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Copyright ©2020 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill
Education.
6-1
Solutions Manual for
Heat and Mass Transfer: Fundamentals & Applications
6th Edition
Yunus A. Çengel, Afshin J. Ghajar
McGraw-Hill Education, 2020
Chapter 6
FUNDAMENTALS OF CONVECTION
PROPRIETARY AND CONFIDENTIAL
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6-2
Mechanism and Types of Convection
6-1C A fluid flow during which the density of the fluid remains nearly constant is called incompressible flow. A fluid whose
6-2C In forced convection, the fluid is forced to flow over a surface or in a tube by external means such as a pump or a fan.
In natural convection, any fluid motion is caused by natural means such as the buoyancy effect that manifests itself as the rise
6-3C If the fluid is forced to flow over a surface, it is called external forced convection. If it is forced to flow in a tube, it is
6-6C Nusselt number is the dimensionless convection heat transfer coefficient, and it represents the enhancement of heat
k
hL
Nu c
=
where Lc is the characteristic length of the surface and k is the thermal conductivity of the fluid.
6-7C Heat transfer through a fluid is conduction in the absence of bulk fluid motion, and convection in the presence of it. The
6-3
6-8 The rate of heat loss from an average man walking in still air is to be determined at different walking velocities.
Assumptions 1 Steady operating conditions exist. 2 Convection heat transfer coefficient is constant over the entire surface.
Analysis The convection heat transfer coefficients and the rate of heat losses at different walking velocities are
(a)
C. W/m956.5m/s) 5.0(6.86.8 20.5353.0 === Vh
6-9 The rate of heat loss from an average man in windy air is to be determined at different wind velocities.
Assumptions 1 Steady operating conditions exist. 2 Convection heat transfer coefficient is constant over the entire surface.
Analysis The convection heat transfer coefficients and the rate of heat losses at
different wind velocities are
(a)
C. W/m174.9m/s) 5.0(8.148.14 20.6969.0 === Vh
W296.3=−=−= C)10)(29m C)(1.7. W/m174.9()( 22
TThAQss
(b)
C. W/m8.14m/s) 0.1(8.148.14 20.6969.0 === Vh
W478.0=−=−= C)10)(29m C)(1.7. W/m8.14()( 22
TThAQss
(c)
C. W/m58.19m/s) 5.1(8.148.14 20.6969.0 === Vh
W632.4=−=−= C)10)(29m C)(1.7. W/m58.19()( 22
TThAQss
Air
V
T = 10C
Ts = 29C
6-4
6-10 Heat transfer coefficients at different air velocities are given during air cooling of potatoes. The initial rate of heat
transfer from a potato and the temperature gradient at the potato surface are to be determined.
Assumptions 1 Steady operating conditions exist. 2 Potato is spherical in shape. 3 Convection heat transfer coefficient is
constant over the entire surface.
Properties The thermal conductivity of the potato at the film temperature of Tf = (Ts + T∞)/2 = (20°C + 5°C)/2 = 12.5°C is
kfluid = 0.02458 W/m∙K (from Table A-15).
Analysis The initial rate of heat transfer from a potato is
222 m 02011.0m) 08.0( ===
DAs
W5.8=−=−= C5))(20m C)(0.02011. W/m1.19()( 22
TThAQss
where the heat transfer coefficient is obtained from the table at 1 m/s velocity. The
initial value of the temperature gradient at the potato surface is
C/m11,666−=
−
−=
−
−=
−=
−==
=
=
C W/m.02458.0
C5)C)(20. W/m1.19(
)(
)(
2
fluidcondconv
k
TTh
r
T
TTh
r
T
kqq
s
Rr
s
Rr
6-11 The upper surface of a solid plate is being cooled by water. The water convection heat transfer coefficient and the water
temperature gradient at the upper plate surface are to be determined.
Assumptions 1 Steady operating conditions exist. 2 Properties are constant. 3 Heat conduction in solid is one-dimensional. 4
No-slip condition at the plate surface.
Properties The thermal conductivity of the solid plate is given as k = 237 W/m∙K. The thermal conductivity of water at the
film temperature of Tf = (Ts,1 + T∞)/2 = (60°C + 20°C)/2 = 40°C is kfluid = 0.631 W/m∙K (from Table A-9).
