Copyright ©2020 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill
Education.
Solutions Manual for
Heat and Mass Transfer: Fundamentals & Applications
6th Edition
Yunus A. Çengel, Afshin J. Ghajar
McGraw-Hill Education, 2020
Chapter 9
NATURAL CONVECTION
PROPRIETARY AND CONFIDENTIAL
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9-2
Physical Mechanisms of Natural Convection
9-1C Natural convection is the mode of heat transfer that occurs between a solid and a fluid which moves under the influence
9-5C The upward force exerted by a fluid on a body completely or partially immersed in it is called the buoyancy or “lifting”
9-9 The volume expansion coefficient is defined as
P
T
−
=
1
. For an ideal gas,
RTP
=
or
RT
P
=
, and thus
( ) ( )
TTRT
P
T
RT
P
T
RTP
P
1111
/
1
2==
=
−−
=
−=
9-3
9-10 The volume expansion coefficient of saturated liquid water at 70°C is to be determined using its definition and the
values tabulated in Table A-9.
Assumptions Density depends on temperature only and not pressure.
Properties The properties of sat. liq. water are listed in the following table:
T, °C
ρ, kg/m3
β, K-1
65
980.4
70
977.5
0.578 × 10−3
75
974.7
Analysis The volume expansion coefficient is defined as
1
9-11 Using the given ρ(T) correlation, the volume expansion coefficient of liquid water at 70°C is to be determined.
Assumptions Density depends on temperature only and not pressure.
Properties The volume expansion coefficient of liquid water at 70°C is 5.78 × 10−4 K-1 (Table A-9).
Analysis The volume expansion coefficient is defined as
111 T
d
9-4
9-12 The Grashof numbers for a plate placed in various fluids are to be determined.
Assumptions 1 Steady operating conditions exist. 2 Properties are constant. 3 Air behaves as an ideal gas.
Properties The properties of air, liq. water, and engine oil are listed in the following table:
Fluid
Tf , °C
ρ, kg/m3
μ, kg/m∙s
β, K-1
Air (Table A-15)
90
0.9718
2.139 × 10−5
2.755 × 10−3
Liq. water (Table A-9)
90
965.3
0.315 × 10−3
0.702 × 10−3
Engine oil (Table A-13)
80
852.0
3.232 × 10−2
0.700 × 10−3
For air (ideal gas) β = 1/Tf
Analysis The Grashof number is given as
3
3
)()(
cscs
LTTgLTTg −
−
9-5
9-13 The Grashof and Rayleigh numbers for a rod submerged in various fluids are to be determined.
Assumptions 1 Steady operating conditions exist. 2 Properties are constant. 3 Air behaves as an ideal gas. 4 The rod is
orientated such that the characteristic length is its diameter.
Properties The properties of air, liq. water, and engine oil are listed in the following table:
Fluid
Tf , °F
ρ, lbm/ft3
μ, lbm/ft∙s
Pr
β, R-1
Liq. water (Table A-9E)
120
61.71
3.744×10−4
3.63
0.246×10−3
Liq. ammonia (Table A-11E)
120
35.26
7.444×10−5
1.313
1.74×10−3
Engine oil (Table A-13E)
125
54.24
7.617×10−2
1607
0.389×10−3
Air (Table A-15E)
120
0.06843
1.316×10−5
0.723
1.72×10−3
For air (ideal gas), β = 1/Tf .
Analysis The Grashof and Rayleigh numbers are given as
3
3
)()(
DTTgDTTg ss
−
−
9-6
Natural Convection over Surfaces
35
L
9-7
9-18 A vertical plate separates the hot water from the cold water. The temperature of the plate surface on the cold water side
is to be determined.
Assumptions 1 Steady operating conditions exist. 2 Temperature of each surface is constant. 3 The plate thermal conductivity
is constant. 4 Radiation heat transfer is negligible.
