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ch05


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using Fig. D.1, with
( )
4 2
-1
2 2
-3
hL t 1.064 10 m / s 8s
Bi or Bi 0 and Fo 0.34,
k L 50 10 m
\u3b1 \u2212× ×
= \u2192\u221e \u2192 = = =
×
find ( ) ( )
i wall
0,8s
P 0,8s 0.5.
\u3b8
\u3b8
= \u2248
\uf8f9\uf8fa\uf8fb
(3)
Combining Eqs. (2) and (3) with Eq. (1), find ( ) ( )
i i
0,0,8s T 0,0,8s T
1 0.5 0.5
T T
\u3b8
\u3b8
\u221e
\u221e
\u2212
= \u2248 × =
\u2212
( ) ( ) ( )iT 0,0,8s T 0.5 T T 500 0.5 20 500 260 C.\u221e \u221e= + \u2212 = + \u2212 = $ <
(b) The parameters controlling transient conduction with convective boundary conditions are the Biot
and Fourier numbers. Since Bi << 0.1 for the cylindrical shape, we can assume radial gradients are
negligible. That is, we need only consider conduction in the x-direction.
PROBLEM 5.90
KNOWN: Cylindrical-shaped meat roast weighing 2.25 kg, initially at 6°C, is placed in an
oven and subjected to convection heating with prescribed (T\u221e,h).
FIND: Time required for the center to reach a done temperature of 80°C.
SCHEMATIC:
ASSUMPTIONS: (1) Two-dimensional conduction in x and r directions, (2) Uniform and
constant properties, (3) Properties approximated as those of water.
PROPERTIES: Table A-6, Water, liquid ( )( )T 80 6 C/2 315K := + \u2248$ \ufffd \u3c1 = 1/vf = 1/1.009 ×
10-3m3/kg = 991.1 kg/m3, cp,f = 4179 J/kg\u22c5K, k = 0.634 W/m\u22c5K, \u3b1 = k/\u3c1c = 1.531 × 10
-7
m
2/s.
ANALYSIS: The dimensions of the roast are determined from the requirement ro = L and
knowledge of its weight and density,
1/31/3
2
o o 3
M 2.25 kgM V 2L r or r L 0.0712m.
2 2 991.1 kg/m
\u3c1 \u3c1 \u3c0
\u3c0\u3c1 \u3c0
\uf8ee \uf8f9\uf8ee \uf8f9
= = \u22c5 \u22c5 = = = =\uf8ef \uf8fa\uf8ef \uf8fa\uf8f0 \uf8fb \uf8ef \uf8fa\uf8f0 \uf8fb
(1)
The roast corresponds to Case (i), Figure 5.11, and the temperature distribution may be
expressed as the product of one-dimensional solutions, ( ) ( ) ( )
i
T x,r,t T
P x,t C r,t ,
T T
\u221e
\u221e
\u2212
= ×
\u2212
 where
P(x,t) and C(r,t) are defined by Eqs. 5.65 and 5.66, respectively. For the center of the
cylinder,
( ) ( )
( )i
T 0,0,t T 80 175 C
0.56.
T T 6 175 C
\u221e
\u221e
\u2212 \u2212
= =
\u2212
\u2212
$
$
(2)
In terms of the product solutions,
( ) ( ) ( )
i i iwall cylinder
T 0,0,t T T 0,t T T 0,t T
0.56
T T T T T T
\u221e \u221e \u221e
\u221e \u221e \u221e
\uf8f9 \uf8f9\u2212 \u2212 \u2212
= = ×\uf8fa \uf8fa
\u2212 \u2212 \u2212\uf8fb \uf8fb
(3)
For each of these shapes, we need to find values of \u3b8o/\u3b8i such that their product satisfies Eq.
(3). For both shapes,
2
-1oh r hL 15 W/m K 0.0712mBi 1.68 or Bi 0.6
k k 0.634 W/m K
\u22c5 ×
= = = = \u2248
\u22c5
( )22 2 7 2 5oFo t/r t/L 1.53 10 m / s t/ 0.0712m 3.020 10 t.\u3b1 \u3b1 \u2212 \u2212= = = × × = ×
Continued \u2026..
PROBLEM 5.90 (Cont.)
A trial-and-error solution is necessary. Begin by assuming a value of Fo; obtain the respective
\u3b8o/\u3b8i values from Figs. D.1 and D.4; test whether their product satisfies Eq. (3). Two trials
are shown as follows:
Trial Fo t(hrs) )o i wall/\u3b8 \u3b8 )o i cyl/\u3b8 \u3b8 o oi iw cyl
\u3b8 \u3b8
\u3b8 \u3b8
\uf8f9 \uf8f9
×\uf8fa \uf8fa\uf8fb \uf8fb
 1 0.4 3.68 0.72 0.50 0.36
 2 0.3 2.75 0.78 0.68 0.53
For Trial 2, the product of 0.53 agrees closely with the value of 0.56 from Eq. (2). Hence, it
will take approximately 2 ¾ hours to roast the meat.
