Course No: ME 5243- Advanced Thermodynamics

Estimating Different Thermodynamic Relations using Redlich- Kwong-Soave Equation of State. Final Project report

Abu Saleh Ahsan,Md. Saimon Islam, Syed Hasib Akhter Faruqui 5-5-2016

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Table of Contents Nomenclature: .............................................................................................................................................. 2

Abstract: ........................................................................................................................................................ 4

Introduction: ................................................................................................................................................. 5

Derivation...................................................................................................................................................... 6

(a) Evaluation of the Constants ................................................................................................................. 6

(b) Equation of State in Reduced Form ..................................................................................................... 9

c) Critical Compressibility Factor ............................................................................................................ 10

d) Express Z in terms TR, vR’: .................................................................................................................. 12

e) Accuracy of EOS from Equation (d) ..................................................................................................... 13

f) Equation for Departure ....................................................................................................................... 15

𝒉 ∗ −𝒉𝑹𝑻𝒄 ......................................................................................................................................... 15

(u*-u)/RTc ........................................................................................................................................... 15

𝒔 ∗ −𝒔𝑹 ............................................................................................................................................... 15

g) Accuracy of EOS for equation (C) ........................................................................................................ 16

h) Derivation of Expressions: .................................................................................................................. 17

𝒂 ∗ − 𝒂𝑹𝑻𝒄: ....................................................................................................................................... 17

𝒈 ∗ − 𝒈𝑹𝑻𝒄: ...................................................................................................................................... 17

i) Speed of sound .................................................................................................................................... 18

(j) Derive the Properties .......................................................................................................................... 19

Cp ........................................................................................................................................................ 19

Cv ........................................................................................................................................................ 19

1/v vp .................................................................................................................................. 19

k = v/p pv .................................................................................................................................... 19

kT ......................................................................................................................................................... 20

J ......................................................................................................................................................... 20

Summary: .................................................................................................................................................... 21

Appendix ..................................................................................................................................................... 22

MATLAB Code ......................................................................................................................................... 22

Reference .................................................................................................................................................... 23

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Nomenclature:

P = Pressure

Pr = Reduced pressure

Pc = Critical pressure

v = Specific volume

vr = Reduced specific volume

vc = Critical volume

𝑣𝑟∗ = Specific volume of ideal gas

vrf = Reduced volume at liquid state

vrg = Reduced volume at gaseous state

T = Temperature

Tr = Reduced temperature

Tc= Critical temperature

R = Molar gas constant

Z = Compressibility factor

Zr = Reduced compressibility factor

Zc = Critical compressibility factor

𝑔 = Gibbs free energy

𝑔0 = Gibbs free energy for ideal gas

ℎ = Specific enthalpy for real gas

ℎ0 = Specific enthalpy for ideal gas

𝑢 = Specific internal energy for real gas

𝑢0 = Specific internal energy for ideal gas

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𝑠 = Specific entropy for real gas

𝑠0 = Specific entropy for ideal gas

𝑘𝑇 = Isothermal expansion exponent

𝑐 = Speed of sound

𝛽 =Volumetric co-efficient of thermal expansion

𝐶𝑝 = Constant pressure specific heat

𝐶𝑣 = Constant volume specific heat

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Abstract: For the project we will be using “Redlich- Kwong-Soave” Equation of State (EOS) to derive to Estimating

Different Thermodynamic Relations. Starting from “Redlich- Kwong-Soave” equation we have calculated

the constants “a(T)” & “b”. The EOS is again represented in its reduced form. Compressibility factor for

the selected EOS is calculated and expressed in terms of TR & VR. By using the EOS different thermodynamic

relations such as departure enthalpy, entropy, and change in internal energy has been evaluated. Again,

we have expressed different important parameters such as speed of sound, isothermal expansion

exponent, CP & CV in reduced form. As for the substance in question we are using “Nitrogen”.

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Introduction: Real gases are different from that of ideal gases. Thus the evaluated properties of ideal gas cannot be

used as the same for real gases. To understand the characteristics of real gases we have to consider the

following-

compressibility effects;

variable specific heat capacity;

van der Waals forces;

non-equilibrium thermodynamic effects;

issues with molecular dissociation and elementary reactions with variable composition.

In our project, we have taken “Redlich- Kwong-Soave” equation of state (EOS) into consideration to

derive the various fundamental relations of thermodynamics. The compressibility, enthalpy departure

and entropy departure, can all be calculated if an equation of state for a fluid is known which is

“Nitrogen” in our case.

Now, the “Redlich- Kwong-Soave” equation of state (EOS) is almost similar to Van Der Walls equation of state. The equation is-

𝑝 =𝑅𝑇

𝑣−𝑏−

𝑎(𝑇)

𝑣(𝑣+𝑏) … …. … … … … … … … … (i)

In thermodynamics, a departure function is defined for any thermodynamic property as the difference between the property as computed for an ideal gas and the property of the species as it exists in the real world, for a specified temperature T and pressure P. Common departure functions include those for enthalpy, entropy, and internal energy.

Departure functions are used to calculate real fluid extensive properties (i.e properties which are computed as a difference between two states). A departure function gives the difference between the real state, at a finite volume or non-zero pressure and temperature, and the ideal state, usually at zero pressure or infinite volume and temperature.

