Ethane enters furnace at 0.9 kg/min and burns with 200% excess air. 50% of fuel

Question

Ethane enters furnace at 0.9 kg/min and burns with 200% excess air. 50% of fuel is consumed and combustion is assumed complete (meaning no CO is produced). Find the flow rate of Oxygen, Nitrogen, and Air at the inlet, the molar flow rate of all species, and the mass flow rate of water vapor in the exit stream.

Explanation / Answer

ethane molecular weight ,M = 28 molar rate of inflow = 900/28 = 225/7 mol/min= 32.143 mol/min since combustion equation C2H6 + 3.5O2 ------> 2CO2 + 3H2O thus 1 mole of Ethane needs 3.5 mol of oxygen thus 32.143 mol needs 112.5 mol of oxygen since 200% excess air thus amount of oxygen entering is 3 times the required i.e oxygen entering = 3*112.5 = 337.5 mol thus flow rate of oxygen = 337.5 mol/min since in 1 mol of air Nitrogen = 0.78084 mol oxygen = 0.209476 mol since 1 mol air contains 0.209476 mol oxygen thus 1611.1631 mol of air contains 337.5 mol of oxygen thus flow rate of air = 1611.1631 mol/min flow rate of nitrogen = 1611.1631*0.78084 = 1258.0606 mol/min flow rate of ethane = 32.143 mol/min since 50% of fuel is burned thus fuel burn rate = 0.5*32.143 = 16.0715 mol/min since burning 1 mol ethane generates 3 mol of water vapor thus rate of water vapor = 3*16.0715 = 48.2145 mol/min = 48.2145*18 g/min = 867.861 g/min = 0.868 kg/min = flow rate of water vapor

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