PROBLEM 5.91
KNOWN: A long alumina rod, initially at a uniform temperature of 850K, is suddenly
exposed to a cooler fluid.
FIND: Temperature of the rod after 30s, at an exposed end, T(0,0,t), and at an axial distance
6mm from the end, T(0, 6mm, t).
SCHEMATIC:
ASSUMPTIONS: (1) Two-dimensional conduction in (r,x) directions, (2) Constant
properties, (3) Convection coefficient is same on end and cylindrical surfaces.
PROPERTIES: Table A-2, Alumina, polycrystalline aluminum oxide (assume
( )T 850 600 K/2 725K):\u2248 + = \u3c1 = 3970 kg/m3, c = 1154 J/kg\u22c5K, k = 12.4 W/m\u22c5K.
ANALYSIS: First, check if system behaves as a lumped capacitance. Find
( ) ( )oc h r / 2 500 W/m K 0.010m/2hLBi 0.202.
k k 12.4 W/m K
\u22c5
= = = =
\u22c5
Since Bi > 0.1, rod does not behave as spacewise isothermal object. Hence, treat rod as a
semi-infinite cylinder, the multi-dimensional system Case (f), Fig. 5.11.
The product solution can be written as
( ) ( ) ( ) ( ) ( ) ( )
i i i
r,x,t r,t x,t
r,x,t C r , t S x , t
\u3b8 \u3b8 \u3b8
\u3b8
\u3b8 \u3b8 \u3b8
\u2217 \u2217 \u2217 \u2217 \u2217
= = × = ×
Infinite cylinder, C(r*,t*). Using the Heisler charts with r* = r = 0 and
11 2
-1 oh r 500 W/m K 0.01mBi 2.48.
k 12.4 W/m K
\u2212
\u2212 \uf8ee \uf8f9
\u22c5 ×\uf8ee \uf8f9
= = =\uf8ef \uf8fa\uf8ef \uf8fa
\u22c5\uf8f0 \uf8fb \uf8ef \uf8fa\uf8f0 \uf8fb
Evaluate \u3b1 = k/\u3c1c = 2.71 × 10-6m2/s, find 2 6 2oFo t/r 2.71 10 m / s\u3b1 \u2212= = × × 30s/(0.01m)2 =
0.812. From the Heisler chart, Fig. D.4, with Bi-1 = 2.48 and Fo = 0.812, read C(0,t*) =
\u3b8(0,t)/\u3b8i = 0.61.
Continued \u2026..
PROBLEM 5.91 (Cont.)
Semi-infinite medium, S(x*,t*). Recognize this as Case (3), Fig. 5.7. From Eq. 5.60, note that
the LHS needs to be transformed as follows,
( )i
i i i
T T T T T T1 S x,t .
T T T T T T
\u221e \u221e
\u221e \u221e \u221e
\u2212 \u2212 \u2212
= \u2212 =
\u2212 \u2212 \u2212
Thus,
( ) ( ) ( )
( )1/ 22
1/ 2 2 1/ 2
h tx hx h t xS x,t 1 erfc exp erfc .
k kk2 t 2 t
\u3b1\u3b1
\u3b1 \u3b1
\uf8f1 \uf8fc\uf8ee \uf8f9\uf8ee \uf8f9 \uf8ee \uf8f9\uf8ee \uf8f9\uf8ee \uf8f9\uf8f4 \uf8f4\uf8ef \uf8fa\uf8ef \uf8fa \uf8ef \uf8fa\uf8ef \uf8fa= \u2212 \u2212 + +\uf8ef \uf8fa\uf8f2 \uf8fd\uf8ef \uf8fa\uf8ef \uf8fa \uf8ef \uf8fa\uf8ef \uf8fa\uf8ef \uf8fa\uf8f4 \uf8f4\uf8f0 \uf8fb\uf8f0 \uf8fb\uf8f0 \uf8fb \uf8f0 \uf8fb\uf8f0 \uf8fb\uf8f3 \uf8fe
Evaluating this expression at the surface (x = 0) and 6mm from the exposed end, find
( ) ( ) ( )( )
22 6 2
2
500 W/m K 2.71 10 m / s 30s
S 0,30s 1 erfc 0 exp 0
12.4 W/m K
\u2212
\uf8f1 \uf8ee \uf8f9\uf8ee \uf8f9
\u22c5 × ×\uf8f4 \uf8ef \uf8fa\uf8ef \uf8fa\uf8f4 \uf8ef \uf8fa= \u2212 \u2212 +\uf8ef \uf8fa\uf8f2 \uf8ef \uf8fa\uf8ef \uf8fa\u22c5\uf8f4 \uf8ef \uf8fa\uf8ef \uf8fa\uf8f4 \uf8f0 \uf8fb\uf8f0 \uf8fb\uf8f3
( )1/ 22 -6 2500 W/m K 2.71 10 m / s 30s
erfc 0
12.4 W/m K
\uf8fc\uf8ee \uf8f9\uf8ee \uf8f9
\u22c5 × × \uf8f4\uf8ef \uf8fa\uf8ef \uf8fa \uf8f4\uf8ef \uf8fa+\uf8ef \uf8fa \uf8fd
\u22c5\uf8ef \uf8fa\uf8ef \uf8fa \uf8f4\uf8ef \uf8fa\uf8ef \uf8fa \uf8f4\uf8f0 \uf8fb\uf8f0 \uf8fb\uf8fe
( ) ( ) ( ){ }S 0,30s 1 1 exp 0.1322 erfc 0.3636 0.693.\uf8ee \uf8f9 \uf8ee \uf8f9= \u2212 \u2212 =\uf8f0 \uf8fb \uf8f0 \uf8fb
Note that Table B.2 was used to evaluate the complementary error function, erfc(w).
