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Chapter 6. Thermodynamic Properties

of Fluids

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Introduction

Numerical values of thermodynamic properties are most important for heat and work calculations

Enthalpy, Internal Energy, Entropy, …..

Measurable Quantities : PVT data and heat capacity

Properties relations are important to calculate one properties from the other.

Generalized correlations are used when experimental values are not available.

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6.1 Properties Relations for Homogeneous Phases

From the 1

st

law and 2

nd

law of Thermodynamics,

rev

rev dW

dQ )

nU (

d   dWrev  Pd(nV) ) nS ( Td dQrev

) nV ( Pd )

nS ( Td )

nU (

d  

Although this eqn. has been derived for the special case of reversible process, it only contain state properties of the system.

Can be applied to any process

from one equilibrium state to another

The only requirements are that the system be closed and that the change occur between equilibrium states.

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6.1 Properties Relations for Homogeneous Phases

PdV TdS

dU  

When the equations are written in molar units;

PV U

H   TS U

A   TS H

G  

VdP PdV

PdV TdS

) PV ( d dU

dH       dH  TdSVdP

SdT TdS

PdV TdS

) TS ( d dU

dA       dA  PdVSdT

SdT TdS

VdP TdS

) TS ( d dH

dG       dG  VdP SdT

(5)

6.1 Properties Relations for Homogeneous Phases

PdV TdS

) V , S (

dU  

Fundamental Properties Relations

VdP TdS

) P , S (

dH  

SdT PdV

) T , V (

dA   

SdT VdP

) P , T (

dG  

These fundamental properties relations are general equation for homogeneous fluid of constant composition

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Exact Differential Equation

If a function F is a property (not depending on the path)

The total differentiation

6.1 Properties Relations for Homogeneous Phases

) y , x ( F F 

y dy dx F

x dF F

y x



 

 



 

 

Ndy Mdx

dF  

y y x

N F x ,

M F 

 

 



 

 

(7)

6.1 Properties Relations for Homogeneous Phases

Exact Differential Equation

Condition for exactness

y dy dx F

x dF F

y x



 

 



 

 

y x

F y

M 2

x  

 



 

Ndy Mdx

dF  

y x

F x

N 2

y  

 



 

x x y

N y

M 

 

 



 

(8)

6.1 Properties Relations for Homogeneous Phases

Maxwell’s Relation

PdV TdS

) V , S (

dU  

VdP TdS

) P , S (

dH  

SdT PdV

) T , V (

dA   

SdT VdP

) P , T (

dG  

V

S S

P V

T 

 

 

 

 

P

S S

V P

T 

 

 



 

T

V V

S T

P 

 

 



 

T

P P

S T

V 

 

 

 

 

x x y

N y

M 

 

 



 

(9)

Maxwell Relation

S H P

U G

V A T

PdV TdS

dU  

VdP TdS

dH  

SdT PdV

dA    SdT VdP

dG  

V

S S

P V

T

P

S S

V P

T

T

V V

S T

P

T

P P

S T

V

(10)

Enthalpy as a Function of T and P

T and P : easily measurable

VdP TdS

) P , S (

dH  

P P

T C

H 

 

dP

? dT

? ) P , T (

dH  

T V T V

P V T S

P H

P T

T

 

 

 

 

 

 



 

T dP T V

V dT

C dH

P

P

 

 

 

 

(11)

Entropy as a Function of T and P

T and P : easily measurable

P P

T C

H 

 

dP

? dT

? ) P , T (

dS  

P P

P T

T S T

C H 

 

 



 

 

T dP V T

C dT dS

P

P

 

 

VdP TdS

) P , S (

dH  

P

T T

V P

S 

 

 

 

 

(12)

Internal Energy as a Function of T and P

T and P : easily measurable

PV H

U  

dP

? dT

? ) P , T (

dU  

P P

P T

P V T

H T

U 

 

 



 

 



 

P dP P V

T T V

T dT P V

C dU

T P

P

P

 

 

 

 



 

 

 

 

 

 

 

P V P V

P H P

U

T T

T

 

 

 



 

 



 

(13)