Properties Assuming the surface temperature on the cold water side is Ts,c = (100 + 7)°C/2=53.5°C, thus Tf,c = (Ts,c + T∞,c)/2 =
(53.5 + 7)°C/2 = 30.25°C. Then, the properties of water at Tf,c are k = 0.6033 W/m∙K, ρ = 995.5 kg/m3, μ = 0.0007935 kg/m∙s,
ν = μ/ρ = 7.971 × 10−7 m2/s, Pr = 5.502, β = 0.0003072 K−1 (Table A-9).
The thermal conductivity of the plate is given as kplate = 15 W/m∙K.
Analysis The Rayleigh number is
9
3-12
3
,, 109.708)502.5(
)m 2.0(K )75.53)(K 0003072.0)(m/s 81.9(
)(
−
−
cccs LTTg
ccscshs TThTT
l
The above solution is repeated iteratively until Ts,c converges to Ts,c = 46°C.
Discussion The results from the iterative solution are listed in the following table:
Iter Ts,c [°C] Ra Nu
1 53.5 9.708 × 109 307.2
2 43.5 5.915 × 109 264.6
3 47.2 7.194 × 109 280.7
4 45.8 6.694 × 109 274.7
5 46.3 6.870 × 109 276.8
6 46.1 6.799 × 109 276.0
9-8
9-19 Reconsider Prob. 9-18. A vertical plate separates the hot water from the cold water. The effect of kplate on Ts,c is to
be investigated.
Analysis The problem is solved using EES, and the solution is given below.
“GIVEN”
L=0.2 [m]
thickness=0.025 [m]
T_infinity_c=7 [C]
T_s_h=100 [C]
“PROPERTIES”
g=9.81 [m/s^2]
Fluid$=’water’
k=Conductivity(Fluid$, T=T_film_c, x=0)
Pr=Prandtl(Fluid$, T=T_film_c, x=0)
rho=Density(Fluid$, T=T_film_c, x=0)
mu=Viscosity(Fluid$, T=T_film_c, x=0)
nu=mu/rho
beta=Volexpcoef(Fluid$, T=T_film_c, x=0)
T_film_c=1/2*(T_s_c+T_infinity_c)
“ANALYSIS”
Ra=(g*beta*(T_s_c-T_infinity_c)*L^3)/nu^2*Pr
Nusselt=(0.825+0.387*Ra^(1/6)/((1+(0.492/Pr)^(9/16))^(8/27)))^2
h=k/L*Nusselt
q_dot=k_plate/thickness*(T_s_h-T_s_c)
q_dot=h*(T_s_c-T_infinity_c)
kplate [W/m∙K] Ts,c [°C]
3 24.89
4 27.88
6 32.71
8 36.57
10 39.80
12 42.59
15 46.16
20 50.97
25 54.82
30 58.01
35 60.72
40 63.05
45 65.09
50 66.90
65 71.27
80 74.57
100 77.90
130 81.49
20
30
40
50
60
70
80
90
100
Ts,c [°C]
9-10
9-21 A vertical ASTM B152 copper plate with one surface subjected to natural convection with hot air. The rate of heat
removal from the plate is 90 W, and the maximum temperature that the air can reach without causing the surface temperature
of the copper plate to increase above 260°C is to be determined.
Assumptions 1 Steady state conditions. 2 Air is an ideal gas. 3 The local atmospheric pressure is 1 atm. 4 Thermal radiation
is negligible.
Properties The properties of air at the film temperature of Tf = 300°C are (Table A–15) Pr = 0.6935, k = 0.04418 W/m∙K, ν =
4.765 × 10−5 m2/s, β = 1/Tf = 1/(300 + 273 K) = 0.001745 K−1
Analysis The characteristic length of the plate is Lc = L = 0.5 m, and the Rayleigh number is
Ra𝐿=𝑔𝛽(𝑇∞−𝑇𝑠)𝐿3
𝐿{0.825+0.387 [𝑔𝛽(𝑇∞−𝑇𝑠)𝐿3Pr 𝜈2
[1+(0.492 Pr
⁄ )916
or
⁄
9-11
9-22 A vertical plate separates the hot water from the cold air. The surface exposed to the cold air is subjected to radiation
heat transfer also. The temperature of the plate surface exposed to the cold air is to be determined.
