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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 mole fraction (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 to be 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

==== mf and mf

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

BBAAm

BBAA

m

mm MyMy

NMNMN

Nm

M +=+

==

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

)1mf/1( BAA

BA MM

My

+−= and 1 AB yy −=

which are the desired relations.

PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

13-2

13-10 The molar fractions of the constituents of moist air are given. The mass fractions of the constituents are to be determined. Assumptions The small amounts of gases in air are ignored, and dry air is assumed to consist of N2 and O2 only. 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% N2 20% O2 2% H2 O

(Mole fractions)

ol

Then the mass fractions of constituent gases are determined to be

0.764===6.280.28)78.0(mf :N 2

22

NNN2 M

My

0.224===6.280.32)20.0(mf :O 2

22

OOO2 M

My

0.013===6.280.18)02.0(mf :OH OH

OHOH22

22 M

My

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 the mixture, 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

( )( )( )( ) kg 1760kg/kmol 44kmol 40kmol 40

kg 1680kg/kmol 28kmol 60kmol 60

2222

2222

COCOCOCO

NNNN

===→=

===→=

MNmN

MNmN

kg 3440kg 1760kg 168022 CON =+=+= mmmm

mole

60% N2 40% CO2

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

51.2%

48.8%

or0.512kg 3440kg 1760

mf

or0.488kg 3440kg 1680

mf

2

2

2

2

COCO

NN

===

===

m

m

m

mm

m

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 the mixture, 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

( )( )( )( ) kg 1760kg/kmol 44kmol 40kmol 40


2222

2222

COCOCOCO

OOOO

===→=

===→=

MNmN

MNmN

kg 3680kg 1760kg 192022 COO =+=+= mmmm

mole

60% O2 40% CO2


%8.47

%2.52

or0.478kg 3680kg 1760

mf

or0.522kg 3680kg 1920

mf

2

2

2

2

COCO

OO

===

===

m

m

m

mm

m


and

Kkg/kJ 226.0

kmol/kg 80.36

⋅=⋅

==

===

kg/kmol 36.8KkJ/kmol 8.314

kmol 100kg 3680

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, the average 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 kg 23kg 10kg 8kg 5

222 CONO =++=++= mmmmm

5 kg O2 8 kg N2

10 kg CO2

Then the mass fraction of each component becomes

0.435

0.348

0.217

===

===

===

kg 23kg 10mf

kg 23kg 8mf

kg 23kg 5

mf

2

2

2

2

2

2

COCO

NN

OO

m

m

m

m

mm

mm

m

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

kmol 0.227kg/kmol 44

kg 10


kg 8


kg 5

2

2

2

2

2

2

2

2

2

CO

COCO

N

NN

O

OO

===

===

===

M

mN

M

mN

M

mN

Thus, kmol 0.669kmol 0.227kmol 0.286kmol 615.0

222 CONO =++=++= NNNN m

and

0.339

0.428

0.233

===

===

===

kmol 0.669kmol 0.227



2

2

2

2

2

2

COCO

NN

OO

m

m

m

N

Ny

N

Ny

N

Ny

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

and

KkJ/kg0.242

kg/kmol34.4

⋅=⋅

==

===

kg/kmol 34.4

KkJ/kmol 8.314

kmol 0.669kg 23

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 and gas 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 component and the total number of moles are

