20–23
m 0429.0
C W/m.02607.0 2
L
k
c
Heat loss by both natural convection and radiation heat can be expressed as
)()(
44
surrssss
TTATThAQ
C42.6
s
T
20–24
20-31 Prob. 20-30 is reconsidered. The effects of the room temperature and the emissivity of the board on the
w=0.15 [m]
T_infinity=20 [C]
Q_dot=8 [W]
epsilon=0.8
g=9.807 [m/s^2] “gravitational acceleration”
“ANALYSIS”
“(a), plate is vertical“
delta_a=L
Ra_a=(g*beta*(T_s_a-T_infinity)*delta_a^3)/nu^2*Pr
20–25
T
[F]
Ts,a
[C]
Ts,b
[C]
Ts,c
[C]
5
32.54
28.93
38.29
7
34.34
30.79
39.97
9
36.14
32.65
41.66
11
37.95
34.51
43.35
13
39.75
36.36
45.04
15
41.55
38.22
46.73
17
43.35
40.07
48.42
19
45.15
41.92
50.12
21
46.95
43.78
51.81
23
48.75
45.63
53.51
25
50.55
47.48
55.21
27
52.35
49.33
56.91
29
54.16
51.19
58.62
31
55.96
53.04
60.32
33
57.76
54.89
62.03
35
59.56
56.74
63.74
510 15 20 25 30 35
25
30
35
40
45
50
55
60
65
T [C]
Ts [C]
Ts,a
Ts,b
Ts,c
20–26
20-32 A vertical plate with length L is placed in a quiescent air, and the expressions, having the form
n
L
CRaNu
, for the
4/1
Ra59.0Nu L
k
hL
→
4/1
Ra59.0 L
L
k
h
Substituting the RaL yields
4/14/138 )/(51.1)10065.1(
02514.0
59.0 LTTL
L
h
104 < RaL < 109
3/1
Ra1.0Nu L
k
hL
→
3/1
Ra1.0 L
L
k
h
Substituting the RaL yields
3/13/138 19.1)10065.1(
02514.0
L
Discussion The average heat transfer coefficient for laminar conditions (104 < RaL < 109) is dependent on ΔT and L. In
turbulent conditions (1010 < RaL < 1013), the average heat transfer coefficient is not influenced by L.
20–28
20-34 Absorber plates whose back side is heavily insulated is placed horizontally outdoors. Solar radiation is incident on the
plate. The equilibrium temperature of the plate is to be determined for two cases.
Assumptions 1 Steady operating conditions exist. 2 Air is an ideal gas with constant
1–
25
K 002915.0
K)27370(
11
7177.0Pr
/sm 10995.1
C W/m.02881.0
f
T
k
Analysis The solution of this problem requires a trial-and-error approach since the determination of the Rayleigh number and
thus the Nusselt number depends on the surface temperature which is unknown. We start the solution process by “guessing”
the surface temperature to be 115C for the evaluation of the properties and h. We will check the accuracy of this guess later
and repeat the calculations if necessary. The characteristic length in this case is
m 24.0
)m 8.0m 2.1(2
)m 8.0)(m 2.1(
p
A
Ls
c
Then,
7
225
3-12
2
3
10414.6)7177.0(
)/sm 10995.1(
)m 24.0)(K 25115)(K 002915.0)(m/s 81.9(
Pr
)(
cs LTTg
Ra
04.60)10414.6(15.015.0 3/173/1 RaNu
m 24.0
C W/m.02881.0 2
L
k
c
2
m 96.0)m 2.1)(m 8.0(
s
A
In steady operation, the heat gain by the plate by absorption of solar radiation must be equal to the heat loss by natural
convection and radiation. Therefore,
Insulation
Air
T = 25C
Absorber plate
s = 0.87
= 0.09
600 W/m2
L = 1.2 m
20–29
)()(
22
44
skyssss
TTATThAQ
This is close to the assumed surface temperature of 95C. Therefore, there is no need to repeat the calculations.
If the absorber plate is made of ordinary aluminum which has a solar absorptivity of 0.28 and an emissivity of 0.07,
the rate of solar gain becomes
20–31
20-36 The required electrical power to maintain a specified surface temperature of a grill is to be determined.
Assumptions 1 Steady operating conditions exist. 2 Air is an ideal gas. 3 Thermal properties are constant.
Analysis Treating the grill as a horizontal circular plate, the characteristic length is
4/
2
D
A
20–34
20–39 Some cans move slowly in a hot water container made of sheet metal. The rate of heat loss from the four side surfaces
of the container and the annual cost of those heat losses are to be determined.
Assumptions 1 Steady operating conditions exist. 2 Air is an ideal gas with constant properties. 3 The local atmospheric
20–35
20–40 Some cans move slowly in a hot water container made of sheet metal. It is proposed to insulate the side and bottom
surfaces of the container for $350. The simple payback period of the insulation to pay for itself from the energy it saves is to
be determined.
