Chapter Three Thermodynamic Relationships and Evaluations ...€¦ · Advanced Thermodynamics, ME...

Advanced Thermodynamics, ME dept, NCHU 頁 1

Chapter Three Thermodynamic Relationships and Evaluations of

Thermodynamic Properties Update on 2013/10/30

(3.1). General relationships Assume x, y, z, and α are all thermodynamic properties. In other words, they are all point functions. (3.1.1). Exact Differential For any point function, ( , )z z x y

y x

z zdz dx dy Mdx Ndy

x y

,y x

z zM N

x y

yx

M N

y x

e.g.

( , )u u s v

v s

u udu ds dv Tds Pdv

s v

s v

T P

v s

(3.1.2). Reciprocal rule

( , )x x y z ， ( , )y y x z

yz

x xdx dy dz

y z


z x

y ydy dx dz

x z

z x yz

z x yz z

x y y xdx dx dz dz

y x z z

x y x y xdx dz

y x y z z

1zz

x y

y x

1

z

z

xyyx

e.g.

1

T

T

vPPv

；1

P

P

vTTv

------------------------------------------------------------------------------------------------------

Example: RT

Pv

2T

v RT v

P P P

2T

P RT P

v v v

------------------------------------------------------------------------------------------------------

Example: 2

RT aP

v b v

2 3

2

( )T

P RT a

v v b v

2 3

1 12

( )T

T

vRT aPP

v b vv

------------------------------------------------------------------------------------------------------


(3.1.3). Cyclic rule

1

x yz

y

x y xzy z zz

1x yz

x y z

y z z

e.g. 1v T P

T P v

P v T

------------------------------------------------------------------------------------------------------

Example: 2

RT aP

v b v

；

2

1( )( )

aT P v b

R v

2 3

2

( )T

P RT a

v v b v

v

T v b

P R

2 3

2[ ]( ) 1

( ) P

RT a v b v

v b v R T

2 3

12

( )P

v RRT aT v b

v b v

------------------------------------------------------------------------------------------------------ (3.1.4). Chain rule

( , )x x y ， ( , )y y z

y

x xdx dy d

y

z

y ydy dz d

z


z y

z y

x y y xdx dz d d

y z

x y x y xdz d

y z y

( , )x x z

z

x xdx dz d

z

1x y xzy z zx

1x y z

y z x

z z y

x x y x

y


(3.2)、Maxwell Relationship

The first law of thermodynamics:

Q dU W

If the process is reversible， Q TdS ， W PdV

TdS dU PdV ------------------------------------------------------------------------------------------------------

( , )U U S V

dU TdS PdV

V S

U UdU dS dV TdS PdV

S V

V

UT

S

，

S

UP

V

V

UT

S

This is an expression of what temperature is. According to this

relationship, temperature could be negative if internal energy increases while entropy decreases. This is called ‘negative temperature’. ------------------------------------------------------------------------------------------------------ H U PV ，Enthalpy

dH dU PdV VdP TdS VdP

( , )H H S P

P S

H HdH dS dP TdS VdP

S P


P

HT

S

，

S

HV

P

------------------------------------------------------------------------------------------------------ A U TS ，Helmholtz Function

dA dU TdS SdT PdV SdT

( , )A A V T

T V

A AdA dV dT PdV SdT

V T

T

AP

V

，

V

AS

T

( )A P PdV This is the effect of pressure on Helmholtz function in an isothermal

process. ------------------------------------------------------------------------------------------------------ G H TS ，Gibbs Function

dG dH TdS SdT VdP SdT

( , )G G P T

T P

G GdG dP dT VdP SdT

P T

T

GV

P

，

P

GS

T

( )G P VdP : This is the effect of pressure on Gibbs function in an isothermal

process. ------------------------------------------------------------------------------------------------------ In summary, we have the following relationships.


Maxwell Relationship:

S V

T P

V S

S P

T V

P S

T V

S P

V T

T P

S V

P T

If the relationship among P, T, and V is known, the terms V

P

T

and P

V

T

can

be evaluated directly, and entropy can then be obtained. The relationship among P, T, and V, called the equation of state, is the most basic relationship for thermodynamic calculations. All other thermodynamic properties can be derived from the state equation.


(3.3)、State equation

The equation of state is the relationship among T, P, and v. This is the fundamentals of the thermodynamic properties.

Fig. 3.3.1 The T, P, v relationship for water, ice, and vapor.

Fig. 3.3.2 The P/T diagram and the P/v diagram of water and vapor.


(3.3.1). The microscopic view of pressure

Pressure is the global effect that results from the momentum change caused by molecules hitting and bouncing from the wall. P ~ collision frequency × momentum change

momentum change ~ molecular velocity collision frequency ~ molecular velocity × molecular density

molecular density ~ N/V P ~V2*N/V V2 ~ molecular kinetic energy ~ T P ~ T*N/V P = nRT /V (3.3.2). Ideal gas Two assumptions of the ideal gas: (1). Molecules do not occupy any volume in the space. (2). Molecules do not attract or repulse each other. P ~ T*N/V

unR TP

V

Pv RT ， uRR

M

uR =8.314 kJ/kmole-K, Universal gas ocnstant

For air, M = 28.97 kg/kmole, R = 0.287 kJ/kg-K The reduced form of the state equation for ideal gas:

rc

TT

T ， r

c

PP

P ， r

c

vv

v

c rr

c c r

T R TP

P v v

Since 1c

c c

T R

P v ，thus r

rr

TP

v


The reduced form of the state equation is universal for all gases. It contains no specific characteristics of any gas. ------------------------------------------------------------------------------------------------------ Example、Find the specific volume of water vapor in the following states. (1). P = 0.01228 bar，T = 10℃，v = 106.445 kg/m3，actual value v = 106.379 kg/m3。 (2). P = 1 bar，T = 100℃，v = 1.723 kg/m3，actual value v = 1.696 kg/m3。(1.6%) (3). P = 10 bar，T = 200℃，v = 0.2185 kg/m3，actual value v = 0.2060 kg/m3。(6.1%) (4). P = 100 bar，T = 320℃，v = 0.02739 kg/m3，actual value v = 0.01925 kg/m3。

(42%) The calculated value deviates from the actual value at high pressure. ------------------------------------------------------------------------------------------------------ Assignment 3.1、Find the specific volume of NH3 in the following states using the

equation of state for ideal gas and find the exact values from the thermodynamic table. (1). P = 0.4086 bar，T = -50℃。 (2). P = 4.2962 bar，T = 0℃。 (3). P = 20 bar，T = 100℃。

------------------------------------------------------------------------------------------------------ It is noted that the higher the pressure, the more remarkable the deviation between the actual value and the value obtained from the ideal gas model. (3.3.3). Size of a molecule The effective size of a diatomic molecule is about 3~4 ×10-10 m.

The averaged space that a molecule occupies is


0 0 0

u uV V nR T R T kTv

N N n PN n PN P

1

33 kT

d vP

For example, at room temperature and atmospheric pressure, the averaged space for a molecule is 4.086×10-26 m3. If this space is a cubic, the averaged distance between two molecules is 3.44×10-9 m, which is about 10 times of the diameter of a molecule. d

10, for air at room temperature and atmospheric pressure.

Mean free path: The distance that a molecule travels between two successive collisions. At room temperature and atmospheric pressure, the mean free path of air is about 2.09×10-9 m, which is about 6 times of the diameter of a molecule.

2 2

1

2 / 2

vl

N V

3 31

2

l d

1

31

2l d

= 0.6083 d

(3.3.4). Molecular force

Neutral atoms and molecules are subject to two distinct forces in the limit of large distance, and short distance. An attractive van der Waals force at long ranges, and a repulsion force, the result of overlapping electron orbitals, referred to as Pauli repulsion. The Lennard-Jones potential is a simple mathematical model that describes this behavior.

LJ is a model with forces that are strongly repulsive at very short inter particle distances, attractive forces at large distances, and weak attractive forces at very large distance. The L-J potential is of the form

12 6

( ) 4V rr r


where ε is the well depth and σ is the hard sphere radius. The term 12( )r

describes

the repulsive force and the 6( )r

term describes the attractive force. The L-J

potential is approximate and the form of the repulsion term has no theoretical justification. The L-J potential is a relatively good approximation and due to its simplicity often used to describe the properties of gases, and to model dispersion and overlap interactions in molecular models. It is particularly accurate for noble gas atoms and is a good approximation at long and short distances for neutral atoms and molecules.

Fig. 3.3.3: The Lennard-Jones potential between molecules

13 7 7 6

4 12 6 24 2 1dV

Fdr r r r r

0F occurs at 6

1

2r

, or 1.1225r , in which repulsive force equals to

attractive force. At this point, the LJ potential is at its minimum. 12 6

( ) 4V rr r

1.1225r , 0F , repulsive force

1.1225r , 0F , attractive force


14 8 8 6

2 224 26 7 24 26 7

dF

dr r r r r

0dF

dr , the maximum attractive force occurs at

67

26r

, 1.2445r , and

max 2.3964F

At the position 2r , max0.1816 0.0758F F

At the position 4r , max0.00146 0.000611F F

The attraction force decays rapidly as the distance increases. The mean free path of gases at room temperature and atmospheric pressure is about 6 times of the diameter of a molecule. At that distance, the inter molecule force is negligible. However, as two molecules heading to each other, the van der Waals force would become significant at the moment of collision. ------------------------------------------------------------------------------------------------------ Example、Find the maximum attractive force between oxygen molecules.

