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(20)1. A wastewater has a BOD5 of 192 mg/L, a reaction rate k of 0.12/day, and a total Kjeldahl nitrogen content (TKN) of 35

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Answer #1

BOD5 = 192 mg/l

Reaction rate constant, k = 0.12/day

Total Kjeldahl nitrogen content (TKN) = 35 mg/l

Ultimate carbonaceous oxygen demand (CBOD) :

Lo = BOD5 /(1-e-kt)

Lo = 192 mg/l /(1-e-0.12x5)

Lo = 192/(1-e-0.6)

Lo = 192/(1-0.549)

Lo = 192/0.451 = 425.72 mg/l

b) Ultimate nitrogenous oxygen demand (NBOD) :

NBOD = 4.57 mg O2/mg N2 ) x TKN = 4.57 x 35 = 159.95 mg/l

c) Total/theoretical oxygen demand (TBOD) : CBOD + NBOD = 425.72 + 159.95 mg/l = 585.67 mg/l

d) Remaining BOD after 5 days have elapsed = Total BOD - BOD5 = 585.67 - 192 mg/l = 393.67 mg/l

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