Question

A gas turbine power plant receives a shipment of hydrocarbon fuel whose composition is uncertain but may be represented by the expression CxHy. The fuel is burned with excess air. An analysis of the product gas gives the following results on a moisture-free basis: 9.80% (v/v) CO2, 5.1% O2, and 85.1% N2. Collapse question part Fuel Composition Determine the molar ratio of hydrogen to carbon in the fuel r = y/x, and the percentage excess air used in the combustion.

Answer 1

Answer:

See explanation..

Explanation:

Equation of reaction;

CxHy + x+y/4(O2 +3.76N2) ------------->xCO2 + y/2 H20 + ( x +y/4 ) O2 + 3.76 ( x + y/4 ) N2

One metre cube(m^3)of CxHy burnt requires = 4.76 ( x+ y/4 ) of air.

The combustion products formed ; yCO2 = x / 4.76 ( x +y/4 ) +y/4 10.5 /100

= x/ 4.76 ( x + y/4 ) +y/4 0.105 ×4.76 ( x + y/4 ) +y/4.

= x 0.5 x + 0.125 y + y/4

Multiply it by 4 -2x+0, we have;

5y + y = 4x 1.5y = 2x

molar ratio of hydrogen/ carbon,r= y/x= 2/1.5= 1.34.

For yN2 = 84.2 /100 = 3.76 ( x +y/4 ) / 4.76 (x +y/4 ) + y/4

For yO2

= 5.3 /100

= x+ y/4 / 4.76 x + 4.76 y/4 +y/4 0.053

= x +y/4 / 4.76x + 5.76 y/4 0.053( 4.76 x + 5.76y/4 )

= x+y/4 1.2 x + 0.07632 y

= x + 0.25 y 0.2 x

= 0.25 y - 0.07632 y

x = 0.174y/ 0.2= 0.87 y

y/x = 1/0.87 = 1.33

Therefore,

CxHy = C2H2 +3 (O2+3.76 N2 ) ------->2CO2 + O + H2O + 3.76N2

1 m^3+ 3m^3 O2 of oxygen= 2 m^3 + 1 m^3 O2

= 3 × 4.76 = 14.76 m^3

The air required = 14.28 m^3

CO2 - 10.5 m^3 O2 = 5.3 m3, N2 = 84.2 m3 ,O2 = 84.2 /3,76 = 22.3