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
L
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)”