3 alzd v = 1.47 x 1 ~ 'm2/s, wlculate: ( i ) Length of plate over which the boundary layer is laminar, and thickness of thk'boundary layer (laminar). (ii) Shear stress at the location where boundary layer ceases to be laminar, andt,,.. : (iii) Total drag force on the both sides on that portion of plate where boundary layer is laminar. i

- If p = 1. Z l k g / d and v = 0.15 x 1

:( i)

m2/s, calculate:

(i) Boundary layer thickness at the end of the plate, (ii) Shear stress at 250 mm from the leading edge, and (iii) Drag force on one side of theptate. L = 0.5 m Solution. Length of plate, B = 0.6 m Width of plate, U = .4 mls Velocity of air,

374

Heat and

Mass Transfer

375

Forced Convection

Density of air, Kinematic viscosity of air, Velocity profile,

p = 1.24 kg/m3

v = 0.15 x

, (f3

m2/s

u = sin (i) Boundary layer thickness at the end of the plate, 6: Reynolds number, Rex = ux = 4 x 0e5 = 1 . 3 3 10s ~ V 0.15 x lo4 Since Rex c 5 x loS, therefore, the boundary layer is laminar over the entire length of the

- '

distribution will be as shown in the Fig. 7.6. The zone or this layer wherein the temperature fiefd exists is called the thermal boundary layer. Due to the exchange of heat between the plate and the fluid, temperature gradient occurs/results. The thermal boundary layer thickness (6,,,) is arbitrarily defined as the distance y from the plate surface at which

- " 7 f

plate. We know,

It !I

f

I

c-4

Temperature profile

= 0.00657 m = 6.57 mm (ii) Shear stress at 250 mm from the leading edge, 5: ...[Eqn. (7.30)J

Cf.= -=-0.654

But,

0.654

0.654

= 0.002533

0.15 x

(iii) Drag force on one side of the plate, F, :

where

CD =

1.31

=

1 FD = 0.003592 x 7 x 1.24 x 0.3 x 4' = 0.01069 N

7.1.5. Thermal Boundary Layer Whenever a flow of fluid takes place past a heated or cold surface, a temperature field is set up in the field next to the surface. If the surface of the plate is hotter than fluid, the temperature Limit of thermal boundary layer

Y

Free stream

t r Plate surface (I,) t --*

Laminar

I+

Turbulent

Fig. 7.6. Thermal boundary layer formed during flow of cool fluid over a warm plate.

warm fluid over a cool plate.

Figure 7.7, shows the shape of the thermal boundary layer when the free stream temperature t, is above the plate surface temperature t,. The thermal boundary layer concept is analogous to hydrodynamic boundary layer; the parameters which affect their growth are however different. Whereas the velocity profile of the hydrodynamic boundary layer depends mainly on the fluid viscosity. the temperature profile of the thermal boundary layer depends upon the viscosity, velocity of flow, specific heat and thermal conducitivity of the fluid. The relative magnitude of 6 and 6thare affected by the thermo-physical properties of the fluid; the governing parameter, however, is h e non-dimensional Prandtel number, pr =

- 0.003592

1.33 x loS and A = area of the plate = L x B = 0.5 x 0.6 = 0.3 m2

..

Fig. 7.7. Thermal boundary layer formed during flow of

7.

( i ) 6th= 6 (iii) 6 , > 6

........ when Pr = 1; ........ when Pr < 1.

(ii)

4, < 6

........P r >

1

7.1.6. Energy Equation of Thermal Boundary Layer Over a Flat Plate Figure 7.8 (a) shows a hot fluid flowing over a cool flat plate, and development of the thermal boundary layer. In order to derive an energy equation, consider control volume (dr x dy x unit depth) in the boundary layer so that end effects are neglected. The enlarged view of this control volume is shown in Fig. 7.8 (b) in which the quantities of energy entering and leaving have been indicated. Involving principle of conservation of energy for the steady state condition, we have: Heat energy convected (E,,,,) through the control volume in X and Y directions + heat energy conducted (Econd,)through the control volume in X and Y directions + heat generated due to fluid friction (viscous heat generation) in the conuol volume = 0. But as the rate of temperature change in the X-directidn is small and can be neglected the conservation of energy becomes: Heat energy convected in X and Y directions + heat energy conducted in Y-direction + viscous S t heat generation = 0. ...(7.41) or d(E), + d(EcOnv.),+ d(ECod), + viscous heat generation = 0

(i)

(ii)

(iii)

(iv)

Heat and Mass Transfer Forced Convection

boundary layer Plate surface (r,>

Viscous heat generation

(iii) The heat conduction in tbe y-direction: - (~cotuf.h+dy d(Ecod), = (E~md.)~ at

(Ec~nv.!v +dy

...(7.44)

. t

the element).

.x distance travelled

= [Shear stress (r) x area upon which it acts)] =

[,$

(dr x l)] x

($dy)

Fig. 7.8. Energies entering and leaving the control volume.

Substituting the values in Eqn. (7.411, we get

(i) The Energy convected in X-direction: (Econ,.)x = mass x specific heat x temperature = [pu(dy x l)] c,t = (pu dy) c,,t

...(7.46) Form the continuity equation for two-dimensional flow, we have

-

au av -+-=0:

-n ' ays + Pp. (,?ay= ---

... neglecting the product of small quantities . n a

%

,,

u- at + v-at =

ax

a~

thus the Eqn. (7.46) reduces to

...(7.47)

a2t

PC,

Equation (7.47) is the requid di#erential energy qUUt0n f0rflDw PaR a fLY p*1te. If heat genemhon is neglected [when the value of U is relatively 10" and difference of temperature swam and the plate is s d l (of the order of 40°c)1, fhe energ Cqudlon reduces bewen the at k a2t 2t ...(7.48) u -at +,-=--=a ax ay PC, a 3

&=

(ii) The energy convected in Y-direction: The net energy cor+vectedin Y-direction,

+

"

. L

fi may noted, the energy equation is similar to the momentum equation. Further the dimensions of kinematic viscosity v and thermal diffusivity a are the same. The auation (7.48) has been derived with the following assumptiom: 1. Steady incompressible flow. ts t, are constant. 2. m e properties of the fluids evaluated at the film temperature $ 2

-

-- -

278

Heat and Mass Transfer

Forced Convection

3. The body forces, viscous heating and conduction in the flow direction are negligible. Pohlhausen solution for the 'Energy equation': at = a-+ v-

By using the following variables the energy equation u at ax into an ordinary differential equation as follows: (Stretching factor) = Y

45,

yf (StreaIn function) =

JY

"1

can be recast

ay2 .

.'

Pr (Randtl number) =

f (ll), and

Also, the values of the &locity components u and v already calculated earlier are:

u = u&

.

4

...[Eqn. (7.15)l

Thus the partial differential equation (7.48) has been converted into ordinary differential equation. The boundary conditions to the satisfied are: At t = t,, y-0 y=m At t = t, ...(7.55) At q = 0, e(q) = values in terms of new variable ~t q = em) = 1 The solution obtained by Pohlhausen for energy equation is given by:

)

...[Eqn. (7.19)J 4

Further, from temperature parameter 0 (non-dimensional) defined above. we have t = t,#+ (t, - t,) e -ax at-- (t- - tJ ae = (tm - t#) ax atl ax at

-

de The factor fr)

-

and

ax -at-- 0,. - ts) ae = (t- - t,) " ,h ay JY atl ay

or

at

represents the dimensionless dope of the temperature profile at the

\Wh=o surface where q = 0; its value can be obtained by applying the boundary condition at r\ = w, 801) = 1. Thus, J7.50)

.

,,

Evidently the dimensionless slope is a function of Prandtl number and the calculations made by Prandtl gave the following result : = 0.332 (PI)'" For 0.6 < Pr 15,

[

=o

Figure 7.9 shows the values of 0 (dimensionless temperature distribution) plotted for various values of Pr (Prandtl number). - The curve for Pr = 0.7 is typical for air and several other gases. - The curve for Pr = 1 is the same as that of curve I in Fig. 7.4. - These curves also enable us to determine the thickness of thermal boundary layer 6,hand, local average heat transfer coefficients h.

Also,

Thermal

-$28 -t - (1..

or

tJ&e - 4) vxdq2

bsening the above values in the energy equation, we get: at

at

u-+v-=aa~ ax

a2t

ay2

= a (t, - t,)

U -Z d28 -

vx 4 After simplification and arrangement of the above equation, we obtaL

...(7.53)

Heat and Mass Trapfpr

Fdrced Convection

Thickness of thermal boundary layer, 6, : Case I: When Pr = 1.

-

-

-

Substituting for Since y = 6, at the outer edge of thermal boundary layer, therefore,

= hx i t s - tJ = 0.332

A

7

in Eqn. (7.63). we obtain

:\'I'.

k 1i2 pr)in ; (t, - tm)(Rex)

k

or or This equation shows that for Pr = 1, the thickness of thermal boundary layer, 6, is equal to hydrodynamic boundary layer, 6. Case 11. When Pr < 1. ,-

-7

...(7.64

hx = 0.332 ; ( ~ e ~ ) l(pr)'" "

-

...(7.65)

112 pr)ln NU, = h2-X = 0.332 (Re,) ( k

... (In non-dimensional form)

hx = local convective heat transfer coefficient; Nu. = local value of Nusselt number (... at a distance x from the leading edge of the plate]. The average heat transfer coefficient is given by

[where

A

This equation shows that for Pr < 1, 6, > 6 Case 111. When Pr > 1. \

/

If we compare-the Eqns. (7.64) and (7.66), we find that h = 2hx This equation shows that for Pr > 1, ath< 6 Pohlhausen has suggested that the following relation is general may be \the thermal and hydrodynamic boundary layers:

between

6

6, = -

...(7.62) (PrV3 . , The local and average heat transfer coefficients: At the surface of the plate, since there is no fluid motion and the heat transfer can occur only through conduction, the heat flux may be written as

valid for Pr > 0e5. ~ 1 the 1 results in Eqns. (7.64), (7.65) and (7.68) 7.1.7. lote& Energy aquation (Approximate ~olutionof e n e r a equation) volume shown in Fig. 7.10. Assume that p, cp and (thermo-plastic Consider a Hydrodynamic 7 boundary layer Qh Thermal boundary I

u,t,

I

B

f I--From the relation 7.63, we may develop

I' I

#"#--

4

(i.e., surface temperature gradient) as

-

4

Control .mlnrne

Qck

~-xoL.__)l

I+----xe

n

-

4

I I

I I

Thin flat plate

Heat and Mass Transfer. ,

382

propertids) of fluid remain constant within the operating range of the temperature, and the heating of the plate commences at a distance xo from the leading edge of the plate (so that the boundary layer initiates at x = xo and develops and grows beyond that). For unit width of the plate we have: H

Mass of fluid entering through face AB = jo pu dy

4 H

Mass of fluid leaving through face CD =

:.

pu dy

...(7.69)

+

a "

[Io pu d y ] d r

...(7.70)

Mass of fluid entering the control volume through face BC =

[jOH Pu dy + $ {joH pu dy) dx] - joH pu dy = $ [joH pu dy] dx

Forced Convection over a flat plate (that has an unheated starting length xO),let US use cubic velocity and temperature distributions in the integral boundary layer energy equation as follows: he cubic velocity profile within the boundary layer is of the fom; ...[Eqn. (7.33)l

U The conditions which are satisfied by the temperature distribution within the boundary layer are: (i) At y = 0, t = ts

-47.71;

Heat influx through the face AB, Q, = mass x specific heat x temperature

Heat efflux through the face CD,

,

r

W e temperature distribution takes the following fom: -=--

Heat (energy) influx through the face BC (which is outside thermal boundary layer and there the temperature is constant at t,),

...(7.79)

B,, putting the proper values of velocity distribution and temperature distribution into the

Heat conducted into the control volume through face AD,

The energy balance for the element is given by

After simplification and rearrangement, we have

Equation (7.76) is the integral equation for the boundary layer for constant properties and constant free stream temperature t,. If the viscous work done within the element is considered, then Eqn. (7.76) becomes

integrind would be zero). After putting

6th a =

and earlying out the integration, Eqn. (7.80) gets reduced

...(7.81) \-,,-

d r dy = viscous work done within the element ...Eqp. (7.8)] ay2 Usually the viscous dissipation term is very small and is neglected (and may be considered only when velocity of flow field becomes very large).

[when

PC,

0

Exp-ion for the convective heat transfer coefficient for laminar flow over a flat blab: In order to derive an expression for convective heat transfer coefficient for laminar flow

c 6, r < 11, we have d

Neglecting the term involving r4 (because 3 = 20 U (t,

-

F u ~ h e r from , Eqn. (7.77). we kave

- t,) 5 (62)

386

Heat and Mass Transfer

.

Forced Convection

,

-

( v ) Thickness of the boundary layer,

(vi) Local convective heat transfer coeflcient, (vii) Average convective heat transfer coeflcient, (viiil Rate of heat transfer by convection, (k) Total drag force on the plate, and (x) Total mass flow rate through the boundary. Solution. Given: U = 3mls. x = 280 mm = 0.28 m, p = 1.1374 kg/m3,k = 0.02732 W/m°C, cP = 1.005 W g K , v = 16.768 x 10d m2/s.

(vii)Average convective heat transfer eoeff~cient,h:

/

(viii) Rate of heat -transfer by convection, Qco,,:

Let us first amxtain the type of the flow, whether laminar or turbulent.

Re, =

ux -

(ix) Total drag force on the plate, FD: FD = q, x area of plate on one side upto 0.28 m = 0.01519 x 0.28 x 0.28 = 0.00119 N (Ans.) (x) Total mass flow 5 rate through the boundary, m :

V 16.768 x lo4 = 5.0 x 104 Since Re, < 5 ' lo5, ~ hence flow in laminar.

(0 Boundary layer thickness at x = 0.28 m, 6:

m = - pU (6, - 61) 8 or

6=

(2) Local friction coefficient,

or

C'.:

5 x 0.28 7 = 0.00626 m or 6.26 mm 5 x lo4

4 = - 50.664 = 0.002969 x lo4

-

Pr = 0.7; k = 0.0364 W / m K.

= 0.005939

Let us first ascertain the type of flow, whether laminar or tubulent.

cf=x=zF (iv) Shearing stress due to friction, To : = 0.002969 x (v) Thickness of thermal boundary layer, &, :

2

32 = 0.01519 N / ~ ~

6 (Pr) 0.00626 -- 0.00705 m or (0.7)'"

...[Eqn. (7.62)

Irn

X

--

s t h = - -8.669 =

( P ~ ) '-/ ~(0.7)"~

7.05 mm

(Ans.)

9.763 mm

(Ans.)

Local heat transfer coefficient, h, :

(vi) Local convective heat transfer coefficient, 4 :

k (Re,)'" (pr)'" hx = a332 -

Since Re, < 5 x Id, hence flow is laminar. Boundary layer thickness at x = 0.5 m, 6 : 6 = 5x = 5 x 0.5 = 8.669 x lo-) m or 8.669 mm fie" 4iZE Thickness of thermal boundary layer, at x = 0.5 m, St.:

6, =

=

x 1.1374 x 3(0.00626 - 0) = 0.01335 kgls (Ans.)

Example 7.11. Air at atmospheric pressure and 200°C flows over a plate with a velocity of 5 d s . The plate is 15 mm wide and is maintained at a temperature of 120°C. Calculate the thicknesses of hydrodynamic and thermal boundaiy layers and the local heat transfer coefficient at a distance of 0.5 m from the leading edge. Assume that /low is on one side of the plate. p = 0.815 kg/rn3; p = 24.5 x 1 r 6iVs/m2, Pr = 0.7. k = 0.0364 W/m K . (AMIE Summer, 1997)

(iii) Average friction coefficient, C'

or

=8

...[Eqn. (7.64)]

= 0.332 x

-.- x (83l63)ln x (0.7)'"

...[Eqn. (7.64)l = 6.189 w/m2 K

(Ans.)

Example 7.12. Air at atmospheric pressure and 40°C flows with a velociry of (I=5 m / s over a 2 m long f k plde ~ whose surface is kept at a uniform temperature of 120°C. Determine

Heat and Mass'Transfer

the average heat transfer coeflcient over the 2 m length of the plate. Also find out the rate of heat transfer between the plate and the air per I m width of the plate. [Air at 1 arm. and 80°C,

v = 2.107 x 10- m2/s,k = 0.03025 W/m K, Pr = 0.69651 (AMIE Winter, 1998) Solution. Given :t, = 40°C; U = 5 m/s; L = 2 m, t, = 120°C; B = 1 m

...assumed Now,

The properties of air at mean bulk temperature of

v = 2.107 x 10- m2/s; k = 0.03025 W/m K ; Pr = 0.6965. Average heat transfer coefficient, h :

Assuming Re,,= 5 x lo5, the flow is laminar.

5 = z x 1.16 x 2 x 0.00857 = 0.01242 k g h

Using exact solution, the average Nusselt number is given by

or

= 0.664 (~e,)'" ( ~ r )

hL k = 0.664 (4.746 x 1@)ln

8

...[Eqn. (7.68)] X,

(0.6965)'" = 405.48

-.

Note, ~f h e mass added in the boundary is to be calculated when h e fluid moves from 10 + along the main flow direction then it is given by S Am =: PU (8, - 8,)

-where 6 , and l&

Rate of heat transfer, Q :

(Ans.)

are the boundary layer thicknesses at X l and q .

(ii) Heat transferred-per hour, Q :

Q = ZA, ( t ,- t,) = 6.133 x (2 x 1 ) (120 - 40) = 981.28 W

(Ans.) Example 7.13. Air at 27°C and I barflows over a plate at a speed of 2 m/s. (i) Calculate the boundary layer thickness at 400 mm from the leading edge of the plate. Find the massflow rate per unit width of the plate. For air p = 19.8 x 10- kg/ms at 27°C. (ii)If the plate is maintained at 60°C, calculate the heat transferred per hour. The properties of air at mean temperature of (27 + 6 0 ) d = 43.5' C are given below :

v = 17.36 x 10- m2/s; k = 0.02749 W/m°C

- 15'76 3600 = 416.74 kl/h

c,=lOOBJ/kgK; R=287Nm/kgmK; Pr=0.7.

Solution. Given :t = 27°C; p = 1 bar, U = 2 m/s; x = 400 rnm = 0.4 m (i) Boundary layer thickness, 6 :

(M.U.)

-

1000

(Ans.)

Example 7.14. Air at 1 bar and at a temperature of 30°C (Y. = 0.06717 kghrn)flows at a speed of I.2ds over a fiat plate. Determine the boundary layer thickness at distance of Somm and 500 mm from the leading edge of the plate. Also, calculate the mass entrainment bemeen these m o sections. Assume the parabolic velocity distribution as: - = -

Solution. Given : t.. = 30°c, )I = 0.06717 kglhm, U = 1.2 mls Boundary layer thicknesses: = 1.15 kg/m3 E-= 1 x lo5 The density of air, = RT 287 x (30 + 273)

*

-

Boundary layer thickness, 6 = 4.64 x

4.64 x 0.4 = 0.00857 m or 8.57 mm 46869 The mass flow rate per metre width is given by,

6=

-===J - ,

(Ans.)

II

Heat and Mass Transfer Boundary layer thickness,

-

6, = 4.64~ 6, =

Re, =

..

Boundary layer thickness,

...[Eqn. (7.36)]

4.64 x 0.25

.I18490 = 0.00853 m or 8.53

1.15 x 1.2 x 0.5 0.067 17

x 3600

mm (Ans.)

= 36981

62 = 4.64 x 0.5 = 0.01206 m = 12.06 mm (Ans.)

Gzi-

Mass entrainment: The mass flow rate at any position in the boundary layer is given by

Forced Convection

-h = -'O0.2 2'

.. ..

x 0.664 x (2.11 x 1 0 ~ ) " ~(0.7)"~= 10.7

w/m2~c(Ans.)

Q = 3; x A , ( t , "-t,) = 10.7 x (0.2 x 0.5) x (100- 20) = 85.6W (Ans.) Example 7.16. In a certain glass making process, a square plate of glass 1 m2 area and 3 mm thick heated uniformly to 90°C is cooled by air at 2U'Cflowing over both sides parallel to the plate at 2 m/s. Calculate the initial rate of cooling the plate. Neglect temperature gradient in the glass plate and consider only forced connection. Take for glass : p = 2500 kg/m3 and cp= 0.67 W / k g K Take the following- .properties of air : p = 1.076 kg/mJ; cp = 1008 JAcg K, k = 0,0286 W/mOC and p = 19.8 x l(T6 N-$/mL. (N.U., 1997) Solution. Given : A = 1 m2; t, = 90°C; t, = 2Q°C, U = 2 d s .

7 Plate at 90°C

-

Fig. 7.11

:.

The mass entrainment between the two sections be., at x = 250 mm and x = 500 mm.

= 3.045 x kg/s = 10.96 kg/h (Ans.) Example 7.15. Air at 20°C is flowing over a flat plate which is 200 mm wide and 500 mm long. The plate is maintained at 100°C. Find the heat loss per hour from the plate if the air is flowing parallel to 500 mm side with 2m4 velocity. What will be the efect on heat transfer if the flow is parallel to-200 mm side. The properties of dir at (100 + 20)D = 60°C are: v = 18.97 x 1@ m2/s, k = 0.025 W/m°C and Pr = 0.7. (M.U.) . --, Solution. U = 2 m/s, v = 18.97 x lo4 m2/s, k = 0.025 W/m°C and Pr = 0.7. Heat loss per hour from the plate, Q: Case I. When the flow is parallel to 500 mm side:

-

N u = -hL -

where,

- 0.664 (Re,)'"

k uL Re, = V

(pr)'"

...[Eqn. (7.68)]

OS5 = 1 2 7 x lo4 18.97 x 10"

Initial rate o f cooling : The average heat transfer - co-efficient for the air flow parallel to the plate is given by hL .[Eqn. 7.68)] Nu = K = 0.664 ( ~ e , ) " ~(pr)'l3

-

(valid for Pr > 0.5)

Substituting the -values in the abovgeqn, we get

hL= k 0.664 x ( 1 .087x 10")"~x (0.698)]/~ = 194.19

The heat flow (Q) from both sides of the platet is given by : ~ = 2 h A ( t , - t J = 2 ~ 5 . 5 5x~( 91 0 - 2 0 ) = 7 7 7 ~ The heat lost by the plate instantaneously is given by : Q=mc,,At=777 where

.. Q = ~ s ( t s - t , ) = 6 . 7 6 7 ~ ( 0 . 5 ~ 0 . 2 ) ( l W - 2 0 ) = 5 4 . 1 4 W(Am.) Case 11. When the flow is parallel to 200 mm side: Re, =

2 x 0.2 = 2.11 x lo4 18.97 x lo6

-

E

m = (area x thicknes$ x p

/

Example 7.17. A flat plate, l m wide and 1.5 m Long is to be maintained at 90°C in air with a free stream temperature of 10°C. Determine the velocity with which air must flow over

Heat and Mass Transfer

392

-

(ii) Using approximate solution: -

f

flat plate along 1.5 m side so that the rate of energy dissipationfrom the plate is 3.75 kW. Take the following properties of air at 50°C:

=

-

p = 1.09 kg/m3, k = 0.028W/m°C, cp = 1.007 kJ/kg°C, p = 2.03 x I @ kg/m-s Pr = 0.7.

or

k

(M.U.) Solution, Given : L = 1.5 m, B = lm, t, = 90°C, t, = 10°C, Q = 3.75 kW Properties of air at (90 + 10)/2 = 50°C : p = 1.09 kg/m3, k = 0.028 W/m°C, cp = 1.007 kT/kg°C, p = 2.03 x lo-' kg/m-s Free stream velocity, U: s ' The heat flow from the plate to air is given by . Q = j; A, (t, - t,) k where h = - x 0.664 (~e,)'" (pr)ln ...[Eqn. (7.68)] L

&& = 0.646 (Re, )'" (pr)lD k

= 0.646 (2.087 x 1 0 ' ) ' ~ (0.696)'" = 261.53

-

261.53 k h=-= L

or

261.53

X 0.02894 = 3.44 Wlm2 2.2

OC

<

r-

-

(i) The average skin friction, CI : Reynolds number,

or U = 100 m/s (Ans.) Example 7.18. Air at 20°C and at atmospheric pressure jlows over aflat plate at a velocity of 1.8 4 s . If the length of the plate is 2.2 m and is maintained at 100°C, calculate the heat transfer rate per unit width using (i) exact and (ii) approximate methods. The properties of air at mean bulk temperature of (100 + 20)D = 60°C are: p = 1.06 kg/m3, cp = 1.OO5 kJ& OC, k = 0.02894 W/m°C, Pr = 0.696 v = 18.97 x 106 m2/s Solution. Given :t, = 20°C, t, = 100°C, U = 1.8 mls, L = 2.2 m, B = 1 m Heat transfer rab per unit width: Reynolds number, Since Reynolds number is less than 5 x 10' hence flow is laminar. (i) Using exact solution: The average Nusselt number is given by Nu = 0.664 (Re, )In ( ~ r ) ' "

-

...[Eqn. (7.68)J

UL Re, = --$

laminar in nature.

(ii) The average shear stress, T,:

1

rw = -2p u 2 =

Heat transfer rate from the plate,

Q = A, (t, - t,) = 3.536 x (2.2 x \l) (100 - 20) = 62234 W (Am.)

X

-Cf

&L x 1.165 x 1.g2 x 0.004572 = 0.008629 ~ l m '(Ans.)

(iii) The ratio of average shear stress to the shear stress at the trailing edge: The skin friction coefficient at the trailing edge (x = L), 0.664 ...[Eqn. (7.24)J Cfx =

;.

Shear stress at the trailing edge, 1

r , = 5 p d Cgi

=

;.

x 0.75 = 84375 -- 1.8 1.6 x lo4

2

x 1.165 x 1.8~x 0.002286 = 0.004314 ~ 1 n - i ~

Heat and Mass TrAhsfer

Example 7.20. Air at 3U°CJlows with a velocity of 2.8 m/s over a plate 1000 mm (length)

x 600 mm (width) x 25 mm (thickness). The top sugace of the plate is maintained at 90°C. I f the thermal conductivity of the plate material is 25W/m°C, calculate : (i) Heat lost by the plate; (ii) Bottom temperature of the plate for the steady state condition. The thermo-physical properties of air at mean film temperature (90 + 30)Q = 60°C are: p = 1.06 kg/m3, c,, = 1.005 kJAg K, k = 0.02894 W/m°C, v = 18.97 x I@ m2/s, Pr = 0.696. Solution. Given :t, = 30°C, ts = 90°C, U = 2.8 d s , kplate= 25 W/m°C, L = 1000 mm = lm, B = 600 mm = 0.6 m, 6 = 25 mm = 0.025m. ( i ) Heat lost by the plate: Reynolds number at the trailing edge,

Since Reynolds number is less than 5 x lo5, hence flow is laminar throughout the length,

h~ Nu = 0.664 (ReL )'I2 (pr)'I3= k

...[Eqn. (7.68)]

-

0.664 ( ~ e , ) " ~(pr)'l3 x k or Ti (average heat transfer coefficient) = Nu x k L L

Forced Convection

(i) Heat transfer rate from first half of the plate:

For first half of the plate,

Since Re < 5 x lo5 hence the flow is laminar. The local Nusselt number is given by, Nu, = 0.332 (~e,)'" ( ~ r ) " '

...IEqn. 7.651

But,

Average heat transfer coefficient, = 2h, = 2 x 4.322 = 8.644 wlm' "C :. Heat transfer rate from first half of the plate, Q = h A, (t, - 2,) = 8.644 x (0.45

x 0.45) (90 - 30) = 105 W (Ans.)

(ii) Heat transfer rate from full plate: For full plate, x = L = 0.9 m

= 6.542 w/m2 "C

:.

Heat lost by the plate,

Q = h A, (t, - t,) or Q = 6.542 x (1.0 x 0.6) (90 - 30) = 235.5 W (Ans.) (ii) Bottom temperature of the plate, t,: Heat lost by the plate Q (calculated above) must be conducted through the plate, hence exchange from top to bottom surface is

0.025 x 235.5 = 90.39"C (Ans.) 25 (1.0 x 0.6) Example 7.21. Air at 30°C and at atmospheric pressure flows at a velocity, of 2.2 m/s over a plate maintained at 90°C. The length and the width of the plate &re 900 mm and 450 mm respectively. Using exact solution, calculate the heat transfer rate from, (i)first half of the plate, (ii)full plate, and (iii) next half of the plate. The properties of air at mean bulk temperature (90 + 30)Q = 60°C are: p = 1.06 kg/m3, y = 7.211 kghm, v = 18.97 x I@ m2/s, Pr = 0.696, k = 0.02894 W/m°C. Solution. Given :t, = 30°Q, U = 2.2 mls, t, = 90°C, L = 900mm = 0.9m,B = 450 mm = 0.45 m. tb = 90 +

The heat transfer rate - from entire plate, Q, = h A, (t, - t,) = 6.1 13 x (0.9 x 0.45) x (90 - 30) = 148.54 W (Ans.) (iii) Heat transfer rate from next half of the plate: Heat transfer rate from the next half of the plate = Q, - Q, = 148.54 - 105 = 43.54 W (Ans.)

Example 7.22. Castor oil at 25°C flows at a velocity of 0.1 m/s past a flat plate, in a certain process. I f the plate is 4.5 m long and is maintained at a un@rm temperature of 9j°C, calculate the following using exact solution: (i) The hydrodynamic and thermal boundary layer thicknesses on one side of the plate, (ii) The total drag force per unit width on one side of the plate, (iii) The local heat transfer coeflcient at the trailing edge, and (iv) The heat transfer rate. The thermo-physical properties of oil at mean film temperature of (95 + 25)R = 60°C are: p = 956.8 kg/m3; a = 7.2 x I@ m2/s; k = 0.213 W/m°C; v = 0.65 x I @ m2/s. Solution. Given : t, = 25"C, t, = 95"C, L = 4.5m, U = 0.1 mls.

&g

Heat and Mass Transfer

Forced Convection

(iii) Local and average convective heat transfer coeflcients; (iv) Heat transfer rate from both sides for unit width of the plate, Iv) Mass entrainment in the boundary layer, (vi) The skin friction coefficient. Assume cubic velocity profile and approximate method. The thermo-physical properties of air at mean film temperature (60 + 2Oyr' = 40°C are: p = 1.128 kg/rn3, v = 16.96 x 1@ m2/s, k = 0.02755 W/m°C, PI = 0.699.

( i ) The hydrodynamic and thermal boundary layer thicknesses, 6, tith:

Reynolds number at the end of the plate,

\

Since Reynolds number is less than 5 x lo5, hence the flow is laminar in nature. The hydrodynamic boundary layer thickness,

,

Solution. Given: t, = 20°C, t, = 60°C, U = 4.5 mls. Ux-

= 0.2704 m or 270.4 mm (Ans.) ...[Eqn. (7.22)] The thermal boundary layer thidkness, according to Pohlhausen, is given by:

x

or = (0'65 l o b 4 = 902.77 a 7.2 x lo4 0'2704 = 0.02798 m or 27.98 mm (Ans.) 6th = (902.77)'" (ii) The total drag force per unit width on one side of the plate FD: The average skin friction coefficient is given by,

where Pr (Prandtl number)

i

-

,

Re, x v 5 x =U =

id x 16.96 x lo4

=

4.5

(where x, = distance from the leading edge at which the flow in the boundary layer changes from laminar to turbulent). (i) Thickness of hydrodynamic layer, 6: m e thickness of hydrodynamic layer for cubic velocity profile is given by 4.64 x, ...[Eqn. (7.36)] 6 = 7

...[Eqn. (7.25)1

The drag force,

or

FD = f

x -1

u2x area of plate (for one side) 2 P

FD = 0.01596 x

1

y x 956.8 x 0.12 x (4.5 x

(ii) Thickness of thermal boundary layer, The thermal boundary layer is given by 0.975 6 ...[Eqn. (7.31)]

1 ) = 0.3436 N per meter width

6th =

(iii) The local heat transfer coefficient at the trailing edge, h, (at x = L): h x = 0.332 (~e,)'" (pr)'I3 Nu, =

or

h, = 266.98 X

- 266'98

\-- I

...[Eqn. (7.65)]

0a213= 12.64 wlm2 OC (Ans,)

4.5

(iv) The heat transfer rate, Q: Q = A, (t, - t,) h = 2h, = 2 x 12.64 = 2 5 . 2 8 ~ l m ~ 0 ~ where Eqn. (7.6711 Q = 25.28 X (4.5 x 1) (95 - 25) = 7963.2 W (Am.) ,. Example 7.23. Air at 20°C and at atmospheric pressure jlows at a velociry of 4.5 m/s past a jlat plate with a sharp leading edge. The entire plate sur$ace is maintained at a temperature of 60°C. Assuming that the transition occurs at a critical Reynolds number of 5 x 16, find the distance from the leading edge at which the flow in the boundaly layer changes from laminar to turbulent, At the location, calculate the following : (i) Thickness of hydrodynumic layer; (ii) Thickness of therml boundary layer;

*

!?.975 0.01234 = 0.01355 m or 13.55 mm (Ans.) (0.699)"~

The Nusselt number at x = x, is given by ln pr)113 NIL = 0.332 (Re,) (

k

= 0.332 x (6923)In (902.77)'" = 266.98

...[Eqn. (7.8911

Put

Nu, =

h Xx, k

h, =

or

...[Eqn. i'1.65)I

Nu, X k XC

z

Average heat transfer coefficient,

x = L f hxak - - xc

h = 2hc = 2 x 3.05 = 6.1 w/m2'C (Am.)

I

(iv) H a t transfer rate from both sides for unit width of the plaki Q : Q = (us) At =. 6.1 ( 2 x 1.88 x 1) (60 - 20) = 917.U W (Am)

z

Heat and MassqTransfer

Forced Convection

(v) Mass entrainment in the boundary layer, in:

u

boundary layer where u = U. Hence for U = 1, we have

"

Here, 6, = 0 at x = 0 and 6, = 0.01234 m at x = x, = 1.88 m 5 . m = x 1.128 X 4.5 (0.01234 - 0 ) = 0.039 kgls or 140.4 kglh (Ans.)

.

-8

(vi) The skin friction coefficient,

U

(Refer Table 7.1)

= 0.86

C': ...[Eqn. (7.24)1

0.646 = 9.136 X lo4 (Ans.) 5 x 10' Example 7.24. A stream of water at 20°C ( p = 1.205 kg/m3, p = 0.06533 kg/hm)Jows at a velocity of 1.8 4 s over a plate 0.6 m long and placed at zero angle of incidence. Using exact solution, calculate: (i) The stream wise velocity component at the midpoint of the boundary layer, (ii) The maximum boundary layer thickness, and (iii) The maximum value of the normal component of velocity at the trailing edge of the plare. Solution. Given t, = 20°C, p = 1.205 kg/m3, p = 0.06533 kghm, L = 0.6 m, U = 1.8 m/s (i) The stream wise velocity component at the midpoint of the boundary layer, u: The boundary layer thickness by exact solution is given by

CB = -d

LAMINAR TUBE FLOW

7.2

7.2.1. Development of Boundary Layer In case of a pipe flow, the development of boundary layer proceeds in a fashion similar to that for flow along a flat plate. A fluid of uniform velocity entering a tube is retmded near the walls and a boundary layer begins to develop as shown in Fig. 7.12 by dotted lines. n e thickness of the boundary layer is limited to the pipe radius because of the flow being within a confined passage. Boundary layers from the pipe walls meet at the centre of the pipe and the entire flow acquires the characteristics of a boundary layer. Once the boundary layer thickness becomes equal to the radius of the tube there will not be any further change in the velocity distribution, this invariant velocity distribution is calledfully developed v e l c p r o ie., poiseulle flow (parabolic distribution). Boundary Layer

1-

~ d l developed v

i 3

Obviously, the midpoint of the boundary layer y = - occurs at

The stream wise velocity component is obtained form the Blasius solution in tabular form (Refer table 7.1).

$

U = 0.736 = 2.5, we get U or u = 0.736 U = 0.736 x 1.8 = 1.325 m/s (Am.) (ii) The maximum boundary layer thickness, 6,: The maximum boundary layer thickness occurs at x = 0.6 m. Thus.

At

11 = y

Re, = puL - 1.205 x 1.8 x 0.6 = 71713, hence flow in laminar. p (0.06533/3600) ' The boundary layer thickness at the trailing edge,

a,=----5L

5 x 0.6

6- - m0.01 -12m or 11.2 mrn

k-

Entrance length (L,)

I

Fig. 7.12. The development of a laminar velarity profile in the intake region Le of a tube.

According to Langhar (1942), the entrance length (L,) is expwscd

:

5 = 0.0575 Re where

D represents the inside diameter of the pipe.

7.2.2

Velocity Distribution Fig. 7.13 shows a horizontal circular pipe of radius R, having laminar flow of fluid through it. Consider a small concentric cylinder (fluid element) of radius r and length dx as a free body. If z is the shear stress, the shear force F is given by F=zx2nr~du Let p be the intensity of pressure at left end and the intensity of pmsum at the dght end

(Am.)

(iii) The inaxilduni dalue of the normal component of velocity at the trailing edge, v : $ t The maximum value of the normal cornfionent of velocity occurs at the outer edge of the !%

c--

<$

'

Thus the forces acting on the fluid element are: 1. The shear force, T x 2zr x dn on the surface of fluid element.

T

4m

Heat and Mass Transfer

2. The pressure force, p x n? on the left-end.

( + -E ]

. dr n? on the right end.

3. The pressure force, p

(

For steady flow, the net force on the cylinder must be zero.

---

Integrating the above equation w.r.t. 'r', we get

@ 1

Pipe

-0--

4 p dX

Where C is the constant of integration and its value is obtained from the boundary condition:

Substituting this. value of C in Eqn. (7.97),we get

it--.------

x2 -------3:

Fig. 7.13. Laminar flow through a circular pipe.

Equation 7.98 shows that the velocity distribution curye is a parabola (see Fig. 7.13).The maximum velocity occurs at the centre and is given by

From equation 7.98 and 7.99, we have

'

II

-

I

Equation (7.95)shows that flow will occur only if pressure gradient exists in the direction of flow. *. The negative sign shows that pressure decreases in the direction of flow. Equation (7.95) indicates that the shear stress varies linearly across the section (see Fig. 7.14). Its value is zero at the eentre of pipe ( r = 0 ) and maximum at the pipe wall given by

I

-

\ , -

Equation 7.100 is the most commonly used equation for the velocity distribution for laminar flow through pipes. This equation can be used to calculate the discharge as follows: The discharge through an elementary ring of thickness dr at radial distances r is given by

... [7.95 (a)]

I

1

a

Ffom Newton's law of viscosity, du 7=p....fi) dv --, In this equation, the distance y is measured fmm the boundary. The radial distance r is related to distance y by the relation y = R - r or d y = - d r The eqn. (i) becomes

Q = ! ~ Q

Total discharge

Shear stress distribution

Velocity distribution curve

Comparing two values of T frm Eqns. 7.95 and 7.96, we have

-

Fig. 7.14. Shear stress and velocity distribution across a section.

..-.

Q

Average velocity of flow, u = A =

"

R2

nR2

u,, =2

402

Heat and Mass Transfer

Forced Convection Elementary rine

Equation (7.101) shows that the average velocity is one-halfthe maximum velocity. Substituting the value of u,, from Eqn. (7.99), we have

ap is usually expressed in terms of a friction factor f, defined as The pressure gradient -

ax

pU where - is dynamic pressure of the mean flow and D is the tube diameter. 2 From Eqn. (7.102) and (7.103), we get the friction factor as a simple function of Reynolds number,

Fig. 7.15. Analysis of energy in the tube flow.

Considering energy balance on the annular element, we obtain (Heat conducted in),,, = (Heat convected out),,, dQr - dQr+dr = (dQmn,)ne,

which is valid for laminar tube flow, Re < 2300 Further, Eqn. (7.102) can be written as,

dt = p (2nrdr)u c - dx

8p U = R2 The pressure difference between two sections 1 and 2 at distances, x, and x2 (see Fig. 7.12), is given by

-ap

~dnular element

Qr + dr

ax

-.ax

at ar

- k (2nr.h) - + k (271 (r

+ dr)} dx

[$ + $dr] = p (2nr.dr)u c -axat dx at + k ( 2 n d r d x ) -a r at

Neglecting second order terms, we get at &.dr = p r u c --dx.dr

ax

Obviously the head loss hL over a length of pipe varies directly as the first power of the rate of discharge Q and inversely as the fourth power of the pipe diameter. 7.2.3. Temperature Distribution In order to estimate the distribution of temperature let us consider the flow of heat through an elementary ring of thickness dr and length dx as shown in Fig. 7.15. Considering the radial conduction (neglecting axial conduction) and axial enthalpy transport in the annular element, we have: Heat conducted into the annular element, at dQr = - k (271r.d~)ar Heat conducted out of the annular element,

1

a ...(7.106)

or Inserting the value of u from eqn. (7.100), we get,

...(7.107) at or

...(7.108)

at

Let us consider the case of uniform heat flux along the wall, where we can take Net heat convected out of the annular element,

,-

constant. Integrating Eqn. (7.108) we have

ax as a

Heat and Mass TrAnsfer

404

405

Forced. Convection

2

tb=sfo

2;

[I

-s]bs-k 2

a

' ak x{ g 16 - < +4 L }16R2 ] r d r

Integrating again, we have 4

(where C; and C2 a& the constants of integration). The boundary conditions are: at =0

At r = 0,

ar At r = R, t = ts Applying the above boundary conditons, we get

Substituting the values of C, and C2 in Eqn. (7.109), we have or

11 tb = t s - - 96

U*mR2-

a

at ax

The heat transfer coeflcient is calculated from the relation

For determining the heat transfer coeficient for fully developed pipe flow, it is imperative to define a characteristic temperature of the fluid. It is the bulk temperature (tb)or the mixing up temperature of the fluid which is an average taken so as to yield the total energy carried by the fluid and is defined as the ratio of flux of enthalpy at a cross-section to the product of the mass flow rate and the specific heat of the fluid. Thus,

From eqn. (7.1 lo), we have

l R p (2nr.dr)ucpt tb =

O

lRp (2 nr . dr) u cp

...(7.111)

0

For an incompressible fluid having constant density and specific heat lRurrdr tb =

O

...(7.112)

I oR u r d r

The averagetmean velocity following definition:

-

u=

where D is the diameter of the tube. The Nusselt number is given by

(i)also known as the bulk mean velocity is calculated from the

2 R 3 In u r d r "

This shows that the Nusselt number for the fully developed laminar tube flow is constant and is independent of the Reynolds number and Prandtl number. The first analytical solution for laminar flow for constant wall temperature was formulated at by Graetz in 1885. Since - is not constant, therefore, the analysis of constant wall temperature

ax

Substituting this value of u in eqn. (7.1 12), we get

is quite cumbersome. The final result comes out to be

Substituting the value of u from eqns. 7.1 10 and 7.101 and that of t from eqn. (7.1 lo), we get

Example 7.25. For laminar flow in a circular tube of 120 mm radius, the velocity and temperature distribution are given by the relations:

.

Heat and Mass Transfer

406

Forced Convection

u = (2.7r - 3.2 ?) ; t = 85 ( I - 2.2r)OC where the distance r is measured from the tube surfnce. Calculate the following: (i) The average velocity and the mean bulk temperature of the fluid; (ii) The heat transfer coeficient based on the bulk mean temperature if the tube su$ace is maintained at a constant miform temperature of 90°C and there occurs a heat loss of 1000 kJ/h per metre length of the tube. Solution. Given : u = (2.7 r - 3.2 3) ... Velocity distribution t = 85 (1 - 2.2 r)OC ... Temperature distribution. ( i ) Average velocity ( Z ) and mean bulk temperature (t,) : The average velocity is obtained by equating the volumetric flow to the integrated flow through an elementary ring of radius r and thickness dr.

Substituting, R = 0.12 m and ii = 0.193 mls, we get 170 (0.9 x 0.12 - 2.285 X 0.12' tb = 0.193

where

Q = 1000 M h per metre =

f 0R

(2.7 r

looo looo = 277.77 Jls ; t, = 90°C 3600

277.77 = 16.97 W / m 2OC (Am.) (271 x 0.12 x 1 ) (90 - 68.29)

- 3.2 ?) rdr

-

Assume flow to be laminar (and fully developed) Nu = 3.657

(AMIE Summer, 1997)

Solution. Given : ti = 60°C, to = 45'C; D = 1 cm = 0.01 m; U = 3 m/s, t, = 40°C; p = 865 kg/m3; k = 0.14 W/m K ; cp= 1.78 kJ/kg°C. Substituting R = 0.12 m, we have

u = 1.8 x 0.12 - 1.6 x 0.12' = 0.193 m/s (Am.) The mean bulk temperature is given by

tb =

lRu t r d r lRu r d r 0

0

Now,

jR0 u t r d r = J~n

40°C

(2.7r - 3 . 2 q x 8 5 ( 1 - 2 . 2 r ) r d r .

Fig. 7.16

Length required, L : Q = m cp(ti- to) = (PA# cp 0,- to)

(where U = average velocity, Af= flow area)

...From eqn. (i)]

...Given

Example 7.26. Lubricating oil at a temperature of 60°C enters I cm diameter tube with a velocity of 3 d s . The tube su$ace is maintained at 40°C. Assuming that the oil has the following average properties calculate the tube length required to cool the oil to 45OC. p = 865 kg/&; k = 0.140 W/m K: cp= 1.78 kJ/kg°C.

