EGR 334 Thermodynamics Chapter 12: Sections 1-4 Lecture 38: Ideal Gas Mixtures Quiz Today?

EGR 334 ThermodynamicsChapter 12: Sections 1-4

Lecture 38: Ideal Gas Mixtures

Quiz Today?

Today’s main concepts:• Be able to describe ideal gas mixture composition in

terms of mass fractions and mole fractions.• explain use of the Dalton model to relate pressure,

volume, and temperature and to calculate changes in U, H, and S for ideal gas mixtures.

• Be able to apply mass, energy, and entropy balances to systems involving ideal gas mixtures, including mixing processes.

Reading Assignment:

Homework Assignment:

Read Chapter 12, Sections 5-8

Problems from Chap 12: 4, 10, 22, 28

3

Thus far we have been working with pure fluids• Water• Air• R22

Sec 12.1 : Describing Mixture Composition

i ofWeight Molecular

i of mass

i

ii M

mn

Number of moles

For gas mixtures, in addition to the normal to parameters (T, p), we also need to know the mixture composition.

mass total

i of mass

T

ii m

mmfMass fraction where

iimf1

moles total

i of moles

T

ii n

nyMole fraction where

iiy1

• Nitrogen• Oxygen• Ammonia

Avg. Molecular Weight total mass

total moles

i iiT

i iiT T

n Mm

M y Mn n

4Sec 12.2 : Relating P, V, and T for Ideal Gas Mixtures

Dalton Model: Assumes that each component behaves as an ideal gas at the specified T and V of the mixture. This assumes that the gases do not interact with each other.

Ideal gas This refers to the overall mixture, on average. How do we describe the relationship between p, V, and T for each component?

pV nRT

V

TRnp ii Partial Pressure

and i ip y pii

T

ii yn

n

VTRn

VTRn

p

p thus

Amagat Model: Assumes that each component behaves as an ideal gas at the specified T and p of the mixture.

ii

n RTV

pPartial Volume VyV ii and

5

Example (12.6): Natural gas at 23°C, 1 bar enters a furnace with the following molar analysis: 40% propane (C3H8), 40% ethane (C2H6), and 20% methane (CH4).

Determine

i C3H8 C2H6 CH4 Total

T (°C) 23

p (bar) 1

yi 0.40 0.40 0.20 1.0

Mi

mi

mfi

mi (kg/s)

(a) The analysis in terms of mass fractions(b) The partial pressure of each component, in bar(c) The mass flow rate, in kg/s, for a volumetric flow rate of 20 m3/s.

yi is the mole fraction of each component


T (°C) 23

p (bar) 1

yi

Mi

mi

mfi

mi (kg/s)

6

(a) The analysis in terms of mass fractions(b) The partial pressure of each component, in bar(c) The mass flow rate, in kg/s, for a volumetric flow rate of 20 m3/s.

iii Mym

636.1709.444.0 Pm

028.1209.304.0 Em

208.304.162.0 Mm


T (°C) 23

p (bar) 1

yi 0.40 0.40 0.20 1.0

Mi 44.09 30.07 16.04

mi 17.636 12.028

3.208 32.872

mfi

mi(kg/s) 32.872total i

i

m m Units are effectively kg/kmol, thus mt = M

Example (12.6): 40% propane (C3H8), 40% ethane (C2H6), and 20% methane (CH4).

Molecular weights of each component are found in Table A-1:

The mass of each component is found

On a 1 kmol basis then


T (°C) 23

p (bar) 1

yi 0.40 0.40 0.20 1.0

Mi 44.09 30.07 16.04

mi 17.636 12.028

3.208

mfi

mi(kg/s)


T (°C) 23

p (bar) 1

yi 0.40 0.40 0.20 1.0

Mi 44.09 30.07 16.04

mi

mfi

mi(kg/s)

7

Example (12.6):

(b) The partial pressure of each component, in bar(c) The mass flow rate, in kg/s, for a volumetric flow rate of 20 m3/s.

i ip y p

barbarpP 4.014.0 i C3H8 C2H6 CH4 Total

T (°C) 23

p (bar) 1

yi 0.40 0.40 0.20 1.0

Mi 44.09 30.07 16.04

mi 17.636

12.028

3.208 32.872

mfi 0.5365

0.3659

0.0976

1.0

mi(kg/s)

barbarpE 4.014.0

barbarpM 2.012.0

To find Partial Pressure:

8

Example (12.6):

(c) The mass flow rate, in kg/s, for a volumetric flow rate of 20 m3/s.


