Chapter 3 Thermodynamics Properties of Fliuds (Part 2)

8/8/2019 Chapter 3 Thermodynamics Properties of Fliuds (Part 2)

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Chapter 3

Thermodynamics Properties

of Fluids

Chemical EngineeringThermodynamics


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3.4 Two Phases Systems

3.5 Tables of ThermodynamicsProperties

Chapter Outline


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FE

GG !E

G FG

6.4 Two Phases Systems

Generally, for two phases and of a pure

species coexisting at equilibrium,

where and are the molar or specific

Gibbs energies of individual phases .The Clapeyron equation follow from this

equality.


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If the temperature of a two-phase system

is changed, then the pressure must also

change accord with the relation between

vapor pressure and temperature if the two

phases continue to coexist in equilibrium .

FE dGdG !

Substituting expressions for and :EdGFdG

dTSdPVdTSdPV

satsat FFEE

!

EF

EF

EF

EF

V

S

VV

SS

dT

dPsat

(

(!

!

Which rearrangement becomes:


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The entropy change S and the volume

change V are the changes which occurwhen a unit amount of a pure chemical

species is transferred from phase to

phase at the equilibrium temperatureand pressure.

EFEF SH (!(

Integration of equation for this change yields

the latent heat of phase transition:


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which is the Clapeyron equation.

lv

lvsat H

dT

dP

(

(!

Thus, substitute in the previous

equation gives:

EF

EF

VT

H

dT

dP sat

(

(!

T

HS

EFEF (!(

For the particularly important case of

phase transition from liquid l to vaporv, it

is written:

or lv

lvsat

ZRT

H

Td

Pd

(

(!

1ln


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A plot ofln Psatvs 1/Tgenerally yields a

line that is nearly straight :

The Antoine equation, which is more

satisfactory for general use, has the form:

CT

BAPsat

!ln

T

BAPsat !ln

The constants A, B and C are readily available

for a large number of species (Table B.2).

Temperature Dependence of the

Vapor Pressure of Liquids

A and B are constants for given species.

(Example 6.6)


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The Wagner equation is one the best

available; it expresses the reduced vapor

pressure as a function of reduced

temperature:

X

XXXX

!

1ln

635.1DCBA

Psatr

where 1- Trand A, B, Cand D areconstant.


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All the relations are represented by the

generic equation :

where is called the quality andrepresented molar fraction of vapor in the

system. Similarly,

M can represents V,U , H , Sand others .

vvlv MxMxM ! 1

lvvl

x (

!

v

x

Two-phases Liquid/Vapor Systems


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Example 6.7

Superheated steam originally at P1

and T1

expands through a nozzle to an exhaust

pressure P2. Assuming the process is reversible

and adiabatic, determine the down stream state

of the steam and Hfor the following conditions:

(a)P1=1,000 kPa, T

1= 250C and P2 = 200 kPa.

(b)P1 =150(psia), T1= 500(F), and P2 = 50(psia).

6.5 Tables of Thermodynamics

Properties


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Solution 6.7

a) Use superheated table (Table F.2) for the

initial temperature of250C at 1,000 kPa to

get the initial value ofHand S.

No entries appear in the SI tables forsuperheated steam. Use interpolation:

12

1int

12

1int

xxxx

yyyy erpolationerpolation

!


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Interpolation between values for 240C

and 260C yields, at1

000 kPa:H1 = 2,942.9 kJ/kg S1= 6.9252 kJ/kg.K

Since the process is both reversible and

adiabatic, the change in entropy of the steamis zero, S=0.

Hence, for the final state at P2=200 kPa,

S2 =S1 = 6.9252 kJ/kg.K


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From the superheated table, the entropy

(S) of saturated vapor at 200 kPa isgreater than S2. The final state is in the

two-phase liquid/vapor region. Thus T2 is

the saturation temperature at 200 kPa.

Given in the superheat tables as T2 =

120.23C.


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Equation

applied to the entropy becomes:

vx2vx2

vvlv SxSxS 22222 )1( !

Hence, 6.9252 = 1.5301(1 - ) + 7.1268

where the values 1.5301 and 7.1268 are

entropies of saturated liquid and saturated

vaporat 200 kPa. Solving,

= 0.9640vx 2

vvlv xx ! 1


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On a mass basis, the mixture is 96.40%

vaporand 3.60% liquid. Its enthalpy isobtained by further application of Eq.

(6.82a):

H2= (0.0360)(504.7) + (0.9640)(2,706.3)

= 2,627.0 kJ/kg

Finally,

H=H2 - H1= 2,627.0 - 2,942.9 = - 315.9 kJ/kg


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(b) For the initial state at P1=150(psia)

and T1=500(F) data from the

superheated tables in English unitsprovide:

H1=1,274.3 (Btu)(Ibm)

-1

S1=

l.6602(Btu)(lbm)-1

(R)-1

At the final state at 50(psia),

S2 =S1 =1.6602(Btu)(Ibm)-1(R)-1

Similarly, the process is both reversible and

adiabatic, the change in entropy of the steam

is zero, S=0.


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AndH =H2- H1

=1,175.3 - 1,274.3 = - 99.0(Btu)(lbm)

Inspection of the superheated table

shows thatS

2 is here greater than theentropy of saturated vapor at 50 (psia).

Hence the final state is in the superheat

region. Interpolation on entropy at

50(psia) yields:T2= 283.28(F)

H2 =1,175.3 (Btu)(lbm)-1


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The End

Chapter 3 Thermodynamics Properties of Fliuds (Part 2)

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