Ideal Gas Mixture and Psychrometric Applications

Min Huang

Chemical Engineering Program

Tongji University

IDEAL GAS MIXTURE

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

yn

n

VnRT

VRTn

p

p

V

RTnp

ii

iiii

ii

/

/

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

• 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

mii

m

mmm

RT

Vpn

RT

Vpn ,

• Therefore,

• Then

• where

i

imiimmm

i

imiimmm pTsypTspTspTs ),(),(or ),(),(

i

i

ig

ii

ig

i

i

i

ig

i

ig

i

yRp

pRpTSpTS

yRp

pRpTSpTS

lnln),(),(

lnln),(),(

pRdpp

Rdp

T

VdS

dnVdpTdSdH

inT

r

i

ii

ln,

1

Obey the 2nd law

• 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

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

• 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

iiii

ii

VV

yn

n

pnRT

pRTn

V

V

p

RTnV

/

/

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

– (b)

component ni × Mi = mi i(%)

CO2 0.08 × 44 = 3.52 12.37

H2O 0.11 × 18 = 1.98 6.96

O2 0.07 × 32 = 2.24 7.87

N2 0.74 × 28 = 20.72 72.8

1.00 28.46 100

• 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

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.

• 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.

• (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

Kp

pTT

k

k

3741

3300

2.01

1

212

k

TTRM

m

k

VpVpW

constpV

pdVW

n

1

)(

1

121122

kJ

KKKmolkJmolkg

kg

W 21.3425.11

)300374)(//314.8(/79.35

)5.0(

• (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

13

13

2222

DDD

Entropy decreases in the process because entropy is transferred from the system accompanying heat transfer.

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.

• 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.

• 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

21

0

2

0

1

21

0

2

0

12

ln)(

)()()()(

0ln)()(

ln)()(

0

22

22222222

222

222

2222

p

pRyy

TsyTsyTsyTsy

p

pRTsTsy

p

pRTsTsy

sysyss

COO

COCOOOCOCOOO

COCOCO

OOO

COCOOO

DD

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

1ln314.8)8.02.0(663.2508.0358.2312.0

)(8.0)(2.0 2

0

2

0

22

TsTs COO

KkmolkJs

p

pRTsTss

O

OOO

D

D

/ 69.3)2.0ln(314.8358.231667.221

ln)()(

2

222

1

21

0

2

0

KkmolkJs

p

pRTsTss

O

COCOCO

D

D

/ 92.0)2.0ln(314.8663.250365.236

ln)()(

2

222

1

21

0

2

0

• (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

20

2

2

2

121

vvhh

2222 2121

2121

1COCOOO hhyhhy

MM

hhhh

kgkJ

hh

/ 7.194

18468271258.015320211842.06.41

121

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.

• 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.

• 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

K

kmolkmol

KkmolKkmolT

cncn

TcnTcnT

TTcnTTcn

KkmolkJ

KkmolkJ

KkmolkJ

KkmolkJ

OvONvN

OOvONNvN

OOvONNvN

261

21.079.0

30021.025079.0

0)()(

99.2082.20

99.2082.20

2

,,

,1,,1,

2

,12,,12,

2222

222222

222222

bars

bars

Kkmol

bars

Kkmol

Kkmol

p

Tn

p

Tn

Tnnp

V

nRTp

p

RTn

p

RTnV

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

22

,1,1

• (b)

• (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

112

T

QSS

0, adiabatic

),(),(22222222 ,1,11 OOOONNNN pTsnpTsnS

),(),( 22222 222222pyTsnpyTsnS OOONNN

• 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

pyR

T

Tcn

p

pyR

T

Tcn

pTspyTsn

pTspyTsn

constant p constant T

KkJ

p

pyR

T

Tcn

p

pyR

T

Tcn

O

O

O

OpO

N

N

N

NpN

/ 0.5

1

62.121.0ln314.8

300

261ln30.2921.0

2

62.179.0ln314.8

250

261ln13.2979.0

lnln

lnln

2

2

2

22

2

2

2

22

2

,1

2,

2

,1

2,

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

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.

