Lecture 31

Department of Mechanical EngineeringME 322 – Mechanical Engineering

Thermodynamics

Lecture 31

Ideal Gas Mixtures

Mixtures in Engineering Applications

• Natural gas– Methane, ethane, propane, butane, nitrogen,

hydrogen, carbon dioxide, and others• Refrigerants

– Zeotropes - True mixture behavior• Example: R407c - R32/125/134a (23/25/52 by mass)

– Azeotropes - Mixtures that behave as a pure fluid

• Example: R507A - R125/143a (50/50 by mass)• Air and water vapor

– Psychrometric analysis• Air conditioning applications

2

Pure Fluid vs. Mixture Behavior

60 80 100 120 140 160 180 200 220101

102

103

h [Btu/lbm]

P [p

sia]

200°F

150°F

100°F

50°F

0°F

-50°F

R22

R22: a pure fluid; a halogenated methane molecule (chlorodiflouromethane)

R407C: a mixture of R32, R125, and R134a

3

-20 0 20 40 60 80 100 120 140 1608x100

101

102

103

h [Btu/lbm]P

[psi

a]

200°F

150°F

100°F

50°F

0°F

-50°F

R407C

Thermodynamic Properties of Mixtures• Real mixture behavior

– Real mixture model• Very complex to describe analytically

– Topic for an advanced course• EES can calculate real-properties of common mixtures!

• Low-pressure, moderate density– Ideal solution model

• Gases are treated as real fluids with idealized mixing– Topic for an advanced course

• Low-pressure, low density– Ideal gas mixing model

• Gases are treated as ideal gases with idealized mixing– ME 322!!

4

Ideal Gas Mixture Models

• Air Conditioning– Water vapor + air mixtures

• Conditions are suitable for ideal gas property estimation – even for water vapor!

• Combustion Analysis– Products of combustion are often at high

temperatures and low pressure

Even though the ideal gas mixing model is simplified, it turns out to be fairly accurate for two important processes that mechanical engineers deal with ...

The rest of ME 322 deals with these two processes

5

An Example – Gas Turbines

6

Air into the combustion

chamberProducts of combustion leaving

the combustion chamber

Combustion products can contain CO2, H2O, O2, N2, CO, NO2, and others!

In order to get a better estimate of the performance of the gas turbine, we need to be able to determine the properties of the mixture passing through the turbine

Properties of Ideal Gas MixturesConsider any property, B (extensive) or b (intensive). For a mixture,

1 1

N N

m k k kk k

B B m b

Mass Basis*

mass fraction

1 1

N N

m k k kk k

B B n b

Molar Basis

mole fraction**

**Note: The text uses ci for mole fractions

*Other common words: weight basis or gravimetric basis

7

1 1

N Nkm

m k k kk km m

mBb b w bm m

kk

m

mw

m

1 1

N Nkm

m k k kk km m

nBb b y b

n n

kk

m

ny

n

8

Mass/Mole Fraction Conversion

ii

i

mMn

In some instances, a conversion between mass fraction and mole fraction is needed. The mass of a substance is related to the number of moles through the molecular mass, Mi ,

Considering the mass fraction,

ii

m

mwm

i i

k kk

n Mn M

ii

m

kk

k m

n Mnn Mn

i i

k kk

y My M

Mass/Mole Fraction Conversion

9

ii

i

mM

n

A similar analysis for the mole fraction reveals,

ii

m

ny

n

Summary of findings ...

i ii

k kk

y Mw

y M

//

i ii

k kk

w Myw M

i

i

k

k k

mMmM

1

1

i

m i

k

k m k

mm Mmm M

/

/i i

k kk

w Mw M

Example

10

Given: A mixture of ideal gases has the following molar composition; Argon (yAr = 0.20), helium (yHe = 0.54), and the balance is carbon monoxide.

Find: (a) mole fraction of carbon monoxide (b) the molecular mass of the mixture

(c) the gravimetric (mass) composition of the mixtureNote: The molecular mass of the mixture can be found by,

mm

m

mM

n

1k

km

mn

1k k

km

n Mn

k kk

y M

Example

11

The mole fractions of the argon and helium are given. Therefore, the mole fraction of carbon monoxide can be found,

m kk

n n

Ar He CO 1kk

y y y y

Now, the molecular mass of the mixture can be found,

Ar Ar He He CO CO

Table C.13a Table C.13a Table C.13a

lbm lbm lbm lbm0.20 39.94 0.54 4.003 0.26 28.01 17.43lbmol lbmol lbmol lbmol

m k kk

m

M y M y M y M y M

M

1k

k m

nn

1kk

y

CO Ar He1 1 0.20 0.54 0.26y y y

Example

12

The mass fraction composition of the mixture can be found by,

i ii

k kk

y Mw

y M

Therefore,

Table C.13a

CO

lbm0.26 28.01lbmol

0.418lbm17.43lbmol

w

i i

m

y MM

Table C.13a

Ar

lbm0.20 39.94lbmol

0.458lbm17.43

lbmol

w

Table C.13a

He

lbm0.54 4.003lbmol

0.124lbm17.43

lbmol

w

Example

13

Comparison of mole fractions and mass fractions for this mixture ...

Component y wAr 0.20 0.458He 0.54 0.124CO 0.26 0.418

1.00 1.00

It is always a good idea to check if the calculated fractions sum up to one!

Ideal Gas Mixture Properties

14

orm k k m k kk k

b w b b y b

We have previously seen that,

Consider the internal energy and enthalpy of an ideal gas mixture. The components of the mixture exist at the same temperature as the mixture. Therefore, according to the expressions above,

orm m k k k m k k k m kk k

u T w u T u y u T T T

orm m k k k m k k k m kk k

h T w h T h y h T T T

Another Example

15

Given: A mixture of ideal gases is contained in a closed, rigid container that has a volume of 2 ft3. The mixture is an equimolar binary mixture of methane and ethane. The mixture is initially at 15 psia, 20°F. Heat is now transferred to the mixture pressure and temperature become 60 psia, 300°F.

Find: The amount of heat transferred in this process.

12Q

1 1

2 2

20 F 15 psia300 F 60 psia

m m

m m

T PT P

32 ftmV