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5-102
5-108 A plane wall with variable heat generation and constant thermal conductivity is subjected to uniform heat flux
0
q
at
the left (node 0) and convection at the right boundary (node 4). The implicit transient finite difference formulation of the
boundary nodes is to be determined.
Assumptions 1 Heat transfer through the wall is given to be transient, and the thermal conductivity to be constant. 2 Heat
transfer is one-dimensional since the wall is large relative to its thickness. 3 Radiation heat transfer is negligible.
Analysis Using the energy balance approach and taking the direction
of all heat transfers to be towards the node under consideration, the
implicit finite difference formulations become
Left boundary node:
TT
x
TT
ii
i
ii
−
−+
+
++
0
1
0
1
1
0
1
1
5-109 A plane wall with variable heat generation and constant thermal conductivity is subjected to insulation at the left (node
0) and radiation at the right boundary (node 5). The explicit transient finite difference formulation of the boundary nodes is to
be determined.
Assumptions 1 Heat transfer through the wall is given to be transient and one-dimensional, and the thermal conductivity to be
constant. 2 Convection heat transfer is negligible.
Analysis Using the energy balance approach and taking the direction
of all heat transfers to be towards the node under consideration, the
explicit transient finite difference formulations become
Left boundary node:
TT
x
x
TT
ii
i
ii
−
−+
0
1
0
01
0
q
),( txe
)(xe
Insulated
5-103
5-110 A plane wall with variable heat generation and constant thermal conductivity is subjected to combined convection,
radiation, and heat flux at the left (node 0) and specified temperature at the right boundary (node 4). The explicit finite
difference formulation of the left boundary and the finite difference formulation for the total amount of heat transfer at the
right boundary are to be determined.
Assumptions 1 Heat transfer through the wall is given to be transient and one-dimensional, and the thermal conductivity to be
constant.
Analysis Using the energy balance approach and taking the direction of all heat transfers to be towards the node under
consideration, the explicit transient finite difference formulations become
Left boundary node:
1
44 1 0 0 0
0 surr 0 0 0
[( ) ( ) ] ( ) 22
i i i i
i i i i i
p
T T T T
xx
q A A T T hA T T kA e A A c
xt
es r
+
¥
--
DD
+ - + - + + =
DD
Heat transfer at right surface:
t
TT
c
x
A
x
Ae
x
TT
kAQ
ii
p
i
ii
i
−
=
+
−
+
+
4
1
4
4
43
surfaceright 22
Noting that
==
i
itQtQQ
the total amount of heat transfer becomes
=
+
=
−
+
−
−
=
=
20
1
4
1
4
4
34
20
1
surfaceright
surfaceright
22
i
ii
p
i
ii
i
i
t
t
TT
c
x
A
x
Ae
x
TT
kA
tQQ
5-111 A composite plane wall consists of two layers A and B in perfect contact at the interface where node 1 is at the
interface. The wall is insulated at the left (node 0) and subjected to radiation at the right boundary (node 2). The complete
transient explicit finite difference formulation of this problem is to be obtained.
Assumptions 1 Heat transfer through the wall is given to be transient and one-dimensional, and the thermal conductivity to be
constant. 2 Convection heat transfer is negligible. 3 There is no heat generation.
Analysis Using the energy balance approach with a unit area A = 1 and taking the direction of all heat transfers to be towards
the node under consideration, the finite difference formulations become
Node 0 (at left boundary):
t
TT
c
x
x
TT
k
ii
ApA
ii
A
−
=
−+
0
1
0
,
01
2
Node 1 (at interface):
t
TT
c
x
c
x
x
TT
k
x
TT
k
ii
BBApA
ii
B
ii
A
−
+
=
−
+
−+
1
1
1
,
12
10
22
Node 2 (at right boundary):
t
TT
c
x
x
TT
kTT
ii
BpB
ii
B
i
−
=
−
+−
+
2
1
2
,
21
4
2
4
surr 2
])([
h, T
x
• • • • •
0 1 2 3 4
Tsurr
TL
),( txe
0
q
Insulated
x
1
•
•
•
0
2
A
Tsurr
Radiation
B
Interface
5-105
5-114 A hot brass plate is having its upper surface cooled by impinging jet while its lower surface is insulated. The implicit
finite difference equations and the nodal temperatures of the brass plate after 10 seconds of cooling are to be determined.
Assumptions 1 Transient heat conduction is one-dimensional. 2 Thermal properties are constant. 3 Convection heat transfer
coefficient is uniform. 4 Heat transfer by radiation is negligible. 5 There is no heat generation.
Properties The properties of the brass plate are given as
= 8530 kg/m3, cp = 380 J/kg∙K, k = 110 W/m∙K, and
= 33.9
10−6 m2/s.
