j = ym – yi = 6 – 2 = 4
m = xj – xi = 10 – 6 = 4
1
ii
xy
1 6 2
Symmetry 4.772
Convection part of [k] = [kh]
Lij = 4.123 ft
2 1 0
ij
hL t
2 1 0
(20)(4.123)(1) 1 2 0
i
N
521
Lim = 8.246 m
i = – 6,
j = 8,
m = – 2
i = – 4,
j = – 2,
m = 6
44
1
44
1
1
44
By (1)
64
10 0 6 8 2
182
0 10 4 2 6
4(22) 26
+
2 0 1
20(8.246) 0 0 0
61 0 2
[k] =
60.88 4.55 26.12
7.73 3.18
Symmetry 59.52
W
°C
T
T
2
0.183 1
{}f
=
1346
54.6
1273
W
13.21
522
13.22 For the square plate in figure P13-22, determine the temperature distribution. Let Kxx = Kyy =
10
W
mC
, and h = 20
2
W
mC
. The temperature along the left side is maintained at
100 °C and that along the top side is maintained at 200 °C.
523
13.23
HEAT—Problem 13–23
KXX = 1.0 KYY = 1.0
SEMI–BANDWIDTH = 4
NEL NODE NUMBER X(1) Y(1)
1 1 2 3 0.0000 2.0000
2 2 5 3 0.0000 0.0000
6 5 e 6 1.5000 0.0000
X(2) Y(2) X(3) Y(3)
0.0000 0.0000 0.7500 1.0000
1.5000 0.0000 1.5000 2.0000
1.5000 0.0000 2.2500 1.0000
3.0000 0.0000 3.0000 2.0000
*PRESCRIBED NODAL TEMPERATURE VALUES*
7 0.00000E+00
8 0.00000E+00
RESULTING NODAL TEMPERATURE VALUES
1 0.10000E+03 2 0.10000E+03
524
ELEMENT RESULTANTS
ELEMENT GRAD (X) GRAD (Y) AVE TEMP
1 –0.3333E+02 0.0000E+00 0.9167E+02
2 –0.3333E+02 0.0000E+00 0.7500E+02
13.24
525
13.25
13.26
13.27
13.29 The temperature distribution of the earth is shown below with a 60 °F oil pipe 15 ft under the
earth’s surface at 50 °F.
13.30
Btu 778 lb ft 1 h
in. lb
hr
60 min
60s 1 Btu
1 ft
s
Heat source
q* = 129.66
lbf in.
s
529
13.32
530
13.35
13.37 Determine the temperature distribution and rate of heat flow through the plain carbon steel
W
mK
while the underside surface is held at 0oC. Assume that no heat is lost from the sides.
532
13.38
13.40 For the basement wall, determine the temperature distribution and the heat transfer through
534
13.41
13.44 The Allen Wrench, shown in Figure 1, is exposed at one end to at temperature of 300 K,
while the other end has a heat flux of 10
W
m^2
. Determine the temperature distribution
536
throughout the wrench. It has a thermal conductivity of 43.6
W
mK
and a specific heat
capacity of 0.000486
J
gK
.
Part dimensions
Figure 3: Temperature Distribution (K)
13.45 Temperature distribution
13.46 The thermal aspect of this component is that the base has an applied temperature of 100 °F.
xx
KA
L
=
2
(1.5 )
4 144
10 12
41
(0.017)
/
= 0.001
Btu
h °F
Small so neglect
Btu
hF
6
hPL
=
3
6
12
1
(1.5 )
10
4
12
1
= 0.04096
Btu
hF
(1.5)
12
10
4
12
Btu
h
[k(1)] =
2.4
2
11
11
+ 0.04096
21
12
[k(1)] =
1.118 1.24091
1.159 1.2818
0 1.159 1.2818 – 1.118 1.2409
0 0 1.159 1.2818
[K] =
0.164 1.241 0 0
1.159 0.164 1.241 0
0 1.159 0.164 1.241
0 0 1.159 1.282
49.087 1.159 (50)
49.087
49.087
24.5435
[K
2
4
5
64.719
77.715
89.747
100.296
t
t
t
t
1(10)(0.24)(50 F)=120 Btu/h
in
q mct
5(10)(0.24)(100.296) 240.7Btu / h
out
q mct