14-121
14-122
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 0131.19090.01041.0
kg/m 9090.0
K) 273+K)(55/kgkPa.m 287.0(
kPa )76.15325.101(
kg/m 1041.0
K) 273+K)(55/kgkPa.m 4615.0(
kPa 76.15
=+=+=
=
−
==
=
==
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
155)726.01052.1(15.0Pr)Gr(15.0Nu 3/193/1 ===
C) W/m02662.0)(155(
Nu 2
k
14-123
W14,209=++=++= 700,1014862023
evapconvradtoptotal, QQQQ
Therefore, if the water bath is heated electrically, a 14 kW resistance heater will be needed just to make up for the heat loss
225
2
)s/m 10(1.702
133)7255.01025.3(1.0Pr)Gr(1.0Nu 3/193/1 ===
C W/m54.3
m 1
C) W/m02662.0)(133(
Nu 2
conv =
== L
k
h
14-124
14-166 The mole fraction of the water vapor at the surface of a lake and the mole fraction of water in the lake are to be
determined and compared.
kPa 705.1
Csat@15vapor ==
PP
Air
92 kPa, 15C
14-125
14-168E The masses of the constituents of a gas mixture at a specified temperature and pressure are given. The partial
pressure of each gas and the volume of the mixture are to be determined.
Assumptions The gas mixture and its constituents are ideal gases.
Properties The molar masses of CO2 and CH4 are 44 and 16 lbm/lbmol, respectively (Table A-1E)
Analysis The mole numbers of each gas and of the mixture are
lbmol 1875.0
lbm/lbmol 16
lbm 3
:CH
lbmol 0227.0
lbm/lbmol 44
lbm 1
:CO
4
4
4
2
2
2
CH
CH
CH4
CO
CO
CO2
===
===
M
m
N
M
m
N
lbmol 2102.01875.00227.0
42 CHCOtotal =+=+= NNN
Using the ideal gas relation for the mixture and for the constituents, the volume of the mixture and the partial pressures of the
constituents are determined to be
3
ft 49.62=
== psia 25
)R 550)(Rlbmol/ftpsia 73lbmol)(10. 2102.0( 3
P
TNRu
V
psia 22.3
psia 2.70
=
==
=
==
3
3
CH
CH
3
3
CO
CO
ft 62.49
R) 550)(Rlbmol/ftpsia 73lbmol)(10. 1875.0(
ft 62.49
R) 550)(Rlbmol/ftpsia 73lbmol)(10. 0227.0(
4
4
2
2
V
V
TRN
P
TRN
P
u
u
1 lbm CO2
3 lbm CH4
550 R
25 psia
14-126
14-169 Dry air flows over a water body at constant pressure and temperature until it is saturated. The molar analysis of the
saturated air and the density of air before and after the process are to be determined.
Assumptions The air and the water vapor are ideal gases.
Properties The molar masses of N2, O2, Ar, and H2O are 28.0, 32.0, 39.95 and 18 kg / kmol, respectively (Table A-1). The
0.0097
0.2025
0.7566
0.0313
====
====
====
===
325.101
kPa) 156.98(01.0
325.101
kPa) 156.98(209.0
325.101
kPa) 156.98(781.0
325.101
169.3
airdry dry,Ar
Ar
Ar
airdry dry,OO
O
airdry dry,NN
N
OH
OH
22
2
22
2
2
2
P
Py
P
P
y
P
Py
P
P
y
P
Py
P
P
y
P
P
y
(b) The molar masses of dry and saturated air are
=++== kg/kmol0.2995.3901.00.32209.00.28781.0
airdry ii MyM
=+++== kg/kmol62.28180313.09.390097.00.322025.00.287566.0
airsat ii MyM
Then the densities of dry and saturated air are determined from the ideal gas relation to be
( )
( )
( )
3
kg/m1.186=
+
== K27325kg/kmol0.29/Km³/kmolkPa8.314
kPa325.101
/airdry
airdry TMR
P
u
( ) ( )
( )
3
kg/m1.170=
+
== K27325kg/kmol62.28/Km³/kmolkPa8.314
kPa325.101
/airsat
airsat TMR
P
u
Discussion We conclude that the density of saturated air is less than that of the dry air, as expected. This is due to the molar
mass of water being less than that of dry air.
20.9% O2
1% Ar
Water
14-127
14-170 Using the relation
)/400,17exp(1067.2 5TDAB −= −
the diffusion coefficient of carbon in steel is to be
14-128
14-129
5
2253
332
2
ave
3
1049.2
)s/m 10](1.493kg/m 2/)2018.12180.1[(
m) )(0.075kg/m 2018.1)(1.21803m/s 81.9()(
Gr =
+
−
=
−
=−
Lg s
Recognizing that this is a natural convection problem with cold horizontal surface facing up, the Nusselt number and the
convection heat transfer coefficients are determined to be (Table 14–13)
58.5)7316.01049.2(27.0Pr)Gr(27.0Nu 4/154/1 ===
and
C) W/m02495.0)(58.5(
Nu 2
k
14-130
14-172 Air is blown over a circular pan filled with water. The rate of evaporation of water, the rate of heat transfer by
convection, and the rate of energy supply to the water to maintain its temperature constant are to be determined.
Assumptions 1 The low mass flux model and thus the 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 15C). 2 The critical Reynolds number for flow over a
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Education.
14-131
3
K 273)+K)(15/kgmkPa 4615.0(
sv
Away from the surface:
3
3
,
,kg/m 00519.0
K 273)+K)(20/kgmkPa 4615.0(
kPa 7017.0 =
==
TR
P
v
v
v
5
evap
(1.089 10 kg/s)(2466 kJ/kg) 0.02685 kW 26.9 W
v fg
Q m h –
= = ´ = =
(c) The net rate of heat transfer to the water needed to maintain its temperature constant at 15C is
net conv
evap 26.9 ( 7.9)Q Q Q+= = + – = 19.0 W
14-132
14-173 Henry’s law is expressed as
H
P
y)0(
)0( side gas i,
sideliquidi, =
14-174 A glass of water is left in a room. The mole fraction of the water vapor in the air at the water surface and far from the
surface as well as the mole fraction of air in the water near the surface are to be determined when the water and the air are at
the same temperature.
kPa637.1)kPa339.2)(7.0(
C20@satair roomv, ===
PP
1.64%)(or 0.0164=== kPa100
kPa637.1
vapor
vapor P
P
y
kPa339.2
C20@interfacev, == sat
PP
. Then the mole fraction of water vapor in the air at the
2.34%)(or 0.0234=== kPa100
kPa339.2
surface v,
surface v, P
P
y
kPa 661.97339.2100
surface v,surface air, =−=−= PPP
0.0015%101.47 5==== −
bar 65,600
bar)325.101/661.97(
side gasair,dry
sideliquidair,dry H
P
y
Air
70% RH
Water
14-133
14-175 A 0.1-mm thick soft rubber membrane separates pure O2 from air. The mass flow rate of O2 through the membrane
per unit area and the direction of flow are to be determined.
Assumptions 1 Steady operating conditions exist. 2 Mass transfer through the membrane is one-dimensional. 3 The
atm 05.1atm) 5(21.0
22
2
2O2,O
2,O
O===→= PyP
P
P
y
O2
Rubber
membrane
14-134
14-135
14-136