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# ch05

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```T 2FoBi T T+ + \u221e+ + \u2212 = +
The general form of the finite-difference equation for any interior node (1 to 9) is given by Eq. (5.89),
( ) ( )p 1 p 1 p 1 pm mm 1 m 11 2 Fo T Fo T T T+ + +\u2212 ++ \u2212 + =
The finite-difference equation for the midplane node may be obtained by applying the symmetry
requirement to Eq. (5.89); that is, p pm 1 m 1T T .+ \u2212= Hence,
( ) p 1 p 1 p0 1 01 2 Fo T 2 Fo T T+ ++ \u2212 =
For the prescribed conditions, Bi = h\u2206x/k = 5000 W/m2\u22c5K (0.010m)/30 W/m\u22c5K = 1.67. If the explicit
method were used, the stability requirement would be given by Eq. (5.79). Hence, for Fo(1 + Bi) \u2264
0.5, Fo \u2264 0.187. With Fo = \u3b1\u2206t/\u2206x2 and \u3b1 = k/\u3c1c = 5.49 × 10-6 m2/s, the corresponding restriction on
the time increment would be \u2206t \u2264 3.40s. Although no such restriction applies for the implicit method,
a value of \u2206t = 1s is chosen, and the set of 11 finite-difference equations is solved using the Tools
option designated as Finite-Difference Equations, One-Dimensional and Transient from the IHT
Toolpad. For T10 (t) = 300°C, the solution yields
t 161s= <
Continued \u2026..
PROBLEM 5.105 (Cont.)
( )0T t 1364 C= ° <
With a casting speed of V = 15 mm/s, the length of the cooling section is
( )csL Vt 0.015m / s 161s 2.42m= = = <
COMMENTS: (1) With Fo = \u3b1t/L2 = 0.088 < 0.2, the one-term approximation to the exact solution
for one-dimensional conduction in a plane wall cannot be used to confirm the foregoing results.
However, using the exact solution from the Models, Transient Conduction, Plane Wall Option of IHT,
values of T0 = 1366°C and Ts = 200.7°C are obtained and are in good agreement with the finite-
difference predictions. The accuracy of these predictions could still be improved by reducing the
value of \u2206x.
(2) Temperature histories for the surface and midplane nodes are plotted for 0 < t < 600s.
While T10 (600s) = 124°C, To (600s) has only dropped to 879°C. The much slower thermal
response at the midplane is attributable to the small value of \u3b1 and the large value of Bi =
16.67.
0 1 0 0 2 0 0 3 0 0 4 0 0 5 0 0 6 0 0
Tim e (s )
1 0 0
3 0 0
5 0 0
7 0 0
9 0 0
1 1 0 0
1 3 0 0
1 5 0 0
Te
m
pe
ra
tu
re
(C
)
M id p la n e
C o o le d s u rfa ce
PROBLEM 5.106
KNOWN: Very thick plate, initially at a uniform temperature, Ti, is suddenly exposed to a
convection cooling process (T¥,h).
FIND: Temperatures at the surface and a 45mm depth after 3 minutes using finite-difference
method with space and time increments of 15mm and 18s.
SCHEMATIC:
ASSUMPTIONS: (1) One-dimensional transient conduction, (2) Plate approximates semi-infinite
medium, (3) Constant properties.
ANALYSIS: The grid network representing the plate is shown above. The finite-difference
equation for node 0 is given by Eq. 5.82 for one-dimensional conditions or Eq. 5.77,
( ) ( )p+1 p p0 1 0T 2 Fo T Bi T 1 2 Fo 2 Bi Fo T .¥= + × + - - × (1)
The numerical values of Fo and Bi are
( )
( )
6 2
2 2
2 -3
t 5.6 10 m /s 18s
Fo 0.448
x 0.015m
100 W/m K 15 10 mh x
Bi 0.075.
k 20 W/m K
a -D ´ ´
= = =
D
× ´ ´D
= = =
×
Recognizing that T¥ = 15°C, Eq. (1) has the form
p+1 p p
0 0 1T 0.0359 T 0.897 T 1.01.= + + (2)
It is important to satisfy the stability criterion, Fo (1+Bi) £ 1/2. Substituting values,
0.448 (1+0.075) = 0.482 £ 1/2, and the criterion is satisfied.
The finite-difference equation for the interior nodes, m = 1, 2\u2026, follows from Eq. 5.73,
( ) ( )p+1 p p pm mm+1 m-1T Fo T T 1 2Fo T .= + + - (3)
Recognizing that the stability criterion, Fo £ 1/2, is satisfied with Fo = 0.448,
( )p+1 p p pm mm+1 m-1T 0.448 T T 0.104T .= + + (4)
Continued \u2026..
PROBLEM 5.106 (Cont.)
