Example Problem with Complete Solution

2F-1 : An Application of Equations of State 10 pts
Estimate the pressure of ammonia at a temperature of 27oC and a specific volume of 0.626 m3/kg.
a.) Ideal Gas EOS
b.) Virial EOS
c.) van der Waal EOS
d.) Soave-Redlich-Kwong EOS
e.) Compressibility Factor EOS
f.) Steam Tables.
Read : Not much to say here.
Given: T 27 oC Find: P ??? kPa
V 0.626 m3/kg
Assumptions: None.
Equations / Data / Solve:
Begin by collecting all of the constants needed for all the Equations of State in this problem.
R 8.314 J/mol-K Tc 405.55 K
MW 17.03 g NH3 / mol NH3 Pc 1.128E+07 Pa
w 0.250
Part a.) Ideal Gas EOS :
Eqn 1
Solve for pressure :
Eqn 2
We must determine the molar volume before we can use Eqn 2 to answer the question.
Use the definition of molar volume:
Eqn 3
Where :
Eqn 4
Therefore :
Eqn 5
V 1.066E-02 m3/mol
Now, plug values back into Eqn 2. T 300.15 K
P 2.341E+05 Pa
Be careful with the units. P 234.1 kPa
Part b.) Truncated Virial EOS :
Eqn 6
We can estimate B using :
Eqn 7
Eqn 8
Eqn 9
Where :
Eqn 10
We can solve Eqn 6 for P :
Eqn 11
Plugging numbers into Eqns 10, 8, 9, 7 and 11 (in that order) yields :
TR 0.740 B -2.14E-04 m3/mol
B0 -0.6000 Z 9.80E-01
B1 -0.4698 P 229.4 kPa
Part c.) van der Waal EOS :
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.
Eqn 13
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 231.2 kPa
Part d.) Soave-Redlich-Kwong EOS :
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.
Eqn 16
Eqn 17
Eqn 18
Eqn 19
Eqn 20
Where : w 0.250
Now, plug values into Eqns 15 - 20 :
TR 0.7401 a 0.43084 Pa-mol2/m6
m 0.8633 b 2.590E-05 m3/mol
a 1.25575
P 229.9 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
Eqn 21 From part c : TR 0.7401
Defiition of the ideal reduced molar volume :
Eqn 22
VRideal 35.67
Read the Generalized Compressibility
for PR = 0 to 1 (not easy!):
PR 0.02
Z 0.983
We can use the definition of PR to calculate P :
Eqn 23
Eqn 24
P 225.6 kPa
Or, we can use Z and its definition to determine P :
Eqn 25
P 230.1 kPa
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)
1066.56 27 1.67E-03 0.12066 308.11 1465.42
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 0.626 m3/kg lies between 200 kPa and 250 kPa and T = 27oC lies between 25oC and 50oC.  This is a tricky multiple interpolation problem !
The Superheated Ammonia Table gives us : P*(kPa) T (oC) V (m3/kg) H (kJ/kg)
200 20 0.6995 1510.1
200 30 0.7255 1532.5
250 25 0.5668 1518.2
250 50 0.6190 1574.7
We can now interpolate on this data to determine values of the specific volume at T = 27oC at BOTH 200 kPa and 250 kPa.  This will help us setup a second interpolation to determine the pressure that corresponds to T = 27oC and V = 0.626 m3/kg.
At 200 kPa : T (oC) V (m3/kg)
20 0.6995
30 0.7255
Eqn 26
Eqn 27
slope 2.600E-03 (m3/kg)/oC V 0.71766 m3/kg
At 250 kPa : T (oC) V (m3/kg)
25 0.5668
50 0.6190
Eqn 28
Eqn 29
slope 2.085E-03 (m3/kg)/oC V 0.57102 m3/kg
Now, we must interpolate one more time to determine the pressure which, at 27oC, yields a spoecific volume of 0.626 m3/kg :
At 27oC : P (kPa) V (m3/kg)
200 0.7177
250 0.5710
Eqn 30
Eqn 31
slope -340.96 (m3/kg)/kPa P 231.3 kPa
Verify: No assumptions to verify.
Answers : a.) P 234.1 kPa d.) P 229.9 kPa
b.) P 229.4 kPa e.) P 230.1 kPa
c.) P 231.2 kPa f.) P 231.3 kPa