Steady State Heat Transfer In
Composite Hollow Cylinders – The Nuclear Fuel Rod
(A Term Project for MEAE 6630 – Conduction Heat Transfer)
Submitted by: Thomas D. Hammel
Date: April 14, 2000
Page 2 of 14
1. Introduction
This project presents the analytical solution to the problem of steady-state heat transfer for
a two-layer composite hollow cylinder with perfect thermal contact and heat generation,
with an applied heat rate at the inner surface and a convection boundary condition at the
outer surface. The problem is similar to the in-class exercise we completed for a composite
solid cylinder with heat generation and a specified temperature at the outer surface.
Following the presentation of an analytical solution to this problem, an example problem
simulating a nuclear fuel rod at steady-state conditions is solved using the analytical
solution derived in Section 2.
Finally, an ANSYS finite element analysis of the same example problem is presented for
comparison purposes. The complete ANSYS input file listing is included in Appendix A.
Page 3 of 14
2. Analytical Solution
Consider a two-layer composite cylinder as illustrated in Figure 1.
a
k2, g2
b
r
o
c
k1, g1
convection
heat flux
f1
T
h
Figure 1 Two-layer hollow cylinder with perfect thermal contact
The system contains an inner cylinder a rb and an outer cylinder b rc that are in
perfect thermal contact. k1 and k2 are the thermal conductivities, and g1 and g2 are the heat
generation rates of the inner and outer cylinders, respectively. Heat is dissipated by
convection from the outside surface at r = c into an environment at T = TThe inside
surface at r = a is subjected to heating at the rate of f1.
The mathematical formulation of the problem is given by
1 d  d T i  gi
0
r

r dr  dr  k i
in
r i  r  r i  1, i  1, 2
(1)
Page 4 of 14
subject to the boundary conditions
d T1
 f1
dr
at
ra
(2)
T1  T 2
at
r b
(3)
d T1
dT2
 k2
dr
dr
at
rb
(4)
d T1
 h  T 2  T   0
dr
at
rc
(5)
 k1
k1
k2
The solution may be found by simply integrating the two equations defined by Equation (1) and
using the four boundary condition relationships in Equations (2) – (5) to evaluate the four
constants resulting from the integrations of Equation (1). The problem is similar to the in-class
example we solved during the semester, except for the non-zero inner radius and the differing
boundary conditions.
For cylinder and material 1 (i = 1), integrating Equation 1 results in the following expression
r
d T 1 g1 2

r  A0
dr
2 k1
(6)
d T 1 g1
A

r 0
dr
r
2 k1
(7)
or
Separating variable, we obtain
d T1  
g1
2 k1
r dr  A
dr
0
r
(8)
Page 5 of 14
Integrating once more, we get the solution for T1 as
T1  
g1 2
r  A ln r   B
4 k1
(9)
Similarly, for cylinder and material 2, the solution for T2 is
T2  
g2 2
r  C ln r   D
4k2
(10)
Equations (9) and (10) constitute a solution of the general problem with four arbitrary constants.
The solution is made particular to the case considered here by evaluating the four constants, A,
B, C, and D, using the four boundary condition relationships in Equations (2) – (5).
We note here that the following are expressions for the first partial derivatives (which will be
needed to evaluate the constants)
g
T1
A
  1r 
r
r
2 k1
(11)
g
T 2
C
  2 r 
r
r
2k2
(12)
From Equation (2), we have
 k1
g a k
 g
d T1
A
 f 1   k1   1 a    1  1 A
dr
a
2
a
 2 k1
(13)
Rearranging, we obtain an expression for A, as
A 

a  a g1

 f 1
k1  2

From Equation (4), we obtain the following
(14)
Page 6 of 14
 g1
 g
A
C
a    k2   2 b  
k1  
a
b
 2 k1
 2k2
(15)
Knowing A, we readily obtain an expression for C, as
C 
g1
2k2
a 2  b 2  2gk2 b 2  ak f 1
2
(16)
2
Equation (5) results in the expression
 g2
 g