Analysis Applying energy balance on the upper surface of the
solid plate (x = 0), we have
1,2,
−TTh
TT
ss
Air
V = 1 m/s
T = 5C
Potato
Ti = 20C
6-5
6-12 Airflow over a plate surface has a given temperature profile. The heat flux on the plate surface and the convection heat
transfer coefficient of the airflow are to be determined.
Assumptions 1 Steady operating conditions exist. 2 Properties are constant. 3 No-slip condition at the plate surface. 4 Heat
transfer by radiation is negligible.
Properties The thermal conductivity and thermal diffusivity of air at the film temperature of Tf = (Ts + T∞)/2 = (220°C +
20°C)/2 = 120°C are kfluid = 0.03235 W/m∙K and
fluid = 3.565 10−5 m2/s (from Table A-15).
6-8
6-17 An ASTM B98 copper-silicon bolt connects two metal plates. Hot gas flows across the bolt. The minimum heat
removal rate required to keep the bolt surface from going above the maximum use temperature set by the ASME Code for
Process Piping is to be determined.
Assumptions 1 Steady operating conditions exist. 2 Uniform surface temperature. 3 Wall effects from the plates are
negligible.
Analysis From the Newton’s law of cooling, the convection heat transfer rate is
6-18 An ASTM A479 904L stainless steel bar connects two metal plates. Hot gas flows across the square bar. The
maximum velocity that the gas can achieve without heating the stainless steel bar above the maximum use temperature set by
the ASME Code for Process Piping is to be determined.
Assumptions 1 Steady operating conditions exist. 2 Uniform surface temperature. 3 Wall effects from the plates are
negligible.
Discussion The maximum velocity that the gas can achieve without heating the stainless steel bar above the maximum use
6-10
6-20 A metal plate surface is being cooled by convection. The ratio of the temperature gradient in the fluid to the
temperature gradient in the plate at the surface is to be determined.
Assumptions 1 Steady operating conditions exist. 2 The thermal conductivities of the plate and the fluid are constant. 3
Convection heat transfer coefficient is constant over the entire surface. 4 No-slip condition at the plate surface.
Properties The thermal conductivities of the plate and the fluid are constant.
Analysis The temperature gradient in the fluid at the plate surface is
=
−
yTk
y0fluid,fluid )/(
TT
h
T
convcond qq =
0plate,
plate
=
y
y
or
TT
h
T
6-12
6-23 Heat transfer coefficient as a function of air velocity is given during air cooling of a flat plate. The effect of air
velocity on the air temperature and plate temperature gradients at the plate surface is to be determined.
Assumptions 1 Steady operating conditions exist. 2 The thermal conductivities or air and plate are constant. 3 Convection
heat transfer coefficient is constant over the entire surface. 4 No-slip condition at the plate surface.
Properties The thermal conductivity of the plate is given as kplate = 1.4 W/m∙K and the thermal conductivity of air is given as
kair = 0.0266 W/m∙K.
Analysis The air temperature gradient at the plate surface is
=
−
yTk
y0air,air )/(
TT
h
T
85.0
V −(∂T/∂y)0,air −(∂T/∂y)0,plate
[m/s] [K/m] [K/m]
0 0 0
0.1 10036 190.7
0.2 18091 343.7
0.3 25535 485.2
0.4 32609 619.6
0.5 39419 749.0
0.6 46027 874.5
0.7 52470 996.9
0.8 58777 1117
0.9 64966 1234
1.0 71053 1350
1.1 77048 1464
1.2 82963 1576
Discussion The air temperature gradient is larger than the temperature gradient for the plate, because the thermal
conductivity of the plate is larger than the thermal conductivity of air (kplate > kair).
0 0.2 0.4 0.6 0.8 1 1.2
0
20000
40000
60000
80000
V [m/s]
-(¶T/¶y)0 [K/m]
Plate
Air
6-14
6-25 Metal plates are being cooled by air. The effect of cooling time on the plates’ temperature gradient at the surface
is to be determined.
Assumptions 1 The thermal properties are constant. 2 The heat transfer coefficient is uniform over the entire surface. 3
Radiation effects are negligible. 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 the metal plate are given as k = 180 W/mK, ρ = 2800 kg/m3, and cp = 880 J/kgK.
