Unlock document.
This document is partially blurred.
Unlock all pages and 1 million more documents.
Get Access
9-41
9-47 A can of engine oil placed vertically in the trunk of a car 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.
Properties The properties of air at Tf = (Ts + T∞)/2 = 30°C are k = 0.02588 W/m∙K,
= 1.608 × 10−5 m2/s, Pr = 0.7282 (from
Table A-15). Also, β = 1/Tf = 0.0033 K-1.
Analysis The Rayleigh number (Lc = L) is
3-12
3
)7282.0(
)m 15.0(K)1743)(K 0033.0)(m/s 81.9(
Pr
)(
Ra
−
=
−
=−
LTTg s
L
9-44
9-50 Water is boiling in a pan that is placed on top of a stove. The rate of heat loss from the cylindrical side surface of the
pan by natural convection and radiation and the ratio of heat lost from the side surfaces of the pan to that by the evaporation
of water are 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.
Properties The properties of air at 1 atm and the film temperature of
(Ts+T)/2 = (98+25)/2 = 61.5C are (Table A-15)
1-
25
K 00299.0
K)2735.61(
11
7198.0Pr
/sm 10910.1
C W/m.02819.0
=
+
==
=
=
=
−
f
T
k
Analysis (a) The characteristic length in this case is the height of the
pan,
m. 12.0== LLc
Then,
6
225
3-12
2
3
10299.7)7198.0(
)/sm 10910.1(
)m 12.0)(K 2598)(K 00299.0)(m/s 81.9(
Pr
)( =
−
=
−
=−
LTTg
Ra s
We can treat this vertical cylinder as a vertical plate since
4/14/164/1
35
thusand 0.25< 07443.0
)7198.0/10299.7(
)12.0(35
35
Gr
L
D
Gr
L=
=
Therefore,
60.28
7198.0
492.0
1
)10299.7(387.0
825.0
Pr
492.0
1
Ra387.0
825.0
2
27/8
16/9
6/16
2
27/8
16/9
6/1
=
+
+=
+
+=
Nu
2
2
m 09425.0)m 12.0)(m 25.0(
C. W/m720.6)60.28(
C W/m.02819.0
===
=
==
DLA
Nu
k
h
s
and
(b) The radiation heat loss from the pan is
−=
44
)( surrssrad TTAQ
Vapor
2 kg/h
Water
100C
Pan
Ts = 98C
= 0.1
Air
T = 25C
9-45
9-51 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
pressure is 1 atm. 3 Heat loss from the top surface is disregarded.
Properties The properties of air at 1 atm and the film temperature of
(Ts+T)/2 = (60+20)/2 = 40C are (Table A-15)
1-
25
K 003195.0
K)27340(
11
7255.0Pr
/sm 10702.1
K W/m02662.0
=
+
==
=
=
=
−
f
T
k
Analysis The characteristic length in this case is the height of the bath,
m. 5.0== LLc
Then,
8
3-12
3
)m 5.0)(K 2060)(K 003195.0)(m/s 81.9(
)( =
−
−
LTTg
2
m 5.4)m 5.3)(m 5.0()m 1)(m 5.0(2
=+=
s
A
and
Water bath
60C
Aerosol can
9-46
9-52 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
pressure is 1 atm. 3 Heat loss from the top surface is disregarded.
Properties Insulation will drop the outer surface temperature to a value close to the ambient temperature. 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 evaluate air properties at a film temperature of (Ts+T)/2 = 23C and 1 atm based on the problem statement.
