(a) Let us consider CH4 combustion in air with an equivalence ratio of 0.9, wher

Question

(a) Let us consider CH4 combustion in air with an equivalence ratio of 0.9, where we have air flow rate of 400 kg/s. (1) calculate the required flow rate of CH4 (in kg/s), the air-to-fuel ratio on mole basis, and the mole and mass fractions of CO2, H2O, O2, N2 in the product stream. (2) If reactants enter the combustion chamber at 790K, and the products are found to leave the chamber at 1900K, calculate the net heat loss from the combustion chamber. Use constant properties. For all product constituents, use properties near the average temperature of 1345K. (b) Through web search, write down the adiabatic flame temperatures for combustion in air with all reactants at STP, for methane, and hydrogen.

Explanation / Answer

(1)flow rate of CH4 =0.9 x 400Kg/s

                          =360Kg/s

mols of air in a second= 400x1000g/29gmol-1(considering oxygen and nitrigen present in air)

                                               =13793.1034mol

mols of CH4 in a second= 360x1000g/16gmol-1

                                               =22500.0000mol

mol ratio= 13793.1034mol:22500.0000mol

number of nitrogen moles=79% x13793.1034mol=10853.89mol

number of oxygen moles=21% x22500.0000mol=4725mol

CH4+2O2----------> CO2 + 2H2O accordingly 11250 mols of methane reacts in a second

OUT PUT STEAM

number of water mols =4725 mol

number of carbondioxide mols=2362.5 mol

number of nitrogen molesl=10853.89mol

remaining methane mols=22500- 2362.5 mol=20137.5mol

toal number of mols in output=38078.89mol

MOL FRACTIONS

WATER=11250/38078.89=0.29

carbondioxide=2362.5/38078.89=0.06

nitrogen=10853.89/38078.89=0.28

methane=20137.5/38078.89=0.52

BY MULTIPLYING EACH MOLS BY MOLECULAR WEIGHT OU CAN CALCULATE THE MASS FRACTIONS

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