7/28/2019 2012s CHE314 L10 HeatGeneration

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CHE 314 Heat Generation J.P. Mmbaga

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General Equation

x y z p

T T T T k k k q C

x x x y z z t

So far we have considered cases where the internal heat generation per unit volume ( q)

was was zero.

We now consider cases where the heat generation is non-zero. This is very common type

of problem in problems such as:

1) joule/resistance heating2) exothermic/endothermic chemical reactions3) Nuclear reactions, etc.

Consider steady state (1D), constant k with uniform energy generation

02

2

k

q

x

T

solving the integral by integrating it twice we obtain the following general equation

212

2CxC

x

k

qxT

GENERAL Solution

Boundary conditions

At x =-L T=Ts1

21

2

1 2 CLC

L

k

q

Ts

and similarly

21

2

22

CLCL

k

qTs

Adding the two equations together results in:

22

21 2 CLk

qTT ss

therefore

22

2

212 L

kqTTC ss

andL

TTC ss2

121

22

12

2112

2

22ssss TT

L

xTT

L

x

k

LqxT

7/28/2019 2012s CHE314 L10 HeatGeneration

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CHE 314 Heat Generation J.P. Mmbaga

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If the surface temperature on both faces is identical then we have the following equation:

sTL

x

k

LqxT

2

22

12

At x = L or

L then T(x) = Ts

The resulting distribution looks like this

The important part about this distribution is that at x = 0 the slope is 0

02

01

2

2

2

22

ss T

k

Lq

xT

Lk

Lq

xxT

x

In other words the flux

00

"

xdxdTkq

Infinite Solid Cylinder with heat generation:

t

TCq

r

Trk

rr

1

01

k

q

r

Tr

rr

The general solution (integrating twice)

212 ln

4CrCr

k

qrT

Assume Fixed surface temperature and solid cylinder

7/28/2019 2012s CHE314 L10 HeatGeneration

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CHE 314 Heat Generation J.P. Mmbaga

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Simple Case

00

rr

T

r

Crk

qCrCrk

q

dr

drTdr

d 1

2ln

421

2

=0 therefore C1 =0

TsrT )( 0 thereforek

rqTC os

4

2

2

Therefore the solution is

so

o Tr

r

k

rqrT

2

22

14

The boundary conditions can be expressed in terms of T

TTh

dr

dTk s

rr o

The general derivative is

rk

qT

r

r

k

rq

drrT

drs

o

o

21

4 2

22

Evaluating r = ro

ork

q

dr

dT

2

Therefore

TThrk

qk so

2

os r

h

qTT

2

The flux at any point can be found

rq

rk

qk

dx

dTkq

22

"

Alternately an energy balance can be conducted

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CHE 314 Heat Generation J.P. Mmbaga

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The solid sphere

t

TCq

Tk

r

Tk

rr

Trk

rr

sin

sin

1

sin

112222

2

2

Steady state, 1D (i.e. in radial direction only) and constant k

01 22

k

q

r

Tr

rr

Intergrating results in

212 ln

4CrCr

k

qrT

Solving for the following boundary conditions

00

rrT and TsrT )( 0

Results in

so

o Tr

r

k

rqrT

2

22

16

The general derivative is

rk

qTrr

krq

drrT

drs

o

o

31

6 2

22

For a convective boundary

TTh

dr

dTk s

rr o

os rh

qTT

3

An energy balance of generation = convective heat loss will give the same result.

A summary of these results may be found in Appendix C of Incropera/Bergman