Example 2F - 1: An Application of Equations of State

2F-1 :	An Application of Equations of State	10 pts

Estimate the pressure of ammonia at a temperature of 27^oC 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:

^oC

Find:

???

kPa

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.

8.314

J/mol-K

T_c

405.55

17.03

g NH₃ / mol NH₃

P_c

1.128E+07

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

1.066E-02

m³/mol

Now, plug values back into Eqn 2.

300.15

2.341E+05

Be careful with the units.

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 :

T_R

0.740

-2.14E-04

m³/mol

B₀

-0.6000

9.80E-01

B₁

-0.4698

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

T_c

405.55

0.4252

Pa-mol²/m⁶

P_c

1.128E+07

3.74E-05

m³/mol

Now, we can plug the constants a and b into Eqn 12 to determine the pressure.

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 :

0.250

Now, plug values into Eqns 15 - 20 :

T_R

0.7401

0.43084

Pa-mol²/m⁶

0.8633

2.590E-05

m³/mol

1.25575

229.9

kPa

Part e.)

Compressibility EOS :

Given T_R and the ideal reduced molar volume, use the compressibility charts to evaluate either P_R or the compressibility, Z

Eqn 21

From part c :

T_R

0.7401

Defiition of the ideal reduced molar volume :

Eqn 22

V_R^ideal

35.67

Read the Generalized Compressibility
Chart for P_R = 0 to 1 (not easy!):

P_R

0.02

0.983

We can use the definition of P_R to calculate P :

Eqn 23

Eqn 24

225.6

kPa

Or, we can use Z and its definition to determine P :

Eqn 25

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 V_{sat vap} and V_{sat liq} at the given temperature.

If :

V > V_{sat vap}

Then :

The system contains a superheated vapor.

If :

V < V_{sat liq}

Then :

The system contains a subcooled liquid.

If :

V_{sat
vap} > V > V_{sat
liq}

Then :

The system contains an equilibrium mixture of saturated liquid and saturated vapor.

Data :

P*(kPa)

T (^oC)

V_{sat
liq} (m³/kg)

V_{sat vap} (m³/kg)

H_{sat liq} (kJ/kg)

H_{sat vap} (kJ/kg)

1066.56

1.67E-03

0.12066

308.11

1465.42

Because V > V_{sat 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 = 27^oC lies between 25^oC and 50^oC. This is a tricky multiple interpolation problem !

The Superheated Ammonia Table gives us :

P*(kPa)

T (^oC)

V (m³/kg)

H (kJ/kg)

200

0.6995

1510.1

200

0.7255

1532.5

250

0.5668

1518.2

250

0.6190

1574.7

We can now interpolate on this data to determine values of the specific volume at T = 27^oC at BOTH 200 kPa and 250 kPa. This will help us setup a second interpolation to determine the pressure that corresponds to T = 27^oC and V = 0.626 m³/kg.

At 200 kPa :

T (^oC)

V (m³/kg)

0.6995

0.7255

Eqn 26

Eqn 27

slope

2.600E-03

(m³/kg)/^oC

0.71766

m³/kg

At 250 kPa :

T (^oC)

V (m³/kg)

0.5668

0.6190

Eqn 28

Eqn 29

slope

2.085E-03

(m³/kg)/^oC

0.57102

m³/kg

Now, we must interpolate one more time to determine the pressure which, at 27^oC, yields a spoecific volume of 0.626 m³/kg :

At 27^oC :

P (kPa)

V (m³/kg)

200

0.7177

250

0.5710

Eqn 30

Eqn 31

slope

-340.96

(m³/kg)/kPa

231.3

kPa

Verify:

No assumptions to verify.

Answers :

a.)

234.1

kPa

d.)

229.9

kPa

b.)

229.4

kPa

e.)

230.1

kPa

c.)

231.2

kPa

f.)

231.3

kPa

Example Problem with Complete Solution