...(i) R2

1.408 x 0.12)) = 68.290C (Ans.)

(ii) Heat transfer coefficient, h: Q- = hA (t, - tb)

h=

=

+

I j

Also,

~=hA0,

where A = heat transfer area = nDL, and

Heat and Mass Transfer Forced Convection

B. TURBULENT FLOW

...(Given) Now,

Q=5441.7=51.2XmLx 10.82

..

L=

5441.7 = 312.7 m (Ans.) 51.2 x n x 0.01 x 10.82 Example 7.27. When 0.5 kg of water per minute is passed through a tube of 20 mm diameter, it is found to be heated from 20°C to 50°C. m e heating is accomplisted by condensing steam on the sulface of the tube and subsequently the surface temperature of the tube is maintained at 85OC.Detennine the length of the tube required for fully developed flow.

Take the thermo-physical properties of water at 60°C as : p = 983.2 kg/m3, cp = 4.178 kJ/kgK, k = 0.659 W/m°C, v = 0.478 x 1@ m2/s Solution. Given :m = 0.5 kglmin, D = 20 mm = 0.02 m, ti = 20°C, to = 50°C

Length of the tube required for fully developed flow, L: 'The mean film temperature,

f /8fi +

rj = , Let us first determine the type of the flow.

2o

+

2

50)

7.3. INTRODUCTION The flow in the boundary layer, in majority of practical applications in the convective heat transfer, is turbulent rather than laminar. In a turbulent flow the irregular velocity fluctuations are mainly responsible for heat as well as momentum transfer. As the mixing in the turbulent flow is on a macroscopic scale with groups of particles transported in a zig-zig path through the fluid, the exchange mechanism is many times more effective than in laminar flow. Consequently. in turbulent flow, the rates of heat and momentum transfer and the associated friction and heat transfer coefficients are several times larger than that in laminar. Since the nature of turbulent flow is complex, therefore, it is difficult to solve the problems relating turbulent flow analytically. The heat transfer data can best be calculated by laboratory experiments; the other method of study is the analogy between heat and momentum transfer. A. Forced Convection-Flow over a Flat Plate 7.3.1. Turbulent Boundary Layer Refer Fig. 7.17. As compared to laminar boundary layers, the turbulent boundary layers are thicker. Further in a turbulent boundary layer the velocity distribution is much more uniform, than in a laminar boundary layer, due to intermingling of fluid particles between different layers of the fluid. The velocity distribution in a turbulent boundary layer follows a logarithmic law i.e. u log y, which u log y can also be represented by a power law of the type 0.99 U

-

= 6OOC

/

-

19) 1 where n = - (approx) for Re < 1o7 7 but > 5 x 105

.

R e = -D - U - 0.02 x 0.0269 = 1125.5 V 0.478 x lov6 Since Re < 2000, hence the flow is laminar. With constant wall temperature having fully developed flow,

Reynolds number,

hD = 3.65 Nu = k

$

..

=

Laminar sublayer

~1"

...(7.120)

Smooth flat plate

This is known as one-seventh power law. ...[Eqn. (7.11 8)]

Fig. 7.17. Turbulent boundary layer.

The Eqn. (7.120). however, cannot be applied at the boundary itself because at y = 0,

- -; [El =

~ 6 - l y4'7 ' ~ = OJ.This difficulty is circumvented by considering the velocity in the viscous

\-' /

The rate of heat transfer, Q = A, h (t, - t,) = m cp (to - ti) t, = 20

Here,

+ 50 = 35°C = t,

laminar sublayer to be linear and tangential to the seventh-root profile at the point, where the laminar sublayer merges with the turbulent part of the boundary layer. Blasius suggested the following relation for viscous shear stress:

2

(for Re ranging from 5 x 10' to or

L

=

--

377.8

lo7)

Let us now find the values of 6, z,, Cfi,

- 2.76 rn (Ans.) i . ,

)

1

in ,

for the velocity distribution given by Eqn. (7.120)

410

Heat and Mass Transfer

Forced Convection

(i) Boundary layer thickness, 6: U

Substituting the value of

in Von Karman integral eqn. [7.28 (a)], we have

6 -- 0.371

or

X

(Re,)'"

(ii) Shear stress, 70 : b

...[Eqn. (7.12111

Substituting the value of 6 from eqn. (7.123), we get

[

P

= 0.0225 pu2

p ux-

371i (Re,) ' I 5

[In the expression, above, the limits have been taken from 0 to 6 instead of 6' to 6 since the laminar sublayer (6') is very thin]

Now equating the eqns. (7.122) and (7.121), we have

or or

.

(iii)'~ocalskin friction (drag) coefficient, CB:

1/4 1 -d6 - 0.0225 [A] x-

PU

72 dx

(cancelling puZon both sides).

@)'I4

d6 = 0.0225 x

c

[Eqn. (7.124)l

We know Also

r0=

cfXx 11.p d

Now equating the eqns. (7.124) and (7.30), we have Integrating both sides, we have

4 6'14 5

J; (-

= 0.2314

(where C = constant of integration)

+c

Let boundary layer be assumed to be turbulent over the entire length of plate. \Hence, at x = 0, 6 = 0 C=0

:.

= 0.0371

[&)

1/5

0.37 1~ x x =(Re,)'"

(iv) Average value of skin friction (drag) coefficient, Cf :

[Eqn. (7.30)l

Heat and Mass Transfer

413 Forced Cohvection

...(ii)

This is valid for 5 x lo5 < Re,- < lo7. For Reynolds number between lo7 and lo9 the following relationship suggested by Prandtl and Schlichting holds good,

1.328 x, +

0.455 L (loglo

~ " 4- .(Re,)"' '~

7.3.2. Total Drag Due to Laminar and Turbulent Boundary Layers When the leading edge is not very rough, the turbulent boundary layer does not begin at the leading edge, it is usually preceded by the laminar boundary layer. The point of transition from laminar to turbulent layer depends upon the intensity of turbulence. The distance xc (Fig. 7.18) of the transition from the leading edge can be obtained from critical Reynolds number which normally ranges from 3 x lo5 to 3 x lo6.

-

I

Assuming that transition occurs at Re, = 5 x 10';

___+

+ u,t, I

Also,

pi-

I

~ a m i layer y

Leading edge

Turbulent layer

LBpu2 c x -21 ~ A =U C-~f x 2 [where cf= average value of skin friction (drag) coefficient] Ftotal=

f

Equating the above two equations, we have 1670 0.455 cf = (loglo~ e , ) ' . ~ Re, ~

b

The above relation in general may be rewritten as

Fig. 7.18. Drage due to laminar and turbulent boundary layers.

Drag force (FD = F) for the turbulent boundary layer can be estimated from the following relation:

= (Fturb.)total- (Fturb.) x, where (Fturb,)total = the drag which would occur if a turbulent boundary extends along the entire length of the plate, and (Ftu,.)xc = the drag due to fictitious turbulent.bundary layer from the leading Fturb.

edge to a distance x,. Let us assume that the pikeis long enough so that Reynolds number is greater than lo7, then the turbulent drag is given by Fturb. =

0.455 Ooglo~

e

pu2 x-x(LxB)-2~ )

0072 ~(Re,) 11~5 x

pu2

2 ~ x (xc~x B)

...( i )

where

L = length of the plate, B = width of the plate, and U = free stream velocity. The laminar boundary layer prevails within the length xc and its contribution to drag force is given by Fig. 7.19. Skin friction (drag) coefficient for smooth flat plates.

...(7.129)

Heat and Mass Transfer

414

where the value of A depends upon the value of critical Reynolds number Re, (at which the laminar boundary layer transforms to turbulent boundary layer). The values of A for Re,, lo5, 5 x lo5 and 10, are 360, 1670, 3300 respectively.

415

Forced Convection

This being very small, contribution to total drag from laminar boundary layer is negligible; hence

cfis given by

cf

versus Reh Fig. 7.19 shows a log-log plot of Example 7.28. A flat plate 5 m long and 0.75 m wide is kept parallel to the flow of water which is flowing at a velocity of 5 d s . If the average drdg coeficient for turbulent flow past this flat plate is expressed as:

cf=

0'455 (log,, Re,)

2,58

:.

Drag force,

A = ?rDL = n x 5 x = 785.4 m2 1 I FD = x 5 p ~ d= 0.001765 x x 1030 x 785.4 x 8'

cf

find the drag force on both sides of the plate.

Take v = 0.011 x I@ m2/s Solution: Given : L = 5 m, B = 0.75 m, U = 5 d s , v = 0.01 1 x 10'' Drag force on both sides of the plate: The Reynolds number at the end of the plate is given by

Hence total power required to overcome boundary friction, m2/s

Example 7-30. Find the ratio of friction drag on the front half and rear half of the f2ot plate kept at zero incidence in a stream of uniform velocity, i f the boundary layer is turbulent over the whole plate. Solution. The average coefficient of drag for turbulent boundary layer is given by

(cf)

This confirms that the nature of flow is turbulent. The average drag coefficient is expressed as

c -- [log,,0.455 ~ e

k

Area,

d ~ . ~ ~

For the entire plate,

UL ReL = v

Ux U L For the first half of the plate, Re, = v =2v

:.

The drag force on both sides of the plate, FD = 2 x c f x [ ; ~ A U ' )

= 2 x 2.642 x

[where A = area of one side]

IT

x 1 x 1000 x (5 x 0.75) x 5'1 = 247.68 N

Drag force per unit width for the entire plate is, - pu2 FD = Cfx 2 x area per unit width

-

(Ans.)

J

L

Example 7.29. A submarine can be assumed to have cylindrical shape with rounded nose. Assuming its length to be 50 m and diameter 5.0 m, determine the total Dower reauired to overcome boundary friction if it cruises at 8 rn/s velocity in sei water at 20°C (p = 1030 kg/m3), v = I x lo4 m2/s. Solution. Length of submarine, L =50m Diameter of submarine, D = 50 m Velocity of submarine, U =8d s Density of sea water, p = 1030 kg/m3 Kinematic viscosity of sea water, v = 1 x lo6 m2/s Total power required to overcome boundary friction, P : Reynolds number.

\

FD, -

:.

Drag force for the rear half portion of the plate is

Hence,

The length over which boundary layer wilf be laminar is given by Ux -=5x1050rx= v

5 x 1 6 ~ ~ '

U

/

Similarly the drag force per unit width for the front halfportion of the plate is

-FD, -FD2

1 ?x(2)l" 1 1 -- 1 2

- 0S74 = 1.347 (Ans.) 5 - 0.574

Example 7.31. A stream-lined train is ZOO m long with a typical cross-section having a perimeter of 9 m above the wheels. If the kinematic viscosity of air at the prevailing tempemture m2/s and density 1.24 kg/m3, determine the su$ace drag (friction drag) of the train is 1.5 x when running at 90 km/h.

417

416

Heat and Mass Transfer

Make allowance for the fact that boundary layer changes from laminar to turbulent on the train surface. Solution. Length of the train, L = 200 m Perimeter of cross-section of the train above wheels, P = 9 m :. Surface area, A = L x P = 2 0 0 x 9 = 1800m2 Kinematic viscosity of air, v = 1.5 x lo-' m2/s Density of air, p = 1.24 kg/m3

Forced Convection

Maximum boundary layer (laminar) thickness, 8:

(ii) The total drag force FD :

U = 90 kmh = 9ox 1000 = 25 ds

Free stream velocity,

3600 Friction drag, FD : The Reynolds number with length of the train as the characteristic length,

The average coefficient of drag, ...[Eqn. (7.1291

-

uL. - 25 200 Re, = v 1.5x10-~= 3.333 x 108 Obviously the boundary layer is turbulent. Assuming that the abrupt transition from laminar to turbulent flow occurs at a Reynolds number of 5 x lo5, the average coefficient of drag,

0.455 1670 [log,, (3.333 x 108)12." - 3.333 x lo8 = 0.001807 - 5.01 x = 0.0018 The approximate friction drag over the train surface,

-

-

1 1 X 1.24 x 1800 x 25'= 1255.5 N (Ans.) FD = Cf x 2 ~ A V 0.0018 ~ = x2 Example 7.32. A barge with a rectangular surfne 30 m long x 10 m wide is travelling down a river with a velocity of 0.6 m/s. A laminar bouna'ary layer exists upto a Reynolds number equivalent to 5 x IO'and subsequently abrupt transition occurs to turbulent boundary layer. Calculate: (i) The maximum distancefrom the leading edge upto which laminar boundary layer persists and the maximum boundary layer thickness at that point. lii) The total drag force on the flat bottom s u ~ a c eof the barge, and (iii) R e power required to push the bottom surfoce through water at the given velocity. For water p = 998 kg/m3 and v = I x I@ m2/s. Solution. Length of the bottom surface, L = 30 m Width of the bottom surface, B = 10 m .'. Area, A = L x B = 3 0 x 10=300m2 Velocity, U = 0.6 d s Density of water, p = 998 k g h 3 Kinematic viscosity, v = 1 x 10" m2/s (i) The maximum distance up to which laminar boundary layer persists, x,:

-- 5 x lo5

( ~ t ? = ) ~ "C~ V

:.

surface of the barge, 1 Drag force on the bottom 1

FD =

Cfx 5 ~

A = U 0.002M6 ~ x

5x 998 x 300 x (0.6)~

(iii) The power required, P: The power required to push the bottom surface through water at the given velocity, P = FD x U 142.6 x 0.6 = 85.56 W (Am.)

Example 7.33. Ambient air at 2oDC flows at a velocity of 10 m h parallel to a wall 5 m wide and 3 m high. Calculate the heat transfer rate if the wall is maintained at 40°C. The critical Reynolds number is equal to 5 x 1@. fie properties of air at the mean film temperature may be taken as :k = 0.0263 W / m K, v = 15.89 x 1@ m2/s and Pr = 0.707. B the entire boundary layer is assumed turbulent, what will be the percentage error in the computation of heat transfer rate ? Comment on the comparable values of the two results. Appropriate correlation from the following may be used :

-

N = 0.0375 R e f 8 pr'13

-

(AMIE, Summer, 1999)

NU = 0.0375 [ ~ e-p232001 ~ pr113

Solution. Given : t- = 20°C; U = 10 m/s; L = 5 m; B = 3 m; t, = 40°C; R,, = 5 X lo5 k = 0.0263 W/m K; v = 15.89 x lom6m2/s; Pr = 0.707.

Wall

Wind

(t, = 40°C)

t, = 2O0C,

U = 10 mfs (+-

I

B=3m

L = 5 m ------*I Fig. 7;2G

I

418

Heat and Mass Transfer

Forced Convection

Percentage error in computation of heat transfer rate :

i.e., > R,, Using the following equation, we get hL k == 0.0375 k

[

dt Q = - c p A 2O du Separating the variables and integrating within the limits: At the plate surface: u = 0 and t = t, At the outer edge of boundary layer; u = U and t = t,

I I

- 23200 PI.'" ...(Combination of laminar and turbulent flow)

..

k

- 23200 pr'l3

= 0.00019725 (157860.4 - 23200) x 0.8908 = 23.66 w/m2 OC

But

Heat transfer rate,

ha (ts-t,)

= 2 3 . 6 6 ~ ( 5~ 3x (40-20) ) =To98 w If entire boundary is assumed turbulent,

and

-

zo = , C x

1

5p d

Making these substitutions in eqn. (7.132), we 1 get

- hL Nu = -= 0.0375 R e p 8 ~ r ' " k

..[Eqn. (7.23)l

e

c

h,= C i , x ~ p d x $ = $ ( P c p U ) ...(7.133)

...dimensionless fom. hx is called the Stanton number SG. It represents the Nusselt number divided by the -

. Percentage error = 8322 - 7098 x 100 = 17.24% (Am.) 7098 Comments. The difference between the two values is nearly 17%. Further the correlation for turbulent heat transfer are empirical, which may have a variation of 25%. Hence, the rate of heat transfer in such cases may be calculated for all purposes, by assuming the boundary layer to be entirely turbulent.

+

7.3.3. Reynolds Analogy Reynolds analogy is the inter-relationsh@ between f l u i d w o n and Newton's law of viscosity. du We know, zo = p ...(i) dy

[Heat flow along Y-direction

... Fourier equation]

--Nu,

Re,. Pr

- st,

=

5 2

Equation (7.134) is called the Reynolds analogy. By using this interrelationship we can infer heat transfer data from shear stress measurement. hxb Note: The physical significance of Stanton number is : St = p cpU At actual heat flux to the fluid heat flux capacity of the fluid flow

-

Further in case of laminar boundary layer on a flat plate, we have hxx Nu, = -= 0.332 ( ~ e , ) " ~(pr)'" k

...[Eqn. (7.94)]

Dividing both sides of the above equation by Re, (pr)lt3,we get

...(iii) when Pr is unity temperature and velocity [profiles are identical (For most of the gases, 0.6 < Pr < 1.0)] By combining eqns. (i), (ii) and (iii), we get

P cpu ~roducto f the Reynolds and Prandlt numbers, i.e.,

Nux ----0.332 Re, (pr)'13 ( ~ e , ) " ~

5

The L.H.S. of the equality can be rewritten as - Nu, . ( P ~=Jq~( ~~ ~ r ) ~ ~ ~ Re, (pr)'13 Re Pr

-

Heat and Mas6 Transfer

420

~ ~ r o eConvection d

...interrelationship bemeen heat and momentum transfer. Equation (7.137) has been designated as Colburn analogy. For Pr = 1 the Reynolds and Colburn analogies are the same. Heat transfer parameters for turbulmt flow: The heat transfer parameters for turbulent flow may be derived, by using Colbum analopv. as follows: From Colburn analogy

Assuming that the transition occurs at critical Reynolds number, Re, = 5 X lo5, we get kh = : ( ~ r ) ' "10.664 (5 x lo5)'" + 0.036 (R~J"' - 0.036 (5 X 10')~.'1

w.,

and

-

Nu, ~ (Re,)-'" Pr Re, ( ~ r= )0.0288 L

IL

velocity of 50 d s . The local skin friction Example 7.34. Airflows over a heated plate at a co-egident at a point on a plate is 0.004. Estimate the local heat transfer coefficient at this point. The following property data for air are given :

Nu, = 0.0288 (ReJ4" (pr)ln or The average value of heat transfer coefficient is given by,

-

1

.L

,

s t prl" =

(U.P.S.C., 1993)

C $

Solution. Given : U = 50 m/s; cp= 0.004; P = 0.88 kg/m3; P = 2.286 x lo-' kg m/s;

-

cP= 1.001W/kg K: k = 0.035 W/m K.

Local heat transfer coefficient, h, : p . C 2.286 x ~randtlnumber, Pr = k =

or

= 0.036

-

...(Given)

(i)

(Re3U5 ( ~ r ) ~

-

NU = hL. = 0.036 (Red" (Prim k Heat transfer parameters for combination of laminar and turbulent flow: An expression for the average heat transfer coefficient over a plate of length L when bnh laminar and turbulent boundary layers are present can be derived by evaluating the following integral: and

X (1.001 X 1000) = o,654 0.035

Example 7.35. The crankcase df an LC. engine measuring 80 cm x 20 crn may be idealised as a flat plate. The engine runs at 90 k m h and the crankcase is cooled by the air flowing past it at the same speed. Calculate the heat loss from the crank surfoce maintained at 8S°C, to the ambient air at 1S0C.Due to road induced vibrafion, the boundary layer becomes turbulent from the leading edge itselJ: 90 x 1000 Solution. Given : = 90 k f i = 3600 = 25 mls, t, = 85OC, t.. = 15OC, L = 80 cm = L -

The properties of air at tf =

85 + 15 = 50°C are: 2

k = 0.02824 Wlm°C, v = 17.95 x lo4 m2is, Pr = 0.698 ... (From tables)

Heat and Mass Transfer

422

Heat loss from the crankcase, Q :

uL = 25x0.8 The Reynolds number, Re, = v 17.95 x lo4 =1.114x1o6 Since Re, > 5 x lo5, the nature of flow is turbulent. For turbulent boundary layer,

- hZ Nu =

Forced Convection The Reynolds number at the end of the plate is UL 70x1.2 = 8 . 8 4 ~ 1 0 5 Re, = - 9 5 x 10-6 v

-

This shows that both laminar and turbulent boundary layers are present and, Hence,

= 1045 i?u = L0.036 (8.84 x 10')'.~ - 8361 (0.625)'.~~~

~ = 0.036 ( ~ e , ) ' .(~~ r ) ' . " ~= 0.036 (1.1 14 x lo6)'.' ( 0 . 6 9 8 ) ' . ~=~ 2196.92

k or h = - x 2196.92 = O 02824 x 2196.92 = 77.55 w/m2'c L 0.8

-

:.

Q = h~ (t, - $3= 77.55 x (0.8 x 0.2) (85 - 15) = 868.56 W

(Am.) Example 7.36. Air at 20°C and 1.013 bar flows over a flat plate at 40 m/s. The plate is I m long and is maintained at 60°C. Assuming unit depth, calculate the heat transfer from the plate. Use the following correlation: (M.U.) NuL = (~r)'.~'[0.037 ( ~ e , ) ' ,-~8501 Solution. Given : t, = 20°C, U = 40 qds, L = Im, B = I m, t, = 60°C, Properties of air at (60 + 20)/2 = 40°C, from the tables: p = 1.128 kg/m3, c, = 1.005 kJkg°C, k = 0.0275 W/m°C, v = 16.96 x 10d m2/s, Pr = 0.699 Heat transfer from the plate, Q: Reynolds number,

Heat loss,

Q = ;A At

= 65.31 x (1.2 x 0.8) x (950 - 280) = 42007.4 W = 42 kW (Ans.) Example 7-38, Air at 20°C and 1.013 bar jlows over a rectangular container, with top suIface 750 mm long in direction of flow and 1 m wide, at 35 nu's. Determine the heat transfer from the top suface maintained at 60°C Use thefollowing properties of air at average temperature 9f 40% p = 1.906 x I@ kg/ms, c,, = 1.007 kJAg0C and k = 0.0272 W/m°C, and the following co-relationsfor finding average heat transfer coeficient: .. (i) ) ~5 5~ x ~I @ i u = 0.664 ( R ~ J( ~ ~* ~rif ReL . .(ii) ~ > 5 X I@ i u = I0.037 ( ~ e f l ' 8501 ( P ~ ) O * ff~ ReL

.

-

.

(P*U

Solution. Given : p = 1.013 bar, L = 750 mm = 0.75 m, B = 1 m U = 35 d s , Mean film 60 + 20 = 4O0C temp. t, = 60°C, t. = 20°C, t( = -

..

Heat transfer from the to6 surface, Q: Q =

i;A, (5 - ).I

= 92.55 x ( 1 x 1 ) (60 - 20)

= 3702 W or 3.702 kW (Ans.) Example 7.37. In a gas turbine system hot gases at 95PCflow at 70 i d s over the sugace of a combustion chamber which is at a uniform temperature of 280°C. Determine the heat loss from the gases to the combustion chamber which can be idealised as a flat plate measuring I20 cm x 80 cm. The flow is parallel to the 120 cm side and transition Reynolds number is equal to 5 x 16. Take the properties of gas as: p = 0.494 kg/m3, k = 0.075 W/m°C, v = 95 X 10d m2/s, Pr = 0.625. Solution. Given: L = 120 cm = 1.20 m, B = 80 cm = 0.8 m, Re, = 5 x lo5, U = 70 m/s. Heat loss from the gases to the combustion chamber: The average Nusselt number for a flow over ?,flat plate when both laminar and turbulent boundary layers are present is given by I f transition occurs at Re, = 5 x lo5, then the above expression reduces to

Eu = [0.036

(wO,' - 8361 (~r)'.'~'

-

__+

Using the gas equation for air, we have p = pRTorp=$

Air flow

Since Re > 5 x lo5, we shall use eqn. (ii) for finding average heat transfer coefficient.

Heat and Mass Transfer

Forced Convection (ii) Drag force on the wing.

..

Q

= 0.03627 (3489.96 - 850) x 0.8915 = 85.36 w / m 2 ~ c = h,0, - t,) = 85.36 x (0.75 x 1) (60- 20) = 2560.8 W (Ans.)

Example 7.39. A flat plate 1 m wide and 1.5 m long is to be maintained at 90°C in air when free stream temperature is ](PC. Determine the velocity at which air must flow over the plate so that the rate of energy dissipation from the plate is 3.75 kW. 1

Use :

-

-

fiu = hL = 0.664 ( ~ e d ' '(~~ r ) " ~ k

-

...for

laminar flow

(i) Heat loss from the wing:

65 The pressure of the air at flight altitude = p = 5x 1.013 = 0.866 bar

From the characteristic gas equation, we have

-

[0.036(ReL)Od- 8361 (pr)'I3 ... for turbulent flow k Take the following air properties at 50°C: p = 1.0877 kg/m3, k = 0.02813 W/m°C, c, = 1007.3 J7kg°C, p = 2.029 x I@' kg/ms and Pr = 0.703. (pou) Solution. Given: L = 1.5 m. B = 1 m, !,= W0C, t, = 10°C, Q = 3.75 kW, Mean film temperature, t, = (90 + 10)/2 = 50°C Velocity o f flow, U: N U = -hL --

m e properties of air at 10°C

= h,(5 - t,) = ( L x B) (t, 3.75 x lo3 = h x (1.5 x 1) x (90 10)

Q

-

- t,)

The Reynolds number for the entire wing is

Assuming that the critical Reynolds number is 5 x lo5, we have Uxc 125xxc Re, = 5 x lo5 = -;- = 14.16 x lo4 or

xc =

5 x 10' x 1 4 . 1 6 ~l a 6 = 0.0566 125

Thus we assume the flow is either a combination of laminar and turbulent or only turbulent. Considering the former, we have

The value indicates that the flow must be turbulent. Considering the flow to be parallel to the length of the plate, we have

;.

Re, = Re, =

(~~~)1'0'8

= 1246790

* C1

= 1246790

.

,

Example 7-40. An aeroplaneflies with a speed of 450 k M at a height where the surrounding air has a temperature of 1°C and pressure of 65 cm- of Hg. The aeroplane Wing idealised as a jlat plate 6m long, 1.2 m wide is maintained at IQC. If thexow is made parallel to the 1.2 m width calculate: (i) Hear loss from the wing;

--

Heat lost, Q = h A At = 263.89 x (6 x 1.2) x(19 - 1) = 34200 W = 343 k W (Ans.) (ii) Drag force on the wing, FD: The average friction coefficient is given by = 2.676 x 0.072 167 0.072 ' f = (=-~e,"] '[(lO.S x l ~ ~ )(10.59 ~ . x ~lo6)

]

Drag force on one side of the wing, 1 FD = x ;iPA@ (where A = one side area of the wing)

cf

Example 7.41. A square plate maintained at95OC experiences a force of 10.5 N when forced & - \ a t2S°C flows over it at a velocity of 30 d s . Assuming the flow to be turbulent and using- Colburn analogy calculate: (i) The heat transfer coeficient; (ii) The heat loss from the plate surface.

Heat and Mass Transfer

42B

Properties of air are: . p = 1.06 kg/m3, cp = 1.005 W/kg K, v = 18.97 x 1@ m2/s, PP,= 0.696. Solution. Given :FD = 10.5 N , t, = 95OC, t, = 25OC, U = 30 m/s (i) The heat transfer coefficient, % : For turbulent flow, the drag force is given by

Forced Convection

427

(ii) The percentage error if the boundary layer is assumed to be of turbulent nature from the very leading edge of the plate. Assume unit width of the plate and critical Reynolds number to be 5 x ld. 20 = 1 6 0 ~as ~ ): Take the properties of air (at

3Wl

\

/

k = 0.03638 W/m°C, v = 30.08 x lo4 m2/s, Pr = 0.682. Solution. Given: L = 800 mm = 0.8 m, U = 45 m/s, t, = 300°C, t, = 20°C. (i) The heat transferred from the entire plate, considering both laminar and turbulent boundary layers: The Reynolds number for the entire plate,

The critical distance xc from the leading edge at which transition occurs is Re, v x, =

or

L =

(g)1'1'8 = 2.478 m

U

Laminar boundary layer region: Average heat transfer coefficient,

The Reynolds number at the end of the plate,

- =0072 Average skin friction coefficient; Cf (Red0.'

The heat transfer from - the laminar portion, Q,,. = hA (2, - t,) = 44.99 x (0.3342 x 1 ) (300 - 20) = 4209.98 W (per metre width) Turbulent boundary layer region: Average heat transfer coefficient,

From Colburn analogy, we have ...[Eqn. (7.137)]

-

pcu h=-x-jf (pr12"

C

= 70.32 w/rnZOc(Ans.) (ii) Heat loss from the plate surface, Q : Q = KA At = 70.32 x (2.478 x 2.478) x (9T- 25) = 30226 W = 30,226 kW (Ans.) Example 7.42. Air at 20°CJ20ws past a 800 mm long plate at a velocity of 45 nds. If the surjace of the plate is maintained at 300°C, determine: (i) The heat rransferred from the entire plate length to &ir taking into consideration both laminar and turbulent portions of the boundqry layer.

= 0.002812 x 36618.7 x 0.88 = 90.6 w / m 2 " c The heat transfer from - the turbulent portion, Qtuh. = hA (ts - t-1 = 90.6 x [(0.8 - 0.3342) x 11 (300 - 20) = 11816.41 W Hence, total heat transferred from the plate, Qtotal= Qlam.+ Qturb.=4209.98 + 11816.41 ~r16026 W

Alternatively : Overall average heat transfer coefficient, kh = l0.036 (Re,)" - 8361 ( ~ r ) ' "

..

Q,

-

= hA(t, - t,) = 7 1 . 9 x (0.8 x 1) X (300 - 20) = 16025 W

428

Heat and Mass Transfer

'

(ii) The percentage error: If the boundary layer is turbulent from the very beginning, then average heat transfer coefficient,

429

Forced Convectibn or

-

NUU= 0.023

The above expressions are valid for

-

(Q,td)tuh, = hA (t, - t,) = 105 x (0.8 x 1) (300 - 20) = 23520 W lm6 .: Percentage error = 23520x 100 = 46.76 % increase (Ano.) 16026 B. Forced Convection Internal Flow

-

7.4. TURBULENT TUBE FLOW Some of the important relations for fully developed turbulent flow (Re > 2300) ~hrough pipes and conduits are:

...I 7.151)

(pr)'"

I

The properties of fluid are evaluated at film temperature. Example 7.43. A tube 5 m long is maintained at 100°C by steam jacketing. A fluid flows through the tube at the rate of 175 kg/h at 30°C. The diameter ofthe tube is em. Find Out average heat transfer coeflcient. Take the following properties of the jZuid: (MJJ) p = 850 kg/m3, c, = 2000 J/kg°C, v = 5.1 x I @ mws and k = 0.12 w'mOC. Solution: Given: L = 5 m, D = 2 cm = 0.02 m Average heat transfer - coefficient h:

(i) The velocity distribution : -= u =

where

local average velocity, and u,, = velocity at centre line, R = radius of the pipe, and y = distance from the wall = (R - r). J d fLii2 (ii) The head loss :hL = P 2g (where f = friction factor, u = average flow velocity] ... equally valid for turbulent flow. In case of turbulent flow through tube, it is difficult to derive simple analytical expressions for heat transfer coefficient and Nusselt number. But such expressions can be easily found out by using empirical relations. Thefricrion factor, for turbulent flow, is well represented by the following empirical relations: f = 0.316 for 2 x lo4 < Re< 8 x lo4 ...(7.146) f = 0.184 for lo4 < Re < lo5 ...(7.147) f = 0.005 + 0.396 (Re)'.' for 2 x lo4 < Re < 2 x lo6 ..(7.148) (iii) The wall shear stress, 2,:

. The mass flow per second is given by

-

2,

=

f 2 8 PUmm

...(7.149)

(iv) From Colburn analogy (0.5 < P, < 100) St

(mY3 =f 8

...(7.150) Substituting the value off from Eqn. (7.147) in Eqn. (7.150), we get following equations for heat transfer coefficients:

Since Re > 2300, the expression (i) holds good. Substituting the values, we get

Heat and Mass Transfer

430

'l'he thermo-physical properties of water at 30°C are : W/m°C, v = 0.805 x lo4 m2/s, p = 995.7 kg/m3, cp = 4.174 kJ/kg°C, k = 61.718 x Pr = 5.42. (i) The heat transfer coefficient from the tube surface to the water: VD 12 ~ 0 . 0 6 - 0.894 x lo6 Reynolds number, Re = - = v 0.805 x lo4 Since Re > 2500, hence nature of flow is turbulent. For the turbulent flow hD ...[Eqn. (7.151)] (~r)'.~~~ Nu = = 0.023 k

Forced Convection

Length of the tube required, L :

...(i)

Q = hAS e m where A, = nD,L, = n x 0.025 x L = 0.0785 L m2

Also,

Q = m cp (to- ti)

1

The properties of water should be taken at

I

25OC

= 23959.6 w/rn2'c (Ans.) (ii) The heat trawferr9, Q:

For find h we have to use the given empirical relation.

he heat transferred,

Fig. 7.22

The mass flow rate is given by, 71

71

4

4

h= 0.8333 = - D~~x V X p = - x 0.025~x V x 977.8 = 4230345 W

(Am.)

(iii) The length of the tube, L: Q = h A (t, - t,) 4230345 = 23959.6 x (71 x 0.06 x L) (70 - 30) or 4230345 = 23.4 m (Ans.) L= 23959.6 x 71 x 0.06 (70-30) Example 7.45. Water I S jlowing at the rate of 50 kg/min through a tube of inner diameter 2.5 cm. The inner surj4ace of the tube is maintained at 100°C. I f the temperature of water increases from 25°C and 55°C find the length of the tube required. The following relation may be used : Nu = 0.023 (~r)'.~. The properties of water can be taken from the following table : t ("c)

GX

pxld

@dm3)

(JAg-K)

(W/m-K)

(kg/m-s)

40

992.2,

4174

63.35

652

50

988.1

4178

64.74

550

60

983.2

4182

65.90

470

70

977.8

4187

66.72

405

80

971.8

4195

67.41

335

P

CP

(B.U. Winter, 1998) 50 Solution. m = 50 kg/min = -= 0.8333 k d s ; Di = 2.5 cm = 0.025 m; 60

x 4187 = 2.54 p r =p 9 = 405 x k 66.72 x 10-

Now substituting the values in the given empiiical formula, we get

Now substituting the values in (i), we get 104671 = 9268.2 x 0.0785 L x 58.7 104671 L= = 2.45 m (Ans.) 9268.2 x 0.0785 x 58.7 Example 7.46. Water at 25°C jlows across a horizontal copper tube 1.5 cm OD with a velocity of 2 d s . Calculate the heat transfer rate per unit length if the wall temperature is maintained at 75°C. Given properties of water : p = 988 kg/m3 k = 0.648 W/m K

Forced Convection

Heat and Mass Transfer

432

Heat transfer per unit length of the tube:

(AMIE Summer, 1997) Solution. Given : Al = 75 - 25 = 50°C; D = 1.5 cm = 0.015 m, U = 2 m/s;

p = 988 kg/m3; k = 0.648 W/m K ; p = 549.2 x

N dm2;

cp= 4.174 M/kg K. Heat transfer rate per unit length, &.

L'

Reynolds number, Re = Prandtl number,

Pr =

Heat transfer per unit length of the tube, Q = Z A , A ~ = Z X ( n x ~1 ) ~t = 62.8 x (n x 0.0254 x 1) x 20 = 100.22 W (Ans.) Bulk temperature increase (At)b, over a 3 m length of the tube:

pUD = 988 x 2 x 0.015 = 53969.4 p 549.2 x pc 549.2 x = k

x 4 . 1 7 4 ~lo3 = 3.5376 0.648

Q = mcp (At), = (PAW cp (At),

I

(3 x 100.22) = (1.493 x

x 0.02,

x 10) x 1025 x (At),

3 x 100.22 x 4 = 38.77"C (Ans.) 1.493 x n x (0.0254)' x 10 x 1025 Example 7.48. A fluid is flowing in a pipe which is 300 mm in diameter, 3.5 m long and whose sulface in maintained at a constant temperature. The temperature of the wall surface at the inlet section of the pipe exceeds the fluid temperature by 40°C. What is the rise in the fluid temperature at the end section of the pipe? or

(At)b =

The Reynolds analogy holds good which is given by St = 8 where friction factor f = 0.022. Solution. Given: D = 300 rnm = 0.3 m, L = 3.5 m, f = 0.022, (t, - ti) = 40°C Rise in fluid temperature at the end section of the pipe (to ti) : The energy balance yields Q = h A (t, - t,) = m cp (to - ti)

-

NOW

Q=7i;xmX~t L = 12938.4 x n x 0.0 15 x 50 = 30485.4 W/m or 30.485 kW/m (Ans.)

Example 7.47. Air entering at 2 bar pressure and bulk temperature of 200°C is heated as it flows through a tube with a diameter of 25.4 mm at a velocity of 10 m/s. Calculate the heat transfer per unit length of the rube if constant h e a t f l u condition is maintained at the wall and wall temperature is 20°C above the air temperature all along the length of the tube. How much would the bulk temperature increase over a 3 metre length of the tube? , Take the properties of air as: p = 1.493 kg&, p = 2.57 x lr5~ s / m ' , k = 0.0386 W/m°C, c,, = 1025 J/kg°C Use the relation : Nu = 0.023 ( ~ e ) (" ~~ r ) " ~ Solution. Given: D = 25.4 mm = 0.0254 m; U = 10 mls

or

h ( n ~ ~ ) [ t , - =y ]

Dividing throughout by npVc, D, we have

-

...substituting, St = f8 Inserting the values, we get

Heat and Mass Transfer

0.0048 [(t, - to) + 401 = 0.075 [40 - (t, - to)] 0.0048 (t, - to) + 0.192 = 3 - 0.075 (t, - to) (t, - to) (0.0048 + 0.075) = 2.808

or or or

:.

Rise in bulk temperature of air = 63.06 - 40 = 23.06OC (Ans.) Q = 0.0157 x (1.009 x lo3)x (63.06 - 40) = 365.3 W/m (Ans.)

(Ans.) Example 7.49. Air at 2 bar and 40°C is heated as it jlows through tube of diameter 30 mm at a velocity of 10 rnk Calculate the heat transfzr per unit length of the tube when wall temperature is maintained at 100°C all along length of the tube. How much would be the bulk temperature increase over one metre length of the tube ? Use the following relation : Nu = 0.023 ~ 8P P. . ~ ~ +

= 70°C are :

~ = 2 0 . 6 x 1 0 N- ~ dm2; Pr =0.694; cp= 1.009kJ/kg°C; k=0.0297kg/m°C; (M.U., 1992) Solution. Given :p = 2 bar = 2 x 16 ~/rn'; ti = 40°C, D = 30 mm = 0.03 m;

v= 1 0 d s ; t,= 1oo0c p = pRT

We know that

or

P 2' ~

to = 63.06OC

or

The rise in the temperature of the fluid at the end, (to - t,) = 0, - ti) - 0, - to) = 40 - 35.18 = 4.82OC

The properties air at loo 2

Forced Convection

~

7.5. EMPRICAL CORRELATIONS FOR FORCED CONVECTION The following dimensionless numbers are used for the usual forced convection problems: hL P LV (ii) Reynolds number, Re = (i) Nusselt number, Nu = -; k P (iii) Prandtl number, Pr =

PC k '

h (iv) Stanton number, St = P cnv r

In order to determine the value of convection coefficient h, the following conventional generalised basic equations are used: Nu = f, (Re, Pr) = C, (Re)m(Pr)" ; St = f2 (Re, Pr) = C2 (Re)a ( ~ r ) ~ The values (numerical) of the constants and exponents are determined through experiments. The properties of thefluid are evaluated on the basis of bulk temperature (unless stated otherwise).

A. Laminar Flow 7.5.1. Laminar Flow over Flat Plates and Walls ( a ) The local value of heat transfer coeficient is given by h .x = 0.332 ( ~ e , ) ' . (~ ~ r ) ' . ~ ~ ~ ...Blasius equation Nu, = k The average value of heat transfer coeficient is given by

...(7.153)

Ux PC Re, = -, ReL = UL and Pr = v v k The above equations are valid for the following: (i) All fluids (Pr 2 0.6) except liquid metals; (ii) Reynolds number Re 2 40000; where,

Pr = 0.696

...(Given)

Nu = 0.023 ( ~ e ) '(,~~r ) ' . ~

t, (= 40°C)

= 0.023 (32417)O.~ (0.696)O.~ = 80.79

Fig. 7.23

(iii) The fluid properties are evaluated at the mean film temperature, (b) For liquid metals, the following correlation has been proposed: Nu, = 0.565 (Pe,) Q = hA (AMTD)= m x cp (to- ti)

71

71

where m = - D ~ = V - x ~0.03~x 10 x 2.226= 0.0157 kg/s 4 4

t, + L tf = 2 . ...(7.155)

where Re, = Re,. Pr The above equation is valid for the following : (i) Pr I 0.05 (ii) The fluid properties are evaluated at the film temperature.

7.5.2. Laminar Flow Inside Tubes (a) For fully developed flow :

64 f =Re

..(7.156)

Heat and Mass Transfer

436 ,

Forced Convection

437

(iii) 0.7 I Pr 5 160 (iv) Fluid properties are evaluated at the mean bulk temperature. 2. Colburn suggested the following correlation : Nu = 0.023 (pr)" ...(7.165) ...(7.166) St = 0.023 ( ~ e ) - ' .(~ ~ r ) ~ ~ ... in terms of staton number It is valid for the following : (i) St is evaluated at the mean bulk temperature. (ii) Re and Pr are evaluated at the mean film temperature. L (iii) 5 2 60, Re 2 lo4 , 0.7 < Pr < 160.

(b) For uniform heat flux : Nu = 4.36 ...(7.157) (c) For constant wall temperature : ...(7.158) Slug flow : Nu = 5.78 ...(7.159) Fully developed flow : Nu = 3.66 (4 For average Nusselt number for flow inside tubes, the following correlations, have been developed:

This equation is valid (i) for constant wall temperature and fully developed flow, (ii) when properties are evaluated "at the bulk temperature, (iii) tube length is much greater than- pameter, (iv) when Re and Nu are calculated on the basis of pipe diameter as the length parameter. (v) when 0.5 < Pr < 100.

,

.

3. The Mc Adams and Colburn correlationsl are fairly accurate for small to moderate temperature difference (lO°C in case of liquids and 50°C for gases). For larger temperature differences, Sieder and Tate proposed the ioll~wingcorrelafion: = 0.023

(pr)"

...(7.167)

This equation is valid for the following:

L

(i) - > 60, Re 2 lo4, 0.7 I Pr I 16700 D (ii) All fluid properties except ps are evaluated at the mean bulk temperature, psis evaluated at the surface temperature. 4. Desman and Sams suggested the following equation for very large temperature difference (t, -"tb) with air: Nu = 0.026 (~r)'.~ ....(7.168) The equation is valid for the following:

This equation is valid for (i) short tubes

[$> 2)

(ii) Re < 2100

ii) 0.48 < Pr < 16700 (iv) 0.0044 < (~'CL,)< 9.75 (v) In this equation all properties except p, are evaluated at bulk temperature; pJis evaluated at surface temperature.

B. Turbulent Flow 7.5.3. Turbulent Flow Over Flat Plate The local and average Nusselt numbers, from Colburn analogy, are given by: Nu, = 0.029 (pr)lt3 = 0.036 (pr)ll3 where the properties are evaluated at the mean film temperature. When the flow lies in the transition range, Nu = 0.036 [ ( ~ e , ) ' .-~

AVP~)"~

...(7.162)

7.5.5. Turbulent Flow Over Cylinders 1. The following empirical correlation is widely used for turbulent flow over cylinders.

-

hD = c ( ~ e ) (pr)'" " Nu = -

...(7.163)

7.5.4. Turbulent Flow in Tubes The following correlations for fully developed turbulent flow in circular tubes have been proposed : 1. Mc Adams has proposed the following general correlation on the basis pf Nu = @ (Re, Pr) for heating and cooling of all -fluids with some limitations as given bdow : Nu = 0.023 (Pr)" ...(7.164) where n = 0.4 ... for heating n = 0.3 ... for cooling

L

1 60

(ii) 1 x lo4 < Re < 12 X 104

...(7.169)

k

A = 18700Lwhen Re, 4 x lo5 A = 23100 when Re, = 5 x lo5

where

(i)

ts (i) - upto 3.55 (ii) Re 2 lo4 tb (iii) Re is evaluated at mean film temperature (iv) Nu and Pr are evaluated at mean bulk temperature.

where C and n are constants and have the values as given in the table 7.2 below:

Table 7.2. Constants for Eqn. (7.169) for Flow across Cylinder (Hilpert, 1933; Knudsen, 1958) c

S.No.

Re

C

n

1.

0.4 to 4

0.989

0.330

2.

4 to 40

0.911

0.385

3.

40 to 4 x lo3

0.683

0.466

4.

4 x lo3 to 4 x

lo4

0.193

0.618

5.

4 x lo4 to 4

lo5

0.027

0.805

All properties are evaluated at the film temperature.

Forced Convection

Heat and Mass Transfer

438

2. Churchill and Bernstein have suggested the following empirical correlation which covers the entire range of Re and wide range of Pr :

The average heat transfer coefficient is given by

This equation is valid for the following:

(i) Re . Pr > 0.2 (ii) All properties are to be evaluated at the film temperature.

