4-81
4-82
4-105 Thick stainless steel and copper slabs are subjected to an energy pulse. The temperatures of both slabs at the
depth of 5 cm from the surface as a function of time are to be determined.
Assumptions 1 The slabs are treated as semi-infinite solids. 2 Thermal properties are constant. 3 Heat transfer by radiation is
negligible. 4 Entire energy from the pulse enters the slabs.
Properties The properties of stainless steel are given as k = 14.9 W/mK and α = 3.95× 10−6 m2/s; the properties of copper are
given as k = 401 W/mK and α = 117 × 10−6 m2/s.
Analysis The problem is solved using EES, and the solution is given below.
“GIVEN”
e_s=5e7 [J/m^2]
T_i=20 [C]
x=0.05 [m]
“PROPERTIES”
“stainless steel”
k_ss=14.9 [W/m-K]
alpha_ss=3.95e-6 [m^2/s]
“copper”
k_cu=401 [W/m-K]
alpha_cu=117e-6 [m^2/s]
“ANALYSIS”
T_ss-T_i=e_s/(k_ss*(pi#*t/alpha_ss)^0.5)*exp(-x^2/(4*alpha_ss*t))
T_cu-T_i=e_s/(k_cu*(pi#*t/alpha_cu)^0.5)*exp(-x^2/(4*alpha_cu*t))
T (x, t) [°C]
Time [s] SS Cu
1 20.0 23.6
2 20.0 57.2
3 20.0 94.0
6 20.0 147.5
8 20.0 158.0
10 20.0 161.0
20 20.3 150.3
30 23.5 136.3
40 31.4 125.3
50 42.5 116.7
60 54.8 109.9
80 78.2 99.6
Discussion The copper slab, having a much higher thermal diffusivity value, diffuses the heat energy from the pulse much
quicker than the stainless steel slab. As shown in the table and figure, the copper temperature increases sharply for the first 10
seconds from an initial temperature of 20°C to a temperature of 161°C, while the temperature of stainless steel stays constant
at its initial temperature of 20°C .
4-83
4-106 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. 4-49 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
4-84
Transient Heat Conduction in Multidimensional Systems
4-107C The product solution enables us to determine the dimensionless temperature of two– or three-dimensional heat
4-109C This short cylinder is physically formed by the intersection of a long cylinder and a plane wall. The dimensionless
4-85
4-86
After 20 minutes
−−− ),0,0(107.0)0931.1()0580.1(
500),0,0( )208.2()8516.0()208.2()5932.0( 22 tTee
tT
4-87
4-112 A cubic block and a cylindrical block are exposed to hot gases on all of their surfaces. The center temperatures of each
geometry in 10, 20, and 60 min are to be determined.
Assumptions 1 Heat conduction in the cubic block is three-dimensional, and thus the temperature varies in all x–, y, and z–
directions. 2 Heat conduction in the cylindrical block is two-dimensional, and thus the temperature varies in both axial x– and
radial r- directions. 3 The thermal properties of the granite are constant. 4 The heat transfer coefficient is constant and
uniform over the entire surface. 5 The Fourier number is > 0.2 so that the one-term approximate solutions are applicable
(this assumption will be verified).
Properties The thermal properties of the granite are k = 2.5 W/m.C and = 1.1510-6 m2/s.
Analysis
Cubic block: This cubic block can physically be formed by the intersection of three infinite plane wall of thickness 2L = 5
cm. Two infinite plane walls are exposed to the hot gases with a heat transfer coefficient of
h= 40 W / m . C
2
and one with
h= 80 W / m . C
2
.
After 10 minutes: The Biot number and the corresponding constants for
C. W/m40 2=h
are
400.0
)C W/m.5.2(
)m 025.0)(C. W/m40(2
=
== k
hL
Bi
0580.1 and 5932.0 11 ==⎯→⎯ A
The Biot number and the corresponding constants for
C. W/m80 2=h
are
800.0
)C W/m.5.2(
)m 025.0)(C. W/m80(2
=
== k
hL
Bi
1016.1 and 7910.0 11 ==⎯→⎯ A
Ti = 20C
5 cm 5 cm 5 cm
4-88
Cylinder: This cylindrical block can physically be formed by the intersection of a long cylinder of radius ro = D/2 = 2.5 cm
exposed to the hot gases with a heat transfer coefficient of
C. W/m40 2=h
and a plane wall of thickness 2L = 5 cm exposed
to the hot gases with
C. W/m80 2=h
.
