Question

A fuel gas containing 86% methane, 8% ethane, and 6% propane by volume flows to a furnace at a rate of 1450 m3/h at 15°C and 150 kPa (gauge), where it is burned with 8% excess air. Calculate the required flow rate of air in SCMH (standard cubic meters per hour).

Answer 1

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