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, vRβ: .................................................................................................................. 12
e) Accuracy of EOS from Equation (d) ..................................................................................................... 13
f) Equation for Departure ....................................................................................................................... 15
π β βππΉπ»π ......................................................................................................................................... 15
(u*-u)/RTc ........................................................................................................................................... 15
π β βππΉ ............................................................................................................................................... 15
g) Accuracy of EOS for equation (C) ........................................................................................................ 16
h) Derivation of Expressions: .................................................................................................................. 17
π β β ππΉπ»π: ....................................................................................................................................... 17
π β β ππΉπ»π: ...................................................................................................................................... 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
π£πβ = 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
π = Gibbs free energy
π0 = Gibbs free energy for ideal gas
β = Specific enthalpy for real gas
β0 = Specific enthalpy for ideal gas
π’ = Specific internal energy for real gas
π’0 = Specific internal energy for ideal gas
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π = Specific entropy for real gas
π 0 = Specific entropy for ideal gas
ππ = Isothermal expansion exponent
π = Speed of sound
π½ =Volumetric co-efficient of thermal expansion
πΆπ = Constant pressure specific heat
πΆπ£ = Constant volume specific heat
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Abstract: For the project we will be using βRedlich- Kwong-Soaveβ Equation of State (EOS) to derive to Estimating
Different Thermodynamic Relations. Starting from βRedlich- Kwong-Soaveβ equation we have calculated
the constants βa(T)β & βbβ. 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 βNitrogenβ.
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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 βRedlich- Kwong-Soaveβ 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
βNitrogenβ in our case.
Now, the βRedlich- Kwong-Soaveβ equation of state (EOS) is almost similar to Van Der Walls equation of state. The equation is-
π =π π
π£βπβ
π(π)
π£(π£+π) β¦ β¦. β¦ β¦ β¦ β¦ β¦ β¦ β¦ β¦ (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 β
(i) =>
πΏπ
πΏπ£)
π= β
π π
(π£βπ)2+
π(π)
π£2(π£+π)+
π(π)
π£(π£+π)2 β¦ β¦ β¦ β¦ β¦ β¦ β¦ β¦ (ii)
And,
πΏ2π
πΏπ£2)π
=2π π
(π£βπ)3+ π(π) [β
1
π£2(π£+π)2β
2
π£3(π£+π)β
1
π£2(π£+π)2β
2
π£(π£+π)3]
πΏ2π
πΏπ£2)π
=2π π
(π£βπ)3+ π(π) [β
2
π£2(π£+π)2β
2
π£3(π£+π)β
2
π£(π£+π)3]
πΏ2π
πΏπ£2)π
=2π π
(π£βπ)3β π(π) [
2
π£2(π£+π)2+
2
(π£+π)+
2
π£(π£+π)3]
πΏ2π
πΏπ£2)π
=2π π
(π£βπ)3β π(π) [
3π£2+3π£π+π2
π£3(π£+π)3] β¦ β¦ β¦. β¦. β¦. β¦. β¦. β¦. (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,
πΏπ
πΏπ£)
π= β
π ππ
(π£πβπ)2+
π(π)
π£π2(π£π+π)
+π(π)
π£π(π£π+π)2= 0
ππ, 0 = βπ ππ
(π£πβπ)2+
π(π)
π£π2(π£π+π)
+π(π)
π£π(π£π+π)2
ππ,π ππ
(π£πβπ)2=
π(π)
π£π2(π£π+π)
+π(π)
π£π(π£π+π)2
ππ,π ππ
(π£πβπ)2=
π(π) (2π£π+π)
π£π2(π£π+π)2
β¦. β¦. .β¦ β¦. β¦.. β¦.. β¦. β¦ (iv)
And,
πΏ2π
πΏπ£2)π
=2π ππ
(π£πβπ)3β π(π) [
3π£π2+3π£ππ+π2
π£π3(π£π+π)3
] = 0
ππ,2π ππ
(π£πβπ)3= π(π) [
3π£π2+3π£ππ+π2
π£π3(π£π+π)3
] β¦. β¦. β¦. β¦. β¦. β¦. β¦. (v)
Now, Dividing Equation (iv) with equation (v) we get,
(π£π β π) =(π£π+π).π£π.(π£π+π)
3π£π2+3π£ππ+π2
ππ, (π£π β π)(3π£π2 + 3π£ππ + π2) = (π£π + π). π£π . (π£π + π)
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ππ, 3π£π3 + 3π£π
2 + π2π£π β 3ππ£π2 β 3π£ππ
2 β π3 = 2π£π3 + 2π£π
2 + π£π2π + π£ππ
2
ππ, π3 + 3π2π£π + 3ππ£π2 β π£π
3 = 0
ππ, (π + π£π)3 = 2π£π
3 = (β23
π£π)3
ππ, π = (β23
β 1)π£π = (β23
β 1)π§ππ ππ
ππ
ππ, π = (β23
β 1).1
3.π ππ
ππ= 0.08664
π ππ
ππ
ππ, π = 0.08664 π ππ
ππ β¦. β¦. β¦. β¦. β¦. β¦. β¦. (vi)
Note: Here, π§π =1
3. As the original equation is from Redlich-Kwong equation, we will use the value of π§π
from the Redlich-Kwong equation.