Heat transfer from both sides of the plate, per metre width, Q = 2 [ Z A , (t,-t,)] = 2l18.88 x (0.35 x 1 ) (605 - 15)] = 7797.44 W (Ans.) Example 7.51. A surface condenser consists of two hundred thin walled circular tubes (each tube is 22.5 mm in diameter and 5 m long) arranged in parallel, through which water flows. I f the mass flow rate of water through the tube bank is 160 kg/s and its inlet and outlet temperatures are known to be 21°C and 29OC respectively, calculate the average heat transfer coeficient associated with flow of water. Solution. Given: D = 22.5 mm = 0.0225 m, L = 5 m, t, = 21°C, to = 2g°C. Average heat transfer coefficient, h:

7.5.6. Turbulent Flow over Spheres 1. For flow of gases over spheres, Mc Adams suggested the following correlation: ...(7.171) ...for 25 < Re < 1 x lo5 Nu = 0.37(~e)O.~ Fluid properties are to be evaluated at the film temperature. 2. Kramers proposed the following correlation for flow of liquids past spheres: ...(7.172) Nu = [0.97 + 0.68 ( ~ e ) ~ . ~ ] ( ~ ... r for ) ~ I. <~ Re < 2000 The fluid properties are to be evaluated at the film temperature. 3. Whitaker has proposed a single equation for flow of gases and liquids past sphere; his correlation is

ti + to - 21 + 29 t, = - 250C 2 2 The thermo-physical properties of water at 25°C are: p = 996.65 kg/m3; p = 0.862 x kglms; k = 0.6079 W/m°C; c, = 4.178 kJkg°C The mean bulk temperature,

for 0.71 < Pr < 300; 3.5 < Re < 7.6 x lo4; 1.0 < (&I&) < 3.2 All properties except c(, are to be evaluated at t,. Example 7.50. Air at a temperature of 15OC flows at a velocity of 6.5 m/s across a flQt plate maintained at a temperature of 605OC. Calculate the amount of heat transferred per metre width from both sides of the plate over a distance of 350 mm from the leading edge. The following relation holds good in the case of large temperature difference between the plate and the fluid:

Nu, = 0.332 (pr)'I3 (Re)u2

Prandtl number, The mass flow rate of water through each tube,

($"."'

where TTF and T, are the absolute temperatures of the plate sugace and free stream respectively and all fluid properties are evaluated at the mean $film temperature. Solution. Given: T, = 15 + 273 = 288 K, T, = 605 + 273 = 878 K, x = 350 mrn = 0.35 m. Amount of heat transferred per unit width : 605 + 15 = 310°C The mean film temperature, tf = 2 The thermo-physical properties, at 3 1O°C are: p = 0.614 kg/m3; c,, = 1.046 kJ/kg°C k = 0.04593 W/m°C; p = 29.7 x lo4 kglms; Pr = 0.675 0.614 x 6.5 x 0.35 Re, = pux = 47032 Reynolds number, CL 29.7 x lo4 The Nusselt number is given by

As Re >> 2500, therefore, the nature of flow is turbulent. Since the water is being heated in condenser, hence we may use McAdams correlation Nu = 0.023 ( ~ e ) ' (. ~ r ) ' . ~ ...[Eqn. (7.164)] = 0.023 (52518.26)O.~ (5.9241O.~ = 279.9 or

-

Nu = hD = 279.9 k

Example 2.52. Water at 20°C with a flow rate of 0.015 kg/s enters a 2.5 cm ID tube which is maintained at a uniform temperature of 90°C. Assuming hydrodynamically and thermally fully developed flow determine the heat transfer coeficient and the tube length required to heat the water to 70°C. [Water properties at 20°C : p = 1000.5 kg / d , cp = 4181.8 J/kg K,

Nu, = 0.332 (pr)'I3(Re)lJ2

i

Forced Convection

Heat and Mass Transfer

442

443

For calculating the tube side heat transfer coeficient use the Dittus-Boelter equation and for the shell side heat transfer coeficient, the average value may be taken as 10760 w/m2 K. Data : Properties of water at 30°C.

The thermo-physical properties o f air at 90°C are : cp = 1.009 kT/kg°C p = 0.972 kg/m3; v = 22.1 x m2/s k = 0.03127 W/m°C; Pr = 0.69 p = 22.14 x lo4 kgtms;

k = 0.659 W / m K;

p = 979.8 kg/m3

cp = 4.180 kJ/kg K;

p=0.4044

XI&'

pa s

Reynolds number,

(AMIE Summer, 1998) Solution. Given : thl = th2= t,,, = 8 0 V , t,, = 20°C; tc2= 40°C;

As Re >> 2500 the flow is turbulent. (i) The heat loss from the duct over its 4.5 m length, Q: Q = m cp At = 0.06 x (1.009 x lo3) (110 - 70) = 2421.6 W (ii) The heat flux and duct surface temperature at a length of 4.5 m: The heat flux at x = 4.5 m is calculated from the relation: n ~t ~t

V, = 1.4 m/s; h, = 10760 w / m 2 K

(Ans.)

kw= 0.659 W/m°C; p, = 979.8 kg/m3; cp, = 4.18 kJ/kg K ; pw = 0.4044 x (i) The number of tubes, N :

Pas

..... at

30°C

I

Q = mwCpw (4.2 - tcl) where, h, = 6.5 W / ~ ~ " C hi (inside convection coefficient)is calculated from the correlation Nu = 0.023 (~r)'.~~~ = 0.023 (19169.5)0.8( 0 . 6 9 ) " ~=~54.22 ~

...(Given)

Also,

= 15.277 x 4.18 x lo3 (40 - 20) = 1277157 W mw= pw x flow area x velocity

.. (where N = no. o f tubes)

15.277 x 4 = 78.97, i.e., 79 979.8 x n x (0.0134)~ x 1.4 (ii) The length of each tube, L : For two through tubes, N=

:.

Thermal resistance,

I

x (0.0134)' x 1.4

1 1 1 1 Z Rth = -+hi ho = 9.42 + 6.5 = 0.26

(Ans.)

Hence, heat flux From the resistance network,

or

Using Dittus-Boelter equation, we have

250 t = 70 - -= 43.46OC (Ans.)

9.42 Example 7.55. A horizontal tubular 1-1 condenser is used to condense saturated steam at 80°C. The condenser is a shell and tube one with brass tubes (k = 110 W/m°C) of 1.59 cm OD and 1.34 cm ID. Steam is outside tubes and cooling water enters the tubes at 20°C with a velocity of 1.4 d s and leaves at 40°C. I f the rate of cooling water supply is 55000 kg/h and the latent heat of condensation of steam at 80°C is 2304 kJ/kg, calculate : ( i ) The number of tubes; (ii) The length of each tube.

4

-- 0'659 x 0.023 (45453)O.' 0.0134

(2.565)'" = 8774 w/m2 K

(Ans.)

Heat and Mass Transfer

444

Forced Convection From (i) and (ii), we get 9869.1 1 L = 4480

- 0.454 m (Ans.) 9869.11 Example 7.57. A square channel of side 20 mm and length 2.5 m, in a heat transfer problem, carries water at a velocity of 45 d s . The mean temperature of the water along the length of the channel is found to be 30°C while the inner channel sur$ace temperature is 70°C. Calculate the heat transfer coeflcient from channel wall to the water. Use the correlation: or

NOW, where

Q = uaoO, 0, =

'1

- 2'

In (@,/€I2)

- (80- 20) - (80 - 40)

in[:

IZ]

20

- 0.4055

- 49.320C, and

\

"'

j

Reynolds number Nusselt number is given by ...(Given)

V )

Nu = 7 + 0.025 (Pr . ~ e ) ' . ~ =7

+ 0.025

(0.0248 x 49997)0.8= 14.46

I

The heat gained by mercury due to convection, Q = X A , At = 6282.87 x (n x 0.02 x L) (50 - 25) = 9869.11 L

J

(where A = cross-sectional area of channel, P = perimeter of the channel).

k h = - x 573.4 = -x 573.4 = 17703.7 w / I ~ ~ "(Ans.) c D, 0.02 Example 7 3 . Water is heated while flowing through a 1.5 cm x 3.5 cm rectangular tube at a velocity of 1.2 m/s. The entering water temperature is 40°C and tube wall is maintained at 85°C. Determine the length of the tube required to raise the temperature of water by 35°C. Use the following properties of water : p = 985.5kg/m3; k = 0.653 W/m K; v = 0.517 x m2/s; c, = 4.19 kJ/kg K. (N.M.U.,1998) Solution. Given : x = 3.5 cm = 0.035 m; y = 1.5 cm = 0.015 m; V = 1.2 m / s ; t, = 40°C; to = 75OC. Length of tube required, L : Q = hA, 8, = mc, (to- ti) ...(i) where A, is the surface area of the tube. or

..

"/,

i'le thermo-physical properties are evaluated at the mean bulk temperature except Pr, . Pr, is evaluated at the channel sur$ace temperature. Take 'epui9alent diameter as the characteristic length of the channel. The thermo-physical properties of water at 309iZ"i&e: * p = 995.7 kg/m3; k = 0.6175 W/m°C; v = 0,805 x I@ rn2/s; Pr = 5.42;PrS at 700C = 2.55. Solution. Given: V = 4.5 rnls, tb = 30°C, r, = 70°C Heat transfer coefficient, E: The equivalent diameter of the channel,

Substituting the values in the above eqn,, we get 1277157 = 4158.7 x (79 x n x,0.0159 x L) x 49.32 1277157 = 1.578 m (Ans.) L= 4158.7 x 79 x n x 0~0139X 49.32 Example 7.56. Liquid mercury flows at a rate of 1.6 kg/s through a copper tube of 20 mm diameter. The mercury enters the tube at 15°C and after getting heated it leaves the tube at 35°C. Calculate the tube length for constant heat flux at the wall which is maintained at an average temperature of 50°C. For liquid metal flowing through a tube, the following empirical correlation is presumed to agree well with experimental results -: Nu = 7 + 0.025 ( ~ e ) ' . ~ where Pe is the Peclet number; Pe = Pr.Re Solution. Given : D = 20 mm = 0.02m, m = 1.6 kgls, ti = 15"C, to = 35°C. Tube length, L: 15 + 35 t, = -= 25°C The bulk temperature, 2 The thermo-physical properties of mercury at 25OC are: p = 13582 kg/m3; k = 8.69 W/m°C m2/s; Pr = 0.0248 c, = 140 Jlkg°C; v = 1.5 x pVD p A V D Re=--------- mD - 4m - 4m The Reynolds number, CL $D2r n : D ~E ~ ( P

L=--

...(i)

Heat and Mass Transfer

446

Forced Convection

0.621 ~ ( 4 . 1 103)x35 9~ = 6.57 m (Ans.) 5946.3 x 0.1 x 23.3 Example 7.59. 3.8 kg of oil per second is heated from 20°C to 40°C by passing through a circular annulus with a velocity of 0.3 m/s. The hot gases at 400°Care passed through the inside tube of 100 mm diameter and are cooled to 100°C. Find the length of the pipe required for the above heat transfer process assuming the gas is flowing in opposite direction to the oil ? Take the following properties of oil and gases at mean temperature : Oil Gases

L=

(where A, = area of cross-section).

l = 3 . 5 cm

P

v

800 kg/m2 8 x ~ O m2/s - ~ 3350 J/kg K

0.8 kg/m3 32.8 x 1 r 6m2/s 1050 W/K

c~ k 0.2 W/m°C 0.035 W/m°C Solution. Given :m, = 3.8 kg/s; t,, = 20°C; tC2= 40°C; V, = 0.3 m/s;

(P.U.)

t,,, = 400°C; th2= 100°C; d = 100 mm = 0.1 m [The suffix c stands for cold (i.e., oil) and suffix h stands for hot (i.e., gases)]. Length of the pipe required, L : The diameter of the annulus is calculated as follows : Fig. 7.26 = 0.035 x 0.015

x 1.2 x 985.5 = 0.621 kg/s

The heat transfer co-efficient h for the given configuration is given by :

Nu =

5 = 0.023 (I?e)O.' (~r)"" k

Characteristic length,

4 4 P

LC= -=

[Eqn. (7.151)]

L

For oil : For annulus,

A

D,, = (D - d)

...[Eqn. [6.28)1

4 ( 1 x b ) -- 21b 2 (1 + b) - (1 + b)

Nu = 0.023 (~e)'.' (pr)'l3 = 0.023 (2764)O.' (107.2)~'~ = 61.9

x (4.19 x lo3)= 3.27 p r =P-Ce =PC A = 985.5x0.517 x k k 0.653 Substituting the values in (ii), we get 0'021 = 0.023 (0.487 x 105)0.8(3.271°." = 191-23 0.653

Now substituting the values in (i), we get

For gases : Heat lost by gases = heat gained by oil mh x I050 x (400 - 100) = 3.8 x 3350 x (40 - 20) mh = 0.808 kg/s .. The velocity of hot gases (Vh)can be calculated as follows :

...[Eqn. [7.151)]

Heat and Mass Transfer

448

Forced Convection

449

Solution. Given : D = 19 mm = 0.019 m; V = 0.061 m/s; L = 1.5 m ( 1 Pas = 1 N-s/m2)

Overall heat transfer coefficient, x, (velocity depth) = 0.05 ReD= 0.05 x 1720.4 x 0.019 = 1.634 m

Log-mean temperature difference,

xt (temperature depth) = x,

x P,

= 1.634 x 6.5 = 10.62 m

The tube length L is less than x,, as well as x,, so both depths are developing.

(Ans.)

Example 7.61. Liquid metal flows at a rate of 270 kg/min through a 5 cm diameter stainless steel tube. It enters at 415OC and is heated to 440°C as it passes through the tube. The tube wall tenzperature is 20°C higher than liquid bulk temperature and a constant heat flow is maintained along the tube. Calculate the length of the tube required to effect the transfer. For constant wall temperature, ...(i) Nu = 5 + 0.025 (pe)08 For constant heat f l u , ...(ii) Nu = 4.82 + 0.0185 (pe)Og2' Use the following fluid properties : p = 1.34 x I O - ~kg/m-s, cp= 149J/kgK, and

(P.U. Winter, 1997)

Pr = 0.013, k = 15.6 W/m K.

Example 7.60. Gasoline at the mean bulk temperature of 27Ocjows inside a circular tube of 19 mm inside diameter. The average bulk velocity is 0.061 i d s and the tube is 1.5 m long. The $ow starts at the heated tube inlet (no upstream developing section) and the tube sur$ace temperature is constant at 38OC. Determine the average heat transfer co-eficient over 1.5 m length of the tube. Are the temperature and velocity profies developing or developed in the 1.5 m length of the tube ? Take the following fluid properties : CL, = 5,223 x 1 F 4 Pas; pb= 5.892 x 1 F 4 PU S; kb= 0.1591 w/mOc;

Solution. Given : h = u 60 0= 4.5 kg/s; D = 5 cm = 0.05 rn; At = 440 - 415 = 25OC Length of the tube required, L : The temperatures of the wall and flowing metal along the tube are shown in Fig. 7.27. We have to use the eqn. (ii) as it is a constant heat flux condition to the tube. Q = h c p A t = 4 . 5 x 149x25

pb = 874.6 kg/m3; cpb= 1757 J/kg K; Ph = 6.5 Use the following formula :

= 16762.5 W

460°C 440°C

435°C 415°C

Fig. 7.27

x, (entry length for fully developed velocity profile) = 0.05 Re.D, and

f

i

x, (entry length for fully developed temperature profile) = x,, . Ph

(M.U., 1997)

Heat and Mass Transfer

450

Now,

Nu = 4.82 + 0.0185 (Pe)0.827 = 4.82 + 0.0185 (1094.6)O.~~~ = 10.85

i.e.,

-hD

k

Further, z.e.,

- 10.85

Q = hA (At)= 16762.5

(calculated above)

16762.5 = 3385.2 x (nx 0.05 x L) x 25

Forced Convection

Nu = 0.44 (~e)"' For 10
Example 7.62. Assuming that a man can be represented by a cylinder 350 mm in diameter and 1.65 m high with a sulface temperature of 28°C. Calculate the heat he would lose while standing in a 30 km/h wind at 12°C. Solution. Given : D = 350 mm = 0.35 m, L = 1.65 m, t, = 28"C, t, = 12°C.

Heat lost by the man, Q : 28 + 12 - 2ooc t - ----f 2 The properties of air at 20°C are: k = 2.59 x W/m°C, v = 15.0 x lo4 m2/s, Pr = 0.707. UD 8.33 0.35 Reynolds number, Re = -= = 1.94 105 v 1 5 . 0 ~lo4 Using Eqn. (7.169), we have The film temperature,

k h = - . C (Re)" (pr)'I3 D where C = 0.027 and n = 0.805 (From table 7.2) Substituting the value in the above equation, we get

:.

h A, (t, - t,) = 32.15 x (n x 0.35 x 1.65) (28 - 12) or Q = 933.26 W (Ans.) Example 7.63. A copper bus bar 25 mm diameter is cooled by air (in cross-flow) at 30°C and flowing past the bus bar with a velocity of 2.5 d s . the sugace temperature of the bar is not to exceed 85OC and resistivity of copper is 0.0175 x 1Qd ohm-m3/m, calculate the following: (i) The heat transfer coeflcient from the sutjGace to the air; (ii) The permissible current intensity for the bus bar. The following empirical correlations may be applicable for a single cylinder placed in cross-flow : Heat lost by the man,

Q =

(ii) The permissible current intensity for the bus bar, I: Heat dissipation to air = h A, (t, - t,) = 33.6 x (n x 0.025 x 1 ) (85 - 30) = 145.14 Wlm The heat generated in the bus bar

From ( 1 ) and (2), we have 35.65 x lo4

P

= 145.14

Example 7.64. Air stream at 24OC is flowing at 0.4 d s across a 100 W bulb at 130°C. & the bulb is approximated by a 65 mm diameter sphere, calculate: (i) The heat transfer rate, and (ii) The percentage of power lost due to convection. Solution. Given: t, = 24°C t, = 130°C, D = 65 min = 0.065 m, U = 0.4 mls (i) The heat transfer rate, Q: The film temperature, The properties of air at 77°C are: k = 0.03 W/m°C; v = 2.08 x lo-' m21s, Pr = 0.697 UD 0.4 x 0.065 = 1250 Reynolds number, Re=-= 2.08 x l o 5

hD % = 0.37 ( ~ e ) ' =. ~T

Equation (7.171) gives the Nusselt number as

Heat and Mass Transfer

452

-- 0'03 x 0.37 (1250)'.~= 12.32 0.065

:.

The heat transfer rate, Q = % A , (t,-t,)

=Lx7

1

w/~"c

,

x ~(tS-t,) ~

Forced Convection

453

layer on the four walls is turbulent, the heat transfer takes place only from the four suq4aces and the wall surjGace of the truck is maintained at 10°C. Neglecting heat transfer from the front and back and assuming the flow to be parallel to I0 m longside, calculate the following: (i) The heat loss from the four surj5aces; (ii) The tonnage of refrigeration; (iii) The power required to overcome the resistance acting on the four sudaces. 50 + 10

The properties of air (at )i = 2 = 30°C) are: p = 1.165 kg/m3, cc = 1.005 kUkg°C, k = 0.02673 w/m2"c, v = 16 x (ii) The percentage of power lost due to convection:

m2/s, Pr = 0.701.

Solution. Given : U = 90 kmlh = 903600 loo0 = 25; d s , tr., = 50°C, ts = 10°C, L = 10 m ,

%age of power lost = 17'33 x 100 = 17.33% (Am) 100 Example 7.65. Compare the heat transfer coeflcients under the following conditions, assuming that there is no change in the temperatures of the liquid and the tube wall and the flow through the tube is turbulent in character. (i) Twofold increase in the diameter of the tube; the flow velocity is mintairied constant by a change in the rate of liquidflow; (ii) Two-fold increase in the flow velocity, by varying mass flow rate. Solution. In case of turbulent tube flow, the average heat transfer coefficient is given by

B = 4m, H = 3m. (i) The heat loss from the four surfaces: The Reynolds number at the end of the side,

Evidently the boundary layer is turbulent for which hL Nu = - = 0.036 (Re)'.' ( ~ r ) ~ . ~ ~ ~ k

-

hD x u = - = 0.023 (Re)08( ~ r ) ' . ~ ~ ~ k

lod

...[Eqn. (7.141)]

...[Eqn. (7.151)]

:.

-

h = 0.023

v

(Pr)0.33

(WO'~

(D)0.2

( i ) When the flow velocity and the fluid properties remain unchanged:

Heat loss from the four surfaces, Q = 5 A At = 48.64 [2(4 + 3) x 101 x (50 - 10) = 272384 W = 272.384 kW (Ans.) (ii) The tonnage of refrigeration (TR) : 272.384 x 3600 = 70 TR (-: 1 TR = 14000 Wlh) The cooling capacity required = 14000 (iii) The power required to overcome the resistance, P: Average skin friction coefficient

This shows that the heat transfer coeflcient decreases to 0.87 when there is a twofold increase in the diameter of the tube. (ii) When the tube diameter and fluid properties remain the same: h

Drag force,

..(uO'

.. This shows that the heat rransfer is increased to 1.74 times when there is a hvo-fold increase in flow velocity. Example 7.66. A refrigerated truck is moving on a high way at 90 km/h in a desert area where the ambient air temperature is 50°C. The body of the truck may be considered as a rectangular box measuring 10 m (lengrh)x 4 m (width) x 3 m (height). Assume that the boundary

F, =

cfx 1 pAu2

1 x ~ 1.165 x x [2 ( 4 + 3) x lo] ~ = 2.62 x = 133.54 N P = FD x U = 133.54 x 25 = 3338.5 W = 3.3385 kW (Ans.)

2

Example 7.67. 800 kg/h of cream cheese at 15°C is pumped through a tube 100 mm in diameter, 1.75 m long and maintained at 95°C. Calculate : ( i ) The temperature of cheese leaving the heated section; (ii) The rate of heat transfer from the tube to the cheese. Use the following correlation for larninarflow inside a tube :

5

~

Heat and Mass Transfer

Forced Convection

455

I f 12 percent of the power dissipation is lost through the insulated end pieces of the cylinder, calculate the experimental value of the convective heat transfer coeflcient. Compare this value with that obtained by using the correlation : L\

A

The thermo-physical properties of cheese are: p = 1150 kg/m3; p = 22.5 kg/ms k = 0.42 W/m°C; cp = 2.75 kJ/kg°C Solution. Given: m = 800 kg&, D = 100 rnm = 0.1 m, L = 1.75 m, t, = 95"C, t, = 15°C. (i) The temperature of cheese leaving the heated section, tz : Velocity of cheese flowing through the tube,

\

/

where all thermo-physical properties except Pr, are evaluated at the mean bulk temperature lfree stream) of air. Pr, is evaluated at the average temperature of cylinder. The thermo-physicAl properties of air at 25°C are: k = 0.0263 W/m°C; v = 15.53 x I@ m2/s; Pr = 0.702 Pr, at 130°C = 0.685.

Solution. Given: t, = 25"C, U = 16.5 m/s, Power dissipation by heater = 100 W, D = 20mm = 0.02 m. (a) Heat flow from the heater to the air flowing past it is given by Q = h ~ At,

Reynolds number

100

or

Obviously the nature of the flow is laminar and the given correlation is valid for the given flow situation.

[1-g) -

=hx

Prandtl number,

(71

x 0.02 x 0.12) x (130 - 25)

= 111.16 W / ~ ~ " (Ans.) C (b) Rcynolds number Using the correlation given, we have

-

Nu = 0.26 ( ~ e ) ' . ~ - t l +tz 15 + t2 Mean bulk temperature of cheese, tb = -- 2 - 2 Now, heat gained by cheese = convective heat flow from the tube to cheese, or or t2 = 29.82"C (ii) The rate of heat transfer from the tube to the cheese, Q : Q = 71 A, (t, - tb) = m cp (t2 - t l )

(Ans.) Example 7.68. The following data relate to a metallic cylinder of 20 mm diameter and 120 mm in length heated internally by an electric heater and subjected to cross flow of air in a low speed wind tunnel: Temperature of free stream = 25°C; Velocity of free stream air = 16.5 m/s; Average temperature of cylinder su$ace = 130°C; Power dissipation by heater = 100 W.

h =

0'0263 x 90.93 = 119.57 w / ~ ~(Ans.) ~ c x 90.93 = D 0.02

HIGHLIGHTS Important Formulae A. Laminar Flow I. Flow over flat plate: If

= 9056.67 W

k

If

v < 5 x los

...boundary layer is laminar (velocity distribution is parabolic)

Ux - > 5 x lo5 ...boundary layer is turbulent on that portion (velocity distribution v

follows log law or a power law)

(i) Displacement thickness,

6* =

(ii) Momentum thickness,

0= 0

Heat and Mass Transfer

462

\

/

The thermo-physical properties are evaluated at the mean bulk temperature except Pr,; Pr, is evaluated at the channel surface temperature. Take equivalent diameter as characteristic length of the channel. The thermo-physical properties of water at 30°C are : p = 995.7 kg/m3,k = 0.6175 W/m°C; v = 0.805 x lo4 m2/s; Pr = 5.42; Pr, at 70°C = 2.55 [Ans. 25456.6 w/m2 OC] 31. A copper bus bar 20 mm diameter is cooled by air (in cross-flow) at 30°C and flowing past the bus bar

with a velocity of 2 m/s. If the surface temperature of the bar is not to exceed 80°C and resistivity of copper is 0.0175 x lo4 ohm-m3/m, calculate the following: ( i ) The heat transfer coefficient from the surface to the air; (ii) The permissible current density for the bus bar. The following empirical correlations may be applicable for a single cylinder placed in cross-flow : Nu = 0.44 ( ~ e ) ' ~ For 10 < Re < lo3 Nu = 0.22 ( ~ e ) ' . ~ For lo3 < R e < 2 x lo5 The thermo-physical properties are evaluated at t, (30°C) and are given as : [Am. ( i ) 32.14 w/m2 OC (ii) 1346 amp.] k = 0.02673 W/m°C v = 16 x lo4 m2/s

Free Convection 8.1. Introduction. 8.2. Characteristic parameters in free convection. 8.3. Momentum and energy equations for laminar free convection heat transfer on a vertical flat plate. 8.4. Integral equations for momentum and energy on a flat plate - velocity and temperature profiles on a vertical flat plate - solution of integral equations for vertical flat plate - free convection heat transfer coefficient for a vertical wall. 8.5. Transition and turbulence in free convection. 8.6. Empirical correlations for free convection. 8.7. Simplified free convection relations for air. 8.8. Combined freeand forced convection. Typical Examples . - Highlights - Theoretical Questions - Unsolved Examples.

8.1. INTRODUCTION When a surface is maintained in still fluid at a temperature higher or lower than that of the fluid, a layer of fluid adjacent to the surface gets heated or cooled. A density difference is created between this layer and the still fluid surrounding it. The density difference introduces a buoyant force causing flow of fluid near the surface. Heat transfer under such conditions is known as free or natural convection. Thus "Free or natural convection is the process of heat transfer which occurs due to movement of the jluid particles by density changes,associated with temperature differential in a jluid." This mode of heat transfer occurs very commonly, some examples are given below: (9 The cooling of transmission lines, electric transformers and rectifiers. (ii) The heating of rooms by use of radiators. (iii) The heat transfer from hot pipes and ovens surrounded by cooler air. (iv) Cooling the reactor core (in nuclear power plants) and carry out the heat generated by nuclear fission etc. - In free convection, the flow velocities encountered are lower compared to flow velocities in forced convection, consequently the value of convection coefficient is lower, generally by one order of magnitude. Hence, for a given rate of heat transfer larger area could be required. As there is no need for additional devices to force the liquid, this mode is used for heat transfer in simple devices which have to be left unattended for long periods. - The rate of heat transfer is calculated using the general convection equation given below: ...( 8. 1) Q = h A (t, - t,) where Q = heat transfer, W h = convection coefficient, w/m2"c A = area, m2, and t, = temperature of fluid at distances well removed from the surface (here the stagnant fluid temperature).

8.2. CHARACTERISTIC PARAMETERS IN FREE CONVECTION It has been observed that during heat transfer from a heated surface to the surrounding fluid, the fluid adjacent to the surface gets heated; this results in thermal expansion of the fluid and

Heat and Mass Transfer

464

reduction in its density (compared to the fluid away from the surface). Subsequently a buoyancy force acts on the fluid causing it to flow up the surface, and in the neighbourhood of the plate, hydrodynamic and thermal boundary layers are set up. Since here the flow velocity is developed due to difference of temperatures, the two boundary layers are of the same order irrespective of Prandtl number. - A property that comes into play in free or natural convection is the coeflcient of thermal expansion of the fluid defined by

Free Convection

In free convection, the additional force is the body force pg and hence the momentum equation gets modified to

...(8.6) In free convection, outside the boundary layer, (y -,00) as p tends to p,, whereas u and v approach zero. So Eqn (8.6) becomes

For an ideal gas and hence -

Since in free convection heat transfer coefficients are low and Reynolds number is not an independent parameter, a new dimensionless grouping plays the major role (in free convection) which incorporates the coefficient of thermal expansion P in the expression. This dimensionless grouping is called the Grashof number, expressed as

(i.e. the change in pressure on a height

ah is equal to the weight per unit area of the fluid element). Subtituting Eqn. (8.7) in Eqn. (8.6), we

Fig. 8.1 : Free-convection boundary layer over heated vertical plate.

get

L = characteristic length A t = (t, - t,), where t, and t, are the surface temperature and temperature of the surrounding fluid respectively. The role of Grashof number is the same in free convection as that of Reynolds number in forced convection. The critical Grashof number for the flow of air over a flat plate has been observed to be 4 x lo8 (approximately). In general, ...(8.4) Nu = f (Gr, Pr) = C(Gr)a(prlb In several cases, the above relation simplifies to the form Nu = C (Gr . Pr)m ...(8.5) Hence, a new dimensionless group is often used called Rayleigh number viz., Ra = Gr . Pr The product is also a criterion of laminar or turbulent character of the flow as determined by its values. Thus lo4 < Gr Pr < lo9 ... for laminar flow G r Pr > lo9 ... for turbulent flow

...(8.8)

where

;

Further the density difference (p, of expansion p

- p) may be expressed in terms of the volume coefficient

Thus so that Eqn. (8.8) beocmes

The energy equation for free convection boundary layer remains the same as forced convection at low velocities, viz.,

8*4* INTEGRAL EQUATIONS FOR MOMENTUM AND ENERGY ON A FLAT PLAm For free convection on a flat plate, the following equations can be derived. Integral momentum equation:

- lw l-) The values of Gr and Pr are evaluated at the mean film temperature t -[f

8.3. MOMENTUM AND ENERGY EQUATIONS FOR LAMINAR FREE CONVECTION HEAT TRANSFER ON A VERTICAL FLAT PLATE Figure 8.1. shows a free convection boundary layer formed on a flat vertical plate when it is heated. Here, the velocity profile is different from that of forced convection boundary layer. At the surface 1 wall the velocity in zero, it increases to some maximum value and then decreases to zero again at the edge of the boundary since the free stream is at rest. Initially the boundary layer is laminar in character and at some distance from the bottom edge, it becomes turbulent. This onset of turbulence depends on the fluid properties and the temperature difference (t, - t,). In order to analyse the problem, let us consider the momentum equation for forced convection.

Integral energy equation:

8-41. Velocity and Temperature Profiles on a Vertical Fht Plate Assuming the velocity and temperature profiles to be similar at any x, the temperature profile

Heat and Mass Transfer

466

467

2m + n [T) c12C2 (x)

may be taken as t-t, t, - t- - c1+ ~2 The following boundary conditions apply: (i) At y = 0 t = t, (ii) At y = 6 t = t,

Free Convection

--

2m+n-1

C1

=3 gp (ts- t,) C# - v -(x)~-"

@)+~3@Jf

The temperature distribution is, therefore, obtained as t-t, ts - tw The velocity profile may be similarly assumed.

...(8.14)

For a similar solution to exist, both sides of these equations must be independent of x. Equating the exponents, we get 2m+n-1=m-n m+n-1 =-n 1 1 This gives m = - and n = - (Thus u1 = clxin, 6 = c2x1I4) 2 4 Inserting these values into the above equations, we get Cl2c2 -- gp (ts - t,)

84

Cl

c 2

T-v-

...(8.21)

c 2

2a -c1C2 - --

where ux is any arbitrary function with the dimension of velocity. The boundary condition are: (i) At y = 0, u = 0 (ii) At y = 6, u = 0 (iii) At y = 6,

...(8.19)

c 2

au - 0 -

...(8.23)

[

>

20 v + ' / 4 gp (t - t ) ~~=3.93[~+--)

ay

The velocity distribution is, therefore, found to be

...(8.22)

40 C2 Solving these two equations for C, and C2, we get

-'I4 [:)-In

...(8.24)

From Eqn. (8.16), the maximum velocity within the boundary layer is found to be 4 4 Urn, = - U , = - c ~ x ~ ~ ~ 27 27 which on substitution for C1 from Eqn. (8.23) gives ;JIA[gB(r;-tw)]l"(x)l. = 0.766 v 0 952 + -

u,,

(.

The resultant expression for boundary layer thickness is

-6 = c22-1 = c 2 x -314

where

X

8.4.2. Solution of the Integral Equations for Vertical Flat Plate Substitution of the velocity and temperature distributions from Eqns. (8.14) and (8.16) into Eqn. (8.12) for momentum integral gives

and substitution into Eqn. (8.13) for energy integral gives 1 d (u, 6 ) = 2a 0, - t J (t, - t-) 30 6 To solve these equations, let us assume u, and 6 as exponential functions of x 6=C# u, = C I P ; (where C,, C2, m and n are constants). Substituting these values in the above equation, we get

-

which on substitution for C2 from Eqn. (8.24) yields

... is dimensionless numbers (where Gr, =

gp ( 4 - tw) 2

v2

v and Pr = - )

a

8.4.3. Free Convection Heat Transfer Coefficient for a Vertical Wall The heat transfer coefficient may be evaluated from = - ka

($)y;O

= hp

0, - t,)

...(8.28)

Heat and Mass Tjansfer

468

Using the temperature distribution of Eqn. (8.14), we obtain

Laminar pow: NUL = 0.59 (Gr .Pr)'" Turbulent flow: = 0.10 (Gr .Pr)'"

Substituting this value in Eqn. (8.28), we get

All the fluid properties are evaluated at the mean film temperature ff=-

for (lo4< Gr . Pr < lo9) for (lo9 < Gr .Pr < 1012)

Churchill and Chu have recommended the following correlations: 0.67 (Gr ~ r ) " ~ NuL = 0.68 + for (Gr .Pr c lo9)

I I":

-

...(8.30)

From Eqns. (8.27) and (8.29) we obtain the heat transfer correlation for nzm a 1 convection for a vertical flat plate as Nu, = 0.508 (pr)ln (0.952 + pr)-'I4 ( ~ r , ) " ~ For a given Prandtl number, Nu, varies as (Gr,)'" or h, a x-"~ The local film heat transfer coefficient decreases with x. It is inversely proportional to the fourth root of x. By integration over the distance, the average heat transfer coefficient is found to be

...(8.37) ...(8.38)

...(8.39)

2

for (Gr. Pr > lo9)

...(8.40)

8.6.2. Horizontal Plates In case of an irregular plate, the characteristic length is defined as the surface area divided the perimeter of the plate. (i) The upper surface heated or the lower surface cooled: Laminar flow: NUL = 0.54 (Gr . Pr)'" for (lo5 c Gr. Pr S 2 x lo7) ...(8.4 1 ) Turbulent flow: NUL = 0.14 (Gr .pr)In for (2 x lo7 c Gr. Pr I3 x 10") ...(8.42) , (ii) The lower sulface heated or upper surface cooled: Laminar flow: NuL = 0.27 (Gr . Pr)'" for (3 x 16 c Gr. Pr 5 3 x 101°) ...(8.43) Turbulent flow: NUL = 0.107 (Gr . ~ r ) ' " for (7 x lo6 c Gr. Pr $ 11 x 10l0)...(8.44) , - I 8.6.3. Horizontal Cylinders ' For such a case, the outside diameter is used as the characteristic dimension. Mc Adam has recommended the following correlations: iarninar flow: Nu = 0.53 (Gr .Pr)'I4 for (lo4 Gr . Pr c 109) ...(8.45) Turbulent flow: Nu = 0.13 (Gr . pr)ln for (lo9 < Gr. Pr c 1012) ...(8.46) The following general correlation has been suggested by Churchill and Chu for use over a wide range of Gr.Pr. 0.387 (Gr .pr)'l6 for (lo-' c Gr.Prc 1012) (8.47) { I + ( O S S ~ / P ~ }) ~ / ' ~ The fluid properties, in all preceding equations, are determined at the mean film temperature 1

'&

-

For air with Pr = 0.7, Eqns. (8.31) and (8.33) simplify to Nu, = 0.378 ( G ~ J " ~

...(8.34)

-

...(8.35) NuL = 0.504 (~t-3"~ These result are only 5 per cent higher than the exact solution of the free convection problem obtained numerically by Schmidt and Beckrnan. Here all properties are evaluated at 'film temperature'.

8.5. TRANSITION AND TURBULENCE IN FREE CONVECTION A transition from laminar to turbulent condition in both forced and free convection is caused by thermal and hydrodynamic instabilities. It has been observed that the transition in free convection boundary layer occurs when the value of Rayleigh number approaches 10'. Thus

The analysis of turbulent free convection is very complex. However, the results obtained . from various experiments have been used to obtain the pertinent empirical relations.

8.6. EMPIRICAL CORRELATIONS FOR FREE CONVECTION The following dimensionless numbers apply for the usual fiee convection problems: hL L3P@ (i)Nusselt number Nu = (ii) Grashof number Gr = --a

k'

P C

(iii) Prandtl number Pr = -f

8.6.1. Vertical Plates and Cylinders The commonly used correlations are:

v2

,,

]

8.6.4. Inclined Plates For this case multiply Grashof number by cos 8, where 8 is the angle of inclination from the vertical and use vertical plate constants. 8.6.5. Spheres Yuge (1959) has recommended the following correlation for free convection from a sphere of diameter D. Nu = 2 + 0.43 (Gr . ~ r ) ' " for (1 c Gr. Pr c 16)and Pr = 1 ) ...(8.48) 8.6.6. Enclosed Spaces In the literature, several correlationsfor heat transfer between surfaces at different temperatures

-

--

Heat and Mass Transfer

separated by enclosed fluids are available. In this section, some correlations are presented that are per- Boundin tinent to the most common geometries. wall 9, (a) Vertical spaces Jacob has suggested the following correlations Qfor vertical enclosed air spaces shown in Fig. 8.2.

Free Convection

8.6.8. Concentric Sphere Spaces Refer Fig. 8.3. Raithby and Hollands have recommended the following correlations: ...(8.55) Q = ke 7~ (Di Do IL) (ti - to) and ke is expressed as

...[8.55 (a)]

[fr*

for (2000 GrL c 2 x 104)' .

- -

ke = Nu = (ii) k

where

...(8.49)

Fig. 8.2 Fne conveaion in vertical enclosure. k = 0.064 (&dl" for ( 2 x 104 c GrL c ' l i x lo6) ...(8.50)

Ra, = (Gr . Pr), =

8.7. SIMPLIFIED FREE CONVECTION RELATIONS FOR AIR The simplified relation for the heat transfer coefficient from different surfaces to air at atmopheric pressure, in general, is given by

ke = effective thermal conductivity,

h =

H = height of the air space, and L = thickness of the air space. Gr is based on thickness of air space, L. (6)Horizontal spaces The following correlation has been suggested by Jacob for hotizontal enclosed air spaces:

-

for (lo4 < GrL < 4 x l d )

ke = ' L = 0.068 (Grdl" (ii) Nu = k k

for (4 x l d < G p

...(8.31)

In the case of liquids contained in horizontal spaces, G1dbe;apd Dropkin have proposed the following correlation:

k

(Pr)00"07

+pry

ke Pr k = 0.386 [0.861

(RaC)lA

Turbulent

Laminar

Surface and its orientation

-h = l

Vertical plate or cylinder

-

A 2 T

h = 1.32 ( ~ t ) ' "

(lo4 c Gr. r < lo9)

(lo9 < Gr. Pr < 1 0 ' ~ )

-

h = 1.25 (At)'"

Horizontal cylinder

-h = 1.32

Horizontal plate: Heated surface facing up

for ( 3 x lo5 c G r . P r c 7 x lo9) ...(8.53)

8.6.7. Concentric Cylinders Spaces Raithby and Hollands have recommended the following correlations for long horizontal concentric cylinders:

...(8.57)

Table 8.1: Simplified Free Convection Relations for Air

...(8.52) ...[8.52 (a)]

c[$J

where C and m are constants depending on geometry and flow conditions; the significant length L is also a function of geometry and flow. Table 8.1 lists the values suggested by Mc Adarns for various geometries, orientations and flow conditions indicated by the magnitude of product (Gr . Pr). -

S.No.

ke hL (i) - = Nu = - = 0.195 (Grd1I4 k k

- 'L - 0.069 (Grdl" NU = -

1'"

L RaL 7/15 5 [(DoDir(Di7/15+DL ) a The above equations are valid ior ( l o 2 I Ra, S lo4) where

($r

(lo5 < Gr. Pr < 2 x lo7)

(lo9 < Gr. Pr < 1012)

-

h = 1.67 (~t)'"

2 x lo7 < Gr. Pr < 3 x 10")

Heated surface facing down

I /

I

I

Spheres

-

h = [2 + 0.392 (Gr)114 k

]D

( 1 c Gr < 10')

...(8.54) [In ( ~ d ~ i ) l ~ Rac = (Gr. Pr)c = p Dr3/5 R ~ L for lo2 I Ra, I lo7

where

T-7 +

and ke is related as

2n ke = ln ( D m (ti to)

-

Fig. 83. Long concentric cylinders (or spheres)

8.8. COMBINED FREE AND FORCED CONVECTION Free convection effects are negligible when Gr cc Re2, at the other extreme, free convection dominates when Gr >> Re2. There are certain situations where the free as well as the forced convection are of comparable magnitude. One such case is when air is flowing over a heated su#ace at a low velocity. A dimensionless group (GrlRe2) is used to delineate the convection regimes. The various flow regimes are characterised as follows: ... Pure free convection (GrlRe2) 1 1

Heat and Mass Transfer

472

... Mixed (free and forced) convection ... Pure forced convection.

(crlRe2)z 1 (GrlRe2)I 1 8.8.1.

Free Convection

Gr Pr = 2.208 x 101° x 0.6975 = 1.54 x 101° For this value-of Gr Pr (tubulent range)

-

External Flows The local Nusselt number Nu, for mixed convection on vertical plates is given by ...(8.58) Nu, = 0.332 (Re,)'/* ( ~ r ) ' "if (Gr,lRe~) I A ...(8.59) Nu, = 0.508 ( ~ r ) ' "(0.952 + ~r)-l"( ~ r , ) " ~if (GrxlRex2)> A and A r 0.6 for Pr I 10 where

Nu - hL = 0.10 (Gr.pr)lB (for lo9 < Gr.Pr < 1012) L- k

:.

and A = 1.0 for Pr = 100 For horizontal plates when (Gr, l~e?') I 0.083 the following equation for forced convection may be used ...(8.60) Nu, = 0.332 ~ e , prln ' ~

8.8.2. Internal Flows (i) For mixed convection in laminar flow, Brown and Gauvin recommended a correlation of the form as

...[Eqn. (8.38)]

Range of heat loss, Q = hA (t, - t,) = 17.283 x (ITX 0.18 x 1'S)~ ( 1 0 0 20)

-

= 1172.8 kYh (Ans.) Example 8.2. A cylindrical body of 300 mm diameter and 1.6 m height is maintained at a constant temperature of 36.PC. The surrounding temperature is 13S°C. Find out the amount of heat to be generated by the body per hour if p= 1.025 kg/m3; cD= 0.96kJAg°C;

I K I . Assume Nu = 0.12 (Gr.pr)lA (the v = 15.06 x 1 r 6m2/s; k = 0.0892 W/m-h-'C and P = -

298 symbols have their usual meanings). (AMIE Winter, 1997) Solution. Given :D = 300 mm = 0.03 m; L = 1.6 m; t, = 36S°C; t, = 13S°C;

(r-3)

where Gz = Graetz number = Re Pr (D/L) and &, CL, = Viscosities of the fluid at the bulk mean temperature and surface temperature respectively. (ii) For mixed convection with turbulent flow in horizontal tubes, Metais and Eckert (1966) suggest ...(8.62) = 4.69 ( ~ e ) ' (pr)O.'l . ~ ~ (Gr)'.07 (D/L)P.~~ Example 8.1. A vertical cylinder 1.5 m high and 180 mm in diameter is maintained at 100°C in an atmosphere environment of 20°C. Calculate heat loss by free convection from the surface of the cylinder. Assume properties of air at mean temperature as, p = 1.06 kg/m3, (AMIE Summer, 2000) V = 18.97 x 10- m2/s, C, = 1.004 k.//kg°C and k = 0.1042 kJ/mh°C.

k = 0.0892 Wm-h-"C;

1 298

= -K-I ; Nu = 0.12 (Gr.pr)113

The amount of heat to be generated : Grashof number, Gr =

L3gP 0 s - t,) ,J

'

Solution. Given :L = 1.5 m; D = 180 mm = 0.18 m, ts = 100°C;

CLc = 2 PVC Prandtl number, Pr = --L = 1.025 x (15.06 x 10- x 3600) x 0.96 k k 0.0892 \

Heat loss by free convection, Q : p = pv = 1.06 x (18.97 x

/

hL = 0.12 (Gr.pr)'I3 Nusselt number, Nu = k x 3600) = 0.07239 kg/rnh

Heat lost from the surface by natural convection,

Q = hA (t, - t,) = 13.478 x (n x 0.3 x 1.6) (36.5 - 13.5) = *467.5 W h

* This is the amount of heat to be generated. A '

(Ans.)