T (°C) 23

p (bar) 0.40 0.40 0.20 1.0

yi 0.40 0.40 0.20 1.0

Mi 44.09 30.07 16.04

mi 17.636

12.028

3.208 32.872

mfi 0.5365

0.3659

0.0976

1.0

mi(kg/s)

First convert volumetric flow rate to mass flow rate:

V V VA A P A pm

v RT R M T

3 2 2(20 / )(10 / )

8.314 /296

32.872 /

m s kN m kJm

kN mkJ kmol KK

kg kmol

26.71kg

s

9

Example (12.6):


T (°C) 23

P (bar) 0.40 0.40 0.20 1.0

yi 0.40 0.40 0.20 1.0

Mi 44.09 30.07 16.04

mi 17.636

12.028

3.208 32.872

mfi 0.5365

0.3659

0.0976

1.0

mi(kg/s)

14.33 9.77 2.61 26.71

Can also find the mass flow rate of each component

i im mf m

0.5365 26.71 /Pm kg s

0.3659 26.71 /Em kg s

0.0976 26.71 /Mm kg s

(c) The mass flow rate, in kg/s, for a volumetric flow rate of 20 m3/s.

2.61kg

s

9.77kg

s

14.33kg

si C3H8 C2H6 CH4 Total

T (°C) 23

P (bar) 0.40 0.40 0.20 1.0

yi 0.40 0.40 0.20 1.0

Mi 44.09 30.07 16.04

mi 17.636

12.028

3.208 32.872

mfi 0.5365

0.3659

0.0976

1.0

mi(kg/s)

26.71

10Sec 12.3 : Evaluating U, H, S and specific heats

The properties of U, H, S, and c of a mixture are additive.

i

iT mmmmm ......321Mass of system:

Internal energy i

iT UUUUU ......321

i

iiT umumumummu ......332211

i

iiT ununununun ......332211

i

iiT uyuyuyuyu ......332211

Where a similar set of equations can be written for H and S

For specific heat i

iPiPPPP cycycycyc ,3,32,21,1 ......

i

iViVVVV cycycycyc ,3,32,21,1 ......

11Sec 12.4 : Analyzing Systems Involving Mixtures

1212 @@ TuTunUU iii

i

12 @@ TuTuyu iii

i

For mixtures with constant composition (no chemical reaction):

12 TTcu v For constant specific heats,

12, TTcu ivi

Equations for enthalpy (H) are similar to those for internal energy (U), but uses cp.

with

12Sec 12.4 : Analyzing Systems Involving Mixtures

2 1 2 2 1 1@ , @ ,i i ii

S S n s T p s T p 2 ,2 1 ,1@ , @ ,i i i i i

i

s y s T p s T p ,2 2

2 1 2 1,1 1

@ @ ln @ @ lni ii i i i

i i

p y ps s T s T R s T s T R

p y p

For constant specific heats,

2 2

1 1

2 2,

1 1

ln ln

ln ln

P

i P i

T ps c R

T p

T pwith s c R

T p

2 2

1 1

2 2,

1 1

ln ln

ln ln

V

i V i

T Vs c R

T V

T Vwith s c R

T V

For entropy, is also dependent upon pressure changes.

or

13

Example (12.17): A mixture of 2 kg of H2 and 4 kg of N2 is compressed in a piston-cylinder assembly in a polytropic process for which n = 1.2. The temperature increases from 22 to 150°C. Using constant values for the specific heats, determine(a) The heat transfer, in kJ.(b) The entropy change, in kJ/K.

