[heat Transfer] Chapter 2

Chapter 2 Heat conduction equation

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Contents Chapter 1 2.1 Introduction

3 page

2.2 One dimensional heat conduction equation 2.3 General heat conduction equation

5 page

2.4 Boundary and initial conditions

9 page

2.5 Solution of Steady One-dimensional Heat Conduction Problems

15 page

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2.1 Introduction Preview 1.

Description of heat conduction equation

2.

Discussion of boundary condition Ex) Steady vs. Transient Heat Transfer Time = 2 pm 15℃

7℃

Time = 2 pm

Time = 5 pm 7℃

15℃

15℃

Q2  Q1

Q1

7℃

 Q 1

(a) Steady

Time = 5 pm 12℃

9℃

Q2  Q1

(b) Transient (Unsteady) ※ steady state: The temperature does not vary with time

One dimensional system: Temperature gradients exist along a single coordinate system, and heat transfer occurs exclusively in that direction Multi Energy Transport (MET) Lab.

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2.1 Introduction Coordinate system z

z

z P( x, y, z )

P( r,  , z ) z

y



x

r

(b) Cylindrical coordinates

Multi Energy Transport (MET) Lab.

r

y

y

 x

x

(a) Rectangular coordinates

P( r,  , )



4

(c) Spherical coordinates

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2.3 General heat conduction equation General heat conduction equation Energy balance:

Ein  Eg  Eout  Est

Qx Qy Qz  Eg Qxdx Qydy Qzdz  Est

Qzdz

Q ydy

Q y

dx

Q z

T Qx  kdydz x T Q y  kdxdz y T Qz  kdxdy z  E g  gdxdydz ,

dz

E g Est

Q x

Rate equations:

Q xdx

dy

Qx   , Qxx  Qx  dx x   Q , Q yy  Q y  y dy y Qz   , Qzz  Qz  dz z  E st = (CdxdydzT ) t Continue

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2.3 General heat conduction equation Substituting and rearranging   T    T    T        k k k g (CT )       x  x  y  y  z  z  t

Or in the vector form  (kT )  g 

 (CT ) t

For constant properties k, ρ, C

 T  T  T g 1 T  2 2   , 2 x y z k  t 2

2



2

where:

k : thermal diffusivity, m2 /sec C

 C : thermal capacitance

Materials of large α(large k and/or low ρC) will respond quickly to changes in their thermal environment, while materials of small α will respond more sluggishly, taking longer to reach a new equilibrium state Multi Energy Transport (MET) Lab.

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2.3 General heat conduction equation General heat conduction equation in other coordinates - Cylindrical coordinate: z r

dz

x  r cos  , y  r sin  , z  z

dr

z

y

x

1   T  kr r r  r

 1   T    T  2 k   k  r     z  z

T     g  C  t 

d



- Spherical coordinate: 

d

r

x



x  r sin  cos  , y  r sin  sin  , z  cos

dr

y d

1   2 T  1   T  1  T  T  kr k k sin g C           r2 r  r  r2 sin2      r2 sin     t

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2.3 General heat conduction equation 3D heat conduction equation   T    T    T  T k  k  k   g   C   x  x  y  y  z  z  t

(1) Steady-state:

T 0 t

  T    T    T  k   k   g  0   k x  x  y  y  z  z  (2) Steady-state, no heat generation:  T  0 t

 0 , g=

  T    T    T  k   k 0   k x  x  y  y  z  z  (3) 1D, transient, constant conductivity: x o n ly ,

 2T T k 2  g   C x t Multi Energy Transport (MET) Lab.

