Upload
syed-hasib-akhter-faruqui
View
130
Download
2
Embed Size (px)
Citation preview
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
1 | P a g e
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, vRโ: .................................................................................................................. 12
e) Accuracy of EOS from Equation (d) ..................................................................................................... 13
f) Equation for Departure ....................................................................................................................... 15
๐ โ โ๐๐น๐ป๐ ......................................................................................................................................... 15
(u*-u)/RTc ........................................................................................................................................... 15
๐ โ โ๐๐น ............................................................................................................................................... 15
g) Accuracy of EOS for equation (C) ........................................................................................................ 16
h) Derivation of Expressions: .................................................................................................................. 17
๐ โ โ ๐๐น๐ป๐: ....................................................................................................................................... 17
๐ โ โ ๐๐น๐ป๐: ...................................................................................................................................... 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
2 | P a g e
Nomenclature:
P = Pressure
Pr = Reduced pressure
Pc = Critical pressure
v = Specific volume
vr = Reduced specific volume
vc = Critical volume
๐ฃ๐โ = 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
๐ = Gibbs free energy
๐0 = Gibbs free energy for ideal gas
โ = Specific enthalpy for real gas
โ0 = Specific enthalpy for ideal gas
๐ข = Specific internal energy for real gas
๐ข0 = Specific internal energy for ideal gas
3 | P a g e
๐ = Specific entropy for real gas
๐ 0 = Specific entropy for ideal gas
๐๐ = Isothermal expansion exponent
๐ = Speed of sound
๐ฝ =Volumetric co-efficient of thermal expansion
๐ถ๐ = Constant pressure specific heat