kmol 0.568

kg/kmol 44kg 25

kg 25

kmol .6884kg/kmol 16

kg 75kg 75

2

2

22

4

4

44

CO

COCOCO

CH

CHCHCH

===→=

===→=

M

mNm

M

mNm

mass

75% CH4 25% CO2

kmol .2565kmol 0.568kmol 688.424 COCH =+=+= NNN m

Then the mole fraction of each component becomes

10.8%

89.2%

or0.108kmol 5.256kmol 0.568

or0.892kmol 5.256kmol 4.688

2

2

4

4

COCO

CHCH

===

===

m

m

N

Ny

N

Ny


and

Kkg/kJ 0.437

/

⋅=⋅

==

===

kg/kmol 19.03KkJ/kmol 8.314

kmolkg 03.19kmol 5.256kg 100

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/kmol 28kmol 2kmol 2

kg/kmol 2.0kmol 8kmol 8

2222

2222

NNNN

HHHH

===→=

===→=

MNmN

MNmN

8 kmol H2 2 kmol N2

The total mass and the total number of moles are

kmol 10kmol 2kmol 8

kg 72kg 56kg 16

22

22

NH

NH

=+=+=

=+=+=

NNN

mmm

m

m


and

KkJ/kg1.155

/

⋅=⋅

==

===

kg/kmol 7.2

KkJ/kmol 8.314

kmolkg 2.7kmol 10

kg 72

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

( )( )( )( ) lbm 112

lbm 10

===→=

===→=

lbm/lbmol 28lbmol 4lbmol 4

lbm/lbmol 2.0lbmol 5lbmol 5

2222

2222

NNNN

HHHH

MNmN

MNmN

5 lbmol H2 4 lbmol N2

The total mass and the total number of moles are

lbmol 9lbmol 4lbmol 5

lbm 122lbm 112lbm 10

22

22

NH

NH

=+=+=

=+=+=

NNN

mmm

m

m


lbmollbm 5613.lbmol 9

lbm 122/===

m

mm N

mM

and Rlbm/Btu 1465.0 ⋅=⋅

==lbm/lbmol 13.56

RBtu/lbmol 1.986

m

um M

RR

13-17 The mass fractions of the constituents of a gas mixture are given. The volumetric analysis of the mixture 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 component and the total number of moles are


kg 50kg 50


kg 30kg 20


kg 20kg 20

2

2

22

2

2

22

2

2

22

CO

COCOCO

N

NNN

O

OOO

===→=

===→=

===→=

M

mNm

M

mNm

M

mNm

mass

20% O2 30% N2

50% CO2

kmol 832.2136.1071.1625.0222 CONO =++=++= NNNN m

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

40.1%

37.8%

22.1%

or0.401kmol 2.832kmol 1.136

or0.378kmol 2.832kmol 1.071

or0.221kmol 2.832kmol 0.625

2

2

2

2

2

2

COCO

NN

OO

===

===

===

m

m

m

N

Ny

N

Ny

N

Ny


kmolkg 31.35kmol 2.832kg 100 /===

m

mm N

mM

and Kkg/kJ 0.235 ⋅=⋅

==kg/kmol 35.31

KkJ/kmol 8.314

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 existed alone at the mixture temperature and volume. This law holds exactly for ideal gas mixtures, but only approximately 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 only approximately 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 equation of state using the properties of the individual component instead of the mixture, Pivi = RiTi. The P-v-T behavior 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 mixture temperature 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 mixture temperature 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 volumes each 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 O2 10 kmol CO2

290 K

150 kPa

Analysis The total number of moles is

Then

3m 289.3=⋅⋅

==

=+=+=

kPa 150K) K)(290/kmolmkPa 4kmol)(8.31 (18

kmol 18kmol 10kmol 8

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 O2 10 kmol CO2

400 K

150 kPa

Analysis The total number of moles is

Then

3m 399.1=⋅⋅

==

=+=+=

kPa 150K) K)(400/kmolmkPa 4kmol)(8.31 (18

kmol 18kmol 10kmol 8

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 the mixture 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

3m23.3 kPa 250

K) K)(280/kmolmkPa 4kmol)(8.31 (2.5

kmol 2.5kmol 2kmol 5.0

3

NAr 2

=⋅⋅

==

=+=+=

m

mumm

m

PTRN

NNN

V

Also,

kPa357.1 )kPa (250K 280K 400

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 mixture as 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 kmol

= → = = =

= → = = =3

1 kg CO2 3 kg CH4

300 K

200 kPa N N Nm = + = + =CO CH2 4

kmol 0.1875 kmol 0.2102 kmol0 0227.

