LECT06. Real Gas Relationships -...

Introduction to Chemical Engineering Calculations

Prof. Manolito E Bambase Jr. Department of Chemical Engineering. University of the Philippines Los Baños

Lecture 6.

Real Gas Relationships

31

Real Gas Relationships6

Prof. Manolito E Bambase Jr. Department of Chemical Engineering. University of the Philippines Los Baños SLIDE

2

PVT Relationship for a Real Gas

For a real gas,

PV nRT

The best method for obtaining the PVT relationship for a real gas is thru experimentation. In the absence of an experimental data, the following methods may be used:

1. Compressibility Chart

2. Equations of State

3. Estimated Properties



3

Compressibility Factor, Z

The compressibility factor is a factor that compensates for the nonideality of the gas.

Using this factor, the ideal gas equation becomes a real gas equation:

PV = nZRT

This is called the generalized equation of state.

The value of Z can be obtained from compressibility charts.



4



5

The Critical and Reduced Properties

Substances have critical properties (critical temperature critical pressure, and critical volume). Values of critical properties for common substances are available.

Reduced properties are PVT conditions normalized by their respective critical conditions:

r r rC C C

T P VT P VT P V



6

Example 6-1. Use of the Compressibility Factor

After liquid ammonia (NH3) fertilizer has been injected into the soil, there is still some ammonia left in the source tank (V = 120 ft3), but in the form of a gas.

Determine the weight of NH3 remaining in the tank if the pressure is 292 psig and temperature of the tank is 1250F.



7


Solve for n using the generalized equation of state,

PVnZRT

As given, values of the P-V-T are

P = 292 psi + 14.7 psi = 306.7 psiT = 12560F + 460 = 5850RV = 120 ft3

And the value of R:

3

0

ft atmR 10.73

lbmol R



8


The value of Z is obtained from the generalized compressibility chart if Tr and Pr are known,

The critical properties for ammonia are:

TC = 405.5K 729.90R

PC = 11.28 Mpa 1636 psia

Solving for the reduced parameters:

0

r r0c c

T 585 R P 306.7psiaT 0.801 and P 0.187

T P 1636psia729.9 R



9


From the compressibility chart,

Z 0.855

Solving for n and m:

3

330

0

3 33

306.7 psia 120ftn 6.858lbmol NH

ft atm0.855 10.73 585 Rlbmol R

17lbmm 6.858lbmol NH 116.58lbm NH1lbmol NH


Prof. Manolito E Bambase Jr. Department of Chemical Engineering. University of the Philippines Los Baños SLIDE10


Liquid oxygen is used in the steel industry, in the chemical industry, in hospitals, as rocket fuel, and for wastewater treatment as well as many other applications.

In a hospital a tank of 0.0284 m3 volume is filled with 3.500 kg of liquid O2 that vaporized at -250C.

Determine whether the pressure in the tank exceed the safety limit of the tank of 104 kPa.




Using the generalized equation of state:

nZRTPV

Given the following values:

n = 3.500 kg O2 (1/32) = 0.109 kmol

T = - 250C + 273 = 248 K

V = 0.0284 m3

R = 8.314 (kPa)(m3)/(kmol)(K)




To determine the value of Z, values of Tr and Pr must be known.

Critical properties of oxygen:

TC = 154.8 K

PC = 5.08 MPa = 5080 kPa

And the reduced temperature is

rc

T 248KT 1.602

T 154.8K




The pressure is determined by iteration

Step P (kPa) Pr Z Pcalc (kPa) e = |P – Pcalc|

1 10000.00 1.97 0.87 6845.20 3154.80

2 6845.20 1.35 0.91 7201.31 356.11

3 7201.31 1.42 0.88 6983.90 237.41

4 6983.90 1.37 0.89 7043.04 79.14

5 7043.04 1.39 0.90 7122.17 79.14



Equations of State – Van der Waals Equation

2

2 2

2 2C

C

C

C

a n aˆP V b RT or P V nb nRTV̂ V

ˆwhere V specific molar volume(V/n)

R T27 a =64 P

RT1b8 P



Equations of State – Redlich-Kwong Equation

1/2

2 2.5C

C

C

C

RT aP ˆ ˆ ˆV b T V V b

R Twhere a = 0.42748P

RTb 0.08664P



Equations of State – Peng-Robinson Equation

2 2C

C

C

C21/2

r

2

RT aP ˆ ˆ ˆ ˆV b V V b b V b

R Twhere a =0.45724P

RTb 0.07780P

1 1 T

0.37464 1.5422 0.26992accentricfactor



Example 6-3. Use of Van der Waals’ Equation

A cylinder 0.150 m3 in volume containing 22.7 kg of propane (C3H8) stands in the hot sun. A pressure gauge shows that the pressure is 4790 kPa gauge.

Using the Van der Waals equation of state, determine the temperature of the propane in the cylinder.




Solving for T using Van der Waals equation

2

2n aP V nbVT

nR

Given the following values:

n = 22.7 kg (1/44) = 0.516 kmol = 516 gmol

V = 0.150 m3 = 0.150 106 cm3

P = 4790 kPa + 101 kPa = 4891 kPa = 48.3 atm

R = 82.06 (atm)(cm3)/(gmol)(K)




The values of a and b can be computed using values for the critical properties or obtained from handbook.

23 36 cm cma 9.24 10 atm and b 90.7

gmol gmol

Solving for T:

2 66

26

516 9.24 1048.3 0.150 10 516 90.7

0.150 10T

516 82.06T 384K



Example 6-4. Solution of Van der Waals Equation for V

Given the values in a vessel of

Solve for the volume of the vessel.

234

3

0

30

ftP 679.7 psia a 3.49 10 psialbmol

psia ftn 1.136 lbmol R 10.73lbmol RftT 683 R b 1.45

lbmol




Since the Van der Waals equation cannot be easily solved explicitly for volume, it is written in cubic form:

Obtaining the coefficients:

2 33 2

3 23 2 1 0

nRT n a n abV nb V V 0P P P

(a x a x a x a 0)

2 3nRT n a n abnb 13.896 ; 66.262; 109.147P P P




Rewriting the equation:

f(V) = V3 – 13.896V2 + 66.262V – 109.147 = 0

Solve for the roots using any suitable technique such as the Newton’s method which is an iterative procedure. Starting with a guess value for V, a new value is calculated using the following formula:

kk 1 k

k

f (V )V Vf '(V )




f’(V) is obtained by differentiating f(V):

f’(V) = 3V2 – 27.792V + 66.262

For the initial guess value, the volume is calculate using the ideal gas law:

30

1.136 10.73 683nRTV 12.26ftP 679.7




Step VK(ft3) f(V) f'(V) VK+1(ft3) E

0 12.26 457.32 176.45 9.67 2.59

1 9.67 136.30 77.99 7.92 1.75

2 7.92 40.81 34.34 6.73 1.19

3 6.73 12.26 15.13 5.92 0.81

4 5.92 3.61 6.89 5.40 0.52

5 5.40 0.92 3.66 5.15 0.25

6 5.15 0.13 2.69 5.10 0.05

7 5.10 0.00 2.55 5.10 0.00

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LECT06. Real Gas Relationships -...

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