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T  0 , k  co n st. t

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2.4 Boundary and initial conditions Specified temperature boundary condition, (Dirichlet boundary condition) T(x,t)

T1 0

L

T2

x  0 : T (0, t )  T1 x  L : T ( L, t )  T2

x

Specified heat flux boundary condition, (Neumann boundary condition) Energy balance:

Heat flux

x

0

q 0  q cond

L

q0

T q 0   k x

qcond x0

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q 0  0 ( insulated ) T x x0

0 x0

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2.4 Boundary and initial conditions Convection boundary condition Energy balance:

T , h 0

x

qconv  qcond

L

h(T  T )  k

q cond

q conv

T x

x0

x0;T

Generalized boundary condition Energy balance on the considered surface which may involve convection, radiation, and specified heat flux simultaneously.

 Heat transfer   Heat transfer   to the surface    from the surface       in all modes   in all modes      Multi Energy Transport (MET) Lab.

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2.4 Boundary and initial conditions Ex 2.1 A long copper bar of rectangular cross section, whose width w is much greater than its thickness L, is maintained in contact with a heat sink at its lower surface, and the temperature throughout the bar is approximately equal to that of the heat sink, T0. Suddenly, an electric current is passed through the bar and an air stream of temperature T∞ is passed over the top surface, while the bottom surface continues to be maintained at T0. Obtain the differential equation and the boundary and initial conditions that could be solved to determine the temperature as a function of position and time in the bar.

Solution 1. Given

Cooper bar initially in thermal equilibrium with a heat sink is suddenly heated by passage of an electric current.

2. Find

Differential equation and boundary and initial conditions needed to determine temperature as a function of position and time within the bar.

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2.4 Boundary and initial conditions 3. Schematic

C opper bar ( k ,  ) T ( x, y, z, t )  T ( x, t ) A ir T , h

z x

w

A ir T , h

q

y

T (L, t)

L

I 

L

x

T 0  T ( 0 .t )

4. Assumption

1. 2. 3. 4.

Since the bar is long and w × L, side effects are negligible Heat transfer within the bar is one dimensional in the x direction There is an uniform volumetric heat generation All properties are constant

5. Properties Continue Multi Energy Transport (MET) Lab.

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2.4 Boundary and initial conditions 6. Solve The temperature distribution is governed by the heat conduction for the onedimensional and constant property conditions of the present problem  x

  T    T  T  T       k k k g C        t  x  y  y  z  z 

 2T g 1 T ,  2   k  t x

(

T T k   0,   ) y  z C

The temperature of bottom surface is maintained at a value of T0

T (0, t )  T0 The convection heat transfer B,C is appropriate for the top surface

k

T x

 h[T ( L, t )  T ] xL

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2.4 Boundary and initial conditions Before the change in conditions, the bar is initially at a uniform temperature T0

T ( x, 0)  T0 7. Comment - The heat sink at x = 0 could be maintained by exposing the surface to an ice bath or by attaching it to a cold plane - The temperature of the top surface T(L, t) will change with time. This temperature is an unknown and may be obtained after finding T(x, t)

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2.5 Solution of Steady One-dimensional Heat Conduction Problems Ex 2.2 Consider the base plate of a 1200 W household iron that has a thickness of L = 0.5 cm, base area of A = 300 cm2, and thermal conductivity of k = 15 W/m·K. The inner surface of the base plate is subjected to uniform heat flux generated by the resistance heaters inside, and the outer surface loses heat to the surroundings at T∞ = 20℃ by convection. Taking the convection heat transfer coefficient to be h = 80 W/m2·K and disregarding heat loss by radiation. Obtain an expression for the variation of temperature in the base plate, and evaluate the temperatures at the inner and the outer surfaces.

Solution 1. Given

The inner surface of the base plate is subjected to uniform heat flux and the outer surface losses heat by convection q0  1200W, A=300m 2 , L  0.5cm, T  20 o C, h = 80 W/m 2 o C

2. Find

Differential equation and temperatures at the inner and the outer surfaces

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2.5 Solution of Steady One-dimensional Heat Conduction Problems 3. Schematic Resistance heater 1200 W

L

Q g  1200 W Q cond Qconv

Base plate

k = 15 W/m o C

T  20 C

300 cm2

h  80 W/m2 o C

A  300 cm2

h  80 W/m2 o C x T  20 C

k = 15 W/m o C

Insulation

L

4. Assumption

1. 2. 3. 4.

