Question

Some wastewater has a BOD5 of 150 mg/L at 20 o C. Nitrification was inhibited. The reaction rate k at that temperature has been determined to be 0.23 /day. (a) Find the ultimate carbonaceous BOD. (b) Find the reaction rate coefficient at 15 o C. (c) Find the BOD5 at 15 o C.

Answer 1

given,

BOD₅ = 150 mg/L at 20°C

k = 0.23/day

using formula

`L = L_0(1-e^(-kt))`

a)
`150= L_0(1-e^(-0.23* 5))\\150=L_0* 0.683\\L_0=219.62\ mg/L`

L₀ = 219.62 mg/L

b)
`k_(15)=k_(20)(1.047)^(T-20)\\k_(15)=0.23* (1.047)^(-5)\\k_(15)=0.183`

k = 0.183

c) BOD₅ at 15⁰C

`L = L_0(1-e^(-kt))`

`L =219.62(1-e^(-0.183* 5))`

L = 131.772 mg/L