Question

A certain crude oil shows the following ultimate analysis: C=87.1% and H2=12.9%. The volumetric analysis of the dry flue gas resulting from the combustion of this oil shows CO2=12%, O2=4.6%, CO=0%, and N2=83.4%

a. Theoretical Air-Fuel Ratio

b. Actual Air Fuel Ratio

c. Excess air in Percent

Answer 1

a) First we need to find the Oxygen(O₂) needed to burn Carbon(C) and Hydrogen(H₂):

For Carbon:

From the above equation we see that 1kg of C requires 2.67kg of O₂

O₂ required for 0.871kg (Given in question) of C = 0.871x2.67 = 2.325kg of O₂

For Hydrogen:

From the above equation we see that 1kg of Hygrogen(H₂) requires 8kg of Oxygen(O₂)

O₂ required for 0.129kg(Given in question) of H₂ = 1.032kg of O₂

Total O₂ required = 2.325 + 1.032 = 3.357kg of O₂

As O₂ present in 100kg of air is 23kg by mass

Theoretical air required = kg = 14.595 kg/kg of fuel

Theoretical air/fuel ratio = 14.595 (Ans)

b) The dry fuel gas composition by volume needs to be converted to by mass before calculating the actual air required. The calculations is showed in t he table below:

Fuel gas Constituents	Volume in 1m3 of fuel gas (a)	Molecular Weight (b)	Proportional Mass in kg (c)=(a)x(b)	Actual mass in kg/kg of fuel gas (d)	Mass of C in kg/kg of fuel gas (e)
CO2	0.12	44	5.28
O2	0.046	32	1.472		--
CO	0.0	28	0.0	0.0	--
N2	0.834	28	23.352		--
Total	1.00	--	30.104	1.00	0.0478

Mass of N₂ in the fuel gas per kg of fuel = 0.775 x 18.22 = 14.12kg

As nitrogen present in 100 kg of air is 77 kg and 23 kg of O₂

Actual mass of air supplied =

Actual air/fuel ratio = 18.34 (Ans)

c)Amount of excess air supplied per kg of fuel burnt = Actual air supplied - Theoretical air supplied

=18.34 - 14.595 = 3.745kg