Ch4 Material Balances-part2-revised

# Wetgas yn2 05 y co2 03 y o2 01 yh2o 01 05

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Unformatted text preview: cess air **Air: 21% oxygen 4.76 mole air / mole O2 “Stack Gas” or “Flue Gas” = product leaving combustion chamber. Theoretical Oxygen: moles or molar flow rate of O2 needed for complete combustion of all fuel to CO2 + H2O Theoretical Air: the amount of air needed to satisfy theoretical O2 **Air: 21% oxygen 4.76 mole air / mole O2 Excess Air: the amount of air that exceeds theoretical air Percent Excess Air = 100x(moles fed – moles theoretical)/(moles theoretical) Balances on Combustion Reactions: • theoretical oxygen is calculated from stoichiometry based on complete oxidation (combustion) of all reactants– regardless of how much is really reacted, or whether it is actually completely oxidized. • Any of the balance methods can be used, but atomic balances are preferred / typically the easiest. Composition on a dry basis: the composition of a gas that contains water, but calculated with the water taken out. WetGas : yN2 0.5 y CO2 0.3 y O2 0.1 yH2O 0.1 0.5 0.56 0.9 0.3 y CO2 0.33 0.9 0.1 0.11 y O2 0.9 1.00 Dry Basis yN2 10% water, 90% dry gas If yN2 0.6 Dry gas, w / 15% Humidity yN2 ,wet 0.6 mole 0.85 mole 0.51 mole dry mole wet Page 146 test yourself CH CH4 + 2 O2 CO2 + 2 H2O CH4 + 3/2 O2 CO + 2 H2O 1) 2x100 = 200 mol/h 2) Still 200 mol/h !!! 3) 200 mol/h x 4.76 mole air / mole O2 = 952 mole air/h 4) 100 = 100*(n – 952)/952 n = 1904 mole/h (= 2x952) 5) % excess = 100*(300‐200)/200 = 50% 100 mol/h CH4...
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