ThermoSolutions CHAPTER13

13-1

Chapter 13 GAS MIXTURES

Composition of Gas Mixtures

13-1C It is the average or the equivalent gas constant of the gas mixture. No.

13-2C No. We can do this only when each gas has the same mole fraction.

13-3C It is the average or the equivalent molar mass of the gas mixture. No.

13-4C The mass fractions will be identical, but the mole fractions will not.

13-5C Yes.

13-6C The ratio of the mass of a component to the mass of the mixture is called the mass fraction (mf),and the ratio of the mole number of a component to the mole number of the mixture is called the molefraction (y).

13-7C From the definition of mass fraction,

m

ii

mm

ii

m

ii M

My

MNMN

mm

mf

13-8C Yes, because both CO2 and N2O has the same molar mass, M = 44 kg/kmol.

13-9 A mixture consists of two gases. Relations for mole fractions when mass fractions are known are tobe obtained . Analysis The mass fractions of A and B are expressed as

m

BBB

m

AA

mm

AA

m

AA M

My

MM

yMNMN

mm

mfandmf

Where m is mass, M is the molar mass, N is the number of moles, and y is the mole fraction. The apparentmolar mass of the mixture is

BBAAm

BBAA

m

mm MyMy

NMNMN

Nm

M

Combining the two equation above and noting that 1BA yy gives the following convenient relationsfor converting mass fractions to mole fractions,

)1mf/1( BAA

BA MM

My and 1 AB yy

which are the desired relations.

13-2

13-10 The molar fractions of the constituents of moist air are given. The mass fractions of the constituentsare to be determined.Assumptions The small amounts of gases in air are ignored, and dry air is assumed to consist of N2 and O2only.Properties The molar masses of N2, O2, and H2O are 28.0, 32.0, and 18.0 kg/kmol, respectively (Table A-1).Analysis The molar mass of moist air is

M y Mi i 0 78 28 0 0 20 32 0 0 02 18 28 6. . . . . . kg / km Moist air 78% N220% O22% H2 O

(Mole fractions)

ol

Then the mass fractions of constituent gases are determined to be

0.7646.280.28)78.0(mf:N 2

22

NNN2 M

My

0.2246.280.32)20.0(mf:O 2

22

OOO2 M

My

0.0136.280.18)02.0(mf:OH OH

OHOH22

22 MM

y

Therefore, the mass fractions of N2, O2, and H2O in the air are 76.4%, 22.4%, and 1.3%, respectively.

13-11 The molar fractions of the constituents of a gas mixture are given. The gravimetric analysis of themixture, its molar mass, and gas constant are to be determined.Properties The molar masses of N2, and CO2 are 28.0 and 44.0 kg/kmol, respectively (Table A-1) Analysis Consider 100 kmol of mixture. Then the mass of each component and the total mass are

kg1760kg/kmol44kmol40kmol40


2222

2222

COCOCOCO

NNNN

MNmN

MNmN

kg3440kg1760kg168022 CON mmmm

mole

60% N240% CO2

Then the mass fraction of each component (gravimetric analysis) becomes

51.2%

48.8%

or0.512kg3440kg1760

mf

or0.488kg3440kg1680

mf

2

2

2

2

COCO

NN

m

m

mmmm

The molar mass and the gas constant of the mixture are determined from their definitions,

and

MmN

RRM

mm

m

mu

m

3,440 kg100 kmol

8.314 kJ / kmol K34.4 kg / kmol

34.40 kg / kmol

0.242 kJ / kg K

13-3

13-12 The molar fractions of the constituents of a gas mixture are given. The gravimetric analysis of themixture, its molar mass, and gas constant are to be determined.Properties The molar masses of O2 and CO2 are 32.0 and 44.0 kg/kmol, respectively (Table A-1) Analysis Consider 100 kmol of mixture. Then the mass of each component and the total mass are



2222

2222

COCOCOCO

OOOO

MNmN

MNmN

kg3680kg1760kg192022 COO mmmm

mole

60% O240% CO2


%8.47

%2.52

or0.478kg3680kg1760

mf

or0.522kg3680kg1920

mf

2

2

2

2

COCO

OO

m

m

mmmm


and

Kkg/kJ226.0

kmol/kg80.36

kg/kmol36.8KkJ/kmol8.314

kmol100kg3680

m

um

m

mm

MR

R

Nm

M

13-4

13-13 The masses of the constituents of a gas mixture are given. The mass fractions, the mole fractions, theaverage molar mass, and gas constant are to be determined.Properties The molar masses of O2, N2, and CO2 are 32.0, 28.0 and 44.0 kg/kmol, respectively (Table A-1) Analysis (a) The total mass of the mixture is

kg23kg10kg8kg5222 CONO mmmmm

5 kg O28 kg N2

10 kg CO2

Then the mass fraction of each component becomes

0.435

0.348

0.217

kg23kg10mf

kg23kg8mf

kg23kg5

mf

2

2

2

2

2

2

COCO

NN

OO

m

m

m

mmmmmm

(b) To find the mole fractions, we need to determine the mole numbers of each component first,

kmol0.227kg/kmol44

kg10

kmol0.286kg/kmol28

kg8

kmol0.156kg/kmol32

kg5

2

2

2

2

2

2

2

2

2

CO

COCO

N

NN

O

OO

Mm

N

Mm

N

Mm

N

Thus,kmol0.669kmol0.227kmol0.286kmol615.0

222 CONO NNNN m

and

0.339

0.428

0.233

kmol0.669kmol0.227

kmol0.669kmol0.286

kmol0.699kmol0.156

2

2

2

2

2

2

COCO

NN

OO

m

m

m

NN

y

NN

y

NN

y

(c) The average molar mass and gas constant of the mixture are determined from their definitions:

and

KkJ/kg0.242

kg/kmol34.4


kmol0.669kg23

m

um

m

mm

MR

R

Nm

M

13-5

13-14 The mass fractions of the constituents of a gas mixture are given. The mole fractions of the gas andgas constant are to be determined.Properties The molar masses of CH4, and CO2 are 16.0 and 44.0 kg/kmol, respectively (Table A-1) Analysis For convenience, consider 100 kg of the mixture. Then the number of moles of each componentand the total number of moles are

kmol0.568kg/kmol44

kg25kg25

kmol.6884kg/kmol16

kg75kg75

2

2

22

4

4

44

CO

COCOCO

CH

CHCHCH

Mm

Nm

Mm

Nm mass

75% CH425% CO2

kmol.2565kmol0.568kmol688.424 COCH NNN m

Then the mole fraction of each component becomes

10.8%

89.2%

or0.108kmol5.256kmol0.568


2

2

4

4

COCO

CHCH

m

m

NN

y

NN

y


and

Kkg/kJ0.437

/


kmolkg03.19kmol5.256kg100

m

um

m

mm

MR

R

Nm

M

13-15 The mole numbers of the constituents of a gas mixture are given. The mass of each gas and the apparent gas constant are to be determined.Properties The molar masses of H2, and N2 are 2.0 and 28.0 kg/kmol, respectively (Table A-1) Analysis The mass of each component is determined from

kg56

kg16

kg/kmol28kmol2kmol2

kg/kmol2.0kmol8kmol8

2222

2222

NNNN

HHHH

MNmN

MNmN

8 kmol H22 kmol N2

The total mass and the total number of moles are

kmol10kmol2kmol8

kg72kg56kg16

22

22

NH

NH

NNN

mmm

m

m


and

KkJ/kg1.155

/


kmolkg2.7kmol10

kg72

m

um

m

mm

MR

R

Nm

M

13-6

13-16E The mole numbers of the constituents of a gas mixture are given. The mass of each gas and the apparent gas constant are to be determined.Properties The molar masses of H2, and N2 are 2.0 and 28.0 lbm/lbmol, respectively (Table A-1E). Analysis The mass of each component is determined from

lbm112

lbm10

lbm/lbmol28lbmol4lbmol4

lbm/lbmol2.0lbmol5lbmol5

2222

2222

NNNN

HHHH

MNmN

MNmN

5 lbmol H24 lbmol N2

The total mass and the total number of moles are

lbmol9lbmol4lbmol5

lbm122lbm112lbm10

22

22

NH

NH

NNN

mmm

m

m


lbmollbm5613.lbmol9

lbm122/

m

mm N

mM

and Rlbm/Btu1465.0lbm/lbmol13.56

RBtu/lbmol1.986

m

um M

RR

13-17 The mass fractions of the constituents of a gas mixture are given. The volumetric analysis of themixture and the apparent gas constant are to be determined.Properties The molar masses of O2, N2 and CO2 are 32.0, 28, and 44.0 kg/kmol, respectively (Table A-1) Analysis For convenience, consider 100 kg of the mixture. Then the number of moles of each componentand the total number of moles are

kmol136.1kg/kmol44

kg50kg50

kmol071.1kg/kmol28

kg30kg20

kmol625.0kg/kmol32

kg20kg20

2

2

22

2

2

22

2

2

22

CO

COCOCO

N

NNN

O

OOO

Mm

Nm

Mm

Nm

Mm

Nm

mass

20% O230% N2

50% CO2

kmol832.2136.1071.1625.0222 CONO NNNN m

Noting that the volume fractions are same as the mole fractions, the volume fraction of each componentbecomes

40.1%

37.8%

22.1%




2

2

2

2

2

2

COCO

NN

OO

m

m

m

NN

y

NN

y

NN

y


kmolkg31.35kmol2.832kg100 /

m

mm N

mM

and Kkg/kJ0.235kg/kmol35.31

KkJ/kmol8.314

m

um M

RR

13-7

P-v-T Behavior of Gas Mixtures

13-18C Normally yes. Air, for example, behaves as an ideal gas in the range of temperatures and pressures at which oxygen and nitrogen behave as ideal gases.

13-19C The pressure of a gas mixture is equal to the sum of the pressures each gas would exert if existedalone at the mixture temperature and volume. This law holds exactly for ideal gas mixtures, but onlyapproximately for real gas mixtures.

13-20C The volume of a gas mixture is equal to the sum of the volumes each gas would occupy if existed alone at the mixture temperature and pressure. This law holds exactly for ideal gas mixtures, but onlyapproximately for real gas mixtures.

13-21C The P-v-T behavior of a component in an ideal gas mixture is expressed by the ideal gas equationof state using the properties of the individual component instead of the mixture, Pivi = RiTi. The P-v-Tbehavior of a component in a real gas mixture is expressed by more complex equations of state, or by Pivi= ZiRiTi, where Zi is the compressibility factor.

13-22C Component pressure is the pressure a component would exert if existed alone at the mixturetemperature and volume. Partial pressure is the quantity yiPm, where yi is the mole fraction of component i.These two are identical for ideal gases.

13-23C Component volume is the volume a component would occupy if existed alone at the mixturetemperature and pressure. Partial volume is the quantity yiVm, where yi is the mole fraction of component i.These two are identical for ideal gases.

13-24C The one with the highest mole number.

13-25C The partial pressures will decrease but the pressure fractions will remain the same.

13-26C The partial pressures will increase but the pressure fractions will remain the same.

13-27C No. The correct expression is “the volume of a gas mixture is equal to the sum of the volumeseach gas would occupy if existed alone at the mixture temperature and pressure.”

13-28C No. The correct expression is “the temperature of a gas mixture is equal to the temperature of the individual gas components.”

13-29C Yes, it is correct.

13-30C With Kay's rule, a real-gas mixture is treated as a pure substance whose critical pressure and temperature are defined in terms of the critical pressures and temperatures of the mixture components as

iimiim TyTPyP ,cr,cr,cr,cr and

The compressibility factor of the mixture (Zm) is then easily determined using these pseudo-critical point values.

13-8

13-31 A tank contains a mixture of two gases of known masses at a specified pressure and temperature.The volume of the tank is to be determined.Assumptions Under specified conditions both O2 and CO2 can be treated as ideal gases, and the mixture as an ideal gas mixture.

8 kmol O210 kmol CO2

290 K 150 kPa

Analysis The total number of moles is

Then

3m289.3kPa150

K)K)(290/kmolmkPa4kmol)(8.31(18

kmol18kmol10kmol8

3

COO 22

m

mumm

m

PTRN

NNN

V

13-32 A tank contains a mixture of two gases of known masses at a specified pressure and temperature.The volume of the tank is to be determined.Assumptions Under specified conditions both O2 and CO2 can be treated as ideal gases, and the mixture as an ideal gas mixture.

8 kmol O210 kmol CO2

400 K 150 kPa

Analysis The total number of moles is

Then

3m399.1kPa150


kmol18kmol10kmol8

3

COO 22

m

mumm

m

PTRN

NNN

V

13-33 A tank contains a mixture of two gases of known masses at a specified pressure and temperature.The mixture is now heated to a specified temperature. The volume of the tank and the final pressure of themixture are to be determined.Assumptions Under specified conditions both Ar and N2 can be treated as ideal gases, and the mixture as an ideal gas mixture.Analysis The total number of moles is

0.5 kmol Ar 2 kmol N2

280 K 250 kPa Q

And

3m23.3kPa250

K)K)(280/kmolmkPa4kmol)(8.31(2.5

kmol2.5kmol2kmol5.0

3

NAr 2

m

mumm

m

PTRN

NNN

V

Also,

kPa357.1)kPa(250K280K400

11

22

1

11

2

22 PTT

PT

PT

P VV

13-9

13-34 The masses of the constituents of a gas mixture at a specified pressure and temperature are given.The partial pressure of each gas and the apparent molar mass of the gas mixture are to be determined.Assumptions Under specified conditions both CO2 and CH4 can be treated as ideal gases, and the mixtureas an ideal gas mixture.Properties The molar masses of CO2 and CH4 are 44.0 and 16.0 kg/kmol, respectively (Table A-1) Analysis The mole numbers of the constituents are

m NmM

m NmM

CO COCO

CO

CH CHCH

CH

2 2

2

2

4 4

4

4

1 kg 1 kg44 kg / kmol

0.0227 kmol

kg 3 kg16 kg / kmol

0.1875 kmol3

1 kg CO23 kg CH4

300 K 200 kPa

N N Nm CO CH2 4kmol 0.1875 kmol 0.2102 kmol0 0227.

yNN

yNN

m

m

COCO

CHCH

2

2

4

4

0.0227 kmol0.2102 kmol

0.1875 kmol0.2102 kmol

0108

0 892

.

.

Then the partial pressures become

kPa4.178

kPa6.21

kPa200892.0

kPa2000.108

44

22

CHCH

COCO

m

m

PyP

PyP

The apparent molar mass of the mixture is

MmNm

m

m

4 kg0.2102 kmol

19.03 kg / kmol

13-35E The masses of the constituents of a gas mixture at a specified pressure and temperature are given.The partial pressure of each gas and the apparent molar mass of the gas mixture are to be determined.Assumptions Under specified conditions both CO2 and CH4 can be treated as ideal gases, and the mixtureas an ideal gas mixture.Properties The molar masses of CO2 and CH4 are 44.0 and 16.0 lbm/lbmol, respectively (Table A-1E) Analysis The mole numbers of gases are

lbmol0.1875lbm/lbmol16

lbm3lbm3

lbmol0.0227lbm/lbmol44

lbm1lbm1

4

4

44

2

2

22

CH

CHCHCH

CO

COCOCO

Mm

Nm

Mm

Nm1 lbm CO23 lbm CH4

600 R 20 psia

lbmol0.2102lbmol0.1875lbmol2702.042 CHCO NNN m

892.0lbmol0.2102lbmol0.1875

108.0lbmol0.2102lbmol0.0227

4

4

2

2

CHCH

COCO

m

m

NN

y

NN

y

Then the partial pressures become

psia17.84psia2.16

psia20892.0psia20108.0

44

22

CHCH

COCO

m

m

PyPPyP

The apparent molar mass of the mixture is

lbm/lbmol19.03lbmol0.2102

lbm4

m

mm N

mM

13-10

13-36 The masses of the constituents of a gas mixture at a specified temperature are given. The partialpressure of each gas and the total pressure of the mixture are to be determined.Assumptions Under specified conditions both N2 and O2 can be treated as ideal gases, and the mixture as an ideal gas mixture.Analysis The partial pressures of constituent gases are

0.3 m3

0.6 kg N20.4 kg O2

300 K kPa103.9

kPa178.1

3

3

OO

3

3

NN

m0.3K)K)(300/kgmkPakg)(0.2598(0.4

m0.3K)K)(300/kgmkPakg)(0.2968(0.6

2

2

2

2

V

V

mRTP

mRTP

andkPa282.0kPa103.9kPa1.178

22 ON PPPm

13-37 The volumetric fractions of the constituents of a gas mixture at a specified pressure and temperatureare given. The mass fraction and partial pressure of each gas are to be determined.Assumptions Under specified conditions all N2, O2 and CO2 can be treated as ideal gases, and the mixtureas an ideal gas mixture.Properties The molar masses of N2, O2 and CO2 are 28.0, 32.0, and 44.0 kg/kmol, respectively (Table A-1) Analysis For convenience, consider 100 kmol of mixture. Then the mass of each component and the total mass are




2222

2222

2222

COCOCOCO

OOOO

NNNN

MNmN

MNmN

MNmN65% N220% O2

15% CO2

350 K 300 kPa kg3120kg660kg640kg8201

222 COON mmmmm


21.2%

20.5%

58.3%

or0.212kg3120kg660mf

or0.205kg3120kg640mf

or0.583kg3120kg1820mf

22

22

22

COCO

OO

NN

m

m

m

mmmmmm

For ideal gases, the partial pressure is proportional to the mole fraction, and is determined from

kPa45kPa60

kPa195

kPa30015.0kPa30020.0kPa30065.0

22

22

22

COCO

OO

NN

m

m

m

PyPPyPPyP

13-11

13-38 The masses, temperatures, and pressures of two gases contained in two tanks connected to each other are given. The valve connecting the tanks is opened and the final temperature is measured. The volume of each tank and the final pressure are to be determined.Assumptions Under specified conditions both N2 and O2 can be treated as ideal gases, and the mixture as an ideal gas mixtureProperties The molar masses of N2 and O2 are 28.0 and 32.0 kg/kmol, respectively (Table A-1) Analysis The volumes of the tanks are