๐ถ๐ฃ = Constant volume specific heat
4 | P a g e
Abstract: For the project we will be using โRedlich- Kwong-Soaveโ Equation of State (EOS) to derive to Estimating
Different Thermodynamic Relations. Starting from โRedlich- Kwong-Soaveโ equation we have calculated
the constants โa(T)โ & โbโ. 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 โNitrogenโ.
5 | P a g e
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 โRedlich- Kwong-Soaveโ 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
โNitrogenโ in our case.
Now, the โRedlich- Kwong-Soaveโ equation of state (EOS) is almost similar to Van Der Walls equation of state. The equation is-
๐ =๐ ๐
๐ฃโ๐โ
๐(๐)
๐ฃ(๐ฃ+๐) โฆ โฆ. โฆ โฆ โฆ โฆ โฆ โฆ โฆ โฆ (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.
6 | P a g e
Derivation
(a) Evaluation of the Constants
Taking the first and second derivative of pressure WRT to volume be โ
(i) =>
๐ฟ๐
๐ฟ๐ฃ)
๐= โ
๐ ๐
(๐ฃโ๐)2+
๐(๐)
๐ฃ2(๐ฃ+๐)+
๐(๐)
๐ฃ(๐ฃ+๐)2 โฆ โฆ โฆ โฆ โฆ โฆ โฆ โฆ (ii)
And,
๐ฟ2๐
๐ฟ๐ฃ2)๐
=2๐ ๐
(๐ฃโ๐)3+ ๐(๐) [โ
1
๐ฃ2(๐ฃ+๐)2โ
2
๐ฃ3(๐ฃ+๐)โ
1
๐ฃ2(๐ฃ+๐)2โ
2
๐ฃ(๐ฃ+๐)3]
๐ฟ2๐
๐ฟ๐ฃ2)๐
=2๐ ๐
(๐ฃโ๐)3+ ๐(๐) [โ
2
๐ฃ2(๐ฃ+๐)2โ
2
๐ฃ3(๐ฃ+๐)โ
2
๐ฃ(๐ฃ+๐)3]
๐ฟ2๐
๐ฟ๐ฃ2)๐
=2๐ ๐
(๐ฃโ๐)3โ ๐(๐) [
2
๐ฃ2(๐ฃ+๐)2+
2
(๐ฃ+๐)+
2
๐ฃ(๐ฃ+๐)3]
๐ฟ2๐
๐ฟ๐ฃ2)๐
=2๐ ๐
(๐ฃโ๐)3โ ๐(๐) [
3๐ฃ2+3๐ฃ๐+๐2
๐ฃ3(๐ฃ+๐)3] โฆ โฆ โฆ. โฆ. โฆ. โฆ. โฆ. โฆ. (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,
๐ฟ๐
๐ฟ๐ฃ)
๐= โ
๐ ๐๐
(๐ฃ๐โ๐)2+
๐(๐)
๐ฃ๐2(๐ฃ๐+๐)
+๐(๐)
๐ฃ๐(๐ฃ๐+๐)2= 0
๐๐, 0 = โ๐ ๐๐
(๐ฃ๐โ๐)2+
๐(๐)
๐ฃ๐2(๐ฃ๐+๐)
+๐(๐)
๐ฃ๐(๐ฃ๐+๐)2
๐๐,๐ ๐๐
(๐ฃ๐โ๐)2=
๐(๐)
๐ฃ๐2(๐ฃ๐+๐)
+๐(๐)
๐ฃ๐(๐ฃ๐+๐)2
๐๐,๐ ๐๐
(๐ฃ๐โ๐)2=
๐(๐) (2๐ฃ๐+๐)
๐ฃ๐2(๐ฃ๐+๐)2
โฆ. โฆ. .โฆ โฆ. โฆ.. โฆ.. โฆ. โฆ (iv)
And,
๐ฟ2๐
๐ฟ๐ฃ2)๐
=2๐ ๐๐
(๐ฃ๐โ๐)3โ ๐(๐) [
3๐ฃ๐2+3๐ฃ๐๐+๐2
๐ฃ๐3(๐ฃ๐+๐)3
] = 0
๐๐,2๐ ๐๐
(๐ฃ๐โ๐)3= ๐(๐) [
3๐ฃ๐2+3๐ฃ๐๐+๐2
๐ฃ๐3(๐ฃ๐+๐)3
] โฆ. โฆ. โฆ. โฆ. โฆ. โฆ. โฆ. (v)
Now, Dividing Equation (iv) with equation (v) we get,
(๐ฃ๐ โ ๐) =(๐ฃ๐+๐).๐ฃ๐.(๐ฃ๐+๐)
3๐ฃ๐2+3๐ฃ๐๐+๐2
๐๐, (๐ฃ๐ โ ๐)(3๐ฃ๐2 + 3๐ฃ๐๐ + ๐2) = (๐ฃ๐ + ๐). ๐ฃ๐ . (๐ฃ๐ + ๐)
7 | P a g e
๐๐, 3๐ฃ๐3 + 3๐ฃ๐
2 + ๐2๐ฃ๐ โ 3๐๐ฃ๐2 โ 3๐ฃ๐๐
2 โ ๐3 = 2๐ฃ๐3 + 2๐ฃ๐
2 + ๐ฃ๐2๐ + ๐ฃ๐๐
2
๐๐, ๐3 + 3๐2๐ฃ๐ + 3๐๐ฃ๐2 โ ๐ฃ๐
3 = 0
๐๐, (๐ + ๐ฃ๐)3 = 2๐ฃ๐
3 = (โ23
๐ฃ๐)3
๐๐, ๐ = (โ23
โ 1)๐ฃ๐ = (โ23
โ 1)๐ง๐๐ ๐๐
๐๐
๐๐, ๐ = (โ23
โ 1).1
3.๐ ๐๐
๐๐= 0.08664
๐ ๐๐
๐๐
๐๐, ๐ = 0.08664 ๐ ๐๐
๐๐ โฆ. โฆ. โฆ. โฆ. โฆ. โฆ. โฆ. (vi)
Note: Here, ๐ง๐ =1
3. As the original equation is from Redlich-Kwong equation, we will use the value of ๐ง๐
from the Redlich-Kwong equation.
