18–77
time
[min]
To,o
[C]
TL,o
[C]
Q
[kJ]
5
10
15
20
25
30
35
40
45
50
55
60
124.5
103.4
86.49
73.03
62.29
53.73
46.9
41.45
37.11
33.65
30.88
28.68
123.2
102.3
85.62
72.33
61.74
53.29
46.55
41.17
36.89
33.47
30.74
28.57
66.03
118.5
160.4
193.7
220.4
241.6
258.5
272
282.8
291.4
298.2
303.7
510 15 20 25 30 35 40 45 50 55 60
0
25
50
75
100
125
0
50
100
150
200
250
300
To,o and TL,o [C]
Q [kJ]
you are a student using this Manual, you are using it without permission.
18–94 A long roll of large 1-Mn manganese steel plate is to be quenched in an oil bath at a specified rate. The temperature of
the sheet metal after quenching and the rate at which heat needs to be removed from the oil in order to keep its temperature
constant are to be determined.
Assumptions 1 The thermal properties of the steel plate are constant. 2 The heat transfer coefficient is constant and uniform
1.0 036.0
C W/m.5.60
)m 0025.0)(C. W/m860(
2
k
hL
Bi
c
Since
0.1<Bi
, the lumped system analysis is applicable. Therefore,
s 36min 6.0
m/min 15
m 9
velocity
length
time
s 10092.0
m) C)(0.0025J/kg. 434)(kg/m (7854
C. W/m860 1–
3
2
cpp
s
Lc
h
c
hA
b
V
Then the temperature of the sheet metal when it leaves the oil bath is determined to be
C65.5
)(
45820
45)(
)( s) 36)(s 10092.0( -1 tTe
tT
e
TT
TtT bt
i
15 m/min
18–80
18–95 Large steel plates are quenched in an oil reservoir. The quench time is to be determined.
Assumptions 1 The thermal properties of the plates are constant. 2 The heat transfer coefficient is constant and uniform over
18–82
18–97E A person shakes a can of drink in an iced water to cool it. The cooling time of the drink is to be determined.
Assumptions 1 The can containing the drink is cylindrical in shape
with a radius of ro = 1.25 in. 2 The thermal properties of the drink are
taken to be the same as those of water. 3 Thermal properties of the
drink are constant at room temperature. 4 The heat transfer coefficient
is constant and uniform over the entire surface. 5 The Biot number in
this case is large (much larger than 0.1). However, the lumped system
analysis is still applicable since the drink is stirred constantly, so that
its temperature remains uniform at all times.
Properties The density and specific heat of water at room temperature
(90F) are = 62.12 lbm/ft3, cp = 0.999 Btu/lbm.°F, k = 0.358
Btu/h.ft.°F (Table A-15E).
Analysis The characteristic length and Biot number for the can of drink are
1.049.3
FftBtu/h 358.0
)ft 04167.0)(FftBtu/h 30(
ft 04167.0
ft) (1.25/122ft) ft)(5/12 (1.25/122
ft) (5/12ft) 12/25.1(
22
2
2
2
2
2
surface
k
hL
Bi
rLr
Lr
A
L
c
oo
o
c
V
For the reason explained above we can use the lumped system analysis to determine how long it will take for the canned
drink to cool to 40°F
s 615
tee
TT
TtT
LC
h
VC
hA
b
tbt
i
cpp
)s 00322.0(
1–1–
3
2
1–
3290
3240
)(
s 00322.0h 599.11
ft) F)(0.04167Btu/lbm. 999.0)(lbm/ft (62.22
F.Btu/h.ft 30
Milk
3C
Water
32F
Drink
90F
18–84
18–99 Ball bearings leaving the oven at a uniform temperature of 900C are exposed to air for a while before they are
dropped into the water for quenching. The time they can stand in the air before their temperature falls below 850C is to be
determined.
Assumptions 1 The bearings are spherical in shape with a radius of ro = 0.6 cm. 2 The thermal properties of the bearings are
0.1< 0166.0
)C W/m.1.15(
)m 002.0)(C. W/m125(
2
k
hL
Bi
c
Therefore, the lumped system analysis is applicable. Then the
allowable time is determined to be
s 3.68
tee
TT
TtT
Lc
h
c
hA
b
tbt
i
cpp
s
)s 0161.0(
1–
3
2
1–
30900
30850
)(
s 01610.0
m) C)(0.002J/kg. 480)(kg/m 8085(
C. W/m125
V
The result indicates that the ball bearing can stay in the air about 4 s before being dropped into the water.
900C
18–87
18-102 The temperature at the center of a spherical glass bead after 3 minutes of cooling is to be determined using (a) Table
18-2 and (b) the Heisler chart (Figure 18-19).
Assumptions 1 Heat conduction is one-dimensional. 2 Thermal properties are constant. 3 Convection heat transfer coefficient
2.0
K W/m7.0
)m 005.0)(K W/m28(2
k
hr
Bi o
18-88
18-103 The water pipes are buried in the ground to prevent freezing. The minimum burial depth at a particular location is to
be determined.
Assumptions 1 The temperature in the soil is affected by the thermal conditions at one surface only, and thus the soil can be
considered to be a semi-infinite medium with a specified surface temperature of –10C. 2 The thermal properties of the soil
are constant.
18–89
18-104 A thick wall is exposed to cold outside air. The wall temperatures at distances 15, 30, and 40 cm from the outer
surface at the end of 2-hour cooling period are to be determined.
18–90
18-105 The temperature at the edge of a steel block after 10 minutes of cooling is to be determined.
Assumptions 1 Two-dimensional heat conduction in x and y directions. 2 Thermal properties are constant. 3 Convection heat
transfer coefficient is constant. 4 Heat transfer by radiation is negligible.
Properties The properties of steel are (
= 7832 kg/m3, cp = 434 J/kg ∙ K, k = 63.9 W/m ∙ K, and
= 18.8 10−6 m2/s.
Analysis For a quarter-infinite medium, at the edge of the steel block (x = y = 0), we have
2
2
2
)K W/m9.63(
k
04155.0
K W/m9.63
)s 6010)(/sm 108.18()K W/m25(262
k
αth
9548.0)s 600 ,0(
infsemi
The temperature at the edge of the steel block after 10 minutes of cooling is
22
)s 600 ,0 ,0(
i
TT