3

3

m0.465

m0.295

kPa500K)K)(298/kgmkPakg)(0.2598(3

kPa300K)K)(298/kgmkPakg)(0.2968(1

3

OO

3

NN

2

2

2

2

PmRT

PmRT

V

V3 kg O2

25 C500 kPa

1 kg N2

25 C300 kPa

333ONtotal m.760m0.465m295.0

22VVV

Also,

kmol0.09375kg/kmol32

kg3kg3


kg1kg1

2

2

22

2

2

22

O

OOO

N

NNN

Mm

Nm

Mm

Nm

kmol0.1295kmol0.09375kmol03571.022 ON NNN m

Thus,

kPa422.23

3

m0.76K)K)(298/kmolmkPa4kmol)(8.31(0.1295

m

um

TNRP

V

13-12

13-39 The volumes, temperatures, and pressures of two gases forming a mixture are given. The volume of the mixture is to be determined using three methods.Analysis (a) Under specified conditions both O2 and N2 will considerably deviate from the ideal gasbehavior. Treating the mixture as an ideal gas,

kmol2.406K)K)(200/kmolmkPa(8.314

)mkPa)(0.5(8000


)mkPa)(0.3(8000

3

3

NN

3

3

OO

2

2

2

2

TRPN

TRPN

u

u

V

V

3m0.8kPa8000


kmol3.849

kmol2.406kmol443.1

3

NO 22

m

mumm

m

PTRN

NNN

V

0.5 m3 N2200 K 8 MPa

0.3 m3 O2200 K 8 MPa N2 + O2

200 K 8 MPa

(b) To use Kay's rule, we need to determine the pseudo-critical temperature and pseudo-critical pressure of the mixture using the critical point properties of O2 and N2 from Table A-1. But we first need to determinethe Z and the mole numbers of each component at the mixture temperature and pressure (Fig. A-15),

O2: 77.0575.1

MPa5.08MPa8

292.1K154.8

K200

2

22

22

O

Ocr,O,

Ocr,O,

Z

PP

P

TT

T

mR

mR

N2: 863.0360.2

MPa3.39MPa8

585.1K126.2

K200

2

22

22

N

Ncr,N,

Ncr,N,

Z

PP

P

TT

T

mR

mR

kmol2.787K)K)(200/kmolmkPa314(0.863)(8.

)mkPa)(0.5(8000

kmol1.874K)K)(200/kmolmkPa14(0.77)(8.3

)mkPa)(0.3(8000

3

3

NN

3

3

OO

2

2

2

2

TZRPN

TZRPN

u

u

V

V

kmol4.661kmol2.787kmol874.122 NO NNN m

The mole fractions are

MPa4.07MPa)39(0.598)(3.MPa)08.5)(402.0(

K137.7K)6.2(0.598)(12K).80.402)(154(

598.0kmol4.661kmol2.787

0.402kmol4.661kmol1.874

2222

2222

2

2

2

2

N,crNO,crO,cr,cr

N,crNO,crO,cr,cr

NN

OO

PyPyPyP

TyTyTyT

NN

y

NN

y

iim

iim

m

m

13-13

Then,

82.0966.1

MPa4.07MPa8

452.1K137.7

K200

'Ocr,

'Ocr,

2

2m

mR

mR

Z

PP

P

TT

T

(Fig. A-15)

Thus, 3m0.79kPa8000

K)K)(200/kmolmkPa4kmol)(8.3161(0.82)(4.6 3

m

mummm P

TRNZV

(c) To use the Amagat’s law for this real gas mixture, we first need the Z of each component at the mixturetemperature and pressure, which are determined in part (b). Then,

83.0863.0598.077.0402.02222 NNOO ZyZyZyZ iim

Thus, 3m0.80kPa8000

K)K)(200/kmolmkPa4kmol)(8.3161(0.83)(4.6 3

m

mummm P

TRNZV

13-14

13-40 [Also solved by EES on enclosed CD] The mole numbers, temperatures, and pressures of two gases forming a mixture are given. The final temperature is also given. The pressure of the mixture is to bedetermined using two methods.Analysis (a) Under specified conditions both Ar and N2 will considerably deviate from the ideal gasbehavior. Treating the mixture as an ideal gas,

MPa18.2)MPa(5K)(1)(220K)(4)(200

:stateFinal:stateInitial

111

222

2222

1111 PTNTNP

TRNPTRNP

u

u

V

V

(b) Initially,

3 kmol N2190 K 8 MPa

1 kmol Ar

220 K 5 MPa

90.00278.1

MPa4.86MPa5

457.1K151.0

K220

Ar

Arcr,

1

Arcr,

1

Z

PP

P

TT

T

R

R

(Fig. A-15)

Then the volume of the tank is

33

Ar m0.33kPa5000

K)K)(220/kmolmkPa4kmol)(8.31(0.90)(1P

TRZN uV

After mixing,

Ar: 90.0

278.1kPa)K)/(4860K)(151.0/kmolmkPa(8.314

kmol))/(1m(0.33

//

/

325.1K151.0

K200

3

3

Arcr,Arcr,

Ar

Arcr,Arcr,

ArAr,

Arcr,A,

Ru

m

uR

mrR

PPTR

NPTR

TT

T

VvV (Fig. A-15)

N2: 75.3

355.0kPa)K)/(3390K)(126.2/kmolmkPa(8.314

kmol))/(3m(0.33

/

/

/

585.1K126.2

K200

3

3

Ncr,Ncr,

N

Ncr,Ncr,

NN,

Ncr,N,

22

2

22

2

2

22

Ru

m

uR

mR

PPTR

NPTR

TT

T

VvV (Fig. A-15)

Thus,

MPa12.7MPa)39.3)(75.3()(MPa4.37MPa)86.4)(90.0()(

22 NcrN

ArcrAr

PPPPPP

R

R

andMPa17.1MPa12.7MPa37.4

2NAr PPPm

13-15

13-41 EES Problem 13-40 is reconsidered. The effect of the moles of nitrogen supplied to the tank on thefinal pressure of the mixture is to be studied using the ideal-gas equation of state and the compressibilitychart with Dalton's law. Analysis The problem is solved using EES, and the solution is given below."Input Data"R_u = 8.314 [kJ/kmol-K] "universal Gas Constant"T_Ar = 220 [K] P_Ar = 5000 [kPa] "Pressure for only Argon in the tank initially."N_Ar = 1 [kmol]{N_N2 = 3 [kmol]}T_mix = 200 [K] T_cr_Ar=151.0 [K] "Critical Constants are found in Table A.1 of the text"P_cr_Ar=4860 [kPa] T_cr_N2=126.2 [K] P_cr_N2=3390 [kPa]

"Ideal-gas Solution:"P_Ar*V_Tank_IG = N_Ar*R_u*T_Ar "Apply the ideal gas law the gas in the tank."P_mix_IG*V_Tank_IG = N_mix*R_u*T_mix "Ideal-gas mixture pressure"N_mix=N_Ar + N_N2 "Moles of mixture"

"Real Gas Solution:"P_Ar*V_Tank_RG = Z_Ar_1*N_Ar*R_u*T_Ar "Real gas volume of tank"T_R=T_Ar/T_cr_Ar "Initial reduced Temp. of Ar"P_R=P_Ar/P_cr_Ar "Initial reduced Press. of Ar"Z_Ar_1=COMPRESS(T_R, P_R ) "Initial compressibility factor for Ar"P_Ar_mix*V_Tank_RG = Z_Ar_mix*N_Ar*R_u*T_mix "Real gas Ar Pressure in mixture"T_R_Ar_mix=T_mix/T_cr_Ar "Reduced Temp. of Ar in mixture"P_R_Ar_mix=P_Ar_mix/P_cr_Ar "Reduced Press. of Ar in mixture"Z_Ar_mix=COMPRESS(T_R_Ar_mix, P_R_Ar_mix ) "Compressibility factor for Ar in mixture"P_N2_mix*V_Tank_RG = Z_N2_mix*N_N2*R_u*T_mix "Real gas N2 Pressure in mixture"T_R_N2_mix=T_mix/T_cr_N2 "Reduced Temp. of N2 in mixture"P_R_N2_mix=P_N2_mix/P_cr_N2 "Reduced Press. of N2 in mixture"Z_N2_mix=COMPRESS(T_R_N2_mix, P_R_N2_mix ) "Compressibility factor for N2 in mixture"P_mix=P_R_Ar_mix*P_cr_Ar +P_R_N2_mix*P_cr_N2 "Mixture pressure by Dalton's law. 23800"

NN2[kmol]

Pmix[kPa]

Pmix,IG[kPa]

1 9009 90912 13276 136363 17793 181824 23254 227275 30565 272736 41067 318187 56970 363648 82372 409099 126040 45455

10 211047 50000

1 2 3 4 5 6 7 8 9 100

45000

90000

135000

180000

225000

NN2 [kmol]

Pm

ix[k

Pa]

Solution MethodChartChartIdeal GasIdeal Gas

13-16

13-42E The mole numbers, temperatures, and pressures of two gases forming a mixture are given. For a specified final temperature, the pressure of the mixture is to be determined using two methods.Properties The critical properties of Ar are Tcr = 272 R and Pcr = 705 psia. The critical properties of N2 are Tcr = 227.1 R and Pcr = 492 psia (Table A-1E). Analysis (a) Under specified conditions both Ar and N2 will considerably deviate from the ideal gasbehavior. Treating the mixture as an ideal gas,

psia2700)psia750(R)(1)(400R)(4)(360

:stateFinal:stateInitial

111

222

2222

1111 PTNTNP

TRNPTRNP

u

u

V

V

(b) Initially,

3 lbmol N2340 R

1200 psia

1 lbmol Ar

400 R 750 psia

90.007.1

psia705psia750

47.1R272R400

Ar

Arcr,

1

Arcr,

1

Z

PP

P

TT

T

R

R

(Fig. A-15)

Then the volume of the tank is

33

Ar ft5.15psia750

R)R)(400/lbmolftpsia73lbmol)(10.(0.90)(1P

TRZN uV

After mixing,

Ar: 82.0

244.1psia)R)/(705R)(272/lbmolftpsia(10.73

lbmol))/(1ft(5.15

//

/

324.1R272R360

3

3

Arcr,Arcr,

Ar

Arcr,Arcr,

ArAr,

Arcr,Ar,

Ru

m

uR

mR

PPTR

NPTR

TT

T

Vvv (Fig. A-15)

N2: 85.3

347.0psia)R)/(492R)(227.1/lbmolftpsia(10.73

lbmol))/(3ft(5.15

/

/

/

585.1R227.1

R360

3

3

Ncr,Ncr,

N

Ncr,Ncr,

NN,

Ncr,N,

22

2

22

2

2

22

Ru

m

uR

mR

PPTR

NPTR

TT

T

Vvv (Fig. A-15)

Thus,

psia1894psia)492)(85.3()(psia578psia)570)(82.0()(

22 NcrN

ArcrArPPPPPP

R

R

andpsia2472psia1894psia578

2NAr PPPm

13-17

Properties of Gas Mixtures

13-43C Yes. Yes (extensive property).

13-44C No (intensive property).

13-45C The answers are the same for entropy.

13-46C Yes. Yes (conservation of energy).

13-47C We have to use the partial pressure.

13-48C No, this is an approximate approach. It assumes a component behaves as if it existed alone at themixture temperature and pressure (i.e., it disregards the influence of dissimilar molecules on each other.)

13-18

13-49 Oxygen, nitrogen, and argon gases are supplied from separate tanks at different temperatures to forma mixture. The total entropy change for the mixing process is to be determined.Assumptions Under specified conditions all N2, O2, and argon can be treated as ideal gases, and themixture as an ideal gas mixture.Properties The molar masses of O2, N2, and Ar are 32.0, 28.0, and 40.0 kg/kmol, respectively (TableA-1).The properties of Argon are R = 0.2081 kJ/kg.K and cp = 0.5203 kJ/kg.K (Table A-2). Analysis Note that volume fractions are equal to mole fractions inideal gas mixtures. The partial pressures in the mixture are

Ar200 C

N260 C

21% O278% N221% Ar

200 kPa

O210 C

kPa2kPa)200)(01.0(

kPa156kPa)200)(78.0(

kPa42kPa)200)(21.0(

Ar2,Ar

N2,N

O2,O

22

22

m

m

m

PyP

PyP

PyP

The molar mass of the mixture is determined to be

kg/kmol96.28(0.01)(40)(0.78)(28)kg/kmol)(0.21)(32ArArNNOO 2222

MyMyMyM m

The mass fractions are

0138.0kg/kmol28.96

kg/kmol40)01.0(mf

7541.0kg/kmol28.96

kg/kmol28)78.0(mf

2320.0kg/kmol28.96

kg/kmol32)21.0(mf

ArArAr

NNN

OOO

2

22

2

22

m

m

m

MM

y

MM

y

MM

y

The final temperature of the mixture is needed. The conservation of energy on a unit mass basis for steadyflow mixing with no heat transfer or work allows calculation of mixture temperature. All components of the exit mixture have the same common temperature, Tm. We obtain the properties of O2 and N2 from EES:

m

TT

mpTTp

outin

T

hh

TchhTchhee

mm

mm

)5203.0)(0138.0(

)7541.0()2320.0()200)(5203.0)(0138.0()47.36)(7541.0()85.13)(2320.0(

mfmfmfmfmfmf

@@

Ar,Ar@N@OAr,1Ar,ArC60@NC10@O 2222

Solving this equation with EES gives Tm = 50.4ºC. The entropies of O2 and N2 are obtained from EES to be

kJ/kg.K7893.6kPa156C,4.50

kJ/kg.K7461.6kPa200C,60

kJ/kg.K7082.6kPa42C,4.50

kJ/kg.K1797.6kPa200C,10

2,N

1,N

2,O

1,O

2

2

2

2

sPT

sPT

sPT

sPT

The entropy changes are

kg/kg.K04321.07461.67893.6

kg/kg.K5284.01797.67082.6

1,N2,NN

1,O2,OO

222

222

sss

sss

kJ/kg.K7605.0200

2ln)2081.0(2732002734.50ln)5203.0(lnln

1

2

1

2Ar P

PR

TT

cs p

The total entropy change is

kJ/kg.K0.1656)7605.0)(0138.0()04321.0)(7541.0()5284.0)(2320.0(

mfmfmf ArArOOOOtotal 2222ssss

13-19

13-50 Volumetric fractions of the constituents of a mixture are given. The mixture undergoes an adiabatic compression process. The makeup of the mixture on a mass basis and the internal energy change per unitmass of mixture are to be determined.Assumptions Under specified conditions all CO2, CO, O2, and N2 can be treated as ideal gases, and themixture as an ideal gas mixture.Properties 1 The molar masses of CO2, CO, O2, and N2 are 44.0, 28.0, 32.0, and 28.0 kg/kmol, respectively(Table A-1). 2 The process is reversible.Analysis Noting that volume fractions are equal to mole fractions in ideal gas mixtures, the molar mass of the mixture is determined to be

kg/kmol80.30(0.70)(28)(0.10)(32)(0.05)(28)(0.15)(44)222222 NNOOCOCOCOCO MyMyMyMyM m


0.6364

0.1039

0.0454

0.2143

kg/kmol30.80kg/kmol28)70.0(mf




2

22

2

22

2

22

NNN

OOO

COCOCO

COCOCO

m

m

m

m

MM

y

MM

y

MM

y

MM

y

15% CO25% CO 10% O270% N2

300 K, 1 bar The final pressure of mixture is expressed from ideal gas relation to be

22

1

212 667.2

K300)8)(kPa100( T

TTT

rPP (Eq. 1)

since the final temperature is not known. We assume that the process is reversible as well being adiabatic(i.e. isentropic). Using Dalton’s law to find partial pressures, the entropies at the initial state are determinedfrom EES to be:

kJ/kg.K0115.7kPa64.63)1006364.0(K,300

kJ/kg.K9485.6kPa39.10)1001039.0(K,300

kJ/kg.K79483kPa55.4)10004545.0(K,300

kJ/kg.K2190.5kPa43.21)1002143.0(K,300

1,O

1,N

1,CO

1,CO

2

2

2

sPT

sPT

sPT

sPT

The final state entropies cannot be determined at this point since the final pressure and temperature are notknown. However, for an isentropic process, the entropy change is zero and the final temperature and thefinal pressure may be determined from

0mfmfmfmf222222 NNOOCOCOCOCOtotal sssss

and using Eq. (1). The solution may be obtained using EES to be T2 = 631.4 K, P2 = 1684 kPa The initial and final internal energies are (from EES)

kJ/kg9.163kJ/kg8.156kJ/kg3780kJ/kg8734

K4.631

kJ/kg,11.87kJ/kg24.76

kJ/kg4033kJ/kg8997

K300

2,N

2,O

2,CO

2,CO

2

1,N

1,O

1,CC

1,CO

1

2

2

2

2

2

2

uuu

u

T

uuu

u

T

The internal energy change per unit mass of mixture is determined from

kJ/kg251.8)11.87(9.1636364.06)24.76(8.1561039.0

)4033()3780(0454.0)8997()8734(2143.0

)(mf)(mf)(mf)(mf 1,N2,NN1,O2,OO1,CO2,COCO1,CO2,COCOmixture 222222222uuuuuuuuu

13-20

13-51 Propane and air mixture is compressed isentropically in an internal combustion engine. The work input is to be determined.Assumptions Under specified conditions propane and air can be treated as ideal gases, and the mixture asan ideal gas mixture.Properties The molar masses of C3H8 and air are 44.0 and 28.97 kg/kmol, respectively (TableA-1).Analysis Given the air-fuel ratio, the mass fractions are determined to be