Example 8.3. A hot plate 1.2 m wide, 0.35 m high and at 115°C is exposed to the ambient still air at 2S°C. Calculate the following: (i) Maximum velocity at 180 mm from the leading edge of the plate; (ii) The boundary layer thickness at 180 mm from the leading edge of the plate;

Heat and Mass Transfer

474

m, = 1.7 p.v

--

The thermo-physical properties of air at 70°C are : p = 1.029 kg/m3; k = 0.0293 w/m°C; v = 20.02 x lo4 m2/s

2 gP 0 s - tw)

Gr, =

and

GrL = 3.745 x lo7 x

v2

(i) Maximum velocity at 180 mm from the leading edge of the plate, u-: urn = 0.766 v 0 952 +

(

-

[""'a-

,

"1'"

= 0.766 x 20.02 x lo4 (0.952 + 0.694)-In x = 0.406 m/s

...[Eqn. (8.25)]

(4'"

9.81 ~0.002915(115-25) (20.02 x 10-6)~ (.; Pr

(Ans.)

475

(v) 1dtal mass flow through the boundary, mf : Thtal mass flow through the boundary can be calculated from the following formula :

(iii) Local heat transfer coefficient at 180 mm from the leading edge of the plate; (iv) Average heat transfer coefficient over the suface of the plate; (v) Total mass flow through the boundary; (vi) Heat loss from the plate; (vii) Rise in temperature of the air passing through the boundary. Use the approximate solution. Solution. Given : t, = 115OC, t, = 25OC, x = 180 mm = 0.18 m 115 + 25 4 +t = 700C tf = The mean film temperature, 2 - * 2

Grashof number

Free ConvecX~on

]

(0.18)'~

=aV = 0.694 ...as given above)

[(P:) (Pr + 0.952)

or m, = 0.00478 kg/s (Ans.) (vi) Heat loss from the plate, & : Heat lost from the plate will be from both the sides, hence . Q = 2 ZA, (t,-t,) = 2 x 5.43 x (0.35 x 1.2) (115 - 25 ) = 410.5 W (Ans.) (vii) Rise in temperature of the air passing through the boundary, At: Heat lost, Q = m, c, At or 410.5 = 0.00478 x (1.005 x lo3) x At 410.5 or At = = 85.4S°C (Ans.) 0.00478 x (1.005 x lo3) Example 8.4. A 350 mm long glass plate is hung vertically in the air at 24OC while its temperature is maintained at 80°C. Calculate the boundry layer thickness at the trailing edge of the plate. I f a similar plate is placed in a wind tunnel and air is blown over it at a velocity of 5 m/s, find the boundary layer thickness at its trailing edge. Also determine the average heat transfer coeflcient, for natural and forced convection for the above mentioned data. Solution. Given: L = 350 mm = 0.35 m, t, = 24OC, ts = 80°C, U = 5 rn/s

ts + t, !f=---80 + 24 2 2

Film temperature,

- s2oC

The properties of air at 52OC are:

(ii) The boundary layer thickness at 180 mm from the leading edge of the plate, 6:

-X6 = 3.93

...[Eqn. (8.27)]

(0.952 + pr)'I4 (~r,)-''~( ~ r ) - l n

= 3.93 (0.952 + 0.694)"~ (3.745 x lo7)-'" (0.694)-ln = 0.0683 6 = 0.0683 x 0.18 = 0.01229 m = 12.29 mm (Ans.)

or (iii) Local heat transfer coefficient at 180 mm from the leading edge of the plate, h,: 2x hxx 0.508 (pr)ln (0.952 + pr)-lI4 ( ~ r ~= ) ' ~ Nu, =: k 6 ...[Eqn. (8.31)]

-

"

h x = - xx- - - - 6-

-

2k ij

0.35

v2

- (0.35)3x 9.81 x 3.07 x

0'02964 = 4.823 wlmtoc (Ans.)

--0.02964 10.677 (0.694)ln (0.952 + 0.694)-I" (27.532 x 107)114]

-

L3 gfl ( 4 - 1-1

1W3 (80 - 24) = 2.133 x lo8 (18.41 x 1 0 ~ ) ~ . Rayleigh number, RaL = Gr,. Pr = 2.133 x lo8 x 0.7 = 1.493 x lo8 This value of the Rayleigh number, according to Eqn. (8.36), indicates a laminar boundary layer. Using the Eqn. (8.27), we get Grashof number, Gr, =

- 0.01229

(iv) Average heat transfer coefficient over the surface of the plate, % : k h =; 10.677 (pr)ln (0.952 + ~r)-'" ( G I J ' ~ ] i

Boundary layer thickness, 6: Free convection:

6 = 0.35 [ 3.93 (0.952 + 0.7 = 15.4 mm (Ans.) Forced convection: or

...[Eqn. (8.33)]

Reynolds number,

)'I4

(2.133 x 10')-'I4 (0.7)-'~*]= 0.0154 m

Heat and Mass Ttansfer '

476

Free Conve~tion

477

So the boundary layer is laminar. 5L a=----x 0'35 = 0.00567 m = 5.67 mm (Ans.) .JRe - d9.505 x 10 Thus the boundary layer thickness in forced convection is less than that in free convection. Heat transfer coefficient h : Free convection : - hL ...[Eqn (8.33)] Nu - - = 0.677 ( ~ r ) ' "(0.952 + pr)-'I4 ( ~ 1 ) ' ' ~ L k

h, H (Nu), = -- 0.6 (16.68 x 1 0 ~ ) =" 38.34 ~~ k

k h = - x 60.378 = 28'15 lo-' x 60.378 = 4.856 W / ~ ' O C (Ans.) 0.35 L Forced convection :-

hvxB (Nu), = -- 0.27 (6.83 x 107)0.25= 24.54 k

Qv = hv x A, x (At)= 49.07 x (2 x 15 x 0.2) x (30 - 10) = 5888.4 W 1

or

-

Qh= hh x A h x (At)= 19.63 x (15 x 0.32) x (30 10) = 1884.5 W

:

28'15 k x 18133 = 14.62 w l m 2 ' ~ ( A m ) , h = - x 181.78 = or 0.35 L Thus it is seen that heat transfer coefficient in forced convect?m is muc in free convection. Example 8.5. A sheet metal air duct carries air-conditioned air at an average tempe+arure of 10°C. The duct size is 320 mm x 200 mm and length of the duct exposed to the surrounding air at 30°C is 15 m long. Find the heat gain by the air in the duct. Assume 200 mm side is vertical and top su$ace of the duct is insulated. Use the following properties : Nu = 0.6 ( G r . ~ r ) " for ~ j vertical sugace. ~for su$ace. Nu = 0 . 2 7 ( ~ r . ~ r ) " ~ horizontal Take the properties of the air at mean temperature of (30 + 10)/2 = 20°C as given below.

c,,=lOO J / k g K ; p=1.204kg/m3; p = 18.2 x 1 r 6 ~ - s / m ~ ;

(M.U. Winter, 1998) v = 15.1 x 10- m2/s; k = 0.256 W/m K and Pr = 0.71. Solution. Refer Fig. 8.4. There will be two vertical surfaces and Air conditioned one horizontal surface from which the heat will be gained by natural convection. There F / air is no heat transfer from the top surface as it is insulated. Now, (Gr), =

g w 3 (At) v2

:.

Total heat gain, Q = Qv

+ Qh = 5888.4 + 1884.5 = 7772.9 W

(Am.) Example 8.6. Airflows through long rectangular heating duct of width and height of 0.75 m and 0.3 m respectively. The outer surjoce temperature of the duct is maintained at 45OC. If the duct is exposed to air at 15OC in a cramp-space beneath a home, what is the heat loss from the duct per metre length ? use' the following correlations : (i)Top sugace : B Take LC=-= 0.375 m - 2 NuL = 0.54 (Ra$25 if 104 5 Rat 5 107 = 0.15 ( ~ a 3 if" ~lo7 ~ 5 Ra, 5 10"

(ii)Bottom surface : B LC-- = 0.375 nz - 2 N U ~ = O . ~ ~ ( R105 ~ JI~R. ~q ~S ,101° (iii) Sides of the duct :

Take

Use the following properties of air : v = 1 6 . 2 ~ 1 0 - ~ r n ~ /as = , 22,9~10-~m~/s

9.81x( 273 + 20 ) x 0 . 2 ~ x ( 3 0 - 1 0 )

k=0.0265 W/mK, p=0.0033 K-' and Pr=0.71. b

Solution. The configuration of the duct is shown in Fig. 8.5

(N.M.U. Winter, 1995)

...[Eqn. (8.36)]

I1 I

Fig. 8.4

I

Heat and Mass Transfer a

inside the

-

Free Convection

479

Example 8.7. A vertical plate measuring 180 mm x 180 mm and at 50°C is exposed to atmosphere at IO°C. Compare the free convection heat transfer from this plate with that which would result due to forced convection over the plate at a velocity equal to twice the maximum velocity which would occur in free convection boundary layer. Solution. Free convection : Velocity profile 50 + 10 The film temperature, tf = -= 30°C. 2 The thermo-physical properties at 30°C are :

-

L

-

1 = (++ 273)

Fig. 8.5 -

Boundary layer

9.81 x 0.0033 X (45 - 15) L: = 2.62 x lo9 (L:) x (16.2 x (22.9 x

For two sides : Lc=H=0.3m

..

I - 303 - 0.0033 K-I

(0.18)3x 9.81 x 0.0033 x (50 - 10) (16 x 1 0 ~ ) ~ = 2.95 x lo7 Since Gr < lo9, hence the flow in laminar.

-

Ra, = 2.62 x lo9 (0.31~ = 7.07 x lo7

For laminar flow, using the following relation, we get,

-

NU, =

-

hL

Fig. 8.6

= 0.677 (~r)'' (0.952 + ~r)-'" (~r,)"'

The heat lost from one side of the plate, QM = h A, At = 5.47 x (0.18 X 0.18) x (50 - 10) = 7.089 W Forced convection :

For top surface :

For bottom surface : \

/

The average heat transfer coefficient with forced convection if velocity is assumed equal to 2u,,

:

EL= 0.664 (Re)'' The rate of heat loss per unit length of the duct is given by,

Q=Qs+Qt+Qb = [2 x h, x (0.3 x 1) + h, x (0.75 x 1) + h,, x (0.75 x 1) (ts - fa)

-

I

= [2 x 4.23 x 0.3 + 4.127 x 0.75 x 1 + 1.75 x 0.75 x 11 (45 - 15) = 208.4 W/m

(Ans.)

(pr)ln

Heat and Mass frmsf&r

480

Heat lost due to forced - convection, Qfirced = h A, At = 7.039 x (0.18 x 0.18) (50 - 10) = 9.12 W Hence

$FreeConvection

481

Solution. Given :A = 1 x 0.5 = 0.5 m2; tp (= t,) = 130°C; t, = 20°C; m = 20 kg; cp= 400 J/kg K

-- 7.089 = 0.777 (Ans.)

9-l2 Vertical Example 8.8. Two vertical plates, each ,,late 120 mm high and at 85OC are placed in a tank ofwuter at 15°C. Calculate the minimum spacing which will prevent interference of the free convection boundary layers. Solution: Let 6 = boundary layer thickness at the trailing edge of the plate. L = 26 = the minimum spacing required. 85 + 15 The film temperature, tf = ----- = 50°C 2 The thermo-physical properties of water at 50°C are: m2/s k = 0.674 Wm°C; v = 0.556 x

-

Plate

Qforced

\ = 1.06 x lo9

:.

Gr.Pr = 1.06 x 109 x 0.709 = 0.75 x lo9 The heat transfer coefficient for the vertical plate is given by

EL= 0.59 ( G r . ~ r ) ' /for ~

(lo4< Gr.Pr < lo9) ...[Eqn. (8.37)] Convection Substituting the - values in the above eqn., we get cuttents - hL Nu - -= 0.59 (0.75 x 1 0 ~ ) " ~ Fig. 8.8 L-

k

Fig. 8.7

(ii) Initial rate of cooling the plate in OCImin. : Heat lost from both sides of the plate is given by Q = 2 h x A (ts- t,) = 2 [5.66 x ( 1 x 0.5) x (130 - 20)] = 622.6 W G r . Pr = 11.88 x lo9 x 3.54 = 42.055 x lo9 Since G r . Pr > lo9, hence the character of the flow is turbylent. For turbulent flow, using the following relation, we get

I

[

Also,

Q = mc,(At); where At is the rate of cooling.

..

At=

622'6 = O.O778OC/s = 4.668OUmin 20 x 400

(Am.)

(iii) Time required for cooling plate from 180°C to 80°C, z : The instantaneous heat lost by the plate is given by

6 = 0.0306 x 0.12 = 0.003672 m = 3.672 mm L = 26 = 2 x 3.672 = 7.344 mrn ( ~ n s . j Example 8.9. A hot plate 1 m x 0.5 m at 130°C is kept vertically in still air at 20°C. Find : (i) Heat transfer coeflcient, (ii) Initial rate of cooling the plate in OC/min. (iii) Time required for cooling plate from 180°C if the heat transfer is due to convection d y . Mass of the plate is 20 kg and cp= 409 J/kg K. Assume 0.5 m side is vertical and that the heat transfer coeflcient calculated in ( i ) above remains constant and convection takes place from both sides of the plate. 130 + 20 Take properties of air at = 75OC as : 2 m2/s (N.M.U., 5998) c, = 1007J/kgOC, p = 1.07 m2/s; k = 0.029 J / k g K; V = 19.1 x /

..

80 - 20 2 x 5 . 6 6 ~(1 ~ 0 . 5 1n(180-20) ) = - 1413.43 x (- 0.9808) = 1386 s or 0.385 h

Z=

- 20 x 400

(Am.)

Example 8.10. A square plate 40 cm x 40 cm maintained at 400 K is suspended vertically in atmospheric air at 300 K. (i) Determine the boundary layer thickness at trailing edge of the plate. (ii) Calculate the average heat transfer coeflcient using a relation

Heat and Mass Transfer

Free Convection

.

Nu = 0.516 (Gr, ~ r ) " ' ~ Take the following properties of air : v = 20.75 x lo4 m2/s; k = 0.03 W/m°C; P = 2.86 x 1 r 3 K-*;Pr = 0.7 (P.U> Solution. Given : t, = 400 K, t, = 300 K (i) Boundary layer thickness, 6: The boundary layer thickness is given by 6 = x 13.93 (0.952 + pr)'I4 (Gr)-'I4 (pr)-lJ2] ...[Eqn. (8.27)] 2 go (At) -- (0.4)'~ 9.81 x (2.86 x x (400 - 300) = 4.17 x lo8 Gr, = v2 (20.75 x 10")~ Substituting the values in the above equation, we get 6 = 0.4 [ 3.93 (0.952 + 0.7)'" (4.17 x 10' )-'I4 (0.7)-'"1 = 0.0149 m = 14.9 mm (Ans.) (ii) Average heat transfer coefficient, h : G r . Pr = (4.17 x 10') (0.7) = 2.919 x lo8

-

Nu =

or

- -13'02 x 0.13 ( 93.12 x 1 0 ' ~ ) " ~= 34500 W I ~ ~ " C

2.2 :. Heat dissipation from both sides of each plate, Q = 2 h A , At = 2 x 34500 x (2.2 x 1.45) x (960 - 340) = 136.47 x lo6 W = 136.47 MW (Ans.) 8.12. Find the convective heat loss from a radiator 0.6 m wide and 1.2 m high Example maintained at a temperature of 90°C in a room at 14°C. Consider the radiator as a vertical plate. Solution. Given : L = 1.2 m, B = 0.6 m, t, = 90°C. t, = 14°C. Convective heat loss, Q : Film temperature The thermo-physical properties of air at 52°C are: k = 28.15 x W/m°C; v = 18.41 x 10" m2/s,

hL = 0.5 16 (Gr . ~ r ) ' . ~ ' k

Pr = 0.7;

. h- = k x 0.516 ( G r . ~ r ) ' . ~ ~ L =

$x 0.5 16 (2.919 x 10')'.~'

= 5.058 W I I ~ " C (Ans.)

Rayleigh number = Ra, = GrL.Pr =

Example 8.11. A nuclear reactor with its core constructed of parallel vertical plates 2.2 m high and 1.4 m wide has been designed on free convection heating of liquid bismith. The maximum temperature of the plate su$aces is limited to 960°C while the lowest allowable temperature of bismith is 340°C. Calculate the maximum possible heat dissipation from both sides of each plate. For the convection coeflcient, the appropriate correlation is Nu = 0.13 ( G r . ~ r ) ' . ~ ~ ~ where diferent parameters are evaluated at the mean film temperature. Solution. The mean film temperature, tf = 960

L3g (3 (4 - t J

v2

. Pr

= 11.69 x lo9

From Eqn. (8.36) it follows that the flow becomes turbulent on the radiator plate. Using Eqn. (8.40), we have

340 = 650°C 2 The thermo-physical properties of bismith are: p = lo4 kg/m3 ; p = 3.12 kglm-h; c, = 150.7 Jkg°C; k = 13.02 W/m°C

Gr =

L

~

cl'

~

+

~ x P( 1 0A~ 9.81 ) ~~ x 1.08 x - ~(2.21~ (3.12/3600)2

= 9.312 x 10'' Gr.Pr = 9.312 x 1015 x 0.01 = 93.12 x lo1* Using the given correlation, we get

x (960

- 340)

The convective heat Kss, Q = h A, At = 6.187 x (1.2 x 0.6) (90 - 14) = 338.55 W

(Ans.)

Example 8.13. A gas at 195°C is flowing through a thin walled vertical duct which is in the form of circular cross-section having diameter of 450 mm. The ambient air at 15"C, which may be considered still, surrounds the duct. Find the rate of heat transfer using the following simplified relation for air for laminar flow.

N U = -hL - - 0.13 ( ~ r . ~ r ) ' ~ ~ ~ ' k

where, L is the length of the duct in metres.

Heat and Mass Transfw

484

A t = 195 - 15 = 180°C, D = 450 mm = 0.45 m

Solution. Given :

\

:.

1

The rate o f heat transfer, Q = A, At = 5.055 x (n x 0.45 x 1 ) x 180 = 1286.34 Whn length. (Ans.) Example 8.14. A horizontal heated plate measuring -1.5 m x 1.1 m and at 215"C, facing upwards, is placed in still air at 25°C. Calculated the heat loss by natural convection. The convective film coeficient for free convection is given by the following empirical relation: h = 3.05 (T,)'" w/m2"c where, T f is the mean jilm temperature in degrees kelvin. Solution. Given: A, = 1.5 x 1.1 = 1.65 m2, t, = 215"C, t, = 25°C. Mean film temperature,

R&e Convection

485

80°C and {he pipe is placed in atmospheric air at 20°C. Considering heat loss both by radiation and natural convection, calculate: (i) The heat loss from 6m length of the pipe; (ii) The overall heat transfer coeficient and the heat transfer coeficient due to radiation

.5

+ 2 x2.5

The mean film temperature,

= 12.5 cm=0.125 m, E =0.9, t,= 80°C, t,= 20°C

20 = 50°C. ' f = 80 + 2

The thermo-physical properties of air at 50°C are: p = 1.092 kg/m3;c, = 1007 J/kg°C; p = 19.57 x 10" kglms; k = 27.81 x

WIm°C

.. h = 3.05 (393)'" = 13.58 w/rnZOc Rate of heat loss, by natural convection, Q = h A, (t, - t,) = 13.58 x 1.65 x (215 - 25) = 4257.33 W (Ans.) Example 8.15. A transfonner is cooled by immersing in an oil bath which is housed in a cylindrical tank which is 0.8 m in diameter and 1.3 m long. If the electrical loss is 1.3 kW, calculate the surface temperature of the tank; the entire loss of electrical energy may be assumed to be due to free convection from the bottom of the tank. For this case the following simplified relations for the boundary layer are applicable :

Using

... For a cylindrical plane; A

0.25

... For n vertical plane.

h = 1.45 Assume the ambient air temperature as 20°C. Solution. Given :D = 0.8 m, L = 1.3 m. The total rate o f heat transfer from the tank, Qtoml = Qsge + Qtop Qside = hside X Aside X At

...(i)

(i) Heat loss from 6 m length of pipe: Heat lost by convection, Q, = h A, At = 6.24 x (n x 0.125 x 6 ) x (80 - 20) = 882.16 W Heat lost by radiation, Q , . ~ ,= E OA ( T -~T ~ ~ ~ ) = 0.9 x (5.67 x lo-') x (n x 0.125 x 6 ) [(80 + 27314 - (20 + 273)4] = 980.81 W Total heat loss, Qt = Q, + Qrd, = 882.16 + 980.81 = 1862.97 W (Ans.) (ii) Heat transfer coefficients, overall (hJ and due to radiation alone (h,,) Q, = h,. A At

Substituting the values in eqn. (i), weTget 1.3 x lo3 = 4.13 ( t - 20)'." + 0.77 (t - 2 0 ) ' ~=~4.9 ~ (t - 2 0 ) l . ~ or

(t - 20) =

[

]

3 1A.25

1.3 x 10 4.9

= 86.89

or t = 86.89 + 20 = 106.89°C (Am.) Example 8.16. A steam pipe 7.5 em in diameter is covered with 2.5 cm thick layer of insulation which has a suface emissivity of 0.9. The surface temperature of the insulation is

..

h,

= ht - h,

= 13.17

- 6.24 = 6.93 W/mZOc

:

(Am)

Example 8.17. Calculate the heat transfer from a 60W incandescent bulb at 115°C to ambient air at 25°C. Assume the bulb as a sphere of 50 mm diameter. Also, find the percentage of power lost by free convection. The correlation is given by: Nu = 0.60 ( ~ r . ~ r ) " Solution. Given : t, = 1 10°C, ,= . . t = 25"C, D = 50 mm = 0.05 m.

l

I

! I' 1

Heat and Mass Transfer

486

Free Convection

487

+ = -<=

+ 115+25=700C 2 2 The thermo-physical properties of air at 70°C are: k = 2.964 x W/m°C, v = 20.02 x lo4, Pr = 0.694,

The film temperature,

...(where L = 1 m)

= 1.514 [368.04 + 76.761 = 441 W/m

=h x

-

(t, t,) where h is calculated by using the given relation. QCon,.

('.' The characteristic length L = D in this case)

Ra = Gr.Pr = Pg (At) (L,)' va

Using given correlation, we get

-

Heat transfer = h A , At = 9.72 x (x x 0.05~)x (1 15 - 25) = 6.87 W (Ans.) %age of power lost by free convection = 6.87 x LOO = 11.45 Z (Ana) 60 Example 8.18. A horizontal high pressure steam pipe of 10 cm outside diameter passes through a large room whose walls and air are at 23°C. The pipe outside suface temperature is 465°C and it emissivity is 0.85. Estimate the heat loss from the pipe per unit length. Use the following correlation for the calculation of film coeflcient

..

- 2.725 x 10-~,x9.81 X (165 - 23) X (0.1)' 22.8 x

x 32.8 x

=

-

Q,,,,, = 7.2 x (x x 0.1 x 1) (165 23) = 321.2 W, when L = 1 m

Q = 441 + 321.2 = 762.2 W/m (Ans.) Example 8.19. A vertical slot of 20 mm thickness is formed by two 2 nr x 2 m square plates. I f the temperatures of the plates are 115OC and 25°C respectively, calculate the following: (i) The effective thermal conductivity; (ii) The rate of heatflow through the slot. Solution. Given: L = 20 mm = 0.02 m; t1 = 115°C; t2 = 25°C (i)The effective thermal conductivity, k, : Hence,

The mean temperature between the plates = 115 + 25 = 700C 2 The properties of air at 70°C are: k = 0.0295 W/m°C; v = 2.0 x lo-' m2/s

where Ra is known as Rayleigh Number and is given by va

Take the following properties of air : The Grashof number based on the gap L, is a = 3 2 . 8 ~ 1 0 - ~ r n ~ Pr=0.697, /s, P=2.725xl~'~~-'. (N.M.U. Wincdr, 1995) Solution. Given : D = 10 cm = 0.1 m; t, = 23OC; t, = 165°C; E = 0.85 m2/s; Properties of air : k = 0.0313 W/m K; v = 22.8 x m2/s, Pr = 0.697, P = 2.725 x a = 32.8 x KHeat loss from the pipe per unit length, Q : Heat lost by convection and radiation from the sphere is given by :

Q = Qconv. + Qrad.

'

= I 5.147 x lo4 The effective thermal conductivity is given by

...[Eqn. (8.50)]

Frpe Co

Heat and ~ a ~iarisfer &

488

489

& (viscosity at the wall) = 25.19 x lo4 ~ s / m ~

(ii) The rate of heat flow through the slot Q : Q = ke . A

.

(F)

Reynolds number Thuy'the nature of flow is laminar. drashof number,

Example 8.20. Two horizontal panels separated by a distance of 30 mm contain air at atmaspheric pressure. The temperatures of lower and upper panels are 55OC and 20.6OC respectively. Calculate the free convection heat transfer per m2 of the panel surface. Solution. Given. L = 30 mm = 0.03 m, tl F S ° C , t2 = 20.6"C

- 038 P 0 s - t,)

Gr =

I

3

The mean temperature of two panels = 55 +2 20.6 = 37.80C The properties of air at 373°C are: p = 1.121 kglm3 ; k = 29.2 x low3W/m°C;

'

= 37.8

+ 273

= 3.22 x

v = 0.171 x 104m2/s Pr = 0.70

t

Since [ G r / ( ~ e )t:~ 1, ] hence it is a case of mired (free and forced) convection and the following equation may be used for calculation of Nu.

e

...[Eqn. (8.61)]

K-'

Free convection heat transfer per m2 of the panel surface, q:

-

where,

a.

L Nu Z - L k 0 1 - t2)

Also,

)::"I-

= 356.68 x 0.688

= 15.34

...[Eqn. (8.51)1

Nu = 0.195 (Gr$" NU = 0.195 (10.03 x lo4)lI4= 3.47

or

Gz = Graetz number = Re.Pr

-h = - kN-u =

3'208 x 7.7 = 9.88 w/m2"c 0.025 Q = h A, At = 9.88 x (n X 0.025 x 0.4) x (180 - 20) = 49.66W

...[Eqn. 8.52 (a)]

(where q = QlA)

D

(Ana.)

TYPICAL EXAMPLES Example 8.22. Derive a relation between Grashof and Reynolds numbers assuming the heat transfer coeficients over vertical plates for pure forced and free convection are equal in laminar flow. Solution. For pure forced convection: Zu = 0.664 (Re)" (pr)ln ...(i ) ' For pure natural convection: Nu = 0.677 (~r)l)l" (0.952 + (Gr)'" ...(ii) Equating equations (i) and (ii), we get 0.664 (Re)ln ( ~ r ) ' "= 0.677 (pr)In (0.952 + pr)-'I4 (Gr)'I4 or ( ~ r= )(pr12 ~ (0.952 + ~r)-l(Gr)

Example 8.21. (Mixed convection) Air at 200C and 1 atmosphere is forced throvgh a 25 mrn diameter tube 400 mm long, at an average velocity of 0.33 d s . Calculate the rate of heat transfer if the tube wall is maintained at 1800C. Solution. Given: tb = 20°C, t, = 180°C, V = 0.33 m/s, L = 400 mm = 0.4 m, D = 25 mm = 0.025 m. Rate of heat transfer : * 180 + 20 = ]OoOC The mean film temperature, tf = 2 t

The properties of air at 100°C are: k = 3.208 x 1w2W/mOG,v.= 23.13 x 1@ ~ s / mPr~ = , 0.688

'

= 100

+ 273 = 2.68 x

1p3

For the bulk temperature of 20°C, we have lb= 18.15 x ~ s / m ~ Again, for the surface temperature of tube wall (i.e., 180°C)

.

"

*

Example 8.23. Two horizontal steam pipes having diameters I00 mm and 300 mm are so laid in a boiler house that the mutual heat transfr m a y be neglected. The surface temperature of each of the steam pipes is 475OC. If the temperature of the ambient air is 35OC, calculate the mtio of heat transfer coeficients and heat losses per metre length of the pipes. Solution. Given :Dl = 100 mm = 0.1 m D2 = 300 mm = 0.3 m, t, = 475OC, t- = 35°C The steam mains are located in the boiler house where the ambient air is stationary, thus his is a case of free convection for which the following relation applies :

Heat and Mass Transfer

Free Convection

Free convection heat transfer rate,

-

Since, the surface temperature of both pipes is the same and both are exposed to the same ambient conditions, therefore, thermo-physical properties are the same. Nu = (D3)lI4= ( D ) ~ ' ~ ..

:.

The ratio of heat transfer coefficients, h1

%= Also.

:.

The ratio of heat losses,

(2) (z) 1/ 4

1/4

=

1.316

(Ans.)

=

Q, - h , Dl

- -= 1.316 x Q2 h2 0 2

Q = h A, (t,,- t_) = 6. = 871.2W (Ans.) Example 8.25. A 2-stroke motor cycle petrol engine cylinder consists of I 5 fins. I f the outside and inside diameters of each fin are 200 mm a d 100 mm respectively, the overage fin su$ace temperature is 475OC and atmosphe;ic air temperature is 2S°C, calculate the heat transfer rate from the fins for the following cases: (i) When the motor cycle is stationary; (ii) When the motor cycle is running at a speed of 60 kmh. The fin may be idealised as single horizontal flat plate of the same area. Solution. Case (i) When the motor cycle is stationary the heat transfer will be due to free convection.

+ 25 = 250°C. 2 The thermo-physical properties of air at 250°C from the tables are: k = 4.266 x 1W2 W/m°C; v = 40.61 x 106 m2/s; Pr = 0.677 The mean film temperature,

= 0.438

(Ans.)

tf

=

475

Example 8.24. A rqom is heated by initiating fire at the fireplace. The exfitration of room air through a chimney ih reduced by the use of a glass-door fire screen of height 0.8 m and width of 1.Im. If the ambient air temperature is 2J°C and the sulface temperature attained by the glass is 175'C, calculate the free convectin heat transfer rate from the fire place to the room. For this case the followibg correlation given by Churchill and Chu is valid:

where significant length L = 0.9 D

Gr.Pr = 2.985 x lo7 x 0.677

Solution. Given: t, = 17S°C, t, = 2S°C, L = 0.8m, B = 1.1 m

-

2 x lo7

Since Gr.Pr c lo9, hence the nature of flow is laminar.

The film temperature, The thermo-physical properties of air at 100°C are: WIIII~~C, v = 23.13 x lo4 m2/s, Pr = 0.688 k = 3.208 x 1 = 2.68 x p = -1=

Tf Grashof number,

(100

+ 273)

Since the heat transfer is from both sides of the fins, therefore, n x (0.2~- 0.13 [475 - 251 A, At = 8.56 2 x 15 x 4 = 2722.82 W (Ans.) Case (ii) :When the motor cycle is running the heat transfer will be due to forced convection. The velocity of air on the fin = speed of the motor cycle

[

Q =

Reynolds number Hence the flow is turbulent. For turbulent flow, for cooling,

-

1

Heat and W S s transfer

:.

Free Convection

493

HIGHLIGHTS 1. Free or natural convection is the process of heat transfer which occurs due to movement of the fluid particles by density changes associated with temperature differential in a fluid. 2. List of formulae : I. Flow over vertical plates (Laminar $ow) (a) Exact solution:

The heat transfer from the cylinder,

(i) Nu, = 0.676 = 18713 W

(Ans.)

Example 8.26. Gas at 350°C is conveyed through 250 mm diameter pipe laid in an atmosphere of quiescent air at 20°C.The convection heat transfer coeficient from ht 0.25a hot cylindrical su$ace w/mZOcwhere At is the freely exposed to still air is prescribed by the relation: h = 1.55

.,

-

-

-

(ii) Nu = 0.902 (b) Approximate solution:

temperature difference and D is the diameter of the cylinder is metres. (i) Calculate the heat loss per metre length of the bare pipe. (ii) Estimate the percentage reduction in heat loss $the pipe is covered with 75 mm thick laver o f material whose thermal conductivi~is 0.072 W/maC. Any temperature drop in thk meial may be neglected. Solution. Convective heat transfer coefficient, 0.25 350 h = 1.55 0.25 20] = 9.34 w l m 2 ~ c

[

(i) Heat loss from the bare pipe: Convective heat loss per metre length of the bare pipe. 0 = h A. At = 9.34 x (n x 0.25 x 1 ) x (350 - 20) = 2420.75 W (Ans.) (ii) Percentage reduction in heat loss when pipe is lagged : 4 = 250 + ( 2 x 75) = 400 mm = 0.4 m Dl = 250 mm = 0.25 m, On attainment of steady state, the heat conducted through the lagging equals the convective heat loss from the outer su$ace of the lagging. Thus 271 k L (tl - t) = h x ( X D2 L) (t - tm) In (r2/r1> where t is the temperature at the outer surface of the lagging. Substituting the appropriate values, we get 2n 0.072 (350 - ') ;1.55 x (n x 0.4 x 1) (t - 20) In (0.2/0.125) Or 0.962 (350 - = 2.449 (t 350 - t = 2.545 (t - 2 0 ) ' . ~ ~ or t z 64°C By trial and error, we get,

-

(27

:.

.

Heat loss from the lagged pipe, 2 n k L ( t l - 1) - --275.28w = ln (r2/r1) In (0.2/0.125) 2420.75 - 275.28 Percentage reduction in loss = 2420.75

v (where Gr, = gP 0 s - L ) X 3 and Pr = )

3

(vi) m =

a

P 6 . u, 12 .

(viii) Total mass, m = 1.7 pv

[

(~r-)' (Pr + 0.952) h x 2x (ix) Nu, = A = 0.508 ( ~ r ) ' "(0.952 + Pr)-'I4 (Grx)'I4= k 6

(x)

-

-

N u L = hkL ---

-

Nu, = 0.677 (Pr)" (0.952 + Pr)-'" (GrL)lM

11. Flow over vertical plates (Turbulent jlow):

6 1 +' 0.494 ( ~ r - ) O"~ ~ ~ ] (xi) - = 0.565 ( ~ r ) - ~ ~ X Grx

[

(xii) Nu, = 0.0295 ( ~ r ) ~ / ' ~ [ I + 0.494 (Pr)"

Heat and Mass Transfer UNSOLVED EXAMPLES

1. A hot plate 1.1 m wide, 300 mm high and at 120°C is exposed to the ambient still air at 20°C. Calculate : (r) Maximum velocity boundary layer thickness and local heat transfer coefficient at 150 mm from the leading edge of the plate, (ii) Total mass flow through the boundary, (iii) Heat loss from the plate and (iv) Rise in temperature of the air passing through the boundary. [Ans. (i) 0.39 mls, 11.4 mm, 5.199 W/m2"C, (ii) 0.004372 kgls,] 2. A 300 rnm long glass plate is hung vertically in the air at 27OC while its temperature is maintained at 77°C. Calculate the boundary layer thickness at the trailing edge of the plate. If a similar plate is placed in a wind tunnel and air is blown over it at a velocity of 4 mls, find the boundary layer thickness at its trailing edge. Also determine the heat transfer coefficient, for natural and forced convection for the above mentioned [Ans. 15.2 mm, 5.67 mm; 4.9 W/m2"C, 14.11 WlmZ0CI data. 3. A nuclear reactor with its core constructed of parallel vertical plates 2.25 m high and 1.5 m wide has been designed on free convection heating of liquid bismuth. The maximum temperature of the plate surfaces is limited to 975°C while the lowest allowable temperature of bismuth is 325°C. Calculate the maximum possible heat dissipation from both sides of each plate. For the convection coefficient, the appropriate correlation is Nu = 0.13 (Gr . pr)' 333 (Ans. 153 MW) where different parameters are evaluated at the mean film temperature. 4. Find the convective heat loss from a radiator 0.5 m wide and 1 m high maintained at a temperature of (Ans. 110.08 W) 84°C in a room at 20°C. Consider the radiator as a vertical plate. 5. A steam pipe 60 mm in diameter is covered with 20 mm thick layer of insulation which has a surface emissivity of 0.92. The surface temperature of the insulation is 75OC and the pipe is placed in atomspheric air at 25OC. Considering heat loss both by radiation and natural convection, calculate : (i) The heat loss from 5 m length of the pipe; (ii) The overall heat transfer coefficient and the heat transfer coefficient due to radiation alone. (Ans. (i) 1049.83 W, (ii) 7.07 W / ~ ~ " C ) 6. Calculate the heat transfer from a 60 W incandescent bulb at 125°C to ambient air at 25°C. Assume the

bulb as a sphere of 50 mm diameter. Also, find the percentage of power lost by free convection. (Ans. 7.71 W; 19.278%) The correlation is given by : Nu = 0.60 (Gr. pr)'I4 7. A vertical slot of 15 mm thickness is formed by two 2m x 2m square plates. If the temperature of the plates are 120°C and 20°C respectively, Calculate the following: (i) The effective thermal conductivity; [Ans. (i) 0.0317 W/m°C, (ii) 845.3 W] (ii) The rate of heat flow through the slot. 8. Two horizontal plates separated by a distance of 25.4 mm contain air at atmospheric pressure. The temperatures of lower and upper panels are 60°C and 15.6"C respectively. CaIculate (Ans. the free167.2 convection w/m2) heat transfer per mZof the panel surface.

Boiling and Condensation (Heat Transfer with Phase Change)

I

-

9.1. Introduction, 9.2. Boiling heat transfer : General aspects Boiling regimes - Bubble Laps and size consideration - Bubble growth and collapse - Critical diameter of bubble - factors affecting nucleate boiling - Boiling correlations. 9.3. Condensation heat transfer: General aspects - Laminar flow condensation on a vertical plate - Turbulent film condensation - Film condensation on horizontal tubes - Film condensation inside horizontal tubes - Influence of the presence of noncondensable gases Highlights - Theoretical Questions - Unsolved Exam~les.

I

I

9.1.

INTRODUCTION

W e have considered in the preceding2hapter b convective heat transfer, homogeneous single phase systems only. However. there are specific convection processes as boiling a n d condensation which a r e associated with change of phase. Whereas boiling involves change from liquid to vapour phase of a fluid substance, condensation refers to a change from the vapour to a liquid phase, The mode of heat transfer with change of phase (i.e. boiling and condensation processes) finds wide applications as mentioned below: (i) Cooling of nuclear reactors and rocket motors; (ii) Steam power plants (Boilers and condensers); (iii) Refrigerating and airconditioning systems (Evaporators and condensers); (iv) Melting of metal in furnaces; (v) Refineries and sugar mills (Heat exchangers); (vi) Process heating and cooling etc. Boiling and condensation processes entail the following unique features: (i) As a consequence of phase change in these processes, the heat transfer to or from the fluid c a n occure without influencing the fluid temperature. (ii) The heat transfer coeficient and rates, due to latent heat associated with phase change, are generally much higher compared with the normal convection process (i.e., without phase change) (iii) High rate of heat transfer is achieved with small temperature direrenee. The phenomena associated with boiling and condensation are much more complex (than the normal convection process) due to the following being very significant: (i) Latent heat effects; (ii) Surface tension; (iii) Surface characteristics and other properties of two phase systems.

9.2.

BOILING H E A T TRANSFER

9.2.1. General Aspects Boiling is the convective heat transfer process that involves a phase change from liquid to

Boiling and Condensation vapour state. Boiling is also defined as evaporation at a solid-liquid surface. This is possible only when the temperature of the surface (t,) exceeds the saturation temperature corresponding to the liquid pressure (tsa,). Heat is transferred from the solid surface to the liquid according to the law ...(9.1) Q = h A, (t, - t,,,) = h A, A t, At, = (t, - tsa$ is known as excess temperature. where The boiling process finds applications in the following cash: (i) Steam production (for generation of power and for indps&al processes and space heating) in steam and nuclear power plants; (ii) Heat absorption in refrigeration and air conditioning systems; (iii) Distillation and refining of liquids; (iv) Concentration, dehydration and drying of foods and materials, (v) Cooling the machines like nuclew reactors and rocket motors where the large quantities of heat are released in relatively small volume (dissipation rates are as high as lo8 w/m2; the maximum heat transfer rate in modem boiler is about 2 x lo5 w/m2). The boiling heat transfer phenomenon may occurs in the following forms: 1. Pool boiling In this case the liquid above the hot surface is essentially stagnant and its motion near the surface is due to free convection and mixing induced by bubble growth and detachment. The pool boiling occurs in steam boilers involving natural convection. 2: Forced convection boiling This refers to a situation where the fluid motion is induced by external means (and also by free convection and bubble induced mixing). The liquid is pumped and forced to flow. This type of boiling occurs in water tube boilers involving forced convection. 3. Sub-cooled or local boiling In this case the liquid temperature is below the saturation temperature and bubbles are formed in the vicinity of heat surface. These bubbles after travelling a short path get condensed in the liquid which has a temperature less than the boiling point. 4 . Saturated boiling Here, the liquid temperature exceeds the saturation temperature. The vapour bubbles formed at the solid surface (liquid-solid interface) are then propelled through the liquid by buoyancy effects and eventually escape from a free surface (liquid-vapour interface).

9.2.2. Boiling Regimes The process of boiling depends upon the nature of the surface, thermo-physical properties of thefluid and vapour bubble dynamics. Due to involvement of large number of variables, general V

Vapour bubbles

Fig. 9.1. Pool boiling with liquid-vapour interface.

Heat and Mass Transfer

497

equations describing the boiling process are not available. Nonetheless, considerable progress has been made in arriving at a physical understanding of the boiling mechanism. Figure 9.1 shows the temperature distribution in saturated pool boiling with a liquid-vapour interface. It is evident from the figure that although there is a sharp decline in the liquid temperature close to the solid surface, the temperature.through most of the liquid remains slightly above saturation. Consequently bubbles generated at liquid-solid interface rise to and are transported acrlpss the liquid-vapour interface. Whether the boiling phenomenon corresponds to pool boiling or forced circulation boiling, there are three definite regimes of boiling (Interface evaporation, nucleate boiling and film boiling ) associated with progressively increasing heat flux, as shown in Fig. 9.2. This specific curve has been obtained from an electrically heated platinum wire submerged in a pool of water (at saturation temperature) by varying its surface temperature and measuring the surface heat flux q,. 1. Interface evaporation Interface evapo&n (evaporation process with no bubble formation) exists in region I, called the free convection zone. Here the excess temperature, A t,, is very small and 5 5°C. In this region the liquid near the surface is superheated slightly, the convection currents circulate the liquid rd evaporation takes place at the liquid surface. I

Interface +evaporation or-+ Free convection

-Nucleate boiling

-

Film boiling

-1

t

h

"E

s

p*

Heat transferred by superheated liquid rising to the liquid-vapour interface where evaporation takes place

Excess temperature At, = t, - t,,

-

Fig. 9.2. The boiling curve for water.

2. Nucleate boiling This type of boiling exists in regions 11and III. With the increase in At, (excess temperature) the formation of bubbles on the surface of the wire at certain localised spots commences. The bubbles condense in the liquid without reaching the liquid surface. In fact it is the region 11where nucleate boiling starts. With further increase in At, the bubbles are formed more rapidly and rise to the surface of the liquid resulting in rapid evaporation, as indicated in the region III. The nucleate b i I k g - i s thus characterised by formation of bubbles at the nucleation sites and the resulting liquid agitation. The bubble agitation induces considerablefluid mixing and that promotes substantial increase in the heatflux and the boiling heat transfer coefficient (The equipment used tor boiling should be designed to operate in this region only). Nucleate-boiling exists upto At, = 50°C. The maximum heat flux, known as the critical hear $m, occurs at point A (see Fig. 9.2) and is of v & d e r of l h 4 ~ 1 m ~ .

Boiling and Condensation

498

3. Film boiling Film boiling comprises of regions N,V and VZ. The trend of increase of heat flux with increase in excess temperature observed upto region ZZZ is reversed in region N (called film boiling region). This is due to the fact that the bubble formation is very rapid and the bubbles blanket the heating surface and prevent the incomingfresh liquid from taking their place. Eventually the bubbles coalesce and form a vapour film which covers the surface completely. Since the thermal conductivity of vapour film is much less than that of the liquid the heat flux drops with growth in At,. Within the temperature range 50°C < At, < 15O0C, conditions oscillate between nucleate and film boiling and the phase is referred to as transition boiling, unstable jilm boiling or partial film boiling (region IV). With further increase in At, #he vapour film is stablised and the heating surface is completely covered by a vapour blanket and the heat flux is the lowest as .shown in region V. The surface temperatures required to maintain a stable film are high and under these conditions a sizeable amount of heat is lost by the surface due to radiation, as indicated in the region VZ. The phenomenon of stable film boiling can be observed when a drop of water falls on a red hot stove. The drop does not evaporate immediately but dances a few times on the stove; this is due to the formation of a stable steam film at the interface between the hot surface and the liquid droplet. Critical heat flux or burnout point: The critical heat flux or burnout point (Point A in Fig. 9.2) is the point of w i m u m heat flux on the boiling curve at which transition from nucleate to film boiling initiates. This point is also called the boiling crisis because the boiling process beyond that .point is unstable unless of course, point B is reached. The temperature at point B is extremely high and normally above the melting of the solid. So if the heating of the metallic surface is not limited to point A, it is possible that the metal may get damaged or it may even melt (For this reason, point A is often termed as boiling crisis or burnout point). Thus we may be interested to operate the equipment close to this value and not beyond it. 9.2.3. Bubble Shape and Size Consideration The heat transfer rate in nucleate boiling is greatly influenced by the nature and condition of the heating suvace and su@ace tension at the solid-liquid interjke (shape, size or inclination of bubbles, however, do not have much effect on the heat transfer rate). The surface tension signifies wetting capability of the surface with the liquid (i.e., low surface tension, highly wetted surface) and that influences the angle of contact between the bubble and the solid surface. If the surface is contaminated, its wetting characteristics are affected which eventually influence the size and shape of the vapour bubbles. If the surface tension of the liquid is low, it tends to wet the surface, so that the bubble is readily pushed by the liquid and rises. The liquid shear-off the bubbles causing them to become globular or oval as shown in Fig. 9.3 (i) (for totally wetted surface). In case of liquids having intermediate surface tension (partially wetted surface) a momentary balance may exist between the bubbles and solid surface so that it is necessary to form larger bubbles before the buoyant force can free them from the surface; the shape of the bubble is shown in Fig. 9.3 (ii). ,

Solid Totally wetted (i)

Q Partially wetted (ii)

/\c Unwetted (iii)

Fig. 9.3. Typical shapes of steam bubbles.

