STEM Problem Detail

Problem #98

Environmental Science Hard

Problem Statement

Prompt

Provide me the solution of The Example 9.3.

Solution Rendering...

**Solution**

DO of stream $$= 90\%$$ of saturation DO
\[
= 0.9 \times 9.17 = 8.253 \text{ mg/l}
\]

DO of the mix
\[
= \frac{8.253 \times 5000 + 0 \times 1200}{5000 + 1200}
= 6.656 \text{ mg/l}
\]

Initial DO deficit,
\[
D_0 = 9.17 - 6.656 = 2.514 \text{ mg/l}
\]

Minimum DO to be maintained in the stream $$= 4$$ mg/l

Maximum permissible saturation deficit is
\[
D_c = 9.17 - 4 = 5.17 \text{ mg/l}
\]

Now, the first stage BOD of mixture of sewage and stream is
\[
\left( \frac{L_0}{f_s D_c} \right)^{f_s-1}
= f_s \left[ 1 - (f_s - 1)\frac{D_0}{L_0} \right]
\qquad (9.12)
\]

where
\[
f_s = \frac{R}{K} = \frac{0.3}{0.1} = 3
\]

\[
D_0 = 2.514 \text{ mg/l}, \quad D_c = 5.17 \text{ mg/l}
\]

\[
\left( \frac{L_0}{3 \times 5.17} \right)^2
= 3 \left[ 1 - (3 - 1)\frac{2.514}{L_0} \right]
\]

\[
\frac{L_0^2}{240.56}
= 3 \left[ 1 - \frac{5.028}{L_0} \right]
\]

Solving this by trial and error, we get
\[
L_0 = 23.86 \text{ mg/l}
\]

Now,
\[
y_5 = L_0 (1 - 10^{-K_1 t})
\]

\[
y_5 = 23.86 \left( 1 - 10^{-0.1 \times 5} \right)
= 16.31 \text{ mg/l}
\]

This is permissible BOD of the mix.

\[
16.31
= \frac{2 \times 5000 + y_e \times 1200}{5000 + 1200}
\]

From which
\[
y_e = 75.94 \text{ mg/l}
\]

Hence permissible BOD of effluent
\[
= 75.94 \text{ mg/l}
\]

Actual BOD of effluent $$= 160$$ mg/l

Degree of treatment required
\[
= \frac{160 - 75.94}{160} \times 100
= 52.56\%
\]

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