05882.0171

1AF1mf

9412.01716

1AFAFmf

83HC

air

PropaneAir

95 kPa 30ºC

The molar mass of the mixture is determined to be

kg/kmol56.29

kg/kmol44.005882.0

kg/kmol28.979412.0

1mfmf

1

83

83

HC

HC

air

air

MM

M m

The mole fractions are

03944.0kg/kmol44.0kg/kmol56.29)05882.0(mf

9606.0kg/kmol28.97kg/kmol56.29)9412.0(mf

838383

HCHCHC

airairair

MM

y

MM

y

m

m

The final pressure is expressed from ideal gas relation to be

22

1

212 977.2

K273.15)(30)5.9)(kPa95( T

TTT

rPP (1)

since the final temperature is not known. Using Dalton’s law to find partial pressures, the entropies at theinitial state are determined from EES to be:

kJ/kg.K7697.6kPa75.3)9503944.0(C,30

kJ/kg.K7417.5kPa26.91)959606.0(C,30

1,HC

1,air

83sPT

sPT

The final state entropies cannot be determined at this point since the final pressure and temperature are notknown. However, for an isentropic process, the entropy change is zero and the final temperature and thefinal pressure may be determined from

08383 HCHCairairtotal smfsmfs

and using Eq. (1). The solution may be obtained using EES to be T2 = 654.9 K, P2 = 1951 kPa

The initial and final internal energies are (from EES)

kJ/kg1607kJ/kg1.477

K9.654kJ/kg2404

kJ/kg5.216C30

2,HC

2,air2

1,HC

1,air1

8383u

uT

uu

T

Noting that the heat transfer is zero, an energy balance on the system gives

mm uwuwq ininin

where )()( 1,HC2,HCHCair,1air,2air 838383uumfuumfum

Substituting, kJ/kg292.2)2404()1607()05882.0()5.2161.477)(9412.0(in muw

13-21

13-52 The moles, temperatures, and pressures of two gases forming a mixture are given. The mixturetemperature and pressure are to be determined.Assumptions 1 Under specified conditions both CO2 and H2 can be treated as ideal gases, and the mixtureas an ideal gas mixture. 2 The tank is insulated and thus there is no heat transfer. 3 There are no other forms of work involved.Properties The molar masses and specific heats of CO2 and H2 are 44.0 kg/kmol, 2.0 kg/kmol, 0.657 kJ/kg. C, and 10.183 kJ/kg. C, respectively. (Tables A-1 and A-2b).Analysis (a) We take both gases as our system. No heat, work, or mass crosses the system boundary,therefore this is a closed system with Q = 0 and W = 0. Then the energy balance for this closed systemreduces to

22

22

H1CO1

HCO

systemoutin

0

0

TTmcTTmc

UUU

EEE

mm vv

H2

7.5 kmol 400 kPa

40 C

CO2

2.5 kmol 200 kPa

27 CUsing cv values at room temperature and noting that m = NM,the final temperature of the mixture is determined to be

K308.80C40CkJ/kg10.183kg27.5C27CkJ/kg0.657kg442.5

C35.8m

mm

TTT

(b) The volume of each tank is determined from

33

H1

1H

33

CO1

1CO

m7948kPa400


m1831kPa200


2

2

2

2

.

.

PTNR

PTNR

u

u

V

V

Thus,

kmol.010kmol5.7kmol5.2

m.9779m.7948m18.31

22

22

HCO

333HCO

NNNm

m VVV

and kPa3213

3

m79.97K)K)(308.8/kmolmkPa4kmol)(8.31(10.0

m

mumm V

TRNP

13-22

13-53 The temperatures and pressures of two gases forming a mixture are given. The final mixturetemperature and pressure are to be determined.Assumptions 1 Under specified conditions both Ne and Ar can be treated as ideal gases, and the mixture asan ideal gas mixture. 2 There are no other forms of work involved.Properties The molar masses and specific heats of Ne and Ar are 20.18 kg/kmol, 39.95 kg/kmol, 0.6179 kJ/kg. C, and 0.3122 kJ/kg. C, respectively. (Tables A-1 and A-2).Analysis The mole number of each gas is

Ar200 kPa

50 C

Ne100 kPa

20 Ckmol0.0335

K)K)(323/kmolmkPa(8.314)mkPa)(0.45(200


)mkPa)(0.45(100

3

3

Ar1

11Ar

3

3

Ne1

11Ne

TRPN

TRPN

u

u

V

V

Thus, 15 kJ N N Nm Ne Ar kmol 0.0335 kmol 0.0520 kmol0 0185.

(a) We take both gases as the system. No work or mass crosses the system boundary, therefore this is aclosed system with W = 0. Then the conservation of energy equation for this closed system reduces to

Ar1Ne1outArNeout

systemoutin

TTmcTTmcQUUUQ

EEE

mm vv

Using cv values at room temperature and noting that m = NM, the final temperature of the mixture is determined to be

K289.2C50CkJ/kg0.3122kg39.950.0335C20CkJ/kg0.6179kg20.180.0185kJ15

C16.2m

m

m

TTT

(b) The final pressure in the tank is determined from

PN R T

Vmm u m

m

(0.052 kmol)(8.314 kPa m / kmol K)(289.2 K)0.9 m

3

3 138.9 kPa

13-23

13-54 The temperatures and pressures of two gases forming a mixture are given. The final mixturetemperature and pressure are to be determined.Assumptions 1 Under specified conditions both Ne and Ar can be treated as ideal gases, and the mixture asan ideal gas mixture. 2 There are no other forms of work involved.Properties The molar masses and specific heats of Ne and Ar are 20.18 kg/kmol, 39.95 kg/kmol, 0.6179 kJ/kg. C, and 0.3122 kJ/kg. C, respectively. (Tables A-1 and A-2b).Analysis The mole number of each gas is


)mkPa)(0.45(200


)mkPa)(0.45(100

3

3

Ar1

11Ar

3

3

Ne1

11Ne

TRPN

TRPN

u

u

V

VAr

200 kPa 50 C

Ne100 kPa

20 C

Thus,8 kJ

N N Nm Ne Ar kmol 0.0335 kmol 0.0520 kmol0 0185.

(a) We take both gases as the system. No work or mass crosses the system boundary, therefore this is aclosed system with W = 0. Then the conservation of energy equation for this closed system reduces to

Ar1Ne1out

ArNeout

systemoutin

TTmcTTmcQUUUQ

EEE

mm vv

Using cv values at room temperature and noting that m = NM, the final temperature of the mixture is determined to be

K300.0C50CkJ/kg0.3122kg39.950.0335C20CkJ/kg0.6179kg20.180.0185kJ8

C0.27m

m

m

TTT

(b) The final pressure in the tank is determined from

kPa1.1443

3

m0.9K)K)(300.0/kmolmkPa4kmol)(8.31(0.052

m

mumm

TRNPV

13-24

13-55 [Also solved by EES on enclosed CD] The temperatures and pressures of two gases forming a mixture in a mixing chamber are given. The mixture temperature and the rate of entropy generation are tobe determined.Assumptions 1 Under specified conditions both C2H6 and CH4 can be treated as ideal gases, and the mixture as an ideal gas mixture. 2 The mixing chamber is insulated and thus there is no heat transfer. 3There are no other forms of work involved. 3 This is a steady-flow process. 4 The kinetic and potentialenergy changes are negligible.Properties The specific heats of C2H6 and CH4 are 1.7662 kJ/kg. C and 2.2537 kJ/kg. C, respectively. (Table A-2b).Analysis (a) The enthalpy of ideal gases is independent ofpressure, and thus the two gases can be treated independentlyeven after mixing. Noting that , the steady-flow energy balance equation reduces to

W Q 0

462

446262

CHHC

CHCHHCHC

outin

(steady)0systemoutin

0

0

0

iepiep

ieieiiee

eeii

TTcmTTcm

hhmhhmhmhm

hmhm

EE

EEE

200 kPa

CH445 C4.5 kg/s

C2H6

20 C9 kg/s

Using cp values at room temperature and substituting, the exit temperature of the mixture becomes

K302.7C45CkJ/kg2.2537kg/s4.5C20CkJ/kg1.7662kg/s90

C29.7m

mm

TTT

(b) The rate of entropy change associated with this process is determined from an entropy balance on themixing chamber,

[ ( )] [ ( )] [ ( )] [ ( )]

S S S S

m s s m s s S S m s s m s sin out gen system

C H CH gen gen C H CH2 6 4 2 6 4

0

1 2 1 2 2 1 2 1

0

0The molar flow rate of the two gases in the mixture is

kmol/s0.2813kg/kmol16

kg/s4.5kmol/s0.3kg/kmol30

kg/s9

4

4

62

62CH

CHHC

HC MmN

MmN

Then the mole fraction of each gas becomes

484.02813.03.0

2813.0516.02813.03.0

3.0462 CHHC yy

Thus,

KkJ/kg0.265)4ln(0.48K)kJ/kg(0.5182K318K302.7

lnK)kJ/kg2.2537(

lnlnlnln)(

KkJ/kg0.240ln(0.516)K)kJ/kg(0.2765K293K302.7

lnK)kJ/kg(1.7662

lnlnlnln)(

44

4

6262

62

CH1

2

CH1

2,

1

2CH12

HC1

2

HC1

2,

1

2HC12

yRTT

cPPy

RTT

css

yRTT

cPPy

RTT

css

pm

p

pm

p

Noting that and substituting,P Pm i, ,2 1 200 kPa

kW/K3.353KkJ/kg0.265kg/s4.5KkJ/kg0.240kg/s9genS

13-25

13-56 EES Problem 13-55 is reconsidered. The effect of the mass fraction of methane in the mixture on the mixture temperature and the rate of exergy destruction is to be investigated.Analysis The problem is solved using EES, and the solution is given below.

"Input from the Diagram Window"{Fluid1$='C2H6'Fluid2$='CH4'm_dot_F1=9 [kg/s] m_dot_F2=m_dot_F1/2T1=20 [C] T2=45 [C] P=200 [kPa]}{mf_F2=0.1}{m_dot_total =13.5 [kg/s] m_dot_F2 =mf_F2*m_dot_total} m_dot_total = m_dot_F1 + m_dot_F2 T_o = 25 [C] "Conservation of energy for this steady-state, steady-flow control volume is"E_dot_in=E_dot_outE_dot_in=m_dot_F1*enthalpy(Fluid1$,T=T1) +m_dot_F2 *enthalpy(Fluid2$,T=T2)E_dot_out=m_dot_F1*enthalpy(Fluid1$,T=T3) +m_dot_F2 *enthalpy(Fluid2$,T=T3)"For entropy calculations the partial pressures are used."Mwt_F1=MOLARMASS(Fluid1$)N_dot_F1=m_dot_F1/Mwt_F1Mwt_F2=MOLARMASS(Fluid2$)N_dot_F2=m_dot_F2 /Mwt_F2 N_dot_tot=N_dot_F1+N_dot_F2y_F1=IF(fluid1$,Fluid2$,N_dot_F1/N_dot_tot,1,N_dot_F1/N_dot_tot)y_F2=IF(fluid1$,Fluid2$,N_dot_F2/N_dot_tot,1,N_dot_F2/N_dot_tot)"If the two fluids are the same, the mole fractions are both 1 .""The entropy change of each fluid is:"DELTAs_F1=entropy(Fluid1$, T=T3, P=y_F1*P)-entropy(Fluid1$, T=T1, P=P) DELTAs_F2=entropy(Fluid2$, T=T3, P=y_F2*P)-entropy(Fluid2$, T=T2, P=P) "And the entropy generation is:"S_dot_gen=m_dot_F1*DELTAs_F1+m_dot_F2*DELTAs_F2"Then the exergy destroyed is:"X_dot_destroyed = (T_o+273)*S_dot_gen

mfF2 T3[C]

Xdestroyed[kW]

0.01 95.93 20.480.1 502.5 24.080.2 761.4 270.3 948.5 29.20.4 1096 30.920.5 1219 32.30.6 1324 33.430.7 1415 34.380.8 1497 35.180.9 1570 35.870.99 1631 36.41

0 0,2 0,4 0,6 0,8 120

22

24

26

28

30

32

34

36

38

0

400

800

1200

1600

2000

mfF2

T3[C]

Xdestroyed

[kW]

F1 = C2H6

F2 = CH4

mtotal = 13.5 kg/s

13-26

13-57 An equimolar mixture of helium and argon gases expands in a turbine. The isentropic work output of the turbine is to be determined.Assumptions 1 Under specified conditions both He and Ar can be treated as ideal gases, and the mixture asan ideal gas mixture. 2 The turbine is insulated and thus there is no heat transfer. 3 This is a steady-flow process. 4 The kinetic and potential energy changes are negligible.Properties The molar masses and specific heats of He and Ar are 4.0 kg/kmol, 40.0 kg/kmol, 5.1926 kJ/kg. C, and 0.5203 kJ/kg. C, respectively. (Table A-1 and Table A-2).Analysis The Cp and k values of this equimolar mixture are determined from

KkJ/kg0.945

KkJ/kg0.5203kg/kmol22

kg/kmol405.0KkJ/kg5.1926

kg/kmol22kg/kmol45.0

Ar,ArAr

He,HeHe

,mf,

mf

kg/kmol22405.045.0ArArHeHe

pcmM

Mypc

mM

Myipcimp

mMiMiy

mMmNiMiN

mmim

i

MyMyiMiymM

c

2.5 MPa 1300 K

He-Arturbine

200 kPa

andkm = 1.667 since k = 1.667 for both gases.

Therefore, the He-Ar mixture can be treated as a single ideal gas with the properties above. For isentropicprocesses,

K2.473kPa2500

kPa200K130070.667/1.66/1

1

212

kk

PP

TT

From an energy balance on the turbine,

kJ/kg781.3K.24731300KkJ/kg0.945

0

21out

21out

out21

outin


TTcwhhwwhh

EE

EEE

p

13-27

13-58E [Also solved by EES on enclosed CD] A gas mixture with known mass fractions is acceleratedthrough a nozzle from a specified state to a specified pressure. For a specified isentropic efficiency, theexit temperature and the exit velocity of the mixture are to be determined.Assumptions 1 Under specified conditions both N2 and CO2 can be treated as ideal gases, and the mixtureas an ideal gas mixture. 2 The nozzle is adiabatic and thus heat transfer is negligible. 3 This is a steady-flow process. 4 Potential energy changes are negligible.Properties The specific heats of N2 and CO2 are cp,N2 = 0.248 Btu/lbm.R, cv,N2 = 0.177 Btu/lbm.R, cp,CO2 = 0.203 Btu/lbm.R, and cv,CO2 = 0.158 Btu/lbm.R. (Table A-2E).Analysis (a) Under specified conditions both N2 and CO2 can be treated as ideal gases, and the mixture as an ideal gas mixture. The cp, cv, and k values of this mixture are determined from

382.1RBtu/lbm0.173RBtu/lbm0.239

RBtu/lbm173.0158.02.0177.08.0

mfmfmf

RBtu/lbm239.0203.02.0248.08.0

mfmfmf

,

,

CO,CON,N,,

CO,CON,N,,

2222

2222

m

mpm

iim

ppipimp

cc

k

cccc

cccc

v

vvvv 80% N220% CO2

90 psia 1800 R 12 psia

Therefore, the N2-CO2 mixture can be treated as a single ideal gas with above properties. Then theisentropic exit temperature can be determined from

R1031.3psia90psia12

R180020.382/1.38/1

1

212

kk

s PP

TT

From the definition of adiabatic efficiency,

R1092.822

21

21

21

21

3.1031800,1800,1

92.0 TT

TTcTTc

hhhh

sp

p

sN

(b) Noting that, q = w = 0, from the steady-flow energy balance relation,

ft/s2,909Btu/lbm1

/sft25,037R1092.81800RBtu/lbm0.23922

20

2/2/

0

22

212

021

22

12

222

211

outin


TTcV

VVTTc

VhVh

EE

EEE

p

p

13-28

13-59E EES Problem 13-58E is reconsidered. The problem is first to be solved and then, for all otherconditions being the same, the problem is to be resolved to determine the composition of the nitrogen andcarbon dioxide that is required to have an exit velocity of 2000 ft/s at the nozzle exit.Analysis The problem is solved using EES, and the solution is given below.

"Input Data"

mf_N2 = 0.8 "Mass fraction for the nitrogen, lbm_N2/lbm_mix"mf_CO2 = 0.2 "Mass fraction for the carbon dioxide, lbm_CO2/lbm_mix"T[1] = 1800 [R] P[1] = 90 [psia] Vel[1] = 0 [ft/s]P[2] = 12 [psia] Eta_N =0.92 "Nozzle adiabatic efficiency"

"Enthalpy property data per unit mass of mixture:"

" Note: EES calculates the enthalpy of ideal gases referenced tothe enthalpy of formation as h = h_f + (h_T - h_537) where h_f is theenthalpy of formation such that the enthalpy of the elements or theirstable compounds is zero at 77 F or 537 R, see Chapter 14. The enthalpyof formation is often negative; thus, the enthalpy of ideal gases canbe negative at a given temperature. This is true for CO2 in this problem."

h[1]= mf_N2* enthalpy(N2, T=T[1]) + mf_CO2* enthalpy(CO2, T=T[1]) h[2]= mf_N2* enthalpy(N2, T=T[2]) + mf_CO2* enthalpy(CO2, T=T[2])

"Conservation of Energy for a unit mass flow of mixture:""E_in - E_out = DELTAE_cv Where DELTAE_cv = 0 for SSSF"h[1]+Vel[1]^2/2*convert(ft^2/s^2,Btu/lbm) - h[2] - Vel[2]^2/2*convert(ft^2/s^2,Btu/lbm) =0 "SSSFenergy balance"

"Nozzle Efficiency Calculation:"

Eta_N=(h[1]-h[2])/(h[1]-h_s2)h_s2= mf_N2* enthalpy(N2, T=T_s2) + mf_CO2* enthalpy(CO2, T=T_s2)

"The mixture isentropic exit temperature, T_s2, is calculated from setting theentropy change per unit mass of mixture equal to zero."