On the unwetted. surface [Fig. 9.3 (iii)], the bubbles spread out; forming a wedge between the water and heating surface, thereby allowing hydrostatic forces to resist the action of buoyancy.

Heat and Mass Transfer

499

The formation of bubble as sgaun in Fig. 9.3 (i) gives high heat transfer rate compared. with the bubble shapes shown in Fig. 9.3 (ii) and (iii). Addition of agents for reducing the sut$ace tension was found to have the same effect as providing of wettable sudace and to give increased rates of heat transfer. 9.2.4. Bubble Growth and Collapse

-

d (pV- p,) or where

Vapour

= 2m.o

20 PV - PI = 7

...(9.2)

-

pv = vapour pressure inside the bubble pl = liquid pressure over the surface of bubble

o = surface tension of vapour-liquid interface. The vapour may be considered as a perfect gas for which the Clayperon equation may be used, which is given below:

where From perfect gas law:

-

spherical bubble Liquid

From Experiments it has been observed that the bubbles are not always in thermodynamic equilibrium with surrotmding liquid. The vapour inside the bubble is not necessarily at the same temperature as the liquid. Consider the forces acting on a sP&cal vapour bubble as shown in Fig. 9.4; the pressure forces on-the bubble must be balanced by the surface tension a t vapour-liquid interface. Thus

u

Fig. 9.4. Force balance on a spherical vapou bubble.

hfg= latent heat of vapourisation.

e-

RT - P v [whers R = gas or vapour constant; pv = density of vapour formed] Substituting the above equation in Eqn. (9.3) and rearranging, we get

hfg P v P . hfg -Tv - Tsar Tsar R Tm2 where Tv= vapour temperature inside the bubble T, = saturation temperature of vapour inside the bubble at p,. From Eqn. (9.2) and (9.4), we get Pv - Pl -=--

.--

The above equation suggests that if (TI - Tsar)> (Tv - T,), the bubble of radius r will grow otherwise it will collapse. Here T, is the temperature surrounding the bubble.

9.2.5. Critical Diameter of Bubble Refer figure 9.5. The maximum diameter of the bubble formed on the heating surface depends on the following parameters: olv= tension between liquid and vapour ols= tension between liquid and solid surface ovs= tension between vapour and solid surface

Boiling and Condensation

500

P = angle formed by the bubble as

Bubble

shown in Fig. 9.5 dc = maximum or critical diameter of bubble. g(pI pv)= buoyancy force.

-

Heat and Mass Transfer

601

9.2.7.1. Nucleate pool boiling (i) For nucleate pool boiling, Rosenhow has recommended the following correlation:

wnere

q,

= surface heat flux, w/m2;

-liquid viscosity, kg/ms; ;4 sc*,
>

"i.'

By the use of the dimensional analysis technique, we get I

dc

-- .

)

i? (PI- 4)

(9.6)

Fig. 9.5. Critical diameter of bubble.

where C is constant_whichis generally calculated by experimental results. ...suggested by Fritz. The value of C = 0.0148 for water bubbles

9.2.6. Factors Affecting Nucleate Boiling The nucleate boiling is affected by the following factors: 1. Material, shape and condition of the heating surface The boiling heat transfer coefficient depends greatly on the material of the heating surface; under identical conditions of pressure and temperature difference, it is different for different metals (viz oopper has high value than steel, zinc and chromium). The heat transfer rates are also influenced by the condi
1.q..

-_____I-=

Table 9.1. Values of C,for Pool Boiling S.No. 1.

2. 3. 4. 5.

6. 7. 8. 9. 10.

1

Liquid-su~ace

I

Water - copper Water - brass Water - platinum Water - ground and polished stainless steel Water mechanically polished stainless steel Benzene - chromium Ethanol - chromium n-pentane - chromium n-butanol copper Isopropyl alcohol - copper

-

-

(ii) Jacob has proposed the following correlation for nucleate boiling at atmosphericpressure on a flat plate and with low heat fluxes Nu = 0.16 (Gr . ...(9.9) (iii) For the nucleate boiling on a vertical flat plate, Jacob correlation is of the form: Nu = 0.6 1 (Gr . ~ r ) ~ . ~ ~ ...(9.10) I

9.2.7.2. Critical heat flux foq nucleate pool boiling On the boiling curve the critical heat flux is an important point. It is always desirable to operate a boiling process close to this point. Zuber (1958) predicted the following expression, for such a case: qsc = 0.18 (pV)lnhfg [go(PI -PV )ill4 ...(9.11) The expression given above is independent of fluid viscosity, conductivity and specific heat.

9.2.7.3. Film pool boiling In stable film boiling, the heat transfer is due to both convection and radiation. Bromley (1950) has suggested the following correlation for film boiling from the outer surface of horizontal tubes: + hd . (h)'" (h)413= (hconv.)4n ...(9.12)

I

Boiling and Condensation The equation (9.12) being tedious to solve, could be written within zk 5% of error as 3 ...(9.13) h = h",. + q hrd The convective coefficient, h,,

(in the absence of radiation), is given by

Haat and Mass Transfer

..

503

L 9.817810x lo4

=A

(ii) The excess temperature, At, : Using the correlation, 1.58

where D is the outer diameter of the tube. The vapouk properties in the above equation are evaluated at the arithmetic mean of the surface and saturation temperatures. Radiative heat transfer coefficient 5.67 x lo4 & (T: hrud = 8 (' - Tsar,) where E is the emissivity of solid. Example 9.1. A wire of 1.2 mm diameter and 200 mm length, is submerged horizontally in water at 7 bar. The wire carries a current of I35 A with an applied voltage of 2.18 V. If the su@ace of the wire is maintained at 200°C, calculate : (i) The heat jlux, and (ii) The boiling heat transfer coeficient. Solution. Given : d = 1.2 mm = 0.0012 m, 1 = 200 mm = 0.2 m, I = 135 A, V = 2.18 V, t, = 200°C. (i) The heat flux, q : The electrical energy input to the wire is given by Q = VI = 2.18 x 135 = 294.3 W Surface area of the wire, A = d l = IG x 0.0012 x 0.2 = 7.54 x lo4 m2

-

J

(ii) The boiling heat transfer coefficient, h: Corresponding to 7 bar, t,,, = 164.97"C, and 9 = h (1s - tsu,)

,/*

Example 9.2. An electric wire of 1.25 mm diameter and 250 mm long is laid horizontally and submerged in water at atmbspheric pressure. The wire has an applied voltage of 18 V and carries a current of 45 amperes. Calculate: (i) he heat jlux, and (ii) The excess temperature. The following correlation for water boiling on horizontal submerged su@ace holds good: h = 1.58

(fr5

= 5.62 ( ~ t , ) w/m2"c ~,

Solution. Given : d = 1.25 mm = 0.00125 m, 1 = 250 mm = 0.25 m, V = 18 V , I = 45 A. (i) The heat flux, q : Electrical energy input to the wire, Q = VI = 18 x 45 = 810 W Surface area of the wire, A, = d l = IG x 0.00125 x 0.25 = 9.817 x lo4 m2

= 0.825 x lo6 w/m2 = 0.825 M W I ~ ~

[fr

= 5.62 (At,)'

At, =

...g iven

[1.58 (0.8255.62x 106)0.75]

0'333

= 19.68OC

ExampIe 9.3. A nickel wire of I mm diameter and 400 mm long, carrying current, is submerged in a water bath which is open to atmospheric pressure. Calculate the voltage at the burnout point if at this point the wire carries a current of 190 A. Solution. Given : d = 1 rnm = 0.001 m; 1 = 400 mm = 0.4 m, I = 190 A The thermo-physical properties of water and vapour at 100°C are: ~jm. pI = (pf) = 958.4 kg/m3, p, = 0.5955 kg/m3, hfg= 2257 Mlkg, a = 58.9 x Voltage at the burnout point, Vb: At burnout i.e., the points of critical heat flux, the correlation is ...[Eqn. (9.1 I)] qsc = 0.18 (pV)lnhB [ go (PI-pV)l1" = 0.18 (0.5955)'~x 2257 x lo3 [9.81 x 58.9 x (958.4 - 0.5955)11" = 1.52 x lo6 w/m2 = 1.52 M W / ~ ~ Electric energy input to the wire, Q=VbxI

or Vb = 10.05 V xample 9.4. Water is boiled at the rate of 25 kg/h in a polished copper pan, 280 mm in r, at atmosphericpressure. Assuming nucleate boiling conditions, calculate the temperature of the bottom surjace of the pan. Solution. Given: m = 25 kgh; D = 280 mm = 0.28 m The properties of water at atmospheric pressure are: tsar= 100°C; p, = 958.4 kg/m3; pv = 0.5955 kg/m3; cpl = 4220 J k g K; p, = 279 x lo4; Pr, = 1.75; hfg= 2257 kJkg; o = 58.9 x lo-' N/m;n = 1 (for water) The temperature of the bottom surface, t,: Excess temperature At, = t, - tsa For nucleate boiling (assumed), the following correlation holds good:

...[Eqn. (9.811 For polished copper pan, Csl = 0.013

Heat and Mass Transfer

505

Boiling and Condensation

50a

(iii) Critical heat flux, q,,: q, = surface heat flux =

Here

2 = --f mh

(where m = rate of water evaporation)

.

At, =

[279

\.

J

254544 x 1@ x 2257 x 1 d

}

0.5]0.333

r 1 3 x 22;i2;

10

At, = t, - tsa = 12.2 i. e. t, = 12.2 + tsat= 12.2 + 100 = 112.2OC or Example 9.5. Water at atmospheric pressure is to be boiled in polished copper pan. The ?/" diameter of the pan is 350 mm and is kept at llS°C. Calculate the following: (i) Power of the burner; (ii) Rate of evaporation in kgh; (iii) Critical heat flux for these conditions. Solution. Given : D = 350 mm = 0.35 m, ts = 115OC. tsa,= 100°C .The thermo-physical properties of water (from table) at 100°C are: p, (= pf) = 958.4 kg/m3; pv = 0.5955 hglm3; cpl (= cpf) = 4220 Jkg K; & = (14) = 279 x 106 Ns/m2 Nlm Pr, (= Prf) = 1.75; hfg = 2257 Wkg; n = 1; o = 58.9 x The excess temperature, At, = t, - tsa = 115 - 100 = 15OC (i) Power of the burner to maintain boiling: As per boiling curve, for At, = 15°C nucleate pml boiling will occur and for this the following correlation holds good:

qSc= 0.18 (P,)'" hfg [ go ( Pl- Pv)I'" ...[Eqn. 9.1 1)3 = 0.18 (0.5955)'" x 2257 x lo3 [9.81 x 58.9 x low3(958.4 - 0.5955)]"~ = 1.52 x lo6 w1m2 = 1.52 M W / ~ ~ '-yA~xample9.6 A metal-clad heating element of 10 mm diameter and of emissivity 0.92 is submerged in a water bath horizontally. Zf the surface temperature of the metal is 260°C under steady boiling conditions, calculate the power dissipation per unit length for the heater. Assume that water is exposed to atmospheric pressure and is at a uniform temperature. Solution. Given = D = 10 mm = 0.01 m, E = 0.92, tS = 260°C The thermo-physical properties of water at 100°C from table art? : pl = pf = 958.4 kg/m3; h,-,- = 2257 kJkg The thermo-physical properties of vapour at 260°C from table are: pv = 4.807 kglm3, Cpv= 2.56 Ulkg K, K = 0.0331 WImK; 'a,pv = pg = 14.85 x lo4 Ns/m2 Power dissipation per unit length for the heater: The excess temperature At, = t, - tsat = 260 - 100 = 160°C As per boiling curve, at At, = 160°C. there exists a film pool boiling conditions. In this case, the heat transfer is due to both convection and radiation. The heat transfer coefficient, h (approximate) is calculated from the equation: 3 h = ~ C O ~+V7 hrd ...[Eqn. (9.13)] The convective heat transfer coefficient,

...[Eqn. (9.14)]

-

...[Eqn (9.8)l For polished copper pan, CsI = 0.013 Substituting the values in the above Eqn. we get 9.8 1 (958.4 - 0.5955) 9s = 279 x 106 x (2257 x 1 ~ ~ )58.9

[

]O1

...Refer table 9.1 4220 x 15 [0.013 x 2257 x lo3x 1.75

= 629.7 x 399.4 x 1.873 = 471.06 x ld w1m2 = 47 1.06 kw1m2 The boiling heat transfer rate (power of the burner) is given by

(ii) Rate of evaporation, m,: Under steady state conditions, all the heat added to the pan will result in evaporation of water. Thus

or h, = 395.84 w / m 2 ~ c The radiation heat transfer coefficient, ):, 5.67 x lo4 E (T: - T brad = (Ts - TS'J 5.67 x lo-* x 0.92 [(260 + 273)4- (100 + 273)4] [(260 + 273) - (lo0 + 273)] or h, = 20 W I ~ ~ O C . h = 395.84 + 20 = 415.84 w / m 2 ~ c Hence the power dissipation per unit length for the heater = h x (xD x 1) x (260 100) = 415.84 x A x 0.01 x 160 = 2090 Wlm = 2.09 kW/m 9.3. CONDENSATION HEAT TRANSFER

...I Eqn.(9. l5)]

.

-

93.1. General Aspects The condensation process is the reverse of boiling process.Thecondensationsets iq, whenever r saturation vapour comes in contact with a surface whose temperature is lower than the saturation

Boiling and Condensation temperature comsponding to the vapour pressure. As the vapour condenses, latent heat is liberated and there is flow of heat to the surface. The liquid condensate may get somewhat sub-cooled by contact with the cooled surface and that may eventually cause more vapour to condense on the exposed surface or upon the previously formed condensate. Depending upon the condition of cool surface, condensation may occur in two possible ways: Film condensation and dropwise condensation. 1. Film Condensation / If the condensate tends to wet the surface and therebyforms a liquidfilm, then the condensation process is known as 'jilm condensation'. Here, the heat from the vapour to the cooling medium is transferred through the film of the condensate formed on the surface. The liquid flows down the cooling surface under the action of gravity and the layer continuously grows in thickness because of newly condensing vapours. The continuous film offers thermal resistance and checks further transfer of heat between the vapour and the surface. Further, the heat transfer from the vapour to the cooling surface takes place through the film formed on the surface. The heat is transferred from the vapour to the condensate formed on the sur$ace by 'convection' and it isfurther transferred from the condensate film to the cooling surfoee by the 'conduction'. This combined mode of heat transfer by conduction and convection reduces the rates of heat transfer considerably (compared with dropwise condensation). That is the reason that heat transfer rates of filmwise condensation are lower than dropwise condensation. Figure 9.6 (i) shows the film condensation on a vertical plate.

Heat and Mass Transfer

507

surfaces become 'wetted' after being exposed to condensing vapours ovkq a period of time. Dropwise condensation can be obtained under controlled conditions with the help of certain additives to the condensate and various surface coatings but its commercial viability has not yet been approved. For this reason the condensing equipment in use is designed on the basis of filmwise condensation. (.

9.3.2. Laminar Film Condensation on a Vertical Plate I An analysis for filmwise condensation on a vertical plate can be made on lines prepared by Nusselt (1916). Unless the velocity of the vapour is very high or the liquid film very thick, the motion of the condensate would be laminar. The thickness of the condensate film will be a function of the rate of condensation of vapour and the rate at which the condensate is removed from the surface. The film thickness on a vertical surface will increase gradually from top to bottom as shown in Fig. 9.7.

Condensate film

Element P$ f b h dy) pig f b h dy)

L

(b)Force balance m

drops

0

0

0

tv

f- Condensate film ( i ) Film pondensation

I

tsar

(a) Film growth, velocity

(c) Mass and heat balance

aFig. 9.7. Film condensation on a flat vertical plate.

and temper ture profiles (ii) Dropwise condensation

Fig. 9.6. Film and dropwise condensations on a vertical surface

2. Dropwise condensation In 'dropwise condensation' the vapour condenses into small liquid droplets of various sizes which fall down the surface in random fashion. The drops form in cracks and pits on the surface. grow in size, break away from the surface, knock off other droplets and eventually run off the surface, without forming a film under the influence of gravity. Fig. 9.6 (ii) shows the dropwise condensation on a vertical plate. In this type of condensation, a large portion of the area of sdid surface is directly exposed to vapour without an insulating film of condensate liquid, consequently higher heat transfer rate (to the order of 750 kwlm2) are achieved. Dropwise condensation has been observed to occur either on highly polished surfaces, or on surfaces contaminated with impurities like fatty acids and organic compounds. This type of condensation gives coefficient of heat transfer generally 5 to 10 times larger than with film condensation. Although dropwise condensation would be prefened to filmwise condensation yet it is extremely difficult to achieve or maintain. This is because most

Nusselt's analysis of film condensation makes the following simplifying assumptions. 1. The film of the liquid formed flows under the action of gravity. 2. The condensate flow is laminar and the fluid properties are constant. 3. The liquid film is in od thermal contact with the cooling surface and, therefore, temperature at the inside o ,the film is taken equal to the surface temperature t,. Further, the temperature at the liquid-ttapour interface is equal to the saturation temperature t,,, at the prevailing pressure. 4. Viscous shear and gravitational rorces are assumed to act on the fluid; thus normal viscous force and inertia forces are neglected. 5. The shear stress at the liquid-vapour interface is negligible. This means there is no velocity

\

gradient at the liquid-vapour interface [ i.e., 6. The heat ansfer across the condensate layer is by pure conduction and temperature distribut6 is linear.

Boiling and Condensation

508

7. The condensing vapour is entirely clean and free from gases, air and non-condensing impurities. 8. Radiation between vapour and liquid film; horizontal component of velocity at any point in the liquid film; and curvature of the film are considered negligibly small. Consider the process of film condensation occuring on the surface of a flat vertical plate as shown in Fig. 9.7. The coordinate system has also been depicted on the figure. The origin '0' is at the upper end of the plate, the X-axis lies along the vertical surface with the positive direction of X measured downward and Y-axis is perpendicular to it. The vertical plate height I, width b, and 6 denotes the thickness of the film at a distance x from the origin. The liquid film thickness which is zero at the upper end of plate gradually increases as further condensation occurs at the liquid-vapour interface and attains its'maximum value at the lower end of the plate. p, = density of liquid film, . Let, pv = density of vapour, hfg = latent heat of condensation, k = conductivity of liquid film, p = absolute viscosity of liquid film, ts = surface temperature, and tSa = saturation temperature of vapour at prevailing pressure. (a) Velocity distribution o find an expression for velocity distribution u as a function of distance y from the wall , let us consider the equilibrium between gravity and viscous forces on an elementary volume (b dx dy) of the liquid film. ...(i) Gravitational force on the element = plg (b dx dy) - pv g (b dx dy) Viscous shear force on the element

...(ii)

du

Heat and Mass Transfer

509

Substituting the values of C,and C2 we get the velocity profile

Equation (9.18) is the required velocity profile. The mean flow velocity urn, of the liquid film at a distance y is given by

(b) Mass flow rate The mass flow rate of condensate through apy x position of the film is given by : Mass flow rate (m) = mean flow velocity (urn)x flow area x density

or

m = (PI - PV) 8

.a2 x b . 6 x p l = ~ l ( ~ l - ~ v ) g . b .

0) 3Y The mass flow is thus a function of x; this is so because the film thickness 6 is essentialfy dependent upon x. As the flow proceeds from x to (x + dx) the film grows from 6 to (6 + d6) because of additional condensate. The mass of condensate added between x and (x + dx) can be worked out by differentiating Eqn. (9,.20) with respect to x (or 6).

3P

Equating (i) and (ii), we get

1

- d2u =

(P~-PV)~

d>"

...(9.16)

c1

Upon integration, we have

Integrating again, we get CL The relevant boundary conditions are:

Using these boundary conditions, we get the following values of C, and C2.

(c) Heat flux The heat flow rate into the film (dQ) equals the rate of energy release due to condensation at the surface. Thus

According to our assumption the heat transfer across the condensate layer is by pure conduction, hence

Combining Eqns. (9.22) and (9.23), We have

510

Boiling and Condensation

Heat and Mass Transfer

Integrating the above equation, we get

1'"

Substitution of the boundary condition : 6 = 0 at x = 0 yields C, = 0.Hence 4k p (tsa - is)x

PI (PI - Pv) g hfg

...(9.24)

The equation (9.24) depicts that the heat film thickness increases as the fourth root of the distance down the surface; the increase is rather rapid at the upper end of the vertical surface and slows thereafter. (d) Film heat transfer coefficient According to Nusselt assumption the heat flow from the vapour to the surface is by conduction through the liquid film. Thus

The heat flow can also be expressed as dQ = hx ( b &) (tsat - ts) where hx is the local heat transfer coefficient From (i) and (ii), we get

...(ii)

, where hL is the local heat transfer coefficient at the lower edge of the plate. . * 4 This shows that the average heat transfer coefficient is 3 times the local heat transfer coefficient at the trailing edge of plate. Equation 9.28 is usually written in the form

The Nusselt solution derived above is an approximate one because experimental results have shown that it yields results which are approximately 20 percent lower than the measured values. McAdams proposed to use a value of 1.13 in place of coefficient 0.943. Hence h = 1.13

L P L (tsar - ts) J

While using above equation, it may be noted that, all liquid properties are to be evaluated at the temperature Equation (9.25)depicts that at a definite point on the heat transfer surface, the film coefficient hx is directly proportional to thermal conductivity k and inversely proportional to thickness offilm 6 at that point. Substituting the value of 6 from equation (9.24), we get

ls) and hfg should be evaluated at tSap

The total heat transfer to the surface, Q = h As (tsat - ts) The total condensation rate,

...(9.31)

Figure 9.8. shows the variation of film thickness and film coefficient with plate height

Local heat transfer coefficient at the lower end of the plate, i.e., x = 1

Evidently the rate of condensation heat transfer is higher at the upper end of the plate than that at the lower end. The average value of heat transfer can be obtained by integrating the local value of coefficient [Eqn (9.26)]as follows :

("'i

Y u

t

Film

5

Y

e

0

.g zi k3 8 .Y

s

3C4 3iZ Plate height

-+

Fig. 9.8. Film thickness and film coefficient

vs. plate height.

Fig. 9.9. Condensation on an inclined surface.

Boiling and

[graphical representation of Eqns. (9.24), (9.26) and (9.28)]. The film thickness increases with the increase of plate height. Heat transfer rate decreases with the increase of plate height since thermal resistance increases with the film thickness. e. Inclined flat plate surface For inclined flat surfaces, the gravitational acceleration g in equation (9.30) is replaced g sine where 0 is the angle between the surface and horizontal (Refer Fig. 9.9). The Eqn. (9.3 is modified as :

1

, I

/~

*

1

; (1

1

I

or hinclined = hwnical X ...(9.34) Equation (9.34) is applicable only for cases where 0 is small; is not at all applicable for horizontal plate. Condensation 9.3.3. Turbulent Film Condensation surface ', When the plate on which condensation occurs is quite long or when the liquid film is vigorous &ugh, the condensate flow may become turbulent. The turbulent results in higher-heat transfer rates because heat is now transferred not only by condensdtion but also by eddydifision. The-transition film criterion may be expressed in terms of Reynolds number defined as 'I.

C.. 4I A

-

Re = PI Urn Dh PI

where

1 I

I

--

Fig. 9.10 -

513

Heat and Mass Transfer

(

v.

For the plate, A = L x B and P = B, where L and B are height and width of plate, respectively. Thus,

When the value of Re exckds 1800 (approximately), the turbulence will appear in the liquid film. For Re > 1800, the following correlation is used:

9.3.4. Film Condensation on Horizontal Tubes NusseIt's analysis for laminar filmwise condensation on horizontal tubes leads to the following relations:

... For single horizonal tube.

"W.W...

1s of film cndens;ation on a vekcal surface.

D,, = hydraulic diameter cross-sectionalareaof fluid flow =4x wetted perimeter urn= mean or average velocity of flow

where m = pA urn For a vertical plate of unit depth, P = 1, the Reynolds number is sometimes expressed in terms of the mass flow rate per unit depth of plate r, so that

where

...For horizontal tube bank with N tubes placed directly over one another in the vertical direction, D = outer diameter of the tube.

9.3.5. Film Condensation Inside Horizontal Tubes Condensation of vapour inside the tubes finds several engineering applications such as condensers used in refrigeration and airconditioning systems and several chemical and petrochemical industries. The phenomena insides tubes are very complicated because the overall flow rate of vapour strongly affects the heat transfer rate and also the rate of condensation on the walls. Chato (1962) has recommended the following correlation for low velocities inside horizontal tubes (condensation of refrigerants):

where Equation (9.43) is restricted to low vapour Reynolds number such that R~~ =

with

r = 0, at the top of the plate and r increasing with x.

-

The Reynolds number may also be related to heat transfer coefficient as follows: Q = h As (tsar- ts) = m hjR

(" ")

< 3500

where Rev is evaluated at inlet conditions to the tubes.

9.3.6. Influence of the Presence of Noncondensable Gases The presence of noncondensable gas such as air in a condensing vapour produces a detrimental effect on the heat transfer coefficient. It has been observed that even with a few percent by volume of air in steam the condensation heat transfer coefficient is reduced by more than fifty percent. This is owing to the fact that when a vapour containing noncondensable gas condenses, the noncondensable gas is left at the surface. Any further condensation at the surface will occur only after incoming vapour has diffused through this noncondensable gas

Heat and Mass Transfer

514

collected in the vicinity of the surface. The noncondensable gas adjacent to the sui$ace acts as a thermal resistance to the condensation process. The rate of condensation decreases greatly when the condensable vapour is contaminated with even very small amounts of noncondensable gases. As the presence of noncondensable gas in a condensing vapour is undesirable, the general practice in the design of a condenser should be to vent the noncondensable gas to the maximum extent possible. Example 9.7. Discuss the diflerent types of processes for condensation of vapours on a solid su$ace. (AMIE Summer, 1998) Whenever a saturated vapour comes in contact with a surface at a lower Solution. temperature condensation occurs. There are two modes of condensation. Filmwise - in which the condensation wets the surface forming a continuous film which covers the entire surface; Dropwise - in which the vapour condenses into small droplets of various sizes which fall down the surface in a random fashion. Filmwise condensation generally occurs on clean uncontanimated sugaces. In this type of condensation the film covering the entire surface grows in thickness as it moves down the surface by gravity. There exists a thermal gradient in the film and so it acts as a resistance to heat transfer. In dropwise condensation a large portion of the area of the plate is directly exposed to the vapour, making heat transfer rates much higher (5 to 10 times) than those infilmwise condensation. Although dropwise condensation would be preferred to filmwise condensation yet it is extremely d@cult to achieve or maintain. This is because most surfaces become "wetted" after being exposed to condensing vapours over a period of time. Dropwise condensation can be obtained under controlled conditions with the help of certain additives to the condensate and various surface coatings, but its commercial viability has not yet been proved. For this reason the condensing equipments in use are designed on the basis of filmwise condemation. Example 9.8. Saturated steam at tsa,= 90°C (p = 70.14~kPa)condenses on the outer surface of a 1.5 m long 2.5 m OD vertical tube maintained at a uniform temperature T, = 70°C. Assuming film condensation, calculate : (i) The local transfer coeficient at the bottom of the tube, and (ii) The average heat transfer coeficient over the entire length of the tube. Properties of water at 80' C are :p, = 974 kg/m3, kt = 0.668 W/m K, p, = 0.335 xld kg/ms, hfg= 2309 kJ/kg, p, << p,. (AMIE Winter, 1998)

Solution. Given :t,,, = 90°C (3 = 70.14 kPa); L = 1.5 rn; D = 2.5 cm= 0.025 m; t, = 70°C.

Properties of water at 80°C

; pl = 974 kg/m3;

k = 0.668 W/m K; y = 0.335 x lo3 kg/ms; hfg= 2309 H/kg {pv<< pl) (i) The local heat transfer coefficient,h, : With usual notations, the local heat transfer coefficient for film condensation is given as

...[Eqn. (9.26)]

Boiling and Condensation

:.

Local heat transfer coefficient at the bottom of the tube, x = 1.5 m, is

(ii) Average heat transfer coefficient, h : - 4 4 h = h, =, x 3552.9 = 4737.2 J

w/&

O

C

(Ans.)

J

Example 9.9. Saturated steam at 120°C condenses on a 2 cm OD vertical tube which is 20 cm long. The tube wall is maintained at a temperature of 119°C. Calculate the average heat transfer coeflcient and the thickness of the condensate film at the base of the tube. Assume Nusselt's solution is valid. Given : I p, = 1.985 bar; pw = 943 kg/m3; hfg= 2202.2 kJ/kg; kw=0.686 W/m K; p = 2 3 7 . 3 x l u 6 Ns/m2,

(AhfIE Summer, 1997)

Solution. From Nusselt's solution, we have

neglecting p,, in comparison to pl = 1.92 x 1013 'I4 = 5. I x 10-5 m or 0.051 nun (An.)

Now,

:.

k hL---=

6,

o'686 0.051 x

= 13451

Average heat transfer coefficient,

Example 9.10. A vertical cooling jin approximating a flat plate 40 cm in height is exposed to saturated steam at atmospheric pressure (t,,, = IOOOC, A,- = 2257 kJ/kg). The fin is maintained at a temperature of 90°C. Estimate the following : (i) Thickness of the film at the bottom of the fin; (ii) Overall heat transfer coeficient; and (iii) Heat transfer rate after incorporating McAdam's correction. The relevant fluid properties are : p, = 965.3 kg/m3 k, = 0.68 W/m°C p, = 3.153 x

N s/m2

The following relations hay be used :

,,

Heat and Mass Transfer

4kA (tsar. - t x 4=[ghfePl (PI- PV)

]

Boiling and Condensation

1/4

or

-

h = 0.943

[

(980.3)~x (66.4 x I O - ~x) ~ 9.81 x (2257 x l d ) 434 x lo4 x 0.5 (loo - 30)

(AMIE, summer, 1999) Solution. Given : L = 60 cm = 0.6 m; tSa,= 400°C; hfg= 2257 kJ/kg; t, = 90°C; p, = 965.3 kg/m3; kl = 0.68 W/m°C;

pl = 3.153 x

N s/m2 (ii) The condensate rate per meter width, m:

(i) Thickness of film at the bottom edge of the fin, 6 L :

534S9 lo3 = 236.86 kg& (Ans.) 2257 Example 9.12. A vertical plate 350 mm high and 420 mm wide, at 40°C is exposed to saturated steam at 1 atm. Calculate the following: (i) The Jilm thickness at the bottom of the plate; (ii) The maximum velocity at the bottom of the plate; (iii) The total heat flux to the plate. Assume vapour density is small compared to that of the condensate. Solution. Given: t, = 40°C; tsar= 100°C, L = 350 mm = 0.35 m, B = 420 mm = 0.42 m. The properties will be evaluated at the film temperature, i.e., the average of tsarand t,;

w=-= hfg

= 0.0001 136 m = 0.1 136 mm

(ii) Overall heat transfer coefficient,

(Ans.)

I; :

t f = loo +40 = 70°C; further hfL is evaluated at 100°C.

(iii) Heat transfer rate with McAdam's correction : With McAdam's correction, the value of 7; is 20 percent higher. Hence heat transfer rate after incorporating McAdam's correction for unit width, is : Q = 1 . 2 ~ 7 9 8 1 . 2 2 ~ ( 0l)x(l00-90) .4~

(Ans.) Example 9.11. A vertical plate 500 mm high and maintained at 30°C is exposed to saturated steam at atmospheric pressure. Calculate the following: ( i ) The rate of heat transfer, and (ii) The condensate rate per hour per metre of the plate width for film condensation. The properties of water film at the mean temperature are: p = 980.3 kg/m3; k = 66.4 x W/nt°C; p = 434 x 1Od kg/ms and hfg = 2257 kmg. Assume vapour density is small compared to that of the condensate. 2 Solution. Given: L = 500 mm = 0.5 m; B = lm; t, = 30°C; ( i ) The rate of heat transfer per metre width, Q:

2 The properties at 70°C are: kgtms; k = 0.667 W/m°C and hfg = 2257 kl/kg. p, = 977.8 kg/m3; P = 0.4 x (i) The film thickness at the bottom of the plate, 6: 6=

= 38309.8 W/m or 38.3098 kW per m width

..[Eqn. (9.29)]

r p,'Pg%

= 0.943 1 *

llA

I

L P L (tsar - J

...neglecting p,[(p,

<< pi ...given)]

...[Eqn. (9.24)l

g P, (PI - P,) hfg

Neglecting p,, pv << PI or

4 x 0.667 x 0.4 x lo-) (100 - 40) x 0.35 9.81 x (977.81~ x 2257 x lo3

I

1/4

(': x

(ii) The maximum velocity at the bottom of the plate, u,,

(Sy-$] - "(.-$] P

...(ghen)

=

104

= 0.18 mm

= 1 = 0.35 m in this case ) :

u = (p1-pv)g P

At y = 6, u = urnax.,therefore, p1gS2 977.8~9.81x(1.8Xl@)~ urn=-=0,38slnls 2P 2 ~ 0 . 4 ~

...[Eqn. (9.17)] ..neglecting p,

Heat and Mass Transfer

518

(iii) The total heat flux to the plate, Q:

...[Eqn. (9.29)]

Boiling and Condensation

519

(iii) The heat transfer rate, Q : Q = h A, (I,,, - t,) = h x ( L x - ty) = 5926.4 x (0.6 x 1) (100 - 60) = 142233.6 W (iv) The condensate mass flow rate, m :

--

142233'6 = 0.063 kgls or 226.8 k g h (Am.) 2257 x lo3

Let us check whether the flow is laminar or not

..[Eqn.(9.35)] The total heat flux is given by Q = % A (tsar- t,) = L x ( L x B) (t,,, - ts) = 4931.35 x 0.35 X 0.42 X (100 - 40) = 43494 W or 43.494 k W (Ans.) Example 9.13. Vertical f i t plate in the form of fin is 600 m in height and is exposed to steam at atmospheric pressure. if surface of the plate is maintained at 60°C, calculate the following: (i) The film thickness at the trailing edge of the film, (ii) The overall heat transfer coeficient, (iii) The heat transfer rate, and (iv) The condensate mass flow rate. Arsume laminarflow conditions and unit width of the plate. Solution. Given :L = 600 mm = 0.6 m; ts = 100°C; The properties of vapour at atmospheric pressure are: tsar= l0O0C, hfg = 2257 H k g ; pv = 0.596 kglm3. loo 6o = 80°C are: The properties of saturated vapour at the mean film temperature tf = 3 +

rr

p, = 971.8 kg/m3, k = 67.413 x lo-' W/m°C, p = 355.3 x lod Ns/m2 or kglms (i) The film thickness at the trailing edge of the plate, 6 (at x = L = 0.6 m):

This shows that the assumption of laminar flow is correct. Example 9.14. A vertical tube of 60 mm outside diameter and 1.2 m long is exposed to steam at atmospheric pressure. The outer surface of the tube 50°C by circulating cold water through the tube. Calculate th (i) The rate of heat transfer to the coolant, and (ii) The rate of condensation of steam. Solution. Given : D = 60 mm = 0.06 m, L = 1.2 m, t, = 50°C Assuming the condensation film is laminar and noncondensable gases in steam are absent. loo+ 50 = 75°C The mean film temperature tf = 2 The thermo-physical properties of water at 75°C are: p, = 975 kg/m3, pl = 375 x lo4 Ns/m2, k = 0.67 W/m°C. The properties of saturated vapour at tsa = 100"C\are : p, = 0.596 kg/m3, hfg = 2257 Hkg. (i) The rate of heat transfer, Q: For laminar condensation on a vertical surface ...[Eqn. (9.30)]

...[Eqn. (9.24)l

(ii) The overall heat transfer coefficient, %:

Using McAdam's correction which is 20% higher than Nusselt's result. h = 4938.68 x 1.2 = 5926.4 w/rn2"c (Ans.)

= 4627.3 w/m2"c Q = h A, (tsar- t,) = h (n DL) (t,,, - ts) = 4627.3 x (71 x 0.06 X 1.2) (100 - 50) = 52333.5 = 52.333 kW (Ans.) (ii) The rate of condensation of steam, m : The condensation rate is given by

Heat and Mass Transfer

520

Let us check the assumption of laminar film condensation by calculating, Re.

...[Eqn. (9.3511

Since Re (= 1312.85) < 1800, hence the flow is laminar. Example 9.15. A horizontal tube of outer diameter 20 mm is exposed to dry steam at 100°C. The tube surfQcetemperature is maintained at 84°C by circulating water through it. Calculate the rate of formation of condensate per metre length of the tube. Solution. Given: D = 20 mm = 0.02 m, ts = 84°C; t,, = 100°C '

The mean film temperature t, = loo + 84 = 9 2 0 ~ 2 The properties of saturated liquid at 92°C are: p, = 963.4 kg/m3, CL, = 306 x lo4 ~ s l mk~=; 0.677 W/m°C The properties of saturated vapour at tsar= 100°C are: I p, = 0.596 kg/m3, hfg = 2257 Wkg Rate of formation of condensate per metre length of the tube, m: The average heat transfer coefficient is given by

Boiling and Condensation

521

p, = 992 kg/m3; j~ = 663 x lo4 ~ s / mk~=; 0.631 W/m°C

Since the tubes are arranged in square array, therefore, the number of horizontal tubes in = 25 vertical column is : N = ( i ) The heat transfer coefficient, The average heat transfer coefficient for steam condensing on bank of horizontal tubes is given by

a

x:

...[Eqn. (9.41)]

(ii) The rate at which steam is condensed per unit length, m The rate of condensation for the single tube of the array per metre length is

...[Eqn. (9.40)]

The rate of condensation for the complete array is m = 625 x m, = 625 x 1.116 x lom3= 0.6975 kg1s.m = 11579.7 w / m 2 ~ c The heat transfer per unit length is

= 11579.7 X 71 x 0.02 x (100 - 84) = 11641.2 W Rate of formation of condensate per metre length of the tube,

"'

- =--

11641'2 = 5.157 x kgls = 18.56 kgh (Ans.) L hfg - EKiiY Example 9.16. A steam condenser consisting of a square array of 625 horizontal tubes,

each 6mm in diameter, is installed at the exhaust hood of a steam turbine. The tubes are exposed to saturated steam at a pressure of 15 kPa. If the tube sudace temperature is maintained at 25°C calculate: (i) The heat transfer coeficient, and (ii) The rate at which steam is condensed per unit length of the tubes. Assume film condensation on the tubes and absence of noncondensable gases. Solution. Given: D = 6 mm = 0.006 m, ts = 25°C. Corresponding to 15 kPa pressure, the properties of vapour (from the table) are: tsar= 54OC, p, = 0.098 kg/m3, hfg = 2373 kJkg. The properties of saturated water at film temperature t f = 54 + 25 = 39.5"C are : 2

-

(Ans.)

TYPICAL EXAMPLES Example 9.17. A 750 mm square plate, maintained at 28OC is exposed to steam at 8.132 kPa. Calculate the following: ( i ) The film thickness, local heat transfer coeficient and mean jlow velocity of condensate at 400 mm from the top of the plate, (ii) The average heat transfer coeficient and total heat transfer from the entire plate, (iii) Total steam condensation rate, and (iv) The heat transfer coeficient if the plate is inclined at 25" with the horizontal plane. Solution. Given: L = B = 750 mm = 0.75 m, t, = 28OC, x = 400 mm = 0.4 m Assme laminar flow film condensation. Properties of saturated vapour at 8.132 kPa (or 0.08132 bar) are: t,,, = 42°C; p, = 0.0561 kg/m3; hfg = 2402 kJkg 42 + 28 The mean film temperature t, = -= 35°C 2 The properties of saturated water at 35°C are: W/m°C, p = 728.15 x lo6 kg/ms. p1 = 993.95 kg/m3; k = 62.53 x (i) 6,, h,, u, at 400 mm from the top of the plate: The film thickness at a distance x from the top edge of the plate is given by: ...[Eqn. (9.24)]

HBat and Mass Transfer

At x = 0.4 m,

6, = 1.819 x 104x (0.4)'"- 1.45 x 1 0 4 m = 0.145 mm (Ans.) SL = 1,819 x (0.75)"~= 1.69 x lo4 m = 0.169 mm The local heat transfer coeficient,

Boiling and Gondensation The mean film temperature tf = 100 + 54 = 7 7 0 ~ 2 The properties of the condensate at 77°C are : W/m°C; & = 365 x 10" ~ s / m ~k ; = 668 x

The properties of saturated vapour at tSat= 100°C pv = 0.596 kg/m3; hfg= 2257 kJkg. Assuming the flow to be turbulent the relevant equations are :

The mean flow velocity of condensate,

Re =

- t,)

...[Eqn. (9.38)]

hfg PI

-

h = 0.0077 1 ' ( " -

and

L

Eliminating

(ii) Average heat transfer coefficient (g):

4 h L (t,,

P?

1/3

lZg] J

(~e)'.'

...[Eqn. (9.39)]

from these equations, we get the condition that the flow will be turbulent if

Substit-lting the values, we get

(where = the film thickness at the bottom of the plate). Using McAdarns correction, h = 1.2 x 4933.33 = 5920 W I ~ ~ O C The total heat transfer from the entire plate, Q: Q = i;A, (tsar- t,) = ( L x B) (tsar- 2,) = 5920 x (0.75 x 0.75) (42 - 28 ) = 46620 W (Ans.) PY (iii) Total steam condensation rate, m:

-

Thus the film is turbulent as assumed and Re = 4144.8

46620 = 0.0194 kgls or 69.87 kglh (Ans.) 2402 x lo3 (iv) f i e heat transfer coefficient when the plate is inclined 25" with the horizontal, m =

hinc~in,:

hind,,,

= hvertica, x (sin 0)'" = 5920 x (sin 250)"~= 4773.2 w / ~ ~ " c(Ans.)

...[Eqn. (9.34)]

Let us check the type of flow [using Eqn. (9.35)],

Hence assumption is correct. Example 9.18. A vertical plate 3.2 m high maintained at 54"C, is exposed to saturated steam at atmospheric pressure. Calculate the heat transfer rate per unit width. Solution. Given: L = 3.2 m; B = 1 m, t, = 54°C; tsa,= 100°C. Heat transfer rate per unit width : In order to determine whether the condensate film is laminar or turbulent; the Reynolds number must be checked.

= 0.0077 x (2.0797 x 10'~)'" x 27.99 = 5866.62 w/m2 "C Heat transfer rate per unit width, Q = h As (t,,, 4 ) = 5866.62 x (3.2 x 1 ) (100 - 54) = 863566 W/m = 863.566 kW/m (Am) Example 9.19. A condenser is to be designed to condense 1800 kg/h of dry and saturated steam at a pressure of 10 kPa. A square array of 400 tubes, each of 8 mm in diameter, is to be used. If the tube su$ace temperature is to be maintained at 24"C, calculate the following: (i) The heat transfer coeficient, and (ii) The length of each tube assuming single pass. Solution. Given: m = 1800 kgh; D = 8 mm = 0.008 m, t, = 24°C. (i) The heat transfer coefficent, h: Corresponding to 10 kPa (0.1 bar), from table, the properties of dry and saturated vapour are :

-

tsar= 458°C; pv =

(J -

= 0.0676 kg/m3; hf, = 2393 Wlkg

The properties of saturated vapour at the mean film temperature,

Heat and Mass Transfer

pl = 993.95 kg/m3; k = 62.53 x W/m°C; JL. = 728.15 x lo4 kglms. , As the tubes are to be arranged in an array, therefore, the number of horizontal tubes in the vertical column is, N = = 20. The average heat transfer coefficient for steam condensing on bank of horizontal tubes is given by :

...[Eqn. (9.41)]

Boiling and Condensation

525

The average heat transfer coefficient is given by

Using McAdam's correlation, we have h = 1.2 x 3432.09 x (L)-'I4 = 4118.5 x (L)-'/~ The heat transfer rate is given by Q = i;A, (tsa,- t,) = m hfg or 41 18.5 x (L)-'" (n x 0.35 x L) (120.2 - 80) = 70 x (2201.6 x lo3) or

(ii) The length of each tube, assuming single pass, L : The heat transfer rate, Q = - 4 (tsa,- ts) mhfg = h x (400 x nDL) (tsa,- t,) or

182046.8 (Ll3I4= 42808.88

(ii) The thickness of condensate layer, 6: 6 = 1.988 x 10-4 x (L)'I4 = 1.988 x lo4 x (0.1452)'" = 1.227 x lo4 m = 0.1227 mm (Am.) Let us check whether flow is lamniar or not.

