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Judging from these two equations, the mole ratio of oxygen to sewage is (1/4)/(1/50) = (25/2). Oxygen is the same as the ultimate carbonaceous biochemical oxygen demand, CBOD. Let fBOD be the mole ratio of BOD5 to CBOD, which would be the same as the mole ratio of BOD5 to oxygen. The mole ratio of BOD5 to sewage is then f BOD (25/2).

mass of cells mass of cells . „ c5h7no2 1 „ mol cells

BOD5mass of BOD, implies YBOD5mass of BOD5 = 113 YBOD5mol BOD

f bod5(32)

1 mol cells

113 Y bod5

/bod5 ["2 J mol sewage

Vf bod5/ mol sewage

Although not correct exactly, for practical purposes, COD may considered equal to CBOD or oxygen:

mass of cells v mass of cells . r v C5H7no2 = 1 mol cells YCODmass of COD implies YCOD;m^Z:oOD = 113Ycoditk.I COD

113 (25) mol sewage

Therefore,

c dc CODmol sewage

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