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Derivation

(a) Evaluation of the Constants

Taking the first and second derivative of pressure WRT to volume be –

(i) =>

𝛿𝑝

𝛿𝑣)

𝑇= −

𝑅𝑇

(𝑣−𝑏)2+

𝑎(𝑇)

𝑣2(𝑣+𝑏)+

𝑎(𝑇)

𝑣(𝑣+𝑏)2 … … … … … … … … (ii)

And,

𝛿2𝑝

𝛿𝑣2)𝑇

=2𝑅𝑇

(𝑣−𝑏)3+ 𝑎(𝑇) [−

1

𝑣2(𝑣+𝑏)2−

2

𝑣3(𝑣+𝑏)−

1

𝑣2(𝑣+𝑏)2−

2

𝑣(𝑣+𝑏)3]

𝛿2𝑝

𝛿𝑣2)𝑇

=2𝑅𝑇

(𝑣−𝑏)3+ 𝑎(𝑇) [−

2

𝑣2(𝑣+𝑏)2−

2

𝑣3(𝑣+𝑏)−

2

𝑣(𝑣+𝑏)3]

𝛿2𝑝

𝛿𝑣2)𝑇

=2𝑅𝑇

(𝑣−𝑏)3− 𝑎(𝑇) [

2

𝑣2(𝑣+𝑏)2+

2

(𝑣+𝑏)+

2

𝑣(𝑣+𝑏)3]

𝛿2𝑝

𝛿𝑣2)𝑇

=2𝑅𝑇

(𝑣−𝑏)3− 𝑎(𝑇) [

3𝑣2+3𝑣𝑏+𝑏2

𝑣3(𝑣+𝑏)3] … … …. …. …. …. …. …. (iii)

Now at critical point first and second derivative of pressure WRT to volume be zero. So from equation

(ii) and (iii) we can write,

𝛿𝑝

𝛿𝑣)

𝑇= −

𝑅𝑇𝑐

(𝑣𝑐−𝑏)2+

𝑎(𝑇)

𝑣𝑐2(𝑣𝑐+𝑏)

+𝑎(𝑇)

𝑣𝑐(𝑣𝑐+𝑏)2= 0

𝑜𝑟, 0 = −𝑅𝑇𝑐

(𝑣𝑐−𝑏)2+

𝑎(𝑇)

𝑣𝑐2(𝑣𝑐+𝑏)

+𝑎(𝑇)

𝑣𝑐(𝑣𝑐+𝑏)2

𝑜𝑟,𝑅𝑇𝑐

(𝑣𝑐−𝑏)2=

𝑎(𝑇)

𝑣𝑐2(𝑣𝑐+𝑏)

+𝑎(𝑇)

𝑣𝑐(𝑣𝑐+𝑏)2

𝑜𝑟,𝑅𝑇𝑐

(𝑣𝑐−𝑏)2=

𝑎(𝑇) (2𝑣𝑐+𝑏)

𝑣𝑐2(𝑣𝑐+𝑏)2

…. …. .… …. ….. ….. …. … (iv)

And,

𝛿2𝑝

𝛿𝑣2)𝑇

=2𝑅𝑇𝑐

(𝑣𝑐−𝑏)3− 𝑎(𝑇) [

3𝑣𝑐2+3𝑣𝑐𝑏+𝑏2

𝑣𝑐3(𝑣𝑐+𝑏)3

] = 0

𝑜𝑟,2𝑅𝑇𝑐

(𝑣𝑐−𝑏)3= 𝑎(𝑇) [

3𝑣𝑐2+3𝑣𝑐𝑏+𝑏2

𝑣𝑐3(𝑣𝑐+𝑏)3

] …. …. …. …. …. …. …. (v)

Now, Dividing Equation (iv) with equation (v) we get,

(𝑣𝑐 − 𝑏) =(𝑣𝑐+𝑏).𝑣𝑐.(𝑣𝑐+𝑏)

3𝑣𝑐2+3𝑣𝑐𝑏+𝑏2

𝑜𝑟, (𝑣𝑐 − 𝑏)(3𝑣𝑐2 + 3𝑣𝑐𝑏 + 𝑏2) = (𝑣𝑐 + 𝑏). 𝑣𝑐 . (𝑣𝑐 + 𝑏)

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𝑜𝑟, 3𝑣𝑐3 + 3𝑣𝑐

2 + 𝑏2𝑣𝑐 − 3𝑏𝑣𝑐2 − 3𝑣𝑐𝑏

2 − 𝑏3 = 2𝑣𝑐3 + 2𝑣𝑐

2 + 𝑣𝑐2𝑏 + 𝑣𝑐𝑏

2

𝑜𝑟, 𝑏3 + 3𝑏2𝑣𝑐 + 3𝑏𝑣𝑐2 − 𝑣𝑐

3 = 0

𝑜𝑟, (𝑏 + 𝑣𝑐)3 = 2𝑣𝑐

3 = (√23

𝑣𝑐)3

𝑜𝑟, 𝑏 = (√23

− 1)𝑣𝑐 = (√23

− 1)𝑧𝑐𝑅𝑇𝑐

𝑝𝑐

𝑜𝑟, 𝑏 = (√23

− 1).1

3.𝑅𝑇𝑐

𝑝𝑐= 0.08664

𝑅𝑇𝑐

𝑝𝑐

𝑜𝑟, 𝑏 = 0.08664 𝑅𝑇𝑐

𝑝𝑐 …. …. …. …. …. …. …. (vi)

Note: Here, 𝑧𝑐 =1

3. As the original equation is from Redlich-Kwong equation, we will use the value of 𝑧𝑐

from the Redlich-Kwong equation.