Now putting the value of βbβ and π£π = π§ππ ππ
ππ=
1
3.π ππ
ππ in equation (iv) we get,
π ππ
(π£πβπ)2=
π(π) (2π£π+π)
π£π2(π£π+π)2
ππ, π(π) = π ππ π£π
2(π£π+π)2
(2π£π+π) (π£πβπ)2
ππ, π(π) = π ππ [π§π
π ππππ
]2(π§π
π ππππ
+0.08664 π ππππ
)2
(2π§ππ ππππ
+0.08664 π ππππ
) (π§ππ ππππ
β0.08664 π ππππ
)2
ππ, π(π) = π ππ [π§π
π ππππ
]2 (
π ππππ
)2 (
1
3+0.08664 )
2
(π ππππ
) (2
3+0.08664) (
π ππππ
)2 (
1
3β0.08664 )
2
ππ, π(π) = [
1
3.π ππππ
]2β (0.176377600711111)
(1
ππ) (
2
3+0.08664) (
1
3β0.08664 )
2
ππ, π(π) = 0.42747π 2ππ
2
ππ β¦ β¦. β¦. β¦. β¦. β¦. β¦. β¦. β¦. (vii)
Now, this is for the critical point only. As for other points for the assigned substance Nitrogen we get
a(T)= 0.42747π 2ππ
2
ππ(1 + [{0.0007T4 - 0.3012T3 + 45.74036T2 - 3068.87T + 76836.9287}]*T)
at critical point T=TR=1 thus we get again,
a(T)= 0.42747π 2ππ
2
ππ
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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
π =π π
π£βπβ
π(π)
π£(π£+π)
ππ, ππππ =π ππππ
π£ππ£πβ0.08664 π ππππ
β0.42747
π 2ππ2
ππ(1+[{0.0007π4 β 0.3012π3 + 45.74036π2 β 3068.87π + 76836.9287}]βπ)
π£ππ£π(π£ππ£π+0.08664 π ππππ
)
ππ, ππ =π ππππ
(π£ππ£πβ0.08664 π ππππ
)ππ
β0.42747
π 2ππ2
ππ(1+[{0.0007π4 β 0.3012π3 + 45.74036π2 β 3068.87π + 76836.9287}]βπ)
ππ π£ππ£π(π£ππ£π+0.08664 π ππππ
)
ππ, ππ =
π ππππππ
(π£ππ£πβ0.08664 π ππππ
)β
0.42747π 2ππ
2
ππ2 (1+[{0.0007π4 β 0.3012π3 + 45.74036π2 β 3068.87π + 76836.9287}]βπ)
π£ππ£π(π£ππ£π+0.08664 π£ππ§π
)
ππ, ππ =ππ (
π£ππ§π
)
[π£ππ£πβ0.08664 (π£ππ§π
)]β
0.42747 (π£ππ§π
)2(1+[{0.0007π4 β 0.3012π3 + 45.74036π2 β 3068.87π + 76836.9287}]βπ)
π£ππ£π(π£ππ£π+0.08664 π£ππ§π
)
ππ, ππ =ππ (
1
π§π)
[π£πβ0.08664 (1
π§π)]
β0.42747 (
1
π§π)2(1+[{0.0007π4 β 0.3012π3 + 45.74036π2 β 3068.87π + 76836.9287}]βπ)
π£π(π£π+0.08664 1
π§π)
ππ, ππ =3 ππ
[π£πβ0.25992]β
0.0474967(1+[{0.0007(ππππ)4 β 0.3012(ππππ)
3 + 45.74036(ππππ)2 β 3068.87(ππππ) + 76836.9287}]β(ππππ)
π£π(π£π+0.25992)
At critical point it becomes,
ππ, ππ =ππ
0.3333 π£πβ0.08664β
1
21.0541 π£π2+5.472384 π£π
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c) Critical Compressibility Factor