y

NN

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

( )( )( )( ) kPa 4.178

kPa 6.21

===

===

kPa 200892.0

kPa 2000.108

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 mixture as 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

lbmol 0.1875lbm/lbmol 16

lbm 3lbm 3

lbmol 0.0227lbm/lbmol 44

lbm 1lbm 1

4

4

44

2

2

22

CH

CHCHCH

CO

COCOCO

===→=

===→=

M

mNm

M

mNm

1 lbm CO2 3 lbm CH4

600 R 20 psia

lbmol 0.2102lbmol 0.1875lbmol 2702.042 CHCO =+=+= NNN m

892.0

lbmol 0.2102lbmol 0.1875

108.0lbmol 0.2102lbmol 0.0227

4

4

2

2

CHCH

COCO

===

===

m

m

N

Ny

N

Ny

Then the partial pressures become

( )( )( )( ) psia 17.84

psia 2.16===

===

psia 20892.0psia 20108.0

44

22

CHCH

COCO

m

m

PyPPyP

The apparent molar mass of the mixture is

lbm/lbmol19.03 lbmol 0.2102

lbm 4===

m

mm N

mM


13-10

13-36 The masses of the constituents of a gas mixture at a specified temperature are given. The partial pressure 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 N2 0.4 kg O2

300 K

kPa 103.9

kPa 178.1

=⋅⋅

=

=

=⋅⋅

=

=

3

3

OO

3

3

NN

m 0.3K) K)(300/kgmkPa kg)(0.2598 (0.4

m 0.3K) K)(300/kgmkPa kg)(0.2968 (0.6

2

2

2

2

V

V

mRTP

mRTP

and kPa282.0 kPa 103.9kPa 1.178

22 ON =+=+= PPPm

13-37 The volumetric fractions of the constituents of a gas mixture at a specified pressure and temperature are 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 mixture as 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

( )( )( )( )

( )( ) kg 660kg/kmol 44kmol 15kmol 15



2222

2222

2222

COCOCOCO

OOOO

NNNN

===→=

===→=

===→=

MNmN

MNmN

MNmN

65% N2 20% O2

15% CO2

350 K 300 kPa kg 3120kg 660kg 640kg 8201

222 COON =++=++= mmmmm


21.2%

20.5%

58.3%

or 0.212kg 3120kg 660mf

or 0.205kg 3120kg 640mf

or 0.583kg 3120kg 1820mf

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

( )( )( )( )( )( ) kPa 45

kPa 60kPa 195

===

===

===

kPa 30015.0kPa 30020.0kPa 30065.0

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 mixture Properties 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

m 0.465

m 0.295

=⋅⋅

=

=

=⋅⋅

=

=

kPa 500K) K)(298/kgmkPa kg)(0.2598 (3

kPa 300K) K)(298/kgmkPa kg)(0.2968 (1

3

OO

3

NN

2

2

2

2

PmRT

PmRT

V

V

3 kg O2 25°C

500 kPa

1 kg N2 25°C

300 kPa

333ONtotal m .760m 0.465m 295.0

22=+=+= VVV

Also,

kmol 0.09375

kg/kmol 32kg 3

kg 3


kg 1kg 1

2

2

22

2

2

22

O

OOO

N

NNN

===→=

===→=

M

mNm

M

mNm

kmol 0.1295kmol 0.09375kmol 03571.022 ON =+=+= NNN m

Thus,

kPa 422.2=⋅⋅

=

=

3

3

m 0.76K) K)(298/kmolmkPa 4kmol)(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 gas behavior. Treating the mixture as an ideal gas,

kmol 2.406K) K)(200/kmolmkPa (8.314

)m kPa)(0.5 (8000

kmol 1.443K) K)(200/kmolmkPa (8.314

)m kPa)(0.3 (8000

3

3

NN

3

3

OO

2

2

2

2

=⋅⋅

=

=

=⋅⋅

=

=

TRPN

TRPN

u

u

V

V

3m0.8 kPa 8000

K) K)(200/kmolmkPa 4kmol)(8.31 (3.849

kmol 3.849


3

NO 22

=⋅⋅

==

=

+=+=

m

mumm

m

PTRN

NNN

V

0.5 m3 N2 200 K 8 MPa

0.3 m3 O2 200 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 determine the Z and the mole numbers of each component at the mixture temperature and pressure (Fig. A-15),

O2: 77.0575.1

MPa 5.08MPa 8

292.1K 154.8

K 200

2

22

22

O

Ocr,O,

Ocr,O,

=

===

===

Z

PP

P

TT

T

mR

mR

N2: 863.0360.2

MPa 3.39MPa 8

585.1K126.2

K200

2

22

22

N

Ncr,N,

Ncr,N,

=

===

===

Z

PP

P

TT

T

mR

mR

kmol 2.787K) K)(200/kmolmkPa 314(0.863)(8.

