14-101
14-148 The liquid layer on the inner surface of a circular pipe is dried by blowing air through it. The mass transfer coefficient
is to be determined.
s/m 10332.2
1
)K 288(
1087.1
1087.1
25
072.2
10
072.2
10
air–OH2
−−
−
==
==
P
T
DD
AB
Analysis The Reynolds number of the flow is
m) m/s)(0.12 3(
VD
Air, 15C
1 atm, 3 m/s
Room, 15C
14-102
14-103
14-150 The inner surface of a pipe for transporting water contains lead. The level of lead in the water from the pipe is to
be determined.
A-9).
Analysis The Reynolds number of the flow is
Re=𝜌𝑉𝐷
𝜇=(998 kg/m3)(0.1 m/s)(0.02 m)
1.002×10−3kg/m∙s =1992<2300
4)= (0.1 m/s)𝜋(0.02 m)2
4=3.1416×10−5m3/s=0.031416 L/s
The level of lead in the water flowing from the pipe is
14-104
14-105
Simultaneous Heat and Mass Transfer
14–152C In steady operation, the mass transfer process does not have to involve heat transfer. However, a mass transfer
14-153C During evaporation from a water body to air, the latent heat of vaporization will be equal to convection heat transfer
14-154C It is possible for a shallow body of water to freeze during a cool and dry night even when the ambient air and
surrounding surface temperatures never drop to 0C. This is because when the air is not saturated (
< 100 percent), there will
14-106
14-107
14-108
14-157 Prob. 14-156 is reconsidered. The water temperature as a function of the relative humidity of air is to be
phi=0.40
“PROPERTIES”
Fluid$=’steam_IAPWS’
h_f=enthalpy(Fluid$, T=T_s, x=0)
h_g=enthalpy(Fluid$, T=T_s, x=1)
14-109
14-110
14-111
14-160 A thin layer of liquid water on a concrete surface is experiencing simultaneous heat and mass transfer. The
conduction heat flux through the concrete is to be determined.
Assumptions 1 The analogy between heat and mass transfer is applicable. 2 Steady state condition exists. 3 Constant
DAB = 2.50×10−5 m2/s (Table 14-4).
Analysis Applying the Chilton-Colburn analogy,
3/2
mass
heat
=
AB
pD
c
h
h
→
3/2
heat
mass
=
AB
p
D
c
h
h
14-112
14-113
 
ScRe06.0Re4.02Sh
4/1
0.43/22/1
++=
s
14-114
14-162 The heating system of a heated swimming pool is being designed. The rates of heat loss from the top surface of the
pool by radiation, natural convection, and evaporation are to be determined.
Assumptions 1 The low mass flux conditions exist so that the Chilton-Colburn analogy between heat and mass transfer is
applicable since the mass fraction of vapor in the air is low (about 2 percent for saturated air at 300 K). 2 Both air and water
14-115
Note that
s
, and thus this corresponds to hot surface facing up. The perimeter of the top surface of the pool is p =
1.5(25+ 25) = 75 m. Therefore, the characteristic length is
m 625 2
A
14-116
14-117
Note that
s
, and thus this corresponds to hot surface facing up. The perimeter of the top surface of the pool is p =
1.5(25+ 25) = 75 m. Therefore, the characteristic length is
m 625 2
A
14-118
14-164 Glass bottles are washed in hot water in an uncovered rectangular glass washing bath. The rates of heat loss from the
top and side surfaces of the bath by radiation, natural convection, and evaporation as well as the rates of heat and water mass
that need to be supplied to the water are to be determined.
Assumptions 1 The low mass flux conditions exist so that the
Chilton-Colburn analogy between heat and mass transfer is
negligible. 5 The air motion around the bath is negligible so that
there are no forced convection effects.
Properties The air-water vapor mixture is assumed to be dilute,
and thus we can use dry air properties for the mixture at the
( )
/sm1073.2
atm1
K5.310
1087.11087.1 25
072.2
10
072.2
10
air–OH2
−−− ==== P
T
DD AB
The saturation pressure of water at 25C is
kPa. 169.3
Csat@25 =
P
Properties of water at 50C are
kPa 35.12 and kJ/kg 2383 == vfg Ph
(Table A-9). The specific heat of water at the average temperature of (15+50)/2 =
32.5C is cBpB = 4.178 kJ/kg.C.
The gas constants of dry air and water are RBairB = 0.287 kPa.mP3P/kg.K and RBwaterB = 0.4615 kPa.mP3P/kg.K (Table
A-1). Also, the emissivities of water and the sheet metal are given to be 0.61 and 0.95, respectively, and the specific heat of
glass is given to be 1.0 kJ/kg.C.
Analysis (a) The mass flow rate of glass bottles through the water bath in steady operation is
kg/s 2=kg/min 120=n)bottles/mi (800kg/bottle) (0.150=rate flow Bottle
bottlebottle = mm
Air, 25C
Water
bath
14-119
Treating the water vapor and the air as ideal gases and noting that the total atmospheric pressure is the sum of the vapor and
3
,,
3
3
,
,
3
3
,
,
kg/m 0427.19598.00829.0
kg/m 9598.0
K) 273+K)(50/kgkPa.m 287.0(
kPa )35.12325.101(
kg/m 0829.0
K) 273+K)(50/kgkPa.m 4615.0(
kPa 35.12
=+=+=
=
−
==
=
==
sasvs
sa
sa
sa
sv
sv
sv
TR
P
TR
P
3
,,
3
3
,
,
3
3
,
,
kg/m 1777.11662.10115.0
kg/m 1662.1
K) 273+K)(25/kgmkPa 287.0(
kPa )585.1325.101(
kg/m 0115.0
K) 273+K)(25/kgmkPa 4615.0(
kPa 585.1
=+=+=
=
−
==
=
==
av
a
a
a
v
v
v
TR
P
TR
P
146)7262.01026.1(15.0Pr)Gr(15.0Nu 3/193/1 ===
C) W/m02643.0)(146(
Nu 2
k
14-120
W10,604=++=++= 772011581726
evapconvradtoptotal, QQQQ
Therefore, if the water bath is heated electrically, a 10.6 kW resistance heater will be needed just to make up for the heat loss
225
2
)s/m 10(1.679
127)7262.01080.2(1.0Pr)Gr(1.0Nu 3/193/1 ===
C W/m36.3
m 1
C) W/m02643.0)(127(
Nu 2
conv =
== L
k
h