Analysis The nodal spacing is given to be x = 2.5 cm. Then the number of nodes becomes M = L/x +1 = 10/2.5 + 1 = 5.
This problem involves 5 unknown nodal temperatures, and thus we need to have 5 equations. The finite difference equation
for node 0 on the top surface subjected to convection is obtained by applying an energy balance on the half volume element
about node 0 and taking the direction of all heat transfers to be towards the node under consideration:
TT
x
TT
ii
ii
i
−
−
+++
+
1
0
1
0
1
1
1
5-106
5-115 A uranium plate initially at a uniform temperature is subjected to insulation on one side and convection on the other.
The transient finite difference formulation of this problem is to be obtained, and the nodal temperatures after 5 min and under
steady conditions are to be determined.
Assumptions 1 Heat transfer is one-dimensional since the plate is large relative to its thickness. 2 Thermal conductivity is
constant. 3 Radiation heat transfer is negligible.
Properties The conductivity and diffusivity are given to be k = 28 W/m°C and
/sm 105.12 26−
=
.
Analysis The nodal spacing is given to be x = 0.015 m. Then the number of nodes becomes
1/ += xLM
= 0.09/0.015+1
= 7. This problem involves 7 unknown nodal temperatures, and thus we need to have 7 equations. Node 0 is on insulated
boundary, and thus we can treat it as an interior note by using the mirror image concept. Nodes 1, 2, and 3 are interior nodes,
and thus for them we can use the general explicit finite difference relation expressed as
i
m
i
m
i
m
i
i
i
TT
xe
+
12
xe
i
m
i
i
i
i
2
1)21()(
+
k
k
k
where
C20 C, W/m35 C, W/m28 , W/m10 m, 015.0 236
0=====
Thkex
, and
6
105.12 −
=
m2/s.
The upper limit of the time step t is determined from the stability criteria that requires all primary coefficients to be greater
than or equal to zero. The coefficient of
i
T4
is smaller in this case, and thus the stability criteria for this problem can be
expressed as
)/1(2
)/1(2
1
0221
2
kxh
x
t
kxhk
xh
+
→
+
→
−−
since
2
/xt =
. Substituting the given quantities, the maximum allowable the time step becomes
s 8.8
C)] W/m.28/(m) 015.0)(C. W/m35(1/s)[m 105.12(2
)m 015.0(
226
2
=
+
−
t
Therefore, any time step less than 8.8 s can be used to solve this problem. For convenience, let us choose the time step to be
t = 7.5 s. Then the mesh Fourier number becomes
4167.0
)m 015.0(
s) /s)(7.5m 105.12(
2
26
2=
=
=
−
x
t
Substituting this value of
and other given quantities, the nodal temperatures after 560/7.5 = 40 time steps (5 min) are
determined to be
5-107
After 5 min:
5-109
Time
[s]
T1
[C]
T2
[C]
T3
[C]
T4
[C]
T5
[C]
T6
[C]
T7
[C]
Row
0
7.5
15
22.5
30
37.5
45
52.5
60
67.5
100
103.3
106.7
110
113.4
116.7
120.1
123.4
126.8
130.1
100
103.3
106.7
110
113.4
116.7
120.1
123.4
126.7
130
100
103.3
106.7
110
113.4
116.7
120
123.3
126.6
129.9
100
103.3
106.7
110
113.3
116.6
119.8
123.1
126.3
129.5
100
103.3
106.7
109.8
113.1
116.2
119.4
122.6
125.7
128.9
100
103.3
106.2
109.3
112.3
115.5
118.5
121.7
124.8
127.9
100
103.3
106.7
109.8
113.1
116.2
119.4
122.6
125.7
128.9
1
2
3
4
5
6
7
8
9
10
…
…
…
…
…
…
…
…
…
…
…
…
…
…
…
…
…
…
3525
3533
3540
3548
3555
3563
3570
3578
3585
3593
3600
1276
1277
1279
1281
1283
1285
1287
1288
1290
1292
1294
1274
1276
1277
1279
1281
1283
1285
1287
1288
1290
1292
1268
1270
1272
1274
1276
1277
1279
1281
1283
1285
1286
1259
1261
1263
1265
1266
1268
1270
1272
1274
1275
1277
1246
1248
1250
1252
1254
1255
1257
1259
1261
1262
1264
1230
1232
1233
1235
1237
1239
1240
1242
1244
1246
1247
1209
1211
1213
1215
1216
1218
1220
1222
1223
1225
1227
471
472
473
474
475
476
477
478
479
480
481
0500 1000 1500 2000 2500 3000 3500
0
200
400
600
800
1000
1200
1400
T [C]
T1
T7
5-110
5-117E A plain window glass initially at a uniform temperature is subjected to convection on both sides. The transient finite
difference formulation of this problem is to be obtained, and it is to be determined how long it will take for the fog on the
windows to clear up (i.e., for the inner surface temperature of the window glass to reach 54F).
