3. Relating p, V, and T for Ideal Gas
Mixtures
• The Dalton model: the Dalton model assumes that
each mixture component behaves as an ideal gas as
if it were alone at the temperature T and volume V of
the mixture. Individual components would not exert
the mixture pressure p but rather a partial pressure.
pyp
y
n
n
VnRT
VRTn
p
p
V
RTn
p
ii
i
iii
i
i
/
/
4. Ideal Gas Mixture Properties
• Consider the internal energy and enthalpy of an ideal
gas mixture. Recall that
• For ideal gas, pv=nRT, The components of the
mixture exist at the same temperature as the
mixture.
r
i
ii
r
i
ii
dnVdpTdSdH
dnpdVTdSdE
1
1
5. • Therefore,
• What about the entropy of an ideal gas?
• The mixture must obey the Dalton model
i
miimm
i
miimm
i
miimm
i
miimm
ThyThThTh
TuyTuTuTu
)()(or)()(
)()(or)()(
i
miimmm TsypTs ?),(),(
m
mi
i
m
mm
m
RT
Vp
n
RT
Vp
n ,
6. • Therefore,
• Then
• where
i
imiimmm
i
imiimmm pTsypTspTspTs ),(),(or),(),(
i
i
ig
ii
ig
i
i
i
ig
i
ig
i
yR
p
p
RpTSpTS
yR
p
p
RpTSpTS
lnln),(),(
lnln),(),(
pRdp
p
R
dp
T
V
dS
dnVdpTdSdH
inT
r
i
ii
ln,
1
Obey the
2nd law
7. • For the Gibbs energy of an ideal-gas mixture
• Gig = Hig – TSig,
i
ig
i
ig
i
ig
i
i
ig
i
ig
i
ig
i
ig
i
ig
i
ig
i
yRTGG
yRTTSHG
STHG
ln
ln
8. summary
Using Dalton’s Law ...
orm m k k k m k k k
k k
u T w u T u y u T
orm m k k k m k k k
k k
h T w h T h y h T
orpm m k pk k pm k pk k
k k
c T w c T c y c T
orvm m k vk k vm k vk k
k k
c T w c T c y c T
, , or , ,m m m k k m k m m m k k m k
k k
s T P w s T P s T P y s T P
9. • Amagat model is that each mixture component
behaves as an ideal gas as if it existed separately at
the pressure p and temperature T of the mixture.
i
i
iii
i
i
VV
y
n
n
pnRT
pRTn
V
V
p
RTn
V
/
/
10. EXAMPLE 1
• Converting Mole Fractions to Mass Fractions
• The molar analysis of the gaseous products of
combustion of a certain hydrocarbon fuel is CO2, 0.08;
H2O, 0.11; O2, 0.07; N2, 0.74.
– (a)Determine the apparent molecular weight of the
mixture.
– (b)Determine the composition in terms of mass fractions.
• Solution
– (a) M = 0.08(44)+ 0.11(18)+ 0.07(32) + 0.74(28)
=28.46 g/mol
12. • Converting Mass Fractions to Mole Fractions
• A gas mixture has the following composition in terms
of mass fractions: H2, 0.10; N2, 0.60; CO2, 0.30.
Determine
– (a) the composition in terms of mole fractions and
– (b)the apparent molecular weight of the mixture.
• Solution
– (b) M = m/n = 100/78.2 = 12.79 g/mol
Component mi ÷ Mi = ni yi(%)
H2 10 ÷ 2 = 5.00 63.9
N2 60 ÷ 28 = 2.14 27.4
CO2 30 ÷ 44 = 0.68 8.7
100 7.82 100
13. EXAMPLE 2
• Compressing an Ideal Gas Mixture
• A mixture of 0.3 kg of carbon dioxide and 0.2 kg of
nitrogen is compressed from p1=1 bar,T1=300 K to
p2=3 bars in a polytropic process for which k=1.25.
Determine
– (a) the final temperature, in K,
– (b) the work, in kJ,
– (c) the heat transfer, in kJ,
– (d) the change in entropy of the mixture, in kJ/K.
14. • Assume
• As shown in the accompanying figure, the system is the mixture of
CO2 and N2. The mixture composition remains constant during the
compression.
• Each mixture component behaves as if it were an ideal gas
occupying the entire system volume at the mixture temperature.
The overall mixture acts as an ideal gas.
• The compression process is a polytropic process for which k=1.25.
• The changes in kinetic and potential energy between the initial and
final states can be ignored.
