Coal with the given ultimate analysis and a heating value of 12,850 Btu/lb is bu
ID: 991498 • Letter: C
Question
Coal with the given ultimate analysis and a heating value of 12,850 Btu/lb is burned in a utility boiler. The dry volumetric stack gas analysis is also given below. Determine the following... a) Excess air percentage, b) total pounds of water vapor up the stack per 100 pounds of coal burned.
Coal Ultimate Analysis- Carbon(C)- 72%, Hydrogen(H2)- 4.4%, Sulfur(S)- 1.6, Oxygen(O2)- 3.6, Nitrogen(N2)- 1.4, Water(H20)- 8%, Ash- 9%
Dry Stack gas analysis (component) (Volume percent)= (CO2)- 13.03177934, (N2a+N2)- 80.75336314, (SO2)- 0.10838367, (O2)- 6.10647385
Explanation / Answer
Coal Ultimate Analysis- Carbon(C) = 72%, Hydrogen(H2) = 4.4%, Sulfur(S) = 1.6, Oxygen(O2) = 3.6, Nitrogen(N2) =1.4, Water(H2O) = 8%, Ash = 9%
Dry Stack gas analysis (component) (Volume percent)= (CO2) = 13.03177934, (N2a+N2) = 80.75336314, (SO2) = 0.10838367, (O2) = 6.10647385
need the realation molar of the componet and we know than this composition is in % w/w so divide any %/MW
Carbon(C) = 72/12 = 6 mol C Hydrogen(H2) = 4.4/1 = 4.4mol Sulfur(S) = 1.6,/32 = 0.05mol Oxygen(O2) = 3.6/32 = 0.1125 mol Nitrogen(N2) =1.4/14 = 0.1mol Water(H2O) = 8/18 = 0.44mol and Ash = 9g
and in the gas the composition to pressure of 1 atm or 760mmHg we have than PV = nRT R = 0.082atm l /k mol T = 25C +273 = 298K and P = 1atm
n = PV/RT =) n = 1atm x V/24.436atm l/mol =) Dry Stack gas analysis (component) in mol = (CO2) = 13.03177934 x1atm/24.436atm l/mol = 0.533mol , (N2a+N2) = 80.75336314x1atm/24.436atm l/mol = 3.30mol , (SO2) = 0.10838367x1atm/24.436atm l/mol = 0.0044mol , (O2) = 6.10647385x1atm/24.436atm l/mol = 0.25mol
a) the nO2 present in the gas are same the n in excess = 0.25mol in relation whit coal = g of O2 /g coal + g of O2 = ) g of O2 = 0.25mol x 18g/mol = 4.5g the exess = 100x 4.5g/104.5g = 4.3%
b) for 100 pound of coal we have 8 pound of water