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Homework Help: Stoichiometry-calculation of mass fraction of water in exhaust

  1. Jan 7, 2014 #1
    1. The problem statement, all variables and given/known data

    An engine is operated with a special fuel mixture solely composed of carbon, hydrogen and oxygen.
    A fuel breakdown showed the following values: mass fraction carbon
    mass fraction hydrogen ,fuel density
    c = 0.81
    h = 0.15
    ρ=0.75 kg/dm³
    fuel mass flow rate m_b=2 g/s
    relative air fuel ratio λ =1
    exhaust gas mass flow rate m_e =30.52 g/s

    1.Please determine the stoichiometric air requirement of the analysed fuel.

    2.The engine is operated with humid air which, as an approximation, is to be treated as a gas mixture of oxygen, nitrogen and water.
    Please calculate the water mass fraction of the humid air [itex]ξ_{H2O,air}[/itex] with the assumption that no other hydrogen compositions, besides water, are to be found in the exhaust gas.

    2. Relevant equations

    Stoichiometric air requirement = (1/0.232)*(2.664.c+7.937.h-o)

    dry air mass flow rate m_a =m_b.(Stoichiometric air requirement).(relative air fuel ratio λ)

    3. The attempt at a solution

    mass proportions of carbon c,hydrogen h and oxygen o in fuel must add-up to 1.
    therefore, o=0.04.
    Stoichiometric air requirement = (1/0.232)*(2.664.c+7.937.h-o)
    ∴Stoichiometric air requirement =14.26

    mass flow rate of dry air= mass flow rate of fuel.(relative air-fuel ratio).(Stoichiometric air requirement)

    ∴mass flow of dry air=28.5 g/s


    I know that,

    mass of water in exhaust=mass of water produced due to combustion of organic fuel(CxHyOz)+mass of water in the humid air)

    ∴MASS FRACTION OF WATER IN HUMID AIR=[itex]\frac{m_{H20,air}}{m_{air,humid}}[/itex]

    ∴MASS FRACTION OF WATER IN HUMID AIR=[itex]\frac{m_{H20,air}}{m_{H20,air}+m_{air,dry}}[/itex]

    $$m_{air,dry}=28.5 g/s$$
    Last edited: Jan 7, 2014
  2. jcsd
  3. Jan 7, 2014 #2


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