ε = 38 k =38 × 1.38 × 10-23 J/molecule, σ = 3.21 × 10-10 m

1.63 × 10-12 N, maxF 3.91 × 10-12 N

------------------------------------------------------------------------------------------------------

Electrostatic force > van der Waals force >> gravitational force

Gravitational force:

1 22

m mF G

r = 9.01 × 10-43 N

G = 6.67 × 10−11 N m2 kg−2.

At 1.2445r = 3.995 × 10-10 m, and m1 = m2 = 4.65 × 10-26 kg

Coulomb's law:

1 22

04

e eF

r = 1.44 × 10-9 N

0 = 8.8542 × 10-12 C2/Nm2

At 1.2445r = 3.995 × 10-10 m, and e1 = e2 = 1.60219 x 10-19 C


(3.3.5). Equation of state for real gases Consider the volume that molecules occupy and the interaction between molecules. Cubic state equation This is a third order equation of v in terms of P and T.

The general form: 2 2

RT aP

v b v cbv db

Van der Waals equation: 2

RT aP

v b v

Redlich-Kwong equation: 1/ 2( )

RT aP

v b v v b T

Soave equation: ( )

RT aP

v b v v b

Berthelot equation: 2

RT aP

v b Tv

Peng-Robinson equation: 2 22

RT aP

v b v bv b

Clausius equation: 2( )

RT aP

v b T v c

Lorentz equation: 2

( )RT v b a

Pv v v

Dieterici equation: exp( )RT a

Pv b vRT

Virials equation

21

RT B CP

v v v

BWR, Benedict-Webb-Rubin equation: 2 4 5

1Pv B C D E

RT v v v v

2 2 3 6 3 2 2 2

1 1( ) ( ) (1 )exp( )

RT C a cP BRT A bRT a

v T v v v v T v v

Compressibility and generalized chart

( , )r r

PvZ Z P T

RT ，

ZRTv

P


(3.3.6). Van der Waals gas Volume of molecules and the inter molecules force are considered. Molecular attraction ~ density Collision frequency ~ density A molecule striking on the surface of a wall is pulled by the molecules behind it due to the inter-molecular attraction force. The result of the pulling force is a reduction in the collision velocity onwards the wall. The pulling force is proportional to gas density. The collision frequency is also proportional to gas density. As a result, the pressure reduction is proportional to the square of the gas density.

2

RT aP

v b v

If temperature is fixed, i.e. in an isothermal process, the relationship between pressure and specific volume is a cubic equation.

Fig. 3.3.4: Van der Waals equation of state


It can be shown that there are at least two regions that are not feasible in the behavior of real gas. The first region is where the pressure is negative ( 0P ), and the

second region is that the slope of pressure is positive ( 0T

P

v

).

The first term of the equation RT

v b is the major term to determine the pressure of

the gas. The second term 2

a

v is a minor modification to correct the effect of inter

molecular force. However, when the volume is reduced, the effect of the second term overwhelms the first term such that pressure becomes negative. As volume is reduced even more to close the value of b, the effect of first term dominates once again, and pressure approaches infinity. The turn-down and turn up of the pressure trace in the vicinity of zero volume are of no physical meaning because the original modification of van der Waals model to ideal gas model is not tended to resolve the problem at highly dense state. As distance between molecules approaches zero, system is no longer in the gas phase. Transition from gas phase to liquid phase may occur during the process. (3.3.6.1). How to obtain the value of a and b for van der Waals gas? At the critical point, we have

0T

P

v

，

2

20

T

P

v

2 32 0

( )T

P RT a

v v b v

，

3

22( )c

c

RTva

v b

2

2 3 42 6 0

( )T

P RT a

v v b v

，4

33( )c

c

RTva

v b

3

( ) 2c

c

v

v b

，

21

3c

b

v ，

1

3 cb v ，9

8 c ca v RT

2

3 9 3

2 8 8c c

cc c c c c

RT a RT RT RTP

v b v v v v

，

3

8c

cc

RTv

P


2 227

64c

c

R Ta

P ，

1

8c

c

RTb

P

------------------------------------------------------------------------------------------------------ Example: Find the values of a and b of van der Waals gas for water vapor.

cT = 647.3 K， cP = 220.9 bar 2 227

64c

c

R Ta

P = 5.531 bar(m3/kmole)2，=1.703 kPa(m3/kg)2

1

8c

c

RTb

P = 0.0305 m3/kmole = 0.001693 m3/kg

------------------------------------------------------------------------------------------------------ Assignment 3.2Find the specific volume of NH3 in the following states using the Van der Waals equation of state and find the exact values from the thermodynamic table. (1). P = 0.4086 bar，T = -50℃。 (2). P = 4.2962 bar，T = 0℃。 (3). P = 20 bar，T = 100℃。

------------------------------------------------------------------------------------------------------

0b N v ， 3

0

bv d

N ，

1133

260

0.0305

6.02 10

bd

N

3.7×10-10 m = 3.7 A

It seems that d is about the size of a molecule, and b represents the volume that molecules occupy.

The van der Waals constants a bar(m3/kmole)2 b m3/kmole Air 1.368 0.0367 C4H10 13.86 0.1162 CO2 3.647 0.0428 CO 1.474 0.0395 CH4 2.293 0.0428 N2 1.366 0.0386 O2 1.369 0.0317 C3H8 9.349 0.0901 R12 10.49 0.0971 SO2 6.883 0.0569 H2O 5.531 0.0305


(3.3.6.2). How to obtain the solution of the state equation? The state equation of van der Waals gas is a third order equation of v in terms of P and T. If T and V are given, and P is to be estimated, or P and v are given, and T is the known, the equation can be solved easily. However, if P and T are given, and v is to be determined, the third order equation could not be solved directly. Some numerical scheme should be involved to obtain the solution. If specific volume and temperature are known, pressure can then be obtained directly.

2

RT aP

v b v

------------------------------------------------------------------------------------------------------ Example: Find the pressure of water vapor. (P = 100 bar，T = 320℃，v = 0.01925 m3/kg。)

2

RT aP

v b v

= 10991. kPa = 109.91 bar

Error = 9.91% ------------------------------------------------------------------------------------------------------ If specific volume and pressure are known, temperature can then be obtained directly.

2

1( )( )

aT P v b

R v

------------------------------------------------------------------------------------------------------ Example: Find the temperature of water vapor. (P = 100 bar，T = 320℃，v = 0.01925 m3/kg。)

2

1( )( )

aT P v b

R v = 555.4 k = 282.3 ℃

Error = -6.3% ------------------------------------------------------------------------------------------------------ If temperature and pressure are known, specific volume can then be obtained with the help of some numerical schemes.

3 2( ) 0RT a ab

v b v vP P P


(i). Iterative method

1

2

i

i

RTv b

aP

v

, where 0

RTv

P = 0.02737 m3/kg

1 2 3 4 5 6 7

0.027366 0.023989 0.02281 0.022311 0.022083 0.021976 0.021925

8 9 10 11 12 13 14

0.0219 0.021888 0.021882 0.02188 0.021878 0.021878

Error = (0.021878 - 0.01925)/ 0.01925= 13.65% Comment: The iterative method is not very efficient. It needs 13 steps to get the converged solution. However, it could be carried out conveniently by means of calculator. (ii). Newton Rapshion method

1

( )ii i

f vv v

dfdv

, where 0

RTv

P

3 2( ) ( )RT a ab

f v v b v vP P P

23 2( )df RT a

v b vdv P P

1 2 3 4 5 6

0.027366 0.02361 0.022133 0.021884 0.021877 0.021976 0.021877

------------------------------------------------------------------------------------------------------ Example. Ammonia is compressed isothermally at 160℃ from the pressure of 1 bar

to the pressure of 100 bar. Calculate the work of the process assuming that ammonia is a van der Waals gas. NH3， cT 406 K， cP 112.8 bar

a 1.467603

2( )m

kPakg

，b 0.0021953m

kg，R= 0.488 kJ/kg-K


Ideal gas： 2

1

lnRT v

W Pdv dv RTv v

1v 2.1130 m3/kg， 2v 0.02113 m3/kg，W -973.10 kJ/kg

van der Waalsgas：2

RT aP

v b v

The real values of specific volumes are: T1=373 K, P1=1 bar, 1v 1.815 m3/kg T2=373 K, P2=100 bar, 2v 0.0039246 m3/kg

22

1 2 1

1 1ln

RT a v bW Pdv dv RT a

v b v v b v v

W -1265.93 + 369.57 = -896.36 kJ/kg It needs less work to compress real gas than that of ideal gas because as the gas density gets higher, the potential energy increases. The difference is 76.74 kJ/kg. ------------------------------------------------------------------------------------------------------ Assignment 3.3 A certain kind of gas is compressed isothermally at 27℃ from the

pressure of 1 bar to the pressure of 100 bar. Calculate the work of the process assuming that this gas is a van der Waals gas. (1). Methane (2). Hydrogen (3). Air ------------------------------------------------------------------------------------------------------


(3.3.7). Redlich-Kwong gas Assume that the attraction force is temperature dependent.

