Chemical Engineering Question

Question

An oxygen tank with a volume of 2.5 ft3 is kept in a room at 50°F. An engineer has used the

ideal gas equation of state to determine that if the tank is first evacuated and then charged with 35.3

lbm of pure oxygen, its rated maximum allowable working pressure (MAWP) will be attained.

Operation at pressures above this value is considered unsafe.

a) What is the maximum allowable working pressure (psig) of the tank?

b) You suspect that at the conditions of the fully charged tank, ideal-gas behavior may not be a good

assumption. Use the SRK equation of state to obtain a better estimate of the maximum mass of oxygen that may be charged into the tank. Did the ideal-gas assumption lead to a conservative estimate (on the safe side) or a nonconservative estimate of the amount of oxygen that could be charged?

c) Suppose the tank is charged and ruptures before the amount of oxygen calculated in Part (b)

enters it. (It should have been able to withstand pressures up to four times the MAWP.) Think of at least five possible explanations for the failure of the tank below its rated pressure limit.

Answer 1

a) We need to use the ideal gas equation: P.V = n.R.T with P in Pa, V in m³, n in mol, R = 8.314 SI, T in K.

We have:

1. V = 2.5 ft³ = 0.07079 m³.
2. n = m/M with m = 35.3 lbm = 35.3x453.592 = 16,011.79 g and M = 2x16.0 = 32.0 g/mol for O₂ so n = 16,011.79/32.0 = 500.36 mol.
3. T = 50°F = (50°F − 32) × 5/9 + 273.15 = 283.15 K.

So: P_max = n.R.T/V = 500.36x8.314x283.15/0.07079 = 16,639,384.5 Pa = 166.3x10⁵ Pa = 164.2 atm = 2,417 psig !!!

b) The SRK equation of state is: P' = R.T/(V_m − b) − a/((T)^1/2.V_m.(V_m + b))

with:

1. V_m = V/n = Molar volume = 0.07079/500.36 = 0.00014148 m³/mol (= 0.14148 L/mol)
2. a = 1.814×10^-6 Pa.K^1/2.m⁶/mol² (for 283.15 K and P = 164 atm - Note: we can find different values for that "constant").
3. b = 3.21×10^-11 m³/mol (for 283.15 K).

So, we calculate: P'_max = 8.314x283.15/(0.00014148 − 3.21×10^-11) − 1.814×10^-6/((283.15)^1/2.0.00014148.(0.00014148 + 3.21×10^-11)) = 16,639,163.7 Pa < P = 16,639,384.5 Pa.

The pressure is less with the SRK equation, that means the maximum mass of oxygen that may be charged into the tank is less:

n = PV/RT = m/M so m_max = P_max.V.M/( R.T) = 16,639,163.7x0.07079x32.0/(8.314x283.15) = 16011.31 g < 16,011.79 g.

c) If the tank breaks before Pmax, it means interactions between molecules are bigger and the SRK equation is not enough to show that, and the temperature is not constant at 50°F (it increases with the pressure)...