ch05
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ch05


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[A]
and RHS matrix [C] are:
Recall that the problem asks for the time required to reach
95% of the difference for steady-state conditions. This
provides information on approximately how long it takes
for the plate to come to a steady operating condition. If
you worked Problem 4.71, you know the steady-state
temperature distribution. Then you can proceed to find the
p
mT values with increasing time until the first node reaches
the required limit. We should not expect the nodes to reach their limit at the same time.
Not knowing the steady-state temperature distribution, use the implicit FDE in matrix form
above to step through time \u2192 \u221e to the steady-state solution; that is, proceed to p \u2192
10,20\u2026100 until the solution matrix [T] does not change. The results of the analysis are
tabulated below. Column 1 labeled Tm(\u221e) is the steady-state distribution. Column 2,
Tm(95%), is the 95% limit being sought as per the graph directly above. The third column is
the temperature distribution at t = to = 248s, Tm(248s); at this elapsed time, Node 1 has
reached its limit. Can you explain why this node was the first to reach this limit? Which
nodes will be the last to reach their limits?
Tm(\u221e) Tm(95%) Tm(248s) <
55.80 54.51 54.51
49.93 48.93 48.64
47.67 46.78 46.38
59.03 57.58 57.64
51.72 50.63 50.32
49.19 48.23 47.79
63.89 62.20 62.42
52.98 51.83 51.52
50.14 49.13 48.68
62.84 61.20 61.35
53.35 52.18 51.86
50.46 49.43 48.98
PROBLEM 5.127
KNOWN: Nodal network and operating conditions for a water-cooled plate.
FIND: Transient temperature response.
SCHEMATIC:
ASSUMPTIONS: (1) Steady-sate conditions, (2) Two-dimensional conduction.
ANALYSIS: The energy balance method must be applied to each nodal region. Grouping
similar regions, the following results are obtained.
Nodes 1 and 5:
p+1 p+1 p+1 p
1 2 6 12 2 2 2
2 t 2 t 2 t 2 t1 T T T T
x y x y
\u3b1 \u3b1 \u3b1 \u3b1\uf8eb \uf8f6\u2206 \u2206 \u2206 \u2206
+ + \u2212 \u2212 =\uf8ec \uf8f7\uf8ec \uf8f7\u2206 \u2206 \u2206 \u2206\uf8ed \uf8f8
p+1 p+1 p+1 p
5 54 102 2 2 2
2 t 2 t 2 t 2 t1 T T T T
x y x y
\u3b1 \u3b1 \u3b1 \u3b1\uf8eb \uf8f6\u2206 \u2206 \u2206 \u2206
+ + \u2212 \u2212 =\uf8ec \uf8f7\uf8ec \uf8f7\u2206 \u2206 \u2206 \u2206\uf8ed \uf8f8
Nodes 2, 3, 4:
p+1 p+1 p+1 p+1 p
m,n m,nm-1,n m+1,n m,n-12 2 2 2 2
2 t 2 t t t 2 t1 T T T T T
x y x x y
\u3b1 \u3b1 \u3b1 \u3b1 \u3b1\uf8eb \uf8f6\u2206 \u2206 \u2206 \u2206 \u2206
+ + \u2212 \u2212 \u2212 =\uf8ec \uf8f7\uf8ec \uf8f7\u2206 \u2206 \u2206 \u2206 \u2206\uf8ed \uf8f8
Nodes 6 and 14:
p+1 p+1 p+1 p
76 1 62 2 2 2
2 t 2 t 2h t 2 t 2 t 2h t1 T T T T +T
k y k yx y y x
\u3b1 \u3b1 \u3b1 \u3b1 \u3b1 \u3b1
\u221e
\uf8eb \uf8f6\u2206 \u2206 \u2206 \u2206 \u2206 \u2206
+ + + \u2212 \u2212 =\uf8ec \uf8f7\uf8ec \uf8f7\u2206 \u2206\u2206 \u2206 \u2206 \u2206\uf8ed \uf8f8
p+1 p+1 p+1 p
14 15 19 142 2 2 2
2 t 2 t 2h t 2 t 2 t 2h t1 T T T T +T
k y k yx y x y
\u3b1 \u3b1 \u3b1 \u3b1 \u3b1 \u3b1
\u221e
\uf8eb \uf8f6\u2206 \u2206 \u2206 \u2206 \u2206 \u2206
+ + + \u2212 \u2212 =\uf8ec \uf8f7\uf8ec \uf8f7\u2206 \u2206\u2206 \u2206 \u2206 \u2206\uf8ed \uf8f8
Continued \u2026..