Ideal-Gas State

T dP T V

V dT

C dH

P

P

 

 

 

 

 dP

T V T

C dT dS

P

P

 

 

P R T

) P / RT ( T

V

P P

ig  

 

 



 

dT C dH  P

P R dP T

C dT

dS  P

(14)

Alternative Form for Liquid

For liquids,

T V V

P

 

 

 T dP T V

V dT

C dH

P

P

 

 

 

 

 dP

T V T

C dT dS

P

P

 

 

1 T

VdP

dT C

dH  P   VdP

T C dT

dS  P 

(15)

Internal Energy as a Function of T and V

Sometimes T and V are more useful than T and P

Equation of States are expressed as P = F(V,T) PdV

TdS )

V , S (

dU  

V V

T C

U 

 

dV

? dT

? ) V , T (

dU  

T P T P

V P T S

V U

V T

T

 

 

 

 

 

 



 

dV T P

T P dT

C dU

V

V

 

  

 

 

(16)

Entropy as a Function of T and V

Sometimes T and V are more useful than T and P

Equation of States are expressed as P = F(V,T) PdV

TdS )

V , S (

dU  

V V

V T

T S T C

U 

 

 

 

 

dV

? dT

? ) V , T (

dS  

V

T T

P V

S 

 

 



 

T dV P T

C dT dS

V

V

 

 

(17)

Gibbs Free Energy as a Generating Function

G = G(T,P)

T, P  Canonical Variables for Gibbs energy

Conforms to a general rule that is both simple and clear

SdT VdP

) P , T (

dG  

   

RT dT dP H

RT V

RT dT dT S

RT dT H

RT dP S

RT V

dT TS RT H

SdT 1 RT VdP

1

RT dT dG G

RT 1 RT

d G

2

2 2 2

 

 

(18)

Gibbs Free Energy as a Generating Function

RT dT dP H

RT V RT

d G    2

 

Fundamental Properties Relationship

 

P T

RT / G RT

V 



 

 

T P

RT / T G

RT

H 



 

Gibbs energy, when given as a function of T and P, serves as a generating function for other thermodynamic properties, and

implicitly represents complete property information

(19)

Example 6.1

Determine the enthalpy and entropy changes of liquid water for a change of state from 1 bar and 25 oC to 1,000 bar and 50 oC. Data for water are given in the following table.

T (oC) P (bar) Cp (J/molK) V (cm3/mol)  (K-1)

25 1 75.305 18.071 256×10-6

25 1,000 …… 10.012 366×10-6

50 1 75.314 18.234 458×10-6

50 1,000 …… 18.174 568×10-6

(20)

Example 6.1

1 T

VdP

dT C

dH P T VdP

C dT

dS P

) P P ( V ) T 1

( ) T T ( C

H P 2 1 2 2 1

) P P ( T V

ln T C

S 2 1

1 2

P

Use mean averaged values from Table

K mol / J 310 . 2 75

314 . 75 305 .

CP 75

For P = 1 bar,

For T = 50 oC,

mol / cm 204 . 2 18

174 . 18 234 .

V 18 3

1 6

6 513 10 K

2 10 568

458

mol / J 400 , 3 ) P P ( V ) T 1

( ) T T ( C

H P 2 1 2 2 1

K mol / J 13 . 5 ) P P ( T V

ln T C

S 2 1

1 2

P

(21)

6.2 Residual Properties

No experimental method to measure G as a function of T and P

Residual Property

Departure function from ideal gas value

 M is any molar property : V, U, H, S, A or G

id

R M M

M  

id

R G G

G  



 

 

 P

V ZRT )

1 Z P ( RT P

V RT V

V

VR id

RT dT dP H

RT V RT

d G    2

 

 dT

RT dP H

RT V RT

d G 2

R R

R   

 

(22)

6.2 Residual Properties

Fundamental residual-property relation RT dT dP H

RT V RT

d G 2

R R

R   

 

 

T R

R

P RT / G RT

V

 

P R

R

T RT / T G

RT

H

P ) dP 1 Z ( J

RTdP V RT

G RT

G P

0 P

0 R

0 P R

R  

P dP T

T Z RT

H

P P

0

R

P ) dP 1 Z P (

dP T

T Z R

S P

P 0 P

0

R

(const T)

(const T)

(const T)

Derive yourself!