Then, for an air temperature of T = 20C, the corresponding surface temperature is Ts = 26C. The properties of air at 1 atm
and 23C are (Table A-15)
1-
25
K 00338.0
K)27323(
11
7301.0Pr
/sm 10543.1
C W/m.02536.0
=
+
==
=
=
=
−
f
T
k
Analysis We start the solution process by “guessing” the outer
surface temperature to be 26C. We will check the accuracy of this
guess later and repeat the calculations if necessary with a better
guess based on the results obtained. The characteristic length in
this case is the height of the tank,
m. 5.0== LLc
Then,
7
3-12
3
)m 5.0)(K 2026)(K 00338.0)(m/s 81.9(
)( =
−
−
LTTg
W5.97
=
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
the insulation by convection and radiation, which must be equal to the heat conducted through the insulation. The second
conditions requires the surface temperature to be
C)(60
tank −
−
T
TT
Aerosol can
Insulation
Water bath, 60C
9-48
)10777.6(387.0
Ra387.0
2
6/19
2
6/1
9-49
9-54 A cylinder with specified length and diameter, the orientation of the cylinder that would achieve higher heat transfer rate
is to be determined.
Assumptions 1 Steady operating conditions exist. 2 Air is an ideal gas. 3 Thermal properties are constant. 4 Radiation heat
transfer is negligible.
Properties The properties of air at Tf = (Ts + T∞)/2 = 30°C are k = 0.02588 W/m∙K,
= 1.608 × 10−5 m2/s, Pr = 0.7282 (from
Table A-15). Also, β = 1/Tf = 0.0033 K-1.
Analysis For vertical orientation, the Rayleigh number (Lc = L) is
225
3-12
2
3
)7282.0(
)m 15.0(K)1743)(K 0033.0)(m/s 81.9(
Pr
)(
Ra
−
=
−
=−
LTTg s
L
9-50
9-55 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.
Properties The properties of air at Tf = (Ts + T∞)/2 = 20°C are k = 0.02514 W/m∙K,
= 1.516 × 10−5 m2/s, Pr = 0.7309 (from
Table A-15). Also, β = 1/Tf = 0.003413 K-1.
Analysis The Rayleigh number (Lc = D) is
3-12
3
)7309.0(
)m 06.0(K)436)(K 003413.0)(m/s 81.9(
Pr
)(
Ra
−
=
−
=−
DTTg s
D
9-51
9-56 A boiler supplies hot water to a dishwasher through a pipe at 10 g/s. The pipe dimensions are given. The water
exits the boiler at 85°C. The pipe section between the boiler and the dishwasher is exposed to natural convection. The water
temperature entering the dishwasher is to be determined whether it meets the ANSI/NSF 3 standard.
Assumptions 1 Steady state conditions. 2 Air is an ideal gas. 3 The local atmospheric pressure is 1 atm. 4 Thermal radiation
is negligible. 5 Surface temperature of the pipe is constant.
Properties The properties of air at the film temperature of Tf = (T + Ts )/2 = (20 +50)/2 = 35°C are (Table A–15) Pr =
0.7268, k = 0.02625 W/m∙K, ν = 1.655 × 10−5 m2/s, β = 1/Tf = = 1/ (35 + 273 K) = 0.003247 K−1. For water cp = 4.20 kJ/kg·K
Analysis The characteristic length of the horizontal pipe is D = 0.02 m, and the Rayleigh number is
Ra𝐷=𝑔𝛽(𝑇𝑠−𝑇∞)𝐷3
The rate of heat loss from the hot water flowing in the pipe is equal to the heat transfer rate by natural convection between the
pipe surface at the ambient air:
𝑄̇=𝑚̇𝑐𝑝(𝑇in −𝑇out)=ℎ𝐴𝑠(𝑇𝑠−𝑇∞)
9-53
9-58 A cylindrical resistance heater is placed horizontally in a fluid. The outer surface temperature of the resistance wire is to
be determined for two different fluids.
Assumptions 1 Steady operating conditions exist. 2 Air is an ideal gas with constant properties. 3 The local atmospheric
pressure is 1 atm. 4 Any heat transfer by radiation is ignored. 5 Properties are evaluated at 500C for air and 40C for water.