Assumptions 1 Steady operating conditions exist. 2 Air is an ideal gas with constant properties. 3 The local atmospheric
1–
25
K 00338.0
K)27323(
11
7301.0Pr
/sm 10543.1
C W/m.02536.0
f
T
k
225
2
)/sm 10543.1(
53.56
7301.0
492.0
1
)10622.7(387.0
825.0
Pr
492.0
1
Ra387.0
825.0Nu
2
27/8
16/9
6/17
2
27/8
16/9
6/1
2
2
m 7.4)m 60.3)(m 5.0()m 10.1)(m 5.0(2
C. W/m868.2)53.56(
m 5.0
C W/m.02536.0
s
A
Nu
L
k
h
W5.97
])K 27320()K 27326)[(.K W/m1067.5)(m 7.4)(1.0(+C)2026)(m 7.4)(C. W/m868.2(
)()(
44428222
44
surrssssradconv TTATThAQQQ
In steady operation, the heat lost by the side surfaces of the tank must be equal to the heat lost from the exposed surface of
m 05.0
L
It gives Ts = 30.4C, which is sufficiently close to the assumed temperature, 26C. Therefore, there is no need to repeat the
calculations. The total amount of heat loss and its cost during one year are
Then money saved during a one-year period due to insulation becomes
1242$6.72$1315$CostCostsavedMoney
insulation
with
insulation
without
Aerosol can
Insulation
Water bath, 60C
20–36
20-41 A room is to be heated by a cylindrical coal-burning stove. The surface temperature of the stove and the amount of coal
burned during a 14-h period are to be determined.
Assumptions 1 Steady operating conditions exist. 2 Air is an ideal gas with constant properties. 3 The local atmospheric
m. 70.0 LLc
Then,
9
3-12
3
)m 70.0)(K 24130)(K 002857.0)(m/s 81.9(
)(
LTTg s
m 0740.0
)10203.1(
m) 70.0(35
Gr
35
m) 32.0( 4/1101/4
L
D
2.145
7161.0
492.0
1
)10709.1(387.0
825.0
Pr
492.0
1
Ra387.0
825.0
2
27/8
16/9
6/19
2
27/8
16/9
6/1
Nu
222
2
m 7841.04/)m 32.0()m 70.0)(m 32.0(4/
C. W/m080.6)2.145(
m 70.0
C W/m.02931.0
DDLA
Nu
L
k
h
s
Then the surface temperature of the stove is determined from
)297)(m C)(0.7841. W/m080.6( W1500
)()(
22
4
surr
4
radconv
s
ssss
T
TTATThAQQQ
kg 3.88
kJ/kg 30,000
kJ)/0.65 (75,600/
kJ 75,600=s/h) 3600h/day kJ/s)(14 5.1(
coal HV
Q
m
tQQ
20–38
20-43 A soda can placed horizontally in a refrigerator compartment and the heat transfer from the ends of the can are
negligible, determine the heat transfer rate from the can surface.
Assumptions 1 Steady operating conditions exist. 2 Air is an ideal gas. 3 Thermal properties are constant. 4 Radiation heat
transfer is negligible.
Analysis The Rayleigh number (Lc = D) is
225
3-12
2
3
)7309.0(
)m 06.0(K)436)(K 003413.0)(m/s 81.9(
Pr
)(
Ra
DTTg s
D
20–39
20-44 Heat generated by the electrical resistance of a bare cable is dissipated to the surrounding air. The surface temperature
of the cable is to be determined.
Assumptions 1 Steady operating conditions exist. 2 Air is an ideal gas with constant properties. 3 The local atmospheric
pressure is 1 atm. 4 The temperature of the surface of the cable is constant.
Properties We evaluate air properties at a film temperature of (Ts+T)/2 = 60C and 1 atm based on the problem statement.
Then, for an air temperature of T = 20C, the corresponding surface temperature is Ts = 100C. The properties of air at 1 atm
and 60C are (Table A-22)
1–
25
K 003003.0
K)27360(
11
7202.0Pr
/sm 10896.1
CW/m.02808.0
f
T
k
2.590)7202.0(
)/sm 10896.1(
)m 005.0)(K 20100)(K 003003.0)(m/s 81.9(
Pr
)(
225
3-12
2
3
DTTg
Ra s
346.2
7202.0/559.01
)2.590(387.0
6.0
Pr/559.01
387.0
6.0
2
27/8
16/9
6/1
2
27/8
16/9
6/1
Ra
Nu
2
2
m 06283.0)m 4)(m 005.0(
C. W/m17.13)346.2(
m 005.0
C W/m.02808.0
DLA
Nu
D
k
h
s
25
K 002915.0
K)27370(
11
7177.0Pr
/sm 10995.1
f
T
k
387.2
7177.0/559.01
)6.644(387.0
6.0
Pr/559.01
387.0
6.0
2
27/8
16/9
6/1
2
27/8
16/9
6/1
Nu
)(
22
ss
TThAQ
Air
T = 20C
Cable
Ts = ?
L=4 m
D = 5 mm
20–40
20–45 A horizontal hot water pipe passes through a large room. The rate of heat loss from the pipe by natural convection and
radiation is to be determined.
Assumptions 1 Steady operating conditions exist. 2 Air is an ideal gas with constant properties. 3 The local atmospheric
1–
25
K 003096.0
K)27350(
11
7228.0Pr
/sm 10798.1
C W/m.02735.0
f
T
k
Analysis (a) The characteristic length in this case is the outer diameter of the pipe,
m. 06.0 DLc
Then,
5
225
3-12
2
3
)/sm 10798.1(
)m 06.0)(K 2773)(K 003096.0)(m/s 81.9(
)(
DTTg
05.13
7228.0/559.01
)10747.6(387.0
6.0
Pr/559.01
387.0
6.0
2
27/8
16/9
6/15
2
27/8
16/9
6/1
Ra
Nu
2
C. W/m949.5)05.13(
C W/m.02735.0
Nu
k
h
Air
T = 27C
Pipe
Ts = 73C
= 0.8
L =10 m
D = 6 cm