Gas Mixtures

The discussions in this chapter are restricted to nonreactive ideal-gas mixtures. Those interested in real-gas mixtures are encouraged to study carefully the material presented in Chapter 12. Many thermodynamic applications involve mixtures of ideal gases. That is, each of the gases in the mixture individually behaves as an ideal gas. In this section, we assume that the gases in the mixture do not react with one another to any significant degree.

Gas Mixtures

We restrict ourselves to a study of only ideal-gas mixtures. An ideal gas is one in which the equation of state is given by

PV = mRT or

PV = NRu T

Air is an example of an ideal gas mixture and has the following approximate composition. Component N2 O2 Argon CO2 + trace elements

% by Volume 78.10 20.95 0.92 0.03 2

Definitions

The composition of a gas mixture is described by specifying either the mass fraction mfi or the mole fraction yi of each component i.

Consider a container having a volume V that is filled with a mixture of k different gases at a pressure P and a temperature T.

mf i =

A mixture of two or more gases of fixed chemical composition is called a nonreacting gas mixture. Consider k gases in a rigid container as shown here. The properties of the mixture may be based on the mass of each component, called gravimetric analysis, or on the moles of each component, called molar analysis. k gases T = Tm P = Pm

k

i =1

yi =

Ni Nm

=1

and

∑y

Note that k

∑ mf

k

i

i =1

i

=1

i =1

mi = N i M i

To find the average molar mass for the mixture Mm , note k

k

i =1

i =1

mm = ∑ mi = ∑ N i M i = N m M m

k

and

and

The mass and mole number for a given component are related through the molar mass (or molecular weight).

V = Vm m = mm

The total mass of the mixture mm and the total moles of mixture Nm are defined as

mm = ∑ mi

mi mm

N m = ∑ Ni i =1

Solving for the average or apparent molar mass Mm Mm = 3

k k mm N = ∑ i M i = ∑ yi M i N m i =1 N m i =1

( kg / kmol ) 4

The apparent (or average) gas constant of a mixture is expressed as Ru Mm

Rm =

Can you show that Rm is given as

Volume fraction (Amagat model)

( kJ / kg ⋅ K )

Divide the container into k subcontainers, such that each subcontainer has only one of the gases in the mixture at the original mixture temperature and pressure.

k

Rm = ∑ mf i Ri i =1

To change from a mole fraction analysis to a mass fraction analysis, we can show that yM mf i = k i i ∑ yi Mi

Amagat's law of additive vol-umes states that the volume of a gas mixture is equal to the sum of the volumes each gas would occupy if it existed alone at the mixture temperature and pressure.

i =1

k

To change from a mass fraction analysis to a mole fraction analysis, we can show that mf / M i yi = k i ∑ mf i / Mi

Vm = ∑ Vi (Tm , Pm )

Amagat's law:

i =1

The volume fraction of the vfi of any component is

vf i =

i =1

and

Vi ( Tm , Pm ) Vm

k

∑ vf

For an ideal gas mixture NRT Vi = i u m Pm

i

=1

i =1

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Now, consider placing each of the k gases in a separate container having the volume of the mixture at the temperature of the mixture. The pressure that results is called the component pressure, Pi' .

N RT and Vm = m u m Pm

Taking the ratio of these two equations gives

vf i =

Vi N = i = yi Vm N m

The volume fraction and the mole fraction of a component in an ideal gas mixture are the same.

Pi ' =

N i Ru Tm Vm

and

Pm =

N m Ru Tm Vm

Note that the ratio of Pi' to Pm is

Partial pressure (Dalton model) The partial pressure of component i is defined as the product of the mole fraction and the mixture pressure according to Dalton’s law. For the component i

Pi ' Vi N = = i = yi Pm Vm N m For ideal-gas mixtures, the partial pressure and the component pressure are the same and are equal to the product of the mole fraction and the mixture pressure.

Pi = yi Pm k

Dalton’s law:

Pm = ∑ Pi ( Tm ,Vm ) i =1

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8

Other properties of ideal-gas mixtures The extensive properties of a gas mixture, in general, can be determined by summing the contributions of each component of the mixture. The evalu-ation of intensive properties of a gas mixture, however, involves averaging in terms of mass or mole fractions: k k k U m = ∑ U i = ∑ mi ui = ∑ N i ui i =1

i =1

i =1

k

k

k

i =1

i =1

i =1

k

k

k

i =1

i =1

i =1

and

k

um = ∑ mf i ui i =1

Cv ,m

=

C p ,m Cv ,m

(kJ / K)

For constant specific heats, the entropy change of any component is

(kJ / kg or kJ / kmol)

k

and

i =1

hm = ∑ yi hi

(kJ / kg or kJ / kmol)

i =1

k

sm = ∑ mf i si

C p ,m

i =1

k

hm = ∑ mf i hi

um = ∑ yi ui

km =

The entropy of a mixture of ideal gases is equal to the sum of the entropies of the component gases as they exist in the mixture. We employ the Gibbs-Dalton law that says each gas behaves as if it alone occupies the volume of the system at the mixture temperature. That is, the pressure of each component is the partial pressure.