PROBLEM 5.127 (Cont.)
Nodes 7 and 15:
p+1 p+1 p+1 p+1 p
7 72 6 82 2 2 2 2
2 t 2 t 2h t 2 t t t 2h t
1 T T T T T +T
k y k yx y y x k x
\u3b1 \u3b1 \u3b1 \u3b1 \u3b1 \u3b1 \u3b1
\u221e
\u2206 \u2206 \u2206 \u2206 \u2206 \u2206 \u2206
+ + + \u2212 \u2212 \u2212 =
\u2206 \u2206\u2206 \u2206 \u2206 \u2206 \u2206
\uf8eb \uf8f6\uf8ec \uf8f7\uf8ec \uf8f7\uf8ed \uf8f8
p+1 p+1 p+1 p+1 p
15 14 16 20 152 2 2 2 2
2 t 2 t 2h t t t 2 t 2h t
1 T T T T T +T
k y k yx y x x y
\u3b1 \u3b1 \u3b1 \u3b1 \u3b1 \u3b1 \u3b1
\u221e
\u2206 \u2206 \u2206 \u2206 \u2206 \u2206 \u2206
+ + + \u2212 \u2212 \u2212 =
\u2206 \u2206\u2206 \u2206 \u2206 \u2206 \u2206
\uf8eb \uf8f6\uf8ec \uf8f7\uf8ec \uf8f7\uf8ed \uf8f8
Nodes 8 and 16:
p+1 p+1 p+1
78 32 2 2 2
p+1 p+1 p
9 11 82 2
2 t 2 t h t t t1 T T T
k xx y y x
4 t 2 t 2 h t 1 1
 T T T T
3 3 3 k x yx y
2 2 h t 4 2
3 3 k y 3 3
\u3b1 \u3b1 \u3b1 \u3b1 \u3b1
\u3b1 \u3b1 \u3b1
\u3b1
\u221e
\uf8eb \uf8f6\u2206 \u2206 \u2206 \u2206 \u2206
+ + + + \u2212 \u2212\uf8ec \uf8f7\uf8ec \uf8f7\u2206\u2206 \u2206 \u2206 \u2206\uf8ed \uf8f8 \uf8eb \uf8f6\u2206 \u2206 \u2206
\u2212 \u2212 = + +\uf8ec \uf8f7\u2206 \u2206\uf8ed \uf8f8\u2206 \u2206
\u2206
\u2206
p+1 p+1 p+1
16 11 152 2 2 2
p+1 p+1 p
17 21 162 2
2 t 2 t h t t t1 T T T
k xx y y x
4 t 4 t 2 h t 1 1
 T T T T
3 3 3 k x yx y
2 2 h t 2 2
3 3 k y 3 3
\u3b1 \u3b1 \u3b1 \u3b1 \u3b1
\u3b1 \u3b1 \u3b1
\u3b1
\u221e
\uf8eb \uf8f6\u2206 \u2206 \u2206 \u2206 \u2206
+ + + + + \u2212 \u2212\uf8ec \uf8f7\uf8ec \uf8f7\u2206\u2206 \u2206 \u2206 \u2206\uf8ed \uf8f8 \uf8eb \uf8f6\u2206 \u2206 \u2206
\u2212 \u2212 = + +\uf8ec \uf8f7\u2206 \u2206\uf8ed \uf8f8\u2206 \u2206
\u2206
\u2206
Node 11:
p+1 p+1 p+1 p+1 p
11 8 12 16 112 2 2 2 2
2 t 2 t 2h t t t t 2h t
1 T T 2 T T T +T
k x k xx y y x y
\u3b1 \u3b1 \u3b1 \u3b1 \u3b1 \u3b1
\u3b1
\u221e
\u2206 \u2206 \u2206 \u2206 \u2206 \u2206 \u2206
+ + + \u2212 \u2212 \u2212 =
\u2206 \u2206\u2206 \u2206 \u2206 \u2206 \u2206
\uf8eb \uf8f6\uf8ec \uf8f7\uf8ec \uf8f7\uf8ed \uf8f8
Nodes 9, 12, 17, 20, 21, 22:
( ) ( )p+1 p+1 p+1 p+1 p+1 pm,n m,nm,n+1 m,n-1 m-1,n m+1,n2 2 2 22 t 2 t t t1 T T T T T Tx y y x\u3b1 \u3b1 \u3b1 \u3b1\uf8eb \uf8f6\u2206 \u2206 \u2206 \u2206+ + \u2212 + \u2212 + =\uf8ec \uf8f7\uf8ec \uf8f7\u2206 \u2206 \u2206 \u2206\uf8ed \uf8f8
Nodes 10, 13, 18, 23:
( )p+1 p+1 p+1 p+1 pm,n m,nm,n+1 m,n-1 m-1,n2 2 2 22 t 2 t t 2 t1 T T T T Tx y y x\u3b1 \u3b1 \u3b1 \u3b1\uf8eb \uf8f6\u2206 \u2206 \u2206 \u2206+ + \u2212 + \u2212 =\uf8ec \uf8f7\uf8ec \uf8f7\u2206 \u2206 \u2206 \u2206\uf8ed \uf8f8
Node 19:
( )p+1 p+1 p+1 p+1 p19 14 24 20 192 2 2 22 t 2 t t 2 t1 T T T T Tx y y x\u3b1 \u3b1 \u3b1 \u3b1\uf8eb \uf8f6\u2206 \u2206 \u2206 \u2206+ + \u2212 + \u2212 =\uf8ec \uf8f7\uf8ec \uf8f7\u2206 \u2206 \u2206 \u2206\uf8ed \uf8f8
Nodes 24, 28:
p+1 p+1 p+1 po
24 19 25 242 2 2 2
2q t2 t 2 t 2 t 2 t1 T T T +T
k yx y y x
\u3b1\u3b1 \u3b1 \u3b1 \u3b1\uf8eb \uf8f6 \u2032\u2032 \u2206\u2206 \u2206 \u2206 \u2206
+ + \u2212 \u2212 =\uf8ec \uf8f7\uf8ec \uf8f7 \u2206\u2206 \u2206 \u2206 \u2206\uf8ed \uf8f8
p+1 p+1 p+1 po
28 23 27 282 2 2 2
2q t2 t 2 t 2 t 2 t1 T T T +T
k yx y y x
\u3b1\u3b1 \u3b1 \u3b1 \u3b1\uf8eb \uf8f6 \u2032\u2032 \u2206\u2206 \u2206 \u2206 \u2206
+ + \u2212 \u2212 =\uf8ec \uf8f7\uf8ec \uf8f7 \u2206\u2206 \u2206 \u2206 \u2206\uf8ed \uf8f8
Continued \u2026..
PROBLEM 5.127 (Cont.)