ππ =π ππ
π£π β 0.08664 π ππππ
β0.42747
π 2ππ2
ππ
π£π (π£π + 0.08664 π ππππ
)
ππ, ππ =π ππ
{π£π β 0.08664 (π£ππ§π
)}β
0.42747π 2ππ
2
ππ
π£π (π£π + 0.08664 (π£ππ§π
))
ππ, 1 =
π ππππ
π£π {1 β 0.08664 (1π§π
)}β
0.42747π 2ππ
2
ππ2
π£π2 (1 + 0.08664 (
1π§π
))
ππ, 1 =(π£ππ§π
)
π£π {1 β 0.08664 (1π§π
)}β
0.42747(π£ππ§π
)2
π£π2 (1 + 0.08664 (
1π§π
))
ππ, 1 =(
1
π§π)
{1β0.08664 (1
π§π)}
β0.42747(
1
π§π)2
(1+0.08664 (1
π§π))
ππ, 1 =(
1
π§π)(1+0.08664 (
1
π§π))β {1β0.08664 (
1
π§π)} {0.42747(
1
π§π)2}
{{1}2β{0.08664 (1
π§π)}
2}
ππ, 1 β {0.08664 (1
π§π)}
2= (
1
π§π) + 0.08664 (
1
π§π)2β 0.42747(
1
π§π)2+ (0.42747 β .08664) (
1
π§π)3
ππ, 0 = (1
π§π) + 0.08664 (
1
π§π)2β 0.42747 (
1
π§π)2+ (0.42747 β .08664) (
1
π§π)3β 1 + {0.08664 (
1
π§π)}
2
ππ, π§π
2 + 0.08664 π§π β 0.42747 π§π + (0.42747 β .08664) β π§π3 + (0.08664)2 π§π
π§π3 = 0
ππ, π§π2 + 0.08664 π§π β 0.42747 π§π + (0.42747 β .08664) β π§π
3 + (0.08664)2 π§π = 0
ππ, π§π3 β π§π
2 + (0.42747 β 0.08664 β 0.086642)π§π β (0.42747 β .08664) = 0
Solving this equation numerically, we get,
π§π = 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, vRβ:
π =ππ£
π π
From equation (i) substituting the value of p we get,
π =π£
π π [
π π
π£βπβ
π(π)
π£(π£+π)]
ππ, π =π£
π π [
π π
π£β0.08664 π ππππ
β
0.42747π 2ππ
2
ππ(1+[{0.0007π4 β 0.3012π3 + 45.74036π2 β 3068.87π + 76836.9287}]βπ
π£(π£+0.08664 π ππππ
)]
ππ, π = [π£
π£β0.08664 π ππππ
β
0.42747π ππ
2
π ππ (1+[{0.0007π4 β 0.3012π3 + 45.74036π2 β 3068.87π + 76836.9287}]βπ
π£(π£+0.08664 π ππππ
)]
ππ, π =
[
π£
π£(1β0.08664 π πππ£ ππ
) β
0.42747πππ
(1+[{0.0007π4 β 0.3012π3 + 45.74036π2 β 3068.87π + 76836.9287}]βπ
(π£
π ππππ
+0.08664 )
]
ππ, π = [1
(1β0.08664 1
π£π β²
)
β
0.427471
ππ (1+[{0.0007(ππππ)
4 β 0.3012(ππππ)3 + 45.74036(ππππ)
2 β 3068.87(ππππ) + 76836.9287}]β(ππππ)
(π£π β² +0.08664 )
]
ππ, π = [1
(1β0.08664 1
π£π β²
)
β
0.427471
ππ (1+[{176433.1632(ππ)
4 β 604113.552(ππ)3 + 726168.24(ππ)
2 β 386568(ππ) + 76836.9287}]β(ππβ126)
(π£π β² +0.08664 )
]
At critical point,
ππ, π = [1
(1β0.08664 1
π£π β²
)
β0.42747
1
ππ
(π£π β² +0.08664 )