Now putting the value of ‘b’ and 𝑣𝑐 = 𝑧𝑐𝑅𝑇𝑐

𝑝𝑐=

1

3.𝑅𝑇𝑐

𝑝𝑐 in equation (iv) we get,

𝑅𝑇𝑐

(𝑣𝑐−𝑏)2=

𝑎(𝑇) (2𝑣𝑐+𝑏)

𝑣𝑐2(𝑣𝑐+𝑏)2

𝑜𝑟, 𝑎(𝑇) = 𝑅𝑇𝑐 𝑣𝑐

2(𝑣𝑐+𝑏)2

(2𝑣𝑐+𝑏) (𝑣𝑐−𝑏)2

𝑜𝑟, 𝑎(𝑇) = 𝑅𝑇𝑐 [𝑧𝑐

𝑅𝑇𝑐𝑝𝑐

]2(𝑧𝑐

𝑅𝑇𝑐𝑝𝑐

+0.08664 𝑅𝑇𝑐𝑝𝑐

)2

(2𝑧𝑐𝑅𝑇𝑐𝑝𝑐

+0.08664 𝑅𝑇𝑐𝑝𝑐

) (𝑧𝑐𝑅𝑇𝑐𝑝𝑐

−0.08664 𝑅𝑇𝑐𝑝𝑐

)2

𝑜𝑟, 𝑎(𝑇) = 𝑅𝑇𝑐 [𝑧𝑐

𝑅𝑇𝑐𝑝𝑐

]2 (

𝑅𝑇𝑐𝑝𝑐

)2 (

1

3+0.08664 )

2

(𝑅𝑇𝑐𝑝𝑐

) (2

3+0.08664) (

𝑅𝑇𝑐𝑝𝑐

)2 (

1

3−0.08664 )

2

𝑜𝑟, 𝑎(𝑇) = [

1

3.𝑅𝑇𝑐𝑝𝑐

]2∗ (0.176377600711111)

(1

𝑝𝑐) (

2

3+0.08664) (

1

3−0.08664 )

2

𝑜𝑟, 𝑎(𝑇) = 0.42747𝑅2𝑇𝑐

2

𝑝𝑐 … …. …. …. …. …. …. …. …. (vii)

Now, this is for the critical point only. As for other points for the assigned substance Nitrogen we get

a(T)= 0.42747𝑅2𝑇𝑐

2

𝑝𝑐(1 + [{0.0007T4 - 0.3012T3 + 45.74036T2 - 3068.87T + 76836.9287}]*T)

at critical point T=TR=1 thus we get again,

a(T)= 0.42747𝑅2𝑇𝑐

2

𝑝𝑐

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Figure-1: a(T) function determination

-50

0

50

100

150

200

250

300

0 20 40 60 80 100 120 140

a(T)

T

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(b) Equation of State in Reduced Form

𝑝 =𝑅𝑇

𝑣−𝑏−

𝑎(𝑇)

𝑣(𝑣+𝑏)

𝑜𝑟, 𝑝𝑟𝑝𝑐 =𝑅𝑇𝑟𝑇𝑐

𝑣𝑟𝑣𝑐−0.08664 𝑅𝑇𝑐𝑝𝑐

−0.42747

𝑅2𝑇𝑐2

𝑝𝑐(1+[{0.0007𝑇4 − 0.3012𝑇3 + 45.74036𝑇2 − 3068.87𝑇 + 76836.9287}]∗𝑇)

𝑣𝑟𝑣𝑐(𝑣𝑟𝑣𝑐+0.08664 𝑅𝑇𝑐𝑝𝑐

)

𝑜𝑟, 𝑝𝑟 =𝑅𝑇𝑟𝑇𝑐

(𝑣𝑟𝑣𝑐−0.08664 𝑅𝑇𝑐𝑝𝑐

)𝑝𝑐

−0.42747

𝑅2𝑇𝑐2

𝑝𝑐(1+[{0.0007𝑇4 − 0.3012𝑇3 + 45.74036𝑇2 − 3068.87𝑇 + 76836.9287}]∗𝑇)

𝑝𝑐 𝑣𝑟𝑣𝑐(𝑣𝑟𝑣𝑐+0.08664 𝑅𝑇𝑐𝑝𝑐

)

𝑜𝑟, 𝑝𝑟 =

𝑅𝑇𝑟𝑇𝑐𝑝𝑐

(𝑣𝑟𝑣𝑐−0.08664 𝑅𝑇𝑐𝑝𝑐

)−

0.42747𝑅2𝑇𝑐

2

𝑝𝑐2 (1+[{0.0007𝑇4 − 0.3012𝑇3 + 45.74036𝑇2 − 3068.87𝑇 + 76836.9287}]∗𝑇)