After 10 minutes: The Biot number and the corresponding constants for the long cylinder are
)m 025.0)(C. W/m40(2
hr
4-89
4-90
4-114E 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
transfer coefficient is constant and uniform over the entire surface. 4 The Fourier number is > 0.2 so that the one-term
approximate solutions are applicable (this assumption will be verified).
Properties The thermal properties of the hot dog are given to be k = 0.44 Btu/h.ft.F, = 61.2 lbm/ft3 cp = 0.93 Btu/lbm.F,
and = 0.0077 ft2/h.
Analysis (a) This hot dog can physically be formed by the intersection of a long cylinder of radius ro = D/2 = (0.4/12) ft and a
plane wall of thickness 2L = (5/12) ft. The distance x is measured from the midplane.
After 5 minutes
First the Biot number is calculated for the plane wall to be
8.56
)FBtu/h.ft. 44.0(
)ft 12/5.2)(F.Btu/h.ft 120(2
=
== k
hL
Bi
ft) 12/4.0(
2
2===
o
r
Water
212F
r
4-91
4-92
4-115 A rectangular ice block is placed on a table. The time the ice block starts melting is to be determined.
Assumptions 1 Heat conduction in the ice block is two-dimensional, and thus the temperature varies in both x– and y–
directions. 2 The thermal properties of the ice block are constant. 3 The heat transfer coefficient is constant and uniform over
the entire surface. 4 The Fourier number is > 0.2 so that the one-term approximate solutions are applicable (this assumption
will be verified).
Properties The thermal properties of the ice are given to be k = 2.22 W/m.C and = 0.12410-7 m2/s.
Analysis This rectangular ice block can be treated as a short rectangular block that
can physically be formed by the intersection of two infinite plane wall of thickness
2L = 4 cm and an infinite plane wall of thickness 2L = 12 cm. We measure x from
the bottom surface of the block since this surface represents the adiabatic center
Air
4-93
4-94
4-117 A cylindrical ice block is placed on a table. The initial temperature of the ice block to avoid melting for 2 h is to be
determined.
Assumptions 1 Heat conduction in the ice block is two-dimensional, and thus the
temperature varies in both x– and r– directions. 2 Heat transfer from the base of
the ice block to the table is negligible. 3 The thermal properties of the ice block
are constant. 4 The heat transfer coefficient is constant and uniform over the
entire surface. 5 The Fourier number is > 0.2 so that the one-term approximate
solutions are applicable (this assumption will be verified).
Properties The thermal properties of the ice are given to be k = 2.22 W/m.C
and = 0.12410-7 m2/s.
Analysis This cylindrical ice block can be treated as a short cylinder that can
physically be formed by the intersection of a long cylinder of diameter D = 2 cm
and an infinite plane wall of thickness 2L = 4 cm. We measure x from the bottom
surface of the block since this surface represents the adiabatic center surface of
Air
T = 24C
x
r
(ro, L)
Ice block
Ti
4-95
4-118 A short cylinder is allowed to cool in atmospheric air. The temperatures at the centers of the cylinder and the top
surface as well as the total heat transfer from the cylinder for 15 min of cooling are to be determined.
Assumptions 1 Heat conduction in the short cylinder is two-dimensional, and thus the temperature varies in both the axial x–
and the radial r– directions. 2 The thermal properties of the cylinder are constant. 3 The heat transfer coefficient is constant
and uniform over the entire surface. 4 The Fourier number is > 0.2 so that the one-term approximate solutions are
applicable (this assumption will be verified).
Properties The thermal properties of brass are given to be
3
kg/m 8530=
,
CkJ/kg 389.0 =
p
c
,
C W/m110 =k
, and
/sm 1039.3 25−
=
.
Analysis This short cylinder can physically be formed by the intersection of a long cylinder of radius D/2 = 2 cm and a plane
wall of thickness 2L = 20 cm. We measure x from the midplane.