(Page 209 – 210)

(23)

Residual Properties as P 0

A gas become ideal as P  0

Then, all residual properties are zero???

- Not generally true…

ig 0

P R

0

P

U 0 , as lim U U

lim  

0 ) PV ( lim U

lim H

lim

R

0 P R

0 P R

0

P

  

0 T P 0

P R

0

P P

lim Z P RT

1 lim Z

RT V

lim 

 

 



 

  

 

RT J G RT

lim G T

RT / T G

RT H

0 P R R

0 P P

R

R  

 

 

 

 

 

See Figure 3.9 (Page 87)

(24)

Enthalpy and Entropy from Residual Properties

R

ig H

H

H   S Sig SR dT

C H

H T

T ig P ig

0 ig

0

0 T

T ig P ig

0 ig

P ln P T R

C dT S

S

0

T R T

ig P ig

0 C dT H

H H

0

R

0 T

T ig P ig

0 S

P ln P T R

C dT S

S

0

(25)

Residual Properties

Advantages of using Residual Properties

Generating Function  Normally function of T, P

Ideal Gas Property

- P = RT/V

- Cp = Cp (T) = dH/dT , Cv = Cv(T)=dU/dT, Function of T only

If we can use other method to evaluate residual properties, we can calculate all the other properties

- Equation of State

- Generalized Correlation

(26)

Example 6.3

Calculate the enthalpy and entropy of saturated isobutane vapor at 360 K from the following information.

1. Table 6.1 gives compressibility-factor data (Z) for isobutane vapor.

2. The vapor pressure of isobutane vapor at 360 K is 15.41 bar.

3. Set for the ideal-gas reference state at 300 K and 1 bar.

4. The ideal gas heat capacity of isobutene vapor is K mol J

S and mol J

H0ig 18,115 / 0ig 295.976 /

) Kelvin in

T ( T 10 037 . 33 7765 . R 1

Cp 3

(27)

Example 6.3

T R T

ig P ig

0 C dT H

H H

0

R

0 T

T ig P ig

0 S

P ln P T R

C dT S

S

0

P dP T

T Z RT

H

P P

0

R

P )dP 1 Z P (

dP T

T Z R

S P

P 0 P

0

R

P )dP 1 Z ( P and

dP T

evalute Z to

Need P

P 0 P

0

From the data in Table 6.1,

Fit the data Z vs. T at different P…

2596 . P 0

)dP 1 Z (

K 10 37 . P 26

dP T

Z

P 0

1 4 P

P 0

Then, you can perform the rest of calculation!

(28)

6.3 Residual Properties by Equations of State

To evaluate above expression, equation of state is to be expressed as a function of P at const T.

Volume explicit: Z (or V) = f(P)

Pressure explicit: Z (or V) = f(P)

However, many cases are pressure explicit!

P dP T

T Z RT

H

P P

0

R

SR T TZ dPP 0P(Z 1)dPP P

P 0

R

(const T)



1 B'P C'P2 Z



2

V C V

1 B Z

(29)

6.3 Residual Properties by Equations of State

For two-term virial equation,

RT 1 BP Z

RT BP P

)dP 1 Z RT (

G P

0

R

2

P P

0 R

T B dT

dB T

1 R

P P

dP T

T Z RT

H

dT dB R

P P

)dP 1 Z P (

dP T

T Z R

S P

P 0 P

0

R

 

(30)

6.3 Residual Properties by Equations of State

Pressure-explicit EOS

In general, EOS can be expressed as Z = PV/RT

By replacing V by r (molar density)  P = ZrRT

Differentiation gives

Z dZ d

P ) dP

dZ Zd

( RT

dP

r

r

r

r

Z ln 1 d Z

) 1 Z P (

) dP 1 Z RT (

G

0 P

0

R

r

r

 

r

1 d Z

T T Z

RT H

0

R

r

r

r

r

r

r r

r

 

r

r

r d

) 1 Z d (

T T Z

Z R ln

S

0 0

R

(31)

Residual Properties as a function of T and P

From the supplement, we can also derive the following alternative expression.