𝑣𝑟𝑣𝑐(𝑣𝑟𝑣𝑐+0.08664 𝑣𝑐𝑧𝑐

)

𝑜𝑟, 𝑝𝑟 =𝑇𝑟 (

𝑣𝑐𝑧𝑐

)

[𝑣𝑟𝑣𝑐−0.08664 (𝑣𝑐𝑧𝑐

)]−

0.42747 (𝑣𝑐𝑧𝑐

)2(1+[{0.0007𝑇4 − 0.3012𝑇3 + 45.74036𝑇2 − 3068.87𝑇 + 76836.9287}]∗𝑇)

𝑣𝑟𝑣𝑐(𝑣𝑟𝑣𝑐+0.08664 𝑣𝑐𝑧𝑐

)

𝑜𝑟, 𝑝𝑟 =𝑇𝑟 (

1

𝑧𝑐)

[𝑣𝑟−0.08664 (1

𝑧𝑐)]

−0.42747 (

1

𝑧𝑐)2(1+[{0.0007𝑇4 − 0.3012𝑇3 + 45.74036𝑇2 − 3068.87𝑇 + 76836.9287}]∗𝑇)

𝑣𝑟(𝑣𝑟+0.08664 1

𝑧𝑐)

𝑜𝑟, 𝑝𝑟 =3 𝑇𝑟

[𝑣𝑟−0.25992]−

0.0474967(1+[{0.0007(𝑇𝑟𝑇𝑐)4 − 0.3012(𝑇𝑟𝑇𝑐)

3 + 45.74036(𝑇𝑟𝑇𝑐)2 − 3068.87(𝑇𝑟𝑇𝑐) + 76836.9287}]∗(𝑇𝑟𝑇𝑐)

𝑣𝑟(𝑣𝑟+0.25992)

At critical point it becomes,

𝑜𝑟, 𝑝𝑟 =𝑇𝑟

0.3333 𝑣𝑟−0.08664−

1

21.0541 𝑣𝑟2+5.472384 𝑣𝑟

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c) Critical Compressibility Factor

𝑝𝑐 =𝑅𝑇𝑐

𝑣𝑐 − 0.08664 𝑅𝑇𝑐𝑝𝑐

−0.42747

𝑅2𝑇𝑐2

𝑝𝑐

𝑣𝑐 (𝑣𝑐 + 0.08664 𝑅𝑇𝑐𝑝𝑐

)

𝑜𝑟, 𝑝𝑐 =𝑅𝑇𝑐

{𝑣𝑐 − 0.08664 (𝑣𝑐𝑧𝑐

)}−

0.42747𝑅2𝑇𝑐

2

𝑝𝑐

𝑣𝑐 (𝑣𝑐 + 0.08664 (𝑣𝑐𝑧𝑐

))

𝑜𝑟, 1 =

𝑅𝑇𝑐𝑝𝑐

𝑣𝑐 {1 − 0.08664 (1𝑧𝑐

)}−

0.42747𝑅2𝑇𝑐

2

𝑝𝑐2

𝑣𝑐2 (1 + 0.08664 (

1𝑧𝑐

))

𝑜𝑟, 1 =(𝑣𝑐𝑧𝑐

)

𝑣𝑐 {1 − 0.08664 (1𝑧𝑐

)}−

0.42747(𝑣𝑐𝑧𝑐

)2

𝑣𝑐2 (1 + 0.08664 (

1𝑧𝑐

))

𝑜𝑟, 1 =(

1

𝑧𝑐)

{1−0.08664 (1

𝑧𝑐)}

−0.42747(

1

𝑧𝑐)2

(1+0.08664 (1

𝑧𝑐))

𝑜𝑟, 1 =(

1

𝑧𝑐)(1+0.08664 (

1

𝑧𝑐))− {1−0.08664 (

1

𝑧𝑐)} {0.42747(

1

𝑧𝑐)2}

{{1}2−{0.08664 (1

𝑧𝑐)}

2}

𝑜𝑟, 1 − {0.08664 (1

𝑧𝑐)}

2= (

1

𝑧𝑐) + 0.08664 (

1

𝑧𝑐)2− 0.42747(

1

𝑧𝑐)2+ (0.42747 ∗ .08664) (

1

𝑧𝑐)3

𝑜𝑟, 0 = (1

𝑧𝑐) + 0.08664 (

1

𝑧𝑐)2− 0.42747 (

1

𝑧𝑐)2+ (0.42747 ∗ .08664) (

1

𝑧𝑐)3− 1 + {0.08664 (

1

𝑧𝑐)}

2

𝑜𝑟, 𝑧𝑐

2 + 0.08664 𝑧𝑐 − 0.42747 𝑧𝑐 + (0.42747 ∗ .08664) − 𝑧𝑐3 + (0.08664)2 𝑧𝑐

𝑧𝑐3 = 0

𝑜𝑟, 𝑧𝑐2 + 0.08664 𝑧𝑐 − 0.42747 𝑧𝑐 + (0.42747 ∗ .08664) − 𝑧𝑐

3 + (0.08664)2 𝑧𝑐 = 0

𝑜𝑟, 𝑧𝑐3 − 𝑧𝑐

2 + (0.42747 − 0.08664 − 0.086642)𝑧𝑐 − (0.42747 ∗ .08664) = 0

Solving this equation numerically, we get,

𝑧𝑐 = 0.3471 (MATLAB Code at Appendix)