Now putting the value of โbโ and ๐ฃ๐ = ๐ง๐๐ ๐๐
๐๐=
1
3.๐ ๐๐
๐๐ in equation (iv) we get,
๐ ๐๐
(๐ฃ๐โ๐)2=
๐(๐) (2๐ฃ๐+๐)
๐ฃ๐2(๐ฃ๐+๐)2
๐๐, ๐(๐) = ๐ ๐๐ ๐ฃ๐
2(๐ฃ๐+๐)2
(2๐ฃ๐+๐) (๐ฃ๐โ๐)2
๐๐, ๐(๐) = ๐ ๐๐ [๐ง๐
๐ ๐๐๐๐
]2(๐ง๐
๐ ๐๐๐๐
+0.08664 ๐ ๐๐๐๐
)2
(2๐ง๐๐ ๐๐๐๐
+0.08664 ๐ ๐๐๐๐
) (๐ง๐๐ ๐๐๐๐
โ0.08664 ๐ ๐๐๐๐
)2
๐๐, ๐(๐) = ๐ ๐๐ [๐ง๐
๐ ๐๐๐๐
]2 (
๐ ๐๐๐๐
)2 (
1
3+0.08664 )
2
(๐ ๐๐๐๐
) (2
3+0.08664) (
๐ ๐๐๐๐
)2 (
1
3โ0.08664 )
2
๐๐, ๐(๐) = [
1
3.๐ ๐๐๐๐
]2โ (0.176377600711111)
(1
๐๐) (
2
3+0.08664) (
1
3โ0.08664 )
2
๐๐, ๐(๐) = 0.42747๐ 2๐๐
2
๐๐ โฆ โฆ. โฆ. โฆ. โฆ. โฆ. โฆ. โฆ. โฆ. (vii)
Now, this is for the critical point only. As for other points for the assigned substance Nitrogen we get
a(T)= 0.42747๐ 2๐๐
2
๐๐(1 + [{0.0007T4 - 0.3012T3 + 45.74036T2 - 3068.87T + 76836.9287}]*T)
at critical point T=TR=1 thus we get again,
a(T)= 0.42747๐ 2๐๐
2
๐๐
8 | P a g e
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
9 | P a g e
(b) Equation of State in Reduced Form
๐ =๐ ๐
๐ฃโ๐โ
๐(๐)
๐ฃ(๐ฃ+๐)
๐๐, ๐๐๐๐ =๐ ๐๐๐๐
๐ฃ๐๐ฃ๐โ0.08664 ๐ ๐๐๐๐
โ0.42747
๐ 2๐๐2
๐๐(1+[{0.0007๐4 โ 0.3012๐3 + 45.74036๐2 โ 3068.87๐ + 76836.9287}]โ๐)
๐ฃ๐๐ฃ๐(๐ฃ๐๐ฃ๐+0.08664 ๐ ๐๐๐๐
)
๐๐, ๐๐ =๐ ๐๐๐๐
(๐ฃ๐๐ฃ๐โ0.08664 ๐ ๐๐๐๐
)๐๐
โ0.42747
๐ 2๐๐2
๐๐(1+[{0.0007๐4 โ 0.3012๐3 + 45.74036๐2 โ 3068.87๐ + 76836.9287}]โ๐)
๐๐ ๐ฃ๐๐ฃ๐(๐ฃ๐๐ฃ๐+0.08664 ๐ ๐๐๐๐
)
๐๐, ๐๐ =
๐ ๐๐๐๐๐๐
(๐ฃ๐๐ฃ๐โ0.08664 ๐ ๐๐๐๐
)โ
0.42747๐ 2๐๐
2
๐๐2 (1+[{0.0007๐4 โ 0.3012๐3 + 45.74036๐2 โ 3068.87๐ + 76836.9287}]โ๐)
๐ฃ๐๐ฃ๐(๐ฃ๐๐ฃ๐+0.08664 ๐ฃ๐๐ง๐
)
๐๐, ๐๐ =๐๐ (
๐ฃ๐๐ง๐
)
[๐ฃ๐๐ฃ๐โ0.08664 (๐ฃ๐๐ง๐
)]โ
0.42747 (๐ฃ๐๐ง๐
)2(1+[{0.0007๐4 โ 0.3012๐3 + 45.74036๐2 โ 3068.87๐ + 76836.9287}]โ๐)
