Ask Question
13 May, 02:55

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 25.0°C and 210.0 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.

+1
Answers (1)
  1. 13 May, 06:23
    0
    V=4373.2m³/hr

    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
Know the Answer?
Not Sure About the Answer?
Get an answer to your question ✅ “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 25.0°C and ...” in 📙 Chemistry if there is no answer or all answers are wrong, use a search bar and try to find the answer among similar questions.
Search for Other Answers