8-1
Solutions Manual for
Heat and Mass Transfer: Fundamentals & Applications
6th Edition
Yunus A. Çengel, Afshin J. Ghajar
McGraw-Hill Education, 2020
Chapter 8
INTERNAL FORCED CONVECTION
PROPRIETARY AND CONFIDENTIAL
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8-2
General Flow Analysis
8-2C In fluid flow, it is convenient to work with an average or mean velocity Vavg and an average or mean temperature Tm
8-4C For flow through non-circular tubes, the Reynolds number as well as the Nusselt number and the friction factor are
A
4
8-3
8-11C The hydrodynamic and thermal entry lengths are given as
DLhRe05.0=
and
L D
t=005.Re Pr
for laminar flow, and
DLL th 10
in turbulent flow. Noting that Pr >> 1 for oils, the thermal entry length is larger than the hydrodynamic entry
length in laminar flow. In turbulent, the hydrodynamic and thermal entry lengths are independent of Re or Pr numbers, and
are comparable in magnitude.
8-12C The hydrodynamic and thermal entry lengths are given as
DLhRe05.0=
and
DL
t
PrRe05.0=
for laminar flow,
8-13C The region of flow over which the thermal boundary layer develops and reaches the tube center is called the thermal
8-17C When the surface temperature of tube is constant, the appropriate temperature difference for use in the Newton’s law
of cooling is logarithmic mean temperature difference that can be expressed as
)/ln(
ie
ie
TT
TT
−
8-18C The number of transfer units NTU is a measure of the heat transfer area and effectiveness of a heat transfer system. A
8-4
8-19 The average velocity and mean temperature are to be determined from the given velocity and temperature profiles.
Assumptions 1 Steady operating conditions exist. 2 Properties are constant.
Analysis The average velocity in a tube with a radius of R = D/2 is
R
2
8-5
8-20 The mass flow rate and the surface heat flux are to be determined from the given velocity and temperature profiles.
Assumptions 1 Steady operating conditions exist. 2 Properties are constant.
Properties The density of 1 atm air at 20°C is
= 1.204 kg/m3 (Table A-15).
Analysis The average velocity in a tube with a radius of R = D/2 is
R
2
8-7
8-22 Liquid water entering at 5°C and flowing at 0.0035 kg/s is heated in a circular tube at a rate of 1000 W. The water
temperature at the tube exit is to be determined if it exceeds 120°C. The inner surface temperature of the tube is to be
determined if it exceeds 79°C.
Assumptions 1 The flow is steady and incompressible. 2 Uniform tube surface temperature. 3 Inner surface of the tube is
smooth. 4 The convection heat transfer coefficient is constant. 5 The PVDC lining is very thin.
Properties The specific heat of water at 40°C is (Table A–9) cp = 4179 J/kg∙K.
Analysis (a) The heat transfer rate to the water is
Discussion The water exiting the tube is at a temperature below 120°C, which complies with the ASME Boiler and Pressure
8-9
8-24 Combustion gases passing through a tube are used to vaporize waste water. The tube length and the rate of evaporation
of water are to be determined.
Assumptions 1 Steady operating conditions exist. 2 The surface temperature of the pipe is constant. 3 The thermal resistance
of the pipe is negligible. 4 Air properties are to be used for exhaust gases.
Properties The properties of air at the average
temperature of (250+150)/2=200C are (Table A-15)
kJ/kg.K 287.0
CJ/kg. 1023
=
=
R
cp
Also, the heat of vaporization of water at 1 atm or
100C is
kJ/kg 2257=
fg
h
(Table A-9).