DELTAs_mix=mf_N2 * DELTAs_N2 + mf_CO2 * DELTAs_CO2DELTAs_N2 = entropy(N2, T=T_s2, P=P_2_N2) - entropy(N2, T=T[1], P=P_1_N2) DELTAs_CO2 = entropy(CO2, T=T_s2, P=P_2_CO2) - entropy(CO2, T=T[1], P=P_1_CO2) DELTAs_mix=0

"By Dalton's Law the partial pressures are:"P_1_N2 = y_N2 * P[1]; P_1_CO2 = y_CO2 * P[1] P_2_N2 = y_N2 * P[2]; P_2_CO2 = y_CO2 * P[2]

"mass fractions, mf, and mole fractions, y, are related by:"M_N2 = molarmass(N2) M_CO2=molarmass(CO2)y_N2=mf_N2/M_N2/(mf_N2/M_N2 + mf_CO2/M_CO2) y_CO2=mf_CO2/M_CO2/(mf_N2/M_N2 + mf_CO2/M_CO2)

13-29

SOLUTION of the stated problem

DELTAs_CO2=-0.04486 [Btu/lbm-R] DELTAs_mix=0 [Btu/lbm-R] DELTAs_N2=0.01122 [Btu/lbm-R] Eta_N=0.92 h[1]=-439.7 [Btu/lbm] h[2]=-613.7 [Btu/lbm]h_s2=-628.8 [Btu/lbm] mf_CO2=0.2 [lbm_CO2/lbm_mix]mf_N2=0.8 [lbm_N2/lbm_mix] M_CO2=44.01 [lbm/lbmol]M_N2=28.01 [lbm/lbmol] P[1]=90 [psia]P[2]=12 [psia] P_1_CO2=12.36 [psia]P_1_N2=77.64 [psia] P_2_CO2=1.647 [psia]P_2_N2=10.35 [psia] T[1]=1800 [R]T[2]=1160 [R] T_s2=1102 [R]Vel[1]=0 [ft/s] Vel[2]=2952 [ft/s] y_CO2=0.1373 [ft/s] y_N2=0.8627 [lbmol_N2/lbmol_mix]

SOLUTION of the problem with exit velocity of 2600 ft/s

DELTAs_CO2=-0.005444 [Btu/lbm-R] DELTAs_mix=0 [Btu/lbm-R] DELTAs_N2=0.05015 [Btu/lbm-R] Eta_N=0.92h[1]=-3142 [Btu/lbm] h[2]=-3277 [Btu/lbm]h_s2=-3288 [Btu/lbm] mf_CO2=0.9021 [lbm_CO2/lbm_mix]mf_N2=0.09793 [lbm_N2/lbm_mix] M_CO2=44.01 [lbm/lbmol] M_N2=28.01 [lbm/lbmol] P[1]=90 [psia]P[2]=12 [psia] P_1_CO2=76.89 [psia]P_1_N2=13.11 [psia] P_2_CO2=10.25 [psia]P_2_N2=1.748 [psia] T[1]=1800 [R]T[2]=1323 [R] T_s2=1279 [R]Vel[1]=0 [ft/s] Vel[2]=2600 [ft/s] y_CO2=0.8543 [ft/s] y_N2=0.1457 [lbmol_N2/lbmol_mix]

13-30

13-60 A piston-cylinder device contains a gas mixture at a given state. Heat is transferred to the mixture.The amount of heat transfer and the entropy change of the mixture are to be determined.Assumptions 1 Under specified conditions both H2 and N2 can be treated as ideal gases, and the mixture as an ideal gas mixture. 2 Kinetic and potential energy changes are negligible.Properties The constant pressure specific heats of H2 and N2 at 450 K are 14.501 kJ/kg.K and 1.049kJ/kg.K, respectively. (Table A-2b).Analysis (a) Noting that P2 = P1 and V2 = 2V1,

Q

0.5 kg H21.6 kg N2100 kPa 300 K

K600K300222

111

12

1

11

2

22 TTTT

PT

PV

VVV

Also P = constant. Then from the closed system energy balance relation, E E E

Q W U Q Hb

in out system

in out in,

since Wb and U combine into H for quasi-equilibrium constant pressure processes.

kJ2679K300600KkJ/kg1.049kg1.6K300600KkJ/kg14.501kg0.5

2222 N12avg,H12avg,NHin TTmcTTmcHHHQ pp

(b) Noting that the total mixture pressure, and thus the partial pressure of each gas, remains constant, the entropy change of the mixture during this process is

kJ/K6.19kJ/K1.163kJ/K.0265

kJ/K1.163K300K600

lnKkJ/kg1.049kg1.6

lnlnln

kJ/K.0265K300K600lnKkJ/kg14.501kg0.5

lnlnln

22

2

2

2

222

2

2

2

222

NHtotal

N1

2N

N

0

1

2

1

2NN12N

H1

2H

H

0

1

2

1

2HH12H

SSS

TT

cmPP

RTT

cmssmS

TT

cmPP

RTT

cmssmS

pp

pp

13-31

13-61 The states of two gases contained in two tanks are given. The gases are allowed to mix to form ahomogeneous mixture. The final pressure, the heat transfer, and the entropy generated are to bedetermined.Assumptions 1 Under specified conditions both O2 and N2 can be treated as ideal gases, and the mixture as an ideal gas mixture. 2 The tank containing oxygen is insulated. 3 There are no other forms of work involved.Properties The constant volume specific heats of O2 and N2 are 0.658 kJ/kg. C and 0.743 kJ/kg. C,respectively. (Table A-2).Analysis (a) The volume of the O2 tank and mass of the nitrogen are

333N1,O1,total

3

3

N1

11N

33

O1

1O1,

m2.25m2.0m52.0

kg10.43K)K)(323/kgmkPa(0.2968

)mkPa)(2(500

m0.25kPa300

K)K)(288/kgmkPakg)(0.2598(1

22

2

2

2

2

VVV

V

V

RTP

m

PmRT

N2

2 m3

50 C500 kPa

O2

1 kg 15 C

300 kPa

Also,Q

kmol0.40375kmol0.03125kmol3725.0

kmol0.3725kg/kmol28

kg10.43kg43.10


kg1kg1

22

2

2

22

2

2

22

ON

N

NNN

O

OOO

NNN

Mm

Nm

Mm

Nm

m

Thus, kPa444.63

3

m2.25K)K)(298/kmolmkPa4kmol)(8.31(0.40375

m

um

TNRP

V

(b) We take both gases as the system. No work or mass crosses the system boundary, and thus this is a closed system with W = 0. Taking the direction of heat transfer to be from the system (will be verified), theenergy balance for this closed system reduces to

2222 N1O1NOout

systemoutin

mmout TTmcTTmcQUUUQ

EEE

vv

Using cv values at room temperature (Table A-2), the heat transfer is determined to be

system thefromC2550CkJ/kg0.743kg10.43C2515CkJ/kg0.658kg1out

kJ187.2Q

(c) For and extended system that involves the tanks and their immediate surroundings such that theboundary temperature is the surroundings temperature, the entropy balance can be expressed as

surr

out12gen

12gensurr,

out

systemgenoutin

)(

)(

TQ

ssmS

ssmSTQ

SSSS

b

13-32

The mole fraction of each gas is

923.040375.03725.0

077.040375.003125.0

22

22

NN

OO

m

m

NN

y

NN

y

Thus,

KkJ/kg0.0251kPa500

kPa)4.6(0.923)(44lnK)kJ/kg(0.2968

K323K298

lnK)kJ/kg(1.039

lnln)(

KkJ/kg0.5952kPa300

kPa)4.6(0.077)(44lnK)kJ/kg(0.2598

K288K298

lnK)kJ/kg(0.918

lnln)(

2

2

2

2

N1

2,

1

2N12

O1

2,

1

2O12

PPy

RTT

css

PPy

RTT

css

mp

mp

Substituting,

kJ/K0.962K298kJ187.2KkJ/kg0.0251kg10.43KkJ/kg0.5952kg1genS

13-33

13-62 EES Problem 13-61 is reconsidered. The results obtained assuming ideal gas behavior with constantspecific heats at the average temperature, and using real gas data obtained from EES by assuming variablespecific heats over the temperature range are to be compared.Analysis The problem is solved using EES, and the solution is given below.

"Input Data:"T_O2[1] =15 [C]; T_N2[1] =50 [C] T[2] =25 [C]; T_o = 25 [C] m_O2 = 1 [kg]; P_O2[1]=300 [kPa] V_N2[1]=2 [m^3]; P_N2[1]=500 [kPa] R_u=8.314 [kJ/kmol-K]; MM_O2=molarmass(O2)MM_N2=molarmass(N2); P_O2[1]*V_O2[1]=m_O2*R_u/MM_O2*(T_O2[1]+273)P_N2[1]*V_N2[1]=m_N2*R_u/MM_N2*(T_N2[1]+273)V_total=V_O2[1]+V_N2[1]; N_O2=m_O2/MM_O2N_N2=m_N2/MM_N2; N_total=N_O2+N_N2P[2]*V_total=N_total*R_u*(T[2]+273); P_Final =P[2] "Conservation of energy for the combined system:"E_in - E_out = DELTAE_sysE_in = 0 [kJ]E_out = Q DELTAE_sys=m_O2*(intenergy(O2,T=T[2]) - intenergy(O2,T=T_O2[1])) + m_N2*(intenergy(N2,T=T[2]) - intenergy(N2,T=T_N2[1]))P_O2[2]=P[2]*N_O2/N_totalP_N2[2]=P[2]*N_N2/N_total"Entropy generation:" - Q/(T_o+273) + S_gen = DELTAS_O2 + DELTAS_N2 DELTAS_O2 = m_O2*(entropy(O2,T=T[2],P=P_O2[2]) - entropy(O2,T=T_O2[1],P=P_O2[1]))DELTAS_N2 = m_N2*(entropy(N2,T=T[2],P=P_N2[2]) - entropy(N2,T=T_N2[1],P=P_N2[1]))"Constant Property (ConstP) Solution:"-Q_ConstP=m_O2*Cv_O2*(T[2]-T_O2[1])+m_N2*Cv_N2*(T[2]-T_N2[1])Tav_O2 =(T[2]+T_O2[1])/2 Cv_O2 = SPECHEAT(O2,T=Tav_O2) - R_u/MM_O2 Tav_N2 =(T[2]+T_N2[1])/2 Cv_N2 = SPECHEAT(N2,T=Tav_N2) - R_u/MM_N2 - Q_ConstP/(T_o+273) + S_gen_ConstP = DELTAS_O2_ConstP + DELTAS_N2_ConstP DELTAS_O2_ConstP = m_O2*( SPECHEAT(O2,T=Tav_O2)*LN((T[2]+273)/(T_O2[1]+273))-R_u/MM_O2*LN(P_O2[2]/P_O2[1]))DELTAS_N2_ConstP = m_N2*( SPECHEAT(N2,T=Tav_N2)*LN((T[2]+273)/(T_N2[1]+273))-R_u/MM_N2*LN(P_N2[2]/P_N2[1]))

SOLUTIONCv_N2=0.7454 [kJ/kg-K] Cv_O2=0.6627 [kJ/kg-K] DELTAE_sys=-187.7[kJ]DELTAS_N2=-0.262 [kJ/K] DELTAS_N2_ConstP=-0.2625 [kJ/K] DELTAS_O2=0.594 [kJ/K] DELTAS_O2_ConstP=0.594 [kJ/K] E_in=0 [kJ] E_out=187.7 [kJ]MM_N2=28.01 [kg/kmol] MM_O2=32 [kg/kmol] m_N2=10.43 [kg] m_O2=1 [kg] N_N2=0.3724 [kmol] N_O2=0.03125 [kmol]N_total=0.4036 [kmol] P[2]=444.6 [kPa] P_Final=444.6 [kPa] P_N2[1]=500 [kPa] P_N2[2]=410.1 [kPa] P_O2[1]=300[kPa]P_O2[2]=34.42 [kPa] Q=187.7 [kJ] Q_ConstP=187.8 [kJ]R_u=8.314 [kJ/kmol-K] S_gen=0.962 [kJ] S_gen_ConstP=0.9616[kJ]Tav_N2=37.5 [C] Tav_O2=20 [C] T[2]=25 [C]T_N2[1]=50 [C] T_o=25 [C] T_O2[1]=15 [C]V_N2[1]=2 [m^3] V_O2[1]=0.2494 [m^3] V_total=2.249 [m^3/kg]

13-34

13-63 Heat is transferred to a gas mixture contained in a piston cylinder device. The initial state and thefinal temperature are given. The heat transfer is to be determined for the ideal gas and non-ideal gas cases. Properties The molar masses of H2 and N2 are 2.0, and 28.0 kg/kmol. (Table A-1).Analysis From the energy balance relation,

222222 N12NH12HNHin

out,in

outin

hhNhhNHHHQ

UWQEEE

b

Q

6 kg H221 kg N25 MPa 160 K

since Wb and U combine into H for quasi-equilibrium constant pressure processes

NmM

NmM

HH

H

NN

N

2

2

2

2

2

2

6 kg2 kg / kmol

3 kmol

21 kg28 kg / kmol

0.75 kmol

(a) Assuming ideal gas behavior, the inlet and exit enthalpies of H2 and N2 are determined from the ideal gas tables to be

H h h h h

N h h h h2 1 2

2 1 2

: ,

: ,@160 K @ 200 K

@160 K @200 K

4,535.4 kJ / kmol 5,669.2 kJ / kmol

4,648 kJ / kmol 5,810 kJ / kmolThus, kJ4273648,4810,575.04.535,42.669,53idealQ(b) Using Amagat's law and the generalized enthalpy departure chart, the enthalpy change of each gas is determined to be

H2:0

0

006.63.33

200

846.330.15

805.43.33

160

2

1

222

22221

221

Hcr,

2,H,

Hcr,H,H,

Hcr,

1,H,

h

h

mR

mRR

mR

Z

Z

TT

T

PP

PP

TT

T

(Fig. A-29)

Thus H2 can be treated as an ideal gas during this process.

N2:7.0

3.1

58.12.126

200

47.139.35

27.12.126

160

2

1

222

22221

221

Ncr,

2,N,

Ncr,N,N,

Ncr,

1,N,

h

h

mR

mRR

mR

Z

Z

TT

T

PP

PP

TT

T

(Fig. A-29)

Therefore,

kJ/kmol1,791.5kJ/kmol4,648)(5,8100.7)K)(1.3K)(126.2/kmolmkPa8.314(

kJ/kmol8.133,14.535,42.669,5

3

ideal12N12

ideal,H12H12

212

22

hhZZTRhh

hhhh

hhcru

Substituting,kJ4745kJ/kmol1,791.5kmol0.75kJ/kmol1,133.8kmol3inQ

13-35

13-64 Heat is transferred to a gas mixture contained in a piston cylinder device discussed in previousproblem. The total entropy change and the exergy destruction are to be determined for two cases. Analysis The entropy generated during this process is determined by applying the entropy balance on anextended system that includes the piston-cylinder device and its immediate surroundings so that theboundary temperature of the extended system is the environment temperature at all times. It gives

S S S S

QT

S S S m s s QT

in out gen system

in

boundarygen water gen

in

surr( )2 1

Then the exergy destroyed during a process can be determined from its definition .X Tdestroyed gen0S

(a) Noting that the total mixture pressure, and thus the partial pressure of each gas, remains constant, the entropy change of a component in the mixture during this process is

1

2,

0

1

2

1

2 lnlnlnTTcm

PPR

TTcmS ipi

ipii

Assuming ideal gas behavior and using cp values at the average temperature, the S of H2 and N2 are determined from

kJ/K4.87K160K200

lnKkJ/kg1.039kg21

kJ/K18.21K160K200

lnKkJ/kg13.60kg6

ideal,N

ideal,H

2

2

S

S

and

kJ2721

kJ/K8.98

kJ/K8.98K303K303kJ4273

kJ/K4.87kJ/K12.18

gen0destroyed

gen

STX

S

(b) Using Amagat's law and the generalized entropy departure chart, the entropy change of each gas is determined to be

H2:1

1

006.63.33

200

846.330.15

805.43.33

160

2

1

222

22221

221

Hcr,

2,H,

Hcr,H,H,

Hcr,

1,H,

s

s

mR

mRR

mR

Z

Z

TT

T

PP

PP

TT

T

(Table A-30)

Thus H2 can be treated as an ideal gas during this process.

