Question

A lake is fed by a polluted stream and a sewage outfall. The stream and sewage wastes have a decay rate coefficient (k) of 0.5/day (1st order units). Assuming complete mixing and no other water losses or gains, what is the steady-state pollutant concentration as mg/L in the lake? Incoming Stream: C = 10 mg/L, Q = 40 m^3/s Sewage Outfall: C = 100 ppm, Q = 0.5 m^3/s Lake: V= 200 m^3

Answer 1

Given :

k = 0.5 per day

`$C_s = 10 \ mg/L \ ; \ \ Q_s= 40 \ m^3/s$`

`$C_(sw) = 100 \ ppm \ ; \ \ Q_(sw)= 0.5 \ m^3/s$`

Volume, V
`$= 200 \ m^3$`

Now, input rate = output rate + KCV ------------- (1)

Input rate
`$= Q_s C_s + Q_(sw)C_(sw)$`

`$=(40 * 10) + (0.5* 100)$`

`$= 2 * 10^5 \ mg/s$`

The output rate
`$= Q_m C_(m)$`

= ( 40 + 0.5 ) x C x 1000

`$=40.5 * 10^3 \ C \ mg/s$`

Decay rate = KCV

∴
`$KCV =(0.5/d * C \ * 200 * 1000)/(24 * 3600)$`

= 1.16 C mg/s

Substituting all values in (1)

`$2 * 10^5 = 40.5 * 10^3 \ C+ 1.16 C$`

C = 4.93 mg/L