One-Dimensional, Steady-State
Conduction with
Thermal Energy Generation
Chapter Three
Section 3.5, Appendix C
Implications
Implications of Energy Generation
• 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):
𝐸𝐸̇𝑔𝑔 𝐼𝐼2 𝑅𝑅𝑒𝑒 (3.43)
𝑞𝑞̇ = =
𝑉𝑉 𝑉𝑉
or it may be non-uniformly distributed, as in the absorption of radiation
passing through a semi-transparent medium. For a plane wall,
q ∝ e −α 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.
The Plane Wall
The Plane Wall
• Consider one-dimensional, steady-state conduction
in a plane wall of constant k, uniform generation,
and asymmetric surface conditions:
• Heat Equation:
d dT d 2T q
k +q =0→ 2 + =0 (3.44)
dx dx dx k
Is the heat flux q′′ independent of x?
• General Solution:
T ( x) =
− ( q / 2k ) x 2 + C1 x + C2 (3.45)
What is the form of the temperature distribution for
q = 0? q > 0? q < 0?
How does the temperature distribution change with increasing q ?
Plane Wall (cont.)
Symmetric Surface Conditions or One Surface Insulated:
• What is the temperature gradient
at the centerline or the insulated
surface?
• Why does the magnitude of the temperature
gradient increase with increasing x?
• Temperature Distribution:
q L2 x2
T ( x )= 1 − 2 + Ts
2k
(3.47)
L
• How do we determine Ts ?
Overall energy balance on the wall →
− E out + E g =
0
−hAs (Ts − T∞ ) + q As L =
0
q L
T=
s
T∞ + (3.51)
h
• How do we determine the heat rate at x = L?
Radial Systems
Radial Systems
Cylindrical (Tube) Wall Spherical Wall (Shell)
Solid Cylinder (Circular Rod) Solid Sphere
• Heat Equations:
Cylindrical Spherical
1 d dT 1 d 2 dT
kr +q=0 kr + q =0
r dr dr r 2 dr dr
Radial Systems (cont.)
• Solution for Uniform Generation in a Solid Sphere of Constant k
with Convection Cooling:
Temperature Distribution Surface Temperature
dT q r 3
kr 2
=
− + C1 Overall energy balance:
dr 3 q r
0 → Ts = T∞ + o
− E out + E g =
q r 2 C1 3h
T= − − + C2
6k r
dT Or from a surface energy balance:
|r = 0 =0 → C1 =0
dr → q ( r ) q
= q ro
0
Ein − E out = cond o conv → T = T +
3h
s ∞
q ro 2
T ( ro ) =Ts → C2 =Ts +
6k
q ro 2 r2
T ( r=
) 1 − + Ts
6k ro 2
• 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.
Problem: Nuclear Fuel Rod
Problem 3.82 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 q = 108 W/m3 . (b) Evaluation of
temperature distributions in the thorium and graphite
for generation rates in the range 108 ≤ q ≤ 5 ×108 W/m3 .
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction, (3) Constant
properties, (4) Negligible contact resistance, (5) Negligible radiation.
=
PROPERTIES: Table A.1, Thorium: Tmp 2023
= K; Table A.2, Graphite: Tmp 2273 K.
Problem: Nuclear Fuel Rod (cont.)
ANALYSIS: (a) The outer surface temperature of the fuel, T2 , may be determined from the rate
equation
T2 − T∞
q′ =
′
Rtot
1n ( r3 / r2 ) 1
′=
where Rtot + = 0.0185 m ⋅ K/W
2π k g 2π r3h
The heat rate may be determined by applying an energy balance to a control surface about the fuel
element,
Eout = Eg
or, per unit length,
• •
E ′out = E ′g
Since the interior surface of the thorium is essentially adiabatic, it follows that
•
q=′ q π ( r22 − r12=
) 17,907 W/m
Hence,
′ + T∞= 17,907 W/m ( 0.0185 m ⋅ K/W ) + 600 K= 931 K
T2= q′Rtot
With zero heat flux at the inner surface of the fuel element, Eq. C.14 yields
q r22 r12 q r12 r2
• •
T1 =T2 + 1− − 1n =931 K + 25 K − 18 K =938 K <
4kt r22 2kt r1
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
q r22 r 2
Tt ( r ) = T2 + 1 −
4kt r22
2 2
q
r r 1n ( r / r )
− 2
1 − 2 + (T2 − T1 ) 1n r2 / r
1 (C.2)
4kt r ( 2 1)
2
Boundary conditions at r1 and r2 are used to determine T1 and T2.
q r 2 r 2
k 2
1 − 2 + (T2 − T1 )
1
(C.14)
qr
1 4kt r2
r= r1 : q1′′= 0= −
2 r11n ( r2 / r1 )
qr
22 r12
1 − 2 + (T2 − T1 )
(C.17)
k
4 k t
r2
q r2
r = r2 : U 2 (T2 − T∞ ) = −
2 r21n ( r2 / r1 )
−1 −1
=U2 (= ′ )
A2′ Rtot ( 2π r2 Rtot
′ ) (3.37)
Problem: Nuclear Fuel Rod (cont.)
The following results are obtained for temperature distributions in the thorium.
2500
2100
Tmp = 2023 K
Temperature, T(K)
1700
1300
900
500
0.008 0.009 0.01 0.011
Radial location in fuel, r(m)
qdot = 5E8
qdot = 3E8
qdot = 1E8
Operation at q= 5 × 108 W/m3 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 q = 3 × 108 W/m3.
The small radial temperature gradients are attributable to the large value of kt .
Problem: Nuclear Fuel Rod (cont.)
Using the value of T2 from the foregoing solution and computing T3 from the surface condition,
2π k g (T2 − T3 )
q′ =
1n ( r3 / r2 )
the temperature distribution in the graphite is
T2 − T3 r
=Tg ( r ) 1n + T3 (3.31)
1n ( r2 / r3 ) r3
2500
Tmp = 2273 K
2100
Temperature, T(K)
1700
1300
900
500
0.011 0.012 0.013 0.014
Radial location in graphite, r(m)
qdot = 5E8
qdot = 3E8
qdot = 1E8
Problem: Nuclear Fuel Rod (cont.)
Operation at q = 5 × 108 W/m 3 is problematic for the graphite. Larger temperature gradients
are due to the small value of k g .
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 ? (ii) Referring
to the schematic, where might radiation effects be significant? What would be the influence of such
effect on temperatures in the fuel element and the maximum allowable value of q ?
