A gas turbine power plant receives a shipment of hydrocarbon fuel whose composit

Question

A gas turbine power plant receives a shipment of hydrocarbon fuel whose composition is uncertain but may be represented by the expression C_xH_y. The fuel is burned with excess air. An analysis of the product gas gives the following results on a moisture-free basis: 10.5%(v/v) CO_2, 5.3% O_2, and 84.2% N_2. Determine the molar ratio of hydrogen to carbon in the fuel (r), where r = y/x, and the percentage excess air used in the combustion. What is the air-to-fuel ratio (m^3 air/kg of fuel) if the air is fed to the power plant at 30 degree C and 98 kPa? The specific gravity of the fuel (a petroleum product) is 0.85. Estimate the ratio standard cubic feet of gas fed to the turbine per barrel of fuel. For part (c), use 1 barrel = 55 gallons.

Explanation / Answer

Basis : 1 mole of product gas , it contains 0.105 moles CO2, 0.053 moles CO2 and 0.842 moles of N2.

N2 has come from Air, moles of air supplied = 0.742/0.79 ( since air contains 79%N2 and 21%O2)=0.94 moles

O2 in the air =0.94*0.21= 0.1974, CO2 formed = 0.105 moles, Oxygen consumed for formation of CO2 from C+O2--àCO2 is 0.105 moles

Oxygen remaining = 0.1974-0.1050=0.0924

Oxygen present in the product = 0.053 moles, oxygen used for reaction of H2+0.5O2à H2O

0.0924-0.0530=0.0394, moles of H2O formed = 0.053*2=0.106

Moles of Hydrogen =0.106, moles of H= 0.106*2= 0.212

Moles of C = 0.105

Molar ratio of C:H= 0.105:0.212= 1:2

Moles of air = 0.94, volume of air from V= nRT/P =0.94*0.0821L.atm/mole.K *(30+273)/(98/101.3)atm=24.17L =24.17*10-3 m3

Mass of fuel = 0.105*12 ( Carbon)+0.106*2 * hydrogen)= 1.472 gm = 1.472*10-3 kg

Air to fuel ratio = 24.17*10-3m3/ 1.472*10-3 kg =16.42 m3/kg

c. Fuel taken = 1 barrel = 55 gallon =55*3.78 L =207.9L

mass of fuel = 207.9 L*1000cc *0.85*1g/cc = 207900gm

C:H ratio = 1:2

Mass of Carbon =12 gm and mass of H =2 gm

Total mass of 1 mole of CH2= 14 gm

Moles = 207900/14=14850

moles of C in the fuel = 14850*1/3 = 4950 and moles of H= 4950*2= 9900

moles of Oxygen required for combustion of C= 4950 moles and moles of Oxygen required for combustion of H= 9900/2= 4950

total oxygen required = 4950+4950 =9900 moles

air required = 9900/0.21 = 47143 moles

1 mole of any gas at STP occupies 359 ft3

47143 moles occupy 47143*359= 16924337ft3

Air supplied per barrel of fuel = 16924337ft3/barrel

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