Why is DO of the mixture (8.06 mg/L) lower than the river DO of 8.5 mg/L?

Because the sewage is putrefied and carries about zero DO

Because temperature rises after mixing

How is the need for treatment decided?

If the calculated critical deficit exceeds the allowable deficit

If the BOD of the sewage exceeds 20 mg/L

If the river flow is less than 10 m3/s

If the initial deficit D0 is greater than zero

What is f here and what does it represent?

f = K2/K1 = 4, the ratio of reaeration to deoxygenation constants

2080 Chaitra · 8 marks

Is treatment necessary? Town wastewater 300 l/s

Short answer

Streeter-Phelps analysis: mixed flow 5.8 m³/s, initial deficit 1.04 mg/L, ultimate BOD 34.04 mg/L, f=4, critical time 1.87 days, critical deficit 5.54 mg/L > allowable 4.5 mg/L, so treatment needed. Required effluent BOD₅ ≤ 362.8 mg/L, corresponding to 19.4% BOD removal efficiency.

The wastewater generated from a newly established town is proposed to dispose in the nearby river. The characteristics of raw wastewater and that of river are as follows: Wastewater discharge = 300 l/s; BOD5 of wastewater at 20°C = 450 mg/l; The sewage is in putrefied state; Minimum river discharge = 5.5 m³/s; BOD5 of river at 20°C = 0 mg/l; DO of river = 8.5 mg/l; Temperature in the river after mixing = 20°C; De-oxygenation constant = 0.1/day (base 10) at 20°C; Re-oxygenation constant = 0.4/day (base 10) at 20°C; Saturation DO at 20°C = 9.1 mg/l; Allowable minimum DO deficit at d/s of disposal = 4.5 mg/l. Is treatment necessary? If so, what will be the required degree of treatment?

The oxygen sag (Streeter-Phelps) analysis is applied to the mixture of river and wastewater. Constants are in base 10.

Step 1: Conditions after mixing

\begin{aligned} Q_{mix}&=0.3+5.5=5.8\ \text{m}^3/\text{s}\\ DO_{mix}&=\frac{0.3(0)+5.5(8.5)}{5.8}=8.06\ \text{mg/L}\quad(\text{putrefied sewage: DO}\approx0)\\ D_0&=9.1-8.06=1.04\ \text{mg/L}\\ BOD_5^{mix}&=\frac{0.3(450)+5.5(0)}{5.8}=23.28\ \text{mg/L}\\ L_0&=\frac{BOD_5}{1-10^{-0.1\times5}}=\frac{23.28}{0.6838}=34.04\ \text{mg/L} \end{aligned}

Step 2: Critical deficit without treatment

f=\frac{K_2}{K_1}=\frac{0.4}{0.1}=4

t_c=\frac{1}{K_2-K_1}\log_{10}\left[f\left(1-\frac{(f-1)D_0}{L_0}\right)\right]=\frac{1}{0.3}\log_{10}\left[4\left(1-\frac{3\times1.04}{34.04}\right)\right]=1.87\ \text{days}

D_c=\frac{L_0}{f}10^{-K_1t_c}=\frac{34.04}{4}\times10^{-0.1\times1.87}=5.54\ \text{mg/L}

Since $D_c=5.54$ mg/L is greater than the allowable deficit of 4.5 mg/L, treatment is necessary.

Step 3: Required degree of treatment

Find the maximum $L_0$ for which $D_c=4.5$ mg/L (with the same $D_0$ , by trial using the same equations): $L_0=27.45$ mg/L. Then

BOD_5^{mix}=27.45\times0.6838=18.77\ \text{mg/L}

The effluent BOD₅ that can be allowed is:

BOD_5^{eff}=18.77\times\frac{5.8}{0.3}=362.8\ \text{mg/L}

\text{Efficiency of treatment}=\frac{450-362.8}{450}\times100=19.4\%

Answer: Treatment is necessary; the plant must remove about 19.4% of the BOD (effluent BOD₅ not more than about 362.8 mg/L).

✓ Answer checked by Claude Opus 5.5 (high effort)

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Is treatment necessary? Town wastewater 300 l/s

Step 1: Conditions after mixing

Step 2: Critical deficit without treatment

Step 3: Required degree of treatment

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