(kJ)

k

and

Ratio of specific heats k is given as

(kJ)

Hm = ∑ Hi = ∑ mi hi = ∑ N i hi Sm = ∑ Si = ∑ mi si = ∑ N i si

These relations are applicable to both ideal- and real-gas mixtures. The properties or property changes of individual components can be determined by using ideal-gas or real-gas relations developed in earlier chapters.

k

and

i =1

sm = ∑ yi si

(kJ / kg ⋅ K or kJ / kmol ⋅ K)

i =1

k

Cv , m = ∑ mf i Cv , i

k

and

i =1

i =1

k

C p , m = ∑ mf i C p , i i =1

Cv , m = ∑ yi Cv , i k

and

C p , m = ∑ yi C p , i

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10

i =1

The entropy change of the mixture per mass of mixture is

In these last two equations, recall that

Pi , 1 = yi , 1 Pm, 1 Pi , 2 = yi , 2 Pm, 2 Example 13-1 An ideal-gas mixture has the following volumetric analysis The entropy change of the mixture per mole of mixture is Component N2 CO2

% by Volume 60 40

(a)Find the analysis on a mass basis. For ideal-gas mixtures, the percent by volume is the volume fraction. Recall

yi = vf i

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12

Comp.

yi

N2 CO2

0.60 0.40

Mi yiMi mfi = yiMi /Mm kg/kmol kg/kmol kgi/kgm 28 16.8 0.488 44 17.6 0.512 Mm = ΣyiMi = 34.4

(c) Find the specific heats at 300 K. Using Table A-2, Cp N2 = 1.039 kJ/kg⋅K and Cp CO2 = 0.846 kJ/kg⋅K 2

C p , m = ∑ mf i C p ,i = (0.488)(1039 . ) + (0.512)(0.846) 1

(b) What is the mass of 1 m3 of this gas when P = 1.5 MPa and T = 30oC? R Rm = u Mm =

mm =

= 0.940

( kJ / kg ⋅ K )

kJ kgm ⋅ K

Cv , m = C p , m − Rm = (0.940 − 0.242)

kJ kmol ⋅ K = 0.242 kJ kg kg ⋅ K 34.4 kmol

8.314

= 0.698

PmVm RmTm

kJ kgm ⋅ K

kJ kgm ⋅ K

. MPa (1m3 ) 15 103 kJ (0.242 kJ / ( kg ⋅ K ))(30 + 273) K m3 MPa = 20.45 kg =

13

(d) This gas is heated in a steady-flow process such that the temperature is increased by 120oC. Find the required heat transfer. The conservation of mass and energy for steady-flow are

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(e) This mixture undergoes an isentropic process from 0.1 MPa, 30oC, to 0.2 MPa. Find T2. The ratio of specific heats for the mixture is k=

Cp,m Cv , m

=

0.940 = 1347 . 0.698

Assuming constant properties for the isentropic process

m& 1 = m& 2 = m& m& 1h1 + Q& in = m& 2 h2 Q& in = m& (h2 − h1 ) & p , m (T2 − T1 ) = mC The heat transfer per unit mass flow is

Q& in = C p , m (T2 − T1 ) m& kJ = 0.940 (120 K ) kgm ⋅ K

(f) Find ΔSm per kg of mixture when the mixture is compressed isothermally from 0.1 MPa to 0.2 MPa.

qin =

= 112.8

kJ kgm

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2

Δsm = ∑ mf i Δsi

But, the compression process is isothermal, T2 = T1. The partial pressures are given by

i =1

Pi = yi Pm

= (0.488

kg N 2

. = −0167

kJ kgm ⋅ K

The entropy change becomes

For this problem the components are already mixed before the compression process. So,

yi , 2 = yi ,1

Then,

kgm

)( −0.206

kgCO2 kJ kJ ) + (0.512 )( −0131 . ) kg N 2 ⋅ K kgm kgCO2 ⋅ K

Why is Δsm negative for this problem? Find the entropy change using the average specific heats of the mixture. Is your result the same as that above? Should it be? (g) Both the N2 and CO2 are supplied in separate lines at 0.2 MPa and 300 K to a mixing chamber and are mixed adiabatically. The resulting mixture has the composition as given in part (a). Determine the entropy change due to the mixing process per unit mass of mixture.

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Take the time to apply the steady-flow conservation of energy and mass to show that the temperature of the mixture at state 3 is 300 K.

18

But here the components are not mixed initially. So,

y N 2 ,1 = 1 yCO2 , 2 = 1 and in the mixture state 3,

y N 2 , 3 = 0.6 yCO2 , 3 = 0.4

But the mixing process is isothermal, T3 = T2 = T1. The partial pressures are given by

Pi = yi Pm

Then,

The entropy change becomes

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Then, 2

Δsm = ∑ mf i Δsi i =1

= (0.488 = 0163 .

kg N 2 kgm

)(0152 .

kgCO2 kJ kJ ) + (0.512 )(0173 . ) kg N 2 ⋅ K kgm kgCO2 ⋅ K

kJ kgm ⋅ K

If the process is adiabatic, why did the entropy increase? Extra Assignment Nitrogen and carbon dioxide are to be mixed and allowed to flow through a convergent nozzle. The exit velocity to the nozzle is to be the speed of sound for the mixture and have a value of 500 m/s when the nozzle exit temperature of the mixture is 500oC. Determine the required mole fractions of the nitrogen and carbon dioxide to produce this mixture. From Chapter 17, the speed of sound is given by

C = kRT

Answer: yN2 = 0.589, yCO2 = 0.411

NOZZLE Mixture N2 and CO2

z z z z z z z z z z z

Q9-18 Q9-19 Q9-23 Q9-32 Q9-35 Q9-47 Q9-57 Q9-76 Q9-80 Q9-92 Q9-94

(b) 0.422kJ (c) 20.4% (b) 0.39kJ (c) 20.9% (a) 46.1 kPa (b) 87.5 kJ/kg (c) 0.533 (d) 0.578 Given in text book Given in text book (a) 1.848 (b) 158.87 Btu/lbm (c) 59.1% Given in text book (a) 52.7% (b) 27.0% Given in text book Given in text book Given in text book

C = 500 m/s T = 500oC

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