Two-Dimensional Conduction:
Finite-Difference Equations
and
Solutions
Chapter 4
Sections 4.4 and 4.5
Finite-Difference Method
The Finite-Difference Method
• An approximate method for determining temperatures at discrete
(nodal) points of the physical system.
• Procedure:
– Represent the physical system by a nodal network.
– Use the energy balance method to obtain a finite-difference
equation for each node of unknown temperature.
– Solve the resulting set of algebraic equations for the unknown
nodal temperatures.
Finite-Difference Approximation
The Nodal Network and Finite-Difference Approximation
• The nodal network identifies discrete
points at which the temperature is
to be determined and uses an
m,n notation to designate their location.
What is represented by the temperature determined at a nodal point,
as for example, Tm,n?
• A finite-difference approximation
is used to represent temperature
gradients in the domain.
How is the accuracy of the solution affected by construction of the nodal
network? What are the trade-offs between selection of a fine or a coarse mesh?
Energy Balance Method
Derivation of the Finite-Difference Equations
- The Energy Balance Method • As a convenience that obviates the need to predetermine the direction of heat
flow, assume all heat flows are into the nodal region of interest, and express all
heat rates accordingly.
Hence, the energy balance becomes:
Ein  E g  0
(4.34)
• Consider application to an interior nodal point (one that exchanges heat by
conduction with four, equidistant nodal points):
4
 q(i ) ( m, n)  q  x  y    0
i 1
where, for example,
q m 1, n   m, n   k  y 

Tm 1, n  Tm, n
x
(4.35)
Is it possible for all heat flows to be into the m,n nodal region?
What feature of the analysis insures a correct form of the energy balance
equation despite the assumption of conditions that are not realizable?
Energy Balance Method (cont.)
• A summary of finite-difference equations for common nodal regions is provided
in Table 4.2. Consider an external corner with convection heat transfer.
q m 1, n  m, n   q m, n 1 m, n   q   m, n   0
 y  Tm 1, n  Tm, n
 x  Tm, n 1  Tm, n
k
 
k
 
x
y
 2 
 2 
 x 
 y 
h 
  T  Tm, n   h 
  T  Tm, n   0
 2 
 2 
or, with x  y,
Tm 1, n  Tm, n 1  2
hx
 hx 
T  2 
 1 Tm, n  0
k
k


(4.47)
Energy Balance Method (cont.)
• Note potential utility of using thermal resistance concepts to express rate
equations. E.g., conduction between adjoining dissimilar materials with
an interfacial contact resistance.
q m, n 1  m, n  
Rtot 
y / 2
kA  x 

Tm, n 1  Tm, n

Rtot
Rt, c
x 

y / 2
kB  x 

(4.50)
Solution Methods
Solutions Methods
• Matrix Inversion: Expression of system of N finite-difference equations for
N unknown nodal temperatures as:
 AT   C 
Coefficient
Matrix (NxN)
Solution
Solution Vector
(T1,T2, …TN)
(4.52)
Right-hand Side Vector of Constants
(C1,C2…CN)
T    A1 C 
(4.53)
Inverse of Coefficient Matrix
• Gauss-Seidel Iteration: Each finite-difference equation is written in explicit
form, such that its unknown nodal temperature appears alone on the lefthand side:
N aij
C i 1 aij k
k
Ti   i   Tj   
Tj( k 1)
(4.55)
aii
j 1 aii
j  i 1 aii
where i =1, 2,…, N and k is the level of iteration.
Iteration proceeds until satisfactory convergence is achieved for all nodes:
k
k 1
Ti   Ti   
• What measures may be taken to insure that the results of a finite-difference
solution provide an accurate prediction of the temperature field?
Problem: Finite-Difference Equations
Problem 4.41: Finite-difference equations for (a) nodal point on a diagonal
surface and (b) tip of a cutting tool.
(a) Diagonal surface
(b) Cutting tool.
Schematic:
ASSUMPTIONS: (1) Steady-state, 2-D conduction, (2) Constant properties
Problem: Finite-Difference Equations (cont.)
ANALYSIS: (a) The control volume about node m,n is triangular with sides x and y and diagonal
(surface) of length 2 x.
The heat rates associated with the control volume are due to conduction, q1 and q2, and to convection,
qc. An energy balance for a unit depth normal to the page yields
Ein  0
q1  q 2  qc  0
Tm,n-1  Tm,n
Tm+1,n  Tm,n
k  x  1
 k  y  1
h
y
x
With x = y, it follows that
Tm,n-1  Tm+1,n  2 