)m kPa)(0.5 (8000

kmol 1.874K) K)(200/kmolmkPa 14(0.77)(8.3

)m kPa)(0.3 (8000

3

3

NN

3

3

OO

2

2

2

2

=⋅⋅

=

=

=⋅⋅

=

=

TZRPN

TZRPN

u

u

V

V

kmol 4.661kmol 2.787kmol 874.122 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

N

Ny

N

Ny

iim

iim

m

m


13-13

Then,

82.0966.1

MPa 4.07MPa 8

452.1K 137.7

K 200

'Ocr,

'Ocr,

2

2 =

===

===

mm

R

mR

Z

PP

P

TT

T

(Fig. A-15)

Thus, 3m0.79 kPa 8000

K) K)(200/kmolmkPa 4kmol)(8.31 61(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 mixture temperature and pressure, which are determined in part (b). Then,

( )( ) ( )( ) 83.0863.0598.077.0402.02222 NNOO =+=+==∑ ZyZyZyZ iim

Thus, 3m0.80 kPa 8000

K) K)(200/kmolmkPa 4kmol)(8.31 61(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 be determined using two methods. Analysis (a) Under specified conditions both Ar and N2 will considerably deviate from the ideal gas behavior. Treating the mixture as an ideal gas,

MPa 18.2===

==

)MPa (5K) (1)(220K) (4)(200

:state Final :state Initial

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

MPa 4.86MPa 5

457.1K 151.0

K 220

Ar

Arcr,

1

Arcr,

1

=

===

===

Z

PP

P

TT

T

R

R

(Fig. A-15)

Then the volume of the tank is

33

Ar m 0.33kPa 5000

K) K)(220/kmolmkPa 4kmol)(8.31 (0.90)(1=

⋅⋅==

PTRZN uV

After mixing,

Ar: 90.0

278.1kPa) K)/(4860 K)(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)/(3390 K)(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

N

PTR

TT

T

VvV (Fig. A-15)

Thus,

MPa 12.7MPa) 39.3)(75.3()(MPa 4.37MPa) 86.4)(90.0()(

22 NcrN

ArcrAr

======

PPPPPP

R

R

and MPa17.1 MPa 12.7MPa 37.4

2NAr =+=+= PPPm


13-15

13-41 EES Problem 13-40 is reconsidered. The effect of the moles of nitrogen supplied to the tank on the final pressure of the mixture is to be studied using the ideal-gas equation of state and the compressibility chart 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 9091 2 13276 13636 3 17793 18182 4 23254 22727 5 30565 27273 6 41067 31818 7 56970 36364 8 82372 40909 9 126040 45455

10 211047 50000

1 2 3 4 5 6 7 8 9 100

45000

90000

135000

180000

225000

NN2 [kmol]

Pm

ix [

kPa]

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 gas behavior. Treating the mixture as an ideal gas,

psia 2700===

==

)psia 750(R) (1)(400R) (4)(360

:state Final :state Initial

111

222

2222

1111 PTNTNP

TRNPTRNP

u

u

V

V

(b) Initially,

3 lbmol N2 340 R

1200 psia

1 lbmol Ar

400 R 750 psia

90.007.1

psia 705psia 750

47.1R 272R 400

Ar

Arcr,

1

Arcr,

1

=

===

===

Z

PP

P

TT

T

R

R

(Fig. A-15)

