A gaseous fuel containing methane and ethane is burned with excess air. The fuel

Question

A gaseous fuel containing methane and ethane is burned with excess air. The fuel enter the furnace at 25 degree C and 1 atm, and the air enter at 200 degree C and 1 atm. The stack gas leaves the furnace at 1000 degree C and 1 atm and contains 5.32 mole % CO_2, 1.60% 7.32% O_2, 12.24% H_2O, and the balance N_2. (a) Calculate the molar percentages of methane and ethane in the fuel gas and the percentage excess air fed to the reactor. (b) Calculate the heat (KW) transferred from the reactor per 10 m^3/s of fuel gas (c) If 90% of the heat generated from (b) above is used to produce a 20 bar steam 400 degree C from its liquid at 20 degree C, determine the volumetric flow rate of the product steam.

Explanation / Answer

assume a ideal gas behaviour

Nitrogen balance

theoretical in air 3.76 mol of nitrogen in present in air

3.76 n3 = (100) (0.7352) mol N2

3.76 n3 = 73.52 mol N2

n3 = 73.53 mol N2 / 3.76

n3 = 19.55 mol O2 fed

Carbon balance

n1 contains 1 carbon atom, n2 contains 2 carbon atom

n1 (1) + n2 (2) = (100) (0.0532)(1) + (100)(0.0160)(1)

n1 (1) + n2 (2) = 6.92

n1 (1) = 6.92 - n2 (2)

Hydrogen balance

n1 (4) + n2 (6) = (100) (0.1224)(2)

(4)(6.92 - 2 n2 ) + 6 n2 = 24.48

27.68 - 8 n2 + 6 n2 = 24.48

- 2 n2 = 24.48 - 27.68

-2 n2 = - 3.2

n2 = 3.2 / 2

n2 = 1.60 mol C2H6

n1 (1) = 6.92 - n2 (2)

n1 (1) = 6.92 - (1.60) (2)

n1 = 6.92 - 3.20

n1 = 3.72 mol CH4

total fuel consumption

n1 + n2

= 3.72 + 1.60 = 5.32

so percentage molar of methane

5.32 = 100 %

3.72 = x

x = 3.72 x 100 / 5.32

x = 372 / 5.32

x = 69.9 mole % CH4

and

mole % CH4 + mole % C2H6 = 100 %

69.9 mole % of CH4 +mole % C2H6 = 100 %

mole % C2H6 = 100 % - 69.9 mole % of CH4

mole % C2H6 = 30.1 mole % C2H6

b) Theoretical oxygen

3.72 mol CH4 * 2 mol O2 / 1 mol CH4 + 1.60 mol C2H6 * 3.5 mol O2 / 1 mol C2H6

7.44 + 5.6 = mol O2

13.04 mol O2

% Excess air

= (19.55 - 13.04) mol O2 in excess / 13.04 mol O2 required * 100%

= 6.51 / 13.04 * 100 % excess air

= 0.50 * 100 % excess air

= 50 % excess air

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