N2:4.0

8.0

585.12.126

200

475.139.35

268.12.126

160

2

1

222

22221

221

Ncr,

2,N,

Ncr,N,N,

Ncr,

1,N,

s

s

mR

mRR

mR

Z

Z

TT

T

PP

PP

TT

T

(Table A-30)

13-36

Therefore,

kJ/K15.66K303

kJ4745

kJ/K7.37)kJ/K(4.870.4)K)(0.8/kmolmkPa4kmol)(8.310.75(

kJ/K.2118

0

surrsurr

3

ideal,NNN

ideal,HH

22122

22

TQ

S

SZZRNS

SS

ssu

and

kJ3006

kJ/K9.92

kJ/K9.92K303K303kJ4745

kJ/K7.37kJ/K8.211

gen0destroyed

gen

STX

S

13-37

13-65 Air is compressed isothermally in a steady-flow device. The power input to the compressor and therate of heat rejection are to be determined for ideal and non-ideal gas cases.Assumptions 1 This is a steady-flow process. 2 The kinetic and potential energy changes are negligible.Properties The molar mass of air is 29.0 kg/kmol. (Table A-1).Analysis The mass flow rate of air can be expressed in terms of the mole numbers as

200 K 8 MPa

79% N221% O2

N mM

2.90 kg / s29.0 kg / kmol

0.10 kmol / s

(a) Assuming ideal gas behavior, the h and s of air during this process is

KkJ/kmol5.763MPa4MPa8

lnKkJ/kg8.314

lnlnln

processisothermal0

1

2

1

20

1

2

PP

RPP

RTT

cs

h

uup

200 K 4 MPa

Disregarding any changes in kinetic and potential energies, the steady-flow energy balance equation for theisothermal process of an ideal gas reduces to

E E E

E E

W Nh Q Nh

W Q N h W Q

in out system (steady)

in out

in out

in out in out

0

1 20

0

0

Also for an isothermal, internally reversible process the heat transfer is related to the entropy change byQ T S NT s ,

kW3115kW3115KkJ/kmol5.763K200kmol/s0.10 out .. QsTNQ

Therefore,

kW115.3outin QW

(b) Using Amagat's law and the generalized charts, the enthalpy and entropy changes of each gas are determined from

h h R T Z Z h h

s s R Z Z s su cr h h

u s s

2 1 2 10

2 1 2 1

1 2

1 2

( ) ( )

( ) ( )ideal

ideal

where

N2:35.0,8.0

2.0,4.0

36.239.38

74.12.126

220

18.139.34

22

11

22

221

21

Ncr,

2,

Ncr,

Ncr,

1,

sh

sh

mR

mRR

mR

ZZ

ZZ

PP

P

TT

TT

PP

P

(Tables A-29 and A-30)

13-38

O2:5.0,0.1

25.0,4.0

575.108.58

421.18.154

220

787.008.54

22

11

22

221

21

Ocr,

2,

Ocr,

Ocr,

1,

sh

sh

mR

mRR

mR

ZZ

ZZ

PP

P

TT

TT

PP

P

(Tables A-29 and A-30)

Then,

KkJ/kmol7.18)763.5()5.025.0)(314.8)(21.0()35.02.0)(314.8)(79.0(

)()(

kJ/kmol4940)0.14.0)(8.154)(314.8)(21.0()8.04.0)(2.126)(314.8)(79.0(

)()(

2222

2222

O12ON12N12

O12ON12N12

ssyssysyss

hhyhhyhyhh

ii

ii

Thus,

kW94.2

kW143.6

kJ/kmol)494kmol/s)(.100(+kW6.143)(

0

KkJ/kmol7.18K200kmol/s0.10

in12outin

2out1in

outin


out

WhhNQW

hNQhNW

EE

EEE

sTNQ

13-39

13-66 EES Problem 13-65 is reconsidered. The results obtained by assuming ideal behavior, real gasbehavior with Amagat's law, and real gas behavior with EES data are to be compared.Analysis The problem is solved using EES, and the solution is given below.

"Input Data:"y_N2 = 0.79 y_O2 = 0.21 T[1]=200 [K] "Inlet temperature"T[2]=200 [K] "Exit temmperature"P[1]=4000 [kPa] P[2]=8000 [kPa] m_dot = 2.9 [kg/s] R_u = 8.314 [kJ/kmol-K]DELTAe_bar_sys = 0 "Steady-flow analysis for all cases"m_dot = N_dot * (y_N2*molarmass(N2)+y_O2*molarmass(O2))"Ideal gas:"e_bar_in_IG - e_bar_out_IG = DELTAe_bar_sys e_bar_in_IG =w_bar_in_IG + h_bar_IG[1] e_bar_out_IG = q_bar_out_IG +h_bar_IG[2] h_bar_IG[1] = y_N2*enthalpy(N2,T=T[1]) + y_O2*enthalpy(O2,T=T[1])h_bar_IG[2] = y_N2*enthalpy(N2,T=T[2]) + y_O2*enthalpy(O2,T=T[2])"The pocess is isothermal so h_bar_IG's are equal. q_bar_IG is found from the entropy change:"q_bar_out_IG = -T[1]*DELTAs_IG s_IG[2]= y_N2*entropy(N2,T=T[2],P=y_N2*P[2]) + y_O2*entropy(O2,T=T[2],P=y_O2*P[2])s_IG[1] =y_N2*entropy(N2,T=T[1],P=y_N2*P[1]) + y_O2*entropy(O2,T=T[1],P=y_O2*P[1]) DELTAs_IG =s_IG[2]-s_IG[1] Q_dot_out_IG=N_dot*q_bar_out_IGW_dot_in_IG=N_dot*w_bar_in_IG"EES:"PN2[1]=y_N2*P[1]PO2[1]=y_O2*P[1]PN2[2]=y_N2*P[2]PO2[2]=y_O2*P[2]e_bar_in_EES - e_bar_out_EES = DELTAe_bar_sys e_bar_in_EES =w_bar_in_EES + h_bar_EES[1] e_bar_out_EES = q_bar_out_EES+h_bar_EES[2] h_bar_EES[1] = y_N2*enthalpy(Nitrogen,T=T[1], P=PN2[1]) +y_O2*enthalpy(Oxygen,T=T[1],P=PO2[1])h_bar_EES[2] = y_N2*enthalpy(Nitrogen,T=T[2],P=PN2[2]) +y_O2*enthalpy(Oxygen,T=T[2],P=PO2[2])q_bar_out_EES = -T[1]*DELTAs_EESDELTAs_EES =y_N2*entropy(Nitrogen,T=T[2],P=PN2[2]) + y_O2*entropy(Oxygen,T=T[2],P=PO2[2]) - y_N2*entropy(Nitrogen,T=T[1],P=PN2[1]) - y_O2*entropy(Oxygen,T=T[1],P=PO2[1]) Q_dot_out_EES=N_dot*q_bar_out_EESW_dot_in_EES=N_dot*w_bar_in_EES"Amagat's Rule:"Tcr_N2=126.2 [K] "Table A.1"Tcr_O2=154.8 [K] Pcr_N2=3390 [kPa] "Table A.1"Pcr_O2=5080 [kPa] e_bar_in_Zchart - e_bar_out_Zchart = DELTAe_bar_sys e_bar_in_Zchart=w_bar_in_Zchart + h_bar_Zchart[1] e_bar_out_Zchart =q_bar_out_Zchart + h_bar_Zchart[2] q_bar_out_Zchart = -T[1]*DELTAs_Zchart Q_dot_out_Zchart=N_dot*q_bar_out_ZchartW_dot_in_Zchart=N_dot*w_bar_in_Zchart"State 1by compressability chart"Tr_N2[1]=T[1]/Tcr_N2Pr_N2[1]=y_N2*P[1]/Pcr_N2

13-40

Tr_O2[1]=T[1]/Tcr_O2Pr_O2[1]=y_O2*P[1]/Pcr_O2DELTAh_bar_1_N2=ENTHDEP(Tr_N2[1], Pr_N2[1])*R_u*Tcr_N2 "Enthalpy departure, N2"DELTAh_bar_1_O2=ENTHDEP(Tr_O2[1], Pr_O2[1])*R_u*Tcr_O2 "Enthalpy departure, O2"h_bar_Zchart[1]=h_bar_IG[1]-(y_N2*DELTAh_bar_1_N2+y_O2*DELTAh_bar_1_O2) "Enthalpy of realgas using charts"DELTAs_N2[1]=ENTRDEP(Tr_N2[1], Pr_N2[1])*R_u "Entropy departure, N2"DELTAs_O2[1]=ENTRDEP(Tr_O2[1], Pr_O2[1])*R_u "Entropy departure, O2"s[1]=s_IG[1]-(y_N2*DELTAs_N2[1]+y_O2*DELTAs_O2[1]) "Entropy of real gas using charts""State 2 by compressability chart"Tr_N2[2]=T[2]/Tcr_N2Pr_N2[2]=y_N2*P[2]/Pcr_N2Tr_O2[2]=T[2]/Tcr_O2Pr_O2[2]=y_O2*P[2]/Pcr_O2DELTAh_bar_2_N2=ENTHDEP(Tr_N2[2], Pr_N2[2])*R_u*Tcr_N2 "Enthalpy departure, N2"DELTAh_bar_2_O2=ENTHDEP(Tr_O2[2], Pr_O2[2])*R_u*Tcr_O2 "Enthalpy departure, O2"h_bar_Zchart[2]=h_bar_IG[2]-(y_N2*DELTAh_bar_2_N2+y_O2*DELTAh_bar_2_O2) "Enthalpy of realgas using charts"DELTAs_N2[2]=ENTRDEP(Tr_N2[2], Pr_N2[2])*R_u "Entropy departure, N2"DELTAs_O2[2]=ENTRDEP(Tr_O2[2], Pr_O2[2])*R_u "Entropy departure, O2"s[2]=s_IG[2]-(y_N2*DELTAs_N2[2]+y_O2*DELTAs_O2[2]) "Entropy of real gas using charts"DELTAs_Zchart = s[2]-s[1] "[kJ/kmol-K]"

SOLUTIONDELTAe_bar_sys=0 [kJ/kmol] DELTAh_bar_1_N2=461.2DELTAh_bar_1_O2=147.6 DELTAh_bar_2_N2=907.8DELTAh_bar_2_O2=299.5 DELTAs_EES=-7.23 [kJ/kmol-K]DELTAs_IG=-5.763 [kJ/kmol-K] DELTAs_N2[1]=1.831DELTAs_N2[2]=3.644 DELTAs_O2[1]=0.5361DELTAs_O2[2]=1.094 DELTAs_Zchart=-7.312 [kJ/kmol-K]e_bar_in_EES=-2173 [kJ/kmol] e_bar_in_IG=-1633 [kJ/kmol]e_bar_in_Zchart=-2103 e_bar_out_EES=-2173 [kJ/kmol]e_bar_out_IG=-1633 [kJ/kmol] e_bar_out_Zchart=-2103h_bar_EES[1]=-3235 h_bar_EES[2]=-3619h_bar_IG[1]=-2785 h_bar_IG[2]=-2785h_bar_Zchart[1]=-3181 h_bar_Zchart[2]=-3565m_dot=2.9 [kg/s] N_dot=0.1005 [kmol/s]Pcr_N2=3390 [kPa] Pcr_O2=5080 [kPa] P[1]=4000 [kPa] P[2]=8000 [kPa] PN2[1]=3160 PN2[2]=6320PO2[1]=840 PO2[2]=1680Pr_N2[1]=0.9322 Pr_N2[2]=1.864Pr_O2[1]=0.1654 Pr_O2[2]=0.3307q_bar_out_EES=1446 [kJ/kmol] q_bar_out_IG=1153 [kJ/kmol]q_bar_out_Zchart=1462 Q_dot_out_EES=145.3 [kW] Q_dot_out_IG=115.9 [kW] Q_dot_out_Zchart=147 [kW] R_u=8.314 [kJ/kmol-K] s[1]=155.1s[2]=147.8 s_IG[1]=156.7s_IG[2]=150.9 Tcr_N2=126.2 [K] Tcr_O2=154.8 [K] T[1]=200 [K] T[2]=200 [K] Tr_N2[1]=1.585Tr_N2[2]=1.585 Tr_O2[1]=1.292Tr_O2[2]=1.292 w_bar_in_EES=1062 [kJ/kmol]w_bar_in_IG=1153 [kJ/kmol] w_bar_in_Zchart=1078 [kJ/kmmol]W_dot_in_EES=106.8 [kW] W_dot_in_IG=115.9 [kW] W_dot_in_Zchart=108.3 [kW] y_N2=0.79y_O2=0.21

13-41

13-67 The volumetric fractions of the constituents of a mixture of products of combustion are given. The average molar mass of the mixture, the average specific heat, and the partial pressure of the water vapor inthe mixture are to be determined.Assumptions Under specified conditions all N2, O2, H2O, and CO2 can be treated as ideal gases, and the mixture as an ideal gas mixture.Properties The molar masses of CO2, H2O, O2, and N2 are 44.0, 18.0, 32.0, and 28.0 kg/kmol, respectively(Table A-1). The specific heats of CO2, H2O, O2, and N2 at 600 K are 1.075, 2.015, 1.003, and 1.075 kJ/kg.K, respectively (Table A-2b). The specific heat of water vapor at 600 K is obtained from EES. Analysis For convenience, consider 100 kmol of mixture. Noting that volume fractions are equal to molefractions in ideal gas mixtures, the average molar mass of the mixture is determined to be

kg/kmol28.62kmol76.41)12.206.50(4.89

)(76.41)(28)(12.20)(32(6.50)(18)kg/kmol)kmol)(44(4.892222

22222222

NOOHCO

NNOOOHOHCOCO

NNNNMNMNMNMN

M m76.41% N212.20% O26.50% H2O4.89% CO2

600 K 200 kPa The average specific heat is determined from

kJ/kmol.K31.59kmol76.41)12.206.50(4.89

075)(28)(76.41)(1.003)(32)(12.20)(1.15)(18)(6.50)(2.0kg/kmol)4kJ/kg.K)(45kmol)(1.07(4.892222

222222222222

NOOHCO

NN,NOO,OOHOH,HCOCO,CO

NNNNMcNMcNMcNMcN

c pppOpp,m

The partial pressure of the water in the mixture is

kPa13.0kPa)200)(0650.0(

0650.0kmol76.41)12.206.50(4.89

kmol6.50

2222

2

NOOHCO

OH

mvv

v

PyP

NNNN

Ny

13-42

13-68 The volumetric fractions of the constituents of a mixture are given. The makeup of the mixture on amass basis and the enthalpy change per unit mass of mixture during a process are to be determined.Assumptions Under specified conditions all N2, O2, and CO2 can be treated as ideal gases, and the mixtureas an ideal gas mixture.Properties The molar masses of CO2, O2, and N2 are 44.0, 32.0, and 28.0 kg/kmol, respectively (Table A-1).Analysis Noting that volume fractions are equal to mole fractions in ideal gas mixtures, the molar mass of the mixture is determined to be

kg/kmol60.31(0.70)(28)(0.10)(32)kg/kmol)(0.20)(44222222 NNOOCOCO MyMyMyM m

The mass fractions are 70% N210% O2

20% CO2

T1 = 300 K T2 = 500 K

0.6203

0.1013

0.2785



kg/kmol31.60kg/kmol44

)20.0(mf

2

22

2

22

2

22

NNN

OOO

COCOCO

m

m

m

MM

y

MM

y

MM

y

The enthalpy change of each gas and the enthalpy change of the mixture are (from Tables A-18-20)

kJ/kg21.209kg/kmol28

kJ/kmol)8723581,14(


kJ/kmol)8736770,14(


kJ/kmol)9431678,17(

22

22

22

N

K300@K500@N

O

K300@K500@O

CO

K300@K500@CO

Mhh

h

Mhh

h

Mhh

h

kJ/kg201.1)21.209)(6203.0()56.188)(1013.0()43.187)(2785.0(

mfmfmf222222 NNOOCOCO hhhhm

13-43

Special Topic: Chemical Potential and the Separation Work of Mixtures

13-69C No, a process that separates a mixture into its components without requiring any work (exergy)input is impossible since such a process would violate the 2nd law of thermodynamics.

13-70C Yes, the volume of the mixture can be more or less than the sum of the initial volumes of themixing liquids because of the attractive or repulsive forces acting between dissimilar molecules.

13-71C The person who claims that the temperature of the mixture can be higher than the temperatures of the components is right since the total enthalpy of the mixture of two components at the same pressure andtemperature, in general, is not equal to the sum of the total enthalpies of the individual components before mixing, the difference being the enthalpy (or heat) of mixing, which is the heat released or absorbed as two or more components are mixed isothermally.

13-72C Mixtures or solutions in which the effects of molecules of different components on each other are negligible are called ideal solutions (or ideal mixtures). The ideal-gas mixture is just one category of ideal solutions. For ideal solutions, the enthalpy change and the volume change due to mixing are zero, but theentropy change is not. The chemical potential of a component of an ideal mixture is independent of theidentity of the other constituents of the mixture. The chemical potential of a component in an ideal mixtureis equal to the Gibbs function of the pure component.