]
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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
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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
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f) Equation for Departure For simplicity of calculation we will consider the reduced equation at critical point,
πβ β π
πΉπ»π
ββ β β
π ππ= β« ππ [(
πππ
πππ) β ππ] ππ£π β ππ(1 β π)
ππ
β
= β« ππ [(π
πππ) (
ππ
0.3333 π£πβ0.08664β
1
21.0541 π£π2+5.472384 π£π
) βππ
0.3333 π£πβ0.08664β
ππ
β
1
21.0541 π£π2+5.472384 π£π
] ππ£π β ππ(1 β π)
= β« ππ [(1
0.3333 π£πβ0.08664β
ππ
0.3333 π£πβ0.08664+
1
21.0541 π£π2+5.472384 π£π
] ππ£π β ππ(1 β π) ππ
β
= ππ [(ln(0.3333 π£πβ0.08664)
0.3333 β
ππ ln(0.3333 π£πβ0.08664)
0.3333 +
ln(5.472384
π£π)+21.0541
5.472384 ] β ππ(1 β π)
= 0.873 ln(0.333π£π β 0.08664) [1 β ππ] + 0.053 ln (5.472384
π£π) + 1.12 β ππ(1 β π)
(u*-u)/RTc Now to derive departure from internal energy
π’β β π’
π ππ= ββ« ππ [ππ (
πππ
πππ) β ππ] ππ£π
ππ
β
= β« ππ [ππ (π
πππ) (
ππ
0.3333 π£πβ0.08664β
1
21.0541 π£π2+5.472384 π£π
) βππ
0.3333 π£πβ0.08664β
ππ
β
1
21.0541 π£π2+5.472384 π£π
] ππ£π
= β« ππ [(ππ
0.3333 π£πβ0.08664β
ππ2
0.3333 π£πβ0.08664+
ππ
21.0541 π£π2+5.472384 π£π
] ππ£π ππ
β
= ππ ππ [(ln(0.3333 π£πβ0.08664)
0.3333 β
ππ ln(0.3333 π£πβ0.08664)
0.3333 +
ln(5.472384
π£π)+21.0541
5.472384 ]
= 0.873 ππ ln(0.333π£π β 0.08664) [1 β ππ] + 0.053Tr ln (5.472384
π£π) + 1.12
πβ β π
πΉ
π β β π
π = β« ππ [(
πππ
πππ) β (
1
ππ)] ππ£π β ln(π§)
ππ
β
=β« ππ [(1
0.3333 π£πβ0.08664β
1
ππ ] ππ£π β ln(π§)
ππ
β
= ππ [(ln(0.3333 π£πβ0.08664)
0.3333 β (
1
lnπ£π) ]β ln (z)
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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=(0.3471β0.291)
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,
πβ β π
πΉπ»π:
We know,
πβ β π
π ππ= ββ« ππ [(
πππ
πππ) β
ππ
ππ]ππ£π + ππππ(π)
ππ
β
ππ,πβ β π
π ππ= ββ« ππ [(
π
πππ(
ππ
0.3333 π£π β 0.08664β
1
21.0541 π£π2 + 5.472384 π£π
)) βππ
ππ] ππ£π
ππ
β
+ ππππ(π)
Or, πββπ
π ππ= ββ« ππ [
1
0.33βππβ0.086β