Analysis The density of air at the inlet and the mass
flow rate of exhaust gases are
3
kg/m 7662.0
K) 273250(kJ/kg.K) 287.0(
kPa 115 =
+
== RT
P
kg/s 0.007522=m/s) (5
4
m) (0.05
)kg/m 7662.0(
4
2
3
avg
2
avg
=
== V
D
VAm c
The rate of heat transfer is
W5.769)C150250)(CJ/kg. 1023)(kg/s 007522.0()( =−=−= eip TTcmQ
The logarithmic mean temperature difference and the surface area are
C82.79
250110
150110
ln
250150
ln
lm =
−
−
−
=
−
−
−
=
is
es
ie
TT
TT
TT
T
2
2
lm
lm m 0.08034
)C82.79)(C. W/m120(
W5.769 =
=
=⎯→⎯= Th
Q
AThAQss
Then the tube length becomes
cm 51.2====⎯→⎯= m 5115.0
m) 05.0(
m 0.08034 2
D
A
LDLAs
s
The rate of evaporation of water is determined from
kg/h 1.23=kg/s 0003409.0
kJ/kg 2257
kW 7695.0 ===⎯→⎯=
fg
evapfgevap h
Q
mhmQ
Ts=110C
L
D = 5 cm
Exh. gases
250C
5 m/s
150C
8-10
8-25 Combustion gases passing through a tube are used to vaporize waste water. The tube length and the rate of evaporation
of water are to be determined.
Assumptions 1 Steady operating conditions exist. 2 The surface temperature of the pipe is constant. 3 The thermal resistance
of the pipe is negligible. 4 Air properties are to be used for exhaust gases.
Properties The properties of air at the average temperature of (250+150)/2=200C are (Table A–15)
kJ/kg.K 287.0
CJ/kg. 1023
=
=
R
cp
Also, the heat of vaporization of water at 1 atm or 100C is
kJ/kg 2257=
fg
h
(Table A-9).
Analysis The density of air at the inlet and the mass flow rate of
exhaust gases are
3
kg/m 7662.0
K) 273250(kJ/kg.K) 287.0(
kPa 115 =
+
== RT
P
kg/s 0.007522=m/s) (5
4
m) (0.05
)kg/m 7662.0(
4
2
3
avg
2
avg
=
== V
D
VAm c
The rate of heat transfer is
W5.769)C150250)(CJ/kg. 1023)(kg/s 007522.0()( =−=−= eip TTcmQ
The logarithmic mean temperature difference and the surface area are
C82.79
250110
150110
ln
250150
ln
lm =
−
−
−
=
−
−
−
=
is
es
ie
TT
TT
TT
T
2
2
lm
lm m 0.2410
)C82.79)(C. W/m40(
W5.769 =
=
=⎯→⎯= Th
Q
AThAQss
Then the tube length becomes
cm 153====⎯→⎯= m 534.1
m) 05.0(
m 0.2410 2
D
A
LDLAs
s
The rate of evaporation of water is determined from
kg/h 1.23=kg/s 0003409.0
kJ/kg 2257
kW 7695.0 ===⎯→⎯=
fg
evapfgevap h
Q
mhmQ
Ts =110C
L
D = 5 cm
Exh. gases
250C
5 m/s
150C
8-11
8-26 Steam is condensed by cooling water flowing inside copper tubes. The average heat transfer coefficient and the number
of tubes needed are to be determined.
Assumptions 1 Steady operating conditions exist. 2 The surface temperature of the pipe is constant. 3 The thermal resistance
of the pipe is negligible.
Properties The properties of water at the average temperature of
(10+24)/2=17C are (Table A–9)
CJ/kg. 8.4183
kg/m 7.998 3
=
=
p
c
Also, the heat of vaporization of water at 30C is
kJ/kg 2431=
fg
h
.
Analysis The mass flow rate of water and the surface area are
kg/s 0.4518=m/s) (4
4
m) (0.012
)kg/m 7.998(
4
2
3
avg
2
avg
=
== V
D
VAm c
The rate of heat transfer for one tube is
W460,26)C1024)(CJ/kg. 8.4183)(kg/s 4518.0()( =−=−= iep TTcmQ
The logarithmic mean temperature difference and the surface area are
C63.11
1030
2430
ln
1024
ln
lm =
−
−
−
=
−
−
−
=
is
es
ie
TT
TT
TT
T
2
m 0.1885=m) m)(5 012.0(
== DLAs
The average heat transfer coefficient is determined from
C.kW/m 12.1 2=
=
=⎯→⎯= W1000
kW 1
)C63.11)(m 1885.0(
W460,26
2
lm
lm TA
Q
hThAQ
s
s
The total rate of heat transfer is determined from
kW 65.364kJ/kg) 2431)(kg/s 15.0( === fgcondtotal hmQ
Then the number of tubes becomes
13.8=== W460,26
W650,364
Q
Q
Ntotal
tube
Steam, 30C
L = 5 m
D = 1.2 cm
Water
10C
4 m/s
24C
8-12
8-27 Steam is condensed by cooling water flowing inside copper tubes. The average heat transfer coefficient and the number
of tubes needed are to be determined.