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Figure-2: Solution for Critical Compressibility Factor

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d) Express Z in terms TR, vR’:

𝑍 =𝑝𝑣

𝑅𝑇

From equation (i) substituting the value of p we get,

𝑍 =𝑣

𝑅𝑇 [

𝑅𝑇

𝑣−𝑏−

𝑎(𝑇)

𝑣(𝑣+𝑏)]

𝑜𝑟, 𝑍 =𝑣

𝑅𝑇 [

𝑅𝑇

𝑣−0.08664 𝑅𝑇𝑐𝑝𝑐

−

0.42747𝑅2𝑇𝑐

2

𝑝𝑐(1+[{0.0007𝑇4 − 0.3012𝑇3 + 45.74036𝑇2 − 3068.87𝑇 + 76836.9287}]∗𝑇

𝑣(𝑣+0.08664 𝑅𝑇𝑐𝑝𝑐

)]

𝑜𝑟, 𝑍 = [𝑣

𝑣−0.08664 𝑅𝑇𝑐𝑝𝑐

−

0.42747𝑅𝑇𝑐

2

𝑇 𝑝𝑐 (1+[{0.0007𝑇4 − 0.3012𝑇3 + 45.74036𝑇2 − 3068.87𝑇 + 76836.9287}]∗𝑇

𝑣(𝑣+0.08664 𝑅𝑇𝑐𝑝𝑐

)]

𝑜𝑟, 𝑍 =

[

𝑣

𝑣(1−0.08664 𝑅𝑇𝑐𝑣 𝑝𝑐

) −

0.42747𝑇𝑐𝑇

(1+[{0.0007𝑇4 − 0.3012𝑇3 + 45.74036𝑇2 − 3068.87𝑇 + 76836.9287}]∗𝑇

(𝑣

𝑅𝑇𝑐𝑝𝑐

+0.08664 )

]

𝑜𝑟, 𝑍 = [1

(1−0.08664 1

𝑣𝑅′

)

−

0.427471

𝑇𝑅 (1+[{0.0007(𝑇𝑟𝑇𝑐)

4 − 0.3012(𝑇𝑟𝑇𝑐)3 + 45.74036(𝑇𝑟𝑇𝑐)

2 − 3068.87(𝑇𝑟𝑇𝑐) + 76836.9287}]∗(𝑇𝑟𝑇𝑐)

(𝑣𝑅′ +0.08664 )

]

𝑜𝑟, 𝑍 = [1

(1−0.08664 1

𝑣𝑅′

)

−

0.427471

𝑇𝑅 (1+[{176433.1632(𝑇𝑟)

4 − 604113.552(𝑇𝑟)3 + 726168.24(𝑇𝑟)

2 − 386568(𝑇𝑟) + 76836.9287}]∗(𝑇𝑟∗126)

(𝑣𝑅′ +0.08664 )

]

At critical point,

𝑜𝑟, 𝑍 = [1

(1−0.08664 1

𝑣𝑅′

)

−0.42747

1

𝑇𝑅

(𝑣𝑅′ +0.08664 )

]

13 | P a g e

e) Accuracy of EOS from Equation (d)

TR = 1

TR VR'

From Table

From Equation % Error

1 0.7 0.58 1.086951409 87.40542

1 0.8 0.635 0.919693232 44.83358

1 0.9 0.675 0.796209555 17.95697

1 1 0.701 0.70147159 0.067274

1 1.2 0.75 0.565944624 24.54072

1 1.4 0.775 0.473864828 38.85615

1 1.6 0.81 0.407336589 49.71153

1 1.8 0.83 0.357071096 56.97939

1 2 0.845 0.317780354 62.39286

TR = 1.05

TR VR'

From Table

From Equation % Error

1.05 0.7 0.64 1.112828194 73.87941

1.05 0.8 0.68 0.942651495 38.62522

1.05 0.9 0.71 0.816840904 15.04801

1.05 1 0.74 0.720204302 2.675094

1.05 1.2 0.77 0.581765455 24.44604

1.05 1.4 0.8 0.487557258 39.05534

1.05 1.6 0.419405385

1.05 1.8 0.367860496

1.05 2 0.327535613

TR = 1.10

TR VR'

From Table

From Equation

1.4 0.7 1.24221212

1.4 0.8 1.05744281

1.4 0.9 0.91999765

1.1 1 0.73723404

1.1 1.2 0.596148029

1.1 1.4 0.500004921

1.1 1.6 0.430377018

1.1 1.8 0.377669042

1.1 2 0.336404031

14 | P a g e

TR = 1.20

TR VR'