15. • (a) For an ideal gas, the temperatures and pressures
at the end states of a polytropic process,
• (b)
• m = 0.3+0.2, n = nCO2 + nN2, M = m/n
K
p
p
TT
k
k
374
1
3
300
2.0
1
1
2
12
k
TTR
M
m
k
VpVp
W
constpV
pdVW
n
1
)(
1
12
1122
kJ
KKKmolkJ
molkg
kg
W 21.34
25.11
)300374)(//314.8(
/79.35
)5.0(
16. • (c) The change in internal energy of the mixture
equals the sum of the internal energy changes of the
components.
• Inserting values for DU and W into the expression for
Q:
• (d)
kJ
U
TuTunTuTunU NNNCOCOCO
3.26
)62297770)(0071.0()69399198)(0068.0(
)()()()( 1212 222222
D
D
from Ideal gas tables
kJQ 91.721.343.26
KkJ
snsnS NNCOCO
/0231.0
314.8682.191105.1980071.0
314.8915.213475.2220068.0
1
3
1
3
2222
DDD
Entropy decreases in the process because entropy is
transferred from the system accompanying heat transfer.
17. EXAMPLE 3
• Gas Mixture Expanding Isentropically through a
Nozzle
• A gas mixture consisting of CO2 and O2 with mole
fractions 0.8 and 0.2, respectively, expands
isentropically and at steady state through a nozzle
from 700 K, 5 bars, 3 m/s to an exit pressure of 1 bar.
Determine
– (a) the temperature at the nozzle exit, in K,
– (b) the entropy changes of the CO2 and O2 from inlet to
exit, in
– (c) the exit velocity, in m/s.
18. • Assume
• The control volume shown by the dashed line on the
accompanying figure operates at steady state.
• The mixture composition remains constant as the mixture
expands isentropically through the nozzle. The overall mixture
and each mixture component act as ideal gases. The state of
each component is defined by the temperature and the
partial pressure of the component.
• The change in potential energy between inlet and exit can be
ignored.
19. • solution
• (a) The temperature at the exit can be determined
using the fact that the expansion occurs
isentropically
therefore
1
2
1
0
1
0
2
0
2
0
1
2
1
0
2
0
1
2
1
0
2
0
12
ln)(
)()()()(
0ln)()(
ln)()(
0
22
22222222
222
222
2222
p
p
Ryy
TsyTsyTsyTsy
p
p
RTsTsy
p
p
RTsTsy
sysyss
COO
COCOOOCOCOOO
COCOCO
OOO
COCOOO
DD
20. T1 = 700oC
Through iteration, T2 = 517.6 K
•(b) The change in the specific entropy for each of the
components can be determined using
5
1
ln314.8)8.02.0(663.2508.0358.2312.0
)(8.0)(2.0 2
0
2
0
22
TsTs COO
KkmolkJs
p
p
RTsTss
O
OOO
D
D
/69.3)2.0ln(314.8358.231667.221
ln)()(
2
222
1
2
1
0
2
0
KkmolkJs
p
p
RTsTss
O
COCOCO
D
D
/92.0)2.0ln(314.8663.250365.236
ln)()(
2
222
1
2
1
0
2
0
21. • (c) The energy rate balance for the one-inlet, one-
exit control volume at steady state
• where apparent molecular weight
M=(0.8)44+0.2(32)=41.6 kg/kmol
• v2 = 624 m/s
2
0
2
2
2
1
21
vv
hh
2222 2121
21
21
1
COCOOO hhyhhy
MM
hh
hh
kgkJ
hh
/7.194
18468271258.015320211842.0
6.41
1
21
22. EXAMPLE 4
• Adiabatic Mixing at Constant Total Volume
• Two rigid, insulated tanks are interconnected by a valve.
Initially 0.79 kmol of N2 at 2 bars and 250 K fills one tank.
The other tank contains 0.21 kmol of O2 at 1 bar and 300
K. The valve is opened and the gases are allowed to mix
until a final equilibrium state is attained. During this
process, there are no heat or work interactions between
the tank contents and the surroundings. Determine
– (a)the final temperature of the mixture, in K,
– (b)the final pressure of the mixture, in atm,
– (c)the amount of entropy produced in the mixing process, in
kJ/K.
23. • Assumptions:
• 1. The system is taken to be
the nitrogen and the oxygen
together.
• 2.When separate, each of the
gases behaves as an ideal gas.
