me201 properties of mixture

7/24/2019 ME201 Properties of Mixture

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Class Lecture Delivered by

Dr. A. K. M. Monjur Morshed

Assistant Professor

Department of Mechanical Engineering

Bangladesh University of Engineering and Technology

Topics: Properties of Mixture


2/16

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


3/16

Pure Fluid vs. Mixture Behavior

60 80 100 120 140 160 180 200 220

101

102

103

h [Btu/lbm]

P[

psia]

200F

150F

100F

50F

0F

-50F

R22

R22: a pure fluid; a

halogenated methane molecule

(chlorodiflouromethane)

R407C: a mixture of R32, R125,

and R134a

3

-20 0 20 40 60 80 100 120 140 160

8x100

101

102

103

h [Btu/lbm]

P

[psia] 200F

150F

100F

50F

0F

-50F

R407C


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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


5/16

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 ...

5


6/16

An Example Gas Turbines

6

Air into the

combustion

chamber

Products of combustion leavingthe combustion chamber

Combustion products can

contain CO2

, H2

O, 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 theproperties of the mixture

passing through the turbine


7/16

Properties of Ideal Gas Mixtures

Consider any property, B (extensive) or b (intensive). For amixture,

1 1

N N

m k k k

k k

B B m b

Mass Basis*

mass fraction

1 1

N N

m k k k k k

B B n b

Molar Basis

mole fraction**

**Note: The text uses cifor

mole fractions*Other common words: weight

basis or gravimetric basis

7

1 1

N Nkm

m k k k

k km m

mBb b w b

m m

kk

m

mwm

1 1

N Nkm

m k k k

k km m

nBb b y b

n n

kk

m

nyn


8/16

Mass/Mole Fraction Conversion

8

ii

i

mM

n

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 k

k

n Mn M

ii

m

kk

k m

nM

nn

Mn

i i

k k

k

y My M


9/16

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 k

k

y Mwy M

//

i ii

k k

k

w Myw M

i

i

k

k k

m

M

m

M

1

1

i

m i

k

k m k

m

m M

m

m M

/

/

i i

k k

k

w M

w M


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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 mixture

Note: The molecular mass of the mixture can be found by,

mm

m

mM

n 1

k

km

mn

1k k

km

n Mn k kk y M


11/16

Example

11

The mole fractions of the argon and helium are given.

Therefore, the mole fraction of carbon monoxide can be found,

m k

k

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 17.43

lbmol lbmol lbmol lbmol

m k k

k

m

M y M y M y M y M

M

1k

k m

n

n 1k

k

y

CO Ar He1 1 0.20 0.54 0.26y y y


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Example

12

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

i ii

k k

k

y Mw

y M

Therefore,

Table C.13a

CO

lbm0.26 28.01

lbmol

0.418lbm

17.43lbmol

w

i i

m

y M

M

Table C.13a

Ar

lbm0.20 39.94

lbmol

0.458lbm

17.43lbmol

w

Table C.13a

He

lbm0.54 4.003

lbmol

0.124lbm

17.43lbmol

w


13/16

Example

13

Comparison of mole fractions and mass fractions for this

mixture ...

Component y w

Ar 0.20 0.458

He 0.54 0.124

CO 0.26 0.418

1.00 1.00

It is always a good idea to check if the

calculated fractions sum up to one!


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Ideal Gas Mixture Properties

14

orm k k m k k k 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 k k ku T w u T u y u T T T

orm m k k k m k k k m k

k k

h T w h T h y h T T T


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Another Example

15

Given: A mixture of ideal gases iscontained 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, 20F. Heat is nowtransferred to the mixture pressure and

temperature become 60 psia, 300F.

Find: The amount of heat transferred inthis process.

12Q

1 1

2 2

20 F 15 psia

300 F 60 psiam m

m m

T P

T P

32 ftmV


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me201 properties of mixture

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