From Table

From Equation % Error

1.4 0.7 1.24221212

1.4 0.8 1.05744281

1.4 0.9 0.91999765

1.2 1 0.767036082

1.2 1.2 0.621317533

1.2 1.4 0.521788333

1.2 1.6 0.449577376

1.2 1.8 0.394833997

1.2 2 0.351923762

TR = 1.40

TR VR' From Table

From Equation % Error

1.4 0.7 1.24221212

1.4 0.8 1.05744281

1.4 0.9 0.91999765

1.4 1 0.813867863

1.4 1.2 0.660869611

1.4 1.4 0.556019408

1.4 1.6 0.479749366

1.4 1.8 0.421807498

1.4 2 0.37631191

15 | P a g e

f) Equation for Departure For simplicity of calculation we will consider the reduced equation at critical point,

𝒉∗ − 𝒉

𝑹𝑻𝒄

ℎ∗ − ℎ

𝑅𝑇𝑐= ∫ 𝑍𝑐 [(

𝜕𝑃𝑟

𝜕𝑇𝑟) − 𝑃𝑟] 𝑑𝑣𝑟 − 𝑇𝑟(1 − 𝑍)

𝑉𝑟

∞

= ∫ 𝑍𝑐 [(𝜕

𝜕𝑇𝑟) (

𝑇𝑟

0.3333 𝑣𝑟−0.08664−

1

21.0541 𝑣𝑟2+5.472384 𝑣𝑟

) −𝑇𝑟

0.3333 𝑣𝑟−0.08664−

𝑉𝑟

∞

1

21.0541 𝑣𝑟2+5.472384 𝑣𝑟

] 𝑑𝑣𝑟 − 𝑇𝑟(1 − 𝑍)

= ∫ 𝑍𝑐 [(1

0.3333 𝑣𝑟−0.08664−

𝑇𝑟

0.3333 𝑣𝑟−0.08664+

1

21.0541 𝑣𝑟2+5.472384 𝑣𝑟

] 𝑑𝑣𝑟 − 𝑇𝑟(1 − 𝑍) 𝑉𝑟

∞

= 𝑍𝑐 [(ln(0.3333 𝑣𝑟−0.08664)

0.3333 −

𝑇𝑟 ln(0.3333 𝑣𝑟−0.08664)

0.3333 +

ln(5.472384

𝑣𝑟)+21.0541

5.472384 ] − 𝑇𝑟(1 − 𝑍)

= 0.873 ln(0.333𝑣𝑟 − 0.08664) [1 − 𝑇𝑟] + 0.053 ln (5.472384

𝑣𝑟) + 1.12 − 𝑇𝑟(1 − 𝑍)

(u*-u)/RTc Now to derive departure from internal energy

𝑢∗ − 𝑢

𝑅𝑇𝑐= −∫ 𝑍𝑐 [𝑇𝑟 (

𝜕𝑃𝑟

𝜕𝑇𝑟) − 𝑃𝑟] 𝑑𝑣𝑟

𝑉𝑟

∞

= ∫ 𝑍𝑐 [𝑇𝑟 (𝜕

𝜕𝑇𝑟) (

𝑇𝑟

0.3333 𝑣𝑟−0.08664−

1

21.0541 𝑣𝑟2+5.472384 𝑣𝑟

) −𝑇𝑟

0.3333 𝑣𝑟−0.08664−

𝑉𝑟

∞

1

21.0541 𝑣𝑟2+5.472384 𝑣𝑟

] 𝑑𝑣𝑟

= ∫ 𝑍𝑐 [(𝑇𝑟

0.3333 𝑣𝑟−0.08664−

𝑇𝑟2

0.3333 𝑣𝑟−0.08664+

𝑇𝑟

21.0541 𝑣𝑟2+5.472384 𝑣𝑟

] 𝑑𝑣𝑟 𝑉𝑟

∞

= 𝑍𝑐 𝑇𝑟 [(ln(0.3333 𝑣𝑟−0.08664)

0.3333 −

𝑇𝑟 ln(0.3333 𝑣𝑟−0.08664)

0.3333 +

ln(5.472384

𝑣𝑟)+21.0541

5.472384 ]

= 0.873 𝑇𝑟 ln(0.333𝑣𝑟 − 0.08664) [1 − 𝑇𝑟] + 0.053Tr ln (5.472384

𝑣𝑟) + 1.12

𝒔∗ − 𝒔

𝑹

𝑠∗ − 𝑠

𝑅= ∫ 𝑍𝑐 [(

𝜕𝑃𝑟

𝜕𝑇𝑟) − (

1

𝑉𝑟)] 𝑑𝑣𝑟 − ln(𝑧)

𝑉𝑟

∞

=∫ 𝑍𝑐 [(1

0.3333 𝑣𝑟−0.08664−

1

𝑉𝑟 ] 𝑑𝑣𝑟 − ln(𝑧)

𝑉𝑟

∞

= 𝑍𝑐 [(ln(0.3333 𝑣𝑟−0.08664)

0.3333 − (

1

ln𝑣𝑟) ]– ln (z)

16 | P a g e

g) Accuracy of EOS for equation (C) From table A1 for Nitrogen we get,

Zc=0.291

And at part (c) we calculated,

Zc=0.3471

Thus, Accuracy=(0.3471−0.291)

0.291=0.192783=19.2783%

17 | P a g e

h) Derivation of Expressions: For simplicity of calculation we will consider the reduced equation at critical point,

