Example Problem with Complete Solution

2F-1 : An Application of Equations of State 10 pts
Estimate the pressure of ammonia at a temperature of 22oC and a specific volume of 0.500 m3/kg.
               
a.) The Ideal Gas EOS          
b.) The Virial EOS          
c.) The van der Waal EOS        
d.) The Soave-Redlich-Kwong EOS      
e.) The Compressibility Factor EOS      
f.) The Steam Tables          
               
Read : Not much to say here.
Given : T 22 oC
V 0.500 m3/kg
Find : P ??? kPa
Data : R 8.314 J/mol-K
MW 17.03 g NH3 / mol NH3
Tc 405.55 K
Pc 1.128E+07 Pa
w 0.250
Part a.) Ideal Gas EOS : Equation 1 Eqn 1
Solve for pressure : Equation 2 Eqn 2
We must determine the molar volume before we can use Eqn 2 to answer the question.
Use the definition of molar volume: Equation 3 Eqn 3
Where : Equation 4 Eqn 4
Therefore : Equation 5 Eqn 5
V 8.52E-03 m3/mol
Now, plug values back into Eqn 2. T 295.15 K
P 2.882E+05 Pa
Be careful with the units. P 288.2 kPa
Part b.) Truncated Virial EOS : Equation 6 Eqn 6
We can estimate B using : Equation 7 Eqn 7
Equation 8 Eqn 8
Equation 9 Eqn 9
Where : Equation 10 Eqn 10
We can solve Eqn 6 for P : Equation 11 Eqn 11
Plugging numbers into Eqns 10, 8, 9, 7 and 11 (in that order) yields :
TR 0.728 B -2.23E-04 m3/mol
B0 -0.6186 Z 9.74E-01
B1 -0.5143 P 280.6 kPa
Part c.) van der Waal EOS : Equation 12 Eqn 12
We can determine the values of a and b, which are constants that depend only on the chemical species in the system, from the following equations.
Equation 13
Eqn 13 Equation 14 Eqn 14
Tc 405.55 K a 0.4252 Pa-mol2/m6
Pc 1.128E+07 Pa b 3.74E-05 m3/mol
Now, we can plug the constants a and b into Eqn 12 to determine the pressure.
P 283.6 kPa
Part d.) Soave-Redlich-Kwong EOS : Equation 15 Eqn 15
We can determine the values of a, b and a, which are constants that depend only on the chemical species in the system, from the following equations.
Equation 16 Eqn 16 Equation 17 Eqn 17
Equation 18 Eqn 18 Equation 19 Eqn 19
Equation 20 Eqn 20
Where : w 0.250
Now, plug values into Eqns 15 - 20 :
TR 0.7278 a 0.43084 Pa-mol2/m6
m 0.8633 b 2.590E-05 m3/mol
a 1.26971 P 281.5 kPa
Part e.) Compressibility EOS :
Given TR and the ideal reduced molar volume, use the compressibility charts to evaluate either PR or the compressibility, Z
Equation 21 Eqn 21 From part c : TR 0.7278
Defiition of the ideal reduced molar volume :
Equation 22 Eqn 22
VRideal 28.49
Read the Generalized Compressibility
Chart for PR = 0 to 1 :
PR 0.025
Z 0.975
We can use the definition of PR to calculate P :
Equation 23 Eqn 23 Equation 24 Eqn 24
P 282.0 kPa
Or, we can use Z and its definition to determine P :
Equation 25 Eqn 25
P 281.0 MPa
Part f.) The Ammonia Tables provide the best available estimate of the pressure.
We begin by determining the state of the system.  In this case, it would be easiest to lookup the vsat vap and vsat liq at the given temperature.
If : v > vsat vap Then : The system contains a superheated vapor.
If : v < vsat liq Then : The system contains a subcooled liquid.
If : vsat vap > v > vsat liq Then : The system contains an equilibrium mixture of saturated liquid and saturated vapor.
Data : P*(kPa) T (oC) vsat liq (m3/kg) vsat vap (m3/kg) hsat liq (kJ/kg) hsat vap (kJ/kg)
913.27 22 0.001647 0.1405 284.4 1462.9
Because V > Vsat vap, the ammonia is superheated in this system.
At this point we can make use of the fact that we have a pretty good idea of what the actual pressure is in the tank (from parts a-d) or we can scan the superheated vapor tables to determine which two pressures bracket our known value of the specific volume.  The given specific volume of 0.500 m3/kg lies between 250 kPa and 300 kPa and T = 22oC lies between 20oC and 30oC.  This is a tricky multiple interpolation problem !
The Superheated Ammonia Table gives us :
P*(kPa) T (oC) v (m3/kg) h (kJ/kg)
250 20 0.5568 1507.6
250 30 0.5780 1530.3
300 20 0.4613 1504.2
300 30 0.4792 1527.4
We can now interpolate on this data to determine values of the specific volume at T = 22oC at BOTH 250 kPa and 300 kPa.  This will help us setup a second interpolation to determine the pressure that corresponds to T = 22oC and V = 0.500 m3/kg.
At 250 kPa :
T (oC) V (m3/kg)
20 0.5568
30 0.5780
Equation 26 Eqn 26
Equation 27 Eqn 27
slope 2.120E-03 (m3/kg)/oC V 0.56104 m3/kg
At 300 kPa :
T (oC) V (m3/kg)
20 0.4613
30 0.4792
Equation 28 Eqn 28
Equation 29 Eqn 29
slope 1.790E-03 (m3/kg)/oC V 0.46488 m3/kg
Now, we must interpolate one more time to determine the pressure which, at 22oC, yields a spoecific volume of 0.500 m3/kg :
At 22oC :
P (kPa) V (m3/kg)
250 0.56104
300 0.46488
Equation 30 Eqn 30
Equation 31 Eqn 31
slope -520.0 (m3/kg)/kPa P 281.7 kPa