๐ฃ๐๐ฃ๐(๐ฃ๐๐ฃ๐+0.08664 ๐ฃ๐๐ง๐
)
๐๐, ๐๐ =๐๐ (
1
๐ง๐)
[๐ฃ๐โ0.08664 (1
๐ง๐)]
โ0.42747 (
1
๐ง๐)2(1+[{0.0007๐4 โ 0.3012๐3 + 45.74036๐2 โ 3068.87๐ + 76836.9287}]โ๐)
๐ฃ๐(๐ฃ๐+0.08664 1
๐ง๐)
๐๐, ๐๐ =3 ๐๐
[๐ฃ๐โ0.25992]โ
0.0474967(1+[{0.0007(๐๐๐๐)4 โ 0.3012(๐๐๐๐)
3 + 45.74036(๐๐๐๐)2 โ 3068.87(๐๐๐๐) + 76836.9287}]โ(๐๐๐๐)
๐ฃ๐(๐ฃ๐+0.25992)
At critical point it becomes,
๐๐, ๐๐ =๐๐
0.3333 ๐ฃ๐โ0.08664โ
1
21.0541 ๐ฃ๐2+5.472384 ๐ฃ๐
10 | P a g e
c) Critical Compressibility Factor
๐๐ =๐ ๐๐
๐ฃ๐ โ 0.08664 ๐ ๐๐๐๐
โ0.42747
๐ 2๐๐2
๐๐
๐ฃ๐ (๐ฃ๐ + 0.08664 ๐ ๐๐๐๐
)
๐๐, ๐๐ =๐ ๐๐
{๐ฃ๐ โ 0.08664 (๐ฃ๐๐ง๐
)}โ
0.42747๐ 2๐๐
2
๐๐
๐ฃ๐ (๐ฃ๐ + 0.08664 (๐ฃ๐๐ง๐
))
๐๐, 1 =
๐ ๐๐๐๐
๐ฃ๐ {1 โ 0.08664 (1๐ง๐
)}โ
0.42747๐ 2๐๐
2
๐๐2
๐ฃ๐2 (1 + 0.08664 (
1๐ง๐
))
๐๐, 1 =(๐ฃ๐๐ง๐
)
๐ฃ๐ {1 โ 0.08664 (1๐ง๐
)}โ
0.42747(๐ฃ๐๐ง๐
)2
๐ฃ๐2 (1 + 0.08664 (
1๐ง๐
))
๐๐, 1 =(
1
๐ง๐)
{1โ0.08664 (1
๐ง๐)}
โ0.42747(
1
๐ง๐)2
(1+0.08664 (1
๐ง๐))
๐๐, 1 =(
1
๐ง๐)(1+0.08664 (
1
๐ง๐))โ {1โ0.08664 (
1
๐ง๐)} {0.42747(
1
๐ง๐)2}
{{1}2โ{0.08664 (1
๐ง๐)}
2}
๐๐, 1 โ {0.08664 (1
๐ง๐)}
2= (
1
๐ง๐) + 0.08664 (
1
๐ง๐)2โ 0.42747(
1
๐ง๐)2+ (0.42747 โ .08664) (
1
๐ง๐)3
๐๐, 0 = (1
๐ง๐) + 0.08664 (
1
๐ง๐)2โ 0.42747 (
1
๐ง๐)2+ (0.42747 โ .08664) (
1
๐ง๐)3โ 1 + {0.08664 (
1
๐ง๐)}
2
๐๐, ๐ง๐
2 + 0.08664 ๐ง๐ โ 0.42747 ๐ง๐ + (0.42747 โ .08664) โ ๐ง๐3 + (0.08664)2 ๐ง๐
๐ง๐3 = 0
๐๐, ๐ง๐2 + 0.08664 ๐ง๐ โ 0.42747 ๐ง๐ + (0.42747 โ .08664) โ ๐ง๐
3 + (0.08664)2 ๐ง๐ = 0
๐๐, ๐ง๐3 โ ๐ง๐
2 + (0.42747 โ 0.08664 โ 0.086642)๐ง๐ โ (0.42747 โ .08664) = 0
Solving this equation numerically, we get,
๐ง๐ = 0.3471 (MATLAB Code at Appendix)