L =

196500 = 1.09 m (Ans.) 1092147.3 Example 9.20. The outer sutface of a cylindrical drum 350 mm diameter is exposed to ' ~ t u r a t e dsteam at 2.0 bar for condensation. If the sutface temperature of the drum is maintained at 80°C calculate the following: (i) The length of the drum; (ii) The thickness of the condensate layer to condense 70 kg/h of steam. Solution. Given : D = 35 mm = 0.35 m; t, = 80°C; m = 70 kg / h Assuming film condensation and laminar flow : Corresponding to 2.0 bar, from the table, the properties of the saturated vapour are:

The properties of saturated water at the mean film temperature 9 = 1202

2

+ 80

= loooC

p, = 956.4 kg/m3; k = 68.23 x W/m°C; p = 283 x lo4 kgtms ( i ) The length of the drum, L: The film at the bottom edge of the drum is given by

...[Eqn. (9.24)]

As Re (= 249.9) < 1800, hence assumuption is correct. HIGHLIGHTS 1. Boiling is the convective heat transfer process that involves a phase change from liquid to vapour state. 2. The boiling heat transfer phenomenon may occur in the follows forms: (i) Pool boiling (ii) Forced convection boiling (iii) Sub-cooled or local boiling (iv) Saturated boiling. 3. The three regimes of boiling are: (i) Interface evaporation (ii) Nucleate boiling (iii) Film boiling. 4. The condensation process is reverse of boiling process. Condensation may occur in two possible ways: (i) Film condensation (ii) Dropwise condensation. If the condensate tends to wet the surface and thereby forms a liquid film, the condensation process is known as 'film condensation'. In 'dropwise condensation' the vapour condenses into small liquid droplets of various sizes which fall down the surface in random fashion. Summary of formulae : A. Boiling

Heat and Mass Transfer

Boiling and Condensation

... for single horizontal tube 13.

'

Nu = 0.16 ( ~ r . P r ) q . ~ ~

... for nucleate boiling at atmospheric pressure

Nu = 0.61 ( ~ r . P r ) ~ . ~ '

on a flat plate with low heat fluxes. ... For the nucleate boilng on a vertical flat plate.

qsc = 0.18 (pv)l0 hfg rg 0 (PI - m (h)"I3= (hconv.)413 + brad . (h)lJ3 h

3

'hem. + 2 hn2d

-

r P, ( P, - pV)pghfg]'I4

h = 0.725 /

L N Pr (tsat - ts)D J

... for horizontal tube bank with N tubes placed directly over one another in the vertical direction.

where D = outer diameter of the tube.

3

1 / 4

where hig = hfg + g c,, (t, - t,)

... critical heat flbx for nucleate pool hoiling

THEORETICAL QUESTIONS

... within f 5% ertor

1. 2. 3. 4. 5. 6. 7. 8. 9. 10.

B. Condensation :

Define the term 'boiling'. Enumerate the applications of boiling heat transfer. Explain briefly the physical mechanism of boiling. Differentiate between pool boiling and forced convection boiling. Explain briefly the various regimes of saturated pool boiling. What is burnout point? Explain briefly the condensation mechanism. Differentiate between the mechanism of filmwise and dropwise condensation. Derive the Nusselt theory of laminar flow film condensation on a vertical palte. Drive the following relation for laminar film condensation on vertical plate.

UNSOLVED EXAMPLES Boiling Heat Transfer

10.

hinclinad = (h)venicalx (sin 8)lJ4

...for Re > 1800

1. Water at atmospheric pressure is to be boiled in polished copper pan. The diameter of the pan is 300 mm and is kept at 111"C. Calculate the following : ( i ) Power of the burner to maintain boiling; (ii) Rate of evaporation in kglh. Take the properties of water at 100°C as follows: pi = 958 kg/m3; pv = 0.597 kg/m3; ~ l=f 278 x lod kg/ms; cfl= 4216 JkgK; hfg = 2257 kJ&g ; Pr = 1.723, o = 58.9 x Nlm [Ans. ( i ) 13.664 kW, (ii) 21.8 kgh] 2. A wire of 1 mm diameter gnd 150 mm length is submerged horizontally in water at 7 bar. The wire carries a current of 131.5 A with an applied voltage of 2.15 V. If the surface of the wire is maintained at 180°C, calculate: ( i )The heat flux, and (ii)The boiling heat transfer coefficient. [Ans. ( i )0.6 MW/m2, (ii) 39920 ~ l r n * O C ] 3. An electric wire of 1.5 mm diameter and 200 mm long is'laid horizontally and submerged in water at atmospheric pressure. The wire has an applied voltage of 16 V and carries a current of 40 amperes. [Ans. ( i )0.679 MWI~', ( i i ) 18.52OC) Calculate: ( i )The heat flux, and (ii)The excess temperature. 4. A nickel wire of 1.5 mm diameter and 500 mm long, carrynig current, is submerged in a water bath which is open to atmospheric pressure. Calculate the voltage at the burnout point if at this point the wire carries a current of 200 A. [Ans. 17.9 V (uppx.)) 5. A metal-clad heating element is of 8 mm diameter and of emissivity 0.95. The element is horizontally immersed in a water bath. The surface temperature of the metal is 260°C under steady state bolllng

Heat and Mass Transfer

528

conditions. Calculate the power dissipation per unit length for the heater if water is exposed to atmospheric [Am. 1.75 kW/m] pressure and is at uniform temperature.

Condensation Heat Transfer 6. A vertical plate 450 mm high and maintained at 30°C is exposed to staturated steam at atmospheric

7.

8. 9.

10.

11.

pressure. Calculate: (i) The rate of heat transfer, and (ii) The condensate rate per hour per metre for plate width film condensation. The properates of water film at the mean temperature are: W/m°C; p = 434 x lo4 kglms; and hf,= 2256.9 kJkg p = 980.3 kg/m3 ; k = 66.4 x [Am. 439.9 x lo3 kJ/h, 218.8 kg/hl A vertical plate in the form of fin is 500 mm in height and is exposed to steam at atmospheric pressure. If the surface of the plate is maintained at 60°C, calculate: (i) The film thickness at the trailing edge of the film, (ii) The overall heat transfer coefficient, (iii) The heat transfer rate, and (iv) The condensate mass flow rate. Assume laminar flow conditions and unit width of the plate. [Ans. (i) 0.1732 mm, (ii) 6227.52 W/m°C (McAdam's), (iii) 124550 W, (iv) 0.055 kg/s] A vertical plate 2.8 m high is maintained at 5 4 T in the presence of saturated steam at atmospheric [Ans. 700 kW/m) pressure. Calculate the heat transfer rate per unit width. A vertical tube of 50 mm outside diameter and 2 m long is exposed to steam at atmospheric pressure. The outer surface of the tube is maintained at a temperature of 84°C by circulating cold water through the tubes. Determine: (i) The rate of heat transfer to the coolant, and (ii) the rate of condensation of [Ans. (i) 179 kW, (ii) 28.6 kgh] steam. A horizontal tube of outer diameter 25 mm is exposed to dry steam at 100°C. The tube surface temperature is maintained at 84°C by circulating water through it. Calculate the rate of formation of condensate per [Ans. 21.94 kg/h] metre length of the tube. A condenser is to be deigned to condense 2250 kg/h of dry and saturated steam at a pressure of 15 kPa. A square array of 400 tubes each of 6 mm in diamter, is to be used. If the tube surface temperature is to be maintained at 2G°C, calculate the heat transfer coefficient and the length of each tube, assuming single [Ans. 5205.3 w/m2"C; 1.35 ml pass.

Heat Exchangers 10.1. Introduction. 10.2. 'Qpes of heat exchangers. 10.3. Heat exchanger analysis. 10.4. Logarithmic mean temperature difference (LMTD). 10.5. Overall heat transfer coefficient. 10.6. Correction factors for multipass arrangements. 10.7. Heat exchanger effectiveness and number of transfer units (NTU). 10.8. Pressure drop and pumping power; rnical Examples-Highhghts-Theoretical Questions-Unsolved Examples.

10.1. INTRODUCTION A 'heat exchanger' may be defined as an equipment which transfers the energy from a hot fluid to a cold fluid, with maximum rate and minimum investment and running costs. In heat exchangers the temperature of each fluid changes as it passes through the exchangers, and hence the temperature of the dividing wall between the fluids also changes along the length of the exchanger. Examples of heat exchangers: (ii) Condensers and boilers in steam plant; (i) Intercoolers and preheaters; (iii) Condensers and evaporators in (iv) Regenerators; refrigeration units; (v) Automobile radiators; (vi) Oil coolers of heat engine; (vii) Milk chiller of a paste(viii) Several other industrial processes. urising plant;

.

I

TYPES OF HEAT EXCHANGERS In order to meet the widely varying applications, several types of heat exchangers have been developed which are classified on the basis of nature of heat exchange process, relative direction of fluid motion, design and constructional features, and physical state of fluids. 1. Nature of heat exchange process Heat exchangers, on the basis of nature of heat exchange process, are classified as follows: (i) Direct contact (or open) heat exchangers. (ii) Indirect contact heat exchangers. (a) Regenerators. (b) Recuperators. (i) Direct contact heat exchangers In a direct contact or open heat exchanger the exchange of heat takes place by direct mixing of hot and cold fluids and transfer of heat and mass takes place simultaneously. The use of such units is made under conditions where mixing of two fluids is either harmless or desirable. Examples: (i) Cooling towers; (ii) Jet condensers; (iii) Direct contact feed heaters. Figure 10.1 shows a direct contact heat exchanger in which steam mixes with cold water, gives its latent heat to water and gets condensed. Hot water and non-condensable gases leave the container as shown in the figure. (ii) Indirect contact heat exchangers 10.2.

Heat and Mass ~ransfer

330

In this type of heat exchanger, the heat transfer between two fluids could be carried out by transmission Hot through wall which separates water the two fluids. This type inContainer cludes the following: (a) Regenerators. (b) Recuperators or surface exchangers. (a) Regenerators: In a regenerator type of heat exchanger the hot and coMfluids Cold pass alternately through a water space containing solid particles (matrix), these particles Fig. 10.1. Direct contact or open heat exchanger. providing alternately a sink and a source for heat flow. Examples : (i) I.C. engines and gas turbines; (ii) Open hearth and glass melting furnaces; (iii) Air heaters of blast furnaces. A regenerator generally operates periodically (the solid matrix alternately stores heat extracted from the hot fluid and then delivers it to the cold fluid). However, in some regenerators the matrix is made to rotate through the fluid passages arranged side by side which makes the heat exchange process continuous. The performance of these regenerators is affected by the following parameters: (i) Heat capacity of regenerating material, (ii) The rate of absorption, and (iii) The release of heat. Advantages 2. Less weight per kW of the plant; 1. Higher heat transfer coefficient; 3. Minimum pressure loss; 4. Quick response to load variation; 5. Small bulk weight; 6. Efficiency quite high. Disadvantages 1. Costlier compared to recuperative heat exchangers. 2. Leakage is the main trouble, therefore, perfect sealing is required. (b) Recuperators: 'Recuperator' is the most important type of heat exchanger in which the flowing fluids exchanging heat are on either side of dividing wall (in the form of pipes or tubes generally). These heat exchangers are used when two fluids cannot be allowed to mix ix., when the mixing is undesirable. Examples: (i) Automobile radiators, (ii) Oil coolers, intercoolers, air preheaters, economisers, superheaters, condensers and surface feed heaters of a steam power plant, (iii) Milk chiller of pasteurising plant, (iv) Evaporator of an ice plant: Advantages 1. Easy construction; 2. More economical; 3. surface area for heat transfer; 4. Much suitable for stationary plants. Disadvantages 1. Less heat transfer coefficient; 2. Less geneuting capacity; 3. Heavy and sooting problems.

,

t

Free Convection

531

The flow through direct heat exchangers and recuperators may be treated as steady state while through regenerators the flow is essentially transient. 2. Relative direction of fluid motion According to the relative directions of two fluid streams the heat exchangers are classified into the following three categories: (i) Parallel flow or unidirection flow (ii) Counte-flow (iii) Cross-flow. (i) Parallel flow heat exchangers In a parallel flow heat exchanger, as the name suggests, the two fluid streams (hot and cold) travel in the same direction. The two streams enter at one end and leave at the other end. The flow arrangement and variation of temperatures of the fluid streams in case of parallel flow heat exchangers, are shown in Fig. 10.2. It is evident from the Fig.l0.2(b) that the temperature diference between the hot and cold fluids goes on decreasing .from inlet to outlet. Since this type of heat exchanger needs a large area of heat transfer, therefore, it is rarely used in practice. r (temperature)

-

L (length)

Fig. 10.2. Parallel flow heat exchanger.

Examples: Oil coolers, oil heaters, water heaters etc. As the two fluids are separated by a wall, this type of heat exchanger may be called parallel flow recuperator or surface heat exchanger. (ii) Counter-flow heat exchangers In a counter-flow heat exchanger, the two fluids flow in opposite directions. The hot and cold fluids enter at the opposite ends. The flow arrangement and temperature distribution for such a heat exchanger are shown schematically in Fig. 10.3. The temperature difference between the

ore

I

I

rc 1

(4

(b) Fig. 10.3. Counter-flow heat exchanger

Heat and Mass Transfer

532

Free Convection

two fluids remains more or less nearly constant. This type of heat exchanger, due to counter flow, gives maximum rate of heat transfer for a given surface area. Hence such heat exchangers are most favoured for heating and cooling of fluids. (iii) Cross-flow heat exchanger In cross-flow heat exchangers, the two fluids (hot and cold) cross one another in space, usually at right angles. Fig. 10.4 shows a schematic diagram of common arrangements of cross-flow heat exchangers. Cold fluid (in) (unmixed stream) Cold fluid (in) Tube(Mixed hot stream)

Hot fluid (out)

E3

e Hot fluid e (in) (Unmixed stream)

Hot

+fluid (out) c3

I

Cold fluid Cold fluid (in) (out) (a) One-shell pass and two-tube pass heat exchanger Cold fluid (out) (a)

Cold fluid (out) (6)

Shell fluid

+

Fig. 10.4. Cross-flow heat exchangers

Refer Fig. 10.4 (a): Hot fluid flows in the the separate tubes and there is no mixing of the fluid streams. The cold fluid is perfectly mixed as it flows through the exchanger. The temperature of this mixed fluid will be uniform across any section and will vary only in the direction of flow. Examples: The cooling unit of refrigeration system etc. Refer Fig. 10.4(b) : In this case each of the fluids follows a prescribed path and is unmixed as it flows through heat exchanger. Hence the temperature of the fluid leaving the heater section is not uniform. Examples: Automobile radiator etc. In yet another arrangement, both the fluids are mixed while they travel through the exchanger; consequently the temperature of both the fluids is uniform across the section and varies only in the direction in which flow takes place. 3. Design and constructional features On the basis of design and constructional features, the heat exchangers are classified as under: ( i ) Concentric tubes In this type, two concentric tubes are used, each carrying one of the fluids. The direction of flow may be parallel or counter as depicted in Fig. 10.2 (a) and Fig. 10.3 (a). The effectiveness of the heat exchanger is increased by using swirling flow. (ii) Shell and tube In this type of heat exchanger one of the fluids flows through a bundle of tubes enclosed by a shell. The other fluid is forced through the shell and it flows over the outside surface of the tubes. Such an arrangement is employed where reliability and heat transfer effectiveness are important. With the use of multiple tubes heat transfer rate is amply improved due to increased surface area. I

A

I

Baffles

I

.-

Tube fluid -b

I

II

* I

\

I

I

.I

\ \A

,

/I

I I I

I

I

\A/

I

(6) Two-shell pass and four-tube pass heat exchanger Fig. 10.5. Shell and tube heat exchangers.

(iii) Multiple shell and tube passes Multiple shell and tube passes are used for enhancing the overall heat transfer. Multiple shell pass is possible where the fluid flowing through the shell is re-routed. The shell side fluid is forced to flow back and forth across the tubes by baffles. Multiple tube pass exchangers are those which re-route the fluid through tubes in the opposite direction. (iv) Compact heat exchangers These are special purpose heat exchangers and have a very large transfer surface area per unit volume of the exchanger. They are generally employed when convective heat transfer coefficient associated with one of the fluids is much smaller than that associated with the other fluid.

Heat and Mass Transfer

534

Example : Plate-fin, flattened fin tube exchangers etc. (4) Physical state of fluids Depending upon the physical state of fluids the heat exchangers are classified~asfollows: (i) Condensers (ii) Evaporators (i) Condensers: In a condenser, the condensing fluid remains at constant temperature throughout the exchanger while the temperature of the colder fluid gradually increases from inlet to outlet. The hot fluid loses latent part of heat which is accepted by the cold fluid (Refer Fig. 10.6).

Fig. 10.6. Temperature distribution in a condenser.

Fig. 10.7. Temperature distribution in an evaporator.

(ii) Evporators: In this case, the boiling fluid (cold fluid) remains at constant temperature while the temperature of hot fluid gradually decreases from inlet to outlet. (Refer Fig. 10.7).

10.3. HEAT EXCHANGER ANALYSIS For designing or predicting the performance of a heat exchanger it is necessary that the total heat transfer may be related with its governing parameters : (i) U (overall heat transfer coefficient due to various modes of heat transfer, (ii) A total surface area of the heat transfer, and (iii) tl, t2 (the inlet and outlet fluid temperatures). Fig. 10.8 shows the overall energy balance in a heat exchanger. m mass flow rate, kgts. Let cp = specific heat of fluid at constant pressure J/kg°C t = temperature of fluid, "C At = temperature drop or rise of a fluid across the heat exchanger. Heat exchanger

Hot fluid , Q

-

-

Heat transfer area Fig. 10.8. Overall energy balance in a heat exchanger. Subscripts h and c refer to the hot and cold fluids respectively; subscripts band 2 correspond to the inlet and outlet conditions respectively.

Free Convection Assuming that there is no heat loss to the surroundings and potential and kinetic energy changes are negligible, from the energy balance in a heat exchanger, we have: ...(10.1) Heat -given up by the hot fluid, Q = mh Cph (th~- th2) ...(10.2) Heat picked up by the cold fluid, Q = mc cpc
10.4. LOGARITHMIC MEAN TEMPERATURE DIFFERENCE (LMTD) Logarithmic mean temperature difference (LMTD)is defined as that temperature difference which, if constant, would give the same rate of heat transfer as actually occurs under variable conditions of temperature difference. In order to derive expression for LMTD for various types of heat exchangers, the following assumptions are made: 1. The overall heat transfer coefficient U is constant. 2. The flow conditions are steady. 3. The specific heats and mass flow rates of both fluids are constant. 4. There is no loss of heat to the surroundings, due to the heat exchanger being perfectly insulated. 5. There is no change of phase either of the fluid during the heat transfer. 6. The changes in potential and kinetic energies are negligible. 7. Axial conduction along the tubes of the heat exchanger is negligible. 10.4.l.Logarithmic Mean Temperature Difference for "Parallel Flow" Refer Fig. 10.9, which shows the flow arrangement and distribution of temperature in a single-pass parallel flow heat exchanger. Let us consider an elementary area dA of the heat exchanger. The rate of flow of heat through this elementary area is given by dQ = UdA (th - tc) = u . d A . A t As a result of heat transfer dQ through the area dA, the hot fluid is cooled by dh whereas the cold fluid is heated up by dtc. The energy balance over a differential area dA may be written as ...(10.4) dQ = - mh Cph . dth = hc Cpc . dtc = U .dA . (th - tc) (Here dth is -ve and dtc is +ve)

and where

dQ = dQ dtc = ;.,cpc c c Ch L ith cph = heat capacity or water equivalent of hot fluid, and Cc = riz, cpC= heat capacity or water equivalent of cold fluid.

?ithand in, are the mass flow rates of fluids and cph and cp, are the respective specific horlr.

Heat and Mass Transfer

Free Convection ...[10.7 (a)]

/---+

----+

Annulus surrounding the pipe

...[10.7 (b)]

II Cold

---+

---+ /Pipe

Hot

4

4

-- --

4

K

Substituting the values of

1 1 and - into Eqn. (10.6), we get Ch

cc

Cold

(a)Flow arrangement

The above equation may be written as Q = UA8,

8, is called the logarithmic mean temperature differencr (LMTD).

Area -+ (b)Temperature distribution

2

Subscri~tsh. c refer to : hot and cold fluids ~ubscriDt1. .2 refer to : inlet and outlet conditions. Fig. 10.9. ~alculaGonbf LMTD for a parallel flow heat exchanger. Substituting the value of dQ from Eqn. (10.4) the above equation becomes

10.4.2. Logarithmic Mean Temperature Difference for "Counter-flow" Refer Fig. 10.10, which shows the flow arrangement and temperature distribution in a single-pass counter-flow heat exchanger. Annulus surrounding the pipe II Pipe t t Cold t

-- -/-

+

Hot

Cold

(a) Flow arrangement

Integrating between inlet and outlet conditions (i.e. from A = 0 to A = A), we get

Now, the total heat transfer rate between the two fluids is given by Q = ch (thl - th2) = Cc (tc2 - tcl)

I

I

Area -+ (b)Temperature distribution

Heat and Mass Transfer

538

Free Convection

Let us consider an elementary area dA of the heat exchanger. The rate of flow of heat through this elementary area is given by ...(10.11) dQ = U . d A (th- tc) = U.dA.At In this case also, due to heat transfer dQ through the area dA, the hot fluid is cooled down by dth whereas the cold fluid is heated by dt,. The energy balance over a differential area dA may be written as ...(10.12) dQ = - mh . Cph . dlh = - mc . cpc.dtc

A special case arises when 0 , = e2 = 0 in case of a counter-flow heat exchanger. In such a case, we have

In a counter-flow system, the temperatures of both the fluids decrease in the direction of heat exchanger length, hence the -ve signs.

This value is indeterminate. The value of 8, for such a case can be found by applying L' Hospital's rule:

and

d t c = - - =dQ -mc Cpc

dQ

cc

Let (8J0,) = R. Therefore, the above expresion can be written as 0 ( R - 1) lim ~ - 1 1 ln(R) Differentiating the numerator and denomenator with respect to R and taking limits, we get 0 lim -8 (R* 1) (1/R) Hence when 0 , = 0, Eqn. (10.3) becomes Q=UA8 0, (LMTD) for a counter-flow unit is always greater than that for a parallel flow unit; hence counter-flow heat exchanger can transfer more heat than parallel-flow one; in other words a counter-flow heat exchanger needs a smaller heating surface for the same rate of heat transfer. For this reason, the counter-flow arrangement is usually used. When the temperature variations of the fluids are relatively small, then temperature variation curves are approximately straight lines and adequately accurate results are obtained by taking the arithmatic mean temperature dijference (AMTD).

-

Inserting the value of dQ from Eqn. (10.11), we get de = - U dA (th - t,)

-- -

[ih

di

Integrating the above equation from A = 0 to A = A, we get

Now, the total heat transfer rate between the-two fluids is given by Q = ch (thl - t h ~=) Cc (tc2 'tcl)

...(10.15)

...[10.15 (a)]

1 1 substituting the values of - and - into Eqn. (10.14), we get Ch cc

However, practical considerations suggest that the logarithmic mean temperature 01 > 1.7. difference (8,) should be invariably used when 82

10.5. OVERALL HEAT TRANSFER COEFFICIENT In a heat exchanger in which two fluids are separated by a plane wall as shown in the Fig. 10.1 1 , the overall heat transfer coefficient is given by 1 U= ...( 10.18) 1 L 1-+-+hi k ho If the fluids are separated by a tubewall as shown in Fig. 10.12 the overall heat transfer coefficient is given by: Inner surface

Since

Q = UA8,

,

Heat and Mass Transfer

Heat ~xchangers*

543

--

Distilled water

0.0001

Treated boiler feed water

0.0001 - 0.0002

Worst water used in heat exchangers

< 0.0002

Fuel oil and crude oil

0.0009

Industrial liquids

0.0002

Transformer or lubricating oil

0.0002

Engine exhaust and fuel gases

0.002

Steam (non-oil bearing)

0.0001

Refrigerant liquids, brine or oil-bearing

0.0002

Table 10.2. Representative values of overall heat transfer coefficient (U) S.No.

Fluid combination

u (w/m2~c)

Water to water

850 - 1170

Water to oil

110 - 350

Steam condensers(water in tubes)

1000 - 6000

Alcohol condensers (water in tubes)

250

Feed water heaters

110 - 8500

Air-condensers

350

Air to various gases

60 - 550

Air to heavy tars and liquids

As low as 45

Air to low viscosity liquids

As high as 600

Finned-tube heat exchanger (water in tubes, air in cross-flow)

25

- 700 - 780

- 50

Fouling processes : 1. Precipitation or crystallization fouling. 2. Sedimentation or particulate fouling. 3. Chemical reaction fouling or polymerisation. 4. Corrosion fouling. 5. Biological fouling. 6. Freeze fouling. Parameters affecting fouling : Velocity

Properties to be censidered for&lectiod:of materials for heat exchaneers : Physical properties ~echanical~operties Climatic properties Chemical environment Quality of surface finish Service life Freedom from noise Reliability. Common failures In heat exchangers : Chocking of tubes either expected or extraordinary. Excessive transfer rates in heat exchanger. Increasing the pump pressure to maintain throughout. Failure to clean tubes at regularly scheduled intervals. Excessive temperatures in heat exchangers. Lack of control of heat exchangers atmosphere to retard scaling. Increased product temperature over a safe design limit. Unexpected radiation from refractory surfaces. Unequal heating around the circumference or along the length of tubes.

01 Example 10.1. For what value of end temperature dflerencis ratio -, is the arithmatic 92 mean temperature difference 5 per cent higher than the lbg-mean temperature diference? Solution: The arithmetic mean temperature difference (8) and log-mean temperature difference (0,) ratio may be written as f0, + 0,)

I

It is given' that 0 is to be 5 percent higher than 0,

Temperature Water chemistry Tube material. Prevention of fouling : The following methods may be used to keep fouling minimum : 1. Design of heat exchanger. 2. Treatment of process system. 3. By using cleaning system.

By hit and trial method, we get

Thus the simple arithmetic mean temperature difference gives results to within 5 percent when end temperature differences vary by no more than a factor of 2.2. Example 10.2. (a)Derive an expressionfor the effectiveness of a parallelflow heat exchanger in terms of the number of transfer units, NTU, and the capacity ratio Cmi,,/Cm,

544

Heat and Mass Transfer

545

Heat Exchangers

(b) h a parallel flow double-pipe heat exchanger water flows through the inner pipe and is heated from 20°C to 70°C. Oil flowing through the annulus is cooled from 200°C to 100°C. It is desired to cool the oil to a lower exit temperature by increasing the length of the heat exchanger. Determine the minimum temperature to which the oil may be cooled. '

(U.P.S.C.. 1995) Solution. (a) Refer Article 10.7. (b) Using subscripts.h and c for oil and water respectively, we have

Let 't' be the lowest temperature to which oil may be cooled and this will be the highest temperature of water too (Refer Fig. 10.13). Hence,

Areallength

-

-

tcl = 20°C

4

Cold water

thl = 75°C

4

~ owater t

4

44 5 ' ~(th2)

tcl = 20°C

4

Cold water

4

4

A

tc2

tc2

(a) Flow arrangement

Fig. 10.13

mhcph(200 - t ) = m,cPc (t - 20)

=2(t-20) 200-t=2t-40 t = 80°C (Ans.) 4' Eample 10.3. The flow rates of hot and cold water streams running through a parallel flow heat exchanger are 0.2 kg/. and 0.5 kg/s respectively. The inlet temperatures on the hot and cold sides are 75°C and 20°C respectively. The exit temperature of hot water is 45°C. If the individual heat transfer coeflcients on both sides are 650 w/m2"c, calculate the area of the heat exchanger.

(b) Temperature distribution

Fig. 10.14. Parallel flow heat exchanger.

Overall heat transfer coefficient U is calculated from the relation 1 1 1 U = -hi+ h, ,

Solution. Given: mh = 0 2 kgls; mc = 0.5 kgls; thl = 750 C ; th2= 4 5 0 ~tc, ; = 2 0 0 ~hi; = h, = 650 w/m2'c. The area of heat exchanger, A: The heat exchanger is shown diagrammatically in Fig. 10.14. The heat transfer rate, Q = 4 cph ( - th2) = 0.2 x 4.187 x (75 - 45) = 25.122 WS Heat lost by hot water = heat gained by cold water mh cph (thl - th2) = me 0.2 x 4.187 X (75 - 45) = 0.5

..

cpc (re- - tcl) X 4.187 x (tc2- 20)

Also,

*

Q = UA9, A=-- Q U 8,

- 25'122

lo* = 2.65 325 x 29.12

' (Ans.)

n-

tC2 =

32°C Logarithmic mean temperature difference (LMTD) is given by

...[Eqn. (10.9)]

&xample 10.4. The following data relate to a parallel flow heat exchanger in which air is heated by hot exhaust gases. ...155450 kJ Heat transferred per hour ...120 w/rn2"c Inside heat transfer coeflcient ...195 w/rn20c Outside heat transfer coeflcient ...450°C and 2.50% respectively Inlet and ouilet temperatures of the hot fluid

Heat and Mass Transfer

546

Inlet and outlet temperatures of the cold fluid ...60°C and 120°C, respectively Inside and outside diameters of the tube ..SO mm and 60 mm, respectively. Calculate the length of the tube required for the necessary heat transfer to occur. Neglect the tube resistance. ~ C = 450°C; ; Solution. Given : Q = 155450 kJ/h; h, = 120 w/m2'c; ho = 195 W I ~ ~ th, = 250°C; tcl = 60°C; tc2 = 120°C; di = 50 mm = 0.05m; do = 60 mm = 0.06 m. th2 Length of each tube, L: Logarithmic mean temperature differenqe (LMTD) is given by

Heat Exchangers

Heat lost by the cold fluid, Q = 0.694 x 400 (tc2- 25) = 277.6 (tc2- 25)

...(ii)

Equating (i) and (ii), we have 5560 (200 - th2)= 277.6 (tc2- 25)

...(iii) Also, heat transferred is given by, Q = UA8, where, Here 8, = thl - tc, = 200 - 25 = 175OC; and

The overall heat transfer coefficient, U is given by

..

e2= tm - tc2

175- ( h 2 - tc2)

"=m th2

-42

Substituting the values in the above equati*,

.

we get

. U = 66.09 w / m 2 ~ c Total heat transfer rate is given by

...(iv)

Q = U A 8,= U x ( n d 0 L ) x 8 , Q L= - 155450 x (1000/3600) = 14.65 m (Ans.) U x 7t do x 8, 66.09 x 7t x 0.06 x 236.66 Example 10.5. A hot fluid at 2OO0C enters a heat exchanger at a mass flow rate of 10* kg/h. Its specijc heat is 2000 Jkg K. It is to be cooled by another fluid entering at 25OC with a mass flow rate 2500 kg/h and specijc heat 400 J/kg K. The overall heat transfer coejficient based on outside urea of 20 m2 is 250 w / m 2 K. Find the exit temperature of the hot fluid when the fluids are in parallel flow. (GATE, 1998)

-

L

J

Substituting the values of tc2from (iii) in (iv),we get 175 - Itu - (4025 - 20 th2)l Q=250~20[

1

-

Solution. Given :thl = 200°C; mh = lWW - 2.78 kg/s; c,,,, = 2000 J/kg K; t,, = 25OC; 3600

t h ~=

2ooo {-

Using- hit and trial method, the value of th2may be found out.

Hot fluid

th2

Cold fluid

t,, = 25OC I

I Fig. 10.15.

1]

Equating (i) and (v), we get 175 (2 1 th, - 4025) 5560 (2W - lh2) = 5000[ 175 21 t h 2 - 4 2 5

Exit temperature of the hot fluid, tm : Heat lost by the hot fluid, Q = mhcph(thl- th2)

-

J Example 10.6. In a certain double pipe heat exchanger hot water flows at a rate of 5VW kgih and gets cooled from 95'C to 65OC. At the,same time 50000 kg/h of cooling water at 30°C enters the heat exchanger. The flow conditions are such that overall heat transfer coejficient remains constant at 2270 w/m2 K. Determine the heat tmsfer area required and the effectiveness, assuming two streams an in pamllel flow. Assume for the both the s t m s cp=4.2 kJAg K. (GATE, 1997)

-

Solution. Given :mh = 50000 3600 - 13.89 k g / ~ ;fhl = 95OC; th2= 6S°C;

Heat and Mass Transfer

cph= cpc= 4.2 kJ/kg or 4200 J/kg K. Q = heat lost by hot water = heat gained by cold water.

Heat Exchangers

Solution: Refer Fig. 10.9, for parallel flow heat exchanger. The heat flow rate through an elementary strip of area dA is ...(i) dQ = - mh cphdt,, = mc cpc dlc = -Ch dth = Cc dtc ...(ii) dQ dQ and dr, = dth = - -

cc

Ch

From eqns. (i) and (ii), we get

d (th - tc) = d (0) = - dQ

[& + -&]

...(iii)

cold fluid

(water) Length ----+

Fig. 10.16. Parallel-flow heat-exchanger. mh Cph

(thl

- t h l ) = lj2c Cpc

(tc2

-t ~ l )

13.89 x 4200 x (95 - 65) = 13.89 x 4200 x (tc2- 30) .. tc2= 60°C Log mean temperature difference, or

LMTD, 0, = (91 - 92) In ('31/92)

d~ = u.dA.0 = (a + be) dA.0 Also Integrating eqn. (iv) within limits 1 and 2, we get 91 - 92 Q= 1

[G

+

...(vi)

$1

-

...(vii)

1 1 = (31 - 92 or -+Q Ch cc From eqns. (iv) and (v) and using eqn. (vi), we have - d0 - -de =- Q d e (a + b e ) dA.0 = 8 , -8, 02-01

...(viii)

[i k] +

Integrating eqn. (viii) between limits 1 and 2, we get

Q = UA9,

Also,

-

or

13.89 x 4200 x (95 65) = 2270 x A x 23.4 Heat transfer area, A = 32.95 m2 Also, Qacrual = mh~ph(thl - th2) and Qm. = mh cph (thl - tcl) :. Effectiveness of the heat exchqger,

:.

Example 10.7. Prove that the rate of heat t r a n g r Q for a double parallel flow heat exchanger in which the overall heat trans$er coeficient varies linearly with temperature direrence i.e., U = a + be, where a and b are constants, is given by

Q= Here su&

u1e2

- u2 01

The constant a may be expressed by solving UI = a + beI and U2 = a + beZ which results in

a=

u192 - u2 0 1 0, - 0 ,

ibstituting this value of a in eqn. (ix), we get

A

ln Wle2/U2 011 I and suffix 2 represent inlet and outlet of the heat exchanger respectively.

402

-

u201

Q = In t(Ule2)/(U2%)I

*

... Proved.

Heat and Mass Transfer

Example 10.8. In a counter-jlow double pipe heat exchunger, water is heated fmm 25°C to 65OC by an oil with a specijic heat of 1.45 W/kg K and mars flow rate of 0.9 kgh. 73e oil is cooled from 230°C to 160DC. If the overall heat tran.$er coescient is 420 W@C, calculate the following: (i) The rate of heat transfer, (ii) The mass flow rate of water, and (iii) The surjGace area of the hat exchanger. Solution. Given: tcl = 25°C; tc2 = 65°C. cph = 1.45 Hlkg K;mh = 0.9 kgls;

cph

(thl

- th2) = 91.35 =

m, =

Q=UAO, A=-- Q

- 91'35 lo3 = 1-45 m2

(Am.)

~J'Example 10.9. An oil cooler for a lubrication system has to cool 1000 kg/h of oil (cp = 2.09 kJ/7cg°C) from 8 P C to 4O"C by using a cooling wafer flow of 1000 kg/h at 30°C. Give your choice for a parallel flow or counter-jlow heat exchanger, with reasons. Calculate the surface area of the heat exchanger, if the overall heat transfer coeflcient is 24 w/m2"c. r Take cp of water = 4.18 kJ/kg°C. 1000 Solution. Given: Ah = 1000 kgls; cph = 2.09 W k g 0 ~ ; ' c = p , 4.18 Wlkg°C; kc = 3600 3600 kgh;

(Ans.)

Heat lost by oil (hot fluid) = heat gained by water (cold fluid) mh

Also,

U 0,,.. 420 x 149.5

thl = 230°C; th2 = 160°C, LI = 420 w/m2"c. (i) The rate of heat transfer, Q: or Q = 0.9 x (1.45) x (230 - 160) = 91.35 kJ1s (ii) The mass flow rate of water, m,':

Heat Exchangers

-

cpc x (tc2- tcl)

ri?, x 4.187 (65 - 25) 91'35 = 0.545 kgls 4 . 1 8 7 ~(65-25)

(Am.)

Surface area of heat exchanger, A: I,@ subscripts h and c stand for hot and cold fluids respectively.

Since tC2> th2,counter-flow arrangement must be used. (a) Flow arrangement

Again,

8, =

(31

- 82

In (81/(32)

(b)Temperature distribution

Fig. 10.17. Counter-flow heat exchanger.

(iii) The surface area of heat exchanger, A: Logerithmic mean temperature difference (LMTD) is given by

Fig. 10.18.

Heat and Mass Transfer

553

Heat Exchangers

(i) When the flow is counter: Also,

Q = UA0,

Om = 01 - 02 In W 0 2 ) In this

0, =

= O2 = 100°C I

I

Example 10.10. Show that in a double-pipe counter flow heat exchanger if &,ch = 4 c , the temperature projiles of the two fluids along its length are parallel straight lines.

nr

t,

Solution. For heat exchanger, dQ = - mh ch dth= mc c, dtc = - Ch dt, =

I

= 30°C I

(b) Parallel flow

The heat transfer rate is given by Q = UA0, mCx cpCx (tc2- tcl) = U A Om or or

F Ch = heat capacity of hot fluid.

or

2.917 x 4.187 x (80 - 30) = 814 x A x 100 2.917 x 4.187 x lo3 (80 - 30) = A = 814 x 100'

m2

...(Given) Length

In counter-jlow system, temperature of both the fluids decrease in the direction of heat exchanger length, therefore

Fig. 10.19.

Again, or

-

dt,, = - dQ and dtc= -dQ cc Ch

Id,'d,l

dth-dtc=dO=-dQ --L

Since Ch= Cc,

'cl

Fig. 10.20.

d Also, C, = heat capacity of cold fluid, E

mhch = mc cC Ch = cc

+

I

(a) Counter-flow

m, = mass flow rate of cold fluid. Due to heat exchanger, temperature of hot fluid decreases by dth and that of cold fluid increases by dt,.

or

tc2= 80°C

t,, = 30°C

ccdt,

where ih,, = mass flow rate of hot fluid, and

and

th, = 130°C

= 80°C

(GATE, 1996)

-1

:.

d0 = 0 or 0 = constant. Thus, both the straight lines showing the variation of temperatures along the length are parallel lines. Proved J' Example 10.11. A counter-jlow double [email protected] exchanger using superheated steam is used to heat water at the rate of 10500 kgh. T k steam enters the hear exchanger at 180°C and lehves at 13O0C The inlet and exit temperatures of water are 30°C and 80°C respectively. I f overall heat transfer coeficient from steam to water is 814 w/mZ0c, calculate the heat transfer area. What would be the increase in area if the fluid flows were parallel?

...

.

Q = UA0, 2.917 x (4.187 x lo3) x (80 - 30) = 814 x A x 91 2.917 x (4.187 x lo3)x (80 - 30) = 8.24 A = 814 x 91

m2

-,7 S = 0.0987 or 9.87% (Ans.) Increase in area = 8.24, 1. . I

Example 10.12. A counter-flow heat exchanger, through which passes 12.5 kg/. of air to be cooled from 540°C to 146OC, contains 4200 tube, %dch having a diameter of 30 mm. The inlet and outlet temperatures of cooling water are 25OC and 75OC respectively. I f the water side resistance to flow is negligible, calculate the tube lehgNt required for this duty. 0.8 p4 For turbulent flow inside tubes : Nu = 0.023 Re P Properties of the air at the average temperature are as follows: p = 1.009 kg/m3; cp = 1.0082 kJ/kg°C; p = 2.075 x I @ kg/ms (lVs/m2)and k = 3.003 x 1 r 2 W/m°C. Solution. Given : A, = 12.5 kgls, thl = 540°C; th2= 146OC; tC1= 25OC; t, = 7S°C; N = 4200.

554

Heat and Mass Transfer

Heat Exchangers

Tube length, L: Re =

Reynolds number,

f?L!.! ..

P Mass flow

mh x cph X (thl- th2) = mc Cpc (tc2 - t c l ) 22.5 x 1 x (650 -th2) = 13.33 x 2.71 x (350 - 180) th2 = 377°C

m = NAVp, therefore,

6

4

Gas

4

4

tc2= 3S0°C t

+

Steam

t

+ tcl (= 180°C)

4

4

Gas

C3

4

fhl

=650T

th, = 6s0°C

Prandtl number, Pr = P c = 2.075 x 10" x 1.0082 x lo3 = 0.6966 k 3.003 x lo-2 h d = 0.023 Nusselt number, Nu = k

I

th2 (= 377OC)

thl (= 377Q

(a) Flow arrangement

Since the water side resistance to flow is negligible

..

-I - - I- -

I or u = 21.22 w / ~ ~ ~ U - h - 21.22 Logarithmic mean temperature difference (LMTD) is given by

em =

c

01 -82 In (el/@,)

I Area ---b (b) Temperature distribution Fig. 10.21. I Counter-flow heat exchanger.

Overall heat transfer doeffici{nt is given by Further, the rate of heat transfer, Q = mh x cphx (th2-thl)= U A

...(g iven)

em= U x(Nn d L ) X%

- th,) --

12.5 x (1.0082 x lo3)x (540 - 146) = 2.31 m (Ans.) Ux N n d x 8 , 21.22 x 4200 x n x 0.03 x 255.5 J Example 10.13. Steam enters a cowter-flow heat exchanger, dry saturated at 10 bar and leaves at 350°C. The man flow of steam is 800 kg/min. The gas enters Ske heat exchanger at " 650°C and mass flow rate is 1350 kgimin. If the tubes are 30 mm diameter and 3 m long, determine the number of tubes required. Neglect the resistance offered by metallic tubes. Use the following data : For steam :t,,, = 18O0C (at 10 bar); cps = 2.71 W/kg°C; hs = 600 w/m2O€ For gas :cpg = 1 kJ/kg°C; h, = 250 w/m2Oc. (P.U) 800 1350 Solution. Given: 4 = mc = -= 13.33 kg/s; ritp t &, = 60 60 - 22.5 k g / ~ ;thl = 650°C; tcl (= tsar)= 180°C; tc2= 350°C; d = 30 mm = 0.03 m; L = 3 m. Number of tubes required, N: Heat lost by gases = heat gained by steam L=

Ah X Cph ( 4 2

.

-

Total heat transfer rate is given by Q = UA8, where A =Nx(~dL)=Nxnx0.03x3=0.2827~m~

Substituting the values in e4n. (i), we get 6142.5 x lo3 = 176.5 x 0.2827 N x 244.9

N =

6142'5 = 503 tubes (Am.) 176.5 x 0.2827 x 244.9

Heat and Mass Transfer

556

eat

Example 10.14. In a shell and tube counter-jlow heat exchanger waterflows through a copper tube 20 mm I.D. (internal diamet8i-) and 23 mm O.D. (outer diameter), while oil flows through the shell. Water enters at 20°C and comes out at 30"C, while oil enters at 75°C and comes out at 60°C. The water and oil side jilm coeflcients are 4500 and I250 w/m2"c respectively. 'The thermal conductivity of the tube wall is 355 W/m°C. 7Xe fouling factors on the water and oil sides may be taken to be 0.0004 and 0.001 respectively. If the length of the tube is 2.4 m, calculate the following: (i) The overall heat transfer coeflcient; (ii) The heat transfer rate. Solution. Given : di = 20 mm = 0.02 m; do = 23 mm = 0.023 m; tcl = 20°C; tc2 = 30°C;

Exchangers

Date given : Nu = 0.023 ( ~ e ) ' (. ~ r ) ' . ~ Fouling factor, water side = 0.0005 m2 K / W Fouling factor, oil side = 0.0008 m2 K / W Water and oil properties : Property P (kg/m3) ( k J h g K)

~p

thl = 75°C; th2 = 60°C, hi = 4500 W/m20C;ho= 1250 W/m2"C;

k (W/m K)

k = 355 W/m°C; Rfi = 0.0004; Rfo = 0.001; L = 2.4 m (i) The overall heat transfer coefficient, U,: The overall heat transfer coefficient based on outer surface of inner pipe is given by, 1 ro 1 ro '0 =-X-+-R, +-In (rdri) + R f o i - -1 Uo ri hi ri f k h~ ...[Eqn. (10.28)]

Oil

Water

850

995

1.89

4.187

0.138

0.615 4.18xlo-'

7.44 x 1 0 - ~

v (m2/s)

(U.P.S.C., 1998) Solution. Given : Inner dia. of the copper tube (di)= 16 m m = 0.016 m; Outer dia. of the copper tube (do),= 19 m m = 0.019 m; Inner dia. of the steel tube, (di)$= 26 mm = 0.026 m;

(Rf stands for fouling factor)

Outer dia. of the steel tube,

tcl = 30°C; 1 . u, = 0.00252 = 396.8 w / m 2 ~ c thl = 7 5 ' ~ (ii) The heat transfer rate, Q: Area, A, = n d o L = n x 0.023 x 2.4 tc2= 30°C = 0.1734 m2 Logarithmic mean temperature difference (LMTD) is given by,

.

fhl

(do),= 30 m m = 0.03 m;

= 65°C; th2= 50°C

n

ic =4 ~ ( 0 . 0 1 6x) ~1.48 x 995 = 0 796 k g s ; mh = 0.4 kg/s

t,z (32

fluid (waer) Area ---b F i g .10.22.

