Question

Propane burns at an equivalence ratio (ER) of 0.6, determine actual air-fuel ratio. If excess air is 5%, what will be the actual air fuel ratio?

Answer 1

Answer:

when 5% excess air is supplied, moles of air supplied/moles of fuel = $23.81\times 1.05 =25$

Explanation:

Equivalence ratio = 0.6

Equivalence ratio = Actual air to fuel ratio (AAFR)/ stoichiometric air to fuel ratio SAFR

combustion reaction of propane is

$C_3H_8+ 5O_2 ----->3CO_2+4H_2O$

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 $= \frac{5}{0.21} = 23.81$

Theoretical air to fuel ratio $= \frac{23.81}{1} = 23.81$

Given$\frac{AFR}{SFR} = 0.6$

Actual Air Fuel Ratio $= 23.81\times 0.6 = 14.3$

when 5% excess air is supplied, moles of air supplied/moles of fuel = $23.81\times 1.05 =25$