12 | P a g e
d) Express Z in terms TR, vRโ:
๐ =๐๐ฃ
๐ ๐
From equation (i) substituting the value of p we get,
๐ =๐ฃ
๐ ๐ [
๐ ๐
๐ฃโ๐โ
๐(๐)
๐ฃ(๐ฃ+๐)]
๐๐, ๐ =๐ฃ
๐ ๐ [
๐ ๐
๐ฃโ0.08664 ๐ ๐๐๐๐
โ
0.42747๐ 2๐๐
2
๐๐(1+[{0.0007๐4 โ 0.3012๐3 + 45.74036๐2 โ 3068.87๐ + 76836.9287}]โ๐
๐ฃ(๐ฃ+0.08664 ๐ ๐๐๐๐
)]
๐๐, ๐ = [๐ฃ
๐ฃโ0.08664 ๐ ๐๐๐๐
โ
0.42747๐ ๐๐
2
๐ ๐๐ (1+[{0.0007๐4 โ 0.3012๐3 + 45.74036๐2 โ 3068.87๐ + 76836.9287}]โ๐
๐ฃ(๐ฃ+0.08664 ๐ ๐๐๐๐
)]
๐๐, ๐ =
[
๐ฃ
๐ฃ(1โ0.08664 ๐ ๐๐๐ฃ ๐๐
) โ
0.42747๐๐๐
(1+[{0.0007๐4 โ 0.3012๐3 + 45.74036๐2 โ 3068.87๐ + 76836.9287}]โ๐
(๐ฃ
๐ ๐๐๐๐
+0.08664 )
]
๐๐, ๐ = [1
(1โ0.08664 1
๐ฃ๐ โฒ
)
โ
0.427471
๐๐ (1+[{0.0007(๐๐๐๐)
4 โ 0.3012(๐๐๐๐)3 + 45.74036(๐๐๐๐)
2 โ 3068.87(๐๐๐๐) + 76836.9287}]โ(๐๐๐๐)
(๐ฃ๐ โฒ +0.08664 )
]
๐๐, ๐ = [1
(1โ0.08664 1
๐ฃ๐ โฒ
)
โ
0.427471
๐๐ (1+[{176433.1632(๐๐)
4 โ 604113.552(๐๐)3 + 726168.24(๐๐)
2 โ 386568(๐๐) + 76836.9287}]โ(๐๐โ126)
(๐ฃ๐ โฒ +0.08664 )
]
At critical point,
๐๐, ๐ = [1
(1โ0.08664 1
๐ฃ๐ โฒ
)
โ0.42747
1
๐๐
(๐ฃ๐ โฒ +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,
๐โ โ ๐
๐น๐ป๐
โโ โ โ
๐ ๐๐= โซ ๐๐ [(
๐๐๐
๐๐๐) โ ๐๐] ๐๐ฃ๐ โ ๐๐(1 โ ๐)
๐๐
โ
= โซ ๐๐ [(๐
๐๐๐) (
๐๐
0.3333 ๐ฃ๐โ0.08664โ
1
21.0541 ๐ฃ๐2+5.472384 ๐ฃ๐
) โ๐๐
0.3333 ๐ฃ๐โ0.08664โ
๐๐
โ
1
21.0541 ๐ฃ๐2+5.472384 ๐ฃ๐
] ๐๐ฃ๐ โ ๐๐(1 โ ๐)
= โซ ๐๐ [(1
0.3333 ๐ฃ๐โ0.08664โ
๐๐
0.3333 ๐ฃ๐โ0.08664+
1
21.0541 ๐ฃ๐2+5.472384 ๐ฃ๐
] ๐๐ฃ๐ โ ๐๐(1 โ ๐) ๐๐
โ
= ๐๐ [(ln(0.3333 ๐ฃ๐โ0.08664)
0.3333 โ
๐๐ ln(0.3333 ๐ฃ๐โ0.08664)
0.3333 +
ln(5.472384
๐ฃ๐)+21.0541
5.472384 ] โ ๐๐(1 โ ๐)
= 0.873 ln(0.333๐ฃ๐ โ 0.08664) [1 โ ๐๐] + 0.053 ln (5.472384
๐ฃ๐) + 1.12 โ ๐๐(1 โ ๐)