Assumptions 1 Steady operating conditions exist. 2 The surface temperature of the pipe is constant. 3 The thermal resistance
of the pipe is negligible.
Properties The properties of water at the average temperature of
(10+24)/2=17C are (Table A–9)
CJ/kg. 8.4183
kg/m 7.998 3
=
=
p
c
Also, the heat of vaporization of water at 30C is
kJ/kg 2431=
fg
h
.
Analysis The mass flow rate of water is
kg/s 0.4518=m/s) (4
4
m) (0.012
)kg/m 7.998(
4
2
3
avg
2
avg
=
== V
D
VAm c
The rate of heat transfer for one tube is
W460,26)C1024)(CJ/kg. 8.4183)(kg/s 4518.0()( =−=−= iep TTcmQ
The logarithmic mean temperature difference and the surface area are
C63.11
1030
2430
ln
1024
ln
lm =
−
−
−
=
−
−
−
=
is
es
ie
TT
TT
TT
T
2
m 0.1885=m) m)(5 012.0(
== DLAs
The average heat transfer coefficient is determined from
C.kW/m 12.1 2=
=
=⎯→⎯= W1000
kW 1
)C63.11)(m 1885.0(
W460,26
2
lm
lm TA
Q
hThAQ
s
s
The total rate of heat transfer is determined from
kW 6.1458kJ/kg) 2431)(kg/s 60.0( === fgcondtotal hmQ
Then the number of tubes becomes
55.1=== W460,26
W600,458,1
Q
Q
Ntotal
tube
Steam, 30C
L = 5 m
D = 1.2 cm
Water
10C
4 m/s
24C
8-13
8-28 Prob. 8–26 is reconsidered. The effect of the cooling water average velocity on the number of tubes needed to
achieve the indicated heat transfer rate in the condenser is to be investigated.
Analysis The problem is solved using EES, and the solution is given below.
“GIVEN”
T_i=10 [C]
T_e=24 [C]
T_s=30 [C]
L=5 [m]
D=1.2e-2 [m]
m_dot_cond=0.15
“PROPERTIES”
Fluid$=’water’
C_p=CP(Fluid$, T=T_ave, x=0)*Convert(kJ/kg-C, J/kg-C)
rho=Density(Fluid$, T=T_ave, x=0)
h_f=enthalpy(Fluid$, T=T_sat, x=0)
h_g=enthalpy(Fluid$, T=T_sat, x=1)
h_fg=h_g-h_f
T_ave=(T_i+T_e)/2 “T_ave = 1/2*(T_i+T_e)”
T_sat=T_s
“ANALYSIS”
A_c=pi#*D^2/4 “Cross-section area”
A_s=pi#*D*L “Surface area”
m_dot=rho*A_c*V_avg
DELTAT_lm=(T_i-T_e)/ln((T_s-T_e)/(T_s-T_i))
Q_dot=m_dot*C_p*(T_e-T_i)
Q_dot=h*A_s*DELTAT_lm
Q_dot_total=m_dot_cond*h_fg*1e3
N_tube=Q_dot_total/Q_dot
Vavg [m/s] Ntubes
0.3 183.6
0.4 137.7
0.5 110.2
0.6 91.81
0.7 78.70
0.8 68.86
0.9 61.21
1.0 55.09
1.2 45.91
1.4 39.35
1.6 34.43
1.8 30.60
2.0 27.54
2.5 22.03
3.0 18.36
3.5 15.74
40
80
120
160
200
Ntube
8-14
8-29 Steam is condensed by cooling water flowing inside copper tubes. The average heat transfer coefficient and the cooling
water mean velocity needed to achieve the indicated heat transfer rate in the condenser are to be determined.