13-73 Brackish water is used to produce fresh water. The minimum power input and the minimum heightthe brackish water must be raised by a pump for reverse osmosis are to be determined.Assumptions 1 The brackish water is an ideal solution since it is dilute. 2 The total dissolved solids inwater can be treated as table salt (NaCl). 3 The environment temperature is also 12 C.Properties The molar masses of water and salt are Mw = 18.0 kg/kmol and Ms = 58.44 kg/kmol. The gas constant of pure water is Rw = 0.4615 kJ/kg K (Table A-1). The density of fresh water is 1000 kg/m3.Analysis First we determine the mole fraction of pure water in brackish water using Eqs. 13-4 and 13-5. Noting that mfs = 0.00078 and mfw = 1- mfs = 0.99922,

kg/kmol01.18

0.180.99922

44.580.00078

1mfmf

1mf1

m

w

w

s

s

i

i

MMM

M

99976.0kg/kmol18.0kg/kmol01.18)99922.0(mfmf

w

mww

i

mii M

MyMMy

The minimum work input required to produce 1 kg of freshwater from brackish water isrfresh watekJ/kg03159.076)ln(1/0.999K)K)(285.15kJ/kg4615.0()/1ln(0inmin, ww yTRw

Therefore, 0.03159 kJ of work is needed to produce 1 kg of fresh water is mixed with seawater reversibly.Therefore, the required power input to produce fresh water at the specified rate is

kW8.85kJ/s1kW1kJ/kg)9/s)(0.0315m280.0)(kg/m1000( 33

inmin,inmin, wVW

The minimum height to which the brackish water must be pumped is

m3.22kJ1N.m1000

N1kg.m/s1

m/s9.81kJ/kg03159.0 2

2inmin,

min gw

z

13-44

13-74 A river is discharging into the ocean at a specified rate. The amount of power that can be generatedis to be determined.Assumptions 1 The seawater is an ideal solution since it is dilute. 2 The total dissolved solids in water canbe treated as table salt (NaCl). 3 The environment temperature is also 15 C.Properties The molar masses of water and salt are Mw = 18.0 kg/kmol and Ms = 58.44 kg/kmol. The gas constant of pure water is Rw = 0.4615 kJ/kg K (Table A-1). The density of river water is 1000 kg/m3.Analysis First we determine the mole fraction of pure water in ocean water using Eqs. 13-4 and 13-5. Noting that mfs = 0.035 and mfw = 1- mfs = 0.965,

kg/kmol45.18

0.180.965

44.580.035

1mfmf

1mf1

m

w

w

s

s

i

i

MMM

M


w

mww

i

mii M

MyMMy

The maximum work output associated with mixing 1 kg of seawater (or the minimum work input requiredto produce 1 kg of freshwater from seawater) is

rfresh watekJ/kg46.1891)K)ln(1/0.9K)(288.15kJ/kg4615.0()/1ln(0outmax, ww yTRw

Therefore, 1.46 kJ of work can be produced as 1 kg of fresh water is mixed with seawater reversibly.Therefore, the power that can be generated as a river with a flow rate of 400,000 m3/s mixes reversiblywith seawater is

kW10582 6

kJ/s1kW1kJ/kg)/s)(1.46m104)(kg/m1000( 353

outmaxoutmax wVW

Discussion This is more power than produced by all nuclear power plants (112 of them) in the U.S., whichshows the tremendous amount of power potential wasted as the rivers discharge into the seas.

13-45

13-75 EES Problem 13-74 is reconsidered. The effect of the salinity of the ocean on the maximum power generated is to be investigated.Analysis The problem is solved using EES, and the solution is given below.

"Properties:"M_w = 18.0 [kg/kmol] "Molar masses of water"M_s = 58.44 [kg/kmol] "Molar masses of salt"R_w = 0.4615 [kJ/kg-K] "Gas constant of pure water"roh_w = 1000 [kg/m^3] "density of river water"V_dot = 4E5 [m^3/s] T_0 = 15 [C]

"Analysis:

First we determine the mole fraction of pure water in oceanwater using Eqs. 13-4 and 13-5. "mf_s = 0.035 "mass fraction of the salt in seawater = salinity"mf_w = 1- mf_s "mass fraction of the water in seawater""Molar mass of the seawater is:"M_m=1/(mf_s/m_s+mf_w/M_w)"Mole fraction of the water is:"y_w=mf_w*M_m/M_w"The maximum work output associated with mixing 1 kg of seawater(or the minimum work input required to produce 1 kg of freshwater fromseawater) is:"w_maxout =R_w*(T_0+273.15)*ln(1/y_w) "[kJ/kg fresh water]""The power that can be generated as a river with a flow rate of 400,000 m^3/s mixesreversibly with seawater is"W_dot_max=roh_w*V_dot*w_maxout"Discussion This is more power than produced by all nuclear power plants (112 of them) in the US., which shows the tremendous amount of power potential wasted as the rivers discharge intothe seas."

mfs Wmax

[kW]0 0

0.01 1.652E+080.02 3.333E+080.03 5.043E+080.04 6.783E+080.05 8.554E+08

0 0.01 0.02 0.03 0.04 0.050

1.000x108

2.000x108

3.000x108

4.000x108

5.000x108

6.000x108

7.000x108

8.000x108

9.000x108

mfs

Wm

ax [

kw]

13-46

13-76E Brackish water is used to produce fresh water. The mole fractions, the minimum work inputsrequired to separate 1 lbm of brackish water and to obtain 1 lbm of fresh water are to be determined.Assumptions 1 The brackish water is an ideal solution since it is dilute. 2 The total dissolved solids inwater can be treated as table salt (NaCl). 3 The environment temperature is equal to the water temperature.Properties The molar masses of water and salt are Mw = 18.0 lbm/lbmol and Ms = 58.44 lbm/lbmol. Thegas constant of pure water is Rw = 0.1102 Btu/lbm R (Table A-1E).Analysis (a) First we determine the mole fraction of pure water in brackish water using Eqs. 13-4 and 13-5. Noting that mfs = 0.0012 and mfw = 1- mfs = 0.9988,

lbm/lbmol015.18

0.180.9988

44.580.0012

1mfmf

1mf1

m

w

w

s

s

i

i

MMM

M

0.99963lbm/lbmol18.0lbm/lbmol015.18)9988.0(mfmf

w

mww

i

mii M

MyMMy

0.0003799963.011 ws yy

(b) The minimum work input required to separate 1 lbmol of brackish water is

aterbrackish w)]00037.0ln(0.00037)99963.0ln(R)[0.99963R)(525Btu/lbmol.1102.0(

)lnln(0inmin,

Btu/lbm0.191

sswww yyyyTRw

(c) The minimum work input required to produce 1 lbm of freshwater from brackish water is waterfreshBtu/lbm0.02149963)R)ln(1/0.9R)(525Btu/lbm1102.0()/1ln(0inmin, ww yTRw

Discussion Note that it takes about 9 times work to separate 1 lbm of brackish water into pure water and salt compared to producing 1 lbm of fresh water from a large body of brackish water.

13-47

13-77 A desalination plant produces fresh water from seawater. The second law efficiency of the plant is tobe determined.Assumptions 1 The seawater is an ideal solution since it is dilute. 2 The total dissolved solids in water canbe treated as table salt (NaCl). 3 The environment temperature is equal to the seawater temperature.Properties The molar masses of water and salt are Mw = 18.0 kg/kmol and Ms = 58.44 kg/kmol. The gas constant of pure water is Rw = 0.4615 kJ/kg K (Table A-1). The density of river water is 1000 kg/m3.Analysis First we determine the mole fraction of pure water in seawater using Eqs. 13-4 and 13-5. Notingthat mfs = 0.032 and mfw = 1- mfs = 0.968,

kg/kmol41.18

0.180.968

44.580.032

1mfmf

1mf1

m

w

w

s

s

i

i

MMM

M


w

mww

i

mii M

MyMMy

The maximum work output associated with mixing 1 kg of seawater (or the minimum work input requiredto produce 1 kg of freshwater from seawater) is

rfresh watekJ/kg313.190)K)ln(1/0.9K)(283.15kJ/kg4615.0()/1ln(0outmax, ww yTRw

The power that can be generated as 1.4 m3/s fresh water mixes reversibly with seawater is

kW1.84kJ/s1kW1kJ/kg)/s)(1.313m4.1)(kg/m1000( 33

outmaxoutmax wVW

Then the second law efficiency of the plant becomes

21.6%216.0MW8.5MW1.83

in

inmin,II W

W

13-78 The power consumption and the second law efficiency of a desalination plant are given. The powerthat can be produced if the fresh water produced is mixed with the seawater reversibly is to be determined.Assumptions 1 This is a steady-flow process. 2 The kinetic and potential energy changes are negligible.Analysis From the definition of the second law efficiency

MW0.594revrev

actual

revII MW3.3

0.18 WWWW

which is the maximum power that can be generated.

13-48

Review Problems

13-79 The molar fractions of constituents of air are given. The gravimetric analysis of air and its molarmass are to be determined.Assumptions All the constituent gases and their mixture are ideal gases.Properties The molar masses of O2, N2, and Ar are 32.0, 28.0, and 40.0 kg/kmol. (Table A-1).Analysis For convenience, consider 100 kmol of air. Then the mass of each component and the total massare




ArArArAr

NNNN

OOOO

2222

2222

MNmN

MNmN

MNmN

AIR

21% O278% N21% Ar m kg2896kg40kg2184kg267ArNO 22

mmmm


1.4%

75.4%

23.2%

or0.014kg2896

kg40mf

or0.754kg2896kg2184

mf

or0.232kg2896

kg672mf

ArAr

NN

OO

2

2

2

2

m

m

m

mmmmmm

The molar mass of the mixture is determined from its definitions,

MmNm

m

m

2,896 kg100 kmol

28.96 kg / kmol

13-80 Using Amagat’s law, it is to be shown that for a real-gas mixture.Z ymi

k

1

Zi i

Analysis Using the compressibility factor, the volume of a component of a real-gas mixture and of the volume of the gas mixture can be expressed as

m

mummm

m

muiii P

TRNZP

TRNZVV and

Amagat's law can be expressed as mmim PT ,VV . Substituting,

Z N R TP

Z N R TP

m m u m

m

i i u m

m

Simplifying, iimm NZNZ

Dividing by Nm, iim ZyZ

where Zi is determined at the mixture temperature and pressure.

13-49

13-81 Using Dalton’s law, it is to be shown that for a real-gas mixture.Z ym ii

k

1

Zi

Analysis Using the compressibility factor, the pressure of a component of a real-gas mixture and of thepressure of the gas mixture can be expressed as

PZ N R T

VP

Z N R TVi

i i u m

mm

m m u m

mand

Dalton's law can be expressed as mmim VTPP , . Substituting,

Z N R TV

Z N R TV

m m u m

m

i i u m

m =

Simplifying, iimm NZNZ

Dividing by Nm, iim ZyZ

where Zi is determined at the mixture temperature and volume.

13-82 A mixture of carbon dioxide and nitrogen flows through a converging nozzle. The required make up of the mixture on a mass basis is to be determined.Assumptions Under specified conditions CO2 and N2 can be treated as ideal gases, and the mixture as an ideal gas mixture.Properties The molar masses of CO2 and N2 are 44.0 and 28.0 kg/kmol, respectively (Table A-1). The specific heat ratios of CO2 and N2 at 500 K are kCO2 = 1.229 and kN2 = 1.391 (Table A-2). Analysis The molar mass of the mixture is determined from

2222 NNCOCO MyMyM m

The molar fractions are related to each other by

CO2N2

500 K 360 m/s

122 NCO yy

The gas constant of the mixture is given by

m

um M

RR

The specific heat ratio of the mixture is expressed as

2222 NNCOCO mfmf kkk


m

m

MM

y

MM

y

2

22

2

22

NNN

COCOCO

mf

mf

The exit velocity equals the speed of sound at 500 K

kJ/kg1/sm1000 22

exit TkRV m

Substituting the given values and known properties and solving the above equations simultaneously using EES, we find

0.1620.83822 NCO mf,mf

13-50

13-83 The volumetric fractions of the constituents of combustion gases are given. The mixture undergoes areversible adiabatic expansion process in a piston-cylinder device. The work done is to be determined.Assumptions Under specified conditions all CO2, H2O, O2, and N2 can be treated as ideal gases, and the mixture as an ideal gas mixture.Properties The molar masses of CO2, H2O, O2, and N2 are 44.0, 18.0, 32.0, and 28.0 kg/kmol, respectively(Table A-1).Analysis Noting that volume fractions are equal to mole fractions in ideal gas mixtures, the molar mass of the mixture is determined to be

kg/kmol63.288)(0.7641)(22)(0.1220)(38)(0.0650)(14)(0.0489)(422222222 NNOOOHOHCOCO MyMyMyMyM m


4.89% CO26.5% H2O12.2% O276.41% N2

1800 K, 1 MPa7476.0


)7641.0(mf

1363.0kg/kmol28.63

kg/kmol32)1220.0(mf

0409.0kg/kmol28.63

kg/kmol18)0650.0(mf

07516.0kg/kmol28.63

kg/kmol44)0489.0(mf

2

22

2

22

2

22

2

22

NNN

OOO

OHOHOH

COCOCO

m

m

m

m

MM

y

MM

y

MM

y

MM

y

Using Dalton’s law to find partial pressures, the entropies at the initial state are determined from EES as

kJ/kg.K2199.8kPa1.764)10007641.0(K,1800

kJ/kg.K2570.8kPa122)10001220.0(K,1800

kJ/kg.K590.14kPa65)10000650.0(K,1800

kJ/kg.K0148.7kPa9.48)10000489.0(K,1800

1,O

1,N

1O,H

1,CO

2

2

2

2

sPT

sPT

sPT

sPT

The final state entropies cannot be determined at this point since the final temperature is not known. However, for an isentropic process, the entropy change is zero and the final temperature may bedetermined from

0mfmfmfmf22222222 NNOOCHOHCOCOtotal sssss

The solution may be obtained using EES to be T2 = 1253 K

The initial and final internal energies are (from EES)

kJ/kg5.696kJ/kg8.662

kJ/kg955,11kJ/kg8102

K1253

kJ/kg,1214kJ/kg1147

kJ/kg779,10kJ/kg7478

K1800

2,N

2,O

2,CH

2,CO

2

1,N

1,O

1,CH

1,CO

1

2

2

2

2

2

2

2

2

uu

uu

T

uu

uu

T


mm uwuwq outoutin

where)(mf)(mf)(mf)(mf 1,N2,NN1,O2,OO1,OH2,OHCO1,CO2,COCO 22222222222

uuuuuuuuum

Substituting,

kJ/kg547.812145.6967476.011478.6621363.0

)779,10()955,11(0409.0)7478()8102(07516.0out muw

13-51

13-84 The mole numbers, pressure, and temperature of the constituents of a gas mixture are given. Thevolume of the tank containing this gas mixture is to be determined using three methods.Analysis (a) Under specified conditions both N2 and CH4 will considerably deviate from the ideal gasbehavior. Treating the mixture as an ideal gas gives

and

3m11.1kPa12,000


kmol8kmol6kmol2

3

CHN 42

m

mumm

m

PTRN

NNN

V

2 kmol N26 kmol CH4

200 K 12 MPa (b) To use Kay's rule, we first need to determine the pseudo-critical

temperature and pseudo-critical pressure of the mixture using the criticalpoint properties of N2 and CH4 from Table A-1,

MPa4.33MPa)4(0.75)(4.6MPa)9(0.25)(3.3

K174.9K).1(0.75)(191K)20.25)(126.(

75.0kmol8kmol6

and0.25kmol8kmol2

4422

4422

4

4

2

2

CH,crCHN,crN,cr,cr

CH,crCHN,crN,cr,cr

CHCH

NN

PyPyPyP

TyTyTyT

NN

yNN

y

iim

iim

mm

Then,

47.077.2

33.412

144.19.174

200

',cr

',cr

m

m

mR

m

mR

Z

PP

P

TT

T

(Fig. A-15)

Thus,

3m0.52)m.111)(47.0( 3idealVV m

m

mummm Z

PTRNZ

(c) To use the Amagat's law for this real gas mixture, we first need to determine the Z of each componentat the mixture temperature and pressure,

N2: 85.054.3

39.312

585.12.126

200

2

22

22

N

Ncr,N,

Ncr,N,

Z

PP

P

TT

T

mR

mR

(Fig. A-15)

CH4: 37.0586.2

64.412

047.11.191

200

4

44

44

CH

CHcr,CH,

CHcr,CH,

Z

PP

P

TT

T

mR

mR

(Fig. A-15)

Mixture:

49.037.075.085.025.04422 CHCHN ZyZyZyZ Niim

Thus,

3m544.0)m11.1)(49.0( 3idealVV m

m

mummm Z

PTRNZ

13-52

13-85 A stream of gas mixture at a given pressure and temperature is to be separated into its constituentssteadily. The minimum work required is to be determined.Assumptions 1 Both the N2 and CO2 gases and their mixture are ideal gases. 2 This is a steady-flow process. 3 The kinetic and potential energy changes are negligible.Properties The molar masses of N2 and CO2 are 28.0 and 44.0 kg/kmol. (Table A-1).Analysis The minimum work required to separate a gas mixture into its components is equal to thereversible work associated with the mixing process, which is equal to the exergy destruction (orirreversibility) associated with the mixing process since

gen0outrev,0

,actoutrev,destroyed STWWWX u

where Sgen is the entropy generation associated with thesteady-flow mixing process. The entropy change associated with a constant pressure and temperature adiabatic mixingprocess is determined from

kJ/kg46.6KkJ/kg0.160K291

KkJ/kg0.160kg/kmol36

KkJ/kmol5.763

kg/kmol36kg/kmol440.5kg/kmol280.5

KkJ/kmol5.763ln(0.5)0.5ln(0.5)0.5KkJ/kmol8.314ln

gen0destroyed

gengen

gen

sTx

Ms

s

MyM

yyRss

m

iim

iiui

100 kPa

CO218 C

N218 C

50% N250% CO2

18 C

13-53

13-86 A gas mixture is heated during a steady-flow process. The heat transfer is to be determined usingtwo approaches.Assumptions 1 Steady flow conditions exist. 2 Kinetic and potential energy changes are negligible.Analysis Noting that there is no work involved, the energy balance for this gas mixture can be written, on a unit mole basis, as

E E E

E E

q h h

q hin

in

in out system (steady)

in out

0

1 2

0

Q

210 K8 MPa

180 K 1O2+3N2

Also, .75.0and52.022 NO yy

(a) Assuming ideal gas behavior, the inlet and exit enthalpies of O2 and N2 are determined from the ideal gas tables to be

kJ/kmol,100.56kJ/kmol,5229:N

kJ/kmol6112.9,kJ/kmol5239.6:O

K210@2K180@12

K210@2K180@12

hhhh

hhhh

Thus,

kJ/kmol872.0229,55.100,675.06.239,59.112,625.0

)()(2222 N12O12Oidealin, hhyhhyhyq Nii

(b) Using the Kay's rule, the gas mixture can be treated as a pseudo-pure substance whose critical temperature and pressure are

MPa3.81MPa)9(0.75)(3.3MPa)8(0.25)(5.0

K133.4K).2(0.75)(126K).8(0.25)(154

2222

2222

N,crNO,crO,cr,cr

N,crNOcr,O,cr,cr

PyPyPyP

TyTyTyT

iim

iim

Then,

1.1

4.1

574.14.133

210

100.281.38

349.14.133

180

2

1

,cr

2,2,

,cr2,1,

,cr

1,1,

h

h

m

mR

m

mRR

m

mR

Z

Z

TT

T

PP

PP

TT

T

(Fig. A-29)

The heat transfer in this case is determined from

kJ/kmol1205kJ/kmol)(8721.1)K)(1.4K)(133.4kJ/kmol(8.314

)(

)()(

idealcr

1212

21

21

qZZTR

hhZZTRhhq

hhu

idealhhcruin

13-54

13-87 EES Problem 13-86 is reconsidered. The effect of the mole fraction of oxygen in the mixture on heattransfer using real gas behavior with EES data is to be investigated.Analysis The problem is solved using EES, and the solution is given below.