The final mixture also acts as
an ideal gas. Each mixture
component occupies the total volume and exhibits the
mixture temperature.
• 3.No heat or work interactions occur with the surroundings,
and there are no changes in kinetic and potential energy.
24. • Solution
• (a) The final temperature of the mixture can be
determined from an energy balance. With
assumption 3, the closed system energy balance
reduces to
• or
012 D WQUUU
)()(
)()(
222
,1,11
2222
222222
TunTunU
TunTunU
OONN
OOONNN
0)()(
0)()()()(
222222
22222222
,12,,12,
,12,12
OOvONNvN
OOOONNNN
TTcnTTcn
TuTunTuTun
26.
bars
bars
Kkmol
bars
Kkmol
Kkmol
p
Tn
p
Tn
Tnn
p
V
nRT
p
p
RTn
p
RTn
V
O
OO
N
NN
ON
O
OO
N
NN
62.1
1
30021.0
2
25079.0
2610.1
2
22
2
22
22
2
22
2
22
,1,1
2
2
2
2
,1,1
• (b)
27. • (c) the closed system form of the entropy balance
• The initial entropy of the system,S1, is the sum of the
entropies of the gases at the respective initial states
• The final entropy of the system,S2, is the sum of the
entropies of the individual components, each
evaluated at the final mixture temperature and the
partial pressure of the component in the mixture
2
1
12
T
Q
SS
0, adiabatic
),(),( 22222222 ,1,11 OOOONNNN pTsnpTsnS
),(),( 22222 222222
pyTsnpyTsnS OOONNN
28. • entropy produced
• or
2
2
2
22
2
2
2
22
222222
222222
2
,1
2
,
2
,1
2
,
,122
,122
lnln
lnln
),(),(
),(),(
O
O
O
OpO
N
N
N
NpN
OOOOOO
NNNNNN
p
py
R
T
T
cn
p
py
R
T
T
cn
pTspyTsn
pTspyTsn
constant p constant T
29.
KkJ
p
py
R
T
T
cn
p
py
R
T
T
cn
O
O
O
OpO
N
N
N
NpN
/0.5
1
62.121.0
ln314.8
300
261
ln30.2921.0
2
62.179.0
ln314.8
250
261
ln13.2979.0
lnln
lnln
2
2
2
22
2
2
2
22
2
,1
2
,
2
,1
2
,
30. EXAMPLE 5
• Adiabatic Mixing of Two Streams
• At steady state, 100 m3/min of dry air at 32oC and 1
bar is mixed adiabatically with a stream of oxygen (O2)
at 127oC and 1 bar to form a mixed stream at 47oC
and 1 bar. Kinetic and potential energy effects can be
ignored. Determine
– (a)the mass flow rates of the dry air and oxygen, in kg/min,
– (b)the mole fractions of the dry air and oxygen in the
exiting mixture, and
– (c)the time rate of entropy production, in kJ/K min
31. Assume
• 1. steady state.
• 2.No heat transfer occurs with the surroundings.
• 3.Kinetic and potential energy effects can be ignored,
• 4.The entering gases can be regarded as ideal gases.
The exiting mixture can be regarded as an ideal gas
mixture.
• 5.The dry air is treated as a pure component.
32. • Solution
• the specific volume of the air at 1 is
• The mass flow rate of the dry air entering is
• From mass balances,
kg
m
mN
K
Kkg
mN
p
RT
M
v
air
air
3
25
1
1
1, 875.0
/10
305
97.28
8314
1
min3
3
1,
1
1, 29.114
/875.0
min/100 kg
air
air
kgm
m
v
AV
m
3,2,
3,1,
22 OO
airair
mm
mm
33. • The enthalpy of the mixture at the exit is evaluated
by summing the contributions of the air and oxygen,
each at the mixture temperature.