C
c    h    2 c 2  C ln c   D  h  T 
k2  
c
 2k2
 4k2

(17)
which leads to an expression for D, as
D 
2
a a

c g2  c

2 

    g1  f 1  b g 2  g1  k 2  ln c   T 

4  k 2 h 
 2k2
 k 2  2

  h  c
(18)
Finally, Equation (3) gives us the expression

g1 2
g2 2
b  A ln b   B  
c  C ln b  D
4 k1
4k2
(19)
which, after substitutions for A, C, and D are made, and much rearranging, we obtain the
following expression
2 g
B  b  1 
4  k1
2
 a

g2 
a  a

  
   g1  f 1  b g 2  g1  ln b 
 2k2
k2 
 k 2 k 1   2

2
 a a
 k
 c g2  c
2 


  T 
   g1  f 1  b g 2  g1  2  ln c  

2
2
c
4


k
k
h
k
h
2


 2
 2
  
(20)
Page 7 of 14
Using these expressions for A, B, C, and D, final expressions for T1(r) and T2(r) are found to be
as follows:
T 1 r   
g1 2 a  a g1

b 2  g1 g 2 
 f 1 ln r  

r  
4  k 1 k 2 
4 k1
k1  2

 a

a  a
 b2

 
   g1  f 1 
g 2  g1  ln b 
 2k2
 k 2 k 1   2

(21)
 a a
 k
 c g2  c
2 
 b2

  T 0

   g 1  f 1 

g 2  g1  2  ln c  
4  k 2 h  
 2k2

 k 2  2
  h  c
for a  r  b
and
T 2 r   


a f 1
c g2  c
g2 2
g2 2
 g1 2
2 



a  b2 
b 
r  
 ln r  
2
2
4
4k2
k
k
k
k
h
2
2 

 2
 2
 a a
 k

 b2

   g 1  f 1 
g 2  g1  2  ln c   T 
 2k2
 k 2  2
  h  c

for b  r  c
(22)
Page 8 of 14
3. Example – Nuclear Fuel Rod at Steady-State Conditions
Nuclear fuel rods are typically manufactured from Zircaloy. These rods typically have a layer of
oxidation on the outer surface of the rod. At full power the fuel pellets within the rods subject the
inside surface of the rods to a constant heat rate, and there tends to be some small amount of heat
generation within the Zircaloy clad. Heat is dissipated by convection from the outside surface of the
oxidation into a forced water flow environment at a uniform temperature.
These conditions are represented by the analytical problem presented and solved in Section 2. To
illustrate the application of the analytical solution, typical input parameters for a nuclear fuel rod are
presented here, and are used in the following two sections to obtain analytical and finite element
analysis results, respectively, for the same problem.
Material Properties (Thermal conductivities assumed constant, at T = 600 F, for this analysis)
Thermal conductivity of clad, k1 = 10.00 Btu/(hr-ft-F)
Thermal conductivity of oxide layer, k2 = 0.02 W/(cm-K)
Geometric Properties
Clad inner radius, a = 0.200 inches
Clad thickness, b - a = 0.030 inches
Oxide thickness, c - b = 0.005 inches
Loading
Heat rate, f1 = 6.00 Kw/ft
Heat generation in clad, g1 = 2% of applied heat rate
Heat transfer coefficient, h = 6000 Btu/(hr-ft2-F)
Bulk coolant temperature, T = 600.0 F
Page 9 of 14
3.1 Analytical Solution to Example
Using the input parameters from Section 3 and Equations (21) and (22) for T1 and T2, respectively, the
steady-state through-wall temperature distribution in the example fuel rod is determined. The results
are calculated at radii chosen to coincide with the nodal spacing developed in the finite element
analysis of the following section.
The results of the analytical solution are presented in Table 1 and Figure 3,
3.2 Finite Element Analysis Solution to Example
An ANSYS finite element analysis was completed using the parameter input of Section 3. A 2-D
axisymmetric model of arbitrary axial length was developed using PLANE55 2-D thermal solid
elements. A radial grid with eight uniformly spaced elements through the clad and two uniformly
spaced elements through the oxidation was used. (See complete ANSYS input listing in Appendix A
for all modeling details.)
A contour plot of the through-wall temperature distribution from the finite element analysis is included
in Figure 2. A tabulation of the same temperature distribution is included in Table 1, along with the
analytical solution developed in Section 3.1.