The time scale is related to p, the number of steps in the calculation procedure, and Dt, the time
increment,
t p t.= D (5)
The finite-difference calculations can now be performed using Eqs. (2) and (4). The results are
tabulated below.
p t(s) T0 T1 T2 T3 T4 T5 T6 T7(K)
0 0 325 325 325 325 325 325 325 325
1 18 304.2 324.7 325 325 325 325 325 325
2 36 303.2 315.3 324.5 325 325 325 325 325
3 54 294.7 313.7 320.3 324.5 325 325 325 325
4 72 293.0 307.8 318.9 322.5 324.5 325 325 325
5 90 287.6 305.8 315.2 321.5 323.5 324.5 325 325
6 108 285.6 301.6 313.5 319.3 322.7 324.0 324.5 325
7 126 281.8 299.5 310.5 317.9 321.4 323.3 324.2
8 144 279.8 296.2 308.6 315.8 320.4 322.5
9 162 276.7 294.1 306.0 314.3 319.0
10 180 274.8 291.3 304.1 312.4
Hence, find
( ) ( )10 100 3T 0, 180s T 275 C T 45mm, 180s T 312 C.= = = =
o o <
COMMENTS: (1) The above results can be readily checked against the analytical solution
represented in Fig. 5.8 (see also Eq. 5.60). For x = 0 and t = 180s, find
( )
( ) ( )
1/2
1/22 -6 21/2
x
0
2 t
100 W/m K 5.60 10 m / s 180sh t
0.16
k 20 W/m K
a
a
=
× ´ ´
= =
×
for which the figure gives
i
i
T T
0.15
T T¥
-
=
-
so that,
( ) ( ) ( )
( )
i iT 0, 180s 0.15 T T T 0.15 15 325 C 325 C
T 0, 180s 278 C.
¥= - + = - +
=
o o
o
For x = 45mm, the procedure yields T(45mm, 180s) = 316°C. The agreement with the numerical
solution is nearly within 1%.
PROBLEM 5.107
KNOWN: Sudden exposure of the surface of a thick slab, initially at a uniform temperature,
to convection and to surroundings at a high temperature.
FIND: (a) Explicit, finite-difference equation for the surface node in terms of Fo, Bi, Bir, (b)
Stability criterion; whether it is more restrictive than that for an interior node and does it
change with time, and (c) Temperature at the surface and at 30mm depth for prescribed
conditions after 1 minute exposure.
SCHEMATIC:
ASSUMPTIONS: (1) One-dimensional transient conduction, (2) Thick slab may be
approximated as semi-infinite medium, (3) Constant properties, (4) Radiation exchange is
between small surface and large surroundings.
ANALYSIS: (a) The explicit form of the FDE for
the surface node may be obtained by applying an
energy balance to a control volume about the node.
( ) ( )
in out conv rad cond st
p p
op p 1
o r sur o
E E q q q E
T T
h T T h T T k 1
x

\u221e
\u2032\u2032 \u2032\u2032 \u2032\u2032 \u2032\u2032 \u2032\u2032 \u2032\u2032
\u2212 = + + =
\u2212
\u2212 + \u2212 + \u22c5 \u22c5
\u2206
\ufffd \ufffd \ufffd
p+1 p
o oT Tx
c 1
2 t
\u3c1 \u2212\u2206\uf8ee \uf8f9= \u22c5\uf8ef \uf8fa \u2206\uf8f0 \uf8fb (1)
where the radiation process has been linearized, Eq. 1.8. (See also Comment 4, Example 5.9),
( ) ( ) 2p p p p 2r r o sur o sur sur0h h T , T T T T T .\u3b5\u3c3 \uf8eb \uf8f6\uf8ee \uf8f9= = + +\uf8ec \uf8f7\uf8ec \uf8f7\uf8ef \uf8fa\uf8f0 \uf8fb\uf8ed \uf8f8 (2)
Divide Eq. (1) by \u3c1c\u2206x/2\u2206t and regroup using these definitions to obtain the FDE:
( ) 2 r rFo k/ c t/ x Bi h x/k Bi h x/k\u3c1\u2261 \u2206 \u2206 \u2261 \u2206 \u2261 \u2206 (3,4,5)
( ) ( )p+1 p po r sur r o1T 2Fo Bi T Bi T T 1 2 Bi Fo 2Bi Fo 2Fo T .\u221e= \u22c5 + \u22c5 + + \u2212 \u22c5 \u2212 \u22c5 \u2212 (6) <
(b) The stability criterion for Eq. (6) requires that the coefficient of poT be positive.
( ) ( )r r1 2Fo Bi Bi 1 0 or Fo 1/2 Bi Bi 1 .\u2212 + + \u2265 \u2264 + + (7) <
The stability criterion for an interior node, Eq. 5.74, is Fo \u2264 1/2. Since Bi + Bir > 0, the
stability criterion of the surface node is more restrictive. Note that Bir is not constant but
depends upon hr which increases with increasing
p
oT (time). Hence, the restriction on Fo
increases with increasing poT (time).
Continued \u2026..
PROBLEM 5.107 (Cont.)
(c) Consider the prescribed conditions with negligible convection (Bi = 0). The FDEs for the
thick slab are:
Surface (0) ( ) ( )pp+1 po r sur r o1T 2Fo Bi Fo Bi T T 1 2Bi Fo 2Bi Fo 2Fo T= \u22c5 + \u22c5 + + \u2212 \u22c5 \u2212 \u22c5 \u2212 (8)
Interior (m\u22651) ( ) ( )p+1 p p pm mm+1 m-1T Fo T T 1 2Fo T= + + \u2212 (9,5,7,3)
The stability```