- #8,506
htf
- 56
- 0
How did you get these figures? I get 2 m^3 / sec.NUCENG said:I calculated a boiloff rate based on adiabatic heating and came up with 20 m^3/sec.
I calculate 1.35 kg/sec boiloff rate: 3.05e6 W/(2.256e6 J/kg) These are ~75 mol, which give 22.4 * 373/273 * 75 l = 2.3e3 l = 2.3 m^3 steam - make it 2 m^3 because it is saturated and not an ideal gas.