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

when 5% excess air is supplied, moles of air supplied/moles of fuel =
`23.81* 1.05 =25`

Step-by-step 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
`= (5)/(0.21) = 23.81`

Theoretical air to fuel ratio
`= (23.81)/(1) = 23.81`

Given
`(AFR)/(SFR) = 0.6`

Actual Air Fuel Ratio
`= 23.81* 0.6 = 14.3`

when 5% excess air is supplied, moles of air supplied/moles of fuel =
`23.81* 1.05 =25`