ch04
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ch04


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= 0. Note that
2h x 50W/m K 0.1m
3.333.
k 1.5 W/m K
D × ´= =
×
Node Finite-Difference Equations
 1 -4T1 + 2T2 + 2T4 = 0
 2 -4T2 + T1 + T3 + 2T5 = 0
 3 -4T3 + 200 + 2T6 + T2 = 0
 4 -4T4 + T1 + 2T5 + T7 = 0
 5 -4T5 + T2 + T6 + T8 + T4 = 0
 6 -4T6 + T5 + T3 + 200 + T9 = 0
 7 -4T7 + T4 + 2T8 + T10 = 0
 8 -4T8 + T7 +T5 + T9 + T11 = 0
 9 -4T9 + T8 + T6 + 200 + T12 = 0
 10 -4T10 + T7 + 2T11 + T13 = 0
 11 -4T11 + T10 + T8 + T12 + T14 = 0
 12 -4T12 + T11 + T9 +200 + T15 = 0
 13 2T10 + T14 + 6.666´30-10.666 T13 = 0
 14 2T11 + T13 + T15 + 6.666´30-2(3.333+2)T14 = 0
 15 2T12 +T14 + 200 + 6.666´30-2(3.333+2) T15 = 0
Using the matrix inversion method, Section 4.5.2, the above equations can be written in the form [A]
[T] = [C] where [A] and [C] are shown on the next page. Using a stock matrix inversion routine,
the temperatures [T] are determined.
Continued \u2026..
PROBLEM 4.60 (Cont.)
-4 2 0 2 0 0 0 0 0 0 0 0 0 0 0
1 -4 1 0 2 0 0 0 0 0 0 0 0 0 0
0 1 -4 0 0 -2 
[A] =
 0 0 0 0 0 0 0 0 0
1 0 0 -4 2 0 1 0 0 0 0 0 0 0 0
0 1 0 1 -4 1 0 1 0 0 0 0 0 0 0
0 0 1 0 1 -4 0 0 1 0 0 0 0 0 0
0 0 0 1 0 0 -4 2 0 1 0 0 0 0 0
0 0 0 0 1 0 1 -4 1 0 1 0 0 0 0
0 0 0 0 0 1 0 1 -4 0 0 1 0 0 0
0 0 0 0 0 0 1 0 0 -4 2 0 1 0 0
0 0 0 0 0 0 0 1 0 -1 -4 1 0 1 0
0 0 0 0 0 0 0 0 1 0 1 -4 0 0 1
0 0 0 0 0 0 0 0 0 2 0 0 -10.66 2 0
0 0 0 0 0 0 0 0 0 0 2 0 1 -10.66 1
0 0 0 0 0 0 0 0 0 0 0 2 0 1 -10.66
é ù
ê ú
ê ú
ê ú
ê ú
ê ú
ê ú
ê ú
ê ú
ê ú
ê ú
ê ú
ê ú
ê ú
ê úê úë û
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
T 0 153.9
T 0 159.7
T-200 176.4
T 0 148.0
T 0 154.4
T-200
T 0
[C] 0 [T] T
-200 T
 0 T
 0 T
-200 T
-200 T
-200 T
-400 T
é ùé ù
ê úê ú
ê úê ú
ê úê ú
ê úê ú
ê úê ú
ê úê ú
ê úê ú= = =ê úê ú
ê úê ú
ê úê ú
ê úê ú
ê úê ú
ê úê ú
ê úê úê ú ê úë û ë û
( )
172.9
129.4
137.0 C
160.7
95.6
103.5
132.8
45.8
48.7
67.0
é ù
ê ú
ê ú
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o
Considering symmetry, the heat transfer rate to the fluid is twice the convection rate from the surfaces of the control volumes
exposed to the fluid. Using Newton\u2019s law of cooling, considering a unit thickness of the bar, find
( ) ( ) ( ) ( )conv 13 14 15
y y
q 2 h T T h y T T h y T T h 200 T
2 2¥ ¥ ¥ ¥
D Dé ù= × × - + ×D × - + ×D - + × -ê úë û
( ) ( ) ( ) ( )conv 13 14 15
1 1
q 2h y T T T T T T 200 T
2 2¥ ¥ ¥ ¥
é ù= ×D - + - + - + -ê úë û
( ) ( ) ( ) ( )conv 2
W 1 1
q 2 50 0.1m 45.8 30 48.7 30 67.0 30 200 30
2 2m K
é ù= ´ ´ - + - + - + -ê úë û×
convq 1487 W/m.= <
PROBLEM 4.61
KNOWN: Upper surface and grooves of a plate are maintained at a uniform temperature T1, while the
lower surface is maintained at T2 or is exposed to a fluid at T\u221e.
FIND: (a) Heat rate per width of groove spacing (w) for isothermal top and bottom surfaces using a
finite-difference method with \u2206x = 40 mm, (b) Effect of grid spacing and convection at bottom surface.
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state, two-dimensional conduction, (2) Constant properties.
ANALYSIS: (a) Using a space increment of \u2206x = 40 mm, the symmetrical section shown in schematic
(a) corresponds to one-half the groove spacing. There exist only two interior nodes for which finite-
difference equations must be written.