Then the volume of the tank is

33

Ar ft 5.15psia 750

R) R)(400/lbmolftpsia 73lbmol)(10. (0.90)(1=

⋅⋅==

PTRZN uV

After mixing,

Ar: 82.0

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

lbmol) )/(1ft (5.15

//

/

324.1R 272R 360

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

lbmol) )/(3ft (5.15

/

/

/

585.1R 227.1

R 360

3

3

Ncr,Ncr,

N

Ncr,Ncr,

NN,

Ncr,N,

22

2

22

2

2

22

=

=⋅⋅

=

==

===

Ru

m

uR

mR

PPTR

N

PTR

TT

T

Vvv (Fig. A-15)

Thus,

psia 1894psia) 492)(85.3()(psia 578psia) 570)(82.0()(

22 NcrN

ArcrAr======

PPPPPP

R

R

and psia 2472=+=+= psia 1894psia 578

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 the mixture 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 form a 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 the mixture 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 in ideal gas mixtures. The partial pressures in the mixture are

Ar 200°C

N2 60°C

21% O2 78% N2 21% Ar

200 kPa

O2 10°C

kPa 2kPa) 200)(01.0(

kPa 156kPa) 200)(78.0(

kPa 42kPa) 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/kmol 96.28(0.01)(40)(0.78)(28)kg/kmol) (0.21)(32ArArNNOO 2222

=++=

++= MyMyMyM m

The mass fractions are

0138.0kg/kmol 28.96

kg/kmol 40)01.0(mf

7541.0kg/kmol 28.96

kg/kmol 28)78.0(mf

2320.0kg/kmol 28.96

kg/kmol 32)21.0(mf

ArArAr

NNN

OOO

2

22

2

22

===

===

===

m

m

m

MM

y

M

My

M

My

The final temperature of the mixture is needed. The conservation of energy on a unit mass basis for steady flow 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.K 7893.6 kPa 156 C,4.50

kJ/kg.K 7461.6 kPa 200 C,60

kJ/kg.K 7082.6 kPa 42 C,4.50

kJ/kg.K 1797.6 kPa 200 C,10

2,N

1,N

2,O

1,O

2

2

2

2

=→=°=

=→=°=

=→=°=

=→=°=

sPT

sPT

sPT

sPT

The entropy changes are

kg/kg.K 04321.07461.67893.6

kg/kg.K 5284.01797.67082.6

1,N2,NN

1,O2,OO

222

222

=−=−=∆

=−=−=∆

sss

sss

kJ/kg.K 7605.0200

2ln)2081.0(2732002734.50ln)5203.0(lnln

1

2

1

2Ar =

−

++

=−=∆PP

RTT

cs p

The total entropy change is

kJ/kg.K 0.1656=++=

∆+∆+∆=∆

)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 unit mass of mixture are to be determined. Assumptions Under specified conditions all CO2, CO, O2, and N2 can be treated as ideal gases, and the mixture 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/kmol 80.30(0.70)(28)(0.10)(32)(0.05)(28)(0.15)(44)222222 NNOOCOCOCOCO

=+++=

+++= MyMyMyMyM m

The mass fractions are

0.6364

0.1039

0.0454

0.2143

===

===

===

===

kg/kmol 30.80kg/kmol 28)70.0(mf




2

22

2

22

2

22

NNN

OOO

COCOCO

COCOCO

m

m

m

m

M

My

M

My

MM

y

M

My

15% CO2 5% CO 10% O2 70% N2

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

22

1

212 667.2

K 300)8)(kPa 100( 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 determined from EES to be:

kJ/kg.K 0115.7 kPa 64.63)1006364.0( K, 300

kJ/kg.K 9485.6 kPa 39.10)1001039.0( K, 300

kJ/kg.K 79483 kPa 55.4)10004545.0( K, 300

kJ/kg.K 2190.5 kPa 43.21)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 not known. However, for an isentropic process, the entropy change is zero and the final temperature and the final pressure may be determined from 0mfmfmfmf

222222 NNOOCOCOCOCOtotal =∆+∆+∆+∆=∆ sssssand 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/kg 9.163kJ/kg 8.156kJ/kg 3780kJ/kg 8734

K 4.631

kJ/kg, 11.87kJ/kg 24.76

kJ/kg 4033kJ/kg 8997

K 300

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/kg 251.8=

−−+−−+−−−+−−−=

−+−+−+−=∆

)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 as an 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

