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Ch4 Material Balances-part2(1)

# 1 05 056 09 03 y co2 033 09 01 011 y o2 09 100 dry

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Unformatted text preview: “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 Example 4.7‐1 (page 131) Assume n1 100moles y1M 0.078 y1O 0.194 y1N2 0.728 fM 0.90 CH4 3 / 2O2 CO 2H2O sD/C 8 M O C W CH4 2O2 CO2 2H2O D At least one input or output stream amount or flow must be given. If not, assume a value as a basis. Use atomic balances (2 reactions) Flow, mol/s Total ni 1 2 100 n2 7.8 19.4 72.8 0 0 0 n2M n2O n2N2 n2C n2D n2w Comp, nij nM nO nN2 nC nD nW nuk 0 6 ndf = nuk – nat bal – nmol bal – nother 6 ‐ 3 ‐ 1 (N2) ‐ 2 = 0 input = output C: 7.8 = n2M + n2C + n2D and n2M = 7.8(1‐ 0.90) = 0.78 moles (fM) n2D = 8n2C (SD/C) so 7.8 = 0....
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