Fig. 10.23. Counter-flow double pipe heat exchanger.

Length of the tube, L : The rate of heat transfer is given by, Q = mh Cph (thl - fh2) = k cpc (tc2 - tcl) 0.4 x 1.89 (65 - 50) = 0.296 X 4.187 (tc2- 30)

The heat transfer rate is given by

Q = u04e, or = 396.8' x 0.1734 x 42.15 = 2900 W (Ans.) Example 10.15. In a counter-flow double pipe heat exchanger waterflows through a copper tube (19 mm O.D. and I6 mm I.D.), at a flow rate of 1.48 m/s. The oil flows through the annulus formed by inner copper tube and outer steel tube (30 mm O.D. and 26 mm I.D.). The steel ttrbe is insulated from outside. The oil enters at 0.4 kg/s and is cooled from 65°C to 500C whereas water enters at 32OC. Neglecting the copper tube wall thermal resistance, calculate the length of the tube required.

(b)

(4

Also, Q = 0 . 4 ~1 . 8 9 ~ ( 6 5 - 5 0 ) =11.34kW Reynolds number for the flow of water through copper tube, Re=-= 4 m 4 x 0.296 = 56826 n(di),p n x 0 . 0 1 6 x ( 9 9 5 x 4 1 8 x 10-3 'i

Now,

Nu = 0.023 (~e)'.' ( ~ r ) ' . ~

(*:

P=Q.V)

...(G h a )

Heat and Mass Transfer

,

(Since, Pr = w and water is heated) k

Now oil flows through annulus diameter, therefore hydraulic diameter,

Dh=(di),-(do),=0.026-0.019=0.007 m Reynolds number through the annulus,

-

=

1790 Since Re c 2500, hence annulus flow is laminar. Heat transfer coefficient at the inner surface of the annulus,

Heat Exchangers

The heat transfer rate is given by, Q = Ua,em = 42.43 x (n x 0.019 x L) = 20.86

*/ Example 10.16. A counter-flow, concentric tube heat exchanger is used to cool the lubricating oil for a large industrial gas turbine engine. The flow rate of cooling water through the inner tube (d, = 20 mm) is 0.18 kg/s, while theflow rate of oil through the outer annulus ld, = 40 mm) is 0.12 kg/s. The inlet and outlet temperatures of oil are 95OC and 65°C respectively. The water enters at 30°C to the exchanger. Neglecting tube wall thermal resistance, fouling factors and heat loss to the surroundings, calculate the length of the tube. Take the following properties at the bulk mean tempeature: Engine oil at 80°C : c, = 2131 J/kg°C; )I = 0.0325 ~ s / m ' ;k = 0.138 W/m°C: Water at 35OC : c, = 41 74 J/kg°C, p = 725 x I@ ~ s / m ' ;k = 0.625 W/m°C, Pr = 4.85 Solution: Given: di = 20 mm = 0.02 m; do = 40 mm = 0.04 m, m, = m, = 0.18 kg/s;

m,, = mh = 0.12 kg/s;

tll,

= 95OC, th2 = 65OC; t,, = 30°C.

Annulus

NU--- h" Dh - 0.023 (~e)'.' ( ~ r ) ' , ~

k

Overall heat transfer coefficient referred to outer diameter of inner tube is given by :

t,,, = 65°C

(Rf stands for fouling factor) r,, = 30°C

-

1

+

0.002203 + 0.000594 + 0.005309 0.0008 + 0.01466

= 42.43 w/m2 K

Logarithmic mean temperature difference is given by,

Fig. 10.24. The counter-flow, concentric tube heal exchanger.

Length of the tube, L: The rate of heat transfer is given by,

Heat and Mass Transfer

560

LMTD is given by,

Heat Exchangers ,/ Example 10.17. Thefollowing data pertain to an oil cooler of the form of thl = 803C tubular heat exchanger, where oil is cooled by a large pool of stagnant -.-- water: -Temperature of stagnant water = 20°C (assumed constant), Inlet and outlet temperatures of oil = 80°C and 30°C respectively, Inside diameter and length of the tube carrying oil = 20 mm and 30 m, respectively, *

The overall heat transfer coefficient U is given by the relation

..(i) Reynolds number for flow of water through the tube is

th2= 3O0c Cold fluid (water)

Fig. 10.25

Specific heat and specific gravity of oil = 2.5 W/kg°C and 0.85 respectively, Average velocity of oil = 0.55 ids. Calculate the overall heat transfer coeflcient obtainable from thk system. Solution. Given : tcl = tC2= 20°C; thl = 80°C; th2 = 30°C; d = 20 mm = 0.02 m; L = 30 m;

Since the flow is turbulent, hence

cph = 2.5 kJ/kg°C; sp.gr = 0.85; V = 0.55 d s . Overall heat transfer coefficient, Ui : The mass flow rate of hot fluid (oil),

= 3088.7 w / m 2 OC The oil flows through the annulus, hence the hydraulic diameter Dh = do - di = 0.02 m Reynolds number through the annulus is Re =

p.umDe - P(do-di) P CL 4 h,

Re =

-

+ d,) y

LMTD is given by,

mh

X

X

4 (d,2 - dl2)P 4 x0.12

= 78.35 0.0325 Since Re c 2300, hence the annular flow is laminar. Assuming uniform temperatun along the inner surface of the annulus the heat transfer coefficient at the inner surface of the annulus, is 7[: (d,

Heat lost by the hot fluid, Q = hhCph (thl- th2)= 0.1468 x (2.5 x LO') x (80 - 30) = 18350 w

x (0.04

+ 0.02) x

The heat transfer rate is given by Q = UiAi 0, = Ui( X di L)0, 18350 = Uix x x 0.02 x 30 x 27.9

...[Eqn. 7.1O8)] Substituting the values in eqn. (i), we get

.. Also,

u=-- I

0.0399

1

- 25.06 w/m2'c

p = 990 kg/m3; c,, = 4.2 kJ/kg°C, k = 0.5418 W/m°C; p = 700 x 106 kg/ms.

Q = hh Cph ( f h l - fh2) = UA 8, = u x (7[: di L) x 8, 0.12 x 2131 x (95-65) = 25.06 x 7[: x 0.02 x L x 44.2 0.12 X 2131 (95 -65) L = 25.06 x ~r: x 0.02 x 44.2 = 110.23 m

/&le 10.18. R e velocity of water flowing through a tube of 22 mm diameter is 2 m/sS Steam condensing at 150°C on the outside su$ace of the tube heats the warerfrom 15°C to 60°C over the length of the tube. Neglecting the tube and steam sidefilm resistance, calculate the following: (i) The heat transfer coeficient, and (ii) The length of the tube. Take the following properties of water at mean temperature :

(Pm

Solution. Given: d = 22 mm = 0.022 m; V = 2 d s , thl = th2= 150°C; tcl = 15OC; tC2= 60.C. (i) The heat transfer coefficient, h : (Ans.)

Heat and Mass Transfer

562

steam

Heat Exchangers

rd=22mm (ii) AMTD =

1

= 62228

-+

Water -+

-+

1

(100 - 30) + (100 - 70) --70 + 30 - 500C 2 - 2

Hot fluid (steam)

(i)Parallel-flow

(ii)Counter-flow Fig. 10.27.

(ii) The length o f the tube, L:

Example 10.20. The amount of F12used in compression refrigeration system is 4 tonnesntour. The brine, flowing at 850 kg/min. with inlet temperature of 12°C is cooled in the evaporator. Assuming F12 entering and leaving the evaporator as saturated liquid and saturated vapour respectively, determine the area of evaporator required. Take the following properties : For FI2: Saturation temperature :- 23°C; cp= 1.1 7 kJ/kdC; hfg= 167.4 kJ/kg (AMIE Summer, 2000) c, (brine)= 6.3 kJ/kg°C; U = 8368 kJ/m2hoc.

I

I

Fig. 10.26

The mass flow rate (water),

Heat gained by water passing through the tube,

Q

mc X Cpc

(tcz-tc1) = 0.753 x (4.2 x lo3) x (60 - 15) = 142317 W Q is also given by, Q = U A en, =

Solution. Making energy balance, we have

X

....( i )

Substituting the values in eqn. (i), we get 142317 = 6771.7 x ( X X 0.022 x L) x 1 1 1 (Here, U = h) or L = 2.74 m (Ans.) Example 10.19. Saturated steam at 100°C is condensing on the shell side of a shell-and-tube heat exchanger. The cooling water enters the tube at 30°C and leaves at 70°C. Calculate the mean temperature d~fferenceif arrangement is (i) parallel flow, (ii) counter flow. (AMIE, Winter, 1997) Solution. The mean temperature difference will be the same for parallel flow or counter flow arrangements as is evident from the diagrams (Fig. 10.27) of temperature variations in both the arrangements. The mean temperature difference may be logarithmic mean (LMTD),or arithmetic mean temperature difference (AMTD). ( i ) LMTD = (100 - 30) - (100 - 70) =-- 40 (100 - 30) ln($) $00-7011

- 4 7 f l o c (Ans.)

.

- 23°C

41= 4 2

w

Cold fluid (F,,)

I I 1

I I I

Fig. 10.28.

:.

Exit temperature of brine, th2= 12 - 850 11160 x 6.3 ,9.90c

LO^ mean temperature, 9,

=

-

01 - 92 - (thl tcl) - (th2 - tcd In ( e 1 / e 2 )Ithl In -

Heat and Mass Transfer

eat Exchangers n

Now,

Q = UAem

(where A = area o f evaporator)

kc = ( -4d ? x V X p) x Np Np = number o f tubes in each pass (N = p x Np)

where

n

60 = 7 x (0.021)~x 2 x 1000 x Np A=

60 = 2.357 I$ (Ans.) 8368 x 33.94 -:' Example 10.21. A heat exchanger is to be designed to condense an organic vapour at a rate of 500 kg/min which is available at its saturation temperature 355 K. Cooling water at 286 K is available at a flow rate of 60 kg/s. The overall heat transfer coeficient is fhl = th2= tmt.= 82°C 475 w/m2'c. Latent heat of condensation Hot fluid (vapour) of the organic vapour is 600 W . g . Calculate: tc2 (= ?I (i) The number of tubes required, if25 mm outer diameter, 2mm thick and 4.87 m long tubes are available, and 'c' (ii) The number of tube passes, if the cooling water velocity (tube (= 130C) side) should not exceed 2 m.4. (M.U.) Fig. 10.29

Solution. Given : do = 25 mm = 0.025 m; di = 25 - 2 x 2 = 21 mm = 0.021 m; L = 4.87 m; tcl = 286 - 273 = 13OC, thl = th2 = t o , = 355 - 273 = 82OC; v = 475 w / m O c , 500 hfg(organicvapour) = 600 kl/kg, = mh = 60 - 8.33 kg/s; m, = rit, = 60 kg/s.

v=2

m,

(i) The number of tubes required, N : Heat lost by vapour = heat gained by water

N 470 Number of passes, p = - = -= 4.91 = 5 (Ans.) N, 95.5 Example 10.22. (a) A steam condenser consists of 3000 brass tubes of 20 mm diameter. Cooling water enters the tubes at 20°C with a mean pow rate of 3000 kgh. The heat transfer coescient on the inner sugace is 11270 w/m2'c, and that for condensation on the outer sutjize is 15500 w/m2 OC. The steam condenses at 50°C, and the condenser load is 230 MW. The latent heat of steam is 2380 kJ/kg. Assuming counter flow arrangement, calculate the tube length per pass if two tube passes are used. (b) Explain why in steam condensers the LMTD is independent of flow arrangement ? (AMIE Summer, 1999) Solution. (a) Given :Np = 3000 per pass; d = 20 rnm = 0.02 m; tcl = 20°C; mc = 3000 kgls; h, = 11270w / m 2 OC; ho = 15500 w / m 2 "C, thl = th2= 50°C; condenser load = 230 MW (or 230 x lo3 kW); hfg= 2380 kJ/kg; number o f passes = 2. Tube length per pass, L : Assuming the tubes to be thin, the overall heat transfer coefficient : 1 ,yo=-- 1 1 1 1 -+-+- 1 hi h, 11270 15500 = 6525.4 w/m2'c

t,, (= 20°C)

Heat exchanger load =

Logarithmic mean temperature difference (LMTD) is given by

F h

Steam

mc cpc(tc2- tcl)

i.e.,

Fig. .10.30

230 x lo3= 3000 x 4.187 (tc2- 20)

:.

Water outlet temperature, tc2= 38.31°C Log-mean temperature difference,

Heat transfer rate is given by,

..

Q = m h x h f g= v A e m = v ( 7 t d o L N ) x 8, 8.33 x 600 x lo3 = 475 x ( 7t x 0.025 x 4.87 x N) x 58.5 N = 470 tubes (Ans.)

In

-

(ii) The number of tube passes,. v : The cold water flow mass passing through each pass (assume p are number of passes) is given by,

Now,

150 - 38.311

Q = VJ9m 230 x lo6 = 6525.4 x (ndL) x (2N,,) x 19.43

t,,, = th2= 50°C

.

566

Heat and Mass Transfer

Heat Exchangers

760-660 x 1.0133 = 0.133 bar = 760 The properties of steam at p, = 0.133 bar, from steam table, are: tsar= 51°C; hh = 2592 kJkg . tc2 = 51 - 4 = 47OC Ps

230x lo6 L= = 4.812 m (Ans.) 6525.4 x n x 0.02 x (2 x 3000) x 19.43 (b) In steam condensers, the temperature o f hot fluid (steam) is the same at inlet and exit. Hence the terminal temperaturedifferencewould not depend upon the arrangement. The effectiveness of heat exchanger for all the arrangements in this case (the heat capacity ratio for condensation being zero) would be E

.

The steam condensed per minute,

= 1 - e-NTU

The temperature variations for parallel-flow and counter-flow are as shown in Fig. 10.31 and Fig. 10.32 respectively.

(i) Mass of cooling water circulated per minute, m, (= m,): Heat lost by steam = heat gained by water mh x ( X . hfg)= mc X cpcX (tc2- tC,)

.Steam r

1250 x (0.9 x 2592) = m, x 4.187 (47 - 24)

..

mc (= m,) = 30280 kglmin (Ans.) (ii) Condenser surface area, A : ms x (x . hfR) = UA9, '= 60

...(i)

I

I

Fig. 10.31. Parallel-flow.

where

Fig. 10.32. Counter-flow.

Example 10.23. A two-pass sudace Steam in (x = 0.9) condenser is required to handle the exhaust from a turbine developing 15 MW with Tho-pass surface specijic steam consumption of 5 kg/kWh. The condenser vacuum is 660 mm of Hg when the barometer reads 760 mm of Hg. The mean velocity of water is 3 d s , water inlet temperature is 24°C. The condensate is saturated water and outlet temperature of cooling water is 4OC less than the Water \ condensate temperature. The quality of exhaust steam is 0.9 dry. The overall heat Condensate out transfer coeflcient based on outer area of tubes is 4000 w/m2"c. The water tubes are 38.4 mm in outer diameter and 29.6 mm in inner diameter. Calculate the following: (i) Mass of cooling water circulated in kg/min, (ii) Condenser sur$ace area, Fig. 10.33. A two-pass surface condenser. (iii) Number of tubes required per pass, and (iv) Tube length. (P.U Solution. Given : di = 29.6 mm = 0.0296 m ; do = 38.4 m m = 0.0384 m; U = 4000 w/m2"c; V = 3 mls; tc, = 24°C; x (dryness fraction) = 0.9. The pressure o f the steam in the condenser,

Substituting the values in eqn. (i), we get

1250 60 x (0.9 x 2592 x

lo3) = 4000 x A x 12.04

or A = 1009.1 m2 (iii) Number of tubes required per pass, N, :

(Ans.)

1

I

30280 x 4 = 244.46 say 245 (Ans.) Np = 60 x n x (0.0296)~ x 3 x 1000 ( Total number o f tubes required, N = 2Np = 2 x 245 = 490) (iv) Tube length, L: A = (n do L) x (2Np) 1009.1 = n x 0.0384 x L x (2 x 245) 1009.1 =. 17.1 m (Ans.) L= n x 0.0384 x 2 x 245 Example 10.24. A feed water heater which supplies hot water to a boiler comprisr 8 ahd and tube heat exchanger with one-shell pass and two-tube passes. One hundred t h i n - ~ l l dCW d each of 20 mm diameter and length of 2m per pass are used. Under normal operating water enters the tubes at 10 kg/s and 17°C and is heated by condensing saturated s1l a on the outer sudace of the tubes. The convection coeflcient of the saturated steam is 10 k

Heat and Mass Transfer

%8

Determine the water exit temperature. Use the following properties of water: c,, = 4.18 kJ/kg°C; p = 0.596 x lod3N S / ~k ~=. 0.635 W/m°C and Pr = 3.93. Solution. Given : p (number of Steam in tube passes) = 2, N (total number of tubes) = 200,d = 20 mm = 0.02 m; L (length per pass) = 2m, m, = riz, = 10 kg/s, tcl = 17OC; Water Water exit temperature, tc2:

Heat Exchangers

AMTD =

(M.U.)

The heat transfer rate is given by Q = m, c, (t, - tcl)= U A, (AMTD) = U x (n d L x N ) (AMTD) or

-

40406 (91.5 - 0.5 tc2 ) = 0.966 (91.5 - 0.5 t,) tc2- 17 = 41800 = 88.39 - 0.483 tc2

t,, = 71°C (Ans.) or Example 10.25. A one ton window airconditioner removes 3.5 W/s from a room and in the process rejects 4.2 U/s in the air cooled condenser. The ambient temperature is 30°C whereas condensing temperature of the refrigerant is 4PC. For the condenser the product of overall heat transfer coe8cient and corresponding area is 350 W/K.Calculate the temperature rise of the (M.U., Winter, 1998) air as it jlows over the condenser tubes. Solution. Given :tcl = 30°C; thl = th2= 45°C; UA = 350 W/K.

in

P

Condensate out n or 10 = - ~ 0 . 0 2 ~vxx 10OOx 100 4 thl = th2= fsat = loo"c 2 10x4 Hot fluid (steam) v = 82 n x 0.02~x 1000 x 100 T tc2= ? = 0.318 m/s Using non-dimensional heat transfer equation to water side, we t,,-.= 17°C get hi d Nlc = - = 0.023 (Re)'.' ( ~ r ) ' , ~ ~ k Fig. 10.34. One-shell pass and two-tube passes condenser.

:.

(where A, = surface area of all the tubes in both passes) 10 x (4.18 x lo3) (t, - 17) = 1607.7 x ( n x 0.02 x 2 x 200) x (91.5 - 0.5 t,) 41800 (tc2- 17) = 40406 (91.5 - 0.5 t,)

or

. n mc=-d2xvxpx~,

4 [where V = velocity of water; N, = N = number of tubes per pass = -

9 , + 02 - 83 + (100 - tc2) 2 = 91.5 2 -

Temperature rise of air, tc2: The arrangement and temperature distribution are shown in Fig. 10.35.

-

Condenser 4.2 kJ/s

refrigerant in

3.5 kJ/s

Substituting the values in eqn. (i), we get

(a)Arrangement

h, = 0'635 x 0.023 (10671)'~~3.93)'~~~ = 1915 w/m2'c 0.02 The overall heat transfer coefficient is given by the relation,

Further,

:.

Fig. 10.35.

Heat lost by the refrigerant in condenser = heat trankferred to air 91 - 92 4.2 x lo00 = UA9, = UA In (01/02)

91=tl,l-tcl=100-17=830C

1

e2 = tIl2-tc2= 100 - tc2

4

Arithmatic mean temperature difference,

(b)Temperature distribution

Heat and Mass Transfer

Heat Exchangers

Substituting the values in eqn. (2),we get Using trial and error method, we get, tn = 35°C

:. Temperature rise of air = 35 - 30 = S°C (Ans.) Example 10.26. A rectangular tube, 30 mm x 50 mm, carries water at a rate of 2 kg/s. Determine the length required to heat water from 30°C to 50°C if the wall temperature is maintained at 90°C. Use the following properties of water at 40°C: p = 992.2 kg/m3; k = 0.634 W/m°C; C, = 4.174 kJ/kg°C; /.k = 6.531 X Ns/m2. (P.U.) 61= 3O0c Solution: Given: h, = rit, = 2 kgts; tcl = 30°C; tc2 = 50°C; thl = th2 = 90°C. Length required, L: Fig. 10.36.

Now, putting the values-in eqn. (I), we have (where A = surface area) Q = UA8, = hA8, 166960 = 5082.4 x [2(0.03 + 0.05) x L] x 49.3 or L = 4.16 m (Ans.) Example 10.27. Carbon dioxide (C02)for a gas cooled reactor is used to generate steam. A flow rate of 90000 k g h of C 0 2 at 4 bar enters the tubes of a shell and tube type steam generator at 500°C. The C 0 2 leaves the generator at 330°C and steam saturation temperature is 250°C. Assume that the steam formed is dry and satwated. Using 25 mm inner diameter (I.D.) copper tubes, 2 mm wall thickness and designed for C 0 2 mass flow rate of 350000 kg/m2h, calculate the length and number of tubes to be used neglecting steam side thermal resistance. Take the following properties of C02: c,, = 1.1 72 kJ/kg°C; p = 0.0000298 Ns/m2: k = 0.043 W/m°C; p = 3.26 kg/m3. (P.U.) Take k (for copper) = 384 W/m°C.

I

Heat gained by cold fluid (water),

Q = m c x C,c X (t,2 - t,l) = 2 x (4.174 x lo3) x (50 - 30) = 166960 W

Water in (cold fluid)

The heat transfer Q is also given by Q = UAe, where

8, = 01- 92 In (8,/82)

- (90 - 30) - (90 - 50) - In [(go - 30)/(90 - 50)]

The heat transfer coefficient 7; (=U is this case) is given by

C 0 2 (hot fluid)

C 0 2 out (th2= 3 3 0 ' ~ )

in (t,, = 5 0 0 ' ~ ) Steam out (tc2= 250°C)

4Ac - 4 X (0003 X 0.05) where de (equivalent diameter) = p 2 (0.03 + 0.05) (A, = cross-sectional area, and P = perimeter)

- 0.0375

P VLC Reynolds number, Re = , where V is the velocity of water flowing through the tube.

CL

m, (= m,) = A,. V.p 2 = (0.03 x 0.05) x V x 992.2

Cold fluid (steam)

n

Fig. 10.37. Gas cooled reactor.

di = 25 mm = 0.025

90000

m h = 3 6 0 0 = 25 kgls; t,, = 500°C; t, m; do = 25 + 2 x 2 = 29 mm = 0.029 m.

Solution: Given: mCC2=

= 330°C, t,, = rc2= 250°C

Heat and Mass Transfer

572

Heat Exchangers

Number of tubes (N)and length of each tube (L): Considering the flow of C02 through tubes : Mass flow, m = A.V.p (where V is the velocity of CO, in the tubes). m G = - (mass flow per unit area per unit time) A

G = ~ v =350000 -3600 (given) v=- 350000 x 1 = 29.82 d s (:. p = 3.26 kg/m3 ...given) 3600 3.26 The mass flow rate of CO, through the tubes is given by

Again,

-

..

P Vd - G.d Reynolds number, Re = P P (where G is in kg/m2.s and p in ~ s l m ~ )

(where N, number of tubes)

or Now I

Prandtl number

P r =PCA = 0.0000298 x (1.172 x 1000) = 0.812 k 0.043

Nusselt number,

N U = -hi - di

k

- 0.023 (Re)'.'

Q = m, x cpl,x

- t,,2)

= 25 x (1.172 x lo3) (500 - 330) = 4.981 x lo6 W Negletcting the steam side thermal resistance, we can write

1 - 1 -

u;- %

or

+-kc,,,' i

In (ro 1 r, )

u.= 0.003 177 = 314.76 w/m2 "C

Also, Q = U;A, 8, :. 4.981 x lo6 = 314.76 x A; x 149.2

L=

(PI)'."

= 3 13.76 w/m2 "C . The temperature distribution during the flow is shown in Fig. 10.37. It is assumed that the water enters into the boiler at saturated condition and comes out as saturated steam, so the temperature during the generation of steam remains constant at 250°C. The logarithmic mean temperature difference (LMTD) is given by

Further,

N = 523.89 say 524 (Am.) Ai = (x diL) x N 106.06 = x x 0.025 x L x 524 106'06

x x 0.025 x 524

2.58 m

(Am.)

Example 10.28. The tubes (k = 106 I4'/m0C) of a single pass condenser are of 30 mm outside diameter and 25 mm inside diameter. The condenser is required to handle 20000 k g h of dry and saturated steam at 50°C. The inlet and outlet temperatures of water are 15°C and 25°C respectively. If the average velocity of water in each tube is 2.5 m/s and steam side plm heat transfer coeficient is 5150 w/m2"c, calculate the outside tube area. The properties of water at mean temperature are given as under: p = 998.2 kg/m3, c,, = 4.182 W/kg K: p = 1004.5 x lod ~ s / m ' ;v = 1.006 x 1od m2/s; and k = 0.598 W/m°C. The latent heat (hh) at 50°C = 2374 W/kg. For turbulent flow inside tubes: Nu = 0.023 Rea8 ~ 9 . ~ . Solution. Given : k = 106 W/m°C; do = 30 mm = 0.03 m; di = 25 mm = 0.025 m; m, = 20000 kglh; h, = 5150 w / ~ ~ " cV; = 2.5 d s . Outside tube area, A: V di 2.5 X 0.025 = 62127 Reynolds number, Re=-- v 1.006 x lom6 4 Prandlt number,

...[Eqn. (10.27)]

Nusselt number, Nu =

hw di - 0.023 Re0.' k

= 0.023 x (62127)O.~x (7.02)'.~ = 282

The overall heat transfer coefficient is given by

h, = heat transfer coefficient on the water side, and h, = heat transfer coefficient on the steam side. Substituting the values, we get where

574

Heat and Mass Transfer

Heat Exchangers

Ui is given by

AMTD (arithmatic mean temperature difference)= (50 - 15) + (50 - 25) = 300c

.(2)

Heat transfer rate is given by,

Q = UA (AMTD) m, hh = UA (AMTD)

where hi, ho= inside and outside heat transfer coefficients respectively, and Rfi = inside thermal resistance due to fouling. In order to find out hi we shall use the following relation: h. d. (~r)'.~~ Nu = k = 0.023

Example 10.29. A single pass shell and tube heat exchanger, consisting of a bundle of 100 tubes (inner diameter = 25 mm and outer diameter = 29 mm) is used for heating 500 kg/min of water from 30°C to 70°C with the help of steam condensing at atmospheric pressure on the shell rhl = rh2 = 100°C side. Calculate the overall heat transfer coeflcient Hot fluid (steam) based on the inner area and length of the tube 70°C bundle if the condensing side heat transfer coefl(tcz) cient is 5000 w / m 2 " ~Take . the fouling factor on the water side to be 0.0002 m2"c7W per tube. Neglect the effect of fouling on the shell side and thermal resistance of the tube wall. 30°C Take the following properties of water at the (tcl) mean temperature of 50°C: Area / length -+ p = 988.1 kg/m3; cp = 4174 J/kg°C; k = 0.6474 W/m°C; p = 550 x I@ kg/ms; v = 0.555 Fig. 10.38. x I@ m2/s, Pr = 3.54. Solution. Given: N = 100; di = 25 mm = 0.025 m: do = 29 rnm = 0.029 m; m, = m, = 500 kglmin; I,., = 30°C; tc2 = 70°C; h, (steam side) = 5000 w / ~ ' o cthl ; = th2 = IOOOC. R~ = 0.0002 ~ * " c / w . Overall heat transfer coefficient (U,); tube length (L): Arrangement o f the heat exchanger is shown in Fig. 10.38 Heat gained by water,

Q = mc X

Cpc X

(tc2- tCl)

The Q transferred from the steam to the water is given by

Q= where

u,A,, 0,

[Eqn. (10.27)l

...( 1 )

U, = overall heat transfer coefficient based on inner surface o f the tubes, A,, = inner surface area o f the tubes. 0, = logarithmic mean temperature difference (LMTD)

where

v di Re = -, v

V being the average velocity of the water flow which can be calculated by using the following equation.

Substituting the values in eqn. (3), we get hi =

-x

0.023 (77481O.~ x (3.54)0.33= 1168 wlm2"C 0.025 Further, inserting the values in eqn. (2), we get

= 0.000856 + 0.0002 + 0.000172 = 0.001228 Ui = 814.3 ~ 1 r n ~ " C . Now, substituting all the values in eqn. ( I ) , we get Q = Uix(Nx~xdixL)xO,

.

1.39 x lo6 = 814.3 x (100 x n; x 0.025 x L) x 47.2 1.39 x lo6 L = = 4.6 m 814.3 x 100 x n x 0.025 x 47.2

(Ans.)

Example 10.30. A multipass heat exchanger has two passes on shell side and four passes on the tube side. The oil is passed through the tubes and cooled from 135°C to 52°C. The cooling water passing through shells enters at 13°C and leaves at 31°C. Calculate the heat tranSfcr mfe using the following data :

Heat and Mass Transfer

Heat Exchangers

h, (oil)= 270 w/m2 "C, ho (water)= 965 w/& "C h (scale on water side) = 2840 w/m2 "C Number of tubes per pass = 120 Length and outer diameter of each tube are 2 m and 2.54 cm respectively. Thickness of the tube = 1.65 mm LMTD correction factor = 0.98 Neglect the tube wall resistance. (A.U. Winter, 1998) Solution. Given : No. of passes : Two on shell side and four on tube side. tcl = 13°C; tc2 = 31°C; thl= 135°C; thz= 52°C; hi (oil) = 270 w/rn2"c;

Substituting the values, we have

r,,

do)

h, (water) = 965 w/m2"c; h (scale on water side) = 2840 w/m2 "C, Np = 120; do = 2.54cm = 0.0254 m; L = 2 m; di= 0.0254 - 2 x 0.00165= 0.0221 m;

:. Rate of heat transfer, Q = U J ,

LMTD correction factor, F = 0.98

(0m)actual = 175.4 x [nD& x (N, X 4)l X 65

Water in

= 175.4 x [n x 0.0254 X 2 X (120 X 4)l X 65 Oil in :

= 873369 W or 873.369 kW

v h

(Ans.)

10.6. CORRECTION FACTORS FOR MULTI-PASS ARRANGEMENTS The expression Om = (01- 02) for LMTD is essentially valid for single-pass heat exchangers. 13 (Ol/O2)

1 T L

Oil out 4

Water out (a) Arrangement

(b) Temperature distribution

Fig. 10.39.

Rate of heat transfer, Q :

The analytical treatment of multiple pass shell and tube heat exchangers and cross-flow heat exchangers is much more difficult than single pass cases; such cases may be analysed by using the following equation: ...(10.31) Q = UAF Om factor; the correction factors have been published in the form of charts where F is the by Bonman, Mueller and Nagle and by TEMA. Correction factors for several common arrangements have been given in Figs. 10.40 to 10.43. The - - data is presented as a function of two non-dimensional variables namely the temperature ratio P and the capacity ratio R. Temperature ratio, P: It is defined as the ratio of the rise in temperature of the cold fluid to the difference in the inlet temperatures of the two fluids. Thus: 4 2 - tCl ...(10.32) *P thl - fel

--

where subscripts h and c denote the hot and cold fluids respectively, and the subscripts 1 and 2 refer to the inlet and outlet conditions respectively. The temperature ratio P indicates cooling or heating effectiveness and it can vary from zero for a constant temperature of one of the fluids to unity for the case when inlet temperatun of the hot fluid equals the outlet temperature of the cold fluid. Capacity ratio R: The ratio of the products of the mass flow rate times the heat capacity of the fluids is termed as capacity ratio R. Thus:

k . Cnc -

66.3 x F = 66.3 x 0.98 = 65°C Theoverall heat transfer coefficient,

...( 10.33)

I

578

Heat and Mass Transfer Heat Exchangers

I

temperature drop inof the thecold hot fluid = ltemoeratursise fluid L

.

For thew correction factor plots it is immaterial whether the hot fluid flows in the shell or the tubes. The value of the correction factor indicates the p e d o m n c e level of a given arrangement for the given terminal fluid temperatures. The correction factor F is always less than univ as no arrangement can be more eflective than the conventional counter flow. Example 10.31. Calculate for the following cases, the surfoce area required for a heat exchanger which is required to cool 3200 k g h of benzene (c, = 1.74 W/kg°C) from 72°C to 42°C The cooling water (c, = 4.18 kUXg°C) at 15°C has a flow rate of 2200 kgh. (i) Single pass counter-flow, (ii) 1-4 exchanger (one-shell pass and four-tube passes), and (iii) Cross flow single pass with water mixed and benzene unmixed. For each configuration, the overall heat transfer coeficient may be taken as 0.28 k~/m'"C.

Sohtion Given: hh= 32W 3600 = 0.889 kgls; cph= 1.74 Wkg°C; thl = 72"C, th2 = 42°C;

Fig. 10.40. Correction factor plot for heat exchanger with one shell pass and two, four or any multiple of tube passes.

Surface area required, A: Using energy balance on both the fluids, we have

0

Fig. 10.41. Correction factor plot for heat exchanger with two shell passes and four, eight or any multiple of tube passes.

- tc1 p='hl - lcl tc2

Fig. 10.42. Correction factor plot for single cmss-flow heat exchanger with both fluids

Heat and Mass Transfer

-

th2) = cpc (t& - tcl) 1.74 (72 - 42) = 0.61 1 X 4.18 (tc2- 15) .. tc2 = 33.2OC An energy balance on the hot fluid yields the total heat transfer, ' h 'ph (lhl

0.889

X

Q = mh cph (thl - th2)= 0.889 (i) Single-pass counter-flow:

:.

Heat Exchangers 3 .

Area of the exchanger, A =

X

Using P = 0.32 and R = 1 .65 the correction factor F from Fig. 10.40 is read as

1.74 (72 - 42) = 46.4 kW

Q = m2 U 8, 0.2846'4 x 32.5 = 5.1

:.

Area of the exchanger,

46.4 e= 0.9 x 0.28 x 32.5 = 5.66 m2

A = F U 8,

(iii) Cross-flow single-pass with water mixed and benzene unmixed: Using P = 0.32 and R = 1.65 the cometion factor F from Fig. 10.43 is read as F 1 0.92 46.4 :. Area of the exchanger, A = F U 8, 0.92 x 0.28 x 32.5 = 5.54 m2 (Ans.)

e=

(Ans.)

(ii) 1-4 exchanger: Since the number of passes is more than one hence 8, ( W D ) needs correction factor, F. To know the constion factor we have to know fvsl P (temperature ratio) and (capacity ratio) R.

Example 10.32. it is required to design a shell-and-tube heat exchanger for heating 2.4 kg/s of water from 20°C to 9PC by hot engine oil (cp = 2.4 W/'kg°C)flowing through the shell of the heat exchanger. The oil makes a single pass entering at 145'C and leaving 9PC with an average heat transfer coeflcient of 380 w/m2'c. The waterflows through 12 thin-walled tubes of 25 mm diameter with each-tube making 8-passes through the shell. The heat transfer coefiient on the water side is 2900 w/m2"c. Calculate the length of the tube required for the heat exchanger to accomplish the required water heating. Solution. Given: m, = mc = 2.4 kgls, tcl = 20°C, tc2= 90°C; thl = 145OC. th2= 90°C, cPh= 2.4 kJ/kg°C, d = 25mm = 0.025 m, N = 12; hi = 2900 Wlm°C, h, = 380 w1m2OC Length of the tube L: The overall heat transfer coefficient, neglecting thermal resistance of the tube, is given by

The parameters required to get the correction factors are:

Form Fig. 10.40, F 1 0.82 For the conventional counter-flow arrangement; el = thl - tc2= 145 - 90 = 55OC 8, = th2- tcl = 90 - 20 = 70°C

The heat transfer rate is given by Fig. 10.43. Comclion factors plot for single-pass cross-flow heat exchanger, one fluid mixed and the other unmixed.

Q = ' , Cpc (tc2 - tcl) = 2.4 x 4.18 x lo3 (90

- 20) = 702240 W

Heat and Mass Transfer

Also,

Q = F U A Om,where A = heating surface

But,

-

The shell length = 26'1 - 3.26 m (Ans.) 8 10.7. HEAT EXCHANGER EFFECTIVENESS AND NUMBER O F TRANSFER UNITS (NTU) A heat exchanger can be designed by the LMTD (logarithmic mean temperature difference) when inlet and outlet conditions are specified. However, when the problem is to determine the inlet or exit temperatures for a particular heat exchanger, the analysis is performed more easily, by using a method based on effectiveness of the heat exchanger (concept first proposed by Nusselt) and number of transfer units (NTU). The heat exchanger effectiveness (E) is defined as the ratio of actual heat transfer to the maximum possible heat transfer. Thus actual heat transfer -2 E = maximum possible heat transfer Q,, Thei actual heat transfer rate Q can be determined by writing an energy balance over either side of tye heat exchanger.

Heal Exchangers

as such it is inconvenient to combine them in a graphical or tabular form. However, by compiling a non-dimensional grouping, E can be expressed as a function of three non-dimensional parameters. This method is known as NTU method. This methodlapproachfacilitates the comparison between the various types of heat exchangers which may be used for a particular application. The effectiveness expresssions for the parallel flow and counter-flow cases can be derived as follows: (i) Effectiveness for the Parallel flow heat exchanger : Refer Fig. 10.8. The heat exchange dQ through an area M of the heat exchanger is given by ...(i) dQ = U.dA (th - tc) = - m.cph.dth = m,.cpc.dtc ...(ii) = - C,.dth = C,.dtc From expression (ii), we have

-

dth = - d Q

cc

Substituting the value of dQ from expression (i) and rearranging, we get (fh

-

Q = mh Cp,l (thl - th2) = mc Cpc (tc2 - tcl) ...( 10.36) The product of mass flow rate and the specific heat, as a matter of convenience, is defined as the fluid capacity rate C: mh cph = Ch = hot fluid capacity rate m, cpc = C, = cold fluid capcity rate ,Cm, = the minimum fluid capacity rate (Chor Cc) C,, = the maximum fluid capacity rate (Ch or Cc). The maximum rate of heat transfer for parallel flow or counter-flow heat exchangers would occur if the outlet temperature of the fluid with smaller value of Ch or Cc Le., Ch were to be equal to the inlet temperature of the otherfluid. The maximum possible temperature change can be achieved by only one ofjuids, depending upon their heat capacity rates. This maximum change cannot be obtained by both the fluids except in the very special case of equal heat capacity rates. Thus :

-

dt, = dQ

and

ch

(th

-

=

- tc)

- u.dA

[i $1 +

Upon integration, we get

From eqn. (10.38), we have the expressions for effectiveness E

Ch (thl

=

(thl

- th2) -'t c l )

Cc (tc2 emin (thl

-

tcl)

'

Hence,

Eliminating th2and tc2from eqn. (10.40) with the help of eqns. (10.41) and (10.42), we get

.Once the effectiveness is known, the heat transfer rate can be very easily calculated by using th<equation

Q=

E

Cm,, (t,,1 -

f, 1 )

...( 10.39)

Number of transfer units (NTU) method : It is obvious from Eqn. (10.38) that effectiveness & is a function of several variables and

\

or

/

E

=

1 - exp [ - (UA / Ch){l + (Ch/ c ~ ) } ] cmin

If Cc > Ch then Cmin= Ch and C,,

( >i +)

= Cc, hence eqn. (10.43) becomes

...(10.43)

Heat and Mqss Transfer

584

Heat Exchangers

If Cc < Ch thed Cm, = Cc and CM, = Ch, hence eqn. (10.43) becomes E

=

- exp [-(UAIC-)I1 + (Cm,/Cm,)}]'

1

...(iii) "

...(10.45)

1 + (Cmin1 Cmm) By rearranging eqns. (10.44) and (10.45), we get a common equation E

=

1

...(iv)

- exP 1- (UA I CmJ (1 +
Substituting these values in e4n. (10.471, we get,

[lhl -

1 + (Cminl Cmm) where Cminand C,, represent the smaller and larger of the two heat capacities Cc and Ch. The grouping of the terms (UA)ICminis a dimensionless expression called the number of transfer units NTU; NTU is a measure of qfectiveness of the heat exchanger. Cmhl C,, isi the second dimensionless parameter and is called the capacity ratio R. The last dimensionless parameter is theflow arrangement, i.e., parallel flow, counter-flow, cross-flow and so on. Thus the effectiveness of a parallel flow heat exchanger is given by E

E

=

1 - exp [- NTU (1+ (Cm, / cmAt]

1 + (Cm, CmA 1 exp [- NTU(1 + R)] = l + R

+

E Cmin (thl cc

..,l10.46 (a)]

cc

=, exp [(UAIC,) I 1

- (CcI Ch)11

= exp [(UNC,) I 1

- (CcI Ch)11

- tcl)]

--E . Cmin

or 1

Ch

. Cmin7 -

1

LC

-y

Substituting these values is eqn. (10.48), we get,

. Cmin , E

d(th - tc) = - dQ [l - l] = dQ [l ch

cc

cc

ch

Substituting the value of dQ from expression (i), we get,

Upon integration, we get

-Cmm

-

From Eqn. (10.38), we have the expressions for effectiveness

or

E

=

...(10.48)

E

Assume Cc < Ch, Cc = Cminand C,, = C., E - Cmin 1 ---;;---

dQ dtc = - -

and

- tcl

...(10.46)

(ii) Counter-flow heat exchanger : Refer Fig. 10.9. The heat exchange dQ through an area dA of the heat exchanger is given by dQ = U.dA (th - tc) ...(i) = - m cPhdth = - m cCpc dtc = - Ch dth = - Cc dt, ...(ii) From expression (ii), we have

..

- kcl

- tcl)]

ch

1

-

dth = - dQ Ch

thl

Cmb ( f h l

exP [(UA/Cmin) ( 1

- (Cmin/cm,)}I

- (Cmin /CrnaJ\I - 1 Cmin exP [(UA/Cmin) ( 1 - ( c m i n /cma>}I - Cm, 1 - exp [(- UA /Cmi,J{ I - (Cmin/c,))] ~ X [(UA P /Cmin) {l

Heat and Mass Transfer

586

= R and UA/C,, = NTU, therefore, 1 - exp [-NTU(1 - R)] E = 1 - R exp [- NTU(1 - R)] We find that effectiveness of parallel flow and counter-flow heat exchangers is given by the following expressions:

Heat Exchangers

-

+ R)] 1 + R - 1 - exp [- NTU ( 1 - R)] (~)counter~ow - 1 - R exp [-NTU(1 - R)], - 1 - exp [- NTU(1

(~)pru/lei flow -

Heat transfer surface

, Hot fluid 3

Since Cmin/ C,,

Y/7////711///7/////1////7/////7n Cold flu~d

...( 1 )

...(2)

-

here R = (C,,,,, 1 C,) Let us discuss two limiting cases of eqns. ( 1 ) and (2) Case I: When R 0 ...Condensers and evaporators (boilers)

By using the above case, we arrive at the following common expression for parallel flow as well as counter-flow heat exchangers E = 1 - exp (- NTU) ...(10.51) Such cases are found is condensers and evaporators in which onejuid remains at constant temperature throughout the exchanger. Here C,,,, =

(29

and thus R = -

0.

Obviously, no matter how large the exchanger is or how large the overall transfer coefficient is the maximum effectiveness for parallel flow heat exchanger is 50%. For counter-flow, this limit is 100%. For this reason, a counter flow is usually more advantageous for a gas turbine heat exchangers. Case I1 : When R = 1 ... Typical regenerators (i) In case of parallelflow heat exchanger using R = 1 , we get 1 - exp (- 2NTU) E =

2 (ii) In case of counter-jlow heat exchanger using R = 1 we get an expression for effectiveness

Number of transfer units, NTU,,

UA

=

Cmin

Fig. 10.44. Effectiveness for parallel flow heat exchanger.

>

I

-

, Hot fluid

r

Heat transfer surface

Cold fluid

which is indeterminate. We can find the value of E by applying L, Hospital's rule : 1 - exp [- NTU ( 1 - R)] lim R . + I 1 - R e x p [ - N T U ( 1 - R)]

e x p [ N T U ( l - R)] - 1 -+ 1 exp [NTU ( 1 - R)] - R Differrentiating the numerator and the denomenator with respect to R and taking the limit, we get, exp [NTU(l - R) (- NTU)] - NTW lim ...( 10.53) R+ 1 exp [NTU(l - R)] (- NTU) - 1 1 + NTU The NTU is a measure of the heat transfer size of the exchanger; the larger the value of NTU, the closer the heat exchanger approaches its thermodynamic limit. The effectiveness of various types of heat exchangers in the form of graphs (prepared by lim

Kays and London) for values of R

(=%I

and NTU are shown in Fig. 10.44 to 10.49. Cmu 10.8. PRESSURE AND PUMPING POWER An important consideration in heat exchanger design, besides the heat transfer requirements,

Number of transfer units, NTU,,-

-ycr Cmin

Fig. 10.45. Effectiveness for counter-flow heat exchanger.

