Islamic Azad University Karaj Branch

Dr. M. Khosravy

Chapter 3 Conduction with Generation

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Implications

•  Involves a local (volumetric) source of thermal energy due to conversion from another form of energy in a conducting medium.

•  The source may be uniformly distributed, as in the conversion from electrical to thermal energy (Ohmic heating):

q•

=Eg

•

!=I 2Re!

(3.38)

or it may be non-uniformly distributed, as in the absorption of radiation passing through a semi-transparent medium.

q•

! exp "! x( )

•  Generation affects the temperature distribution in the medium and causes the heat rate to vary with location, thereby precluding inclusion of the medium in a thermal circuit.

For a plane wall,

Implications of Energy Generation

Dr. M. Khosravy

The Plane Wall Consider one-dimensional, steady-state conduction in a plane wall of

constant k, with uniform generation, and asymmetric surface conditions: •  Heat diffusion equation (eq. 2.3) :

02

2=+

kq

dxTd !

•  Boundary Conditions:

2,1, )(,)( ss TLTTLT ==!

•  General Solution:

212

2CxCx

kq

T ++!=!

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Dr. M. Khosravy

Temperature Profile

22)(

12

)( 2,1,1,2,2

22ssss TT

LxTT

Lx

kLq

xT+

+!

+""#

$%%&

'!=

!

!  Profile is parabolic. !  Heat flux not independent of x

?  What happens when:

(3.3)

?0 increases, ,0 <= qqq !!!

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Dr. M. Khosravy

Symmetrical Distribution •  When both surfaces are maintained

at a common temperature, Ts,1= Ts,2 = Ts

sTLx

kLq

xT +!!"

#$$%

&'= 2

221

2)(!

?  What is the location of the maximum temperature?

2

max

max)(!"#$

%&=

''(

Lx

TTTxT

s

(3.4a)

(3.4b)

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Dr. M. Khosravy

Symmetrical Distribution

"  Note that at the plane of symmetry:

0q" 0 00

=!="#$

%&'

==

xxdx

dT

!  Equivalent to adiabatic surface

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Dr. M. Khosravy

True or False?

•  In one-dimensional problems involving generation the heat rate remains constant in the direction of heat flow

•  We can apply the concept of a conduction resistance when generation is present

•  The following two are represented by the same equations: –  A plane wall of thickness 2L experiencing uniform volumetric

heating and convection conditions at both surfaces –  A plane wall of thickness L, experiencing the same volumetric

heating and convection conditions at one surface, but having the opposite surface well insulated.

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Dr. M. Khosravy

Example The steady-state temperature distribution in a composite plane wall of three different materials, each of constant thermal conductivity, is shown in the schematic below.

a)  Does heat generation occur in any of sections A,B, or C? b)  Based on the schematic, what is the boundary condition at location (4)? c)  Comment on the relative magnitudes of q2” and q3”. d)  Comment on the relative magnitudes of kA and kB.

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Dr. M. Khosravy

Calculation of surface temperature Ts

In equations (3.4a) and (3.4b) the surface temperature, Ts is needed. "  Boundary condition at the wall:

)( !=

"=" TThdxdTk s

Lx

Substituting (dT/dx)x=L from equation (3.4a):

hLqTTs!

+= !(3.5)

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Dr. M. Khosravy

Example (Problem 3.72 textbook) A plane wall of thickness 0.1 m and thermal conductivity 25 W/m.K having uniform volumetric heat generation of 0.3 MW/m3 is insulated on one side, while the other side is exposed to a fluid at 92°C. The convection heat transfer coefficient between the wall and the fluid is 500 W/m2.K. Determine the maximum temperature in the wall.

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Dr. M. Khosravy

Example: Conduction and Metabolic Energy Production

To maintain the body temperature at a constant value, metabolic heat production must balance heat removal. Let’s investigate what happens when conduction is the only mechanism for heat removal from the body.

" Consider a three layer model, consisting of the body core, with thickness L1=0.22 m and k1=0.42 W/m.K, surrounded by a fat and skin layer of thickness L2=0.08 m and effective conductivity k2=0.235 W/m.K. Metabolic heat production within the body equals 880 W/m3 and heat is removed by convection is air. Determine the mean steady-state temperature in the core region.

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Dr. M. Khosravy

Radial Systems Cylindrical (Tube) Wall Spherical Wall (Shell)

Solid Cylinder (Circular Rod) Solid Sphere

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•  Heat Equations: Cylindrical

1rddr

kr dTdr

!"#

$%&+ q

•

= 0

Spherical 1r 2ddr

kr 2 dTdr

!"#

$%&+ q

•

= 0

Dr. M. Khosravy

Radial Systems •  Heat diffusion equation

in the r-direction for steady-state conditions: 01 =+!

"#$

%&

kq

drdTr

drd

r!

•  Boundary Conditions: sor

TrTdrdT ==

=)( ,0

0•  Temperature profile:

so

o Trr

krq

rT +!!"

#$$%

&'= 2

221

4)(!

L

hT ,!

))(2()( 2!"#=# TTLrhLrq soo!

hrq

TT os 2

!+= !

•  Calculation of surface temperature:

and

(3.6)

(3.7)

•  General Solution: 212 ln

4)( CrCr

kqrT ++!=!

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Temperature Distribution

32

13dT q rkr Cdr

= ! +•

21

26CqrT C

k r= ! ! +

•

0 10 0rdT Cdr = = ! =|

( )2

2 6o

o s s

q rT r T C T

k= ! = +

•

( )2 2

21

6o

s

o

q r rT r T

k r! "

= # +$ %& '

•

Surface Temperature

Overall energy balance:

Or from a surface energy balance:

0 in outE E! =• • ( )cond o convq r q! =

3o

s

q rT T

h!" = +•

•  Solution for Uniform Generation in a Solid Sphere of Constant k with Convection Cooling:

•  A summary of temperature distributions is provided in Appendix C for plane, cylindrical and spherical walls, as well as for solid cylinders and spheres. Note how boundary conditions are specified and how they are used to obtain surface temperatures.