P dP T

T Z T dP

T V V P

T H P

T H P

T H

P P

P

P ig

R( , ) ( , ) ( , )

0

0

P Z dP

P dP T

T Z P dP

R T

P V T S P T S P

T

S P

P P

P

P ig

R( , ) ( , ) ( , )

0

0

0 ( 1)

P Z dP

V dP V RT

P T G P

T G P

T

GR( , ) ( , ) ig( , )

0P  

0P( 1)

(32)

Residual Properties as a function of T,V

However, the application of the previous expression is not straightforward, especially for complicate E.O.S.

From the supplement, we derived the alternative expressions, which is much more useful to use for any pressure explicit E.O.S.

) , ( )

1 (

) ,

( P dV RT Z H T P0

T T P V

T

H V ig

V

) , ( ln

ln )

,

( 0 S T P0

P R P

Z R V dV

R T

V P T

S V ig

V

) , ( ln

) 1 (

ln

) , ( )

, ( )

, (

0

0 G T P

P RT P

Z RT Z

RT V dV

P RT

V T S T V

T H V

T G

V ig





 

(33)

6.4 Two-Phase Systems

Vapor-Liquid Transition

Specific Volume : large change

U, H, S : also large change

G ?

At constant T and P

P

T

Solid Phase

Liquid Phase

Vapor Phase Pc

Tc triple

point

Gas Vapor-Liquid Boundary

SdT VdP

) P , T (

dG  

0 ) nG (

d 

 G G

(34)

6.4 Two-Phase Systems

Derivation of Clapeyron Equation

Phase Transition   

 dG dG

SdT VdP

) P , T (

dG  

dT S dP

V dT

S dP

V sat sat





 

 

V S V

V

S S

dT dPsat



  

H T S





 

V T

H dT

dPsat

(35)

6.4 Two-Phase Systems

Derivation of Clapeyron Equation

Especially for vapor liquid transition

vl vl sat

V T

H dT

dP

 

vl sat

vl Z

P

V  RT 

vl 2

vl sat

Z RT

H dT

P ln d

 

 

vl

vl sat

Z R

H T

/ 1 d

P ln d

 

Relation between

Vapor pressure and heat of vaporization

(36)

Example 6.5

The Clapeyron equation for vaporization may be simplified by introduction of reasonable approximations, namely, that the vapor phase is an ideal gas and that the molar volume of the liquid is negligible compared with the molar volume of the vapor. How do these assumptions alter the Clapeyron equation?

 

vl

vl sat

Z R

H T

/ 1 d

P ln d

From the assumption,

1 Z P or

V RT

Vvl v sat lv

The Clapeyron equation becomes

 

1/T

d P ln R d H

sat vl

the Calusius/Calpeyron equation

T A B

P

ln sat  

(37)

6.4 Two-Phase Systems

Temperature Dependence of Vapor Pressure

 Empirical equation for P vs. T

 

vl

vl sat

Z R

H T

/ 1 d

P ln d

 

T A B

P

ln sat   Simple Expression

C T A B P

ln sat

 

Antoine Equation

 

1

D C

B P A

ln

6 3

5 . 1 sat

Wagner Equation

Tr

1

Specific Temperature Range

(38)

6.4 Two-Phase Systems

Two phase Liquid/Vapor System

 When system consist of saturated liquid and saturated vapor

 Other extensive thermodynamic properties, M (M=U, H, S, A, G,..)

v v l

lV n V

n

nV  

v v l

lV x V

x

V  

v v l

v)V x V

x 1 (

V   

v v l

v)M x M

x 1 (

M   

lv v

l x M

M

M   

(39)

6.5 Thermodynamic Diagrams

Graph showing for a particular substance a set of properties (T, P, V, H, S,…)

TS Diagram

PH Diagram (ln P vs. H)

HS Diagram (Mollier Diagram)

(40)

6.6 Tables of Thermodynamic Properties

U, H, S, V,…  Thermodynamic function

Tables of Thermodynamic Data

Tabulation of values of thermodynamic functions (U, H, V,..) at various condition (T

and P)

It is impossible to know the absolute values of U , H for process materials  Only

changes are important ( U, H,…)

 Reference state

- Choose a Tand Pas a reference state and measure changes of Uand Hfrom this reference state

 tabulation

(41)

Steam Tables

Compilation of physical properties of water (H, U, V)

Reference state: liquid water at triple point (0.01 oC, 0.00611 bar)

Properties of Superheated Steam

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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 downstream state of the steam and H for the following condition.

