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

Answer:

given,

BOD₅ = 150 mg/L          at 20°C

k = 0.23/day

using formula

$L = L_0(1-e^{-kt})$

a) $150= L_0(1-e^{-0.23\times 5})\\150=L_0\times 0.683\\L_0=219.62\ mg/L$

L₀ = 219.62 mg/L

b) $k_{15}=k_{20}(1.047)^{T-20}\\k_{15}=0.23\times (1.047)^{-5}\\k_{15}=0.183$

k = 0.183

c) BOD₅       at       15⁰C

 $L = L_0(1-e^{-kt})$

 $L =219.62(1-e^{-0.183\times 5})$

     L = 131.772 mg/L