0out gE E! + =• •

3o

s

q rT T

h!" = +•

Dr. M. Khosravy

Example (Problem 3.84 textbook) A cylindrical shell of inner and outer radii ri and ro, respectively, is filled with a heat-generating material that provides a uniform volumetric generation rate. The inner surface is insulated, while the outer surface of the shell is exposed to a fluid with a convection coefficient h.

a)  Obtain an expression for the steady-state temperature distribution T(r) in the shell.

b)  Determine an expression for the heat rate q’(ro) at the outer radius of the shell in terms of the heat generation rate and the shell dimensions

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Problem: Nuclear Fuel Rod

Problem 3.91 Thermal conditions in a gas-cooled nuclear reactor with a tubular thorium fuel rod and a concentric graphite sheath: (a) Assessment of thermal integrity for a generation rate of . (b) Evaluation of temperature distributions in the thorium and graphite for generation rates in the range .

8 310 W/mq =!

8 810 5x10q! !•

Assumptions: (1) Steady-state conditions, (2) One-dimensional conduction, (3) Constant properties, (4) Negligible contact resistance, (5) Negligible radiation, (6) Adiabatic surface at r1.

Properties: Table A.1, Thorium: 2000K; Table A.2, Graphite: 2300K.mp mpT T! !

Schematic:

Analysis: (a) The outer surface temperature of the fuel, T2 , may be determined from the rate equation

!q =T2 "T#

!Rtotwhere !Rtot =

1n r3 / r2( )2!kg

+ 12!r3h

= 0.0185 m "K/W

The heat rate may be determined by applying an energy balance to a control surface about the fuel element,

Since the interior surface of the element is essentially adiabatic, it follows that

!q = q•

! r22 ! r1

2( ) =17,907 W/m

Hence, T2 = !q !Rtot +T! =17,907 W/m 0.0185 miK/W( )+ 600K = 931K

With zero heat flux at the inner surface of the fuel element, Eq. C.14 yields

T1 = T2 +q•

r22

4kt1!r1

2

r22

"

#$$

%

&''!q•

r12

2kt1n

r2

r1

"

#$

%

&' = 931K + 25K !18K = 938K <

E•

out = E•

gor, per unit length,

!E•

out = !E•

g

Problem: Nuclear Fuel Rod (cont)

Since T1 and T2 are well below the melting points of thorium and graphite, the prescribed operating condition is acceptable.

(b) The solution for the temperature distribution in a cylindrical wall with generation is

Tt r( ) = T2 + q•

r22

4kt1! r

2

r22

"

#$%

&'

!q•r22

4kt1!r12

r22

"

#$

%

&' + T2 !T1( )

(

)

***

+

,

---

1n r2 /r( )1n r2 /r1( )

(C.2)

Boundary conditions at r1 and r2 are used to determine T1 and T 2 .

r = r1 : !!q1 = 0 =qr1•

2"

k q•

r22

4kt1"r12

r22

#$%

&'(+ T2 "T1( )

)

*

++

,

-

.

.

r11n r2 / r1( )

(C.14)

r = r2 : U 2 T2 !T"( ) = q•

r22

!

k qr22

•

4kt1!r12

r22

#$%

&'(+ T2 !T1( )

)

*

++

,

-

.

.

r21n r2 / r1( )

(C.17)

U 2 = !A2 !Rtot( )"1 = 2!r2 !Rtot( )"1 (3.32)

Problem: Nuclear Fuel Rod (cont.)

The following results are obtained for temperature distributions in the graphite.

Operation at is clearly unacceptable since the melting point of thorium would be exceeded. To prevent softening of the material, which would occur below the melting point, the reactor should not be operated much above . The small radial temperature gradients are attributable to the large value of .

8 35x10 W/mq =•

tk

8 33x10 W/mq =•

0.008 0.009 0.01 0.011

Radial location in fuel, r(m)

500

900

1300

1700

2100

2500

Tem

pera

ture

, T(K

)

qdot = 5E8qdot = 3E8qdot = 1E8

Problem: Nuclear Fuel Rod (cont.)

the temperature distribution in the graphite is

Tg r( ) = T2 !T31n r2 / r3( )1n

rr3

"

#$%

&'+T3 (3.26)

0.011 0.012 0.013 0.014

Radial location in graphite, r(m)

500

900

1300

1700

2100

2500

Tem

pera

ture

, T(K

)

qdot = 5E8qdot = 3E8qdot = 1E8

Using the value of T2 from the foregoing solution and computing T3 from the surface condition,

!q =2!kg T2 "T3( )1n r3 / r2( ) (3.27)

Problem: Nuclear Fuel Rod (cont.)

Comments: (i) What effect would a contact resistance at the thorium/graphite interface have on temperatures in the fuel element and on the maximum allowable value of ? q

•

Operation at is problematic for the graphite. Larger temperature gradients are due to the small value of .

8 35x10 W/mq =•

gk

What would be the influence of such effect on temperatures in the fuel element and the maximum allowable value of ?

q•

(ii) Referring to the schematic, where might radiation effects be significant?

Problem: Nuclear Fuel Rod (cont.)

Dr. M. Khosravy

Summary

•  We obtained temperature distributions for problems involving steady-state, one-dimensional conduction with energy generation

•  The heat flux and heat rate are no longer independent of x –  The concept of a thermal resistance cannot be applied

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