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20-1
Solutions Manual
for
Fundamentals of Thermal Fluid Sciences
5th Edition
Yunus A. Çengel, John M. Cimbala, Robert H. Turner
McGraw-Hill, 2017
Chapter 20
NATURAL CONVECTION
PROPRIETARY AND CONFIDENTIAL
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20-2
Physical Mechanisms of Natural Convection
influence of natural means. Natural convection differs from forced convection in that fluid motion in natural convection is
caused by natural effects such as buoyancy.
involved.
water. Therefore, the hull of a ship will sink deeper in fresh water because of the smaller buoyancy force acting upwards.
“lifting” force. The buoyancy force is proportional to the density of the medium. Therefore, the buoyancy force is the largest
in mercury, followed by in water, air, and the evacuated chamber. Note that in an evacuated chamber there will be no
buoyancy force because of absence of any fluid in the medium.
forces in Reynolds number is replaced by the buoyancy forces in Grashof number.
P
20-3
20-9 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-15.
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
Analysis The volume expansion coefficient is defined as
P
T
For density varying with temperature at constant pressure, we can approximate
Hence, the volume expansion coefficient is calculated to be
K-1).
20-10 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-15).
dT
TP
Hence, at T = 70°C the volume expansion coefficient is
20-4
20-11 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
3-132
107.76
GrL
106.68
GrL
20-12E 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:
3-132
105.79 )63.3)(1059.1(Ra 8
water,D
(b)
109.31
GrD
and
101.22
Ra D
(c)
104.41
GrD
and
107.09
Ra D
(d)
101.11
GrD
and
108.02
Ra D
20-13C No, a hot surface will cool slower when facing down since the warmer air in this position cannot rise and escape
easily.
20-14C The heat flux will be higher at the bottom of the plate since the thickness of the boundary layer which is a measure of
thermal resistance is the lowest there.
Gr
20-16 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
307.2
502.5
)10705.9(387.0
825.0
Pr
Ra387.0
825.0Nu
m 2.0
Thus, the rate of heat transfer balance for conduction through the plate
thickness and natural convection is
k
1 53.5 9.708 × 109 307.2
3 47.2 7.194 × 109 280.7
5 46.3 6.870 × 109 276.8
20-7
20-17 Reconsider Prob. 20-16. A vertical plate separates the hot water from the cold water. The effect of kplate on Ts,c is
L=0.2 [m] "Plate length"
thickness=0.025 [m] "Plate thickness"
g=9.81 [m/s^2] "gravitational acceleration"
Fluid$='water'
k=Conductivity(Fluid$, T=T_film_c, x=0)
Pr=Prandtl(Fluid$, T=T_film_c, x=0)
4 27.88
8 36.57
12 42.59
20 50.97
30 58.01
40 63.05
50 66.90
80 74.57
130 81.49
200 86.53
040 80 120 160 200
70
80
90
Ts,c [°C]
20-8
20-18 A vertical plate separates the hot water from the cold air. The surface exposed to the cold air is subjected to radiation
heat transfer also. The temperature of the plate surface exposed to the cold air is to be determined.
0.7277 (Table A-22), and β = 1/Tf,c = 1/ 299.5 K.
The thermal conductivity and the emissivity of the plate are given as kplate = 1.5 W/m∙K and εplate = 0.73, respectively.
Analysis The Rayleigh number is
45.89
492.0
1
825.0
492.0
1
825.0Nu
m 2.0
Thus, the rate of heat transfer balance for conduction through the
plate thickness l, natural convection and radiation is
k
2 84.6 4.934 × 107 49.66 6.670
4 83.72 4.913 × 107 49.60 6.654
20-9
20-19 Reconsider Prob. 20-18. A vertical plate separates the hot water from the cold air. The surface exposed to the
L=0.2 [m]
g=9.81 [m/s^2] "gravitational acceleration"
Fluid$='air'
"Cold air"
k=Conductivity(Fluid$, T=T_film_c)
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]
0.02 86.40
0.04 76.79
0.08 63.74
0.09 61.27
0.10 59.03
0 0.02 0.04 0.06 0.08 0.1
60
100
20-20 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-22), and β = 1/Tf
= 1/303 K = 0.0033 K−1.
The Nusselt number relation for vertical plate is
7.103
7282.0
492.0
1
)10699.5(387.0
825.0
Pr
492.0
1
Ra387.0
825.0Nu
2
6/18
2
6/1
m 5.0
Thus, the heat flux on the plate surface can be determined from the heat loss by
natural convection and radiation on the other side of the plate:
20-21 A thin vertical plate is subjected to uniform heat flux on one side and exposed to hydrogen gas on the other side. The
plate midpoint temperature is to be determined.
Assumptions 1 Steady operating conditions exist. 2 Constant heat flux on the plate surface. 3 Thermal resistance in the plate
Analysis With the assumption that Tf = 50°C, the plate midpoint temperature is estimated as
The Rayleigh number is
)/sm 10240.1(
The Nusselt number relation for vertical plate is
74.35
7191.0
)10598.1(387.0
825.0
Pr
Ra387.0
825.0Nu
2
6/17
2
6/1
m 5.0
Thus, the plate midpoint temperature from the heat flux as
20-12
20-22 A thin vertical plate is subjected to uniform heat flux on one side and exposed to air on the other side. The plate
midpoint temperatures for (a) a highly polish copper surface and (b) a black oxidized copper surface are to be determined.
1.608 × 10−5 m2/s, Pr = 0.7282 (Table A-22), and β = 1/Tf = 1/ 303 K = 0.0033 K−1.