𝒂∗ − 𝒂

𝑹𝑻𝒄:

We know,

𝑎∗ − 𝑎

𝑅𝑇𝑐= −∫ 𝑍𝑐 [(

𝜕𝑃𝑟

𝜕𝑇𝑟) −

𝑇𝑟

𝑉𝑟]𝑑𝑣𝑟 + 𝑇𝑟𝑙𝑛(𝑍)

𝑉𝑟

∞

𝑜𝑟,𝑎∗ − 𝑎

𝑅𝑇𝑐= −∫ 𝑍𝑐 [(

𝜕

𝜕𝑇𝑟(

𝑇𝑟

0.3333 𝑣𝑟 − 0.08664−

1

21.0541 𝑣𝑟2 + 5.472384 𝑣𝑟

)) −𝑇𝑟

𝑉𝑟] 𝑑𝑣𝑟

𝑉𝑟

∞

+ 𝑇𝑟𝑙𝑛(𝑍)

Or, 𝑎∗−𝑎

𝑅𝑇𝑐= −∫ 𝑍𝑐 [

1

0.33∗𝑉𝑟−0.086−

𝑇𝑟

𝑉𝑟] 𝑑𝑉𝑟 + 𝑇𝑟𝑙𝑛(𝑍)

𝑉𝑟

∞

or, 𝑎∗−𝑎

𝑅𝑇𝑐= −𝑍𝑐[(

ln(|165𝑉𝑟−43|)

0.33− 𝑇𝑟𝑙𝑛|𝑉𝑟|] + 𝑇𝑟𝑙𝑛(𝑍)

After putting the value of Zc we get,

𝑎∗ − 𝑎

𝑅𝑇𝑐= −0.291[(

ln(|165𝑉𝑟 − 43|)

0.33− 𝑇𝑟𝑙𝑛|𝑉𝑟|] + 𝑇𝑟𝑙𝑛(𝑍)

𝒈∗ − 𝒈

𝑹𝑻𝒄:

We know,

𝑔∗ − 𝑔

𝑅𝑇𝑐= ∫ [𝑍𝑐𝑃𝑟 −

𝑇𝑟

𝑉𝑟]𝑑𝑣𝑟 + 𝑇𝑟(𝑙𝑛𝑧 + 1 − 𝑧)

𝑉𝑟

∞

𝑔∗ − 𝑔

𝑅𝑇𝑐= ∫ [𝑍𝑐 (

𝑇𝑟

0.3333 𝑣𝑟 − 0.08664−

1

21.0541 𝑣𝑟2 + 5.472384 𝑣𝑟

) −𝑇𝑟

𝑉𝑟]𝑑𝑉𝑟 + 𝑇𝑟(𝑙𝑛𝑧 + 1

𝑉𝑟

∞

− 𝑧)

𝑔∗ − 𝑔

𝑅𝑇𝑐= −

𝑇𝑟𝑙𝑛(|1375𝑉𝑟 − 361|)𝑍𝑐

0.33+

ln (|1368096

𝑉𝑟 + 5263525|)𝑍𝑐

5.47+ 𝑇𝑟(𝑙𝑛𝑧 + 1 − 𝑧)

After putting the value of Zc we get,

𝑔∗ − 𝑔

𝑅𝑇𝑐= −

𝑇𝑟𝑙𝑛(|1375𝑉𝑟 − 361|) ∗ .291

0.33+

ln (|1368096

𝑉𝑟 + 5263525|) ∗ .291

5.47+ 𝑇𝑟(𝑙𝑛𝑧 + 1 − 𝑧)

18 | P a g e

i) Speed of sound

𝑐 = √−𝑣2 √ (𝜕𝑝

𝜕𝑣)𝑠

(𝜕𝑝

𝜕𝑣)𝑠=

𝜕

𝜕𝑣{

𝑅𝑇

𝑣 − 𝑏−

𝑎(𝑇)

𝑣(𝑣 + 𝑏) }

=𝑅𝑇(−1)

(𝑣 − 𝑏)2+

𝑎(𝑇) (2𝑏 + 𝑏)

(𝑣 + 𝑏)2𝑣2

= −𝑅𝑇

(𝑣 − 𝑏)2+

𝑎(𝑇) (2𝑏 + 𝑏)

(𝑣 + 𝑏)2𝑣2

𝒄 = √−𝒗𝟐{−𝑹𝑻

(𝒗−𝒃)𝟐+

𝒂(𝑻) (𝟐𝒃+𝒃)

(𝒗+𝒃)𝟐𝒗𝟐 }

19 | P a g e

(j) Derive the Properties

Cp We can directly derive Cp and Cv from Uj. Now we know Cp and Cv,

𝐶𝑝 = −1

𝑈𝑗[𝑇 [

𝑅𝑇𝑣 − 𝑏

−𝑎(𝑇)

𝑣(𝑣 + 𝑏)

−𝑅𝑇

(𝑣 − 𝑏)2 +𝑎(𝑇)

𝑣2(𝑣 + 𝑏)+

𝑎(𝑇)𝑣(𝑣 + 𝑏)2

] + 𝑣]