"Input Data:"y_N2/y_O2 =3 T[1]=180 [K] "Inlet temperature"T[2]=210 [K] "Exit temmperature"P=8000 [kPa] R_u = 8.34 [kJ/kmol-K]"Solution is done on a unit mole of mixture basis:"y_N2 + y_O2 =1 DELTAe_bar_sys = 0 "Steady-flow analysis for all cases"

"Ideal gas:"e_bar_in_IG - e_bar_out_IG = DELTAe_bar_syse_bar_in_IG =q_bar_in_IG + h_bar_1_IG e_bar_out_IG = h_bar_2_IG h_bar_1_IG = y_N2*enthalpy(N2,T=T[1]) + y_O2*enthalpy(O2,T=T[1])h_bar_2_IG = y_N2*enthalpy(N2,T=T[2]) + y_O2*enthalpy(O2,T=T[2])

"EES:"P_N2 = y_N2*P P_O2 = y_O2*P e_bar_in_EES - e_bar_out_EES = DELTAe_bar_sys e_bar_in_EES =q_bar_in_EES + h_bar_1_EES e_bar_out_EES = h_bar_2_EES h_bar_1_EES = y_N2*enthalpy(Nitrogen,T=T[1], P=P_N2) +y_O2*enthalpy(Oxygen,T=T[1],P=P_O2)h_bar_2_EES = y_N2*enthalpy(Nitrogen,T=T[2],P=P_N2) +y_O2*enthalpy(Oxygen,T=T[2],P=P_O2)

"Kay's Rule:"Tcr_N2=126.2 [K] "Table A.1"Tcr_O2=154.8 [K] Pcr_N2=3390 [kPa] "Table A.1"Pcr_O2=5080 [kPa] Tcr_mix=y_N2*Tcr_N2+y_O2*Tcr_O2Pcr_mix=y_N2*Pcr_N2+y_O2*Pcr_O2e_bar_in_Zchart - e_bar_out_Zchart = DELTAe_bar_syse_bar_in_Zchart=q_bar_in_Zchart + h_bar_1_Zchart e_bar_out_Zchart = h_bar_2_Zchart

"State 1by compressability chart"Tr[1]=T[1]/Tcr_mixPr[1]=P/Pcr_mixDELTAh_bar_1=ENTHDEP(Tr[1], Pr[1])*R_u*Tcr_mix "Enthalpy departure"h_bar_1_Zchart=h_bar_1_IG-DELTAh_bar_1 "Enthalpy of real gas using charts"

"State 2 by compressability chart"Tr[2]=T[2]/Tcr_mixPr[2]=Pr[1]DELTAh_bar_2=ENTHDEP(Tr[2], Pr[2])*R_u*Tcr_mix "Enthalpy departure"h_bar_2_Zchart=h_bar_2_IG-DELTAh_bar_2 "Enthalpy of real gas using charts"

13-55

qinEES

[kJ/kmol]qinIG

[kJ/kmol]qinZchart

[kJ/kmol]yO2

1320 659.6 1139 0.011241 693.3 1193 0.11171 730.7 1255 0.21144 749.4 1287 0.251123 768.1 1320 0.31099 805.5 1387 0.41105 842.9 1459 0.51144 880.3 1534 0.61221 917.7 1615 0.71343 955.2 1702 0.81518 992.6 1797 0.91722 1026 1892 0.99

0 0.2 0.4 0.6 0.8 1600

800

1000

1200

1400

1600

1800

2000

yO2

q in

[kJ/

kmol

]

Ideal GasIdeal GasEESEESComp. ChartComp. Chart

13-56

13-88 A gas mixture is heated during a steady-flow process, as discussed in the previous problem. Thetotal entropy change and the exergy destruction are to be determined using two methods.Analysis The entropy generated during thisprocess is determined by applying the entropybalance on an extended system that includes thepiston-cylinder device and its immediatesurroundings so that the boundary temperature of the extended system is the environmenttemperature at all times. It gives

S S S S

QT

S S S m s s QT

in out gen system

in

boundarygen system gen

in

surr( )2 1

Q

210 K 8 MPa

180 K 1O2+3N2

Then the exergy destroyed during a process can be determined from its definition .X Tdestroyed gen0S

(a) Noting that the total mixture pressure, and thus the partial pressure of each gas, remains constant, the entropy change of a component in the mixture during this process is

1

20

1

2

1

2 lnlnlnTTMc

PPR

TTcs p

iupi

Assuming ideal gas behavior and cp values at room temperature (Table A-2), the s of O2 and N2 are determined from

KkJ/kmol4.49KkJ/kmol4.480.75KkJ/kmol4.520.25

KkJ/kmol4.48K180K210

lnKkJ/kg1.039kg/kmol28


lnKkJ/kg0.918kg/kmol32

2222

2

2

idealsys,

ideal,N

ideal,O

NNOOii sysysys

s

s

and

kJ/kmol488

KkJ/kmol1.61


kJ/kmol872KkJ/kmol.494

0destroyed

gen

gensTx

s

(b) Using the Kay's rule, the gas mixture can be treated as a pseudo-pure substance whose critical temperature and pressure are

MPa3.81MPa)9(0.75)(3.3MPa)8(0.25)(5.0

K133.4K).2(0.75)(126K).8(0.25)(154

2222

2222

N,crNO,crO,cr,cr

N,crNO,crO,cr,cr

PyPyPyP

TyTyTyT

iim

iim

13-57

Then,

45.0

8.0

574.14.133

210

100.281.38

349.14.133

180

2

1

,cr

2,2,

,cr2,1,

,cr

1,1,

s

s

m

mR

m

mRR

m

mR

Z

Z

TT

T

PP

PP

TT

T

(Fig. A-30)

Thus,

KkJ/kmol7.40K)kJ/kmol(4.490.45)K)(0.8kJ/kmol(8.314

)( idealsys,sys 21sZZRs ssu

and

kJ/kmol1034

KkJ/kmol3.41


kJ/kmol1204.7KkJ/kmol7.40

gen0destroyed sTx

sgen

13-58

13-89 The masses, pressures, and temperatures of the constituents of a gas mixture in a tank are given.Heat is transferred to the tank. The final pressure of the mixture and the heat transfer are to be determined.Assumptions He is an ideal gas and O2 is a nonideal gas. Properties The molar masses of He and O2 are 4.0 and 32.0 kg/kmol. (Table A-1) Analysis (a) The number of moles of each gas is

kmol1.25kmol0.25kmol1

kmol0.25kg/kmol32

kg8

kmol1kg/kmol4.0

kg4

2

2

2

2

OHe

O

OO

He

HeHe

NNN

Mm

N

Mm

N

mQ

4 kg He 8 kg O2

170 K 7 MPa

Then the partial volume of each gas and the volume of the tank are

He: 33

1,

1HeHe m0.202

kPa7000K)K)(170/kmolmkPa4kmol)(8.31(1

m

u

PTRN

V

O2: 53.010.1

8.154170

38.108.57

1

O,cr

1

O,cr

1,

21

21

Z

TT

T

PP

P

R

mR

(Fig. A-15)

333OHetank

33

1,

1OO

m0.229m0.027m0.202

m0.027kPa7000

K)K)(170/kgmkPa4kmol)(8.315(0.53)(0.2

2

2

2

VVV

Vm

u

PTRZN

The partial pressure of each gas and the total final pressure is

He: kPa7987m0.229


3

tank

2He2He, V

TRNP u

O2: 39.0

616.3kPa)K)/(5080K)(154.8/kmolmkPa(8.314

kmol))/(0.25m(0.229

/

/

/

42.18.154

220

3

3Ocr,Ocr,

O

Ocr,Ocr,

OO,

Ocr,

2

22

2

22

2

2

22

Ru

m

uR

R

PPTR

NPTR

TT

T

Vvv (Fig. A-15)

MPa9.97MPa1.981MPa7.987MPa1.981kPa1981kPa50800.39

2

22

OHe2m,

OcrO

PPPPPP R

(b) We take both gases as the system. No work or mass crosses the system boundary, therefore this is aclosed system with no work interactions. Then the energy balance for this closed system reduces to

E E E

Q U U Uin out system

in He O2

He: kJ623.1K170220KkJ/kg3.1156kg41He TTmcU mv

13-59

O2:

2.1963.1

08.597.942.1

2.238.110.1

22

2

11

1

hR

R

hR

R

ZP

T

ZPT

(Fig. A-29)

kJ/kmol2742kJ/kmol4949)(6404.2)1K)(2.2K)(154.8kJ/kmol(8.314)()( ideal12cr12 21

hhZZTRhh hhu

Also,

kPa828kPa6172kPa7000

kPa6,172m0.229

K)K)(170/kgmkPa4kmol)(8.31(1

1He,1,1,O

3

3

tank

1He1He,

2PPP

TRNP

m

u

V

Thus,

kJ5.421mkPa)828)(0.229(1981kJ/kmol)kmol)(2742(0.25

)()()()(3

tank1,O2,O12O112212OO 22222VVV PPhhNPPhhNU

Substituting,kJ1045kJ5.421kJ623.1inQ

13-60

13-90 A mixture of carbon dioxide and methane expands through a turbine. The power produced by themixture is to be determined using ideal gas approximation and Kay’s rule.Assumptions The expansion process is reversible and adiabatic (isentropic).Properties The molar masses of CO2 and CH4 are 44.0 and 16.0 kg/kmol and respectively. The criticalproperties are 304.2 K, 7390 kPa for CO2 and 191.1 K and 4640 kPa for CH4 (Table A-1).Analysis The molar mass of the mixture is determined to be

kg/kmol80.32(0.40)(16)(0.60)(44)4422 CHCHCOCO MyMyM m

The gas constant is

60% CO240% CH4

100 kPa

1600 K 800 kPa 10 L/s kJ/kg.K2533.0

kg/kmol32.8kJ/kmol.K314.8

m

u

MR

R


1951.0kg/kmol32.8

kg/kmol16)40.0(mf

8049.0kg/kmol32.8

kg/kmol44)60.0(mf

4

44

2

22

CHCHCH

COCOCO

m

m

MM

y

MM

y

Ideal gas solution:Using Dalton’s law to find partial pressures, the entropies at the initial state are determined from EES tobe:

kJ/kg.K188.17kPa320)80040.0(K,1600

kJ/kg.K424.6kPa480)80060.0(K,1600

1,CH

1,CO

4

2

sPT

sPT


)()(0 1,CH2,CHCH1,CO2,COCO

CHCHCOCOtotal

444222

4422

ssmfssmf

smfsmfs

The solution is obtained using EES to be T2 = 1243 K

The initial and final enthalpies and the changes in enthalpy are (from EES)

kJ/kg1136kJ/kg7877

K1243kJ/kg4.747kJ/kg7408

K16002,CH

2,CO2

1,CH

1,CO1

4

2

4

2

uh

Tuh

T


mm hmWhmWQ outoutin

where

kJ/kg9.745)4.747()1136()1951.0()7408()7877()8049.0(

)(mf)(mf 1,CH2,CHCH,1CO,2COCO 444222hhhhhm

The mass flow rate is

kg/s01974.0K)600kJ/kg.K)(1(0.2533

/s)mkPa)(0.010800( 3

1

11

RTP

mV

Substituting, kW14.72kJ/kg)9.745)(01974.0(out mhmW

13-61

Kay’s rule solution:The critical temperature and pressure of the mixture is

kPa6290kPa)0(0.40)(464kPa)0(0.60)(739

K0.259K).1(0.40)(191K).2(0.60)(304

4422

4422

CHcr,CHCOcr,COcr

CHcr,CHCOcr,COcr

PyPyP

TyTyT

State 1 properties:

EES)(from0001277.0

01025.0002.1

127.0kPa6290kPa800

178.6K259.0K1600

1

1

1

cr

11

cr

11

s

h

R

R

ZZZ

PP

P

TT

T

kJ/kg6714.0K)59.0kJ/kg.K)(22533.0)(01025.0(cr11 RTZh h

kJ/kg5813)6714.0()1.747)(1951.0()7408)(8049.0(11,CHCH,1COCO1 4422

hhmfhmfh

kJ/kg.K00003234.0kJ/kg.K)2533.0)(0001277.0(11 RZs s

kJ/kg.K529.8)00003234.0()188.17)(1951.0()424.6)(8049.0(

mfmf 11,CHCH1,COCO1 4422ssss


)(mf)(mf0

mfmf

1,CH2,CHCH1,CO2,COCO

CHCHCOCOtotal

444222

4422

ssss

sss

The solution is obtained using EES to be T2 = 1243 K

The initial and final enthalpies and the changes in enthalpy are

EES)(from0001171.0

00007368.0

016.0kPa6290

kPa100

80.4K259.0K1243

2

2

cr

22

cr

22

s

h

R

R

ZZ

PP

P

TT

T

kJ/kg04828.0K)59.0kJ/kg.K)(22533.0)(000007368.0(22 crh RTZh

kJ/kg6559)4828.0()1136)(1951.0()7877)(8049.0(

mfmf 22,CHCH,2COCO2 4422hhhh


)( 12outoutin hhmWhmWQ m

where the mass flow rate is

kg/s01970.0K)600kJ/kg.K)(12533(1.002)(0.

/s)mkPa)(0.010800( 3

11

11

RTZP

mV

Substituting,

kW14.71kJ/kg)5813()6559()kg/s01970.0(outW

13-62

13-91 Carbon dioxide and oxygen contained in one tank and nitrogen contained in another tank are allowed to mix during which heat is supplied to the gases. The final pressure and temperature of themixture and the total volume of the mixture are to be determined.Assumptions Under specified conditions CO2, N2, and O2 can be treated as ideal gases, and the mixture as an ideal gas mixture.Properties The molar masses of CO2, N2, and O2 are 44.0, 28.0, and 32.0 kg/kmol, respectively (TableA-1). The gas constants of CO2, N2, and O2 are 0.1889, 0.2968, 2598 kJ/kg.K, respectively (Table A-2). Analysis The molar mass of the mixture in tank 1 are

kg/kmol5.39)(0.375)(32)(0.625)(442222 OOCOCO MyMyM m

The gas constant in tank 1 is

kJ/kg.K2104.0kg/kmol39.5kJ/kmol.K314.8

1m

u

MR

R

The volumes of the tanks and the total volume are

3m6.828276.4551.2

m276.4kPa200

K)27315(kJ/kg.K)2968.0(kg)10(

m551.2kPa125

K)27330(kJ/kg.K)2104.0(kg)5(

21total

3

2

2222

3

1

1111

VVV

V

V

PTRm

PTRm N2

10 kg 15 C

200 kPa

Qin = 100 kJ

62.5% CO237.5% O2

5 kg 30 C

125 kPa

The mass fractions in tank 1 are

3037.0kg/kmol39.5

kg/kmol32)375.0(mf6963.0


)625.0(mf 2

22

2

22

OO1,O

COCO1,CO

mm MM

yM

My

The masses in tank 1 and the total mass after mixing are

kg519.1kg)5)(3037.0(mf

kg481.3kg)5)(6963.0(mf

11,O1,O

11,CO1,CO

22

22

mm

mmkg1510521total mmm

The mass fractions of the combined mixture are

6667.01510mf1012.0

15519.1mf2321.0

15481.3mf

total

22,N

total

1,O2,O

total

1,CO2,CO 2

2

2

2

2 mm

mm

mm

The initial internal energies are

kJ/kg77.84C15kJ/kg16.74

kJ/kg89952C30 1,N1

1,O

1,CO1 2

2

2 uTu

uT

Noting that there is no work interaction, an energy balance gives

)77.84()6667.0()16.74()1012.0()8995()2321.0(kg)kJ)/(15100(

)(mf)(mf)(mf/

,2N,2O,2CO

,1N,2N,2N,1O,2O,2O,1CO,2CO,2COtotalin

totalin

222

222222222

uuu

uuuuuumQumQ m

The internal energies after the mixing are a function of mixture temperature only. Using EES, the finaltemperature of the mixture is determined to be

Tmix = 312.4 K The gas constant of the final mixture is

kg/kmol2680.0.2968)(0.6667)(0.2598)(0.1012)(0.1889)(0.2321)(0

mfmfmf222222 N2,NO2,OCO2,COmix RRRR

The final pressure is determined from ideal gas relation to be

kPa1843total

mixmixtotalmix m6.828

K)4.312(kJ/kg.K)2680.0(kg)15(V

TRmP

13-63

13-92 EES A program is to be written to determine the mole fractions of the components of a mixture of three gases with known molar masses when the mass fractions are given, and to determine the massfractions of the components when the mole fractions are given. Also, the program is to be run for a samplecase.Analysis The problem is solved using EES, and the solution is given below.