Principles to be applied:

Closed Ideal Gas system:

1st Law of Thermodynamics:

2nd Law of thermodynamics:

2 1 12 12U U U Q W

pV nRT

2

2 1

1

QS S S

T

2 2 112 1 1

mR T TW pdV

n

where for polytropic process

i ipV n RTand

14


Find the moles of each component and the Mixtures Molecular Weight:

2

2

2

2

2.018 /H

HH

m kgn

M kg kmol 2

2

2

4

28.01 /N

NN

m kgn

M kg kmol

2

2

0.991

(0.991 0.143)H

Htotal

n kmoly

n kmol

2

2

0.143

(0.991 0.143)N

Ntotal

n kmoly

n kmol

2 2 2 2

0.874 2.018 0.126 28.01 5.29H H N NM y M y M

0.991kmol

0.874

0.143kmol

0.126

Molecular Weight:

15


To Evaluate work for a polytropic process.

n

TTMRm

n

TTmRW

111212

12

8.314 /2 4 150 22

5.29 /

1 1.2

kJ kmol Kkg K

kg kmol

6035.1kJ

16

Example (12.17):

(a) The heat transfer, in kJ.(b) The entropy change, in kJ/K.

Evaluate change in internal energy

2 2 2 2, 2 1 , 2 1H v H N v NU m c T T m c T T

2 10.311 / 150 22

4 0.745 / 150 22

kg kJ kg K K

kg kJ kg K K

Assuming constant heat capacity from Table A-20 at average temperature (359 K)

3021.1kJ

Then using the energy balance to evaluate Q

12 12 12Q U W

3021.1 6035.1 3014kJ

17

Example (12.17):

(b) The entropy change, in kJ/K. 1 1

2 2 2 2

1 1 1 1

ln ln ln lnn

V V

T V T Ts c R c R

T V T T

2 2 2

1 1 1

ln ln ln1 1V V

T T TR Rc c

T n T n T

Using M = 5.29 kg/kmol,The mixture’s cv (avg. heat capacity) needs to be found.

2 2 2 2, , ,V i V i H V H N V Ni

c mf c mf c mf c 2 410.311 / 0.745 / 3.933 /

6 6

kg kgkJ kg K kJ kg K kJ kg K

kg kg

V

V

cc

M

(3.933 / )(5.29 / )V Vc c M kJ kg K kg kmol 20.8 /kJ kmol K

so

18

Example (12.17):

(b) The entropy change, in kJ/K.Therefore:

2

1

ln1V

TRs c

n T

8.314 / 42320.8 / ln

(1.2 1) 395

kJ kmol K KkJ kmol K

K

1.422 /kJ kmol K 1.422 /kJ kmol K

1.422 /0.269 /

5.29 /

s kJ kmol Ks kJ kg K

M kg kmol

(6 )( 0.269 / ) 1.623 /S m s kg kJ kg K kJ K

19Sec 12.4.2 : Mixing of Ideal gases

The previous example considered a mixture that had already been formed.

How is a process different when a mixture is formed from two individual gases which might originally be a different temperatures and pressures?

Whenever two highly ordered substances are mixed, entropy is expected to increase. This is because

-- Gases are initially at different temperature.-- Gases are initially at different pressures-- Gases are distinguishable from one another.

irrev.process


Example 3: Consider a canister that is initially divided into two sections. One side contains 2 lbmol of Nitrogen (N2) at 500°R and 2 atm

One side contains 3 lbmol of Oxygen (O2) at 300°R and 1 atm.

Determine final temperature, pressure, and entropy production when mixed.

0U Q W Start with Energy Balance: Assume no heat lost from system No change of total volume no work

2 2 2 2 2 21 ( ) ( )N N N O O OU n u T n u T

2 2 2 22 ( ) ( )N N f O O fU n u T n u T

where

2 2 2 2 2 2 2 2 2 2

0 ( ) ( ) ( ) ( )N N f O O f N N N O O On u T n u T n u T n u T so

2 2 2 2 2 2 2 20 ( ) ( ) ( ) ( )N N f N N O O f O On u T u T n u T u T

or

2 2 2 2 2 2_ _0 ( ) ( )N v N f N O v O f On c T T n c T T

Assuming constant specific heat, 2 2_ _ and v N v Oc c


Example 3: Consider a canister that is initially divided into two equal sized sections. One side contains 2 lbmol of Nitrogen (N2) at 500°R and 2 atm



