Chapter 1

Question 11: Propane Tank

Propane is pressurized to liquify in order to store in propane cans.

a) what is the minimum pressure to ensure liquid propane forming at room temperature?

b) a pressure guage on a \(1\:L\) propane tank reads \(P_g=1\:MPa\) guage pressure. How much propane is contained in this tank?

Solution Approach for a)

Propane gas starts to liquify at its saturation pressure corresponding to room temperature \(T=250\:^{\circ}C\).

#importing the required library
import CoolProp.CoolProp as CP

#define variables
T_r = 25 + 273.15   #room temperature in K
fluid = 'propane'

P_s = CP.PropsSI("P", "T", T_r, "Q", 0 , fluid)   #saturation pressure of Propane in room temperature in Pa

print('The minimum pressure to liquify propane at room temperature is:', f"{P_s/1000:.1f}", 'kPa')

The minimum pressure to liquify propane at room temperature is: 952.1 kPa

Solution Approach for b)

from density

\(D=m/V\)

so

\(m=D\times V\)

V = 1E-3   #tank volume in m3
P_a = 101325   #atmospheric pressure in Pa
P_g = 1E+6   #guage pressure in Pa

P = P_g + P_a   #propane absolute pressure in Pa
D = CP.PropsSI("D", "T", T_r, "P", P , fluid)   #propane density in kg/m3

m = D * V   #propane mass in kg

print('The amount of propane stored in the tank is:', f"{m:.2f}", 'kg')

The amount of propane stored in the tank is: 0.49 kg