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20 May, 18:27

Fuel cells have been proposed for use in cars and for power generation as part of a hydrogen economy. They offer the advantages of higher efficiency and cleaner fuel with insignificant pollutants. The reaction is 2H2 + O2 - > 2H2O Where the oxygen comes from the air (21% O2 and 79% N2). Hydrogen flows into a fuel cell at a rate of 27 gmol/min. Air (consisting of oxygen and nitrogen) enters the fuel cell in a separate stream. The amount of oxygen entering the fuel cell is 50% more than that needed to react stoichiometrically with all of the entering hydrogen. The conversion of hydrogen in the fuel cell is 85%. Assume that only a single stream exits the fuel cell.

What is the flow rate of each of the species leaving the fuel cell? Note that the flow rates given correspond to a fuel cell rated at approximately 50 kW using 2004 technology.

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  1. 20 May, 21:25
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    The reaction in question is as follows;

    2H2 + O2 - > 2H2O

    from the reaction, 2 moles of hydrogen react with 1 mole of O2 to give 2 moles of water. given the flow rate of hydrogen = 27 gmol/min gmol is defined = mass in grams / molecular weight molecular weight of H2 = 2 mass of hydrogen = 27 gmol x 2 = 54 g of H2 flows / min

    from the reaction; 2moles of H2 react with 1 mole of O2, then the number of moles of O2 that will be needed to react with 27moles of Hydrogen will be 13.5moles which implies the amount of oxygen that needs to flow. As such, the amount of oxygen is 50% more than needed, this makes oxygen the excess reactants.

    As the amount needed is 13.5 moles which is 50% more = 13.5 x 50/100 = 6.75 moles.

    Therefore, total moles = 13.5 + 6.75 = 20.25 moles

    Since the oxygen comes from air which has a component of 21 mole % oxygen and 79 mole% N2. it shows that since we have 20.25 moles of oxygen, then 76.17 moles of nitrogen also will be required.

    Given that the conversion of H2 is 85%, hence the remaining hydrogen is unburnt which is 0.15 x 27 moles = 4.05 moles of hydrogen remaining. Amount of hydrogen burning is 0.85 x 27 = 22.95 moles

    This will consume 22.95/2 = 11.475 moles of oxygen so the amount of O2 unreacted will be = 20.25 - 11.475 = 8.775 moles of oxygen. from the reaction, since 2moles of H2 produce 2moles of water, hence 22.95 moles of hydrogen that is burning it will also produce 22.95 moles of water.

    Therefore, the flow rate of each of the specie will be;

    8.775 gmol/min of oxygen, 22.95 g mol/min of water, 4.05 gmol/min of hydrogen, 76.17 gmol/min of nitrogen.
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