Assumptions 1 Steady operating conditions exist. 2 The surface temperature of the pipe is constant. 3 The thermal resistance
of the tubes is negligible.
Properties The properties of liquid water at the bulk mean fluid temperature of Tb = (Ti + Te)/2 = (60C + 5C)/2 = 32.5C are
(Table A-9):
cp = 4178 J/kg∙K and ρ = 994.8 kg/m3
Also, the heat of vaporization of water at 68C is hfg = 2338 kJ/kg
Analysis The total rate of heat transfer from the condensation is
8-16
8-31 Hot air at 1 atm passing through a tube is used to boil water. The average heat transfer coefficient and the rate of
evaporation of water are to be determined.
Assumptions 1 Steady operating conditions exist. 2 The surface temperature of the pipe is constant. 3 The thermal resistance
of the tube is negligible.
Properties The properties of air at the bulk mean fluid temperature of Tb = (Ti + Te)/2 = (300C + 150C)/2 = 225C are
(Table A-15):
cp = 1029 J/kg∙K and ρ = 0.7085 kg/m3
Also, the heat of vaporization of water at 120C is hfg = 2203 kJ/kg (Table A-9).
Analysis The density of air at the inlet and the mass flow rate of exhaust gases are
kg/s 0.009738=m/s) (7
4
m) (0.05
)kg/m 7085.0(
4
2
3
avg
2
avg
=
== V
D
VAm c
8-19
Laminar and Turbulent Flow in Tubes
8-34C Yes, the volume flow rate in a circular pipe with laminar flow can be determined by measuring the velocity at the
8-36C The friction factor for flow in a tube is proportional to the pressure drop. Since the pressure drop along the flow is
D
2
8-39C In fully developed laminar flow in a circular pipe, the pressure drop is given by
2
avg
2
avg 328
D
LV
R
LV
P
==
The mean velocity can be expressed in terms of the flow rate as
4/
2
avg D
A
V
c
VV
==
. Substituting,
4222
avg
2
avg 128
4/
32
328
D
L
DD
L
D
LV
R
LV
P
V
V
====
Therefore, at constant flow rate and pipe length, the pressure drop is inversely proportional to the 4th power of diameter, and
thus reducing the pipe diameter by half will increase the pressure drop by a factor of 16 .
8-40C In fully developed laminar flow in a circular pipe, the pressure drop is given by
2
avg
2
avg 328
D
LV
R
LV
P
==
When the flow rate and thus mean velocity are held constant, the pressure drop becomes proportional to viscosity. Therefore,
pressure drop will be reduced by half when the viscosity is reduced by half.
8-20
8-41 In fully developed laminar flow in a circular pipe, the velocity at r = R/2 is measured. The velocity at the center of the
pipe (r = 0) is to be determined.
Assumptions The flow is steady, laminar, and fully developed.
Analysis The velocity profile in fully developed laminar flow
in a circular pipe is given by
−= 2
2
max 1)(
R
r
VrV
where Vmax is the maximum velocity which occurs at pipe
center, r = 0. At r =R/2,
4
3
4
1
1
)2/(
1)2/( max
max
2
2
max
V
V
R
R
VRV =
−=
−=
Solving for Vmax and substituting,
m/s 8=== 3
m/s) 6(4
3
)2/(4
max
RV
V
which is the velocity at the pipe center.
8-42 The velocity profile in fully developed laminar flow in a circular pipe is given. The mean and maximum velocities and
the volume flow rate are to be determined.
Assumptions The flow is steady, laminar, and fully developed.
Analysis The velocity profile in fully developed
laminar flow in a circular pipe is given by
−= 2
2
max 1)(
R
r
VrV
The velocity profile in this case is given by
)/1(4)( 22 RrrV −=
Comparing the two relations above gives the maximum velocity to
be Vmax = 4 m/s. Then the mean velocity and volume flow rate
become
m/s 2=== 2
m/s 4
2
max
avg
V
V
/sm 0.0628 3
==== ]m) (0.10m/s)[ 2()( 22
avgavg
RVAV c
V
R
r
0
Vmax
V(r)=Vmax(1–r2/R2)
R
r
0
Vmax
V(r)=Vmax(1–r2/R2)