Solving for final temperature Tf

A good choice to use for the temperature to find the cv is the average temperature of the initial gases (Tave = 400 R). From Table A-20E.

so

and

2 2 2 2 2 2

2 2 2 2

_ _

_ _

N v N N O v O Of

N v N O v O

n c T n c TT

n c n c

2_ (@ 400 ) 0.180 Btu /ov N mc R lb R

2_ (@ 400 ) 0.168 Btu /ov O mc R lb R

228.01 /N m molM lb lb

232.0 /O m molM lb lb

2 2 2_ _ (0.180 Btu / )(28.01 / ) 5.042Btu /v N v N N m m mol molc c M lb R lb lb lb R

2 2 2_ _ (0.168 Btu / )(32.0 / ) 5.376Btu /v O v O O m m mol molc c M lb R lb lb lb R





therefore:

2 2 2 2 2 2

2 2 2 2

_ _

_ _

N v N N O v O Of

N v N O v O

n c T n c TT

n c n c

(2 )(5.042 / )(500 ) (3 )(5.376 / )(300 )

(2 )(5.042 / ) (3 )(5.376 / )mol mol mol mol

mol mol mol mol

lb Btu lb R R lb Btu lb R R

lb Btu lb R lb Btu lb R

377 oR





2 2N OV V V

2 2 2 2N N N Np V n RT

To find the final pressure first find the volume of the total original gases.

where

2 2 2 2O O O Op V n RT

2 2

2

2

N NN

N

n RTV

p 2 2

2

2

O OO

O

n RTV

p

2

2 2

(2 )(1.986 / )(500 ) 1 778

29.4 / 144mol mol

f

lb Btu lb R R ft lbf ft

lb in in Btu

3365 ft 3657 ft

2

2 2

(3 )(1.986 / )(300 ) 1 778

14.7 / 144mol mol

f

lb Btu lb R R ft lbf ft

lb in in Btu

2 2

3365 657 1022N OV V V ft total volume is then


Example 3: Consider a canister that is initially divided into two equal sized sections. One side contains 2 lbmol of Nitrogen (N2) at 500°R and 2 atm



2 2 2 2N N O OM y M y M

2

3 2

778(5 )(1.986 / )(377 ) 1

1022 144t f fmol mol

f

n RT lb ftlb Btu lb R R ftp

V ft Btu in

The mixed gases have a combined Molecular weight given by

using Ideal Gas equation:

2 3(28.01 / ) (32.00 / ) 30.40 /

5 5mol mol

m mol m mol m molmol mol

lb lblb lb lb lb lb lb

lb lb

19.79 1.346psi atm





To find the change in entropy:

where

1

1

f

f

QS S S

T

0

2 2 2 2 2 2 2 21 ( , ) ( , )N N N N O O O NS n s T p n s T p

2 2 2 2 2 2 2 2 2 2 2 2( , ) ( , ) ( , ) ( , )N N f N f N N N O O f O f O O Nn s T y p s T p n s T y p s T p

therefore2 2 2 2 2 2

( , ) ( , )f N N f N f O O f O fS n s T y p n s T y p

then using the form based on ideal gas behavior with constant specific heat

2 2

2 2 2 2

2 2 2 2

_ _ln ln ln lnN f O ff fN p N O p O

N N O O

y p y pT Tn c R n c R

T p T p





To find the change in entropy:

2 2

2 2 2 2

2 2 2 2

_ _ln ln ln lnN f O ff fN p N O p O

N N O O

y p y pT Tn c R n c R

T p T p

377 (0.4)(1.346 )2 (5.042 / ) ln (1.986 / ) ln

500 2

377 (0.6)(1.346 )3 (4.376 / ) ln (1.986 / ) ln

300 1

mol mol mol

mol mol mol

atmlb Btu lb R Btu lb R

atm

K atmlb Btu lb R Btu lb R

K atm

2.365 4.272 6.637 /Btu R

27

end of slides Lecture 38

EGR 334 Thermodynamics Chapter 12: Sections 1-4 Lecture 38: Ideal Gas Mixtures Quiz Today?

Documents

Transcript of EGR 334 Thermodynamics Chapter 12: Sections 1-4 Lecture 38: Ideal Gas Mixtures Quiz Today?