Properties The properties of air at 1 atm and 500C are (Table A-15)
1-
25
K 001294.0
K)273500(
11
,6986.0Pr
/sm 10804.7
C W/m.05572.0
=
+
==
=
=
=
−
f
T
k
The properties of water at 40C are (Table A-9)
1-
26
K 000377.0
32.4Pr
/sm 106582.0/
C W/m.631.0
=
=
==
=
−
k
Analysis (a) 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 1200C for the calculation of h. We will check the accuracy of this guess later and
repeat the calculations if necessary. The characteristic length in this case is the outer diameter of the wire,
m. 005.0== DLc
Then,
)m 005.0(C)201200)(K 001294.0)(m/s 81.9(
)(
3-12
3
−
−
DTTg
Air
T = 20C
Resistance
heater, Ts
300 W
L = 0.75 m
D = 0.5 cm
9-55
9-60 A thick fluid flows through a pipe in calm ambient air. The pipe is heated electrically. The power rating of the electric
resistance heater and the cost of electricity during a 15-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
pressure is 1 atm.
Properties The properties of air at 1 atm and the film temperature
of (Ts+T)/2 = (25+0)/2 = 12.5C are (Table A-15)
1-
25
K 003503.0
K)2735.12(
11
7330.0Pr
/sm 10448.1
C W/m.02458.0
=
+
==
=
=
=
−
f
T
k
Analysis The characteristic length in this case is the outer diameter of the pipe,
m. 3.0== DLc
Then,
7
3-12
3
)m 3.0)(K 025)(K 003503.0)(m/s 81.9(
)( =
−
−
cs LTTg
Asphalt
L = 100 m
D =30 cm
Ts = 25C
= 0.8
Tsky = -30C
T = 0C
9-56
9-61 A fluid flows through a pipe in calm ambient air. The pipe is heated electrically. The thickness of the insulation needed
to reduce the losses by 85% and the money saved during 15-h are 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.
Properties Insulation will drop the outer surface temperature to a value close to the ambient temperature, and possible below
it because of the very low sky temperature for radiation heat loss. For convenience, we use the properties of air at 1 atm and
5C (the anticipated film temperature) (Table A-15),
1-
25
K 003597.0
K)2735(
11
7350.0Pr
/sm 10382.1
C W/m.02401.0
=
+
==
=
=
=
−
f
T
k
Analysis The rate of heat loss in the previous problem was
obtained to be 29,094 W. Noting that insulation will cut down the
heat losses by 85%, the rate of heat loss will be
Asphalt
L = 100 m
D + 2tins
25C
Insulation
= 0.1
Tsky = -30C
T = 0C
9-57
9-62 A boiler supplies hot water to an equipment through a pipe at 10 g/s. The pipe dimensions are given. The water
enters the equipment at 98°C. The pipe section that is exposed to natural convection is 30 m long. The water temperature
exiting the boiler is to be determined whether it meets the ASME Boiler and Pressure Vessel Code.
Assumptions 1 Steady state conditions. 2 Air is an ideal gas. 3 The local atmospheric pressure is 1 atm. 4 Surface
temperature of the pipe is constant.
Properties The properties of air at the film temperature of Tf = (T + Ts )/2 = (20 +80)/2 = 50°C are (Table A–15) Pr =
0.7228, k = 0.02735 W/m∙K, ν = 1.798 × 10−5 m2/s, β = 1/Tf = 1/ (50 + 273 K) = 0.003096 K−1. For water cp = 4.20 kJ/kg·K
Analysis The characteristic length of the horizontal pipe is D = 0.02 m, and the Rayleigh number is
9-60
9-65 Prob. 9-64 is reconsidered. The effect of the surface temperature of the steam pipe on the rate of heat loss from
the pipe and the annual cost of this heat loss is to be investigated.
D=0.0603 [m]
T_s=170 [C]
T_infinity=20 [C]
epsilon=0.7
g=9.807 [m/s^2] “gravitational acceleration"
"ANALYSIS"
delta=D
Ra=(g*beta*(T_s-T_infinity)*delta^3)/nu^2*Pr
Nusselt=(0.6+(0.387*Ra^(1/6))/(1+(0.559/Pr)^(9/16))^(8/27))^2
140
145
150
155
15005
15876
16768
17680
5873
6214
6563
6920
12000
15000
5000
6000