3.3 Comparison of Analytical and Finite Element Analysis Solutions
The analytical solution and the finite element solution are in excellent agreement. All temperatures
agree within 0.02%.
Page 10 of 14
1
MX
MN
ANSYS 5.5.2
APR 12 2000
21:57:25
PLOT NO. 1
NODAL SOLUTION
STEP=1
SUB =1
TIME=1
TEMP
(AVG)
RSYS=0
PowerGraphics
EFACET=1
AVRES=Mat
SMN =628.351
SMX =736.326
628.351
640.348
652.345
664.342
676.34
688.337
700.334
712.331
724.329
736.326
Figure 2 ANSYS Through-Wall Temperature Distribution
for Example Problem
Page 11 of 14
Table 1 Input and Output for Example Fuel Rod Problem
Geometric Properties:
a = inner radius =
t1 = clad thickness =
b = interface radius =
t2 = oxidation thickness =
c = outer radius =
Material Properties:
cylinder
k
Btu/(hr-ft-F)
1
10.0
2
1.16
0.2000
0.0300
0.2300
0.0050
0.2350
k
Btu/(sec-in-F)
2.315E-04
2.675E-05
Boundary Conditions:
inside:
heat flux =
outside:
in.
in.
in.
in.
in.
g
Btu/(sec-in3)
2.514E-01
0
2
0.3771 Btu/(sec-in
6 Kw/in
h
6000
T
600.0 F
2
Btu/(hr-ft -F)
Analytical Solution
r
(in)
0.2
0.20375
0.2075
0.21125
0.215
0.21875
0.2225
0.22625
0.23
0.2325
0.235
T1
(°F)
736.20
730.14
724.18
718.30
712.52
706.82
701.20
695.66
690.20
-----
T2
(°F)
----------------690.20
659.06
628.26
0.0116
2
ANSYS Solution
T
(°F)
736.20
730.14
724.18
718.30
712.52
706.82
701.20
695.66
690.20
659.06
628.26
X
(in.)
0.2
0.20375
0.2075
0.21125
0.215
0.21875
0.2225
0.22625
0.23
0.2325
0.235
)
Btu/(sec-in -F)
TEMP
(°F)
736.33
730.27
724.30
718.43
712.64
706.95
701.33
695.79
690.33
659.18
628.35
Page 12 of 14
Temperature in Composite Cylinder
Analytical
ANSYS
750
Temperature (°F)
725
700
675
650
625
600
0.200
0.210
0.220
0.230
0.240
Radius (in.)
Figure 3 Comparison of Analytical and Finite Element Analysis
Solutions to Example Problem
Page 13 of 14
APPENDIX A - ANSYS Input File Listing
! file = example.inp
! created by T. D. Hammel (April 2000)
/title,Nuclear Fuel Rod Example Problem
et,1,plane55,,,1
! use 2-D thermal solid, axisymmetric
! material properties
! clad
kcld=10.00/(3600*12) ! Btu/(sec-in-°F)
mp,kxx,1,kcld
! Btu/(sec-in-°F)
! oxide layer
Kox = 0.0200
Kox = Kox*57.79
Kox = Kox/(3600*12)
mp,kxx,2,Kox
! W/(cm-°K)
! Btu/(hr-ft-°F)
! Btu/(sec-in-°F)
! geometric input data
a=0.200
! clad inner radius (in)
t1=0.030
! clad thickness (in)
b=a+t1
! clad outer radius = oxidation inner radius (in)
t2=0.0050
! oxidation thickness (in)
c=b+t2
! oxidation outer thickness (in)
! model creation
n_clad=8
n_ox=2
! no. of elements through clad thickness
! no. of elements through oxidation thickness
k,1,a
k,2,b
k,3,c
kgen,2,1,3,,,t2
a,1,2,5,4
a,2,3,6,5
! area 1 = clad
! area 2 = oxidation
lesize,all,,,1
lesize,1,,,n_clad
lesize,3,,,n_clad
lesize,5,,,n_ox
lesize,7,,,n_ox
allsel
mat,1
amesh,1
mat,2
amesh,2
finish
/solu
antype,static
pi = 3.14159
! determine heat rate
bc_flux = 6.00/12
bc_flux = bc_flux/(pi*2*a)
bc_flux= bc_flux*1000/(0.2931*3600)
! Kw/in
! Kw/in^2 - divide by circumference
! Btu/(in^2-sec)
Page 14 of 14
! determine heat generation rate
bc_gen = 0.02*bc_flux/t1
! Btu/(in^3-sec), divide by clad thickness
! determine convection boundary condition input
bc_conv = 6000/(3600*144)
! Btu/(in^2-sec-°F)
bc_Tbulk = 600.0
! °F
! apply convection on outer surface
nsel,s,loc,x,c-.001,c+.001
sf,all,conv,bc_conv,bc_Tbulk
! select oxidation outer nodes (x=c)
! apply convection to oxidation outer nodes
! apply heat rate at inner surface
nsel,s,loc,x,a-.001,a+.001
sf,all,hflux,bc_flux
! select cladding inner nodes (x=a)
! apply heat flux to clad inner nodes
! apply heat generation in cladding
esel,s,mat,,1
nsle
bf,all,hgen,bc_gen
! select cladding elements
! select nodes in cladding elements
! apply heat generation to clad nodes
allsel
outpr,basic,all
solve