Assumptions 1 Steady operating conditions exist. 2 Temperature on each surface is constant. 3 The plate thermal
conductivity is constant. 4 Local atmospheric pressure is 1 atm. 5 The Tsurr is the same as the cold air temperature.
Properties Assuming the surface temperature on the cold air side is Ts,c = (100 + 2)°C/2=51°C, thus Tf,c = (Ts,c + T∞,c)/2 = (51
+ 2)°C/2 = 26.5°C. Then, the properties of air at Tf,c and 1 atm pressure are k = 0.02562 W/m∙K, ν = 1.576 × 10−5 m2/s, Pr =
0.7277 (Table A-15), and β = 1/Tf,c = 1/ 299.5 K.
The thermal conductivity and the emissivity of the plate are given as kplate = 1.5 W/m∙K and εplate = 0.73, respectively.
Analysis The Rayleigh number is
7
312
3
,, 10762.3)7277.0(
)m 2.0(K )251()K 2735.26)(m/s 81.9(
)(
−+
−
−
cccs LTTg
9-12
9-23 Reconsider Prob. 9-22. A vertical plate separates the hot water from the cold air. The surface exposed to the cold
air is subjected to radiation heat transfer also. The effect of the plate thickness on Ts,c is to be evaluated.
Analysis The problem is solved using EES, and the solution is given below.
“GIVEN”
L=0.2 [m]
T_infinity_c=2 [C]
T_s_h=100 [C]
“PROPERTIES”
g=9.81 [m/s^2] “gravitational acceleration”
Fluid$=’air’
“Cold air”
k=Conductivity(Fluid$, T=T_film_c)
Pr=Prandtl(Fluid$, T=T_film_c)
rho=Density(Fluid$, T=T_film_c, P=101.3)
mu=Viscosity(Fluid$, T=T_film_c)
nu=mu/rho
beta=Volexpcoef(Fluid$, T=T_film_c)
T_film_c=1/2*(T_s_c+T_infinity_c)
“Plate”
k_plate=1.5 [W/m-K]
epsilon_plate=0.73
“ANALYSIS”
Ra=(g*beta*(T_s_c-T_infinity_c)*L^3)/nu^2*Pr
Nusselt=(0.825+0.387*Ra^(1/6)/((1+(0.492/Pr)^(9/16))^(8/27)))^2
h=k/L*Nusselt
q_dot=k_plate/thickness*(T_s_h-T_s_c)
q_dot=h*(T_s_c-T_infinity_c)+sigma#*epsilon_plate*((T_s_c+273)^4-(T_infinity_c+273)^4)
Thickness [m] Ts,c [°C]
0.01 92.54
0.02 86.40
0.03 81.23
0.04 76.79
0.05 72.92
0.06 69.51
0.07 66.47
0.08 63.74
0.09 61.27
0.10 59.03
Discussion As the plate thickness
increases, the thermal resistance of the
plate increases, thus reducing the surface
temperature on the cold air side.
0 0.02 0.04 0.06 0.08 0.1
50
60
70
80
90
100
Plate thickness [m]
Ts,c [°C]
9-13
9-24 A CPVC tube is embedded in a vertical concrete wall, where the tube surface is 3 cm from the wall surface. The
surface of the wall is subjected to convection with hot quiescence air at 140°C. Would the tube comply with the ASME Code
for Process Piping?
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 No heat generation in the wall. 6 One-dimensional conduction. 7 Constant properties.
Properties The properties of air at the film temperature of Tf = (T + Ts )/2 = (140 +100)/2 = 120°C are (Table A–15) Pr =
0.7073, k = 0.03235 W/m∙K, ν = 2.522 × 10−5 m2/s, β = 1/Tf = 1/(120 + 273 K) = 0.002545 K−1
Analysis The characteristic length of the vertical wall is Lc = L = 1 m, and the Rayleigh number is
Ra𝐿=𝑔𝛽(𝑇∞−𝑇𝑠)𝐿3
9-15
9-26 A thin vertical plate is subjected to uniform heat flux on one side and exposed to cool air on the other side. The heat flux
on the plate is to be determined.