ππ
ππ] πππ + ππππ(π)
ππ
β
or, πββπ
π ππ= βππ[(
ln(|165ππβ43|)
0.33β ππππ|ππ|] + ππππ(π)
After putting the value of Zc we get,
πβ β π
π ππ= β0.291[(
ln(|165ππ β 43|)
0.33β ππππ|ππ|] + ππππ(π)
πβ β π
πΉπ»π:
We know,
πβ β π
π ππ= β« [ππππ β
ππ
ππ]ππ£π + ππ(πππ§ + 1 β π§)
ππ
β
πβ β π
π ππ= β« [ππ (
ππ
0.3333 π£π β 0.08664β
1
21.0541 π£π2 + 5.472384 π£π
) βππ
ππ]πππ + ππ(πππ§ + 1
ππ
β
β π§)
πβ β π
π ππ= β
ππππ(|1375ππ β 361|)ππ
0.33+
ln (|1368096
ππ + 5263525|)ππ
5.47+ ππ(πππ§ + 1 β π§)
After putting the value of Zc we get,
πβ β π
π ππ= β
ππππ(|1375ππ β 361|) β .291
0.33+
ln (|1368096
ππ + 5263525|) β .291
5.47+ ππ(πππ§ + 1 β π§)
18 | P a g e
i) Speed of sound
π = ββπ£2 β (ππ
ππ£)π
(ππ
ππ£)π =
π
ππ£{
π π
π£ β πβ
π(π)
π£(π£ + π) }
=π π(β1)
(π£ β π)2+
π(π) (2π + π)
(π£ + π)2π£2
= βπ π
(π£ β π)2+
π(π) (2π + π)
(π£ + π)2π£2
π = ββππ{βπΉπ»
(πβπ)π+
π(π») (ππ+π)
(π+π)πππ }
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,
πΆπ = β1
ππ[π [
π ππ£ β π
βπ(π)
π£(π£ + π)
βπ π
(π£ β π)2 +π(π)
π£2(π£ + π)+
π(π)π£(π£ + π)2
] + π£]
Cv Also the relation between Cp and Cv is,
ππ =πΆπ
π
So,
πΆπ£ = β1
πππΎ[π [
π ππ£ β π
βπ(π)
π£(π£ + π)
βπ π
(π£ β π)2 +π(π)
π£2(π£ + π)+
π(π)π£(π£ + π)2
] + π£]
1/v vp
Ξ² = (1/π£)(ππ£
ππ)π
(ππ£
ππ‘)π = β
(ππ
ππ)π£
(ππ
ππ£)π
=
π
π£βπβ
π(π)
π£(π£+π)
βπ π
(π£βπ)2+
π(π)
π£2(π£+π)+
π(π)
π£(π£+π)2
Ξ² = β1
π£[
π
π£βπβ
π(π)
π£(π£+π)
βπ π
(π£βπ)2+
π(π)
π£2(π£+π)+
π(π)
π£(π£+π)2
]
k = v/p pvIsentropic expansion coefficient:
π = βπ£
π (
ππ
ππ£) π
= -π£
π [β
π π
(π£βπ)2+
π(π)
π£2(π£+π)+
π(π)
π£(π£+π)2]
π =βπ£
(π π
π£βπβ
π(π)
π£(π£+π)) [
βπ π
(π£βπ)2+
2π
π (π£+π)3]
20 | P a g e
kT
We know,
πΎπ = β1
π£
πΏπ
πΏπ£)
π
= -1
π£ [β
π π
(π£βπ)2+
π(π)
π£2(π£+π)+
π(π)
π£(π£+π)2]
J
We know,ππ = (ππ
ππ) π£
Also,
πβ = πΆπππ + [π£ β π(ππ£
ππ)π]
From isentropic process,
h= constant
so, πβ = 0
(ππ
ππ) β =
π (ππ£
ππ)πβπ£
ππ
Now, (ππ£
ππ) π = -
(ππ
ππ)π£
(ππ
ππ£)π
ππ = (ππ
ππ) β
=
π[β(ππππ
)π£
(ππππ£
)π]βπ£
ππ
ππ =
βπ
π π£βπ
βπ(π)
π£(π£+π)
βπ π
(π£βπ)2+
π(π)
π£2(π£+π)+
π(π)
π£(π£+π)2
βπ£
ππ
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.