Analysis The characteristic length and the Biot number of the metal plates are (note: plate thickness is 2L = 10 mm),
mm 10
2===== L
LA
V
)m 105)(K W/m30(
32
−
hLc
6-15
6-26 A stainless steel strip is heat treated as it moves through a furnace. The surface temperature gradient of the strip at mid-
length of the furnace is to be determined.
Assumptions 1 The thermal properties are constant. 2 The heat transfer coefficient is uniform over the entire surface. 3
Radiation effects are negligible. 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 stainless steel are given as k = 21 W/mK, ρ = 8000 kg/m3, and cp = 570 J/kgK.
Analysis The characteristic length and the Biot number of the stainless steel strip
mm 5.2
2
mm 5
2
2===== L
A
LA
A
L
s
c
V
(Note: the strip thickness is 2L = 5 mm)
)m 105.2)(K W/m80(
32
−
hLc
6-16
6-27 A steel strip is heat treated as it moves through a furnace. The surface temperature gradient of the strip as a
function of the furnace location is to be determined.
Assumptions 1 The thermal properties are constant. 2 The heat transfer coefficient is uniform over the entire surface. 3
Radiation effects are negligible. 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 stainless steel are given as k = 21 W/mK, ρ = 8000 kg/m3, and cp = 570 J/kgK.
Analysis The characteristic length and the Biot number of the steel strip are (note: plate thickness is 2L = 5 mm),
mm 5
2===== L
LA
V
)m 105.2)(K W/m80(
32
−
hLc
6-17
Boundary Layers and Flow Regimes
6-30C Viscosity is a measure of the “stickiness” or “resistance to deformation” of a fluid. It is due to the internal frictional
6-36C Reynolds number is the ratio of the inertial forces to viscous forces, and it serves as a criterion for determining the
6-18
6-40C Turbulent viscosity
t is caused by turbulent eddies, and it accounts for momentum transport by turbulent eddies. It is
u
6-19
6-43 Using the given velocity and temperature profiles, the expressions for friction coefficient and convection heat transfer
coefficient are to be determined.
Assumptions 1 The fluid is Newtonian. 2 Properties are constant. 3 No-slip condition at the plate surface.
Analysis The shear stress at the wall surface is
2
u
6-20
6-45 For air flowing over a flat plate, the effect of air velocity on the wall shear stress at x = 0.5 m and 1 m is to be
determined.
Assumptions 1 Isothermal condition exists between the flat plate and fluid flow. 2 Properties are constant. 3 Edge effects are
negligible.
Properties The properties of air at 20°C are ν = 1.516 10−5 m2/s and ρ = 1.204 kg/m3 (Table A-15).
Analysis The friction coefficient and the wall shear stress can be determined with
5.0
−
Vx
2
2
V
The problem is solved using EES, and the solution is given below:
"GIVEN"
x_1=0.5 [m]
x_2=1.0 [m]
"PROPERTIES"
nu=1.516e-5 [m^2/s]
rho=1.204 [kg/m^3]
"ANALYSIS"
C_f1=0.664*(V*x_1/nu)^(-0.5)
C_f2=0.664*(V*x_2/nu)^(-0.5)
tau_w1=C_f1*rho*V^2/2 "Wall shear stress at x = 0.5 m"
tau_w2=C_f2*rho*V^2/2 "Wall shear stress at x = 1.0 m"
V τw (x = 0.5 m) τw (x = 1 m)
[m/s] [N/m2] [N/m2]
0.5 0.0007782 0.0005503
1.0 0.002201 0.001556
1.5 0.004044 0.002859
2.0 0.006225 0.004402
2.5 0.008700 0.006152
3.0 0.01144 0.008087
3.5 0.01441 0.01019
4.0 0.01761 0.01245
4.5 0.02101 0.01486
5.0 0.02461 0.01740
5.5 0.02839 0.02008
6.0 0.03235 0.02287
Discussion As the air velocity increases, the wall shear stress also increases.
1 2 3 4 5 6
0
0.005
0.01
0.015
0.02
0.025
0.03
0.035
V [m/s]
tw [N/m2]
x = 0.5 m
x = 1.0 m
