evaluation of pure component fugacity

7/25/2019 Evaluation of Pure Component Fugacity

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Evaluation of Pure Component Fugacity (f), and Coefficient Fugacity ()

Find pure component fugacity (f) and coefficient fugacity () of Hydrogen at 0oC temperature and 960

atm pressure

1. Evaluation of Pure Component Fugacity (f) and Coefficient Fugacity () from compressibility data

Data requirement for this case is compressibility data at least from 0 pressure until 960 atm pressure

which the compressibility data must evaluate at same temperature (0oC)

P (atm) Z P (atm) Z

100 1.069 600 1.431

200 1.139 700 1.504300 1.209 800 1.577

400 1.283 900 1.649

500 1.356 1000 1.72

Coefficient Fugacity () can evaluate from equation

ln= 1

=

lnis area of bellow curve from P=0 until P=P (P=960 for this case). The y axis curve is

and x axiscurve is P.

(atm) Almost 0 Almost 1 Equal to 0.000000

100 1.069 0.000690

200 1.139 0.000695

300 1.209 0.000697

400 1.283 0.000708

500 1.356 0.000712600 1.431 0.000718

700 1.504 0.000720

800 1.577 0.000721

900 1.649 0.000721

1000 1.72 0.000720


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nox axis

lengthy axis length Area

1 50 0.0002 0.01

2 50 0.00055 0.0275

3 50 0.0007 0.035

4 50 0.0007 0.035

5 50 0.0007 0.035

6 50 0.0007 0.035

7 50 0.0007 0.035

8 50 0.00071 0.03559 50 0.00071 0.0355

10 50 0.00071 0.0355

11 50 0.00071 0.0355

12 50 0.00071 0.0355

13 50 0.00072 0.036

14 50 0.00072 0.036

15 50 0.00072 0.036

16 50 0.00072 0.036

17 50 0.00072 0.036

18 50 0.00072 0.03619 60 0.00072 0.0432

0.000000

0.000100

0.000200

0.000300

0.000400

0.000500

0.000600

0.000700

0.000800

0 100 200 300 400 500 600 700 800 900 1000 1100

(Z-

1)/P

P

1

2

3 4 5 6 7 8 91

0

1

1

1

2

1

3

1

4

1

5

1

6

1

7

1

8

1

9

= 0.6492

ln=

ln= 0.6492 = . = .

= = 1.9140 960 =


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2. Evaluation of Pure Component Fugacity (f) and Coefficient Fugacity () from virial equation

Data requirement for this case is acentric factor (), Critical Temperature (, and Critical Pressure (.For Hydrogen:

= 33.2 = 13 = 0.22= =

960 13 =

960 12,83 = 74.8246

= 0.083 0.422. = 0.083 0.4228.2229.= 0.068

= = 033.2=

273 33.2 = 8.2229

= 0.139 0.172. = 0.139 0.172. = 0.1390

ln=+ ln=74.82468.22290.0685 + 0.220.1390ln= 0.6286 = . = .

= = 1.8750 960 =

3. Evaluation of Pure Component Fugacity (f) and Coefficient Fugacity () from qubic equation

Eq of State (Tr) Zc

vdW (1873) 1 0 0 1/8 27/64 3/8

RK (1949) Tr-1/2 1 0 0.08664 0.42748 1/3

SRK (1972) Srk(Tr;) 1 0 0.08664 0.42748 1/3

PR (1976) PR(Tr;) 1 + 2 1 2 0.07779 0.45724 0.30740

,= [1 + 0.480 + 1.574 0.176 ( 1 )]

,= [1 + 0.37464 +1.542260.26992 ( 1 )]


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3.a Van der Waals (vdw) Equation state

Step :

- Determine and from literature data and calculate and For hydrogen

= 33.2 = 13 = 0.22= =

960 13 = 960 12,83 = 74.8246 = = 033.2= 273 33.2 = 8.2229

-

Determine (Tr); ; ; ; from table

= 1 ; = 0 ; = 0 ; = 1/8 ; = 27/64-

Calculate =

=1

8(74.8246

8.2229)= 1.1374

= = 27/64 11/8 8.2229= 0.4104

- Calculate Z

= 1 + + + = 1+1.1374 0.41041.1374 1.1374 +01.1374 +01.1374 = 2.1374 0.46679 1.1374

= 2.0362

- Choose equation I and calculate I

Because = , so equation I is = + =

1.13742.0362+01.1374= 0.5586

- Pure Component Fugacity (f) and Coefficient Fugacity ()

ln= 1 ln ln= 2.0362 1 ln2.03621.1374 0.41040.5586ln= 0.91353

= .

