Answer:
Basis: Hour
From the question, we will have the following reactions;
CH4 + 2O2 -------- CO2 + 2H20 (Methane with O2)
C2H6 + 3.5O2 -------- 2CO2 + 2H2O (Butane with O2)
C3H8 + 5O2 ---------- 2CO2 + 4H2O (Propane with O2)
But we are also given this,
R=8.314J/mol.K, V=1450m3/h, P=150kPa gauge, Pt=150+101kPa=251kPa, T=15C= 288K
Assuming they are all ideal gases, we can find the no of moles of the gases.
n=PV/RT,
n = 251x103 x 1450 /8.314 x 288
n = 151, 999mols = 152kmols
However from the input and complete reactions stoichiometries above, we will have,
1. Methane 86% = 0.86 x 152kmols = 130kmols, required O2 = 2 x 130.7 = 261.44kmols
2. Ethane 8% = 0.08 x 152kmols = 12.2kmols, required O2 = 3.5 x 12.2 = 42.56kmols
3. Propane 6% = 0.06 x 152 kmols = 9.2kmols, required O2 = 5 x 9.1 = 45.5kmols
O2 = 349.5kmols, with 8% excess, Total O2 = 349.5+ (0.08x349.5) = 377.46kmols
But Air:O2 = 21%: 100%
inflow Air = 377.46x 100/21= 1797.5kmols, at standard pressure and temperature.
From PV =nRT
V (M3/H) = nRT/P
===== 1797.5mol x 8.314Nm/mol.K x 273K/101325Nm-2 x 1000
Hence, the required flow rate of air in SCMH = 40,265m³/h