(a) The Biot number is calculated for the plane wall to be
)m 10.0)(C. W/m40(2
hL
D0 = 4 cm
4-96
===
−
−
1944.02191.08871.0),(
),0,(
,
tL
TT
TtLT
cylowall
short
i
4-97
4-119 Prob. 4-118 is reconsidered. The effect of the cooling time on the center temperature of the cylinder, the center
temperature of the top surface of the cylinder, and the total heat transfer is to be investigated.
Analysis The problem is solved using EES, and the solution is given below.
“GIVEN”
D=0.04 [m]
r_o=D/2
height=0.20 [m]
L=height/2
T_i=150 [C]
T_infinity=20 [C]
h=40 [W/m^2-C]
time=15 [min]
“PROPERTIES”
k=110 [W/m-C]
rho=8530 [kg/m^3]
c_p=0.389 [kJ/kg–C]
alpha=3.39E-5 [m^2/s]
“ANALYSIS”
“(a)”
“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 4-2 corresponding to this Bi number, we read”
lambda_1_w=0.1882 “w stands for wall”
A_1_w=1.0060
tau_w=(alpha*time*Convert(min, s))/L^2
theta_o_w=A_1_w*exp(-lambda_1_w^2*tau_w) “theta_o_w=(T_o_w-T_infinity)/(T_i-T_infinity)”
“For long cylinder”
Bi_c=(h*r_o)/k “c stands for cylinder”
“From Table 4-2 corresponding to this Bi number, we read”
lambda_1_c=0.1412
A_1_c=1.0025
tau_c=(alpha*time*Convert(min, s))/r_o^2
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)”
(T_L_o-T_infinity)/(T_i-T_infinity)=theta_L_w*theta_o_c “center temperature of the top surface”
“(c)”
V=pi*r_o^2*(2*L)
m=rho*V
Q_max=m*c_p*(T_i-T_infinity)
Q_w=1-theta_o_w*Sin(lambda_1_w)/lambda_1_w “Q_w=(Q/Q_max)_w”
Q_c=1-2*theta_o_c*J_1/lambda_1_c “Q_c=(Q/Q_max)_c”
J_1=0.07034 “From Table 4-3, at lambda_1_c”
Q/Q_max=Q_w+Q_c*(1-Q_w) “total heat transfer”
4-98
time
[min]
To,o
[C]
TL,o
[C]
Q
[kJ]
5
10
15
20
25
30
35
40
45
50
55
60
96.18
64.26
45.72
34.94
28.68
25.05
22.93
21.7
20.99
20.58
20.33
20.19
94.83
63.48
45.26
34.68
28.53
24.96
22.88
21.67
20.97
20.56
20.33
20.19
45.49
71.85
87.17
96.07
101.2
104.2
106
107
107.6
107.9
108.1
108.3
010 20 30 40 50 60
20
40
60
80
100
40
50
60
70
80
90
100
110
120
To,o and TL,o [C]
Q [kJ]
4-99
4-120 A semi-infinite aluminum cylinder is cooled by water. The temperature at the center of the cylinder 5 cm from the end
surface is to be determined.
Assumptions 1 Heat conduction in the semi-infinite cylinder is two-dimensional, and thus the temperature varies in both the
axial x– and the radial r– directions. 2 The thermal properties of the cylinder are constant. 3 The heat transfer coefficient is
constant and uniform over the entire surface. 4 The Fourier number is > 0.2 so that the one-term approximate solutions are
applicable (this assumption will be verified).
Properties The thermal properties of aluminum are given to be k = 237 W/m.C and = 9.7110-5m2/s.
Analysis This semi-infinite cylinder can physically be formed by the intersection of a long cylinder of radius ro = D/2 = 7.5
cm and a semi-infinite medium. The dimensionless temperature 5 cm from the surface of a semi-infinite medium is first
determined from
)237(
)608)(1071.9()140(
237
)05.0)(140(
exp
)608)(1071.9(2
05.0
2
exp
),(
5
2
52
5
2
2
+−
=
+
+−
=
−
−
−
−
−
erfc
k
th
t
x
erfc
k
th
k
hx
t
x
erfc
TT
TtxT
i
i
4-100