• (b) the mole fractions of the dry air and oxygen in
the exiting mixture
min/1.23
)()(
)()(
)()()()(
32
13
31,32,
33,33,22,11,
2
22
kg
ThTh
ThTh
mm
ThmThmThmThm
OO
aa
airO
OOaairOOaair
154.0846.0
min/67.472.095.3
min/72.0
32
1.23
min/95.3
97.28
29.114
2
2
2
2
2
n
n
yand
n
n
y
kmolnnn
kmol
M
m
n
kmol
M
m
n
O
air
a
air
Oa
O
O
O
air
air
a
34. • (c) The specific entropy of each component in the
exiting ideal gas mixture is evaluated at its partial
pressure in the mixture and at the mixture
temperature
• Since p1 = p3, the specific entropy change of the dry
air is
• since p2 = p3, the specific entropy change of the
oxygen is
0,,,, 333,333,222,111, 22
pTsmpTsmpTsmpTsm OOaairOOaair
113332,113331, ,,,, 2
pTspTsmpTspTsm OOOaaair
1
3
1
0
3
0
1133 ln,,
p
py
M
R
TsTspTspyTs a
air
aaaaa
2nd law
O
O
OOOOO y
M
R
TsTspTspyTs ln,,
2
2
0
3
0
2233
35. • The rate of entropy production becomes
min
42.17
lnln
2
2 2
0
3
0
1
0
3
0
K
kJ
y
M
R
TsTsmy
M
R
TsTsm O
O
OOOa
air
aaair
37. Psychrometrics
• Psychrometrics is the determination of physical and
thermodynamic properties of gas-vapor mixtures,
study of systems involving mixtures of dry air and
water vapor. A condensed water phase also may be
present.
• Such systems is essential for the analysis and design
of air-conditioning devices, cooling towers, and
industrial processes requiring close control of the
vapor content in air.
38. Moist Air
• The term moist air refers to a mixture of dry air and
water vapor in which the dry air is treated as if it
were a pure component.
• As can be verified by reference to appropriate
property data, the overall mixture and each mixture
component behave as ideal gases at the states under
present consideration.
• Accordingly, for the applications to be considered,
the ideal gas mixture concepts introduced previously
apply directly.
39. Humidity Ratio, Relative Humidity, and
Mixture Enthalpy
• Humidity ratio
• Relative humidity
• Mixture Enthalpy
aa
vv
aa
vv
air
vapor
pM
pM
RTVpM
RTVpM
m
m
/
/
pTg
v
pTsatv
v
p
p
y
y
,,,
vav
a
v
a
a
vvaava
hhh
m
m
h
m
H
hmhmHHH
40. • partial condensation of the water vapor can occur
when the temperature is reduced. Water vapor
would cool at constant pv from state 1 to state d,
called the dew point.
• the system would be cooled below the dew point
temperature, some of the water vapor would
condense. The vapor that remains can be regarded
as saturated at the final temperature, state 2.
41. Example 1
• Cooling Moist Air at Constant Pressure
• A 1 kg sample of moist air initially at 21oC, 1 bar, and
70% relative humidity is cooled to 5oC while keeping
the pressure constant. Determine
• (a)the initial humidity ratio,
• (b)the dew point temperature, in oC, and
• (c)the amount of water vapor that condenses, in kg.
42. • (a) The partial pressure of the water vapor, pv1 can be
found from the given relative humidity and pg from
Table at 21oC, 1 bar.
bar0.01741bar0.024870.7φpp gv1
011.0622.0
1
1
1
v
v
aa
vv
pp
p
pM
pM
43. • (b) The dew point temperature is the saturation
temperature corresponding to the partial pressure,
pv1. Interpolation in Table gives T=15.3oC.
• (c) The amount of condensate, mw, equals the
difference between the initial amount of water vapor
in the sample, mv1, and the final amount of water
vapor, mv2 ,
kgm
kg
kg
m
kmmm
mm
lbmm
mmm
a
v
vvv
av
va
vvw
9891.00109.01
0109.0
1011.0/1
1
g11/1/
/
1
1
11111
11
1
21
44. • the partial pressure of the water vapor remaining in
the system at the final state is the saturation
pressure corresponding to 5oC: pg=0.00872 bar
• The total condensate
0054.0622.02
g
g
pp
p
kgmm av 0053.09891.00054.022
kgmmm vvw 0056.00053.00109.021
45. Example 2
• Cooling Moist Air at Constant Volume
• An air–water vapor mixture is contained in a rigid, closed
vessel with a volume of 35 m3 at 1.5 bar, 120oC, and =
10%. The mixture is cooled at constant volume until its
temperature is reduced to 22oC. Determine
• (a)the dew point temperature corresponding to the
initial state, in oC,
• (b)the temperature at which condensation actually
begins, in oC, and
• (c)the amount of water condensed, in kg.
46. • The dew point temperature at the initial state is the
saturation temperature corresponding to the partial
pressure pv1.