The emissivity of highly polished copper is ε = 0.02 and of black oxidized copper is ε = 0.78 (TableA-18).
Analysis With the assumption that Tf = 30°C, the plate midpoint temperature is estimated as
7.103
7282.0
1
825.0
Pr
1
825.0Nu
m 5.0
(a) With the known surface heat flux of 1000 W/m2 and ε = 0.02 (highly polished copper), the plate midpoint temperature can
be determined as
3 141.6 8.423 × 108 116.7 6.778
5 147.0 8.452 × 108 116.8 6.830
6 147.2 8.453 × 108 116.8 6.832
(a) (b)
(b) With the known surface heat flux of 1000 W/m2 and
ε = 0.78 (black oxidized copper), the plate midpoint
3 88.1 7.401 × 108 112.3 6.085
5 88.5 7.415 × 108 112.3 6.092
Discussion For part (a), the highly polished copper surface has a low emissivity, which limits the heat loss on the surface by
20-13
20-23 Reconsider Prob. 20-22. 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
L=0.5 [m]
g=9.81 [m/s^2] "gravitational acceleration"
Fluid$='air'
"Air"
k=Conductivity(Fluid$, T=T_film)
Pr=Prandtl(Fluid$, T=T_film)
(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
500 53.77 5.598E+08
600 61.34 6.106E+08
700 68.55 6.527E+08
for the highly polished copper surface is
higher than that of the black oxidized copper
surface. The highly polished copper surface
140
180
Highly polished
20-24 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
7323.0Pr
9
3-12
3
)m 2.1)(K 525)(K 34720.0)(m/s 81.9(
)(
0
cs LTTg
7.189
7323.0
492.0
1
825.0
Pr
492.0
1
825.0
Nu
m 2.1
L
convection ss TThAQ
W422.2 3.2349.187
)m 4.2)(C W/m.78.0(
s
kA
tQ
t
kA
C. W/m20.35 2
W2.422
or
total
Q
h
-5C
Glass
20-25E A hot plate with an insulated back is considered. The rate of heat loss by natural convection is to be determined for
different orientations.
Assumptions 1 Steady operating conditions exist. 2 Air is an ideal gas with constant
R)4605.102(
f
T
3-12
3
)(
LTTg
6.102
7256.0
)10503.5(387.0
825.0
Pr
Ra387.0
825.0
Nu
2
F.Btu/h.ft 7869.0)6.102(
FBtu/h.ft. 01535.0
Nu
k
h
3-12
3
)(
24.29)10598.8(54.054.0 4/164/1 RaNu
ft 5.0
L
c
number is
10598.8 Ra
. Then,
62.14)10598.8(27.027.0 4/164/1 RaNu
Insulation
20-16
20-26E Prob. 20-25E is reconsidered. The rate of natural convection heat transfer for different orientations of the plate
as a function of the plate temperature is to be plotted.
Analysis The problem is solved using EES, and the solution is given below.
"GIVEN"
g=32.2 [ft/s^2]
"ANALYSIS"
"(a), plate is vertical"
delta_a=L
Ra_a=(g*beta*(T_s-T_infinity)*delta_a^3)/nu^2*Pr
80
7.714
9.985
4.993
90
30.38
39.32
19.66
100
57.37
74.26
37.13
110
87.15
112.8
56.4
125
135.6
175.5
87.75
135
169.9
219.9
109.9
145
205.4
265.9
132.9
155
242.1
313.4
156.7
165
279.9
362.2
181.1
175
318.5
412.2
206.1
180
338.1
437.6
218.8
400
450
500
20-27 It is proposed that the side surfaces of a cubic industrial furnace be insulated for $550 in order to reduce the heat loss
by 90 percent. The thickness of the insulation and the payback period of the insulation to pay for itself from the energy it
saves are to be determined.
Assumptions 1 Steady operating conditions exist. 2 Air is an ideal gas
7177.0Pr
11
3-12
3
)m 3)(K 30110)(K 002915.0)(m/s 81.9(
)(
LTTg
1.545
7177.0
1
825.0
Pr
1
825.0
Nu
2
C.W/m235.5)1.545(
m 3
C W/m.02881.0
c
Nu
L
k
h
kJ 105,500
78.0
$12,816)therm/10.1)($ therms651,11(=energy) ofcost t saved)(UniEnergy (savedMoney
Hot gases
20-19
7255.0Pr
10
)m 3)(K 3050)(K 003195.0)(m/s 81.9(
)(
LTTg
5.400
7255.0
)10239.4(387.0
825.0
Pr
Ra387.0
825.0
Nu
m )2m)(3 3(4 insuls tA
Ts = 41.2C and tinsul = 0.0284 m = 2.84 cm
20-28 A circuit board containing square chips is mounted on a vertical wall in a room. The surface temperature of the chips is
side of the circuit board is negligible.
7282.0Pr
and repeat the calculations if necessary. The characteristic length in this case is the height of the board,
m. 5.0 LLc
Then,
8
)m 5.0)(K 2535)(K 0033.0)(m/s 81.9(
)(
LTTg
72.63
7282.0
)10140.1(387.0
825.0
Pr
Ra387.0
825.0
Nu
22
m 25.0m) 5.0(
m 5.0
s
A
L
Ts = 36.2C
which is sufficiently close to the assumed value in the evaluation of properties and h. Therefore, there is no need to repeat
Air
T = 25C
Tsurr = 25C
= 0.7
