Heat Exchanger Lecture

Heat Transfer/Heat Exchanger • • • • • • • • •

How is the heat transfer? Mechanism of Convection Applications . Mean fluid Velocity and Boundary and their effect on the rate of heat transfer. Fundamental equation of heat transfer Logarithmic-mean temperature difference. Heat transfer Coefficients. Heat flux and Nusselt correlation Simulation program for Heat Exchanger

How is the heat transfer? • Heat can transfer between the surface of a solid conductor and the surrounding medium whenever temperature gradient exists. Conduction Convection Natural convection Forced Convection

Natural and forced Convection

Natural convection occurs whenever heat flows between a solid and fluid, or between fluid layers. As a result of heat exchange Change in density of effective fluid layers taken place, which causes upward flow of heated fluid. If this motion is associated with heat transfer mechanism only, then it is called Natural Convection

Forced Convection  If this motion is associated by mechanical means such as pumps, gravity or fans, the movement of the fluid is enforced.  And in this case, we then speak of Forced convection.

Heat Exchangers • A device whose primary purpose is the transfer of energy between two fluids is named a Heat Exchanger.

Applications of Heat Exchangers Heat Exchangers prevent car engine overheating and increase efficiency Heat exchangers are used in Industry for heat transfer

Heat exchangers are used in AC and furnaces

• The closed-type exchanger is the most popular one. • One example of this type is the Double pipe exchanger.

• In this type, the hot and cold fluid streams do not come into direct contact with each other. They are separated by a tube wall or flat plate.

Principle of Heat Exchanger •

First Law of Thermodynamic: “Energy is conserved.” 0

0 0 0 dE     & & &+ w & & = ∑m hin − ∑ m hout  + q s + e generated dt out  in 

Qh = Am&h C ph ∆Th Qc = Am&c C pc ∆Tc

•Control Volume

∑ m&.hö = −∑ m&.hö in

out

COLD Qh HOT

Cross Section Area

Thermal Boundary Layer

THERMAL

Region III: Solid – Cold Liquid Convection

BOUNDARY LAYER

Energy moves from hot fluid to a surface by convection, through the wall by conduction, and then by convection from the surface to the cold fluid.

NEWTON’S LAW OF CCOLING

dqx = hc .(Tow - Tc ).dA Th

Ti,wall To,wall Tc

Region I : Hot LiquidSolid Convection

Q hot

Q cold

NEWTON’S LAW OF CCOLING

dqx = hh .(Th - Tiw ).dA

Region II : Conduction Across Copper Wall FOURIER’S LAW

dT dqx = - k. dr

• Velocity distribution and boundary layer When fluid flow through a circular tube of uniform crosssuction and fully developed, The velocity distribution depend on the type of the flow. In laminar flow the volumetric flowrate is a function of the radius.

V = volumetric flowrate u = average mean velocity

 In turbulent flow, there is no such distribution. • The molecule of the flowing fluid which adjacent to the surface have zero velocity because of mass-attractive forces. Other fluid particles in the vicinity of this layer, when attempting to slid over it, are slow down by viscous forces. Boundary  layer

r

• Accordingly the temperature gradient is larger at the wall and through the viscous sub-layer, and small in the turbulent core.

qx = hADT qx = hA(Tw - T)

Tube wall heating



Metal wall δ Twh 

cooling Tc

Twc

h

k qx = A(Tw - T) d

• The reason for this is 1) Heat must transfer through the boundary layer by conduction. 2) Most of the fluid have a low thermal conductivity (k) 3) While in the turbulent core there are a rapid moving eddies, which they are equalizing the temperature.

U = The Overall Heat Transfer Coefficient [W/m.K] Region I : Hot Liquid – Solid Convection Region II : Conduction Across Copper Wall Region III : Solid – Cold Liquid Convection

Th - Tc =

qx = hhot .(Th - Tiw ).A qx =

qx = hc (To,wall - Tc ) Ao

qx R1 + R2 + R3

qx = U.A.(Th - Tc )

U=

kcopper .2pL r ln o ri

1 A.SR

Th - Tiw =

To,wall - Ti,wall

qx hh .Ai

æro ö qx .lnç ÷ è ri ø = kcopper .2pL

qx To,wall - Tc = hc .Ao

é ù æro ö lnç ÷ ê ú r 1 1 è iø ú Th - Tc = qx ê + + êhh .Ai k copper .2pL hc .Ao ú ê ú ë û

      . ln      1   +   =  +  . .      

