Stoichiometry-calculation of mass fraction of water in exhaust

[marellasunny]

Homework Statement

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 $ξ_{H2O,air}$ with the assumption that no other hydrogen compositions, besides water, are to be found in the exhaust gas.

Homework 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)

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/sMASS FRACTION OF WATER IN HUMID AIR= ?

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=$\frac{m_{H20,air}}{m_{air,humid}}$

∴MASS FRACTION OF WATER IN HUMID AIR=$\frac{m_{H20,air}}{m_{H20,air}+m_{air,dry}}$

$$m_{air,dry}=28.5 g/s$$

 

[SteamKing]

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