4-101
1
−
+
=
Q
Q
Q
Q
4-102
4-122 A cylindrical aluminum block is heated in a furnace. The length of time the block should be kept in the furnace and the
amount of heat transferred to the block are to be determined.
Assumptions 1 Heat conduction in the cylindrical block is two-dimensional, and thus the temperature varies in both axial x–
and radial r- directions. 2 Heat transfer from the bottom surface of the block is negligible. 3 The thermal properties of the
aluminum 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 (or the transient temperature charts) are applicable (this assumption will
be verified).
Properties The thermal properties of the aluminum block are given to be k = 236 W/m.C, = 2702 kg/m3, cp = 0.896
kJ/kg.C, and = 9.7510–5 m2/s.
Analysis This cylindrical aluminum block can physically be formed by the
intersection of an infinite plane wall of thickness 2L = 60 cm and a long cylinder of
radius ro = D/2 = 7.5 cm. Note that the height of the short cylinder represents the half
thickness of the infinite plane wall where the bottom surface of the short cylinder is
adiabatic. The Biot numbers and corresponding constants are first determined to be
102.0
)C W/m.236(
)m 3.0)(C. W/m80(2
=
== k
hL
Bi
0164.1 and 3135.0 11 ==→ A
0254.0
)C W/m.236(
)m 075.0)(C. W/m80(2
=
== k
hr
Bi o
0063.1 and 2217.0 11 ==→ A
Noting that
2
/Lt
=
and assuming > 0.2 in all dimensions and thus the one-term
approximate solution for transient heat conduction is applicable, the product solution
for this problem can be written as
7627.0
)075.0(
)1075.9(
)2217.0(exp)0063.1(
)3.0(
)1075.9(
)3135.0(exp)0164.1(
120020
1200300
),0(),0(),0,0(
2
5
2
2
5
2
cyl
1
wall
1cylwallblock
2
1
2
1
=
−
−=
−
−
==
−−
−−
tt
eAeAttt
Solving for the time t gives
t = 306 s = 5.1 min
We note that
2.03317.0
m) 3.0(
s) /s)(306m 1075.9(
2
25
2
wall
=
==
−
L
t
L
z
Cylinder
Ti = 20C
r0
Furnace
T = 1200C
L
4-103
4-104
4-123 Prob. 4-121 is reconsidered. The effect of the final center temperature of the block on the heating time and the
amount of heat transfer is to be investigated.
Analysis The problem is solved using EES, and the solution is given below.
“GIVEN”
2*L=0.30 [m]
2*r_o=0.15 [m]
T_i=20 [C]
T_infinity=1200 [C]
T_o_o=300 [C]
h=80 [W/m^2-C]
“PROPERTIES”
k=236 [W/m-C]
rho=2702 [kg/m^3]
c_p=0.896 [kJ/kg–C]
alpha=9.75E-5 [m^2/s]
“ANALYSIS”
“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 4-1 corresponding to this Bi number, we read”
lambda_1_w=0.2224 “w stands for wall”
A_1_w=1.0083
tau_w=(alpha*time)/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=0.2217
A_1_c=1.0063
tau_c=(alpha*time)/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 cylinder”
V=pi*r_o^2*(2*L)
m=rho*V
Q_max=m*c_p*(T_infinity-T_i)
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=0.1101 “From Table 4-3, at lambda_1_c”
Q/Q_max=Q_w+Q_c*(1-Q_w) “total heat transfer”
4-105
To,o
[C]
time
[s]
Q
[kJ]
50
100
150
200
250
300
350
400
450
500
550
600
650
700
750
800
850
900
950
1000
37.79
79.48
123.1
168.9
217
267.7
321.3
378.1
438.7
503.4
572.9
647.9
729.5
818.9
917.7
1028
1153
1298
1469
1678
605.5
1238
1870
2502
3134
3766
4398
5031
5663
6295
6927
7559
8191
8823
9456
10088
10720
11352
11984
12616
0200 400 600 800 1000
0
200
400
600
800
1000
1200
1400
1600
1800
0
2000
4000
6000
8000
10000
12000
14000
To,o [C]
time [s]
Q [kJ]
4-106
Special Topic: Refrigeration and Freezing of Foods
4-124C The common kinds of microorganisms are bacteria, yeasts, molds, and viruses. The undesirable changes caused by
4-125C Microorganisms are the prime cause for the spoilage of foods. Refrigeration prevents or delays the spoilage of foods
4-127C Cooking kills the microorganisms in foods, and thus prevents spoilage of foods. It is important to raise the internal
4-128C The contamination of foods with microorganisms can be prevented or minimized by (1) preventing contamination
by following strict sanitation practices such as washing hands and using fine filters in ventilation systems, (2) inhibiting
4-129C (a) High air motion retards the growth of microorganisms in foods by keeping the food surfaces dry, and creating an
undesirable environment for the microorganisms. (b) Low relative humidity (dry) environments also retard the growth of
4-130C Cooling the carcass with refrigerated air is at -10ºC would certainly reduce the cooling time, but this proposal should
4-133C This claim is reasonable since the lower the storage temperature, the longer the storage life of beef. This is because
4-107
4-134C A refrigerated shipping dock is a refrigerated space where the orders are assembled and shipped out. Such docks
4-135C (a) The heat transfer coefficient during immersion cooling is much higher, and thus the cooling time during
immersion chilling is much lower than that during forced air chilling. (b) The cool air chilling can cause a moisture loss of 1
4-138 The chilling room of a meat plant with a capacity of 350 beef carcasses is considered. The cooling load and the air
flow rate are to be determined.
