Question

# A fuel gas containing 74.0% methane, 12.0% ethane, and 14.0% propane by volume flows to a...

A fuel gas containing 74.0% methane, 12.0% ethane, and 14.0% propane by volume flows to a furnace at a rate of 1450 m3/h at 15.0°C and 180.0 kPa (gauge), where it is burned with 16.0% excess air.

Calculate the required flow rate of air in SCMH (standard cubic meters per hour).

You may take the composition of air to be 21.0 mol% O2 and the balance N2.

Solution:

Explanation: To calculate volumetric flow rate of the gases Methane = (74/100)×1450 = 1073m³/hr

Ethane = (12/100)×1450 = 174m³/hr

Propane = (14/100)×1450 = 203m³/hr

Stichiometric equation for gas to complete combustion CH4 + 2 O2 = CO2 +2 H2O

1m³/hr CH4 requires 2m³/hr O2

1073m³/hr CH4 requires (2×1073)m³/hr O2

1073m³/hr CH4 requires 2146m³/hr O2

C2H6 + 7/2 O2 = 2 CO2 + 3H2O

1m³/hr C2H6 requires 7/2m³/hr O2

174m³/hr C2H6 requires (174×7/2)m³/hr O2

174m³/hr C2H6 requires 609m³/hr O2

C3H8 + 5 O2 = 3 CO2 + 4 H2O

1m³/hr C3H8 requires 5m³/hr O2

203m³/hr C3H8 requires (203×5)m³/hr O2

203m³/hr C3H8 requires 1015m³/hr O2

Total O2 required = 2146+609+1015

Total O2 required = 3770m³/hr

Excess air = 16% × 3770 = 603.2m³/hr

Total flow rate of air = 3770 + 603.2 = 4373.2m³/hr (Ans)

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