05882.0

171

1AF1mf

9412.01716

1AFAFmf

83HC

air

==+

=

==+

=

Propane Air

95 kPa 30ºC

The molar mass of the mixture is determined to be

kg/kmol 56.29

kg/kmol 44.005882.0

kg/kmol 28.979412.0

1mfmf

1

83

83

HC

HC

air

air=

+=

+

=

MM

M m

The mole fractions are

03944.0

kg/kmol 44.0kg/kmol 56.29)05882.0(mf

9606.0kg/kmol 28.97kg/kmol 56.29)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

K 273.15)(30)5.9)(kPa 95( T

TTT

rPP =+

== (1)

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

kJ/kg.K 7697.6 kPa 75.3)9503944.0( C,30

kJ/kg.K 7417.5 kPa 26.91)959606.0( C,30

1,HC

1,air

83=→=×=°=

=→=×=°=

sPT

sPT

The final state entropies cannot be determined at this point since the final pressure and temperature are not known. However, for an isentropic process, the entropy change is zero and the final temperature and the final pressure may be determined from 0

8383 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/kg 1607

kJ/kg 1.477K 9.654

kJ/kg 2404kJ/kg 5.216

C302,HC

2,air2

1,HC

1,air1

8383−=

=→=

−==

→°=u

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/kg 292.2=−−−+−=∆= )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 mixture temperature and pressure are to be determined. Assumptions 1 Under specified conditions both CO2 and H2 can be treated as ideal gases, and the mixture as 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 system reduces 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°C Using cv values at room temperature and noting that m = NM, the final temperature of the mixture is determined to be

( )( )( ) ( )( )( )( )K 308.8

0C40CkJ/kg 10.183kg 27.5C27CkJ/kg 0.657kg 442.5C35.8°=

=°−°⋅×+°−°⋅×

m

mm

TTT

(b) The volume of each tank is determined from

3

3

H1

1H

33

CO1

1CO

m 7948kPa 400


m 1831kPa 200


2

2

2

2

.

.

=⋅⋅

=

=

=⋅⋅

=

=

PTNR

PTNR

u

u

V

V

Thus,

kmol .010kmol 5.7kmol 5.2

m .9779m .7948m 18.31

22

22

HCO

333HCO

=+=+=

=+=+=

NNNm

m VVV

and kPa 321=⋅⋅

== 3

3

m 79.97K) K)(308.8/kmolmkPa 4kmol)(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 mixture temperature and pressure are to be determined. Assumptions 1 Under specified conditions both Ne and Ar can be treated as ideal gases, and the mixture as an 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

Ar 200 kPa

50°C

Ne 100 kPa

20°C

kmol 0.0335K) K)(323/kmolmkPa (8.314

)m kPa)(0.45 (200

kmol 0.0185K) K)(293/kmolmkPa (8.314

)m kPa)(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 a closed 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

( )( )( )( )( )(

( )K 289.2C50CkJ/kg 0.3122kg 39.950.0335C20CkJ/kg 0.6179kg 20.180.0185kJ 15

C16.2°=°−°⋅×+°

)−°⋅×=−

m

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 mixture temperature and pressure are to be determined. Assumptions 1 Under specified conditions both Ne and Ar can be treated as ideal gases, and the mixture as an 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

kmol 0.0335K) K)(323/kmolmkPa (8.314

)m kPa)(0.45 (200

kmol 0.0185K) K)(293/kmolmkPa (8.314

)m kPa)(0.45 (100

3

3

Ar1

11Ar

3

3

Ne1

11Ne

=⋅⋅

=

=

=⋅⋅

=

=

TRPN

TRPN

u

u

V

V

Ar 200 kPa

50°C

Ne 100 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 a closed 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

( )( )( )( )( )(

( )K 300.0C50CkJ/kg 0.3122kg 39.950.0335C20CkJ/kg 0.6179kg 20.180.0185kJ 8

C0.27 °=°−°⋅×+°

)−°⋅×=−

m

m

m

TTT

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

kPa 1.144=⋅⋅

== 3

3

m 0.9K) K)(300.0/kmolmkPa 4kmol)(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 to be 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. 3 There are no other forms of work involved. 3 This is a steady-flow process. 4 The kinetic and potential energy 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 of pressure, and thus the two gases can be treated independently even 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°C 4.5 kg/s

