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18-61
4–77 Thick stainless steel and copper slabs are subjected to uniform heat flux. The temperatures of both slabs at the depth of
1 cm from the surface after 60 s are to be determined.
Assumptions 1 The slabs are treated as a semi-infinite medium with a specified surface heat flux. 2 The thermal properties of
the slabs are constant.
Analysis This is a transient conduction problem in a semi-infinite medium subjected to constant surface heat flux, and the
tα
t
k
4
For stainless steel,
m 01.0
x
18-62
18-78 A thick wood slab is exposed to hot gases for a period of 5 minutes. It is to be determined whether the wood will
ignite.
Assumptions 1 The wood slab is treated as a semi-infinite medium subjected to convection at the exposed surface. 2 The
thermal properties of the wood slab are constant. 3 The heat transfer coefficient is constant and uniform over the entire
18-64
18-80 Prob. 18-79 is reconsidered. The soil temperature as a function of the distance from the earth’s surface is to be
plotted.
k=0.9 [W/m-C]
alpha=1.6E-5 [m^2/s]
18-65
4–81 A spherical watermelon that is cut into two equal parts is put into a freezer. The time it will take for the center of the
exposed cut surface to cool from 25C to 3C is to be determined.
Assumptions 1 The temperature of the exposed surfaces of the watermelon is affected by the convection heat transfer at those
surfaces only. Therefore, the watermelon can be considered to be a semi-infinite medium 2 The thermal properties of the
watermelon are constant.
/s)m 10146.0(). W/m22(
26-22
2
K
h
18-82 The contact surface temperatures when a bare footed person steps on aluminum and wood blocks are to be determined.
Assumptions 1 Both bodies can be treated as the semi-infinite solids. 2 Heat loss from the solids is disregarded. 3 The
properties of the solids are constant.
Properties The
p
ck
value is 24 kJ/m2°C for aluminum, 0.38 kJ/m2°C for wood, and 1.1 kJ/m2°C for the human flesh.
Analysis The surface temperature is determined from Eq. 18-49 to be
C)20(C)kJ/m 24(C)32(C)kJ/m 1.1(
)()(
22
AlAlhumanhuman
pp
TckTck
C28.9
C)kJ/m 38.0(C)kJ/m 1.1(
22
18-66
Transient Heat Conduction in Multidimensional Systems
18-83C The product solution enables us to determine the dimensionless temperature of two- or three-dimensional heat
18-85C This short cylinder is physically formed by the intersection of a long cylinder and a plane wall. The dimensionless
18-86C The heat transfer in this short cylinder is one-dimensional since there is no heat transfer in the axial direction. The
temperature will vary in the radial direction only.
18-68
C449
),0,0(107.0)0931.1()0580.1(
50020
500),0,0( )208.2()8516.0()208.2()5932.0( 22 tTee
tT
18-70
Cylinder: This cylindrical block can physically be formed by the intersection of a long cylinder of radius ro = D/2 = 2.5 cm
400.0
)C W/m.5.2(
)m 025.0)(C. W/m40(2
k
hr
Bi o
0931.1 and 8516.0 11 A
To determine the center temperature, the product solution method can be written as
),0,0(
),0(),0(),0,0(
1
wall
1
wallblock
2
1
2
1
eAeA
TT
TtT
ttt
cyl
i
cyl
18-71
18-89E A hot dog is dropped into boiling water. The center temperature of the hot dog is do be determined by treating hot
dog as a finite cylinder and also as an infinitely long cylinder.
Assumptions 1 When treating hot dog as a finite cylinder, heat conduction in the hot dog is two-dimensional, and thus the
temperature varies in both the axial x- and the radial r- directions. When treating hot dog as an infinitely long cylinder, heat
conduction is one-dimensional in the radial r- direction. 2 The thermal properties of the hot dog are constant. 3 The heat
2728.1 and 5421.1 11 A
2.0015.0
ft) 12/5.2(
h) /h)(5/60ft 0077.0(
2
2
2
L
t
(Be cautious!)
Then the dimensionless temperature at the center of the plane wall is determined from
0
2
2
1
eeA
TT
)FBtu/h.ft. 44.0(
5618.1 and 1589.2 11 A
2.0578.0
ft) 12/4.0(
h) /h)(5/60ft 0077.0(
2
2
2
o
r
t
2
2
1
eeA
TT
o
Hot dog
18-72
0071.0)5618.1( )156.1()1589.2(
1,
2
2
1
eeA
TT
TT
i
o
cylo
F211
),0,0(0084.0
21240
212),0,0(
0084.00071.0185.1
),0,0(
,,
cylinder
short
tT
tT
TT
TtT
cylowallo
i
After 15 minutes
2.0045.0
ft) 12/5.2(
h) /h)(15/60ft 0077.0(
2
2
2
L
t
(Be cautious!)
0
2
2
1
eeA
TT
TT
i
2.0734.1
ft) 12/4.0(
h) /h)(15/60ft 0077.0(
2
2
2
o
r
t
0
2
2
1
eeA
TT
18-75
C85.6
),0,(5047.0
20150
20),0,(
5047.05870.08598.0),(
),0,(
,
tLT
tLT
tL
TT
TtLT
cylowall
cylinder
short
i
(c) We first need to determine the maximum heat can be transferred from the cylinder
kJ 2.325C)20150)(CkJ/kg. 389.0)(kg 431.6()(
kg 431.6m) 15.0(m) 04.0()kg/m 8530(
max
2.32
TTmcQ
Lrm
ip
o
V
18-76
18-92 Prob. 18-91 is reconsidered. The effect of the cooling time on the center temperature of the cylinder, the center
D=0.08 [m]
r_o=D/2
height=0.15 [m]
L=height/2
T_i=150 [C]
"This short cylinder can physically be formed by the intersection of a long cylinder of radius r_o and a plane wall
of thickness 2L"
"For plane wall"
Bi_w=(h*L)/k
"From Table 18-2 corresponding to this Bi number, we read"
theta_o_c=A_1_c*exp(-lambda_1_c^2*tau_c) "theta_o_c=(T_o_c-T_infinity)/(T_i-T_infinity)"
(T_o_o-T_infinity)/(T_i-T_infinity)=theta_o_w*theta_o_c "center temperature of short cylinder"
"(b)"
theta_L_w=A_1_w*exp(-lambda_1_w^2*tau_w)*Cos(lambda_1_w*L/L) "theta_L_w=(T_L_w-T_infinity)/(T_i-
T_infinity)"