1/ 2 1/ 2

1 1 1( )

( )

RT a RT aP

v b v v b T v b b T v v b

2 1/ 2 2 2

1 1 1( ) 0

( ) ( )T

P RT a

v v b b T v v b

2

2 3 1/ 2 3 3

1 2 22 ( ) 0

( ) ( )T

P RT a

v v b b T v v b

2 2 3 3

1 1 2 2 2( ) ( ) 0

( ) ( )v v b v b v v b

2 3 2 3

1 1 1 1 1 10

( ) ( )v v b v v b v b v b

3 3

1 1 20

( )

b b

v v b v b v b

3

3

( )2

v b

v

1/ 32 1 0.259921

b

v

0.259921 cb v

2 1/ 2 2 2

1 1 1( )

( 0.259921 ) 0.259921 (1.259921 )c c c c c c

RT a

v v v T v v

2 3 1/ 2

1

0.5477169 0.702414c

c c c

RT a

v v T


3 / 21.2824325 c ca RT v 3 / 2

2 1/ 2

1.28243250.333338

0.259921 1.259921c c c c

cc c c c c

RT RT v RTP

v v v T v

1

3c c

cc

P vZ

RT ，

1

3c

cc

RTv

P

2 5 / 23/ 21.2824325 0.333338 0.42748c c

cc c

RT R Ta RT

P P

0.259921 0.333338 0.08664c c

c c

RT RTb

P P

------------------------------------------------------------------------------------------------------ Example: Find the values of a and b of Redlich-Kwong gas for water vapor.

cT = 647.3 K， cP = 220.9 bar， 2 5 / 2

0.42748 c

c

R Ta

P = 142.59 bar(m3/kmole)2K0.5=43.912 kPa(m3/kg)2 K0.5

0.08664 c

c

RTb

P = 0.02111 m3/kmole = 0.0011171 m3/kg

------------------------------------------------------------------------------------------------------

0b N v ， 3

0

bv d

N ，

1133

260

0.0211

6.02 10

bd

N

3.3×10-10 m = 3.3 A

The Redlich-Kwong constants a (m3/kmole)2 K0.5 b m3/kmole Air 15.989 0.02541 C4H10 289.55 0.08060 CO2 64.43 0.02963 CO 17.22 0.02737 CH4 32.11 0.02965 N2 15.53 0.02677 O2 17.22 0.02197 C3H8 182.23 0.06242 R12 208.59 0.06731 SO2 144.80 0.03945 H2O 142.59 0.02111

------------------------------------------------------------------------------------------------------ Example. Ammonia is compressed isothermally at 100℃ from the pressure of 1 bar


to the pressure of 100 bar. Calculate the work of the process assuming that ammonia is a Redlich-Kwong gas. NH3， cT 406 K， cP 112.8 bar，R= 0.488 kJ/kg-K

2 5/ 2

0.42748 c

c

R Ta

P = 87.00 bar(m3/kmole)2K0.5=29.964 kPa(m3/kg)2 K0.5

0.08664 c

c

RTb

P = 0.02593 m3/kmole = 0.0015215 m3/kg

1/ 2( )

RT aP

v b v v b T

T1=373 K, P1=1 bar, 1v 1.814 m3/kg T2=373 K, P2=100 bar, 2v 0.0029438 m3/kg

Ideal gas： 2

1

lnRT v

W Pdv dv RTv v

1v 1.824 m3/kg， 2v 0.01824 m3/kg，W -838.25 kJ/kg

2 2 11/ 2 1/ 2

1 1 2

1 1 1 1 ( )( ) ln ln

( )

RT a v b a v b vW dv RT

v b b T v v b v b b T v b v

W -1301.50 +423.99 = -877.51 kJ/kg ------------------------------------------------------------------------------------------------------ Assignment 3.4 A certain kind of gas is compressed isothermally at 27℃ from the

pressure of 1 bar to the pressure of 100 bar. Calculate the work of the process assuming that this gas is a Redlich-Kwong gas. (1). Methane (2). Hydrogen (3). Air ------------------------------------------------------------------------------------------------------


(3.3.8). Other State Equations Berthelot Model

2

RT aP

v b Tv

Soave Model

( )

RT aP

v b v v b

2 2

0.42747 c

c

R Ta

P ， 0.086640 c

c

RTb

P

20.51 1 rm T ， 20.48505 1.5517 0.1563m

where ω is the acentric factor for the species.

10 1satrlog R at 0.7rT

The Peng-Robinson equation of state

2 22

RT aP

v b v bv b

2 2

0.42724 c

c

R Ta

P ， 0.07780 c

c

RTb

P

20.48505 1.5517 0.1563m

22 0.51 (0.37464 1.54226 0.26992 ) 1 rT

Clausius

2( )

RT aP

v b T v c

Lorentz

2 2( )

RT aP v b

v v

Dieterici


exp( )RT a

Pv b vRT

The general form of Cubic state equation

2 2

RT aP

v b v cbv db

2 2

0c

c

R Ta a

P , 0

c

c

RTb b

P

c d 0b 0a

Ideal gas 0 0 0 0 Van der Waals 0 0 1/8 27/64 Redlich Kwong 1 0 0.08664 0.50.42748 rT

Soave 1 0 0.08664 20.50.42748 1 (1 )rf T Peng Robinson 2 -1 0.0778 20.50.42748 1 (1 )rf T Berthelot 0 0 1/8 127 / 64 rT

20.48 1.57 0.17f for Soave

20.37464 1.54226 0.26992f for Peng Robinson

0.7

ln1

ln10 r

satr

atT

P

------------------------------------------------------------------------------------------------------ Assignment 3.5. The Berthelot equation of state can be expressed as the following.

2

RT aP

v b Tv

(i). Show that the constants in the Berthelot equation can be obtained as 2 327

64c

c

R Ta

P ，

1

8c

c

RTb

P

(ii). Find the values of the constants a and b for air. (iii). One kg of air is contained in a chamber of 1 m3 at 300K. Air is then compressed isothermally to a volume of 0.2 m3. Calculate the work of compression. (iv). Find the specific volume of air at 25℃ and 100 bars using the Berthelot

equation. ------------------------------------------------------------------------------------------------------


BWR, Benedict-Webb-Rubin equation

2 2 3 6 3 2 2 2

1 1( ) ( ) (1 )exp( )

RT C a cP BRT A bRT a

v T v v v v T v v

2 4 51

Pv B C D EZ

RT v v v v

0 00 3

A CB B

RT RT

2

3

vc eD

RT

2

3

va ceC b

RT RT

aE

RT


(3.3.9). Compressibility and Generalized Charts (3.3.9.1). Compressibility

PvZ

RT

For ideal gas, 1Z . For real gas,

0lim 1p

Z

.

ZRTv

P ， ideal

RTv

P

ideal

vZ

v

Compressibility is the ratio of the specific volume of real gas to that of ideal gas at the same temperature and pressure. ------------------------------------------------------------------------------------------------------ Example、Find the compressibility of water vapor in the following states. (1). P = 0.01228 bar，T = 10℃，Z = 0.9994 (2). P = 1 bar，T = 100℃，Z=0.9843 (3). P = 10 bar，T = 200℃，Z=0.9428 (4). P = 100 bar，T = 320℃，Z=0.7028

------------------------------------------------------------------------------------------------------ Example、Find the compressibility of NH3 in the following states. (1). P = 0.4086 bar，T = -50℃，Z=0.9863 (2). P = 4.2962 bar，T = 0℃，Z=0.9339 (3). P = 20 bar，T = 100℃，Z=0.9064

------------------------------------------------------------------------------------------------------ (33.9.2). Generalized Chart

rc

TT

T ， r

c

PP

P ， r

c

vv

v

( , )r r rP f T v

c c r r r rc

c r r

Pv P v P v P vZ Z

RT RT T T

( , , )c r rZ Z Z T P


If cZ is a constant, then ( , )r rZ Z T P is an universal relationship. In actual, cZ is not a constant. However, it varies in a narrow range, cZ =0.230~0.307.

Fig. 3.3.5 The compressibility as a function of reduced temperature and reduced

pressure.

