18–22
18-32 The heating times of a sphere, a cube, and a rectangular prism with similar dimensions are to be determined.
Assumptions 1 The thermal properties of the geometries are constant. 2 The heat transfer coefficient is constant and uniform
over the entire surface.
Properties The properties of silver are given to be k = 429 W/mºC, ρ = 10,500 kg/m3, and cp = 0.235 kJ/kgºC.
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18–34C A cylinder whose diameter is small relative to its length can be treated as an infinitely long cylinder. When the
18–35C The Fourier number is a measure of heat conducted through a body relative to the heat stored. Thus a large value of
18–36C The solution for determination of the one-dimensional transient temperature distribution involves many variables that
18–37C Yes. A plane wall whose one side is insulated is equivalent to a plane wall that is twice as thick and is exposed to
18–39C When the Biot number is less than 0.1, the temperature of the sphere will be nearly uniform at all times. Therefore, it
is more convenient to use the lumped system analysis in this case.
18–40C The maximum possible amount of heat transfer will occur when the temperature of the body reaches the temperature
max ip TTmcQ
18–25
18–42 The temperature at the center plane of an aluminum plate with
TTs
, after 15 seconds of heating, is to be
= 97.1 10−6 m2/s.
Analysis For
TTs
, it implies that
h
. Thus,
The Fourier number is
5826.0
)m 05.0(
)s 15)(/sm 101.97(
2
26
2
L
t
The temperature at the center plane after 15 seconds of heating is
2
1
0
TT
TT
i
TeATTT i
2
1
10 )(
C 356 C 500)2732.1)(C 500C 25()5826.0()5708.1(
0
2
eT
Discussion Since
> 0.2, the one-term approximate solution is applicable for this problem.
Alternative solution: This problem can also be solved using transient temperature charts as follows:
The center temperature is determined from
29.0
5826.0
)m 05.0(
)s 15)(/sm 101.97(
0
1
Bi
1
0
2
26
2
TT
TT
L
ti
(Fig. 18-13a)
Therefore,
C362)50025)(29.0(500)(29.0
0 TTTT i
18–27
18–44 Prob. 18–43 is reconsidered. The effects of the temperature of the oven and the heating time on the final surface
L=(0.03/2) [m]
T_i=25 [C]
T_infinity=700 [C]
time=10 [min]
18–28
time
[min]
TL
[C]
2
146.7
4
244.8
6
325.5
8
391.9
10
446.5
12
491.5
14
528.5
16
558.9
18
583.9
20
604.5
22
621.4
24
635.4
26
646.8
28
656.2
30
664
200
300
400
500
600
700
time [min]
TL [C]
18–32
18–48 A long cylindrical wood log is exposed to hot gases in a fireplace. The time for the ignition of the wood is to be
determined.
00.4
)C W/m.17.0(
)m 05.0)(C. W/m6.13(2
k
hr
Bi o
4698.1 and 9081.1 11 A
Once the constant
J0
is determined from Table 18-3 corresponding to
1419.0)2771.0()4698.1(
55015
550420
)/(
),(
2
2
1
)9081.1(
101
e
rrJeA
TT
TtrT
oo
i
o
which is not above the value of 0.2. We use one-term approximate solution (or the transient temperature charts) knowing that
the result may be somewhat in error. Then the length of time before the log ignites is
min 46.2
s 2771
/sm 1028.1
m) 05.0)(1419.0(
27
2
2
o
r
t
250.
),(
1
25.0
m) C)(0.05 W/m(13.6
C W/m17.0
Bi
1
0
2
TT
TtrT
r
x
hr
k
o
o
o
(Fig. 18-14b)
550420
),(
TtrT
10 cm
Wood log, 15C
Hot gases
550C
18–33
18–49E Long cylindrical steel rods are heat-treated in an oven. Their centerline temperature when they leave the oven is to be
determined.
Assumptions 1 Heat conduction in the rods is one-dimensional since the rods are long and they have thermal symmetry about
the center line. 2 The thermal properties of the rod are constant. 3 The heat transfer coefficient is constant and uniform over
0996.1 and 8790.0 11 A
243.0
ft) 12/2(
h) /h)(3/60ft 135.0(
2
2
2
o
r
t
18–36
Alternative solution: This problem can also be solved using transient temperature and heat transfer charts as follows:
270.
1
376.0
m) C)(0.15 W/m(14
C W/m79.0
Bi
1
0
2
TT
TT
L
x
hr
k
s
o
(Fig. 18-14b)
18–37
18–52 A long cylindrical shaft at 400C is allowed to cool slowly. The center temperature and the heat transfer per unit length
of the cylinder are to be determined.
Assumptions 1 Heat conduction in the shaft is one-dimensional since it is long and it has thermal symmetry about the center
705.0
)C W/m.9.14(
)m 175.0)(C. W/m60(2
k
hr
Bi o
1548.1 and 0904.1 11 A
The Fourier number is
s) 60/s)(20m 1095.3(
26
t
Ti = 400C
Air
T = 150C
18–38
18–53 Prob. 18–52 is reconsidered. The effect of the cooling time on the final center temperature of the shaft and the
amount of heat transfer is to be investigated.
Analysis The problem is solved using EES, and the solution is given below.
k=14.9 [W/m-C]
rho=7900 [kg/m^3]
c_p=477 [J/kg-C]
alpha=3.95E-6 [m^2/s]
“ANALYSIS”
18–39
18–54 The convection heat transfer coefficient for a plastic rod being cooled is to be determined.
Assumptions 1 Heat conduction is one-dimensional. 2 Thermal properties are constant. 3 Convection heat transfer coefficient
18–40
18–55 The temperature at the center of a Pyroceram rod after 3 minutes of cooling is to be determined using (a) Table 18–2
and (b) the Heisler chart (Figure 18-18).
10.0
K W/m98.3
)m 005.0)(K W/m80(2
k
hr
Bi o