Assumptions 1 Steady operating conditions exist. 2 Specific heats of beef carcass and air are constant.
Properties The density and specific heat of air at 0C are given to be 1.28 kg/m3 and 1.0 kJ/kgC. The specific heat of beef
carcass is given to be 3.14 kJ/kgC.
Analysis (a) The amount of beef mass that needs to be cooled per
unit time is
time)oolingcooled)/(cmassbeef(Total
=
beef
m
Lights, 2 kW
4-108
4-139 Chickens are to be cooled by chilled water in an immersion chiller. The rate of heat removal from the chicken and the
mass flow rate of water are to be determined.
Assumptions 1 Steady operating conditions exist. 2 The thermal properties of chickens are constant.
Properties The specific heat of chicken are given to be 3.54 kJ/kg.°C. The specific heat of water is 4.18 kJ/kg.C (Table A-
9).
Analysis (a) Chickens are dropped into the chiller at a rate of 500 per hour. Therefore, chickens can be considered to flow
steadily through the chiller at a mass flow rate of
kg/s0.3056 =kg/h 1100)kg/chicken (2.2chicken/h) (500
chicken ==m
Then the rate of heat removal from the chickens as they are cooled from 15C to 3ºC at this rate becomes
kW13.0 C3)ºC)(15kJ/kg.º kg/s)(3.54 (0.3056)( chickenchicken =−== TcmQ p
(b) The chiller gains heat from the surroundings as a rate of 210 kJ/min = 3.5 kJ/s. Then the total rate of heat gain by the
water is
4-140E Chickens are to be frozen by refrigerated air. The cooling time of the chicken is to be determined for the cases of
cooling air being at –40°F and -80°F.
Assumptions 1 Steady operating conditions exist. 2 The thermal properties of chickens are constant.
Analysis The time required to reduce the inner surface temperature of the chickens from 32ºF to 25ºF with refrigerated air at
-40ºF is determined from Fig. 4-51 to be
Immersion chilling, 0.5C
3C
15C
210 kJ/min
4-109
4-141 Turkeys are to be frozen by submerging them into brine at -29°C. The time it will take to reduce the temperature of
turkey breast at a depth of 3.8 cm to -18°C and the amount of heat transfer per turkey are to be determined.
Assumptions 1 Steady operating conditions exist. 2 The thermal properties of
turkeys are constant.
Properties It is given that the specific heats of turkey are 2.98 and 1.65
kJ/kg.°C above and below the freezing point of -2.8°C, respectively, and
the latent heat of fusion of turkey is 214 kJ/kg.