C2H6 20°C 9 kg/s

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

( )( )( ) ( )( )( )

( )K 302.7C45CkJ/kg 2.2537kg/s 4.5C20CkJ/kg 1.7662kg/s 90

C29.7°=°−°⋅+°−°⋅=

m

mm

TTT

(b) The rate of entropy change associated with this process is determined from an entropy balance on the mixing 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

0

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

kmol/s 0.2813kg/kmol 16

kg/s 4.5 kmol/s 0.3kg/kmol 30

kg/s 9

4

4

62

62CH

CHHC

HC ==

===

=

MmN

MmN

&&&&

Then the mole fraction of each gas becomes

484.02813.03.0

2813.0 516.02813.03.0

3.0462 CHHC =

+==

+= yy

Thus,

KkJ/kg 0.265)4ln(0.48 K)kJ/kg (0.5182K 318K 302.7

ln K)kJ/kg 2.2537(

lnlnlnln)(

KkJ/kg 0.240ln(0.516) K)kJ/kg (0.2765K 293K 302.7

ln K)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/K 3.353=⋅+⋅= KkJ/kg 0.265kg/s 4.5KkJ/kg 0.240kg/s 9genS&


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_F1/2 T1=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_out E_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$) N_dot_F1=m_dot_F1/Mwt_F1 Mwt_F2=MOLARMASS(Fluid2$) N_dot_F2=m_dot_F2 /Mwt_F2 N_dot_tot=N_dot_F1+N_dot_F2 y_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 20.48 0.1 502.5 24.08 0.2 761.4 27 0.3 948.5 29.2 0.4 1096 30.92 0.5 1219 32.3 0.6 1324 33.43 0.7 1415 34.38 0.8 1497 35.18 0.9 1570 35.87 0.99 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 as an 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/kg 0.945

KkJ/kg 0.5203kg/kmol 22

kg/kmol 405.0KkJ/kg 5.1926

kg/kmol 22kg/kmol 45.0

Ar,ArAr

He,HeHe

,mf,

mf

kg/kmol 22405.045.0ArArHeHe

⋅=

⋅×

+⋅×

=

+=∑=

===

=×+×=+=∑=

pcmM

Mypc

mM

Myipcimp

mMiMiy

mMmNiMiN

mmim

i

MyMyiMiymM

c

2.5 MPa 1300 K

He-Ar turbine

200 kPa

and km = 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 isentropic processes,

( )

( ) K 2.473kPa 2500

kPa 200K 130070.667/1.66/1

1

212 =

=

=

− kk

PP

TT

From an energy balance on the turbine,

( ) ( )( ) kJ/kg 781.3=−⋅=−=−=+=

=

=∆=−

K.24731300KkJ/kg 0.945

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 accelerated through a nozzle from a specified state to a specified pressure. For a specified isentropic efficiency, the exit 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 mixture as 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/lbm 0.173RBtu/lbm 0.239

RBtu/lbm 173.0158.02.0177.08.0

mfmfmf

RBtu/lbm 239.0203.02.0248.08.0

mfmfmf

,

,

CO,CON,N,,

CO,CON,N,,

2222

2222

=⋅⋅

==

⋅=+=

+==

⋅=+=

+==

∑

∑

m

mpm

iim

ppipimp

cc

k

cccc

cccc

v

vvvv 80% N2 20% CO2

90 psia 1800 R 12 psia

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

( )

( ) R 1031.3psia 90psia 12

R 180020.382/1.38/1

1

212 =

=

=

− kk

s PP

TT

From the definition of adiabatic efficiency,

( )( ) R 1092.8=→

−−

=→−

−=

−−

= 22

21

21

21

21

3.1031800,1800,1

92.0 TT

TTcTTc

hhhh

sp

p

sNη

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

( )