5. Properties

The thermal conductivity of household iron: k = 15 W/m  o C

Steady-state and All properties are constant. Heat transfer is one dimensional in the x-direction. There is no heat generation in the base plate and radiation is negligible. The upper part of the iron is well insulated.

Continue Multi Energy Transport (MET) Lab.

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2.5 Solution of Steady One-dimensional Heat Conduction Problems 6. Solve Inner surface of the base plate is subjected to uniform heat flux as Q 0 1200 W q0    40, 000 W/m 2 2 Abase 0.03 m

Heat conduction equation 0   T    T    T k  k k  x  x  y  y  z  z

  T k x  x

   0, 



0 0 T 0    g   C t  T T   0  1-D y z

g  0  No heat generation T =0  Steady-state t

d2T  2  0 ( since k is const. and x only ) dx

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2.5 Solution of Steady One-dimensional Heat Conduction Problems At x = 0,

q0  40,000  k

dT dx

x 0

x 0

At x = L, k

dT dx

 h(T x L

x L

qcond

q0

qconv

qcond

 T )

xL

The general solution of the differential eq. is

T ( x )  C1 x  C 2 Applying boundary condition dT (0) q  q0   kC1 =q0  C1   0 dx k dT (L) k  h[T (L)  T ]   kC1  h[(C1L  C2 )  T ] dx k

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2.5 Solution of Steady One-dimensional Heat Conduction Problems Substituting and solving for C2

C2  T 

C1k q q  C1 L  T  0  0 L h h k

So, general solution is

 Lx 1 T ( x)  C1 x  C2  T  q0    h  k Temperature is  0.005 m  1  L 1  T (0)  T  q0     20 o C  (40,000 W/m2 )  o 2o  15 W/m C 80 W/m C  k h   533 o C 1 40, 000 W/m 2  o T ( L )  T  q0  0    20 C  h 80 W/m 2 o C   520 o C

7. Comment Multi Energy Transport (MET) Lab.

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2.5 Solution of Steady One-dimensional Heat Conduction Problems Ex 2.3 The temperature distribution across a wall 1-m-thick at a certain instant of time is given as

T  900  300 x  50x2

( o C)

Determine the rate of heat transfer entering the wall (x=0) and the rate of change of energy stored in the wall.

Solution 1. Given

Temperature distribution T(x) at an instant of time t in a one-dimensional wall with uniform heat generation g  1,000 W/m 3 , A  10 m 2 , L  1 m

2. Find

- Heat rate entering, qin (x = 0), the wall - The rate of change of energy storage in the wall, E st

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2.5 Solution of Steady One-dimensional Heat Conduction Problems 3. Schematic

E g E

qin

qout

st

0

L 1 m

x

4. Assumption

1. One-dimensional conduction in the x - direction 2. Isotropic medium with constant properties 3. Uniform internal heat generation, g  W/m3 

5. Properties

- The conductivity of a wall: k = 40 W/m  o C 3 - The density of a wall:   1,600 kg/m - The heat capacity of a wall: C  4 kJ/kg  K Continue

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2.5 Solution of Steady One-dimensional Heat Conduction Problems 6. Solve Since heat is conducted through the solid wall, we can apply the Fourier’s Law of Conduction to this problem,

T Q in  Q cond (0)   kA x

x 0

Qin

 (40)(10)( 300  100 x x 0 )

 120 kW

Qcond x 0

Also the rate of change of energy storage in the wall may be determined by using an energy balance to the wall E in  E g  E out  E st   E out E st  E in  gAL

Ein

 120  1000  10  1  160

E g

Est

Eout

  30 kW

7. Comment Fourier’s law can always be used to compute the conduction heat rate from knowledge of the temperature distribution. Multi Energy Transport (MET) Lab.

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