Procedure Fractions(Type$,A$,B$,C$,A,B,C:mf_A,mf_B,mf_C,y_A,y_B,y_C){If Type$ <> ('mass fraction' OR 'mole fraction' ) then Call ERROR('Type$ must be set equal to "mass fraction" or "mole fraction".')GOTO 10 endif}Sum = A+B+C If ABS(Sum - 1) > 0 then goto 20 MM_A = molarmass(A$) MM_B = molarmass(B$) MM_C = molarmass(C$) If Type$ = 'mass fraction' then mf_A = A mf_B = B mf_C = C sumM_mix = mf_A/MM_A+ mf_B/MM_B+ mf_C/MM_C y_A = mf_A/MM_A/sumM_mix y_B = mf_B/MM_B/sumM_mix y_C = mf_C/MM_C/sumM_mix GOTO 10 endifif Type$ = 'mole fraction' then y_A = A y_B = B y_C = C MM_mix = y_A*MM_A+ y_B*MM_B+ y_C*MM_C mf_A = y_A*MM_A/MM_mix mf_B = y_B*MM_B/MM_mix mf_C = y_C*MM_C/MM_mix GOTO 10 EndifCall ERROR('Type$ must be either mass fraction or mole fraction.')GOTO 10 20:Call ERROR('The sum of the mass or mole fractions must be 1')10:END

"Either the mole fraction y_i or the mass fraction mf_i may be given by setting the parameter Type$='molefraction' when the mole fractions are given or Type$='mass fraction' is given"{Input Data in the Diagram Window}{Type$='mole fraction'A$ = 'N2' B$ = 'O2' C$ = 'Argon'A = 0.71 "When Type$='mole fraction' A, B, C are the mole fractions" B = 0.28 "When Type$='mass fraction' A, B, C are the mass fractions" C = 0.01}Call Fractions(Type$,A$,B$,C$,A,B,C:mf_A,mf_B,mf_C,y_A,y_B,y_C)

SOLUTIONA=0.71 A$='N2' B=0.28 B$='O2'C=0.01 C$='Argon' mf_A=0.680 mf_B=0.306mf_C=0.014 Type$='mole fraction' y_A=0.710y_B=0.280 y_C=0.010

13-64

13-93 EES A program is to be written to determine the apparent gas constant, constant volume specificheat, and internal energy of a mixture of 3 ideal gases when the mass fractions and other properties of theconstituent gases are given. Also, the program is to be run for a sample case. Analysis The problem is solved using EES, and the solution is given below.

T=300 [K] A$ = 'N2' B$ = 'O2' C$ = 'CO2' mf_A = 0.71 mf_B = 0.28 mf_C = 0.01 R_u = 8.314 [kJ/kmol-K]MM_A = molarmass(A$)MM_B = molarmass(B$)MM_C = molarmass(C$)SumM_mix = mf_A/MM_A+ mf_B/MM_B+ mf_C/MM_C y_A = mf_A/MM_A/SumM_mixy_B = mf_B/MM_B/SumM_mixy_C = mf_C/MM_C/SumM_mixMM_mix = y_A*MM_A+ y_B*MM_B+ y_C*MM_C

R_mix = R_u/MM_mixC_P_mix=mf_A*specheat(A$,T=T)+mf_B*specheat(B$,T=T)+mf_C*specheat(C$,T=T)C_V_mix=C_P_mix - R_mixu_mix=C_V_mix*Th_mix=C_P_mix*T

SOLUTIONA$='N2'B$='O2'C$='CO2'C_P_mix=1.006 [kJ/kg-K] C_V_mix=0.7206 [kJ/kg-K] h_mix=301.8 [kJ/kg] mf_A=0.71mf_B=0.28mf_C=0.01MM_A=28.01 [kg/kmol]MM_B=32 [kg/kmol]MM_C=44.01 [kg/kmol]MM_mix=29.14 [kg/kmol]R_mix=0.2854 [kJ/kg-K] R_u=8.314 [kJ/kmol-K]SumM_mix=0.03432T=300 [K] u_mix=216.2 [kJ/kg] y_A=0.7384y_B=0.2549y_C=0.00662

13-65

13-94 EES A program is to be written to determine the entropy change of a mixture of 3 ideal gases whenthe mole fractions and other properties of the constituent gases are given. Also, the program is to be run fora sample case. Analysis The problem is solved using EES, and the solution is given below.

T1=300 [K] T2=600 [K] P1=100 [kPa] P2=500 [kPa] A$ = 'N2' B$ = 'O2' C$ = 'Argon' y_A = 0.71 y_B = 0.28 y_C = 0.01 MM_A = molarmass(A$)MM_B = molarmass(B$)MM_C = molarmass(C$)MM_mix = y_A*MM_A+ y_B*MM_B+ y_C*MM_C mf_A = y_A*MM_A/MM_mixmf_B = y_B*MM_B/MM_mixmf_C = y_C*MM_C/MM_mixDELTAs_mix=mf_A*(entropy(A$,T=T2,P=y_B*P2)-entropy(A$,T=T1,P=y_A*P1))+mf_B*(entropy(B$,T=T2,P=y_B*P2)-entropy(B$,T=T1,P=y_B*P1))+mf_C*(entropy(C$,T=T2,P=y_C*P2)-entropy(C$,T=T1,P=y_C*P1))

SOLUTIONA$='N2'B$='O2'C$='Argon'DELTAs_mix=12.41 [kJ/kg-K] mf_A=0.68mf_B=0.3063mf_C=0.01366MM_A=28.01 [kg/kmol]MM_B=32 [kg/kmol]MM_C=39.95 [kg/kmol]MM_mix=29.25 [kJ/kmol]P1=100 [kPa] P2=500 [kPa] T1=300 [K] T2=600 [K] y_A=0.71y_B=0.28y_C=0.01

13-66

Fundamentals of Engineering (FE) Exam Problems

13-95 An ideal gas mixture whose apparent molar mass is 36 kg/kmol consists of nitrogen N2 and three other gases. If the mole fraction of nitrogen is 0.30, its mass fraction is(a) 0.15 (b) 0.23 (c) 0.30 (d) 0.39 (e) 0.70

Answer (b) 0.23

Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numericalvalues).

M_mix=36 "kg/kmol"M_N2=28 "kg/kmol"y_N2=0.3mf_N2=(M_N2/M_mix)*y_N2

"Some Wrong Solutions with Common Mistakes:"W1_mf = y_N2 "Taking mass fraction to be equal to mole fraction"W2_mf= y_N2*(M_mix/M_N2) "Using the molar mass ratio backwords"W3_mf= 1-mf_N2 "Taking the complement of the mass fraction"

13-96 An ideal gas mixture consists of 2 kmol of N2 and 6 kmol of CO2. The mass fraction of CO2 in themixture is(a) 0.175 (b) 0.250 (c) 0.500 (d) 0.750 (e) 0.825

Answer (e) 0.825


N1=2 "kmol"N2=6 "kmol"N_mix=N1+N2MM1=28 "kg/kmol"MM2=44 "kg/kmol"m_mix=N1*MM1+N2*MM2mf2=N2*MM2/m_mix

"Some Wrong Solutions with Common Mistakes:"W1_mf = N2/N_mix "Using mole fraction"W2_mf = 1-mf2 "The wrong mass fraction"

13-67

13-97 An ideal gas mixture consists of 2 kmol of N2 and 4 kmol of CO2. The apparent gas constant of themixture is(a) 0.215 kJ/kg K (b) 0.225 kJ/kg K (c) 0.243 kJ/kg K (d) 0.875 kJ/kg K(e) 1.24 kJ/kg K

Answer (a) 0.215 kJ/kg K


Ru=8.314 "kJ/kmol.K"N1=2 "kmol"N2=4 "kmol"MM1=28 "kg/kmol"MM2=44 "kg/kmol"R1=Ru/MM1R2=Ru/MM2N_mix=N1+N2y1=N1/N_mixy2=N2/N_mixMM_mix=y1*MM1+y2*MM2R_mix=Ru/MM_mix

"Some Wrong Solutions with Common Mistakes:"W1_Rmix =(R1+R2)/2 "Taking the arithmetic average of gas constants"W2_Rmix= y1*R1+y2*R2 "Using wrong relation for Rmixture"

13-98 A rigid tank is divided into two compartments by a partition. One compartment contains 3 kmol ofN2 at 600 kPa pressure and the other compartment contains 7 kmol of CO2 at 200 kPa. Now the partition is removed, and the two gases form a homogeneous mixture at 300 kPa. The partial pressure of N2 in themixture is(a) 75 kPa (b) 90 kPa (c) 150 kPa (d) 175 kPa (e) 225 kPa

Answer (b) 90 kPa

Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numericalvalues).P1 = 600 "kPa"P2 = 200 "kPa"P_mix=300 "kPa"N1=3 "kmol"N2=7 "kmol"MM1=28 "kg/kmol"MM2=44 "kg/kmol"N_mix=N1+N2y1=N1/N_mixy2=N2/N_mixP_N2=y1*P_mix

"Some Wrong Solutions with Common Mistakes:"W1_P1= P_mix/2 "Assuming equal partial pressures"W2_P1= mf1*P_mix; mf1=N1*MM1/(N1*MM1+N2*MM2) "Using mass fractions"W3_P1 = P_mix*N1*P1/(N1*P1+N2*P2) "Using some kind of weighed averaging"

13-68

13-99 An 80-L rigid tank contains an ideal gas mixture of 5 g of N2 and 5 g of CO2 at a specified pressure and temperature. If N2 were separated from the mixture and stored at mixture temperature and pressure, itsvolume would be(a) 32 L (b) 36 L (c) 40 L (d) 49 L (e) 80 L

Answer (d) 49 L


V_mix=80 "L"m1=5 "g"m2=5 "g"MM1=28 "kg/kmol"MM2=44 "kg/kmol"N1=m1/MM1N2=m2/MM2N_mix=N1+N2y1=N1/N_mixV1=y1*V_mix "L"

"Some Wrong Solutions with Common Mistakes:"W1_V1=V_mix*m1/(m1+m2) "Using mass fractions"W2_V1= V_mix "Assuming the volume to be the mixture volume"

13-100 An ideal gas mixture consists of 3 kg of Ar and 6 kg of CO2 gases. The mixture is now heated at constant volume from 250 K to 350 K. The amount of heat transfer is(a) 374 kJ (b) 436 kJ (c) 488 kJ (d) 525 kJ (e) 664 kJ

Answer (c) 488 kJ


T1=250 "K"T2=350 "K"Cv1=0.3122; Cp1=0.5203 "kJ/kg.K"Cv2=0.657; Cp2=0.846 "kJ/kg.K"m1=3 "kg"m2=6 "kg"MM1=39.95 "kg/kmol"MM2=44 "kg/kmol""Applying Energy balance gives Q=DeltaU=DeltaU_Ar+DeltaU_CO2"Q=(m1*Cv1+m2*Cv2)*(T2-T1)

"Some Wrong Solutions with Common Mistakes:"W1_Q = (m1+m2)*(Cv1+Cv2)/2*(T2-T1) "Using arithmetic average of properties"W2_Q = (m1*Cp1+m2*Cp2)*(T2-T1)"Using Cp instead of Cv"W3_Q = (m1*Cv1+m2*Cv2)*T2 "Using T2 instead of T2-T1"

13-69

13-101 An ideal gas mixture consists of 30% helium and 70% argon gases by mass. The mixture is nowexpanded isentropically in a turbine from 400 C and 1.2 MPa to a pressure of 200 kPa. The mixturetemperature at turbine exit is(a) 195 C (b) 56 C (c) 112 C (d) 130 C (e) 400 C

Answer (b) 56 C


T1=400+273"K"P1=1200 "kPa"P2=200 "kPa"mf_He=0.3mf_Ar=0.7k1=1.667k2=1.667"The specific heat ratio k of the mixture is also 1.667 since k=1.667 for all componet gases"k_mix=1.667T2=T1*(P2/P1)^((k_mix-1)/k_mix)-273

"Some Wrong Solutions with Common Mistakes:"W1_T2 = (T1-273)*(P2/P1)^((k_mix-1)/k_mix) "Using C for T1 instead of K"W2_T2 = T1*(P2/P1)^((k_air-1)/k_air)-273; k_air=1.4 "Using k value for air"W3_T2 = T1*P2/P1 "Assuming T to be proportional to P"

13-102 One compartment of an insulated rigid tank contains 2 kmol of CO2 at 20 C and 150 kPa while theother compartment contains 5 kmol of H2 gas at 35 C and 300 kPa. Now the partition between the twogases is removed, and the two gases form a homogeneous ideal gas mixture. The temperature of themixture is(a) 25 C (b) 29 C (c) 22 C (d) 32 C (e) 34 C

Answer (b) 29 C


N_H2=5 "kmol"T1_H2=35 "C"P1_H2=300 "kPa"N_CO2=2 "kmol"T1_CO2=20 "C"P1_CO2=150 "kPa"Cv_H2=10.183; Cp_H2=14.307 "kJ/kg.K"Cv_CO2=0.657; Cp_CO2=0.846 "kJ/kg.K"MM_H2=2 "kg/kmol"MM_CO2=44 "kg/kmol"m_H2=N_H2*MM_H2m_CO2=N_CO2*MM_CO2"Applying Energy balance gives 0=DeltaU=DeltaU_H2+DeltaU_CO2"0=m_H2*Cv_H2*(T2-T1_H2)+m_CO2*Cv_CO2*(T2-T1_CO2)

13-70

"Some Wrong Solutions with Common Mistakes:"0=m_H2*Cp_H2*(W1_T2-T1_H2)+m_CO2*Cp_CO2*(W1_T2-T1_CO2) "Using Cp instead of Cv"0=N_H2*Cv_H2*(W2_T2-T1_H2)+N_CO2*Cv_CO2*(W2_T2-T1_CO2) "Using N instead of mass"W3_T2 = (T1_H2+T1_CO2)/2 "Assuming averate temperature"

13-103 A piston-cylinder device contains an ideal gas mixture of 3 kmol of He gas and 7 kmol of Ar gas at50 C and 400 kPa. Now the gas expands at constant pressure until its volume doubles. The amount of heattransfer to the gas mixture is (a) 6.2 MJ (b) 42 MJ (c) 27 MJ (d) 10 MJ (e) 67 MJ

Answer (e) 67 MJ


N_He=3 "kmol"N_Ar=7 "kmol"T1=50+273 "C"P1=400 "kPa"P2=P1"T2=2T1 since PV/T=const for ideal gases and it is given that P=constant"T2=2*T1 "K"MM_He=4 "kg/kmol"MM_Ar=39.95 "kg/kmol"m_He=N_He*MM_Hem_Ar=N_Ar*MM_ArCp_Ar=0.5203; Cv_Ar = 3122 "kJ/kg.C"Cp_He=5.1926; Cv_He = 3.1156 "kJ/kg.K""For a P=const process, Q=DeltaH since DeltaU+Wb is DeltaH"Q=m_Ar*Cp_Ar*(T2-T1)+m_He*Cp_He*(T2-T1)

"Some Wrong Solutions with Common Mistakes:"W1_Q =m_Ar*Cv_Ar*(T2-T1)+m_He*Cv_He*(T2-T1) "Using Cv instead of Cp"W2_Q=N_Ar*Cp_Ar*(T2-T1)+N_He*Cp_He*(T2-T1) "Using N instead of mass"W3_Q=m_Ar*Cp_Ar*(T22-T1)+m_He*Cp_He*(T22-T1); T22=2*(T1-273)+273 "Using C for T1"W4_Q=(m_Ar+m_He)*0.5*(Cp_Ar+Cp_He)*(T2-T1) "Using arithmetic averate of Cp"

13-71

13-104 An ideal gas mixture of helium and argon gases with identical mass fractions enters a turbine at1200 K and 1 MPa at a rate of 0.3 kg/s, and expands isentropically to 100 kPa. The power output of theturbine is (a) 478 kW (b) 619 kW (c) 926 kW (d) 729 kW (e) 564 kW

Answer (b) 619 kW


m=0.3 "kg/s"T1=1200 "K"P1=1000 "kPa"P2=100 "kPa"mf_He=0.5mf_Ar=0.5k_He=1.667k_Ar=1.667Cp_Ar=0.5203Cp_He=5.1926Cp_mix=mf_He*Cp_He+mf_Ar*Cp_Ar"The specific heat ratio k of the mixture is also 1.667 since k=1.667 for all componet gases"k_mix=1.667T2=T1*(P2/P1)^((k_mix-1)/k_mix)-W_out=m*Cp_mix*(T2-T1)

"Some Wrong Solutions with Common Mistakes:"W1_Wout= - m*Cp_mix*(T22-T1); T22 = (T1-273)*(P2/P1)^((k_mix-1)/k_mix)+273 "Using C for T1 instead of K"W2_Wout= - m*Cp_mix*(T222-T1); T222 = T1*(P2/P1)^((k_air-1)/k_air)-273; k_air=1.4 "Using k value for air"W3_Wout= - m*Cp_mix*(T2222-T1); T2222 = T1*P2/P1 "Assuming T to be proportional to P"W4_Wout= - m*0.5*(Cp_Ar+Cp_He)*(T2-T1) "Using arithmetic average for Cp"

13-105 Design and Essay Problem

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ThermoSolutions CHAPTER13