Ideal gas： rr

r

TP

v

Van der Waals gas： 3

8c c

cc

P vZ

RT

2 227

64c

c

R Ta

P ，

1

8c

c

RTb

P

rc

TT

T ， r

c

PP

P ，

38

rcc

c

v vv

RTvP

2

8 3

3 1r

rr r

TP

v v


2

3 8 3 3 1( , )

8 3 1 3 1 8r r r r r

c r rr r r r r r r

P v v T vZ Z Z T v

T T v v v T v

This is a universal form of the van der Waals equation. It can be applied to any gas directly without any specified constants. Redlich-Kwong gas：

rc

TT

T ， r

c

PP

P ， r

c

vv

v

1

3c c

cc

P vZ

RT

3/ 2

1/ 2 1/ 2

1.2824325

0.259921 ( 0.259921 )r c c c

r cr c c r c r c c r c

RT T RT vP P

v v v v v v v v T T

1/ 2

1.2824325[ ]

0.259921 ( 0.259921)c r

rc c r r r r

RT TP

P v v v v T

1/ 2

1.28243253[ ]

0.259921 ( 0.259921)r

rr r r r

TP

v v v T

3 / 2

1.2824325( , )

0.259921 ( 0.259921)r r r

c r rr r r r

P v vZ Z Z v T

T v v T

The value of rv can be obtained once rT and rP are given. This is a universal

relationship, regardless what kind of gas is applied. The general relationship for all real gases.

r c r c rr

r c r c r

Pv P v P P vP PZ v

RT RT T T RT T ， c

rc

vPv

RT

------------------------------------------------------------------------------------------------------ Example: Find the specific volume of water vapor. (P = 100 bar，T = 320℃，v = 0.01925 kg/m3。)

cT = 647.3 K， cP = 220.9 bar

rP = 0.4527， rT = 0.9161，Z= 0.76

idealv Zv = 0.0208 kg/m3

------------------------------------------------------------------------------------------------------


Fig. 3.3.6 Generalized compressibility chart(0< Pr <1)

Fig. 3.3.7 Generalized compressibility chart (1< Pr <10)


(3.4). Thermodynamic properties at phase change (3.4.1). Clapeyron Equation This is the relationship between saturation pressure and saturation temperature during phase transition.

Fig. 3.4.1 The saturation pressure and saturation temperature

This is the relationship between T and P during the process of phase transition. dh Tds vdp

During the process of phase change, the pressure does not vary as temperature is fixed.

dh Tds

h T s

hs

T

h u P v

T s u P v

s uP T

v v

，

sat

dP s h

dT v T v


(3.4.1.1). Melting: the solid liquid transition

melth h constant, meltv v constant

1melt

sat melt

dP h

dT v T

1melt

melt

hdP dT

v T

00

lnmelt

melt

h TP P

v T

0melt l sh h h

melt l sv v v

0meltv , 0melt

melt

h

v

, 0T T as 0P P

Melting point will increase at elevated pressure.

0meltv , 0melt

melt

h

v

, 0T T as 0P P

Melting point will decrease at elevated pressure.

0meltv , 0T T as 0P P

Melting point will remain the same at different pressure.

As 0T T , 0

0 0

lnT T T

T T

00

0

melt

melt

h T TP P

v T

P 0meltv 0meltv

0 0

0 0 0

melt

melt

P P h T T

P P v T

T


------------------------------------------------------------------------------------------------------ Example. Find the melting temperature of ice at 1500 bar. Triple point: 0.611 kPa at 273.16 K melth 6.01 kJ/mole, meltv -1.7 cm3/mole T 1500 × 100 × (-1.7 × 10-6) × 273.16 / 6.01 = -11.3 K = -11.3 ℃ Comment: Since >0melth , if 0meltv , i.e., the specific volume of liquid is less than that of solid, the melting point would be depressed under pressurization. That is the case for ice. However, if the specific volume of liquid is greater than that of solid, the melting point would be raised under pressurization. ------------------------------------------------------------------------------------------------------

Melting point of ice at different pressure

(3.4.2). Vaporization: the liquid vapor transition

( )g f g f

sat g f g

h h h hdP

dT T v v Tv

g

RTv

P

2

g f

sat

h hdPP

dT RT

2

ln g f

sat

h hd P

dT RT


If the latent heat of phase change is a constant, fg g fh h h = constant,

1ln g fh hP c

RT

1 00

ln g fh hc P

RT

0 0

ln g f g fh h h hP

P RT RT

00

exp g f g fh h h hP P

RT RT

------------------------------------------------------------------------------------------------------ Example. Predict the saturation pressure of water vapor at 60℃ based on the data at

30℃。

g fh h = 2430.5 kJ/kg

R = 0.461 kJ/kg-K

T0 = 303 K, T = 333 K

P0 = 4.246 kPa

00

exp g f g fh h h hP P

RT RT

= 20.36 kPa (19.94 kPa, error: 2.1%)

------------------------------------------------------------------------------------------------------ Example. The critical temperature of R22 is 369K. However, the saturated

properties of R22 are offered up to 60℃ only in the text book of Thermodynamics.

Please complete the Ps data up to the critical temperature.

------------------------------------------------------------------------------------------------------ Assignment 3.5. An unknown liquid has a vapor pressure of 11.73 kPa at 45oC and

5.20 kPa at 25oC. What is its heat of vaporization?

------------------------------------------------------------------------------------------------------


(3.4.3). Sublimation: the solid vapor transition

------------------------------------------------------------------------------------------------------ Example. Predict the saturation pressure of water vapor at -60℃ based on the data at

triple point。

Triple point: 0.611 kPa at 273.16 K

subh 2837 kJ/mole

0

1 1( )subh

R T T

-6.277

P = 1.15×10-3 kPa

------------------------------------------------------------------------------------------------------ Example. A manufacturer of air dryer claimed that they may produce air with the

dew point at -60℃. Find the relative humidity of air at 25℃ after being treated by

the dryer.

Ps = 3.169 kPa

φ= Pv/Ps = 0.036%

------------------------------------------------------------------------------------------------------


(3.4. 5). Empirical equation

Antoine Equation:

10logB

P AT C

where A, B, and C are experimentally derived parameters. The Antoine equation is

generally reasonably good over a moderate range of temperatures. It doesn't do

very well near the critical point.

formula species A B C Tmin Tmax

CH4 methane 6.69561 405.420 267.777 -181 -152

C2H6 ethane 6.83452 663.700 256.470 -143 -75

C3H8 propane 6.80398 803.810 246.990 -108 -25

C4H10 butane 6.80896 935.860 238.730 -78 19

C5H12 pentane 6.87632 1075.780 233.205 -50 58

Extension reading:

M. Edalat, R.B. Bozar-Jomehri, and G.A. Mansoori, “Generalized Equation Predicts

Vapor Pressure of Hydrocarbons”, Oil & Gas Journal, Feb. 1, 1993.


(3.5). Thermodynamic properties calculations for real gases (3.5.1). u calculations Internal energy is composed of kinetic energy and potential energy. Kinetic energy originates from the motion of molecules, and is the dominant content of total energy. Temperature is an index of kinetic energy. Potential energy stems from the intermolecular force which depends on the distance between molecules. As pressure is raised, the space that molecules may travel is reduced. As a result, the potential energy is reduced too.

The internal energy of water vapor

0.06 bar 1.0 bar 10 bar 100 bar 120℃ 2544.7 2537.3 320℃ 2843.0 2841.5 2826.1 2588.8 500℃ 3132.2 3131.6 3124.4 3045.5 Choose T and v as independent variables.

( , )s s T v

v T

s sds dT dv

T v

From the Maxwell’s relationship, it is known that T V

S P

V T

v v

s Pds dT dv

T T

( , )u u T v

v T

u udu dT dv

T v

vv

uc

T

， v

T

udu c dT dv

v

v v v v

s P s Pdu Tds Pdv T dT dv Pdv T dT T P dv

T T T T


vv

sc T

T

，

T v

u PT P

v T

vv

Pdu c dT T P dv

T

Calculation of u

2 1 vv

Pu u c dT T P dv

T

vc dT : Effect of temperature on internal energy.

v

PT P dv

T

: Effect of pressure on internal energy.

The calculation of internal energy change can be decomposed into three steps.

2 0 1

* * * *2 1 2 2 2 1 1 1[ ] [ ] [ ]T T v v T Tu u u u u u u u

2

0

*2 2 [ ]

v

vv

Pu u T P dv

T

2

1

* *2 1 0

T

v

T

u u c dT

1

0

*1 1 [ ]

v

vv

Pu u T P dv

T

2 2 1

2 1

1

2 1 2 0 1[ ] [ ]v T v

T T v T Tv vT

P Pu u T P dv c dT T P dv

T T

(3.5.1.1). Ideal gas For ideal gas, the effect of pressure on internal energy is zero. Pv RT

0v

P RT P T P

T v

0vdu c dT


------------------------------------------------------------------------------------------------- Example, An insulated rigid chamber is divided into two parts with a diaphragm. The diaphragm neither moves nor conducts heat. The left part is 5L, and is filled with air at 100 bars and 25℃. The right part is 45L, and is kept at vacuum. If the

diaphragm ruptures and the whole chamber is filled with air, find the final pressure and temperature of the air in the chamber. Assume that air is an ideal gas with constant heat capacity. Q U W

0Q , 0W

2 1U U

2 1T T = 298K

12 1

2

VP P

V = 10 bars

------------------------------------------------------------------------------------------------- Example. Ammonia is compressed isothermally at 100℃ from the pressure of 1 bar

to the pressure of 100 bar. Calculate work and the associated heat transfer of the process assuming that ammonia is an ideal gas.

Ideal gas： 2

1

lnRT v

W Pdv dv RTv v

1v 1.824 m3/kg， 2v 0.01824 m3/kg，W -838.25 kJ/kg，Q -838.25 kJ/kg

------------------------------------------------------------------------------------------------------ (3.5.1.2). Van der Waals For van der Waals gas, the effect of pressure on internal energy is no longer zero.