• 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

KKkg

mN

p

RT

Mv

air

air

3

25

1

11, 875.0

/10

30597.28

8314

1

min3

3

1,

11, 29.114

/875.0

min/100 kg

air

airkgm

m

v

AVm

3,2,

3,1,

22 OO

airair

mm

mm

• 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

1331,32,

33,33,22,11,

2

22

kgThTh

ThThmm

ThmThmThmThm

OO

aaairO

OOaairOOaair

154.0 846.0

min/ 67.472.095.3

min/ 72.032

1.23

min/ 95.397.28

29.114

2

2

2

2

2

n

nyand

n

ny

kmolnnn

kmolM

mn

kmolM

mn

O

aira

air

Oa

O

O

O

air

aira

• (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

31

0

3

0

1133 ln,,p

py

M

RTsTspTspyTs a

air

aaaaa 2nd law

O

O

OOOOO yM

RTsTspTspyTs ln,,

2

2

0

3

0

2233

• The rate of entropy production becomes

min42.17

lnln

2

2 2

0

3

0

1

0

3

0

K

kJ

yM

RTsTsmy

M

RTsTsm O

O

OOOa

air

aaair

PSYCHROMETRIC APPLICATIONS

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.

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.

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

va

a

vvaava

hhhm

mh

m

H

hmhmHHH

• 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.

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.

• (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.

bar 0.01741 bar 0.024870.7φpp gv1

011.0622.01

11

v

v

aa

vv

pp

p

pM

pM

• (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

kgkg

m

kmmm

mm

lbmm

mmm

a

v

vvv

av

va

vvw

9891.00109.01

0109.01011.0/1

1

g 11/1/

/

1

1

11111

11

1

21

• 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

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.

• 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.

bar 1985.0985.110.0111 gv pp

• (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.

• 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

TMRv

v

vv

3

5

1

11 145.9

101985.0

393

18

8314/

kg 827.3145.9

35

1

1 v

vv

Vm

• 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

178.0100022.1447.51

100022.1145.93

3

22

22

2

fg

fv

vv

vvx

kg 681.0827.3178.02 vm

kg 146.3681.0827.3212 vvw mmm

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.

• 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

• 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

kg 398.4039397.28/8314

35101985.05.1

//

5

1

1

1

1

TMR

Vpp

TMR

Vpm

a

v

a

aa

Modeling an Adiabatic Saturation Process

• 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.

• 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

airmoistasvvasaa

watermakeup

aswvventeringairmoist

vvaa

ThmThm

ThmmThmThm

'

'

exiting

airmoistasgasa

watermakeupasf

enteringairmoistga

ThTh

ThThTh

'

'

avav mmandmm /' / '

• therefore

asfg

asfasgaasa

ThTh

ThThThTh

'

asg

asg

Tpp

Tp

622.0'

• 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

• 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

• 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.

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.

• (a) Mass balance

• Energy balance

watermm

airdrymm

vv

aa

21

21

2221110 vvaavvaa hmhmhmhmQ

• Solving for Q

1212

1212

vvaaa

vvvaaa

hhhhm

hhmhhmQ

barpp

barbarpp

p

TMRv

v

AVm

va

gv

a

a

a

a

9902.01

0098.0 01228.08.0

/

11

111

1

11

1

1

kgmva / 82.0

109902.0

28397.28

8314

3

51

airdrykg

vaporkg

pp

p

kgm

v

v

a

00616.0

0098.01

0098.0622.0622.0

min/ 9.18282.0

150

1

1

• Finally

• (b)

min/ 3717

8.25193.255600616.01.2832.3039.182

1212

kJ

hhhhmQ vvaaa

231.004246.0

0098.0

2

22

g

v

p

p

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.

• Solution

• (a) The humidity ratio 3 can be found by means of mass rate balances for the dry air and water vapor, respectively

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• The values of va1, and va2, and 2 are readily found from the psychrometric chart.

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• 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,

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• 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.

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Cooling Towers

• Forced-convection, counterflow cooling tower

Example

• Determine the mass flow rates of the dry air and the makeup water, in kg/h.

• 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,

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• 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,

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