−1

r

r

i

o

+

Calculating U using Log Mean Temperature  dqh dqc   d (∆T ) = −  m .C h m .C c  c p   h p

Hot Stream : Cold Stream:

 1 1   d (∆T ) = −U .∆T .dA. +  m .C h m .C c  c p   h p

∫

∆T2

∆T1

∫

∆T2

∆T1

 ∆Th ∆Tc  A2 d ( ∆T ) .∫ dA = −U . + ∆T qc  A1  qh

 1 d (∆T ) 1 = −U . +  m .C h m .C c ∆T c p  h p

[(

 A2 . dA  ∫A1 

)(

æDT ö U . A. U . A in out in out ( ) lnçç 2 ÷ = D T + D T = T T T T h c h h c c ÷ D T q q è 1ø q = U .A Log Mean Temperature

)]

∆T2 − ∆T1  ∆T2  ln   ∆T   1  

Log Mean Temperature evaluation ú h .(T - T ) m  ú c .(T - T ) ú h .C  ú  m  . C  ∆T2 − ∆T1 p 3 6 ∆TLn = U= = c p 7 10  ∆T  A.DTLn A.DTLn ln 2   ∆T1  1 CON CURRENT FLOW 2

1

COUNTER CURRENT FLOW 2

T3

T4

T6

T1

∆ T1 ∆ T2

T6

Wall

T7

T2

T8

T9

T10

∆ A

A

T1

T4

A

T10 T5

T2

T10 T1

T6

T3

T4

T2

T5

T3

T7

T9

T8 Parallel Flow

∆T1 = T − T in h

T6

in c

= T3 − T7

∆T2 = Thout − Tcout = T6 − T10

T8

T7

T9

Counter ­ Current Flow

∆T1 = T − Tcout = T3 − T7 in h

∆T2 = Thout − Tcin = T6 − T10

q = hh  Ai  DTlm (T - T ) - (T6 - T2 ) ∆Tlm = 3 1 (T - T ) ln 3 1 (T6 - T2 )

1

2 T3

T4

T1

T7

T6

Wall

T6 T2

T8 T9

T10

q = hc  Ao  DTlm (T1 - T7 ) - (T2 - T10 ) ∆Tlm = (T1 - T7 ) ln (T2 - T10 )

A

DIMENSIONLESS ANALYSIS TO CHARACTERIZE A HEAT EXCHANGER

Nu = f (Re,Pr, L / D,mb /mo )

•Further Simplification:

Nu = a.Re b .Pr c

Can Be Obtained from 2 set of experiments One set, run for constant Pr And second set, run for constant Re

h

•Empirical Correlation •For laminar flow Nu = 1.62 (Re*Pr*L/D) •For turbulent flow

 = 0.026. Re . Pr 0.8

1/ 3

 µ  .   µ 

•Good To Predict within 20% •Conditions: L/D > 10 0.6 < Pr < 16,700 Re > 20,000

0.14

Experimental Apparatus Switch for concurrent and countercurrent flow

Temperature Indicator

Hot Flow Rotameters

Cold Flow rotameter

Heat Temperature Controller Controller

• Two copper concentric pipes •Inner pipe (ID = 7.9 mm, OD = 9.5 mm, L = 1.05 m) •Outer pipe (ID = 11.1 mm, OD = 12.7 mm)

•Thermocouples placed at 10 locations along exchanger, T1 through T10

Theoretical trend

Examples of Exp. Results

y = 0.8002x – 3.0841

Theoretical trend y = 0.026x

6

Experimental trend

5.5 5

y = 0.0175x – 4.049

4.5 4 3.5

ln (Nu)

Experimental trend

3

2.5 2 9.8

10

10.2

10.4

y = 0.7966x – 3.5415

250

10.6

200

10.8

11

ln (Re)

150

Nus 100

Theoretical trend

50

y = 0.3317x + 4.2533

0 150

4.8

2150

4150

6150

8150

10150

12150

Pr^X Re^Y

4.6 4.4

Experimental Nu = 0.0175Re0.7966Pr0.4622

ln4.2(Nu)

Theoretical

4 0.6

0.8

1

ln (Pr)

1.2

1.4

Experimental trend y = 0.4622x – 3.8097

Nu = 0.026Re0.8Pr0.33

Effect of core tube velocity on the local and over all Heat Transfer coefficients 35000 -

K 30000

-2

25000 hi (W/m2K) ho (W/m2K) U (W/m2K)

20000 15000 10000 5000 Heat Transfer Coefficient Wm 0 0

1

2

3

4 -1

Velocity in the core tube (ms )