Answer :
Answer:
when 5% excess air is supplied, moles of air supplied/moles of fuel = [tex]23.81\times 1.05 =25[/tex]
Explanation:
Equivalence ratio = 0.6
Equivalence ratio = Actual air to fuel ratio (AAFR)/ stoichiometric air to fuel ratio SAFR
combustion reaction of propane is
[tex]C_3H_8+ 5O_2 ----->3CO_2+4H_2O[/tex]
From above reaction, 1 mole of propane, from the reaction, 5 moles of oxygen required,
we know that air contains 21% O_2 and 79% N_2,
Therefore, moles of air based on stoichiometry [tex]= \frac{5}{0.21} = 23.81[/tex]
Theoretical air to fuel ratio [tex]= \frac{23.81}{1} = 23.81[/tex]
Given[tex] \frac{AFR}{SFR} = 0.6[/tex]
Actual Air Fuel Ratio [tex]= 23.81\times 0.6 = 14.3[/tex]
when 5% excess air is supplied, moles of air supplied/moles of fuel = [tex]23.81\times 1.05 =25[/tex]