a If there is 100 mgL of sewage ie C10H19O3N then what is th

a) If there is 100 mg/L of sewage (i.e. C10H19O3N), then what is the COD of the wastewater? b) If the G0’ for that reaction is 30 kJ/e- eq, then what cell yield, in g cell/g COD, would result during aerobic degradation of domestic wastewater? [Note COD specified, so you will need to use the COD’/weight value you determined in part a.]

Solution

(a) solution:-

                             2C₁₀H₁₉O₃N + 25O₂ =    20 CO₂ + 16 H₂O + 2 NH₃

2(12*10 + 1*19 + 16*3 + 14)      25*32

402 g/l                                               800 g/l

Then for 1 g/l CO₂H₁₉O₃N     needed    1.99 g/l COD

Then for 100 mg/l CO₂H₁₉O₃N   needed   100*1.99 = 199 mg/l COD     Answer

(b)

Given

G₀ = 30 kJ/e-eq

For this

f_s = 0.59 g cell

Yield = 0.59 g cell/(1.99 g COD)

         = 0.296 g cell/(g COD)     Answer