Node a: ( )a 1 b 2 14T T T T T 0\u2212 + + + =
( )a b4T 200 T 20 200 0\u2212 + + + = or a b4T T 420\u2212 = (1)
Node b: ( )b 1 a 2 a4T T T T T 0\u2212 + + + =
( )b a4T 200 2T 20 0\u2212 + + = or a b2T 4T 220\u2212 + = (2)
Multiply Eq. (2) by 2 and subtract from Eq. (1) to obtain
7Tb = 860 or Tb = 122.9°C
From Eq. (1),
4Ta - 122.9 = 420 or Ta = (420 + 122.9)/4 = 135.7°C.
The heat transfer through the symmetrical section is equal to the sum of heat flows through control
volumes adjacent to the lower surface. From the schematic,
( ) a 2 b 21 21 2 3 T T T Tx T T xq q q q k k x k2 y y 2 y
\u2212 \u2212\u2206 \u2212 \u2206\uf8eb \uf8f6 \uf8eb \uf8f6
\u2032 \u2032 \u2032 \u2032= + + = + \u2206 +\uf8ec \uf8f7 \uf8ec \uf8f7\u2206 \u2206 \u2206\uf8ed \uf8f8 \uf8ed \uf8f8 .
Continued...
PROBLEM 4.61 (Cont.)
Noting that \u2206x = \u2206y, regrouping and substituting numerical values, find
( ) ( ) ( )1 2 a 2 b 21 1q k T T T T T T2 2
\uf8ee \uf8f9
\u2032 = \u2212 + \u2212 + \u2212\uf8ef \uf8fa\uf8f0 \uf8fb
( ) ( ) ( )1 1q 15 W m K 200 20 135.7 20 122.9 20 3.86 kW m
2 2
\uf8ee \uf8f9
\u2032 = \u22c5 \u2212 + \u2212 + \u2212 =\uf8ef \uf8fa\uf8f0 \uf8fb .
For the full groove spacing, totalq\u2032 = 2 × 3.86 kW/m = 7.72 kW/m. <
(b) Using the Finite-Difference Equations option from the Tools portion of the IHT menu, the following
two-dimensional temperature field was computed for the grid shown in schematic (b), where x and y are
in mm and the nodal temperatures are in °C. Nodes 2-54 are interior nodes, with those along the
symmetry adiabats characterized by Tm-1,n = Tm+1,n, while nodes 55-63 lie on a plane surface.
y\x 0 10 20 30 40 50 60 70 80
0 200 200 200 200 200
10 200 191 186.6 184.3 183.1 182.8
20 200 186.7 177.2 171.2 167.5 165.5 164.8
30 200 182.4 169.5 160.1 153.4 149.0 146.4 145.5
40 200 175.4 160.3 148.9 140.1 133.5 128.7 125.7 124.4
50 141.4 134.3 125.7 118.0 111.6 106.7 103.1 100.9 100.1
60 97.09 94.62 90.27 85.73 81.73 78.51 76.17 74.73 74.24
70 57.69 56.83 55.01 52.95 51.04 49.46 48.31 47.60 47.36
80 20 20 20 20 20 20 20 20 20
The foregoing results were computed for h = 107 W/m2\u22c5K (h \u2192 \u221e) and T\u221e = 20°C, which is tantamount
to prescribing an isothermal bottom surface at 20°C. Agreement between corresponding results for the
coarse and fine grids is surprisingly good (Ta = 135.7°C \u2194 T23 = 140.1°C; Tb = 122.9°C \u2194 T27 =
124.4°C). The heat rate is
( ) ( ) ( ) ( ) ( )46 55 47 56 48 57 49 58 50 59q 2 k T T 2 T T T T T T T T\u2032 \uf8ee= × \u2212 + \u2212 + \u2212 + \u2212 + \u2212\uf8f0
( ) ( ) ( ) ( )51 60 52 61 53 62 54 63T T T T T T T T 2\uf8f9+ \u2212 + \u2212 + \u2212 + \u2212 \uf8fb
[q 2 15 W m K 18.84 36.82 35.00 32.95 31.04 29.46\u2032 = × \u22c5 + + + + + <
]28.31 27.6 13.68 C 7.61kW m+ + + =$
The agreement with 
„
q
 = 7.72 kW/m from the coarse grid of part (a) is excellent and a fortuitous
consequence of compensating errors. With reductions in the convection coefficient from h \u2192 \u221e to h =
1000, 200 and 5 W/m2\u22c5K, the corresponding increase in the thermal resistance reduces the heat rate to
values of 6.03, 3.28 and 0.14 kW/m, respectively. With decreasing h, there is an overall increase in
nodal temperatures, as, for example, from 191°C to 199.8°C for T2 and from 20°C to 196.9°C for T55.
NOTE TO INSTRUCTOR:
 To reduce computational effort, while achieving the same educational
objectives, the problem statement has been changed to allow for convection at the bottom, rather than the
top, surface.
PROBLEM 4.62
KNOWN: Rectangular plate subjected to uniform temperature boundaries.
FIND: Temperature at the midpoint using a finite-difference method with space increment of 0.25m
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) Two-dimensional conduction, (3) Constant
properties.
ANALYSIS: For the nodal network above, 12 finite-difference equations must be written. It follows
that node 8 represents the midpoint of the rectangle. Since all nodes are interior nodes, Eq. 4.33 is
appropriate and is written in the form
m neighbors4T