( ) ( )( ) ft/s 2,909=

−⋅=−=

−+−=

+=+

=

=∆=−

Btu/lbm 1/sft 25,037

R 1092.81800RBtu/lbm 0.23922

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 other conditions being the same, the problem is to be resolved to determine the composition of the nitrogen and carbon 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 to the enthalpy of formation as h = h_f + (h_T - h_537) where h_f is the enthalpy of formation such that the enthalpy of the elements or their stable compounds is zero at 77 F or 537 R, see Chapter 14. The enthalpy of formation is often negative; thus, the enthalpy of ideal gases can be 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 "SSSF energy 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 the entropy change per unit mass of mixture equal to zero." DELTAs_mix=mf_N2 * DELTAs_N2 + mf_CO2 * DELTAs_CO2 DELTAs_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=0.92 h[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 1.049 kJ/kg.K, respectively. (Table A-2b). Analysis (a) Noting that P2 = P1 and V2 = 2V1,

Q

0.5 kg H2 1.6 kg N2 100 kPa 300 K

( )( ) K 600K 300222

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.

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

kJ 2679=−⋅+−⋅=

−+−=∆+∆=∆=

K300600KkJ/kg 1.049kg 1.6K300600KkJ/kg 14.501kg 0.52222 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/K 6.19=+=∆+∆=∆

=

⋅=

=

−=−=∆

=

⋅=

=

−=−=∆

kJ/K 1.163kJ/K .0265

kJ/K 1.163K 300K 600

lnKkJ/kg 1.049kg 1.6

lnlnln

kJ/K .0265K 300K 600lnKkJ/kg 14.501kg 0.5

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 a homogeneous mixture. The final pressure, the heat transfer, and the entropy generated are to be determined. 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,

m 2.25m 2.0m 52.0

kg 10.43K) K)(323/kgmkPa (0.2968

)m kPa)(2 (500

m 0.25kPa 300

K) K)(288/kgmkPa kg)(0.2598 (1

22

2

2

2

2

=+=+=

=⋅⋅

=

=

=⋅⋅

=

=

VVV

V

V

RTP

m

PmRT

N2 2 m3 50°C

500 kPa

O2 1 kg 15°C

300 kPa

Also, Q

kmol 0.40375kmol 0.03125kmol 3725.0


kg 10.43kg 43.10


kg 1kg 1

22

2

2

22

2

2

22

ON

N

NNN

O

OOO

=+=+=

===→=

===→=

NNN

M

mNm

M

mNm

m

Thus, kPa 444.6=⋅⋅

=

=

3

3

m 2.25K) K)(298/kmolmkPa 4kmol)(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), the energy 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/kg 0.743kg 10.43C2515CkJ/kg 0.658kg 1out

kJ 187.2=°−°⋅+°−°⋅=Q

(c) For and extended system that involves the tanks and their immediate surroundings such that the boundary 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

923.0

40375.03725.0

077.040375.003125.0

22

22

NN

OO

===

===

m

m

NN

y

NN

y

Thus,

KkJ/kg 0.0251kPa 500

kPa) 4.6(0.923)(44ln K)kJ/kg (0.2968

K 323K 298

ln K)kJ/kg (1.039

lnln)(

KkJ/kg 0.5952kPa 300

kPa) 4.6(0.077)(44ln K)kJ/kg (0.2598

K 288K 298

ln K)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/K 0.962=+⋅−+⋅=K 298kJ 187.2KkJ/kg 0.0251kg 10.43KkJ/kg 0.5952kg 1genS


13-33

13-62 EES Problem 13-61 is reconsidered. The results obtained assuming ideal gas behavior with constant specific heats at the average temperature, and using real gas data obtained from EES by assuming variable specific 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_O2 N_N2=m_N2/MM_N2; N_total=N_O2+N_N2 P[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_sys E_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_total P_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])) SOLUTION Cv_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 the final 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 H2 21 kg N2 5 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 / kmol

Thus, ( ) ( ) kJ 4273=−×+−×= 648,4810,575.04.535,42.669,53idealQ (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, ( )( ) ( )( ) kJ4745 kJ/kmol 1,791.5kmol 0.75kJ/kmol 1,133.8kmol 3in =+=Q


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