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Critical DO deficit, 300 l/s into river

Short answer

Critical DO deficit is 18.56 mg/l at 1.88 days (90.2 km downstream). Mixed BOD5=78.25 mg/l, ultimate BOD=114.4 mg/l, initial deficit=3.21 mg/l. Minimum DO negative, stream would become anaerobic without treatment.

A town is discharging sewage of 300 liters/second in the river having discharge of 1300 liters/second and a velocity of 48 km/day. The BOD5 of sewage and river water are 400 mg/l and 4 mg/l respectively. The DO of sewage sample is nil. D.O. in the river water is 80% of the saturation value. The temperature of both sewage and river water is 20°C. The saturation DO at 20°C is 9.17 mg/l. Assume K1 = 0.1/day and K2 = 0.4/day (both with base 10). Calculate the critical DO deficit.

River DO =0.8×9.17=7.336=0.8\times 9.17=7.336 mg/l. Both temperatures are 20°C, so no correction is needed. K1=0.1K_1=0.1, K2=0.4K_2=0.4 (base 10), f=4f=4. Flows are in l/s.

Step 1: Conditions just after mixing

Qmix=Qs+Qr=300+1300=1600 l/sBOD5mix=QsBs+QrBrQs+Qr=300×400+1300×41600=78.25 mg/lDOmix=QsDOs+QrDOrQs+Qr=300×0+1300×7.3361600=5.96 mg/lD0=DOsat−DOmix=9.17−5.96=3.21 mg/l\begin{aligned} Q_{mix} &= Q_s+Q_r = 300 + 1300 = 1600\ l/s\\ BOD_5^{mix} &= \frac{Q_sB_s+Q_rB_r}{Q_s+Q_r} = \frac{300\times 400 + 1300\times 4}{1600} = 78.25\ \text{mg/l}\\ DO_{mix} &= \frac{Q_s DO_s+Q_r DO_r}{Q_s+Q_r} = \frac{300\times 0 + 1300\times 7.336}{1600} = 5.96\ \text{mg/l}\\ D_0 &= DO_{sat}-DO_{mix} = 9.17 - 5.96 = 3.21\ \text{mg/l} \end{aligned}

Step 2: Ultimate BOD and self-purification constant

L0=BOD51−10−5K=78.251−10−5×0.1=114.44 mg/lf=RK=0.40.1=4.00\begin{aligned} L_0 &= \frac{BOD_5}{1-10^{-5K}} = \frac{78.25}{1-10^{-5\times 0.1}} = 114.44\ \text{mg/l}\\ f &= \frac{R}{K} = \frac{0.4}{0.1} = 4.00 \end{aligned}

Step 3: Critical time and critical deficit

tc=1K(f−1)log⁡10[f{1−(f−1)D0L0}]=10.1(4.00−1)log⁡10[4.00{1−(4.00−1)3.21114.44}]=1.880 daysDc=L0f 10−Ktc=114.444.00×10−0.1×1.880=18.56 mg/l\begin{aligned} t_c &= \frac{1}{K(f-1)}\log_{10}\left[f\left\{1-(f-1)\frac{D_0}{L_0}\right\}\right]\\ &= \frac{1}{0.1(4.00-1)}\log_{10}\left[4.00\left\{1-(4.00-1)\frac{3.21}{114.44}\right\}\right] = 1.880\ \text{days}\\ D_c &= \frac{L_0}{f}\,10^{-K t_c} = \frac{114.44}{4.00}\times 10^{-0.1\times 1.880} = 18.56\ \text{mg/l} \end{aligned}

Answer: critical DO deficit Dc=18.56D_c = 18.56 mg/l, reached after tc=1.880t_c=1.880 days (about 90.2 km downstream at 48 km/day). Minimum DO =9.17−18.56=−9.39=9.17-18.56=-9.39 mg/l.

Note: the computed deficit is larger than the saturation DO (9.17 mg/l). In reality DO cannot fall below zero, so the stream would go anaerobic near the critical point and the effluent needs treatment before discharge. The negative value is simply what the formula gives for these data.

✓ Answer checked by Claude Opus 5.5 (high effort)

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