Cv Also the relation between Cp and Cv is,

𝑐𝑉 =𝐶𝑝

𝑘

So,

𝐶𝑣 = −1

𝑈𝑗𝐾[𝑇 [

𝑅𝑇𝑣 − 𝑏

−𝑎(𝑇)

𝑣(𝑣 + 𝑏)

−𝑅𝑇

(𝑣 − 𝑏)2 +𝑎(𝑇)

𝑣2(𝑣 + 𝑏)+

𝑎(𝑇)𝑣(𝑣 + 𝑏)2

] + 𝑣]

1/v vp

β = (1/𝑣)(𝜕𝑣

𝜕𝑇)𝑝

(𝜕𝑣

𝜕𝑡)𝑝 = −

(𝜕𝑃

𝜕𝑇)𝑣

(𝜕𝑃

𝜕𝑣)𝑇

=

𝑅

𝑣−𝑏−

𝑎(𝑇)

𝑣(𝑣+𝑏)

−𝑅𝑇

(𝑣−𝑏)2+

𝑎(𝑇)

𝑣2(𝑣+𝑏)+

𝑎(𝑇)

𝑣(𝑣+𝑏)2

β = −1

𝑣[

𝑅

𝑣−𝑏−

𝑎(𝑇)

𝑣(𝑣+𝑏)

−𝑅𝑇

(𝑣−𝑏)2+

𝑎(𝑇)

𝑣2(𝑣+𝑏)+

𝑎(𝑇)

𝑣(𝑣+𝑏)2

]

k = v/p pvIsentropic expansion coefficient:

𝑘 = −𝑣

𝑃 (

𝜕𝑝

𝜕𝑣) 𝑠

= -𝑣

𝑃 [−

𝑅𝑇

(𝑣−𝑏)2+

𝑎(𝑇)

𝑣2(𝑣+𝑏)+

𝑎(𝑇)

𝑣(𝑣+𝑏)2]

𝑘 =−𝑣

(𝑅𝑇

𝑣−𝑏−

𝑎(𝑇)

𝑣(𝑣+𝑏)) [

−𝑅𝑇

(𝑣−𝑏)2+

2𝑎

𝑇 (𝑣+𝑐)3]

20 | P a g e

kT

We know,

𝐾𝑇 = −1

𝑣

𝛿𝑝

𝛿𝑣)

𝑇

= -1

𝑣 [−

𝑅𝑇

(𝑣−𝑏)2+

𝑎(𝑇)

𝑣2(𝑣+𝑏)+

𝑎(𝑇)

𝑣(𝑣+𝑏)2]

J

We know,𝑈𝑗 = (𝜕𝑃

𝜕𝑇) 𝑣

Also,

𝑑ℎ = 𝐶𝑝𝑑𝑇 + [𝑣 − 𝑇(𝜕𝑣

𝜕𝑇)𝑝]

From isentropic process,

h= constant

so, 𝑑ℎ = 0

(𝜕𝑇

𝜕𝑃) ℎ =

𝑇 (𝜕𝑣

𝜕𝑇)𝑝−𝑣

𝑐𝑝

Now, (𝜕𝑣

𝜕𝑇) 𝑝 = -

(𝜕𝑃

𝜕𝑇)𝑣

(𝜕𝑃

𝜕𝑣)𝑇

𝑈𝑗 = (𝜕𝑇

𝜕𝑃) ℎ

=

𝑇[−(𝜕𝑃𝜕𝑇

)𝑣

(𝜕𝑃𝜕𝑣

)𝑇]−𝑣

𝑐𝑝

𝑈𝑗 =

−𝑇

𝑅𝑣−𝑏

−𝑎(𝑇)

𝑣(𝑣+𝑏)

−𝑅𝑇

(𝑣−𝑏)2+

𝑎(𝑇)

𝑣2(𝑣+𝑏)+

𝑎(𝑇)

𝑣(𝑣+𝑏)2

−𝑣

𝑐𝑝

21 | P a g e

Summary:

In our project we firstly evaluated the Two parameters of the Redlich- Kwong-Soave equation. Then converted the equation into reduced form. With the help of MATLAB and Excel we estimated tabulated data and calculations.

22 | P a g e

Appendix

MATLAB Code

clc; clear; x = 1; zc = 1;

while x v2 = zc^3 - zc^2 + (.42747-.08664-.08664^2)*zc + (-.42747*.08664);

if abs(v2) <= 0.00000025 x = 0; clc; fprintf('%d',zc); else zc = zc - 0.000025; end

end

23 | P a g e

Reference 1) http://en.wikipedia.org/

2) http://webbook.nist.gov/chemistry/fluid/

3) http://www.boulder.nist.gov/div838/theory/refprop/MINIREF/MINIREF.HTM

4) https://www.bnl.gov/magnets/staff/gupta/cryogenic-data-handbook/Section6.pdf

5) http://www.swinburne.edu.au/ict/success/cms/documents/disertations/yswChap3.pdf

6) https://www.e-education.psu.edu/png520/m10_p5.html

7) Advanced Engineering Thermodynamics-Adrian Bejan.

8) Thermodynamics, An Engineering Approach- Yunus A Cengel and M.B Boles

9) Provided class lectures and notes.