Question

A 60,000 USWG LPG tank is to be filled. The safety constraint is that the tank cannot be charged with an amount of fuel than would exert 75% of the maximum bursting pressure at 125°F. a) How many kg of liquified alkane fuel can be safely loaded into the tank? b) calculate the energy release in tons of TNT if a mishap occurred with a fully loaded tank. P_max = 320 PSI. The alkane is propane

Answer 1

Psafe =0.75*Pmax =0.75*320 =240 psi =1.6547E+06Pa

60000 US Fluid Gallons = 227.1247068 Cubic Meters

Ideal gas equation: P*V =n*R*T => m= M*P*V/(R*T)   (since n =m/M)

=> mass that can be loaded at 125 oF is m =44.1 *1.6547*10^6* 227.12/(8.314*324.8) =6137439.09 g =6137.439kg

b) HHV of Propane =50.35 MJ/kg

=> Total energy of propane =50.35*10^6*6137.439 =309020053650J =3.09*10^11 J

Energy per ton of TNT     = 4.18 x 10⁹ Joules

=> ton of TNT =3.09*10^11 J/(4.18*10^9 J/ton) =73.92 Ton