Nodes 25, 26, 27:
( )p+1 p+1 p+1 p+1 p+1om,n m,nm,n+1 m-1,n m+1,n2 2 2 2 2q t2 t 2 t 2 t t1 T T T T +Tk yx y y x \u3b1\u3b1 \u3b1 \u3b1 \u3b1\uf8eb \uf8f6 \u2032\u2032 \u2206\u2206 \u2206 \u2206 \u2206+ + \u2212 \u2212 + =\uf8ec \uf8f7\uf8ec \uf8f7 \u2206\u2206 \u2206 \u2206 \u2206\uf8ed \uf8f8
The convection heat rate is
( )( ) ( ) ( )( ) ( ) ([
)( ) ( ) ( )( )conv 6 7 8 1116 15 14 out.
q h x/2 T T x T T x y T T / 2 y T T x
y T T / 2 x T T x/2 T T q
\u221e \u221e \u221e \u221e
\u221e \u221e \u221e
\u2032 = \u2206 \u2212 + \u2206 \u2212 + \u2206 + \u2206 \u2212 + \u2206 \u2212 + \u2206
+\u2206 \u2212 + \u2206 \u2212 + \u2206 \u2212 =
The heat input is
( )in oq q 4 x\u2032 \u2032\u2032= \u2206
and, on a percentage basis, the ratio is
( )conv inn q / q 100.\u2032 \u2032\u2261 ×
Results of the calculations (in °C) are as follows:
Time: 5.00 sec; n = 60.57% Time: 10.00 sec; n = 85.80%
19.612 19.712 19.974 20.206 20.292 22.269 22.394 22.723 23.025 23.137
19.446 19.597 20.105 20.490 20.609 21.981 22.167 22.791 23.302 23.461
 21.370 21.647 21.730 24.143 24.548 24.673
24.217 24.074 23.558 23.494 23.483 27.216 27.075 26.569 26.583 26.598
25.658 25.608 25.485 25.417 25.396 28.898 28.851 28.738 28.690 28.677
27.581 27.554 27.493 27.446 27.429 30.901 30.877 30.823 30.786 30.773
Time: 15.0 sec; n = 94.89% Time: 20.00 sec; n = 98.16%
23.228 23.363 23.716 24.042 24.165 23.574 23.712 24.073 24.409 24.535
22.896 23.096 23.761 24.317 24.491 23.226 23.430 24.110 24.682 24.861
 25.142 25.594 25.733 25.502 25.970 26.115
28.294 28.155 27.652 27.694 27.719 28.682 28.543 28.042 28.094 28.122
30.063 30.018 29.908 29.867 29.857 30.483 30.438 30.330 30.291 30.282
32.095 32.072 32.021 31.987 31.976 32.525 32.502 32.452 32.419 32.409
Time: 23.00 sec; n = 99.00%
23.663 23.802 24.165 24.503 24.630
23.311 23.516 24.200 24.776 24.957
 25.595 26.067 26.214
28.782 28.644 28.143 28.198 28.226
30.591 30.546 30.438 30.400 30.392
32.636 32.613 32.563 32.531 32.520
COMMENTS: Temperatures at t = 23 s are everywhere within 0.13°C of the final steady-
state values.
PROBLEM 5.128
KNOWN: Cubic-shaped furnace, with prescribed operating temperature and convection heat transfer
on the exterior surfaces.
FIND: Time required for the furnace to cool to a safe working temperature corresponding to an inner
wall temperature of 35°C considering convection cooling on (a) the exterior surfaces and (b) on both
the exterior and interior surfaces.
SCHEMATIC:
ASSUMPTIONS: (1) Two-dimensional conduction through the furnace walls and (2) Constant
properties.
ANALYSIS: Assuming two-dimensional conduction through the walls and taking advantage of
symmetry for the cubical shape, the analysis considers the quarter section shown in the schematic
above. For part (a), with no cooling on the interior during the cool-down process, the inner surface
boundary condition is adiabatic. For part (b), with cooling on both the exterior and interior, the
boundary conditions are prescribed by the convection process. The boundaries through the centerline
of the wall and the diagonal through the corner are symmetry