Assumptions 1 Heat transfer is one-dimensional since the window is large relative to its thickness. 2 Thermal conductivity is
constant. 3 Radiation heat transfer is negligible.
Properties The conductivity and diffusivity are given to be k = 0.48 Btu/h.ft°F and
/sft 102.4 26−
=
.
Analysis The nodal spacing is given to be x = 0.125 in. Then the number of nodes becomes
1/ += xLM
=
0.375/0.125+1 = 4. This problem involves 4 unknown nodal temperatures, and thus we need to have 4 equations. Nodes 2
and 3 are interior nodes, and thus for them we can use the general explicit finite difference relation expressed as
i
m
i
m
i
m
i
i
i
TT
xe
+
12
i
i
i
i
1
+
5-112
𝛼 = 𝑘
𝜌𝑐𝑝=16.3
(8000)(500)= 4.075×10−6m2/s
𝜏 = 𝛼∆𝑡
∆𝑥2=(4.075×10−6)(10)
(0.01)2= 0.4075
The numerical results can be solved iteratively with the initial nodal temperatures, Ti = 10°C, and time step, Δt = 10 s. The
following table tabulates the iterated nodal temperatures at every minute:
x [m] =
0
0.01
0.02
0.03
0.04
t [s]
T0 [°C]
T1 [°C]
T2 [°C]
T3 [°C]
T4 [°C]
0
10.0
10.0
10.0
10.0
10.0
60
-10.7
-1.6
4.7
7.7
8.8
120
-17.0
-8.3
-2.0
1.8
3.1
180
-21.8
-13.8
-7.9
-4.2
-2.9
240
-25.9
-18.6
-13.2
-9.8
-8.6
300
-29.7
-23.0
-18.0
-14.9
-13.8
360
-33.1
-27.0
-22.4
-19.6
-18.5
420
-36.2
-30.6
-26.4
-23.8
-22.7
480
-39.1
-33.9
-30.1
-27.6
-26.6
540
-41.7
-36.9
-33.4
-31.1
-30.2
600
-44.0
-39.7
-36.4
-34.3
-33.4
The nodal temperatures, as a function of t, are plotted in the following figure:
-50
-40
-30
-20
-10
0
10
20
0100 200 300 400 500 600
T [°C]
t [s]
T
0
T
1
T
2
5-113
5-119 ASTM F441 CPVC tube is embedded in a metal plate, while the plate’s upper surface is subjected to convection
with hot fluid at T∞ = 300°C and h = 200 W/m2·K. The plate’s bottom surface is well insulated. The plate has an initial
temperature of 20°C. Would the tube comply with the ASME Code for Process Piping if the plate’s upper surface is exposed
to the hot fluid for 5 minutes?
Assumptions 1 Transient heat conduction is one-
dimensional. 2 Thermal properties are constant. 3 Heat
transfer by radiation is negligible. 4 There is no heat
generation. 5 The convection heat transfer coefficients
are constant.
Properties The thermal properties of the metal plate are
cp = 460 J/kg·K, k = 26.9 W/m·K, and ρ = 7730 kg/m3.
Analysis The nodal spacing is given to be Δx = 1 cm.
So, the number of nodes is
𝑚 = 1 to 3: 𝑇𝑚
Node 4 is an insulation boundary,
𝑚 = 4: 𝑇4
𝑖+1 =(1−2𝜏)𝑇4
𝑖+𝜏(2𝑇3
𝑖)
The upper limit of the time step t is determined from the stability criteria that requires all primary coefficients to be greater
than or equal to zero. The coefficient 𝑇0
𝑖 is smaller in this case, and thus the stability criterion for this problem can be
expressed as
𝜏 = 𝛼∆𝑡
∆𝑥2=(7.5651×10−6)(5)
(0.01)2= 0.37826
5-114
The numerical results can be solved iteratively with the initial nodal temperatures, Ti = 20°C, and time step, Δt = 5 s. The
following table tabulates the iterated nodal temperatures at every minute:
x [m] =
0
0.01
0.02
0.03
0.04
t [s]
T0 [°C]
T1 [°C]
T2 [°C]
T3 [°C]
T4 [°C]
0
20.0
20.0
20.0
20.0
20.0
60
64.1
48.7
37.9
31.6
29.5
120
82.1
67.8
57.4
51.0
48.9
180
98.3
85.0
75.3
69.5
67.5
240
113.2
100.9
92.0
86.5
84.7
300
127.1
115.7
107.4
102.4
100.7
The nodal temperatures, as a function of t, are plotted in the following figure:
0
20
40
60
80
100
120
140
050 100 150 200 250 300
T [°C]
t [s]
T0
T1
T2
T3
T4
5-115
5-120 The roof of a house initially at a uniform temperature is subjected to convection and radiation on both sides. The
temperatures of the inner and outer surfaces of the roof at 6 am in the morning as well as the average rate of heat transfer
through the roof during that night are to be determined.
