PVT Data from Compressibility Factor Chart

Home -> ChE Learning Resources -> Solved Problems -> Thermodynamics->

Nitrogen gas at a pressure of 100 Bar and -70oC is contained in a tank of 0.25 m3. Heat is added until the temperature is 37oC. Determine approximately, through the use of Z factor,

(a) Specific volume of gas, in m3/kmol
(b) The final pressure, in Bar.

Solution:

Data for N2:

Pc = 33.9 Bar
Tc = 126.2 K

The value of universal gas constant, consistent with the system of units in this problem is,

R = 1.01325 x 22.4/273 = 0.0831 Bar.m3/(kmol.K)

(a) Molar Volume of Gas:

Initial conditions:

P1 = 100 Bar
T1 = (273 - 70) K
V1 = 0.3 m3

Pr1 = P1/Pc = 100/33.9 = 2.9499
Tr1 = T1/Tc = 203/126.3 = 1.6082

Refer to the Z table 1 and 2.

Interpolating the value of Z for the Tr1 and pr1 values, Z = 0.846

Molar volume of gas (v1) is calculated as follows: v1 = Z x R x T1/P1 = 0.8460 x 0.0831 x 203 /100 = 0.1428 m3/kmol

(a) Molar Volume of Gas:

Final conditions:

T2 = (273 + 37) K
v2 = v1

P2v2/(RT2) = Z2.

In the above equation, both pressure and Z are unknowns.

To calculate Z at the final conditions, the following iterative procedure is used:

(i) Assume a value of pressure (Passumed). For the first iteration the assumption shall be the ideal gas value (Pideal) obtained from the relation Pv = RT.

(ii) Using the assumed pressure of step (i), calculate Tr, Pr and the corresponding value of Z from tables/charts.

(iii) Evaluate P from Pcalculated = Z x Pideal.

(iv) If the calculated pressure of step (iii), is equal to that of that assumed in step (i), the problem is essentially solved, (i.e., the assumed pressure is the answer to the problem), else repeat steps (i) to (iii) till Passumed is nearly equal to Pcalculated.

The above procedure is applied to this problem and the calculations are as below:

Iteration 1:

Passumed = Pideal = R x T2/v2 = 180.5 Bar

Pr = 180.5/33.9 = 5.3246; Tr = 310/126.2 = 2.4564.

For these Pr and Tr values, Z = 1.0432.

Therefore, Pcalculated = 1.0432 x 180.5 = 188.3 Bar.

Since Pcalculated <> Passumed, let us assume another pressure value say, 185 Bar.

Iteration 2:

Passumed = 185 Bar

Z = 1.0464

Pcalculated = 188.88 Bar

Iteration 3:

Passumed = 190Bar

Z = 1.0499

Pcalculated = 189.51 Bar

Iteration 4:

Passumed = 189.5Bar

Z = 1.0495

Pcalculated = 189.4 Bar

In the fourth iteration, 'assumed' and 'calculated' pressures are almost equal. So the pressure at the final conditions when T = 37oc = 189.5 Bar

[Index]

Last Modified on: 01-May-2024

Chemical Engineering Learning Resources - msubbu
e-mail: learn[AT]msubbu.academy
www.msubbu.in