5-61
5-65 A square cross section is undergoing a steady two-dimensional heat transfer. The finite difference equations and the
nodal temperatures are to be determined.
Assumptions 1 Steady heat conduction is two-dimensional. 2 Thermal properties are constant. 3 There is no heat generation
in the body.
Analysis (a) There are 4 unknown nodal temperatures, thus we need to have 4 equations to determine them uniquely. For
nodes 1 to 4, we can use the general finite difference relation expressed as
0
22
2
1 , ,1 ,
2
,1 , ,1 =
+−
+
+− +−+−
y
TTT
x
TTT nmnmnmnmnmnm
Since
yx =
, we have
(b) By letting the initial guesses as T1 = 300°C, T2 = 150°C, T3 = 400°C, and T4 = 250°C the results obtained from successive
iterations are listed in the following table:
Nodal temperature,°C
T1
T2
T3
T4
287.5
209.4
334.4
260.9
285.9
211.7
336.7
262.1
287.1
212.3
337.3
262.4
287.4
212.5
337.5
262.5
287.5
212.5
337.5
262.5
287.5
212.5
337.5
262.5
Discussion The finite difference equations can also be calculated using the EES. Copy the following lines and paste on a
blank EES screen to solve the above equations:
T_1=0.25*(100+T_2+T_3+500)
5-62
5-66 A long solid body is subjected to steady two-dimensional heat transfer. The unknown nodal temperatures are to be
determined.
Assumptions 1 Heat transfer through the body is given to be steady and two-dimensional. 2 There is no heat generation in the
body.
Properties The thermal conductivity is given to be k = 45 W/m°C.
Analysis The nodal spacing is given to be x = x = l = 0.02 m, and the general finite difference form of an interior node for
steady two-dimensional heat conduction for the case of no heat generation is expressed as
8642
TTTT
===
Therefore, there are there are only 3 unknown nodal temperatures,
521 and , , TTT
, and thus we need only 3 equations to determine them
uniquely. Also, we can replace the symmetry lines by insulation and
utilize the mirror-image concept when writing the finite difference
5-64
5-68 A square cross section geometry is subjected to four different boundary conditions. The temperature distribution within
the geometry is to be determined using Gauss-Seidel iteration method.
Assumptions 1 Steady heat conduction is two-dimensional without internal heat generation. 2 Thermal conductivity is
constant.
Properties Thermal conductivity is given as k=20 W/m·K.
Analysis: (a) There are 4 internal nodes (node 2, 2, 6 and
7) and 4 boundary nodes (node 1, 4, 5 and 8). Thus we
need to have 8 equations for 8 unknown temperatures. For
2
2
y
x
511 25.05.025.25 TTT ++=
Node 2: (Internal node)
( )
6312 10025.0 TTTT +++=
Node 3: (Internal node)
( )
7423 10025.0 TTTT +++=
Node 4: (Right boundary node)
( ) ( ) ( ) ( )
0
2
100
24
4348
4=−+
−
+
−
+
−
TTyh
x
TT
yk
y
TT
x
k
y
T
x
k
( )
384 5.045.504944.0 TTT ++=
Node 5: (Left boundary node)
( ) ( ) ( )
0
300
22
1
5156 =
−
+
−
+
−
+y
T
x
k
y
TT
x
k
x
TT
ykyq
( )
615 5.1505.05.0 TTT ++=
Node 6: (Internal node)
( )
30025.0 7256 +++= TTTT
Node 7: (Internal node
( )
30025.0 8367 +++= TTTT
Node 8: (Right boundary node)
( ) ( ) ( ) ( )
0
2
300
28
87848 =−+
−
+
−
+
−
TTyh
x
TT
yk
y
TT
x
k
y
T
x
k
( )
748 5.045.1504944.0 TTT ++=
5-65
(b) By letting the initial guess as 200oC at each node, the temperature distribution obtained using Gauss-Seidel iteration
method is as follows
Nodal temperature, oC
Iteration
T1
T2
T3
T4
T5
T6
T7
T8
0
200.0
200.0
200.0
200.0
200.0
200.0
200.0
200.0
1
175.3
168.8
167.2
157.0
219.1
222.0
222.3
223.1
2
167.6
164.2
160.9
159.6
228.1
228.7
228.2
226.6
3
166.1
163.9
162.9
161.5
231.1
230.8
230.1
228.1
4
166.1
165.0
164.1
162.5
232.2
231.8
231.0
228.8
5
166.3
165.6
164.8
162.9
232.7
232.3
231.5
229.1
6
166.6
165.9
165.1
163.2
233.1
232.6
231.7
229.3
7
166.8
166.1
165.3
163.3
233.3
232.8
231.8
229.4
8
167.0
166.2
165.3
163.4
233.4
232.9
231.9
229.4
9
167.1
166.3
165.4
163.4
233.5
232.9
231.9
229.5
10
167.2
166.4
165.4
163.5
233.5
233.0
232.0
229.5
11
167.2
166.4
165.5
163.5
233.5
233.0
232.0
229.5
5-67
5-70 A rectangular metal block is subjected to specified boundary conditions. The finite difference equations and the nodal
temperatures are to be determined.