2

RT aP

v b v

v

P R

T v b

2 2T v

u P RT RT a aT P

v T v b v b v v

2v

adu c dT dv

v

2 1

( )v

a au c dT

v v


2

2 0 1

1

* * * *2 1 2 2 2 1 1 1 0

2 1

[ ] [ ] [ ]T

T T v v T T v

T

a au u u u u u u u c dT

v v


diaphragm ruptures and the whole chamber is filled with air, find the final pressure and temperature of the air in the chamber. Assume that air is a van der Waals gas with constant heat capacity. Q U W

0Q , 0W

2 1U U 2

1

0 022 1

1 1( ) 0

v

v v

v

au c T dv c T a

v v v

0 2 1

1 1( )

v

aT

c v v

cP 37.7 bar, cT 133 K 2 227

64c

c

R Ta

P 0.163 kPa(m3/kg)2，

1

8c

c

RTb

P =0.001266 m3/kg

2

RT aP

v b v

P1= 100 bar，T1 = 298 K，v1= 0.0081246 m3/kg

v2= 0.081246 m3/kg

0 2 1

1 1( )

v

aT

c v v = -25.2 ℃

T2 = 273 K，P2= 9.549 bar

0 2 1

1 1( )

v

aT

c v v <0

Comment: The volume that gas occupies has been increased but the number of


molecules remains the same. The average distance between molecules is thus increased, and so is the intra molecular potential energy. However, since the internal energy keeps the same, the kinetic energy should be reduced to compensate the increase in potential energy. As a result, temperature would become lower. ------------------------------------------------------------------------------------------------- Example. Ammonia is compressed isothermally at 100℃ from the pressure of 1 bar

to the pressure of 100 bar. Calculate the work of the process assuming that ammonia is a van der Waals gas with constant heat capacity. NH3， cT 406 K， cP 112.8 bar，R= 0.488 kJ/kg-K

a 1.467603

2( )m

kPakg

，b 0.0021953m

kg

van der Waals gas：2

RT aP

v b v

T1=373 K, P1=1 bar, 1v 1.815 m3/kg T2=373 K, P2=100 bar, 2v 0.0039246 m3/kg

22

1 2 1

1 1ln

RT a v bw Pdv dv RT a

v b v v b v v

w -1265.93 + 369.57 = -896.36 kJ/kg 2

1

2 1 02 1 1 2

1 1( )

T

v

T

a au u c dT a

v v v v = -373.14 kJ/kg

It is noted that the variation in internal energy at the same temperature is very much.

2 22 1

1 2 1 2 1 1

1 1 1 1( ) ln ln

v b v bq u u w a RT a RT

v v v b v v v b

=-1265.93 kJ/kg

------------------------------------------------------------------------------------------------------ (3.5.1.3). Redlich-Kwong gas For Redlich-Kwong gas, the effect of pressure on internal energy is as following.

1/ 2 1/ 2

1 1 1( )

( )

RT a RT aP


3 / 2

1

2 ( )v

P R a

T v b v v b T

1/ 2 1/ 2 1/ 2

1 3

2 ( ) ( ) 2 ( )T v

u P RT a RT a aT P

v T v b v v b T v b v v b T v v b T


1/ 2

3

2 ( )v

adu c dT dv

v v b T

1/ 2 1/ 2

3 3[ ] ln

2 ( ) 2

v v

v

P a a vT P dv dv

T v v b T bT v b

*1/ 2

3ln

2

a vu u

bT v b

2

1

2 12 1 01/ 2 1/ 2

2 2 1 1

3 3ln ln

2 2

T

v

T

a v a vu u c dT

bT v b bT v b


diaphragm ruptures and the whole chamber is filled with air, find the final temperature of the air in the chamber. Assume that air is a Redlich-Kwong gas.

cP 37.7 bar, cT 133 K 2 227

64c

c

R Ta

P 0.163 kPa(m3/kg)2，

1

8c

c

RTb

P =0.001266 m3/kg

2

RT aP

v b v

P1= 100 bar，T1 = 298 K，v1= 0.0081246 m3/kg

v2= 0.081246 m3/kg

v

P R

T v b

，

2v

P aT P

T v

2

1

0 022 1

1 1( ) 0

v

v v

v

au c T dv c T a

v v v

0 2 1

1 1( )

v

aT

c v v = -25.2 ℃

T2 = 273 K，P2= 9.549 bar

0 2 1

1 1( )

v

aT

c v v <0

Comment: The volume that gas occupies has been increased but the number of molecules remains the same. The average distance between molecules is thus


increased, and so is the intra molecular potential energy. However, since the internal energy keeps the same, the kinetic energy should be reduced to compensate the increase in potential energy. As a result, temperature would become lower. ------------------------------------------------------------------------------------------------- Assignment 3.6. An insulated rigid chamber is divided into two parts with a diaphragm. The diaphragm neither moves nor conducts heat. The left part is 5L, and is filled with air at 100 bars and 25℃. The right part is 45L, and is kept at vacuum. If the diaphragm ruptures and the whole chamber is filled with air, find the final temperature of the air in the chamber. Assume that air is a Redlich-Kwong gas. ------------------------------------------------------------------------------------------------------ Example. Ammonia is compressed isothermally at 100℃ from the pressure of 1 bar

to the pressure of 100 bar. Calculate the work of the process assuming that ammonia is a Redlich-Kwong gas.

2 2 11/ 2

1 1 2

1 ( )ln ln

( )

v b a v b vw RT

v b b T v b v

2 1 2 12 1 1/ 2 1/ 2 1/ 2

2 2 1 1 2 1

3 3 3 ( )ln ln ln

2 2 2 ( )

a v a v a v v bu u

T v b T v b T v b v

2 1 2 2 1 2 1 21/ 2 1/ 2 1/ 2

2 1 1 1 2 2 1 1

3 ( ) 1 ( ) 1 ( )ln ln ln ln ln

2 ( ) ( ) 2 ( )

a v v b v b a v b v a v v b v bq RT RT

T v b v v b b T v b v T v b v v b

=-1301.50 +423.99/2 = -1089.51 kJ/kg ------------------------------------------------------------------------------------------------------ (3.5.1.4). Berthelot gas For Berthelot gas, the effect of pressure on internal energy is as following.

2

RT aP

v b Tv

2 2v

P R a

T v b T v

2 2 2

2

v

P RT a RT a aT P

T v b Tv v b Tv Tv

2

2v v

v

P adu c dT T P dv c dT dv

T Tv

2v

au c dT

Tv


* * * *2 1 2 2 2 1 1 1 0

2 2 1 1

2 2( ) v

a au u u u u u u u c dT

T v T v


diaphragm ruptures and the whole chamber is filled with air, find the final temperature of the air in the chamber. Assume that air is a Berthelot gas with constant heat capacity.

cP 37.7 bar, cT 133 K 2 227

64c

c

R Ta

P 0.163 kPa(m3/kg)2，

1

8c

c

RTb

P =0.001266 m3/kg

2

RT aP

v b v

P1= 100 bar，T1 = 298 K，v1= 0.0081246 m3/kg

v2= 0.081246 m3/kg

v

P R

T v b

，

2v

P aT P

T v

02 2 1 1

2 20v

a au c T

T v T v

1 1 1 2

0 2 2 1 1 0 1 2 1 1 0 1 1 1 2

2 1 1 2 1 1 2 ( )( ) ( )

( ) ( )v v v

a a a T v T T vT

c T v T v c T T v T v c T T T v v

1 1 1 2 1 1 1 20 0

2 2( ) ( ) 0

v v

a aT T T T v v T v T T v

c c

21

0 1 1 0 1 2

2 1 2 1 1( ) ( ) ( ) 0

v v

a aT T T

c v T c v v

21 1

0 1 1 0 1 1 0 1 2

1 2 1 2 1 2 1 1( ) ( ) 4 ( )

2 v v v

a a aT T T

c v T c v T c v v

------------------------------------------------------------------------------------------------------ Assignment 3.7. An insulated rigid chamber is divided into two parts with a diaphragm. The diaphragm neither moves nor conducts heat. The left part is 5L,


and is filled with air at 100 bars and 25℃. The right part is 45L, and is kept at vacuum. If the diaphragm ruptures and the whole chamber is filled with air, find the final temperature of the air in the chamber. Assume that air is a Berthelot gas. ------------------------------------------------------------------------------------------------------


(3.5.2). h calculations Choose T and P as independent variables.

( , )s s T P

P T

s sds dT dP

T P

From the Maxwell’s relationship, it is known that T P

S v

P T

P P

s vds dT dP

T T

( , )h h T P

P T

h hdh dT dP

T P

pP

hc

T

， p

T

hdh c dT dP

P

P P P P

s v s vdh Tds vdP T dT dP vdP T dT v T dP

T T T T

pP

sc T

T

，

T P

h vv T

P T

pP

vdh c dT v T dP

T

(1). For ideal gas, the effect of pressure on enthalpy is zero. Pv RT

0P

v Rv T v T

T P

0pdh c dT

(2). For van der Waals gas, the effect of pressure on enthalpy is no longer zero.