Assumptions 1 Heat transfer is one-dimensional. 2 Thermal properties, heat transfer coefficients, and the indoor and outdoor
temperatures are constant. 3 Radiation heat transfer is significant.
Properties The conductivity and diffusivity are given to be k = 1.4 W/m.C and
/sm 1069.0 26−
=
. The emissivity of
both surfaces of the concrete roof is 0.9.
Analysis The nodal spacing is given to be x = 0.03 m. Then the
number of nodes becomes
1/ += xLM
= 0.15/0.03+1 = 6.
This problem involves 6 unknown nodal temperatures, and thus
we need to have 6 equations. Nodes 2, 3, 4, and 5 are interior
nodes, and thus for them we can use the general explicit finite
difference relation expressed as
i
m
i
m
i
m
i
i
i
TT
xe
+
12
Convection
ho, To
Concrete
roof
Tsky
Radiation
•
•
•
6
5
4
5-118
Time [s]
T1 [C]
T2 [C]
T3 [C]
T4 [C]
T5 [C]
Row
0
3
3
3
3
3
1
60
35.9
3
3
3
3
2
120
5.389
10.11
3
3
3
3
180
36.75
7.552
4.535
3
3
4
240
6.563
13.21
4.855
3.663
3
5
300
37
9.968
6.402
3.517
3.024
6
360
7.374
15.04
6.549
4.272
3.042
7
420
37.04
11.55
7.891
4.03
3.087
8
480
8.021
16.27
7.847
4.758
3.122
9
540
36.97
12.67
8.998
4.461
3.182
10
…
…
…
…
…
…
…
…
…
…
…
…
…
…
35460
24.85
24.23
23.65
23.09
22.86
592
35520
24.81
24.24
23.65
23.1
22.87
593
35580
24.85
24.23
23.66
23.11
22.88
594
35640
24.81
24.24
23.67
23.12
22.88
595
35700
24.85
24.24
23.67
23.12
22.89
596
35760
24.81
24.25
23.68
23.13
22.9
597
35820
24.85
24.25
23.68
23.14
22.91
598
35880
24.81
24.26
23.69
23.15
22.92
599
35940
24.85
24.25
23.69
23.15
22.93
600
36000
24.82
24.26
23.7
23.16
22.94
601
0 5000 10000 15000 20000 25000 30000 35000 40000
2.5
7
11.5
16
20.5
25
T5 [C]
5-119
5-123 Heat conduction through a long L-shaped solid bar with specified boundary conditions is considered. The temperature
at the top corner (node #3) of the body after 2, 5, and 30 min is to be determined with the transient explicit finite difference
method.
Assumptions 1 Heat transfer through the body is given to be
transient and two-dimensional. 2 Thermal conductivity is
constant. 3 Heat generation is uniform.
Properties The conductivity and diffusivity are given to
be k = 15 W/m°C and
/sm 102.3 26−
=
.
Analysis The nodal spacing is given to be x=x=l=0.015 m.
The explicit finite difference equations are determined on the
basis of the energy balance for the transient case expressed as
TT
i
m
i
m
i
−
+
i
t
l
l
22
2
where
, W/m8000 , W/m102237
0== L
qe
l = 0.015 m, k =15 W/mC, h = 80 W/m2C, and T =25C.
The upper limit of the time step t is determined from the stability criteria that requires the coefficient of
i
m
T
in the
1+i
m
T
4 1 4 1
khl ktl
hl k
( / ) ( / )
• • •
3
1
2
4 5 6 7 8
• • • • •
h, T
Insulated
L
q
5-120
since
2
/lt=
. Substituting the given quantities, the maximum allowable value of the time step is determined to be
s 3.16
C)] W/m.15/(m) 015.0)(C. W/m80(1/s)[m 102.3(4
)m 015.0(
226
2
=
+
−
t
Therefore, any time step less than 16.3 s can be used to solve this problem. For convenience, we choose the time step to be t
= 15 s. Then the mesh Fourier number becomes
2133.0
)m 015.0(
s) /s)(15m 102.3(
2
26
2=
=
=
−
l
t
(for t = 15 s)
Using the specified initial condition as the solution at time t = 0 (for i = 0), sweeping through the 9 equations above will give
the solution at intervals of 15 s. Using a computer, the solution at the upper corner node (node 3) is determined to be 441,
520, and 529°C at 2, 5, and 30 min, respectively. It can be shown that the steady state solution at node 3 is 531C.