P1 = 1,000 kPa, T1 = 250 oC, and P2 = 200 kPa

Solution

Since it is adiabatic process, S = 0  S1 = S2 To calculate H, find H1 and H2.

At the initial state, find the H1 and S1 at P1 = 1,000 kPa and T1 = 250 oC

 From steam table, interpolation between 240 and 260 oC gives H1 = 2942 kJ/kg, S1 = 6.9252 kJ/kgK = S2

Since the entropy of saturated vapor at 200 kPa is greater than S2, the final state is the two phase liquid/vapor region!

From page 731

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Steam Table (page 731)

sat.

liq.

sat.

vap.

175 (448.15)

200 (473.15)

220 (493.15)

240 (513.15)

260 (533.15)

280 (553.15)

300 (573.15)

325 (598.15)

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Steam Table (page 725)

sat.

liq.

sat.

vap.

300 (573.15)

350 (623.15)

400 (673.15)

450 (723.15)

500 (773.15)

550 (823.15)

600 (873.15)

650 (923.15)

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Example 6.7

From the steam table (page 725), the final T2 is the saturation temperature at 200 kPa,

 T2 = 120.23 oC At T2 = 120.23 oC,

Since the entropy of saturated vapor at 200 kPa is greater than S2, the final state is the two phase liquid/vapor region!

, S x S ) x 1 ( S

From v l v v

kg / kJ 3 . 706 , 2 H

, kg / kJ 7 . 504 H

K kg / kJ 1268 .

7 S

, K kg / kJ 5301 .

1 S

v 2 l

2

v 2 l

2

9640 .

0 x

1268 .

7 x 5301 .

1 ) x 1 ( 9252 .

6

v

v v

kg / kJ 9 . 315 9

. 942 , 2 0 . 627 , 2 H

H H

kg / kJ 0 . 627 , 2 3 . 706 , 2 9640 .

0 7 . 504 )

9640 .

0 1 ( H

x H

) x 1 ( H

1 2

v 2 v l

2 v 2

From page 725

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Calculation of Enthalpy and Entropy for Real Fluids

Using residual properties, HR and SR, and ideal gas heat capacities, we can now calculate the enthalpy and entropy for real fluids at any T and P!

For a change from state 1 to state 2,

R 1 T

T ig P ig

0

1 H C dT H

H

0

TT R2

ig P ig

0

2 H C dT H

H

0

R 1 R

2 T

T

ig P 1

2 H C dT H H

H

H 2

1

R 1 R

2 1

T 2 T

ig

P S S

P ln P T R

C dT S 2

1

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Calculation of Enthalpy and Entropy for Real Fluids

Step 1  1ig

Step 1ig  2ig

Step 2ig  2

R 1 1

ig H H

H1 Sig S1 S1R

1

dT C H

H

H 2

1 1 2

T T

ig P ig

ig

ig

1 T 2

T ig P ig

1 ig

2 ig

P ln P T R

C dT S

S

S 2

1

R 2 ig

2

2 H H

H S2 Sig2 SR2

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Homework

Problems

6.3, 6.11, 6.23, 6.48, 6.54, 6.74

For 6.54

- use the heat capacity of n-butane from Table C.1.

- For HRand SR, use Pitzer correlations for the Compressibility factor (Z).

Due:

Recommend Problems

6.2, 6.4, 6.9, 6.12, 6.17, 6.30, 6.31, 6.49, 6.57, 6.58, 6.71, 6.72, 6.76, 6.80, 6.92, 6.93, 6.94

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