(u*-u)/RTc Now to derive departure from internal energy
๐ขโ โ ๐ข
๐ ๐๐= โโซ ๐๐ [๐๐ (
๐๐๐
๐๐๐) โ ๐๐] ๐๐ฃ๐
๐๐
โ
= โซ ๐๐ [๐๐ (๐
๐๐๐) (
๐๐
0.3333 ๐ฃ๐โ0.08664โ
1
21.0541 ๐ฃ๐2+5.472384 ๐ฃ๐
) โ๐๐
0.3333 ๐ฃ๐โ0.08664โ
๐๐
โ
1
21.0541 ๐ฃ๐2+5.472384 ๐ฃ๐
] ๐๐ฃ๐
= โซ ๐๐ [(๐๐
0.3333 ๐ฃ๐โ0.08664โ
๐๐2
0.3333 ๐ฃ๐โ0.08664+
๐๐
21.0541 ๐ฃ๐2+5.472384 ๐ฃ๐
] ๐๐ฃ๐ ๐๐
โ
= ๐๐ ๐๐ [(ln(0.3333 ๐ฃ๐โ0.08664)
0.3333 โ
๐๐ ln(0.3333 ๐ฃ๐โ0.08664)
0.3333 +
ln(5.472384
๐ฃ๐)+21.0541
5.472384 ]
= 0.873 ๐๐ ln(0.333๐ฃ๐ โ 0.08664) [1 โ ๐๐] + 0.053Tr ln (5.472384
๐ฃ๐) + 1.12
๐โ โ ๐
๐น
๐ โ โ ๐
๐ = โซ ๐๐ [(
๐๐๐
๐๐๐) โ (
1
๐๐)] ๐๐ฃ๐ โ ln(๐ง)
๐๐
โ
=โซ ๐๐ [(1
0.3333 ๐ฃ๐โ0.08664โ
1
๐๐ ] ๐๐ฃ๐ โ ln(๐ง)
๐๐
โ
= ๐๐ [(ln(0.3333 ๐ฃ๐โ0.08664)
0.3333 โ (
1
ln๐ฃ๐) ]โ 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=(0.3471โ0.291)
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,
๐โ โ ๐
๐น๐ป๐:
We know,
๐โ โ ๐
๐ ๐๐= โโซ ๐๐ [(
๐๐๐
๐๐๐) โ
๐๐
๐๐]๐๐ฃ๐ + ๐๐๐๐(๐)
๐๐
โ
๐๐,๐โ โ ๐
๐ ๐๐= โโซ ๐๐ [(
๐
๐๐๐(
๐๐
0.3333 ๐ฃ๐ โ 0.08664โ
1
21.0541 ๐ฃ๐2 + 5.472384 ๐ฃ๐
)) โ๐๐
๐๐] ๐๐ฃ๐
๐๐
โ
+ ๐๐๐๐(๐)
Or, ๐โโ๐
๐ ๐๐= โโซ ๐๐ [
1
0.33โ๐๐โ0.086โ
๐๐
๐๐] ๐๐๐ + ๐๐๐๐(๐)
๐๐
โ
or, ๐โโ๐
๐ ๐๐= โ๐๐[(
ln(|165๐๐โ43|)
0.33โ ๐๐๐๐|๐๐|] + ๐๐๐๐(๐)
After putting the value of Zc we get,
๐โ โ ๐
๐ ๐๐= โ0.291[(
ln(|165๐๐ โ 43|)
0.33โ ๐๐๐๐|๐๐|] + ๐๐๐๐(๐)
๐โ โ ๐
๐น๐ป๐:
We know,
๐โ โ ๐
๐ ๐๐= โซ [๐๐๐๐ โ
๐๐
๐๐]๐๐ฃ๐ + ๐๐(๐๐๐ง + 1 โ ๐ง)
๐๐
โ
๐โ โ ๐
๐ ๐๐= โซ [๐๐ (
๐๐
0.3333 ๐ฃ๐ โ 0.08664โ
1
21.0541 ๐ฃ๐2 + 5.472384 ๐ฃ๐
) โ๐๐
๐๐]๐๐๐ + ๐๐(๐๐๐ง + 1
๐๐
โ
โ ๐ง)
๐โ โ ๐
๐ ๐๐= โ
๐๐๐๐(|1375๐๐ โ 361|)๐๐
0.33+
ln (|1368096
๐๐ + 5263525|)๐๐
5.47+ ๐๐(๐๐๐ง + 1 โ ๐ง)