( ) ( )1/ 2-6 2
0.006mS 6mm,30s 1 erfc
2 2.71 10 m / s 30s
\uf8f1 \uf8ee \uf8f9\uf8f4 \uf8ef \uf8fa\uf8f4
= \u2212 \u2212\uf8ef \uf8fa\uf8f2 \uf8ef \uf8fa\uf8f4 × ×\uf8ef \uf8fa\uf8f4 \uf8f0 \uf8fb\uf8f3
( )
2500 W/m K 0.006m
exp 0.1322 erfc 0.3327 0.3636 0.835.
12.4 W/m K
\uf8fc\uf8ee \uf8f9\uf8ee \uf8f9
\u22c5 × \uf8f4\uf8ef \uf8fa \uf8ee \uf8f9+ + =\uf8ef \uf8fa \uf8fd\uf8f0 \uf8fb
\u22c5\uf8ef \uf8fa\uf8ef \uf8fa \uf8f4\uf8f0 \uf8fb\uf8f0 \uf8fb \uf8fe
The product solution can now be evaluated for each location. At (0,0),
( ) ( ) ( ) ( )
i
T 0,0,30s T
0,0, t C 0,t S 0,t 0.61 0.693 0.423.
T T
\u3b8 \u221e\u2217 \u2217 \u2217
\u221e
\u2212
= = × = × =
\u2212
Hence, ( ) ( ) ( )iT 0,0,30s T 0.423 T T 350K 0.423 850 350 K 561K.\u221e \u221e= + \u2212 = + \u2212 = <
At (0,6mm),
( ) ( ) ( )0,6mm,t C 0,t S 6mm,t 0.61 0.835 0.509\u3b8\u2217 \u2217 \u2217= × = × =
( )T 0,6mm,30s 604K.= <
COMMENTS: Note that the temperature at which the properties were evaluated was a good
estimate.
PROBLEM 5.92
KNOWN: Stainless steel cylinder of Ex. 5.7, 80-mm diameter by 60-mm length, initially at 600 K,
suddenly quenched in an oil bath at 300 K with h = 500 W/m2\u22c5K. Use the Transient Conduction,
Plane Wall and Cylinder models of IHT to obtain the following solutions.
FIND: (a) Calculate the temperatures T(r,x,t) after 3 min: at the cylinder center, T(0, 0, 3 min), at the
center of a circular face, T(0, L, 3 min), and at the midheight of the side, T(ro, 0, 3 min); compare your
results with those in the example; (b) Calculate and plot temperature histories at the cylinder center,
T(0, 0, t), the mid-height of the side, T(ro, 0, t), for 0 \u2264 t \u2264 10 min; comment on the gradients and what
effect they might have on phase transformations and thermal stresses; and (c) For 0 \u2264 t \u2264 10 min,
calculate and plot the temperature histories at the cylinder center, T(0, 0, t), for convection coefficients
of 500 and 1000 W/m2\u22c5K.
SCHEMATIC:
ASSUMPTIONS: (1) Two-dimensional conduction in r- and x-coordinates, (2) Constant properties.
PROPERTIES: Stainless steel (Example 5.7): \u3c1 = 7900 kg/m3, c = 526 J/kg\u22c5K, k = 17.4 W/m\u22c5K.
ANALYSIS: The following results were obtained using the Transient Conduction models for the
Plane Wall and Cylinder of IHT. Salient portions of the code are provided in the Comments.
(a) Following the methodology for a product solution outlined in Example 5.7, the following results
were obtained at t = to = 3 min
(r, x, t) P(x, t) C(r, t) T(r, x, t)-IHT T(r, x, t)-Ex
 (K) (K)
0, 0, to 0.6357 0.5388 402.7 405
0, L, to 0.4365 0.5388 370.5