• Interpolating, gives the dew point temperature as
60oC, which is the temperature condensation would
begin if the moist air were cooled at constant
pressure.
bar1985.0985.110.0111 gv pp
47. • (b) In the process from state 1 to state 1’, the water
exists as a vapor only. For the process from state 1’
to state 2, the water exists as a two-phase liquid–
vapor mixture. Note that pressure does not remain
constant during the cooling process from state 1 to
state 2.
• State 1’on the T–v diagram denotes the state where
the water vapor first becomes saturated. The
saturation temperature at this state is denoted as T’.
• Cooling to a temperature less than T’ would result in
condensation of some of the water vapor present.
Since state 1’is a saturated vapor state, the
temperature T’ can be found by interpolating.
48. • The specific volume of the vapor at state 1’equals
the specific volume of the vapor at state 1, which can
be evaluated from the ideal gas equation
• Interpolation with vv1 = vg gives T = 56oC.
• (c) The amount of condensate equals the difference
between the initial and final amounts of water vapor
present. The mass of the water vapor present initially
is
kg
mK
p
TMR
v
v
v
v
3
5
1
1
1 145.9
101985.0
393
18
8314/
kg827.3
145.9
35
1
1
v
v
v
V
m
49. • At the final state, the water forms a two-phase liquid–
vapor mixture having a specific volume of 9.145 m3/kg.
• the quality x2 of the liquid–vapor mixture can is
• where vf2 and vg2 are the saturated liquid and saturated
vapor specific volumes at T2=22oC, respectively.
• the mass of the water vapor contained in the system at
the final state is
• The mass of the condensate,mw2, is then
178.0
100022.1447.51
100022.1145.9
3
3
22
22
2
fg
fv
vv
vv
x
kg681.0827.3178.02 vm
kg146.3681.0827.3212 vvw mmm
50. Example 3
• Evaluating Heat Transfer for Moist Air Cooling at
Constant Volume
• An air–water vapor mixture is contained in a rigid,
closed vessel with a volume of 35 m3 at 1.5 bar,
120oC, and = 10%. The mixture is cooled until its
temperature is reduced to 22oC. Determine the heat
transfer during the process, in kJ.
51. • Solution
• closed system energy balance
• where
• The specific internal energy of the water vapor at the
initial state can be approximated as the saturated
vapor value at T1. At the final state, the water vapor
is assumed to exist as a saturated vapor, so its
specific internal energy is ug at T2. The liquid water at
the final state is saturated, so its specific internal
energy is uf at T2.
12 UUQ
WQU
D
22222222222
1111111
fwgvaawwvvaa
gvaavvaa
umumumumumumU
umumumumU
52. • Therefore,
• The mass of dry air, ma, can be found using the ideal
gas equation with the partial pressure of the dry air
at the initial state obtained using pv1=0.1985 bar.
• Q = -10,603 kJ
11222212 gvfwgvaaa umumumuumQ
kg398.40
39397.28/8314
35101985.05.1
//
5
1
1
1
1
TMR
Vpp
TMR
Vp
m
a
v
a
a
a
54. • The device is assumed to operate at steady state and
adiabatic.
• An air–water vapor mixture of unknown humidity ratio enters
the device at a known pressure p and temperature T.
• As the mixture passes through the device, it comes into
contact with a pool of water. If the entering mixture is not
saturated ( < 100%), some of the water would evaporate.
• The energy required to evaporate the water would come from
the moist air, so the mixture temperature would decrease as
the air passes through the duct.
• For a sufficiently long duct, the mixture would be saturated as
it exits ( = 100%).
• The temperature of the exiting mixture is the adiabatic-
saturation temperature.
55. • A steady flow of makeup water at temperature Tas is
added at the same rate at which water is evaporated.
• Dividing by the mass flow rate of the dry air, the
energy rate balance can be written on the basis of a
unit mass of dry air passing through the device as,
• where
exiting
airmoist
asvvasaa
water
makeup
aswvventering
airmoist
vvaa
ThmThm
ThmmThmThm
'
'
exiting
airmoistasgasa
water
makeupasf
entering
airmoistga
ThTh
ThThTh
'
'
avav mmandmm /'/ '
57. • Applying Mass and Energy Balances to Air-
Conditioning Systems
• Mass balance
• Or
• where
watermmm
airdrymm
vwv
aa
21
21
watermm aw 12
and 2211 avav mmmm
58. • energy balance,
• with
• Substitute mass balance
• into, we have,
• where
2221110 vvaawwvvaa hmhmhmhmhmQ
and 2211 avav mmmm
2221110 gaawwgaa hhmhmhhmQ
221211210 gwgaaa hhhhhmQ
watermm aw 12
2121 TTchh paaa
Ideal gas mixture Using steam table
59. • Evaluating the enthalpies of the water vapor as the
saturated vapor values at the respective
temperatures and the enthalpy of each liquid stream
as the saturated liquid enthalpy at the respective
temperature.