Heat and Mass Transfer Shell fluid

141

Heat Exchangers Cold fluid

/-Tube fluid

4

I

One shell One Shell Pass

Number of transfer units, NTU-=-

UA

Number of transfer units, mu-=

Cmin

CIA Cmin

Fig. 10.46. Effectiveness for 1-2 parallel counter-flow heat exchanger. Fig. 10.48. Effbctiveness for cross-flow heat max exchanger with both fluids unmixed. Shell fluid

1

2 Shells

Mixed fluid

n o shell passes 4,8, 12 etc, Tube passes

Number of transfer units, NTU,,

) Unmixed fluid

=

UA -

Number of transfer units, NTU-

Cmin

Fig. 10.47.

c

=

em&#

Fig. 10.49. Effectiveness for cross-flow heat exchanger with one fluid mixed and other &ud,

Heat and Mass Transfer

590

is the pressure drop pumping cost. The heat exchanger size can be reduced by forcing the fluid through it at higher velocities thereby increasing the overall heat transfer coefficient. But due to higher velocities there will be larger pressure drops resulting in larger pumping costs. The smaller diameter pipe, for a given flow rate, may involve less initial capital cost but definitely higher pumping costs for the life of the exchanger. We know that, Ap = m2 ...(i) where Ap = pressure drop of an incompressible fluid flowing through the pipes, and m = mass flow rate. In order to pump fluid in a steady state, the power requirement is given by m power = v d p = -dp z h3 ...(ii) P This indicates that the power requirement is proportional to the cube of the mass flow rate of the fluid and it may be further increased by dividing it by the pump (fan or compressor) efficiency. Thus, we find that the pumping cost increases greatly with higher velocities, hence, a compromise will have to be made between the larger overall heat transfer coefficient and corresponding velocities. Example 10.33. Steam condenses at atmospheric pressure on the external sulface of the tubes of a steam condenser. The tubes are 12 in number and each is 30 mm in diameter and 10 m long. The inlet and outlet temperatures of cooling water flowing inside the tubes are 25°C and 60°C respectively. I f the flow rate is 1.1 kgh, calculate the following: (i) The rate of condensation of steam, (ii) The mean overall heat transfer coefficient based on the inner sur- t,l = 2 5 0 ~ face area, (iii) The number of transfer units, and Fig. 10.50. (iv) The ejfectiveness of the condenser. Solution. Given : N = 12; di = 30 mm = 0.03 m; L = 10 m; tcl = 25°C tc2= 60°C; thl = th2 = 100°C; m, = m, = 1.1 kgh (i) The rate of condensation of steam, & (= &) Heat lost by steam = heat gained by water liz, x hf, = mc x cPc(tc2 - tcl) where hf, (latent heat of steam) at atmospheric pressure = 2257 kJ/kg. Substituting the values, we get, riz, x 2257 = 1.1 x 4.187 x (60 - 25) or iit, = 0.0714 kgls = 257 kglh (ii) The mean overall heat transfer coefficient, U: Total heat transfer rate is given by

Q

=

m, x

cp, x (tc2 - t,l)

(Ans.)

,

Heat Exchangers 591

where and

A = N x ( n d L ) = 12 X 7~ 11'0.03x 10 = 11.31 Substituting the values in the above equation, we get 161 199.5 = U x 11.31 x 55.68

.

'V,

m2

U = 255.9 w/m2"c (Ans.)

or

(iv) The effectiveness of the condenser, E: E = 1 - exp (- NTU) E = 1 - e-0.628= 0.47 (Ans.) or

Solution. Given:

d = 25 mm = 0.025 m; m, = & =

u = 230 w/m2Oc ; thl = 100°C.

...[Eqn. (10.51)]

0.05 Ws, tc, = 1s0c, tc2= 70°C;

(i) The effectiveness of the heat exchanger, E : Throughout the condenser the hot fluid (i.e., steam), remains at

cm,is infinity and thus Cminis obviously for cold fluid (i.e., water) ~h .

When Ch > C,, then effectiveness is given by

US7;-zO. L m ~

Cmin= mc cW = 0.05 x 4.18 = 0.209 For

or

temperature. Hence Cmin

ulK

cmtn (= R) z 0

ctn~

e = 1 - exp (-NTU) 0.647 = 1 - e-NTU

...[Eqn (10.51)]

Heat and Mass Transfer

9 2

or or

e-NTu= 1 - 0.647 = 0.353 - NTU = In (0.353) = - 1.04

..

5@3

Heat Exchangers

The thermal capacity of hot stream (oil), Ch = mh cph= 0.55 x 2.45

=

1.347 kW

Since Cc > Ch, hence the effectiveness of the heat exchanger is given by

NTU = 1.04

NTU = UA - U X n d L Cmin c min N T u X Cmin 1.04 X (0.209 x 1000) L = = 12 m (Ans.) Und 230 x n x 0.025 (iii) The rate of steam condensation, m,,: Using the overall energy balance, we get

E =

But

&, = 0.00509 kgls or 18.32 k@ (Ans.) or Example 10.35. A counter-jl~wheat exchanger is employed to cool 0.55 kg/s (c, = 2.45 kJLkg°C) of oil from llJ°C to 40°C by the use of water. The intel and outlet temperatures of cooling water are I f C and 7j°C, respectively. The oyerag heat transfer co@cient is expected to be 1450 w / ~ ~ ~Using c . NTU method, calculate the following: (i) The mass flow rate of water; (ii) The effectiveness of the heat exchanger; (iii) The sulfQce area required.

/

actual heat transfer = & = thl - th2 maximum heat transfer Qm, thl - tcl

...[Eqn. (10.35)]

115 - 40 = 0.75 (Ans.) 115 - 15 (iii) The surface area required, A: Here em,,= Ch = 1.347 kW; Cm, = Cc = 1.672 kW, hence -Cmin - R =-= 347 0.806 1.672 cm, For counter-flow heat exchanger, 1 - exp [- NTU(1- R)] E = ...[Eqn. (10.50)] 1 -Rexp [-NTU(1 - R ) After rearrangement, we get E-1 = exp [- NTU (1 - R)] (a- 1 ) 0.75 - 1 or = exp '[- N W (1 - 0.806)] (0.75 x 0.806 - 1 ) or 0.632 = exp [- NTU x 0.1941 or In 0.632 = - 0.194 NTU or NTU = 2.365 E =

[This value of NTU may also be obtained h m Fig. 10.45 for R = NTU

Also

=

% = 0.806 and E = 0.751 em,

UA cmin

5 Fig. 10.51.

Solution. Given: kt,,,, = 4, = 0.55 kgls; cph= 2.45 15OC, tc2 = 75OC;

ILJAC~OC;

thI = I lS0C, th2= 40°C; tcl =

( i ) The mass flow rate of water, mc (= m,): The mass flow rate of water can be found by using the overall energy balance

0.55

..

X

- th2)

= m, Cpc (tc2 - tcl) 2.45 (115 - 40) = mc x 4.18 (75 - 15)

mh Cph (thl

i+ = 0.4 kg/s

(Ans.)

(ii) The effectiveness of the heat exchanger, E: The thermal capacity of cold stream (water), C, = in, cpc= 0.4 x 4.18 = 1.672 kW -

Example 10.36. 16.5 kgh of the product at 650°C (cp = 3.55 kl/kg°C), in a chemical plant, are to be used to heat 20.5 kgh of the incoming fluid from IOO°C (cp = 4.2 kl/kgO~).' If the overall heat transfer coefficient is 0.95 k ~ / m ' "and ~ the installed heat transfer surface is 44 m2, calculate the fluid outlet temperatures for the counter-flow and parallel flow arrangements. Solution. Given: tih = 16.5 kgls, th, = 650°C; cph = 3.55 kT/kg°C; m, = 20.5 kgfs; t,, = 100";

c, = 4.2 kJkg°C; U = 1.2 kw/m2Oc; A = 44 m2.

Fluid outlet temperatures: Case I. Counter-flow arrangement: Thermal capacity of hot fluid, Ch = ljZh x cph= 16.5 x 3.55 = 58.6 k W K 86.1 k W K Thermal capacity of cold fluid, C, = m, x c, = 20.5 x 4.The cold fluid is the maximum fluid, whereas the hot fluid is the minimum fluid. Therefore,

Heat and Mass Transfer

594

i::

n = - =CmiR

c,

Number of transfer units, The value of

E

Also,

30000 Solution. Given: moil= mh = - = 8.333 kgls; cph= 3.6 kJ/kg°C; thl = 100°C; m ,,, 3600 . 50000 = m, = -= 13.89 kgls, cp, = 4.2 kJ/kg°C, t,, = 10°C; U = 1000 w / m 2 ~ cA; = 10 m2. 3600 (i) The outlet temperature of oil and water, th2.tc2:

- 0.68

UA - 0.95 x 44 = 0.71 NTU = -Cmin 58.6

...[Eqn. (10.50)] = 1 - exp [- NTU(1 - R)] 1 - R ~ X ~ ' [ - N TU R)] (~ 1 - e-0.2272 1 - exp [- 0.71 (1 - 0.68)] 1 - 0.68 X exp [- 0.71 (1 - 0.68)] 1 - 0.68 x e-0.2272

(thl - 'h2) Cmin (fh~- tcl) Because, the hot jluid is minimum, we have

or

,595

(effectiveness) for counter-flow arrangement is given by

E

Further,

Heat Exchangers

E

=

ch

...[Eqn. (10.38)]

thz = 650 - 0.443(650 - 100) = 406.3S°C (Ans.) E

=

UA - 1000 X 10 = 0.33 NTU = - cmin 3 0 x 1 0 ~ th = IOOOC

t,, = loOc

- tcl) Cmin (ih~- tc~) c c (tc2

'

...[Eqn. (10.38)]

(ii) I

Fig. 10.52. Lmin

For the calculated values of -= 0.514 and NTU = 0.33, from the Fig. 10.44, we get cmax

..

tc2 = 2658°C (Ans.) Case 11. Parallel jlow arrangement: The value of & for parallel flow arrangement is given by 1 - exp [-NTU(1 + R)] E = l + R

- 1

- exp [- 0.71 (1 + 0.68)J -(1

+ 0.68)

E

= 0.32

... [Eqn. 10.46 (a)]

1 - e-1.1928 = 0.415 1.68 ...[Eqn. (10.28)]

or tc2 = 255.4OC (Ans.) Example 10.37. Oil (cp = 3.6 kJ/kg°C) at 100°Cjlows at the rate of 30000 kgh and enters into a parallel flow heat exchanger. Cooling water (c,, = 4.2 kJ/kg°C) enters the heat exchanger at 10°C at the rate of 50000 kgh. The heat transfer area is 10 m2 and U = 1000 w/rn2"c. Calculate the following: (i) The outlet temperatures of oil, and water; (ii) The maximum possible outlet temperature of water. (P.U.)

..

t,,, = 100 - 0.32 (100 - 10) = 71.2OC

and

42

(Ans.)

= 0'32 (loo - lo) + 10 = ZQ8OC

(Am.) 1.945 (ii) The maximum possible outlet temperature of water, td: When maximum possible outlet temperature of water exists then, th2 = tC2and under this case or or

cpc (tc2 - tcl) cph (thl - tc2) = 30 x lo3 (100 - tc2) = 58.34 x lo3 (tc2- 10) 100 - tc2 = 1.945 (tc2- 10) = l.ecw'tc2 - 19.45 tcz = 40.S°C (Ans.)

Heat and Mass Transfer

Example 10.38. The following data is given for counter-flow heat exchanger: m, = 0.25 kg/s mh = 1 kg/s; cp, = 4.18 kJ/kg°C cph= 1.045 kJ/kg°C; th, = 1000°C ; tC2= 8500C; U = 88.5 w/m2"c; A = 10 in2 Calculate th2and t,,. (P.U.) Solution. Ch = mh cph= 1 x (1.045 x lo3) = 1045 Cc =

m, c,,

Heat Exchangers

The total heat transfer rate, Q: As outlet temperatures of both fluids are not known, we have to use NTU method for solving the problem.

= 0.25 x (4.18 x lo3) = 1045

..

Cmin= C,, = Ch = C, = 1045 The effectiveness E is given by the relation: E

=

- th2) Cmin (thl c c (thl

Fig. 10.54. Counter-flow heat exchanger.

C,

t,,, = 10oo0c

The effectiveness E

E

=

of heat exchanger is given by ch(thl

-

th2)

-

cc

0c2

-

tcl)

Cmin (thl - tcl) Cmin (thl - tcl) Ch = mh cph= 0.152 x 1880 = 258.8 = Cmin

L

UA - 88.5 x 10 NTU = -= 0.85 Cmin 1045

...(i)

C, = m, cp, = 0.076 x 4200 = 319.2 = C,,

Fig. 1053.

For the known values of CmiJCm,and NTU for the counter-flow, we get from Fig. 10.45, E = 0.47 Substituting this value in eqn. (i), we get

cn, cm,

For the calculated values of -= 0.895 and NTU = 1.19, from the Fig. 10.45, we get E rr

0.53

Substituting the values in eqn (i), we get

0.47(1000 - t,,) = 850 - tCl or 470 tC1- 0.47 tcl = 850 - 470

- 0.47

tCl= 850 - t,,

and The total rate of heat transfer is given by Q = UA8,

Example 10.39. Water (cpc= 4200 J/kg°C) enters a counter-flow double pipe heat exhanger at 3PCflowing at 0.076 kg/s. It is heated by oil (c, = 1880 J/kg°C)flowing at the rate of 0.152 kg/s from an inlet temperature of 116°C. For an area of 1m2 and U=340 w/mZOc,determine the total heat transfer rate, Solution: Given: m, = lit, = 0.076 kgls, cDC= 4200 J/kg°C; t,, = 38°C; lit,,,, = lith = 0.152 kgls, cph= 1880 J/kg°C; thl = 116°C; U = 340 W/m2"c;A = 1 m2.

,/hxample 10.40. The overall temperature rise of the coldfludin a cross-jlow heat exchanger is 20°C and overall temperature drop of hot-fluid is 30°C, The effectiveness of heat exchanger

Heat and Mass Transfer

is 0.6. The heat exchanger area is lm2 and overall heat transfer coeflcient is 60 W / ~ ~ OFind C. out the rate of heat transfer. Assume both jluids are unmixed. Solution. Given: tc2- tcl = 20°C; thl - th2= 30°C; E = 0.6; A = lm2; u = 60 W / m 2 ~ ~ ,

Rate of heat transfer, Q: Heat lost by hot fluid = heat gained by water mh cph (thl - th2) = cp, (tc2 - tcl)

..

mc cpc = C,,

and

mh cph = Cmin

I - 0.67 Cm, 1.5 Now from the graph, for the given values of E = 0.6 and Cmin/ C,, = 0.67, we get NTU = 1.4 (From Fig. 10.48) UA mu=Cmin Cmin= UA = 60 x 1 NTU 1.4 - 42.86 = Ch; Cmin 42.86 c,,=-=---- 63.97 = Cc 0.67 0.67

- -

..

Now,

...[Eqn. (10.39)]

Q = E Cmin (thl - tcl) = 0.75 x mh cph(thl- tcl) = 0.75 x 1 x 1 x (420 - 20) = 3200 kJ (Ans.)

(ii) The exit temperature of fluid B, tc2:

Q = mc cpc x ( 4 2 - 41) = 1 x 4 (tc2- 20) 300 or 300 + 20 = 9S°C (Ans.) tc2= .. 4 Example 10.43. Water at the rate of 0.5 kg/s is forced through a smooth 25 mm ID tube of 15 m length. The inlet water temperature is 10°C and the tube wall is at a constant temperature of 40°C. What is the exit water temperature ? Average properties of water are : (AMIE Summer, 1998) c, = 4180 J/kg°C; p = 0.8 X 10- Pa S; k = 0.57 W/m 'C.

-

-'mi'- ---

But

Heat Exchangers

= mh cph (thl - th2) = ch (thl - th2) = 42.86 x 30 = 1285.8 W (Ans.) Example 10.41. Dejhe the terms NTUand effectiveness. Derive an expressionfor effectiveness of a counter-flow heat exchanger in terms of NTU and capacity ratio. (U.P.S.C., 1996) Solution. Refer Article 10.7.

Solution. Given :m, = 0.5 kg/s; D = 25 mm = 0.025 m; L = 15 m Pas;

ti = 10°C; t*= 40°C; cp= 4180 J/kg°C; p = 0.8 x k = 0.57 W/m°C

Exit water temperature, to : m, = pVA W e know that,

Q

J

Example 10.42. Two fluids, A and B exchange heat in a counter-current heat exchanger. Fluid A enters at 4200C and has a mass flow rate of I kg/s. Fluid B enters at 20°C and has a mass flow rate of I kg/.. Effectiveness of heat exchanger is 75%. Determine : (i) The heat transfer rate; (ii) The exit temperature of Jluid B. Specz3c heat of Jluid A is I kJ/kg K and that of fluid B is 4 kJkg K. (GATE, 1999) Solution. Given :tAl= thl= 420°C; mh = 1 kg/s;

= 1 kg/s;

:.

Reynolds number,

Re =

pVD 1000 X 1.086 X 0.025 = 3,183 = P 0.8 x low3

i.e., flow is turbulent (since Re > 2300) Using the relation, = 0.023 ( ~ e ) '(. ~ r ) " ~'

104

...[Eqn. (7.15)] (for t, > tf)

Substituting the values in the above - eqn., we get

cpB= cpc= 4 kJ/kg K . (i) The heat transfer rate, Q : Heat transfer area, A = nDL = x 0.025 x 15 = 1.1781 m2 Since the surface temperature is constant,

-

UA hA 3785 x 1.1781 = 2.133 NTu=-=-2090 Cmin Cmin -

Heat and

690

Effectiveness,

E=

1 - e-NTU

Mass Transfer

...[Eqn. (10.51)]

= 1 - (e)- 2.133 = 0.8815

Now,

.. to = 0.88 15 (40 - 10) + 10 = 3644°C (Ans.) Example 10.44. A counterflow heat exchanger is to heat air entering Hot fluid (Gas) at 400°C with aflow rate of 6 kg/s by the exhaust gas entering at 800°C with a flow rate of 4 kg/s. The overall heat *c2 transfer coefficient is 100 w/m2 K and the outlet temperature of the air is 551S°C. Specific heat at constantpresI sure for both air and exhaust gas can be taken as 1100 Jkg K. Calculate : Fig. 10.55. (i) The heat transfer area needed;

Now,

601

Exchangers

(ii) The number o f transfer units, (NTU) : UA 100 x 48 NTU = -= = 1.09 Cmin 4400

(Ans.)

L Example 10.45. A chemical having specijk heat of 3.3 kJ/kg Kflowing at the rate of 20000 k g h enters a parallel flow heat exchanger at 120°C. The flow rate of cooling water is 50000 k g h with an inlet temperature of 20°C. The heat transfer area is 10 m2 and the overall heat transfer coefficient is 1050 w/m2 K. Find : (i) The effectiveness of the heat exchanger. (ii) The outlet temperature of water and chemical. (U.P.S.C., 1992) Take for water, specijk heat = 4.186 kJ/kg K.

,/

(ii) The nhmber of transfer units. Solution. Given : m, = 6 kg/s; tcl = 400°C; tc2= 551 S°C; thl= 800°C;

eat

thz

Solution. Given : cph= 3.3 kJ/kg K; h --= - 3600

5.56 kg/s;

tCl

(= 400°C)

(GATE, 1995)

t,, = 20°C; A = 10 m2; U = 1050 w / m 2 K. (i) The effectiveness o f the heat exchanger, E : ~ o fluid t capacity rate, Ch = 6th cph = 5.56 ~ 3 . = 3 18.35

i

C, = m, c,, = 6 x 1 100 = 6600, and C , = m,, c,,,, = 4 x 1 100 = 4400

Hence,

Ch < C,

..

Cmln= C,, = 4400 Now, heat transferred to cold air = heat transferred from hot gases

--

( i ) The heat transfer area needed, A :

---*

Water

+

Q = UA 8, Fig. 10.56.

where Ltth2

- 'cl)J

75.75 - (800 - 551.5) - (572.75 - 400) --o.3636 (800 - 55 1.5) (572.75 - 400) Substituting the various values, we get 999900 = 100 x A x 208.33 . A 2 48 m2 (Ans.)

.

Cold fluid capacity rate, Cc = m, c, = 13.89 x 4.186 = 58.14

- 208.33"C

We find Ch c Cc Heat lost by hot fluid = heat gained by cold fluid .'. 5.56 x 3.3 x (120 - th2)= 13.89 x 4.186 x (tc2- 20) or Now,

( 120 - th2)= 3.17 (tC2- 20)

...(i)

Heat and Mass Transfer

Effectiveness,

E=

Heat Exchangers

603

1 - exp [- NTU(l+ R)] l+R

Cmin 18.35 - o.316 where R (capacity ratio) = = -C, 58.14

1 - exp [- 0.572 (1 + 0.316)] - 1 - 0.471 = 0.402 (1 + 0.316) 1.316 (ii) The outlet temperatures of water (tc2)and chemical, (th2): E=

(120 - t,) (120 - 20) or th2= 120 - 0.402 (120 - 20) = 79.S°C Substituting the value of th2= 79J°C in eqn, (i), we get

(Ans.)

(ii) The exit temperatures of hot and cold fluids,Lh2,tc2: When mh is increased from 10 kglmin to 20 kglmin hi will become h,' (considering hot fluid is inside).

0.402 =

(Ans.)

(120-79.8) = 3.17 (tc2- 20)

h,' = 60 x 1.74 = 104.4 w/m2 OC

+ 20 = 32.70, (Ans.) 3.17 Example 10.46. A parallel flow heat exchange has hot and cold water streams running through it and has the following data: mh = I0 kg. / min ;mc = 25 kg / min; Cc2=

#',.

I

cph = cpc = 4.18 kJ/kg OC; thl = 70°C; thl = 7ooc th2 50°c8 fclX 25OC Individual heat transfer coeficent on both sides = 60 w/m2 OC. Calculate the following: (i) The area of the heat exchanger; (ii) The exit temperatures of hot and cold fluids if hot water pow rate is doubled. Fig. 10.57 (P.U.) Solution: (i) The area of heat exchanger, A: Heat lost by hot fluid = heat gained by cold fluid

..

th2= 50°c

UA 38.1 x 0.0161 NTU = - = 0.44 1.39 cmin Cmin = 0.799 and NTU = 44, from the fig. 10.44, we get For the calculated value of -

cm,

Also,

mit, cph (thl - th2) = mc cpc (tc2 - tc,) 10 x 4.18 (70 - 50) = 25 x 4.18 (tc2- 25)

Log-mean temperature difference,

The overall heat transfer coefficient is given by,

t,, = 70 - 0.31 (70 - 25) = 56.0S°C 0.31 (70 - 25) + 25 = 36.16OC tc2= 1.25

(Ans.) (Ans.)

Example 10.47. An oil is cooled to 375 K in concurrent heat exchanger by transferring its heat to the cooling water that leaves the cooler at 300 K. However, it is required that oil must be cooled down to 350 K by lengthening the cooler while the oil and waterpow rates, their inlet temperatures and other dimensions of the cooler remaining unchanged. The inlet temperatures of the cooling water and oil being 288 K and 425 K respectively.

TI-

Heat and Mass Transfer

604

If the length of the original cooler was I m, calculate the following: (i) The outlet temperature of cooling water of the new cooler, and (ii) The length of the new cooler. (M.U.) Solution: Given: thl = 425 - 273 = 152°C; th2= 375 - 2'73 = 102°C; tcl = 288 - 273 = 15°C; tc2 = 300 - 273 = 27°C; L, = lm ... Case I

4,

= 350 - 273 = 77OC; tc2 = ?

... Case I1

Heat Exchangers

605

Case I1 : (After increasing the length) The effectiveness of the parallel flow for the second case is given by 1 - exp [-(NTU), (1 + R)] E = l+R 1 - exp [-0.486 L2 (1 + 0.24)] - 1 - (e)-0.6Lz E = 1 + 0.24 1.24

Cmin where R = -

cm

...(i)

The effectiveness E for the second case is also given by

Lmin

Cc = Cm, and R = -= 0.241

['.' Ch = C,

Cm,

...(ii) 152 - 77 x 0.24 (152 - 15) + 15 = 33OC = (152 - 15) (ii) The length of the new cooler, L, : Equating eqns. ( i ) and (ii), we get. 1 - (e)-0.6% 152 - 77 1.24 152 - 15

'''

Fig. 10.58.

The outlet temperature of the cooling water of the new cooler, 4, : Case I : (Before increasing the length)

..

C,

(Ans.)

Cm in - R = 0.24 = hhcp, and -

cm,

Heat transfer rate is given by L,=-- 134 - 1.89 m (Ans.) 0.6 This indicates 89% increase in length of the cooler. Example 10.48.A simple counter-flow heat exchanger operates under thefollowing conditions: Fluid-A, inlet and outlet temperatures 80°C and 40°C; Fluid-B, inlet and outlet temperatures 20°C and 40°C. The exchanger is cleaned, causing an increase in the overall heat transfer coeflcient by 10% and inlet temperature offluid B is changed to 30°C. What will be new outlet temperatures of fluid A and of fluid B. Assume heat transfer coeflcients and capacity rates are unalered by temperature changes. ... Case I Solution. Given: thl = 80°C. th2 = 40°C. tcl = 20°C tc2 = 40°C ... Case I1 thl = 80°C, th2 = ?, tcl = 30°C, tc2 = ? U2 = 1.1 U, As outlet temperatures of both fluids are to be calculated, so we have to use NTU method to find th2and tc2 for the new inlet condition of the cold fluid after cleaning the exchanger. Further the area of heat exchanger and mass flow rates in both cases remain same. or

or (152 - 102) = (NTO), Or

..

= (NTO), x 102.9

- lo2) = 0.486 102.9 UA, UCL, (NTU), = - = -, where C = n d Cmin cmin WTUh =

UC 0.486 = 0.486 [': --

Cmin

LI

L, = lm (given)]

Heat and Mass Transfer

Heat Exchangers

Again

=

Cmin R = capacity ratio = 7,

where It is obvious that

mh

cph =

cmin

I

I

Case I ( U l = U )

Case I ( U2= 1.1 U 1 ) Fig. 10.59

We can also write

Case 11: as U2 = 1.1 U1 and A and Cm, both the variables remain same in both cases. The effectiveness of the counter-flow heat exchanger for the case I1 is given by E

=

The effectiveness

E

1 - exp [- (NTU), ( 1 - R)] 1 - R exp [- (NTU)2( 1 - R)]

where R = 0.5 (calculated earlier)

1 - e- 0.76 1 - exp [(-I .52) ( 1 - OS)] 1 - 0.5 exp [(-I S 2 ) ( 1 - 0.5)] 1 - 0.5 x e-0.76 = 0.695

-

is also given by

= 47.4"C (Ans.) Example 10.49. In a large steam power plant, a shell and tube lype steam condenser is employed which has the following data: ...2100 M W Heat exchange rate ...one Number of shell passes ...31500, each executing two passes Number of tubes (thin walled) ...25 mm Diameter of each tube ...3.4 x 104 kg/s Mass jlow rate of water through the tubes ...50°C. The condensation temperature of steam (The steam condenses on the outer suface of the tubes) ...11400 w/~**c The heat transfer coeflcient on the steam side The inlet temperature of water Using LMTD correction factor method Steam (mh) I and NTU method, calculate: (i) The outlet temperature of cooling water, and (ii) The length of tube per pass. Take the following properties of water (at tbulk= 27'C): cp = 4.18 Wkg°C, p = 855 x 1@ ~ s / m k~ =, 0.613 W/m°C and Pr = 5.83 (m,) The thermal resistance of tube material (a) Flow arrangement and fouling effects may be neglected. thl = th2 = 5O0C Solution. Given : Q = 2300 x lo6 W ; N, = 31500; d = 25 mm = 0.025 m; thl = th2= 50°C; hw= hc= 3.4 X lo4 tcl = 20°C; h, = 11400 ~ l m ~ ~ c . (i) The water outlet temperature, tC1= 20°C 42

...[Eqn. (10.381 E

or

E

=

- 'h2) -- ',in (thl - th2)---fhl - th2 Cmin (th~- t c ~ ) Cmin(thl- tcl) thl - tcl 'h

(*h~

= 80 - 0.696 (80 - 30) = 452°C

:

In order to obtain water outlet temperature, using overall energy balance, we get Q = mc Cpc ($2 - tCl)

.. (Ans.)

(b)Temperature distribution

Fig. 10.60. Shell and tube type condenser.

2100 x lo6 = 3.4 x lo4 x (4.18 X 1000) (tc2- 20) tc2 = 34.77OC (Ans.)

(ii) The length of the tube per pass, L: (1) LMTD correction factor method:

I

Heat and Mass Transfer

608

The total rate of heat transfer is given by, Q = FUAO, where F = correction factor, U = overall heat transfer coefficient, A = 2Np n d L, where Np is the number of tubes per pass, and Om = log-mean temperature difference. Tn find F we have to find P(temperature ratio) and R (capacity ratio).

...(i)

R=--'hl - th2 - 50 - 50 = 0 37.44 - 20 tc2- t,, With the values of P = 0.492 and R = 0, we get F = 1 To find the value of U, hi has to be calculated first.

*

-

Now,

Cmin = 0 .-.

But

NTU = U A - U(2NpndL) Cmin Cmin

L =

0.677 x 14212 x 104 = 4.131 m 4707.2 x (2 x 31500 x n x 0.025)

(Ans.)

Example 10.50. In a double pipe heat exchanger ih cph= 0.5 inc cpc,The inlet temperatures of hot and cold fluids are thl and tcl. Deduce an expression in t e n s of thl, tcl and th2for the ratio of the area of the counter-flow heat exchanger to that of parallel flow heat exchanger which will give the same hot fluid outlet temperature th2. Find this ratio if thl = 150°C, tcl = 30°C and (U.P.S.C., 1994) th2 = 90°C. I

1 -1 = -+-

U

Here

..

Here, mass flow rate through each tube = riz = 3.4 x 104 = 1.079 kgls 31500 4 X 1.079 Reynokis number, Re = PDV = p n d p n x 0 . 0 2 5 ~ 8 5 5lo6 ~ = 6.43 x 104 Since Re > 2300, hence the flow is turbulent.

:.

Heat Exchangers

h,

1 hi

...neglecting thermal resistance of tube material and fouling effects.

I (b) Counter-flow arrangement

(a) Parallel flow arrangement

Further, LMTD is given by,

Substituting the values in eqn. (i), we get Q = FU(2NPndL)8, 2100 x lo6 = 1 x 4707.2 [ 2 x 31500 x n x 0.025 x L] x 21.8 . L = 4.136 m (Ans.) (2) NTU method: Since the heat exchanger is a condenser, hence Ch = Cmm= and C,,,,, = Cc = m,cpc = 3.4 x lo4 x (4.18 x 1000) = 14212 x lo4 Since Ch > Cc, hence

.

The effectiveness of the heat exchanger & is given by Ch (thl - th2) E = .

Cmin (thl

In this case Cmi, = Ch and Ch = 0.5 Cc and Cc = C,,

= 2Ch

.

i

nh

-c

Cc (tc2

- 'cl)

- tcl)

-

This equation is true for counter flow as well as parallel flow. A, = area of parallel flow heat exchanger, and Let A, = area of counter-flow heat exchanger. As (th2), = (th2),therefore, the heat lost by hot fluid in both the cases is same.

610

Heat and Mass Transfer

.. ..

Q = uAp (ern),= u A c (ern),,since U is not dependent on the direction of flow. Ac -=-

From eqn. (I), we get tc2in terms of thl, th2and tc1as tc2 = tcl + 0 3 (thl (a) Parallel flow:

- th2)

611

Heat Exchangers

A x a m p l e 10.51. 8000 k g h of air at l0O0C is cooled by passing it through a single-pass cross-flow heat exchanger. To what temperature is the air cooled if water entering at 15OCflows through the tubes unmixed at the rate of 7500 kgh? Take : t.J = 500 k.l/h-mZ0c and A = 20 mZ c ( a ) = 1 kr/kg°C and cp (water) rcz = ? = 4.2 kJkg°C Treat both fluids are unmixed.

=

Solution: Given: rhh = 3600 = 2.22 kgls; cph= 1

(b) Counter-flow:

I

1.5 th2- 0.5 thl - tcl

...(i)

NTU =

UA - 1 3 8 9 x 2 0 = Cmin 2220

For the calculated values of

L (th2 - tcl) 1 Substituting the values of (i) and (ii) in eqn. (2), we get

The given data is : thl : 150°C, tc, = 30°C and th2 = 90°C

Cmin

= 0.254 Cmm

and NTU = 1.25, from the Fig. 10.48, we get E = 0.63

..

th2 = 100 - 0.63 (100 -15) = 46.45"C

0'63 (loo - 15) 15 = 28.60C 3.935 The air is cooled to a minimum temperature of 46.4S°C. (Ans.) dExample 10.52. It is required to design a finned tube, cross-flow heat exchanger to heat pressurized water by means of hot exhaust gases, which enter the exchanger at 310°C and leave it at llO°C, respectively. The water flowing at the rate of 1.4 kg/s enters the exchanger at 3oOC and leaves at 130°C. The hot exhaust gases heat tranCr coeficient based on the gas side is 105 w/rn2'c. Using NTU method, calculate the following: and

+

t"2 =

Heat and Mass Transfer

(i) The efectiveness, and (ii) The gas side surface area. Take the following properties: Exhaust gas: cp = 1 kJ/kg°C: water (at tbUIk= 80°C) :cp = 4.2 U/kg°C Solution. Given: m, = i t c = 1.4 kgh; tcl = 30°C; m = 130°C; cpc = 4.2 Wlkg°C; thl = 3 lo°C; th2 = 1 10°C; cPh = 1 kl/kg°C; Uh = 105 w / m 2 OC. 'hI

Heat Exchangers r,, = 4 5 0 " ~

from the hot gases leaving the turbine at 450°C to the air leavhg the'c~m~ressor at 170°C. The air flow rate is 5000 k g h and fuel-air ratio is 0.015 kg&. The overall heat transj2r coeficient fhr the heat exchanger is 52.33 w/m2 OC. The surface area is 50 m2 and arrangement is cross-flow (both fluids unmixed). Calculate the following: (i) The exit tempeatures on the air and gas sides, and (ii) The rate of heat transfer in the exchanger.

tC2= !'

thZ = ?

Fig. 10.64

Take cph = cpc = 1.05 W k g OC Solution: Given : th, = 450°C; tcl = 170°C; A = 50 m2. (i) The exit temperatures on the air and gas sides tc2,th2:

Fig. 10.63. Fixed tube, cross-flow heat exchanger.

mh =

(i) The effectiveness, & : The he'at capacity o f cold fluid, The value o f Ch (heat capacity o f hot fluid) may be obtained by writing overall energy balance (since mh is not given). Hence

& = -lhl -

(ii) The gas side surface area, Ah :

-

Corresponding to Cmin = 0.5 and

- th2

-

thl-tcl

E

310 - 110 = 0.714 310-30

But

NTU =

= 0.714 from Fig. 10.48, we get

41Ah

cmin "fl

The effectiveness & is given by

and

,

1.8 x 2.94 x 1000 = 50.4 m2 (Ans.) 105 Example 10.53. In a gas turbine power plant heat is being transferred in a heat exchanger

0.52 =

tc2 - tcl - tc2 - 170 thl - tcl 450 - 170

tc2 = 0.52 (450 - 170) + 170 = 315.6OC (ii) The rate of heat transfer in exchanger, Q: Q = UA(0m)coun,er= FUA (0m)counter where F = correction factor

..

Cmin

A, =

= 1.4 1 kgls as 1.015 kg of gases are formed per kg of air

(Ans.)

~ m ,

-

3600

50 = 1.795 1457.4 Cmin = 0.984 and NTU = 1.795, From the Fig. 10.48, we get For the calculated values o f

NTU =

Hence Cm, = Cc = 5.88 kWI0C and Cmin= Ch = 2.94 kWI0C When Cc > Ch, then effectiveness E is given by

50M)

(Ans.)

Heat and Mass Transfer

Heat Exchangers 0

-

0

Temperature ratio, P = tc2-tcl - 315.6-170 = 0.52 thl - tCl 450 - 170

41- 4 2 - tcl

B. On the basis of the method of evaporation : 1. Single effect evaporators :

450 - 306.6 = 0.985 315.6 170 Using the values of P and R, from Fig. 10.48, we get

Capacity ratio, R =

th2

-

F = 0.76 Now substituting the values in eqn. (i), we get Q = 0.76 x 52.33 x 50 x 135.5 = 269447 W or 269.45 10.9. EVAPORATORS

Basket vertical tube Long tube vertical.

kW (Ans.)

10.9.1. Introduction Evaporation is a process in which the solvent is removed from a solution of a non-volatile solute and a volatile solvent. Examples. Brine when heated with water (solvent) is evaporated and salt (solute) is produced. Other typical examples of evaporation of water are concentration aqueous solutions of sodium hydroxide: glycerol, sugar, juices and glue. In certain cases where the primary object is to get crystals by evaporation, this special process is termed crystallisatlon. On account of changes ili concentration and properties of the solution the following special problems may creep up : 1. Concentration of liquid 2. Solubility and crystal formation 3. Foaming 4. Pressure and temperature 5. Temperature sensitivity of materials 6. Entrainment of drops 7. Scale deposition and materials of construction. 10.9.2. Classification of Evaporators Evaporators may be classified as follows : A. On the basis of the fluids to be evaporated : 1. Power plant evaporators : (i) Make up evaporators for boiler feed (ii)Process evaporarors (iii) Salt water evaporators (iv) Heat transfer evaporators. In all above evaporators water is the solvent. 2. Chemical evaporators : Chemical evaporators may be natural circulation type or forced circulation type. Natural circulation may fall into the following four main classes : Horizontal tube Vertical tube

(i) Batch evaporators. Here filling, evaporating and emptying are consecutive steps. (ii) Semi-batch. Here feed is continuously added to maintain a constant level until entire change reaches a final density. (iii) Continuous evaporators. Here feed and discharge are continuous, and concentration of both feed and discharge remain substantially constant. 2. Multiple effect evaporators : (I).Forward feed Here raw feed is introduced in the first effect, and passed from effect to effect in parallel to the steam flow. The product is withdrawn from the last feed. (ii) Backward feed, Here the raw feed enters the last (coldest) effect, the discharge from this effect becomes the feed to next to the last effect and so on, until the product is discharged from this first effect, The following characteristics are important in design and operation of evaporators ' 1. The boiling heat transfer coefficient. 2. The boiling point. The heat transfer coefficient (i.e,, boiling film coefficient) depends upon viscosity, and surface tension; p (boiling coefficient) increases with increase in viscosity and varies inversely with surface tension. h increases with increase in boiling point. h .;pa,where n lies between 0.3 to 0.78 depending upon the solution.

-

-

Types of Evaporators : The general types of evaporators are : 1. Open kettle or pan. 2. Horizontal tube natural circulation evaporator. 3. Vertical type natural circulation evaporator. 4. Long tube vertical type natural circulation evaporator. 5. Falling film type vertical natural circulation evaporator. 6. Forced circulation type evaporator. 7. Agitated film evaporator. Example 10.54. (a) Between counter-current and co- current heat exchangers which is more preferable, and why ? (b) A triple effect evaporator is concentrating a liquid that has no appreciable elevation in boiling point. The temperature of the steam to the jirst effect is 108OC, the boiling point of the solution in the last effect is 52OC The overall heat transfer co-eflcients in w/m2'c 2500 in whal tempemtun will the fint effect, 2000 in the second effect and 1000 in the third effect. At(AMIE Winter, 1996) the liquidbil in the first and second effects ? Solution. (a) Counter-current heat exchanger is more preferable to co-current heat exchanger because of the following reasons : 1. The counter-current heat exchanger is thermodynamically the most eficient. 2. For given heat flow rates, and given inlet and outlet fluid temperatures, a counter-current heat exchanger requires the minimum heat transfer area and a cwcurrent heat exchanger requires the maximum heat transfer area.

616

Heat and Mass Transfer

3. In co-current flow (i.e.. parallel flow), the lowest temperature theoretically attainable by the hot fluid is that of the outlet temperature of the cold fluid. If this temperature were attained, the WMTD would be zero, requiring infinite heat transfer surface. Further, inability of the hot fluid in a co-current flow to fall below the outlet temperature of the cold fluid has a marked effect upon the ability of the co-current apparatus to recover heat. Hence, co-current heat flow heat exchanger are rarely used. (b) Assuming that the multiple effect evaporator is forward feed type, the arrangement is drawn below : Steam to ejector --3 I

Feed

Heat Exchangers

617

space pressure is 100 mm Hg (absolute). Find out the steam consumption, steam economy and the heat transfer area if the following data is available. Feed temperature = 37'C; BPE = 52OC; (cP)feed= 0.92

-

O' 75 Overall heat transfer co-eflcient, Uo= 1200 w/m20c. Solution. (a) Refer Article 10.9.2. mv (vapour at 100 mrn Hg) (cp)product=

I I I I

--

----Steam 108°C

(mf= 15000 kg/h, 20% solid) Product

Fig. 10.65. Multiple effect evaporator (Forward feed type).

For the above arrangement, if equal surfaces are employed in each effect (as is the usual practice), the difference in pressures between tbeffects, will be nearly equal. From steam .. tables, Correspondingto 108"C, pressure = 1.3390 bar, Corresponding to 52"C, pressure = 0.1360 bar;

&I,

Average pressure difference = 1.3390 - 0.1360 = 0.401 bar 3 Break-up of the total pressure difference is given in the table below : I

s Heating steam

.

Drip

(AMIE Winter, 1997)

I

1

Pressure (bar)

AP (bar)

Steam or vapour "C

...

108

Steam chest, first effect

1.3390

Steam chest, second effect

0.9380

0.401

98

Steam chest, third effect

0.5370

0.401

83

Vaccuum to ejector

0.1360

0.401

52

(

So, the boiling temperatures of liquid are : First effect = 98°C (Ans.) Second effect = 83°C (Ans.) Example 10.55. (a) How are evaporators dasszj?ed ? What are the characteristics of the solution which are important in design and operation of evaporators and how do these characteristics effect the evaporation process ? (b)A single effect evaporator is used to concentrate 15000 kg/h of a 20% solution of caustic soda to 60% concentration. Heating medium is dry and saturated steam at 125°C. The vapour

4

(m,

Product

= 60%)

Fig. 10.66. Single effect evaporator.

(b) Caustic soda balance : 15000 x 0.2 = mpx 0.6 :. Mass of product, mp : 0.6 0e2= 5000,@l 1500 = m, + mv = 5080 + m (where mv = mass of vapour) Total mass balance, m, = 15000- 5000 = 10000 k g h .. hh of heating steam (at 25°C) from steam tables = 2188 kJ/kg Vapour pressure = 100 mm Hg = (13.6 x 1000) x 9.81 x Eloo0 x

bar = 0.13342 bar

h, = enthalpy of vapour = 2595 Wkg (from steam tables) Assuming that only latent heat of steam is used for heating, an energy balance of the evaporator is made as : mf x ( c ~x 37 ) +~m,~x hl, ~ =~mvx hv+ mpx ( ~ ~ )x 52 ~ ~ d ~ ~ t = mass of feed, m, mf = mass of steam

[ZY

15000 x 0.92 x 37 + m, x 2188 = 10000 x 2595 + 5000 x 0.75 x 52

:.

The steam consumption, m, = 11716 kg/h

(Ans.)

I

Heat and Mass Transfer

618

Steam economy is defined as kg of solvent evaporated per kg of steam used :

I

-

Heat Exchangers

125°C ...(iii)

Steam economy = mv = loooo - 0.8535 Am. m, 11716

...from (ii) and (i)

dB = dQ.Z = U.dA.0.Z

= 11716 x 2188

3600

I

Again integrating, we get

LMTD

Fig. 10.67.

.,,from (iii)

in (02/81)= U.A.Z = UA = U A 8, = U A (LMT.03

I

I

(Am.)

--'-

(b) Given : d = 25 mm = 0.025 m; L = 35 m; tc = 25OC; thl = 85°C; th2= 35°C; V = 0.6 m/s; cph= 2.51 kJ/kg K; specific gravity of oil = 0.8

Heat transfer area = Q - 7121 x 1000 = 73.93 d (Ans.) U8, 1200 x 80.27 Example 10.56. (a) Derive an expression for LMTD of an evaporator. (b) The tube of an oil cooler is submerged in a large pool of stagnant water at temperature of 25°C. The inside diameter of the tube is 25 mm and its length is 35 m. Estimate the overall heat transfer coeficient of this system if the temperature of oil drops from 85°C to 35°C and the average velocity of oil is 0.6 m/s. Assume for oil specific heat = 2.51 HAE K and th, spec@ gravity = 0.8. (AMIE Summer, 1995) Solution. (a)Derivation of expression for LMTD of an evaporator : Let the temperature in the evapoth2 rator be t,, the hot fluid (heat capacity 3I rate mh cph)enters the evaporator at thl tc 4 and leaves at th2. iI The temperature distribution along W-dA I the heat exchanger area is as shown in Area, A --+ Fig. 10.68. Fig. 10.68. Temperature distribution of an evaporator. For an elementary area, dA

Overall heat transfer co-efficient, U :

tc (= 25°C)

- - - -_ --_ --- ------ --_ -- -_ --- -Water -I

25 mm 4

_ - - - _ -_ - - - - - _

0

Fig. 10.69.

(where A!=

m = pAf V

area of flow)

&

where Z = -

1 = a constant. mh Cph

8 = t , - t, d8 = dt, = dQ.2 On integration, we get 2

2

...(ii)

9

(..a

= 385.53

w/m2'c

(Ans.)

Surface area, A, = d L )