2

RT aP

v b v


2

1( )( )

aT P v b

v R

2 3 3

1

( ) 2 ( ) 2( )P

P

v R Ra a RT aTT P v b v bv v v b vv

2

3 3

2( )

2( ) 2( )P

v aRT v b RTv RT v b vv T v

RT a RT aT v b v bv b v v b v

2

3

2( )

2( )

b aRT v b

v b vRT a

v bv b v

2

3

2( )

2( )p

b aRT v b

v b vdh c dT dPRT a

v bv b v

------------------------------------------------------------------------------------------------- Example：Air at 300 K and 10 bars flows through an adiabatic throttle and expands

to atmospheric pressure. Find the temperature behind the throttle assuming that air is a van der Waals gas.

2 1 0h h h 2

1

2 12 1 0

2 2 1 1

2 20

T

p

T

a bRT a bRTh h c dT

v v b v v b

0 2 0 12 1 2 1

2 2( ) ( )p p

bR bR a ac T c T

v b v b v v

0 11 2 1

2

02

2 2( )p

p

bR a ac T

v b v vT

bRc

v b

For air, cP 37.7 bar, cT 133 K

2 227

64c

c

R Ta

P 0.163 kPa(m3/kg)2，

1

8c

c

RTb

P =0.001266 m3/kg


0pc 1.0045 kJ/kg-K

11 2

1 1

RT aP

v b v

P1= 10 bar，T1 = 300 K，v1= 0.08549 m3/kg

Assume v2= 0.861 m3/kg, T2 = 297.7 K P2= 1 bar，T2 = 297.7 K， v2= 0.8538 m3/kg

T -2.3 K If the pressure is raised to 100 bars, P1= 100 bar，T1 = 300 K，v1= 0.008194 m3/kg

T2 = 262.7.7 K, T -37.3 K ------------------------------------------------------------------------------------------------- (3). For Redlich-Kwong gas, the effect of pressure on internal energy is as following.

1/ 2 1/ 2

1 1 1( )

( )

RT a RT aP


2 3/ 2 1/ 2 2 2

1 1 1 1 1 1( ) ( ) 0

( ) 2 ( )

RdT RTdv a dT adP dv

v b v b b T v v b b T v v b

3 / 2 2 1/ 2 2 2

1 1 1 1 1 1 1( ) ( )

2 ( ) ( )

R a RTdv adT dv

v b b T v v b v b b T v v b

3 / 2

2 1/ 2 2 2

1 1 1 1( )

21 1 1

( )( ) ( )

P

R av v b b T v v b

RT aTv b b T v v b

2 1/ 2 21/ 2

2 1/ 2 2 2 2 1/ 2 2 2

1 1 3 3 11 1 1 1 ( )( )( ) 2 ( ) 22

1 1 1 1 1 1( ) ( )

( ) ( ) ( ) ( )P

bRT a v bRT av v b b T v b vv b b T v v bv T v

RT a RT aTv b b T v v b v b b T v v b

2 1/ 2 2

2 1/ 2 2 2

1 1 3 3 1( )

( ) 2 ( ) 21 1 1

( )( ) ( )

p

bRT a v bv b b T v b v

dh c dT dPRT a

v b b T v v b

Calculation of h :


2 1 pP

vh h c dT v T dP

T

2 0 1

* * * *2 1 2 2 2 1 1 1[ ] [ ] [ ]T T P P T Th h h h h h h h

2

*2 2 2

0

[ ]P

P

vh h v T dP

T

2

1

* *2 1 0

T

p

T

h h c dT

1

*1 1 1

0

[ ]P

P

vh h v T dP

T

2 2 1

1

2 1 2 0 1

0 0

[ ] [ ]P T P

pP PT

v vh h v T dP c dT v T dP

T T

For ideal gas, the effect of pressure on enthalpy is zero.

2

1

2 1 0

T

p

T

h h c dT

For van der Waals gas, the effect of pressure on enthalpy is no longer zero.

2

3

2( )

2( )p

b aRT v b

v b vdh c dT dPRT a

v bv b v

For the isothermal process, we have 0dT

2 3 3 2

2 2

( ) ( )

RdT RTdv adv a RTdP dv

v b v b v v v b

2

3 2 2 2

3

2( ) 22

( ) ( )2( )

b aRT v b a RT a bv b vdh dv RT dv

RT a v v b v v bv bv b v

*2 2

0

2[ ] 2

( )

P v

P

v a b a bRTh h v T dP RT dv

T v v b v v b

2

1

2 12 1 0

2 2 1 1

2 2T

p

T

a bRT a bRTh h c dT

v v b v v b


The other way of derivation is as following. 2

1

2 2 1 12 1 2 2 2 1 1 1 0

2 1 2 2 1 1

( ) ( ) ( ) ( )T

v

T

a a v RT a v RT ah h u P v u Pv c dT

v v v b v v b v

2

1

2 10 2 1

2 1 2 2 1 1

T

v

T

a a bRT a bRT ac dT RT RT

v v v b v v b v

2

1

2 10

2 2 1 1

2 2T

p

T

a bRT a bRTc dT

v v b v v b

For Redlich-Kwong gas, the effect of pressure on enthalpy is as following.

2 1/ 2 2

2 1/ 2 2 2

1 1 3 3 1( )

( ) 2 ( ) 21 1 1

( )( ) ( )

p

bRT a v bv b b T v b v

dh c dT dPRT a

v b b T v v b


2 1/ 2 2 2 2 1/ 2 2 2

1 1 1 1 1 1( ) ( )

( ) ( ) ( ) ( )

RTdv a RT adP dv dv


2 1/ 2 2

2 1/ 2 2

2 1/ 2 2 2

1 1 3 3 1( )

1 1 3 3 1( ) 2 ( ) 2( )

1 1 1 ( ) 2 ( ) 2( )( ) ( )

bRT a v bbRT a v bv b b T v b v

dh dP dvRT a v b b T v b v

v b b T v v b

*2 1/ 2 2 1/ 2

1 1 3 3 1 1 3( ) ( ln )

( ) 2 ( ) 2 2

v bRT a v b bRT a v b bh h dv

v b b T v b v v b b T v v b

2

1

2 2 1 12 1 01/ 2 1/ 2

1 2 2 2 1 1 1 1

1 3 1 3( ln ) ( ln )2 2

T

p

T

bRT a v b b bRT a v b bh h c dT


The other way of derivation is as following.

2

1

2 12 1 01/ 2 1/ 2

2 2 1 1

3 3ln ln

2 2

T

v

T

a v a vu u c dT

bT v b bT v b

2

1

2 12 1 0 2 1 11/ 2 1/ 2

2 2 1 1

3 3ln ln

2 2

T

v v

T

a v a vh h c dT P v Pv

bT v b bT v b

2 2 1 12 1 1 1/ 2 1/ 2

2 2 2 1 1 1

1 1(1 ) (1 )v

RTv a v RTv a vP v Pv

v b b T v b v b b T v b


2 12 1 1/ 2 1/ 2

2 1 2 2 1 1

1 1RT b RT b a b a bRT RT

v b v b b T v b b T v b

2

1

2 2 1 12 1 01/ 2 1/ 2

2 2 2 2 1 1 1 1

1 3 1 3( ln ) ( ln )2 2

T

p

T

RT b a v b RT b a v bh h c dT

v b b T v b v b v b b T v b v b

------------------------------------------------------------------------------------------------- Assignment 3.8. Air at 300 K and 10 bars flows through an adiabatic throttle and expands to atmospheric pressure. Find the temperature behind the throttle assuming that air is a Redlich-Kwong gas. ------------------------------------------------------------------------------------------------------ (3.5.3). s calculations (1). T and v as independent variables This is used to calculate the entropy change in a positive displacement type compressor in which compression process occurs in a confined chamber.

v v

v v

c P c P R Rds dT dv dT dv dv

T T T T v v

2 1 lnv

v

c P Rs s dT dv R v

T T v

2 0 1

* * * *2 1 2 2 2 1 1 1[ ] [ ] [ ]T T v v T Ts s s s s s s s

2

0

* 22 2

0

lnv

vv

P R vs s dv R

T v v

2

1

* * 02 1

T

v

T

cs s dT

T

1

0

* 11 1

0

lnv

vv

P R vs s dv R

T v v

2 2 1

0 1 0

0 22 1

1

lnv T v

v

v vv T v

P R c v P Rs s dv dT R dv

T v T v T v

For ideal gas, the entropy change is as following.


RTP

v

0v

P R R R

T v v v

2

1

0 22 1

1

lnT

videal

T

c vs s dT R s

T v

For van der Waals gas, the entropy change is as follows.

2

RT aP

v b v

v

P R

T v b

v

P R R R

T v v b v

1

1 0 1

0 1 1

( ) ( )[ ] ln ln

( )

v

v

P R v b v v bdv R R

T v v b v v

2

1

0 2 2 1 2 12 1

1 1 2 1 2

( ) ( )ln ln ln

( ) ( )

T

videal

T

c v v b v v b vs s dT R R s R

T v v b v v b v

------------------------------------------------------------------------------------------------- Example：An adiabatic piston and cylinder system is filled with air at 25℃, 1 bar

with the volume of 20 L. Air is compressed isentropically to the volume of 1 L. Find the work of the process assuming that air is a van der Waals gas. The specific heat of air is expressed as the following.