Assumptions 1 Steady operating conditions exist. 2 Constant heat flux on the plate surface. 3 Thermal resistance in the plate
is negligible. 4 Local atmospheric pressure is 1 atm. 5 The Tsurr is the same as the cool air temperature.
Properties The film temperature is determined with the plate midpoint temperature, Tf = (TL/2 + T∞)/2 = (55 + 5)°C/2 = 30°C.
Then, the properties of air at Tf = 30°C are k = 0.02588 W/m∙K, ν = 1.608 × 10−5 m2/s, Pr = 0.7282 (Table A-15), and β = 1/Tf
= 1/303 K = 0.0033 K−1.
Analysis The Rayleigh number is
8
312
3
2/ 10699.5)7282.0(
)m 5.0(K )555()K 27330)(m/s 81.9(
)(
−+
−
−
cL LTTg
9-18
9-29 Reconsider Prob. 9-28. A thin vertical copper plate is subjected to uniform heat flux on one side and exposed to
air on the other side. The effect of the heat flux on the plate midpoint temperature for (a) a highly polished copper surface
and (b) a black oxidized copper surface is to be determined.
Analysis The problem is solved using EES, and the solution is given below.
“GIVEN”
L=0.5 [m]
T_infinity=5 [C]
h=k/L*Nusselt
q_dot=h*(T_0.5L-T_infinity)+sigma#*epsilon*((T_0.5L+273)^4-(T_infinity+273)^4)
(a) Highly polished surface (ε = 0.02) (b) Black oxidized surface (ε = 0.78)
q
̇s [W/m2] Ts,c [°C] Ra
q
̇s [W/m2] Ts,c [°C] Ra
500 85.74 7.314E+08
600 98.57 7.735E+08
700 111.1 8.036E+08
800 123.3 8.243E+08
900 135.3 8.377E+08
1000 147.2 8.453E+08
1100 158.9 8.481E+08
1200 170.5 8.473E+08
1300 181.9 8.434E+08
1400 193.3 8.371E+08
1500 204.5 8.288E+08
500 53.77 5.598E+08
600 61.34 6.106E+08
700 68.55 6.527E+08
800 75.46 6.877E+08
900 82.11 7.170E+08
1000 88.52 7.416E+08
1100 94.71 7.621E+08
1200 100.7 7.794E+08
1300 106.5 7.938E+08
1400 112.2 8.058E+08
1500 117.7 8.157E+08
Discussion The plate midpoint temperature
for the highly polished copper surface is
higher than that of the black oxidized copper
surface. The highly polished copper surface
140
180
220
Highly polished
9-19
9-30 A street sign surface is subjected to radiation, the surface temperature of the street sign is to be determined.
Assumptions 1 Steady operating conditions exist. 2 Properties are constant. 3 The surface temperature is constant. 4 The
street sign is treated as a vertical plate. 5 Air is an ideal gas.
Properties The properties of air (1 atm) at 30°C are given in Table A-15: k = 0.02588 W/m∙K, ν = 1.608 × 10−5 m2/s, and Pr =
0.7282. Also, β = 1/Tf = 0.0033 K-1.
Analysis The Rayleigh number is
)m 2.0(K)293)(K 0033.0)(m/s 81.9(
)(
3-12
3
−
−
TLTTg
9-20
9-31 A glass window is considered. The convection heat transfer coefficient on the inner side of the window, the rate of total
heat transfer through the window, and the combined natural convection and radiation heat transfer coefficient on the outer
surface of the window 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 = (5+25)/2 = 15C are (Table A–15)
1–
25
K 003472.0
K)27315(
11
7323.0Pr
/sm 10470.1
C W/m.02476.0
=
+
==
=
=
=
−
f
T
k
Analysis (a) The characteristic length in this case is the height of the window,
m. 2.1== LLc
Then,
9
3-12
3
)m 2.1)(K 525)(K 34720.0)(m/s 81.9(
)( =
−
−
0
cs LTTg
Q
Outdoors
-5C
Glass
Ts = 5C
= 0.9
L = 1.2 m
Room
T = 25C