= . = = 2.4931 960 =


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3.b Redlich Kwong (RK) Equation state

Step :


= 33.2 = 13 = 0.22= =

960 13 = 960 12,83 = 74.8246 = = 033.2= 273 33.2 = 8.2229

-


= 1 ; = 0 ; = 0.08664 ; = 0.62748= .= 8,2229.= 2.8676

- Calculate

=

= 0.0866 (74.82468.2229)= 0.7884 = =0.62748 2.8676

0.0866 8.2229 = 1.7206

- Calculate Z

= 1 + + + = 1+0.7880 1.72060.7880 0.7880 +00.7880 +10.7880 = 1.5066

-

Choose equation I and calculate I

Because , so equation I is = 1 ln (

+ + )=

11 0 ln (

1.5066+10.78841.5066+00.7884)= 0.4209


ln= 1 ln ln= 1.5066 1 ln1.50660.7884 1.72060.4209ln= 0.1134 = . = . = = 1.1201 960 =


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3.c Soave Redlich Kwong (SRK) Equation state

Step :


= 33.2 = 13 = 0.22= =960 13 = 960 12,83 = 74.8246 = = 033.2= 273 33.2 = 8.2229

-


= 1 ; = 0 ; = 0.08664 ; = 0.62748,=[1 + 0.480 + 1.574 0.176 (1 )]

,=1 + 0.480+1.5740.220.1760.22 18.2229,= 0.2780

- Calculate = = 0.0866 (

74.82468.2229)= 0.7884

= =0.62748 0.2780

0.0866 8.2229 = 0.1668

- Calculate Z

= 1 + + +

= 1+0.7880 0.24480.7880 0.7880

+00.7880 +10.7880

= 1.7599- Choose equation I and calculate I


+ + )=

11 0 ln (

1.7599+10.78841.7599+00.7884)= 0.3702


ln= 1 ln

ln= 1.7599 1 ln1.75990.7884 0.16680.3702

ln= 0.7271 = . = . = = 2.0690 960 =


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3.d Peng Robinson (PR) Equation state

Step :


= 33.2 = 13 = 0.22= =960 13 = 960 12,83 = 74.8246 = = 033.2= 273 33.2 = 8.2229

-


= 1 + 2 ; = 1 2 ; = 0.07779 ; = 0.45724,=[1 + 0.37464 +1.54226 0.26992 (1 )]

,=1 + 0.37464+1.542260.220.269920.22 18.2229,= 0.0954

- Calculate = = 0.07779 (

74.82468.2229) = 0.7079

= =0.45724 0.09540.07779 8.2229= 0.0682

-

Calculate Z

= 1 + +

= 1+0.7079 0.06820.7079 0.7079 + 1 20.7079 +1+ 20.7079 = 1.6961- Choose equation I and calculate I


+ + ) =

11 + 2 1 2 ln

1.6961+1+ 20.70791.6961+1 20.7079 = 0.3488


ln= 1 ln ln= 1.7466 1 ln1.74660.7884 0.06820.3488ln= 0.6841 = . = .

= = 1.9820 960 =


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Comparations Pure Component Fugacity (f) and Coefficient Fugacity () of Hydrogen at 0oC

temperature and 960 atm pressure from several Equation.

Equation Coefficient fugacity, Fugacity, f (atm)

From Compressibility Data 1.9140 1837

Virial Equation 1.8750 1800

Van der Waals (VdW) 2.4934 2394

Redlich Kwong (RK) 1.1201 1075

Soave Redlich Kwong (SRK) 2.0690 1962

Peng Robinson (PR) 1.9820 1903

evaluation of pure component fugacity

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