2 3 4pcT T T T

R

where 3.653 , 31.337 10 , 63.294 10 , 91.913 10 , 120.276 10

2 3 4( 1 )v pc c R T T T T R

2

1

2 2 3 3 4 40 22 1 2 1 2 1 2 1

1

( 1) ln ( ) ( ) ( ) ( ))2 3 4

T

v

T

c TdT R T T T T T T T T

T T


2 2 3 3 4 421 2 2 2

1

( 1) ln ( 1) ( 1) ( 1) ( 1))2 3 4

TR T T T T

T

2 12 1

1 2

( )ln 0

( )ideal

v b vs s s R

v b v

2 2 3 3 4 42 1 21 1 1 1

1 2 1

( )ln ( 1)ln ( 1) ( 1) ( 1) ( 1) 0

( ) 2 3 4

v b v TT T T T

v b v T

------------------------------------------------------------------------------------------------- For Redlich-Kwong gas, the entropy change is as follows.

1/ 2( )

RT aP

v b v v b T

3 / 2

1

2 ( )v

P R a

T v b v v b T

3 / 2

1 1 1 1( )

2v

P R R R a

T v v b v b T v v b

1

1 0 0 1 1 13/ 2 3/ 2

0 1 1 0 1 1

( ) 1 1 ( ) ( ) 1 1[ ] ln ln ln ln

( ) 2 ( ) 2 ( )

v

v

P R v b v a v b v v b a vdv R R

T v v b v b T v b v v b T v b

2

1

0 2 2 1 2 12 1 3/ 2 3/ 2

1 1 2 2 2 1 1

( ) 1 1 1 1ln ln ln ln

( ) 2 ( ) 2 ( )

T

v

T

c v v b v a v a vs s dT R R

T v v b v b T v b b T v b

2 1 2 13/ 2 3/ 2

1 2 2 2 1 1

( ) 1 1 1 1ln ln ln

( ) 2 ( ) 2 ( )ideal

v b v a v a vs R

v b v b T v b b T v b

------------------------------------------------------------------------------------------------- Example：An adiabatic piston and cylinder system is filled with air at 25℃, 1 bar

with the volume of 20 L. Air is compressed isentropically to the volume of 1 L. Find the work of the process assuming that air is a Redlich Kwonggas. The specific heat of air is expressed as the following.

2 3 4pcT T T T

R

where 3.653 , 31.337 10 , 63.294 10 , 91.913 10 , 120.276 10


2 3 4( 1 )v pc c R T T T T R 2

1

2 2 3 3 4 40 22 1 2 1 2 1 2 1

1

( 1) ln ( ) ( ) ( ) ( ))2 3 4

T

v

T

c TdT R T T T T T T T T

T T

2 2 3 3 4 421 2 2 2

1

( 1) ln ( 1) ( 1) ( 1) ( 1))2 3 4

TR T T T T

T

2

1

2 1 2 1 0 22 1 3/ 2 3/ 2

1 2 2 2 1 1 1

( ) ( ) ( )ln ln ln ln 0

( ) 2 2

T

v

T

v b v a v b a v b c vs s R dT R

v b v bT v bT v T v

2 2 13/ 2 3/ 2

1 1 2 1

( ) 1 1 1 ( ) ( )ln ln ln

( ) 2

v b a v b v b

v b b RT v v

2 2 3 3 4 421 1 1 1

1

( 1) ln ( 1) ( 1) ( 1) ( 1) 02 3 4

TT T T T

T

-------------------------------------------------------------------------------------------------


(2). T and P as independent variables This is used to calculate the entropy change in an axial type compressor in which compression process occurs in a continuous flow.

P

P

c vds dT dP

T T

2 1p

P

c vs s dT dP

T T

2 0 1

* * * *2 1 2 2 2 1 1 1[ ] [ ] [ ]T T p p T Ts s s s s s s s

2

0

*2 2 [ ]

P

PP

R vs s dP

P T

2

1

0* * 22 1

1

lnT

p

T

c Ps s dT R

T P

1

0

*1 1 [ ]

P

PP

R vs s dP

P T

2 2 1

0 1 0

0 22 1

1

[ ] ln [ ]P T P

p

P PP T P

cR v P R vs s dP dT R dP

P T T P P T

For ideal gas, the entropy change is as following.

RTP

v

0P

R v R R

P T P P

2

1

0 22 1

1

lnT

pideal

T

c Ps s dT R s

T P

For van der Waals gas, the entropy change is as follows.

2

RT aP

v b v

2( )

v b aT P

R v

2 3 2 3

1 2( ) 2 2

P

T a v b a T a b aP

v R v R v v b R v R v


2 3

2 ( )P

R v R RRT a RT a v bP T

v b v v b v

2 3

2

( )T

P RT a

v v b v

2 3

2

( )

RT adP dv

v b v

2 3

2 3

2[ ] [ ][ ]

2 ( ) ( )P

R v R R RT adP dv

RT a RT a v bP T v b vv b v v b v

=2 3

2

2[ ]

( )[ ]

RT aR

Rv b vdv

RT a v bv b v

=3 2

3

2 ( )[1 ]

( )

R RTv a v bdv

v b RTv a v b v

=2 2 2

1 2 12

( ) ( ) ( )

a v b a dv dvdv b

a a aT v T vv v b v v b v v bRT RT RT

How to evaluate the integration of 2 ( )

dva

v v bRT

and

2

1

( )

dvav v v b

RT

?

2X a bx cx

24q ac b

12 2tan

dx cx b

X q q

, ( 0)q

22ln

2

cx b qdx

X q cx b q

, ( 0)q

21ln

2 2

dx x b dx

xX a X a X

2 2( )a a ab

v v b v vRT RT RT


24 4 1 4

1 14

qTab a ab a abq

RT RT RT bR T RT T

2 227

64c

c

R Ta

P ,

1

8c

c

RTb

P

2 22764 2714 3248

c

cq c

c

c

R Ta P

T TRTbR RP

The van der Waals constants

a bar(m3/kmole)2 b m3/kmole qT (K)

Air 1.368 0.0367 112 C4H10 13.86 0.1162 358 CO2 3.647 0.0428 256 CO 1.474 0.0395 112 CH4 2.293 0.0428 162 N2 1.366 0.0386 106 O2 1.369 0.0317 130 C3H8 9.349 0.0901 312 R12 10.49 0.0971 324 SO2 6.883 0.0569 364 H2O 5.531 0.0305 546

It is noted that qT is very low for air such that the value of q is positive for most

applications. In the case that the value of q is positive, the integration can be displaced as the

following.

qT T , 0q

12 2tan

dx cx b

X q q


1

2

22tan

4 4( ) 1 1q q

avdv RT

a T Tab abv v bRT RT T RT T

1

2

212

tan( ) 1 1q

q

vRTa dv R a

aT T Tv v bRT T T

21

2 2

121 1 2ln tan

2 4 4( ) ( ) 1 1q q

avdv v b RT

a ab av T Tab abv v b v v bRT RT RT RT T RT T

21

2 2

212 1

ln tan( ) ( ) 1 1q q

vRTab dv v R aR

a aT v T Tv v b v v bRT RT T T

2 2

1[ ] 2

( ) ( )P

R v a dv dvdP b

a aP T T vv v b v v bRT RT

21 1

2

2 21 12

tan ln tan( )1 1 1 1q q q

q

vRT vRTR v Ra aR

aT T T Tv v bRTT T T T

2

2ln

( )

vR

av v b

RT

2 22 1

2 12 22 2 1 1

2 1

ln ln( ) ( )

ideal

v vs s s R R

a av v b v v b

RT RT

2

1

21 12

0 2 2 12

21 12 2

2

( )ln ln

( )

Tp

T

av v b

c P v RTdT R R

aT P v v v bRT

qT T , 0q


2 ( )( )I II

a abv v v v v v

RT RT

1 41 1

2I

a bv RT

RT a

,

1 41 1

2II

a bv RT

RT a

41I II

a bv v RT

RT a

2

1 1ln

( )( )( )

I

I II I II I II I II II

dv dv dv dv v va v v v v v v v v v v v v v vv v b

RT

1ln

41

I

II

v v

v va bRT

RT a

2

1 1 1( )

( ) I II I II

dv dv dvdv

av v v v v v v vv v bRT

1 1 1 1 1 1 1( ) ( )

I II I I II II

dvv v v v v v v v v v

1 1 1

ln ln )I II

I II I II

v v v v

v v v v v v

1 1 1ln ln )

41

I II

I II

v v v v

v v v va bRT

RT a

2

[ ] ln ln ln )4 4

1 1

I I II

P II I II

R v R v v R b v v b v vdP

P T v v v v v vb bRT RT

a a

= ln 2 ln ln )4

1

I I II

II I II

R v v b v v b v v

v v v v v vbRT

a

2 2 2 2 2 22 1

2 2 2 2 2 22

ln 2 ln ln )4

1

I I IIideal

II I II

R v v b v v b v vs s s

v v v v v vbRT

a


1 1 1 1 1 1

1 1 1 1 1 11

ln 2 ln ln )4

1

I I II

II I II

R v v b v v b v v

v v v v v vbRT

a

For Redlich-Kwong gas, the entropy change is as follows.