Analysis The time required to freeze the turkeys from 1C to -18ºC with
brine at -29ºC can be determined directly from Fig. 4-52 to be
t 180 min. 3 hours
(a) Assuming the entire water content of turkey is frozen, the amount of heat
that needs to be removed from the turkey as it is cooled from 1C to –18C is
Cooling to -2.8ºC:
kJ 79.3C](-2.8)–C)[1kJ/kg kg)(2.98 7()( freshfreshcooling, === TcmQ p
Freezing at -2.8ºC:
kJ 1498kJ/kg) kg)(214 7(
latentfreezing === hmQ
Cooling –18ºC:
kJ 175.6C18)](2.8C)[kJ/kg. kg)(1.65 (7)( frozenpfrozencooling, =−−−== TcmQ
Therefore, the total amount of heat removal per turkey is
kJ1753 6.17514983.79
frozencooling,freezingfreshcooling,total ++=++= QQQQ
(b) Assuming only 90 percent of the water content of turkey is frozen, the amount of heat that needs to be removed from the
turkey as it is cooled from 1C to –18C is
Cooling to -2.8ºC:
kJ 79.3C](-2.98)–C)[1kJ/kg kg)(2.98 7()( freshpfreshcooling, === TcmQ
Freezing at -2.8ºC:
kJ 1348kJ/kg) kg)(214 0.9(7
latentfreezing === hmQ
Cooling –18ºC:
kJ 158C18)](2.8C)[kJ/kg.kg)(1.65 0.9(7)( frozenfrozencooling, =−−−== TcmQ p
kJ 7.31]C18)º(-2.8)[CkJ/kg.º 2.98)(kg 0.17()( freshunfrozencooling, =−−== TcmQ p
Therefore, the total amount of heat removal per turkey is
kJ 1617=+++=++= 7.3115813483.79
unfrozen&frozencooling,freezingfreshcooling,total QQQQ
Brine
–29C
Turkey
Ti = 1C
4-110
Review Problems
4-142 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
the entire surface.
Properties The properties of steel plates are given to be k = 45 W/mK,
= 7800 kg/m3, and cp = 470 J/kgK.
Analysis For sphere, the characteristic length and the Biot number are
m 005.0
2
m 01.0
2
surface
====
L
A
L
c
V
4-111
4-112
4-145 Long copper wires are extruded and exposed to atmospheric air. The time it will take for the wire to cool, the distance
the wire travels, and the rate of heat transfer from the wire are to be determined.
Assumptions 1 Heat conduction in the wires is one-dimensional in the radial direction. 2 The thermal properties of the copper
are constant. 3 The heat transfer coefficient is constant and uniform over the entire surface. 4 The Biot number is Bi < 0.1 so
that the lumped system analysis is applicable (this assumption will be verified).
Properties The properties of copper are given to be k = 386 W/m.C, = 8950 kg/m3, cp = 0.383 kJ/kg.C, and = 1.1310-4
m2/s.
Analysis (a) The characteristic length of the wire and the Biot
number are
m 00075.0
2
m 0015.0
22
2
=====
r
Lr
Lr
A
L
o
o
o
s
c
V
350C
10 m/min
Air
30C
4-113
4-146 Aluminum wires leaving the extruder at a specified rate are cooled in air. The necessary length of the wire is to be
determined.
Assumptions 1 The thermal properties of the geometry are constant. 2 The heat transfer coefficient is constant and uniform
over the entire surface.
Properties The properties of aluminum are k = 237 W/mºC, ρ = 2702 kg/m3, and cp = 0.903 kJ/kgºC (Table A-3).
Analysis For a long cylinder, the characteristic length and the Biot number are
1.0 000211.0
C W/m.237
)m 001.0)(C. W/m50(
m 001.0
4
m 004.0
4
)4/(
2
2
surface
=
==
=====
k
hL
Bi
D
DL
LD
A
L
c
c
V
Since
0.1< Bi
, the lumped system analysis is applicable. Then the cooling time is determined from
s 02049.0
m) C)(0.001J/kg. 903)(kg/m (2702
C. W/m50
1–
3
2
=
===
Lc
h
c
hA
b
cpp
V
Ti = 350ºC
D = 4 mm
4-114
4-115
4-149 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
constant. 3 The heat transfer coefficient is constant and uniform over the entire surface. 4 The Biot number is Bi < 0.1 so that
the lumped system analysis is applicable (this assumption will be verified).
Properties The thermal conductivity, density, and specific heat of the bearings are given to be k = 15.1 W/m.C, = 8085
kg/m3, and cp = 0.480 kJ/kg.C.
Analysis The characteristic length of the steel ball bearings and Biot number are
m 002.0
6
m 012.0
6
6/
2
3
=====
D
D
D
A
L
s
c
V
Furnace
4-116
4-151 The trunks of some dry oak trees are exposed to hot gases. The time for the ignition of the trunks is to be determined.
Assumptions 1 Heat conduction in the trunks is one-dimensional since it is long and it has thermal symmetry about the center
line. 2 The thermal properties of the trunks 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 properties of the trunks are given to be k = 0.17 W/m.C and = 1.2810-7 m2/s.
Analysis We treat the trunks of the trees as an infinite cylinder since heat
transfer is primarily in the radial direction. Then the Biot number becomes
)m 1.0)(C. W/m65(2
hr
Tree
Ti = 30C
Hot
4-117
4-118
4-119
4-120