Below a discharge from a wasterwater treament plant , an 8.6km stream has a reox

Question

Below a discharge from a wasterwater treament plant , an 8.6km stream has a reoxygenation constant of .4 d^-1, a velocity of .15m/s, a dissolved oxygen concentration of 6mg/L, and an ultimate demand of 25mg/L. The stream is 15C, the deoxygenation constant is estimated at .25d^-1. What will be the impact on fish population in the stream

Explanation / Answer

Kr = 0.4 day^-1 v = 0.15 m/s D.O. of mixture of sewage & water = 6 mg/L Ultimate BOD of the mixture(sewage+water), L = 25 mg/L Kd = 0.25 day^-1 T = 15 degree celsius length of stream = 8600 m time required for the sewage to cross the stream, t = Length of stream/velocity of discharge => t = 8600/0.15 = 57333.33 s = 15.93 hours From Streeter-Phelep's equation; Oxygen deficit at any time t, D(t) = [Kd*L/(Kr- Kd)]*[10^(-Kd*t) - 10^(-Kr*t)] + D0*10^(-Kr*t) => D(t= 57333.33 s) = [0.25*25/(0.4-0.25)]*[10^(-0.25*57333.33) - 10^(-0.4*57333.33)] + D0*10^(-0.4*57333.33) from this equation, we will find the value for D0. Now, D0 = initial oxygen deficit = saturated D.O. - D.O. of mixture = saturated D.O. - 6 mg/L from this, we will get the value for saturated D.O. The minimum dissolved oxygen in water for the survival of fishes must be atleast 4 mg/L. So, Dc = saturated D.O. - 4 mg/L            ----->(1) Also, [L/(Dc*f)]^(f-1) = f*[1-(f-1)*(D0*L)] where, L = 25 mg/L f = self purification constant = Kr/Kd = 1.6 D0 = saturated D.O. - 6 mg/L => from here also, we will get the value of Dc, which should not be greater than Dc value calculated in equation (1), otherwise the fishes will die.

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