1/ 2( )

RT aP

v b v v b T

2 3/ 2 1/ 2 2 2

1 1 1 1 1 1( ) ( ) 0

( ) 2 ( )

RdT RTdv a dT adP dv

v b v b b T v v b b T v v b

3 / 2

2 1/ 2 2 2

1 1 1 1( )

21 1 1

( )( ) ( )

P

R av v b b T v v b

RT aTv b b T v v b

3 / 2

1/ 2 2 1/ 2 2 2

1 1 1 1( )

21 1 1

( )( ) ( ) ( )

P

R aR v R v b b T v v b

RT a RT aP Tv b v v b T v b b T v v b


2 1/ 2 2 2 2 1/ 2 2 2

1 1 1 1 1 1( ) ( )

( ) ( ) ( ) ( )

RTdv a RT adP dv dv


3 / 2

1/ 2 2 1/ 2 2 2

1 1 1 1( )

21 1 1

( )( ) ( ) ( )

P

R aR v R v b b T v v bdP dP

RT a RT aP Tv b v v b T v b b T v v b

=

3/ 2

2 1/ 2 2 2

1/ 2 2 1/ 2 2 2

1 1 1 1( ) 1 1 12 ( )

1 1 1 ( ) ( )( )( ) ( ) ( )

R aR RT av b b T v v b dv

RT a RT a v b b T v v bv b v v b T v b b T v v b

2 1/ 2 2 2

3/ 2

1/ 2

1 1 1( )

1 1 1 1( ) ( )( )

2( )

RT aR av b b T v v b

R dvRT a v b b T v v b

v b v v b T


------------------------------------------------------------------------------------------------- Example：Air is compressed with an adiabatic compressor from 25℃, 1 bar to 20

bars. Find the reversible work of the compressor. (1). Assume that air is an ideal gas. (2). Assume that air is a Van der Waals gas. For ideal gas:

0pc dT Rdpds

T P

2

1

ln 0pc dT Ps ds R

T P

The final temperature can be obtained by solving the equation above. For van der Waals gas:

2 2 2 2 2 22 1

2 2 2 2 2 22

ln 2 ln ln )4

1

I I IIideal

II I II

R v v b v v b v vs s s

v v v v v vbRT

a

1 1 1 1 1 1

1 1 1 1 1 11

ln 2 ln ln )4

1

I I II

II I II

R v v b v v b v v

v v v v v vbRT

a

=0

2 2 3 3 4 42 21 2 2 2

1 1

ln ( 1) ( 1) ( 1) ( 1)) ln2 3 4

T PR T T T T R

T P

2 2 2 2 2 2

2 2 2 2 2 22

ln 2 ln ln )4

1

I I II

II I II

R v v b v v b v v

v v v v v vbRT

a

1 1 1 1 1 1

1 1 1 1 1 11

ln 2 ln ln )4

1

I I II

II I II

R v v b v v b v v

v v v v v vbRT

a

=0

------------------------------------------------------------------------------------------------- Assignment 3.9. Methane is compressed isentropically from 1 bar and 27℃ to the pressure of 80 bars, and cooled isobarically to the temperature of 27℃. The

compressed methane then flows through an expansion valve to the pressure of 1 bar.


Find the work of compressor and the final temperature of methane assuming that methane is a van der Waals gas. The specific heat of methane is expressed as the following.

2 3 4pcT T T T

R

where 3.826 , 33.979 10 , 624.558 10 , 922.733 10 , 126.963 10

Ideal gas Real gas Van der Waals gas T1 300 K P1 1 bar H1 627.58 T2 P2 80 bars H2 T3 300 K P3 80 bars h3 548.15 T4 263.8 K P4 1 bar h4 548.15 Wc Qc ------------------------------------------------------------------------------------------------------


(3.6). Joule-Thomson Effect There are two ways to reduce temperature with mechanical device, the constant enthalpy expansion and the constant enthalpy expansion.

Expansion process is the critical process in a refrigeration system. Refrigerant may cool down through expansion. Without the expansion process, cooling effect may not be achieved. However, expansion process does not guarantee a decrease in temperature. It really depends on the operating conditions.

The enthalpy remains the same for the expansion process. For ideal gas,

temperature would not change as enthalpy remains the same. As a result, ideal gas could never be used as a refrigerant.

( )h h T

h1 T h2 h3 P (3.6.1). Joule-Thomson Coefficient The effectiveness of a refrigerant is judged by the temperature change during a


process in which enthalpy remains the same. The Joule-Thomson Coefficient is defined as

Jh

T

P

0J ， 0P ， 0T ，temperature would decrease by expansion

0J ， 0P ， 0T ，temperature would increase by expansion 0J ，the range that refrigerant would work

0pP

vdh c dT v T dP

T

1

Pp

vdT v T dP

c T

1J

h Pp

T vT v

P c T

ZRTv

P

2 2

[ ]r

c r

p p p c r r p

v ZRT ZR RT Z RT Z RT RT Zv T T

T P P P T P T P P T

2 21

r

c rJ

p c r r p

R T T Z

c P P T

0, 0r

Jr p

Z

T

(3.4.3.2). van der Waals gas

2

RT aP

v b v

2 3

20

( )

RdT RTdv adP dv

v b v b v

2 3

2[ ]( )

R RT adT dv

v b v b v


2 3

2( )

P

Rv v b

RT aTv b v

2 2

2 3 2 3

21 1 1 ( )

2 2( ) ( )

JPp p p

a bRTRTv v v bv bT v v

RT a RT ac T c cv b v v b v

2 2

2 2

1 2 ( )2

( )J

p

a v b bRTvac RTv v bv

------------------------------------------------------------------------------------------------- Example：Air at 300 K and 10 bars flows through an adiabatic throttle and expands

to atmospheric pressure. Find the Joule-Thomson Coefficient assuming that air is a van der Waals gas. ------------------------------------------------------------------------------------------------- Assignment 3.9. Air at 300 K and 10 bars flows through an adiabatic throttle and expands to atmospheric pressure. Find the Joule-Thomson Coefficient assuming that air is a Redlich-Kwong gas. ------------------------------------------------------------------------------------------------------


(3.4.3.3). Joule-Thomson inversion temperature

For all real gases, the Joule Thomson coefficient will equal zero at some point called the inversion point. At the inversion point, the Joule-Thomson coefficient changes sign. The Joule-Thomson inversion temperature depends on the pressure of the gas before expansion.

Helium and hydrogen are two gases whose Joule-Thomson inversion temperatures at one atmosphere are very low (e.g., about −222 °C for helium). Thus, helium and


hydrogen will warm when expanded at constant enthalpy at typical room temperatures.

In order to liquefy a gas by a Joule-Thompson expansion the gas must first be cooled to below the J-T inversion temperature. Some inversion temperatures are:

He: 40 K , N2: 621 K, O2: 764 K, Ne: 231 K

We see that N2 and O2 will cool upon expansion at room temperature, but He and Ne will warm upon expansion at room temperature.

The locus of inversion curve

10J

h Pp

T vT v

P c T

P

vT v

T

For van der Waals gas

2 2

2 2

1 2 ( )2

( )J

p

a v b bRTvac RTv v bv

0J ，2 2

2

( )

bRT a

v b v

22(1 )

a bRT

b v

22 2

1 2(1 ) (2 3 )

RT a a b a a bP

v b v v b b v v vb v

223( ) 2( ) 0

b b b P

v v a

2 227

64c

c

R Ta

P ，

1

8c

c

RTb

P

22

2 2

1( )8

27 2764

cr c

c r

c

c

RTP P

b P P PR Ta

P


23( ) 2( ) 027

rb b P

v v ,

1 19

3

rPb

v

2 2

2 2 2

272

2 64 27(1 ) (1 ) (1 )

1 48

c

cc

c

c

R Ta b P b b

T TRTbR v v vRP

2 2 21 1

27 27 39(1 ) (1 ) (2 1 )4 4 3 4 9

r

rr

Pb P

Tv

rP rHT rLT

9 3.0 3.0 8 4.08 2.08 7 4.58 2.85 6 4.98 1.52 5 5.33 1.33 4 5.65 1.18 3 5.95 1.05 2 6.23 0.94 1 6.49 0.84 0 6.75 0.75

9rP , 3.0rT . This is the limit that refrigerant may work.

0rP , 3 27

,4 4rT . This is the limit that refrigerant may work.

He: 40 K , 3 cT 7.8 K

N2: 621 K , 3 cT 378.6 K

O2: 764 K , 3 cT 463.8 K

Ne: 231 K , 3 cT 133.2 K

Chapter Three Thermodynamic Relationships and Evaluations ...€¦ · Advanced Thermodynamics, ME...

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Transcript of Chapter Three Thermodynamic Relationships and Evaluations ...€¦ · Advanced Thermodynamics, ME...