After putting the value of Zc we get,
๐โ โ ๐
๐ ๐๐= โ
๐๐๐๐(|1375๐๐ โ 361|) โ .291
0.33+
ln (|1368096
๐๐ + 5263525|) โ .291
5.47+ ๐๐(๐๐๐ง + 1 โ ๐ง)
18 | P a g e
i) Speed of sound
๐ = โโ๐ฃ2 โ (๐๐
๐๐ฃ)๐
(๐๐
๐๐ฃ)๐ =
๐
๐๐ฃ{
๐ ๐
๐ฃ โ ๐โ
๐(๐)
๐ฃ(๐ฃ + ๐) }
=๐ ๐(โ1)
(๐ฃ โ ๐)2+
๐(๐) (2๐ + ๐)
(๐ฃ + ๐)2๐ฃ2
= โ๐ ๐
(๐ฃ โ ๐)2+
๐(๐) (2๐ + ๐)
(๐ฃ + ๐)2๐ฃ2
๐ = โโ๐๐{โ๐น๐ป
(๐โ๐)๐+
๐(๐ป) (๐๐+๐)
(๐+๐)๐๐๐ }
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,
๐ถ๐ = โ1
๐๐[๐ [
๐ ๐๐ฃ โ ๐
โ๐(๐)
๐ฃ(๐ฃ + ๐)
โ๐ ๐
(๐ฃ โ ๐)2 +๐(๐)
๐ฃ2(๐ฃ + ๐)+
๐(๐)๐ฃ(๐ฃ + ๐)2
] + ๐ฃ]
Cv Also the relation between Cp and Cv is,
๐๐ =๐ถ๐
๐
So,
๐ถ๐ฃ = โ1
๐๐๐พ[๐ [
๐ ๐๐ฃ โ ๐
โ๐(๐)
๐ฃ(๐ฃ + ๐)
โ๐ ๐
(๐ฃ โ ๐)2 +๐(๐)
๐ฃ2(๐ฃ + ๐)+
๐(๐)๐ฃ(๐ฃ + ๐)2
] + ๐ฃ]
1/v vp
ฮฒ = (1/๐ฃ)(๐๐ฃ
๐๐)๐
(๐๐ฃ
๐๐ก)๐ = โ
(๐๐
๐๐)๐ฃ
(๐๐
๐๐ฃ)๐
=
๐
๐ฃโ๐โ
๐(๐)
๐ฃ(๐ฃ+๐)
โ๐ ๐
(๐ฃโ๐)2+
๐(๐)
๐ฃ2(๐ฃ+๐)+
๐(๐)
๐ฃ(๐ฃ+๐)2
ฮฒ = โ1
๐ฃ[
๐
๐ฃโ๐โ
๐(๐)
๐ฃ(๐ฃ+๐)
โ๐ ๐
(๐ฃโ๐)2+
๐(๐)
๐ฃ2(๐ฃ+๐)+
๐(๐)
๐ฃ(๐ฃ+๐)2
]
k = v/p pvIsentropic expansion coefficient:
๐ = โ๐ฃ
๐ (
๐๐
๐๐ฃ) ๐
= -๐ฃ
๐ [โ
๐ ๐
(๐ฃโ๐)2+
๐(๐)
๐ฃ2(๐ฃ+๐)+
๐(๐)
๐ฃ(๐ฃ+๐)2]
๐ =โ๐ฃ
(๐ ๐
๐ฃโ๐โ
๐(๐)
๐ฃ(๐ฃ+๐)) [
โ๐ ๐
(๐ฃโ๐)2+
2๐
๐ (๐ฃ+๐)3]
20 | P a g e
kT
We know,
๐พ๐ = โ1
๐ฃ
๐ฟ๐
๐ฟ๐ฃ)
๐
= -1
๐ฃ [โ
๐ ๐
(๐ฃโ๐)2+
๐(๐)
๐ฃ2(๐ฃ+๐)+
๐(๐)
๐ฃ(๐ฃ+๐)2]
J
We know,๐๐ = (๐๐
๐๐) ๐ฃ
Also,
๐โ = ๐ถ๐๐๐ + [๐ฃ โ ๐(๐๐ฃ
๐๐)๐]
From isentropic process,
h= constant
so, ๐โ = 0
(๐๐
๐๐) โ =
๐ (๐๐ฃ
๐๐)๐โ๐ฃ
๐๐
Now, (๐๐ฃ
๐๐) ๐ = -
(๐๐
๐๐)๐ฃ
(๐๐
๐๐ฃ)๐
๐๐ = (๐๐
๐๐) โ
=
๐[โ(๐๐๐๐
)๐ฃ
(๐๐๐๐ฃ
)๐]โ๐ฃ
๐๐
๐๐ =
โ๐
๐ ๐ฃโ๐
โ๐(๐)
๐ฃ(๐ฃ+๐)
โ๐ ๐
(๐ฃโ๐)2+
๐(๐)
๐ฃ2(๐ฃ+๐)+
๐(๐)
๐ฃ(๐ฃ+๐)2
โ๐ฃ
๐๐
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.