Assumptions 1Two-dimensional steady state heat conduction. 2 No internal heat generation. 3 Thermal conductivity is
constant.
Properties Thermal conductivity of the metal block is given as k = 35 W/m·K.
Analysis (a) There are 15 unknown temperatures while the temperatures on the top surface of the block are to be determined
from the given sinusoidal temperature distribution. Given that
cmyx 25==
, the number of nodes in x and y directions are
5 and 4, respectively. For internal nodes i.e., 2, 3, 4, 7, 8 and 9 we can use the general form of the finite difference equations
without heat generation expressed as
0
22
2
1 , ,1 ,
2
,1 , ,1 =
+−
+
+− +−+−
y
TTT
x
TTT nmnmnmnmnmnm
Since
yx =
, we have
)(25.0 ,11 , ,11 , , nmnmnmnmnm TTTTT +−−+ +++=
For the boundary (external nodes) the finite
difference formulation is obtained using energy
balance and considering all heat transfer towards
these nodes.
Thus the finite difference equations at each node are expressed as follows
Node 1:
( ) ( ) ( )
0
)0sin(100
22
)( 1
16
12
1=
−
+
−
+
−
+−y
T
x
k
y
TT
x
k
x
TT
ykTTyh
Node 2:
04)4/sin(100 2371 =−+++ TTTT
5-68
(b) The nodal temperatures at these different nodes can be determined by solving the equations simultaneously in equation
solver. Solving these equations simultaneously in EES or any other software gives
5-70
5-72 A long solid body is subjected to steady two-dimensional heat transfer. The unknown nodal temperatures and the rate of
heat loss from the top surface are to be determined.
Assumptions 1 Heat transfer through the body is given to be steady and two-dimensional. 2 Heat is generated uniformly in
the body.
Properties The thermal conductivity is given to be k = 150 W/m°C.
Analysis (a) The nodal spacing is given to be x=x=l=0.1 m, and the general finite difference form of an interior node
equation for steady two-dimensional heat conduction for the case of constant heat generation is expressed as
2
node
le
C1159
C1126
==
==
43
21
TT
TT
(b) The total rate of heat transfer from the top surface
top
Q
can be determined from an energy balance on a volume element
at the top surface whose height is l/2, length 0.3 m, and depth 1 m:
depth) (per
C100)–m)(1126 1(C)100120(
2
m 1
)C W/m150(2m)2/1.03.0)( W/m103(
0
100
12
100120
2
1
2)2/13.0(
337
top
1
0top
m
Q
l
T
kl
l
l
kleQ
W760,900
=
+−−−=
=
−
+
−
++
5-72
5-74 Prob. 5-72 is reconsidered. The effects of the thermal conductivity and the heat generation rate on the
temperatures at nodes 1 and 3, and the rate of heat loss from the top surface are to be investigated.
Analysis The problem is solved using EES, and the solution is given below.