60. Example 1
• Heating Moist Air in a Duct
• Moist air enters a duct at 10oC, 80% relative humidity,
and a volumetric flow rate of 150 m3/min. The mixture is
heated as it flows through the duct and exits at 30oC. No
moisture is added or removed, and the mixture pressure
remains approximately constant at 1 bar. For steady-
state operation, determine
• (a)the rate of heat transfer, in kJ/min, and
• (b) the relative humidity at the exit. Changes in kinetic
and potential energy can be ignored.
61. • (a) Mass balance
• Energy balance
watermm
airdrymm
vv
aa
21
21
2221110 vvaavvaa hmhmhmhmQ
62. • Solving for Q
1212
1212
vvaaa
vvvaaa
hhhhm
hhmhhmQ
barpp
barbarpp
p
TMR
v
v
AV
m
va
gv
a
a
a
a
9902.01
0098.001228.08.0
/
11
111
1
1
1
1
1
kgmva /82.0
109902.0
283
97.28
8314
3
51
64. Example 2
• Adiabatic Mixing of Moist Streams
• A stream consisting of 142 m3/min of moist air at a
temperature of 5oC and a humidity ratio of 0.002
kg(vapor)/kg(dry air) is mixed adiabatically with a second
stream consisting of 425 m3/min of moist air at 24oC and
50% relative humidity. The pressure is constant
throughout at 1 bar. Using the psychrometric chart,
determine
• (a)the humidity ratio and
• (b)the temperature of the exiting mixed stream, in oC.
65. • Solution
• (a) The humidity ratio 3 can be found by means of
mass rate balances for the dry air and water vapor,
respectively
watermmm
airdrymmm
vvv
aaa
321
321
66. • Let
• Then
• Since
• Then
• The values of va1, and va2, and 2 are readily found
from the psychrometric chart.
333222111 , avavav mmandmmmm
watermmm aaa 332211
3
2211
3
a
aa
m
mm
213 aaa mmm
21
2211
3
aa
aa
mm
mm
2
2
2
1
1
1 ,
a
a
a
a
v
AV
m
v
AV
m
67. • Thus, at 1 = 0.002 and T1 = 5oC, va1 = 0.79 m3/kg(dry
air). At 2 = 50% and T2 = 24oC, va2 = 0.855 m3/kg(dry
air) and 2 = 0.0094.
• The mass flow rates of the dry air are then
kg(dry air)/min and kg(dry air)/min,
• (b) The temperature T3 of the exiting mixed stream
can be found from an energy rate balance.
• From the table,
1801 am
4972 am
)(
)(
0074.0
497180
4970094.0180002.0
3
airdrykg
vaporkg
332211 vaavaavaa hhmhhmhhm
)(/8.47
)(/10
1
1
airdrykgkJhh
airdrykgkJhh
va
va
68. • Therefore
• Using this value for the enthalpy of the moist air at
the exit, together with the previously determined
value for 3, fixes the state of the exiting moist air.
From inspection of table, T3 = 19oC.
)(
7.37
497180
)8.47(497)10(180
21
2211
3
airdrykg
kJ
mm
hhmhhm
hh
aa
vaavaa
va
71. • Mass balance
• Let
• Energy balance
• Evaluating the enthalpies of the water vapor as the
saturated vapor values at the respective
temperatures and the enthalpy of each liquid stream
as the saturated liquid enthalpy at the respective
temperature,
watermmmmm
airdrymm
vv
aa
42351
43
avav mmandmm 4433
)( 345 amm
4442255333110 vvaawwvvaaw hmhmhmhmhmhmhm
72. • the energy rate equation becomes,
• Let , ,
• Then
• The humidity ratios 3 and 4 can be determined
using the partial pressure of the water vapor
obtained with the respective relative humidity
3 = 0.00688 and 4 = 0.0327,
• Therefore,
4442255333110 gvaaffgvaaf hmhmhmhmhmhmhm
avav mmandmm 4433 )( 345 amm 21 mm
534334434
111
fggaa
ff
a
hhhhh
hhm
m
hkgm
hkgma
/1024.500688.00327.01003.2
/1003.2
57
5
7
