Ch4 Material Balances-part2-revised

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

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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...
View Full Document

Ask a homework question - tutors are online