“GIVEN”
k=150 [W/m-C]
e_dot=3E7 [W/m^3]
DELTAx=0.10 [m]
DELTAy=0.10 [m]
d=1 [m] “depth”
“Temperatures at the selected nodes are also given in the figure”
“ANALYSIS”
“(a)”
l=DELTAx
T_1=T_2 “due to symmetry”
T_3=T_4 “due to symmetry”
“Using the finite difference method, the two equations for the two unknown temperatures are determined to be”
100+120+T_2+T_3-4*T_1+(e_dot*l^2)/k=0
150+200+T_1+T_4-4*T_3+(e_dot*l^2)/k=0
“(b)”
“The rate of heat loss from the top surface can be determined from an energy balance on a volume element
whose height is l/2, length 3*l, and depth d=1 m”
-Q_dot_top+e_dot*(3*l*d*l/2)+2*(k*(l*d)/2*(120-100)/l+k*l*d*(T_1-100)/l)=0
k
[W/m.C]
T1
[C]
T3
[C]
top
Q
[W]
10
30.53
51.05
71.58
92.11
112.6
133.2
153.7
174.2
194.7
15126
5040
3064
2222
1755
1458
1253
1102
987.3
896.5
15159
5073
3097
2254
1787
1491
1285
1135
1020
929
750725
752213
753701
755189
756678
758166
759654
761142
762630
764118
050 100 150 200 250 300 350 400
0
2000
4000
6000
8000
10000
12000
14000
16000
750000
755000
760000
765000
770000
775000
780000
k [W/m-C]
T1 and T3 [C]
Qtop [W]
5-73
e
[W/m3]
T1
[C]
T3
[C]
top
Q
[W]
100000
5.358E+06
1.061E+07
1.587E+07
2.113E+07
2.639E+07
3.165E+07
3.691E+07
4.216E+07
4.742E+07
5.268E+07
5.794E+07
6.319E+07
6.845E+07
7.371E+07
7.897E+07
8.423E+07
8.948E+07
9.474E+07
1.000E+08
129.6
304.8
480.1
655.4
830.6
1006
1181
1356
1532
1707
1882
2057
2233
2408
2583
2759
2934
3109
3284
3460
162.1
337.3
512.6
687.9
863.1
1038
1214
1389
1564
1739
1915
2090
2265
2441
2616
2791
2966
3142
3317
3492
13375
144822
276270
407717
539164
670612
802059
933507
1.065E+06
1.196E+06
1.328E+06
1.459E+06
1.591E+06
1.722E+06
1.854E+06
1.985E+06
2.117E+06
2.248E+06
2.379E+06
2.511E+06
500
1000
1500
2000
2500
3000
3500
5.0x105
1.0x106
1.5x106
2.0x106
2.5x106
3.0x106
Qtop [W]
T1 and T3 [C]
5-75
5-76 A long solid body is subjected to steady two-dimensional heat transfer. The unknown nodal temperatures and the rate of
heat loss from the bottom surface through a 1-m long section are to be determined.
Assumptions 1 Heat transfer through the body is given to be steady and two-dimensional. 2 Heat is generated uniformly in
the body. 3 Radiation heat transfer is negligible.
Properties The thermal conductivity is given to be k = 45 W/m°C.
Analysis The nodal spacing is given to be x = x = l = 0.04
m, and the general finite difference form of an interior node for
steady two-dimensional heat conduction is expressed as
04
2
node
nodebottomrighttopleft =+−+++ k
le
TTTTT
where
C2.142
C W/m45
)m 04.0)( W/m104( 236
2
0
2
node =
== k
le
k
le
The finite difference equations for boundary nodes are obtained by
applying an energy balance on the volume elements and taking the
direction of all heat transfers to be towards the node under
consideration:
04–200240290260 :(interior) 3 Node
0 4–290325290350 :(interior) 2 Node
0
2
)(
325
2
290240
2
:)convection ( 1 Node
2
0
3
2
0
2
2
0
1
111
=++++
=++++
=+−+
−
+
−
+
−
k
le
T
k
le
T
k
le
TThl
l
T
l
k
l
T
kl
l
T
l
k
where
C20 , W/m104 C,. W/m50 C, W/m.45 362 ====
Tehk
Substituting,
T1 = 281.2°C, T2 = 349.3°C, T3 = 283.1°C,
(b) The rate of heat loss from the bottom surface through a 1-m long section is
−==
)(
surface, element,
TThAQQ
m
mm
m
m
e
h, T
Insulated
• • • •
• • • •
• • • •
• • • •
240
200°C
350
260
305
290
3
1
2
4 cm
325
Convection
5-77
5-78 A rectangular block is subjected to uniform heat flux at the top, and iced water at 0C at the sides. The steady finite
difference formulation of the problem is to be obtained, and the unknown nodal temperatures as well as the rate of heat
transfer to the iced water are to be determined.