2 x  1 T  Tm,n  0.
hx
hx 

T   2  2 
Tm,n  0.
k
k 

(b) The control volume about node m,n is triangular with sides x/2 and y/2 and a lower diagonal
surface of length 2  x/2  .
The heat rates associated with the control volume are due to the uniform heat flux, qa, conduction, qb,
and convection qc. An energy balance for a unit depth yields
Ein =0
qa  q b  qc  0
x 
 x 
 y  Tm+1,n  Tm,n

qo    1  k    1
h 2
T  Tm,n  0.
x
2 
 2 
 2 


or, with x = y,
Tm+1,n  2 
hx
x 
hx 
 T  q o 
 1  2 
 Tm,n  0.
k
k 
k 

Problem: Cold Plate
Problem 4.78: Analysis of cold plate used to thermally control IBM multi-chip,
thermal conduction module.
Features:
• Heat dissipated in the chips is transferred
by conduction through spring-loaded
aluminum pistons to an aluminum cold
plate.
• Nominal operating conditions may be
assumed to provide a uniformly
distributed heat flux of qo  105 W/m2
at the base of the cold plate.
• Heat is transferred from the cold
plate by water flowing through
channels in the cold plate.
Find: (a) Cold plate temperature distribution
for the prescribed conditions. (b) Options
for operating at larger power levels while
remaining within a maximum cold plate
temperature of 40C.
Problem: Cold Plate (cont.)
Schematic:
ASSUMPTIONS: (1) Steady-state conditions, (2) Two-dimensional conduction, (3) Constant properties
Problem: Cold Plate (cont.)
ANALYSIS: Finite-difference equations must be obtained for each of the 28 nodes. Applying the energy
balance method to regions 1 and 5, which are similar, it follows that
 y
x  T2   x y  T6   y x    x y   T1  0
Node 5:  y x  T4   x y  T10   y x    x y   T5  0
Node 1:
Nodal regions 2, 3 and 4 are similar, and the energy balance method yields a finite-difference equation of
the form
Nodes 2,3,4:
 y


x  Tm 1, n  Tm 1, n  2  x y  Tm, n 1  2  y x    x y  Tm, n  0
Energy balances applied to the remaining combinations of similar nodes yield the following finite-difference
equations.
Nodes 6, 14:
Nodes 7, 15:
 x
 x
y  T1   y x  T7   x y    y x    hx k T6    hx k T
 y
x  T6  T8   2  x y  T2  2  y x    x y    hx k T7    2hx k  T
 y
y  T19   y x  T15   x y    y x    hx k T14    hx k  T
x   T14  T16   2  x y  T20  2  y x    x y    hx k  T15    2hx k  T
Problem: Cold Plate (cont.)
Nodes 8, 16:
 x
Node 11:
 y
x  T7  2  y x  T9   x y  T11  2  x y  T3  3  y x   3  x y 
 y
x  T15  2  y x  T17   x y  T11  2  x y  T21  3  y x   3  x y 
  h k  x  y T8    h k  x  y  T
  h k  x  y T16    h k  x  y  T
y  T8   x y  T16  2  y x  T12  2  x y    y x    hy k  T11    2hy k  T
Nodes 9, 12, 17, 20, 21, 22:
x  Tm 1, n   y x  Tm  1, n   x y  Tm, n  1   x y  Tm, n 1  2  x y    y x  Tm, n  0
 y
Nodes 10, 13, 18, 23:
 x
 x
Node 19:
Nodes 24, 28:
 x
 x
Nodes 25, 26, 27:
 y
y  Tn 1, m   x y  Tn 1, m  2  y x  Tm 1, n  2  x y    y x  Tm, n  0
y  T14   x y  T24  2  y x  T20  2  x y    y x T19  0
y  T19   y x  T25   x y    y x T24    qo x k 
y  T23   y x  T27   x y    y x T28    qo x k 
x  Tm 1, n   y x  Tm  1, n  2  x y  Tm, n  1  2  x y    y x  Tm, n    2qo x k 
Problem: Cold Plate (cont.)
Evaluating the coefficients and solving the equations simultaneously, the steady-state temperature
distribution (C), tabulated according to the node locations, is:
23.77
23.41
23.91
23.62
28.90
30.72
32.77
28.76
30.67
32.74
24.27
24.31
25.70
28.26
30.57
32.69
24.61
24.89
26.18
28.32
30.53
32.66
24.74
25.07
26.33
28.35
30.52
32.65
(b) For the prescribed conditions, the maximum allowable temperature (T 24 = 40C) is reached when
qo = 1.407  105 W/m2 (14.07 W/cm2).
Options for extending this limit could include use of a copper cold plate (k  400 W/mK) and/or increasing
the convection coefficient associated with the coolant.
With k = 400 W/mK, a value of qo = 17.37 W/cm2 may be maintained.
. With k = 400 W/mK and h = 10,000 W/m2K (a practical upper limit), qo = 28.65 W/cm2.
Additional, albeit small, improvements may be realized by relocating the coolant channels closer to the base
of the cold plate.