A gaseous fuel is being fired under a boiler. The fuel contains only 64 mol % Th

Question

A gaseous fuel is being fired under a boiler. The fuel contains only 64 mol % The dry (or Orsat) flue gas analysis is: 11.2 % CO_2, 4.00 % O_2, and rest N_2. corrected to standard conditions: 1 atm and 15 degree C. Combustion is complete. Using a basis of 100 of gaseous fuel: (a) Determine the atom ratio of hydrogen to carbon r, r = y x. (b) Estimate the percent excess of air used in the combustion. (c) Calculate the air to fuel ratio (m63 air/kg fuel) if the air enter at 30 degree C and 103 kPa.

Explanation / Answer

A gaseous fuel is being fired under a boiler. the fuel contains only 64 mol% hydrocarbons with an average molecular weight of CxHy and 36 mol% nitrogen. The dry (or Orsat) flue gas analysis is 11.2% CO2. 4% O2 and rest N2.

The volume of the dry flue gases is 24 times the volume of the fuel gas fired with each volume corrected to standard conditions:

1 atm and 15o C.

Combustion is complete.

Using a basis of 100 mol of gaseous fuel:

a) Determine the atom ratio of hydrogen to carbon, r (r = y/x

b) Estimate the percent excess of air used in the combustion

c) Calculate the air to fuel ratio (m3 air/kg fuel) if the air enters at 30o C and 103 kPa

Fuel comprises:

Hydrocarbons ---> 64 mol%

Average Molecular Weight----> CxHy

Nitrogen ------> 36 mol%

Dry Gas Comprises:

CO2 ----> 11.2%

O2 ----> 4%

N2----> ? will be 85%

Volume of dry fuel = 24 x (Volume of gaseous fuel

Pressure, P = 1 atm, and Temperature, T = 15o C

Considering 100mol of gaseous fuel, the equation can be framed as:

xCxHy + y N2 + ath (O2 + 3.76N2) -------> aCO2 + bO2 + cN2 + dH2O

C = x (reactant), a (product), so x = a

H = y (reactant), 2d (product), y = 2d or d = y/2

O =2ath (reactant), 2a + 2b  (product), ath = a+b

N =(2 * 3.76)ath (reactant), 2c (product), c = 3.76ath

xCxHy + y N2 + ath (O2 + 3.76N2) -------> aCO2 + bO2 + cN2

CxHy + 2N2 + ath (O2 + 3.76N2) -------> CO2 + O2 + 7.52 N2

64CxHy + 36N2 + ath (O2 + 3.76N2) ------->11.2CO2 + 4O2 + 84.8N2 + dH2O

NItrogen, N= 84.8/3.76 = 22.5

C = 11.2/ 62 = 0.175

H =32

Atomic Ratio of Hydrgen to carbon, r = y/x

= 0.175

Percent excess of air, %excess air = (molesAir Fed - molesTheoritical)/molesTheoritical] x 100

Oxygen = 22.4 + 8/2

= 30.4/2 = 15.2

%excess air = 36/15.2 = 2.36 * 100 = 236%

Air Fuel Ratio, AF = mair/ mfuel

   100 moles = 0.1 kmol

Therefore, AF = 34 (3.6)/ 0.1 x 24

= 122.4 / 2.4 = 51 m3 air/kg fuel

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