Assumptions 1 Heat transfer through the body is given to be steady and two-dimensional. 2 There is no heat generation
within the block. 3 The heat transfer coefficient is very high so that the temperatures on both sides of the block can be taken
to be 0°C. 4 Heat transfer through the bottom surface is negligible.
Properties The thermal conductivity is given to be k = 23 W/m°C.
Analysis The nodal spacing is given to be x=x=l=0.1 m,
and the general finite difference form of an interior node
equation for steady 2-D heat conduction is expressed as
• • •
• • •
• • •
04
2
node
nodebottomrighttopleft
=+−+++
le
TTTTT
where
l = 0.1 m, k = 23 W/mC, T0 =0C, and
22
00 W/m3200)m 5.0 W)/(58000(/ === AQq
This system of 8 equations with 8 unknowns constitutes the finite difference formulation of the problem.
(b) The 8 nodal temperatures under steady conditions are determined by solving the 8 equations above simultaneously with
8 kW heater
Insulated
• • •
1 5 5
5-79
5-80E A long solid bar is subjected to steady two-dimensional heat transfer. The unknown nodal temperatures and the rate of
heat loss from the bar through a 1-ft long section are to be determined.
Assumptions 1 Heat transfer through the body is given to be steady and two-dimensional. 2 Heat is generated uniformly in
the body. 3 The heat transfer coefficient also includes the radiation effects.
Properties The thermal conductivity is given to be k = 16 Btu/h.ft°C.
Analysis The nodal spacing is given to be x=y=l=0.25 ft, and the general finite difference form of an interior node for
steady two-dimensional heat conduction is expressed as
04
2
node
nodebottomrighttopleft =+−+++ k
le
TTTTT
(a) There is symmetry about the vertical, horizontal, and diagonal lines passing
through the center. Therefore,
9731 TTTT ===
and
8642 TTTT ===
, and
521 and ,, TTT
are the only 3 unknown nodal temperatures, and thus we need only
3 equations to determine them uniquely. Also, we can replace the symmetry
h, T
7 8 9
h, T
9731 TTTT ===
8642 TTTT ===
(b) The rate of heat loss from the bar through a 1-ft long section is determined from an energy balance on one-eight section of
the bar, and multiplying the result by 8:
 
ft) 1()2
2
8ft) 1()(
2
)(
2
88
2
2121conv section,eight one
−+=
−+−== − TTT
l
hTT
l
hTT
l
hQQ
• • •
4 5 6
• • •
1 2 3
h, T
e
5-80
5-81 Heat transfer through a square chimney is considered. The nodal temperatures and the rate of heat loss per unit length
are to be determined with the finite difference method.
Assumptions 1 Heat transfer is given to be steady and two–dimensional since the height of the chimney is large relative to its
cross-section, and thus heat conduction through the chimney in the axial direction is negligible. It is tempting to simplify the
problem further by considering heat transfer in each wall to be one dimensional which would be the case if the walls were
thin and thus the corner effects were negligible. This assumption cannot be justified in this case since the walls are very thick
and the corner sections constitute a considerable portion of the chimney structure. 2 There is no heat generation in the
chimney. 3 Thermal conductivity is constant.
Properties The thermal conductivity and emissivity are given
to be k = 1.4 W/m°C and = 0.9.
Analysis (a) The most striking aspect of this problem is the
apparent symmetry about the horizontal and vertical lines
passing through the midpoint of the chimney. Therefore, we
need to consider only one-fourth of the geometry in the
solution whose nodal network consists of 10 equally spaced
nodes. No heat can cross a symmetry line, and thus symmetry
222
2
l
l
where l = 0.1 m, k = 1.4 W/mC, hi = 75 W/m2C, Ti =280C, ho = 18 W/m2C, T0 =15C, Tsurr =250 K, = 0.9, and =
5.6710-8 W/m2.K4. This system of 10 equations with 10 unknowns constitutes the finite difference formulation of the
problem.
(b) The 10 nodal temperatures under steady conditions are determined by solving the 10 equations above simultaneously with
an equation solver to be
• • • •
1 2 3 4
Tsky
ho, To
• • • •
5 6 7 8
Insulated