How is the DO of the river-effluent mixture computed?

Flow-weighted average: (1.8x0 + 22x8.7)/23.8 = 8.04 mg/l

Equal to the river DO, 8.7 mg/l

Simple average of 0 and 8.7 mg/l

How is the distance to the critical deficit found?

x_c = v x t_c = 0.20 m/s x 5.77 d, about 99.7 km

Why is the DO after 1 and 3 days reported as zero rather than -1.23 and -11.42 mg/l?

The formula gives deficits larger than DO_sat (9.1), but DO cannot be negative so the river goes anaerobic

The negative values are correct and DO can be negative

Temperature correction removes the negatives

kr is larger than kd, so DO is always zero

2070 Magh · 8 marks

Critical DO location, industry 1.8 m³/s

Short answer

Critical deficit occurs at 99.7 km (5.77 days) using base-e constants (kd=0.25, kr=0.11). Mixed BOD5=32.0 mg/l, ultimate BOD=44.9 mg/l, initial deficit=1.06 mg/l. DO after 1 and 3 days is effectively zero (formula gives negative), river turns anaerobic.

An industry is going to be established in an urban area near the river side. The river water and industrial effluent characteristics are as follows:

	Industrial effluent	River water
Flow (m³/s)	1.8	22
DO (mg/l)	0	8.7
BOD5, 20°	350	6.0

kd, 20° = 0.25 d⁻¹; kr, 20° = 0.11 d⁻¹; DO saturation = 9.1 mg/l. At what location in the river critical DO deficit would occur if the flow velocity in the river is 0.20 m/s? Also find out DO at the end of 1 and 3 days.

Rates are taken as base $e$ : $k_d=0.25$ /day, $k_r=0.11$ /day (20°C, so no temperature correction). Since $k_r<k_d$ , the ratio $f=k_r/k_d=0.44<1$ . $DO_{sat}=9.1$ mg/l.

Step 1: Mixing

\begin{aligned} Q &= 1.8+22 = 23.8\ \text{m}^3/\text{s}\\ BOD_5^{mix} &= \frac{1.8\times 350+22\times 6}{23.8} = 32.02\ \text{mg/l}\\ DO_{mix} &= \frac{1.8\times 0+22\times 8.7}{23.8} = 8.04\ \text{mg/l}\\ D_0 &= 9.1-8.04 = 1.06\ \text{mg/l}\\ L_0 &= \frac{BOD_5}{1-e^{-5k_d}} = \frac{32.02}{1-e^{-1.25}} = 44.87\ \text{mg/l} \end{aligned}

Step 2: Critical time and location

t_c = \frac{1}{k_r-k_d}\ln\left[\frac{k_r}{k_d}\left\{1-\frac{D_0(k_r-k_d)}{k_dL_0}\right\}\right] = 5.770\ \text{days}

\begin{aligned} D_c &= \frac{k_dL_0}{k_r}e^{-k_dt_c} = 24.10\ \text{mg/l}\\ x_c &= v\,t_c = 0.20\times 86400\times 5.770/1000 = 99.7\ \text{km} \end{aligned}

Step 3: DO after 1 and 3 days

D_t = \frac{k_dL_0}{k_r-k_d}\left(e^{-k_dt}-e^{-k_rt}\right)+D_0e^{-k_rt}

t (days)	$D_t$ (mg/l)	$DO = 9.1 - D_t$ (mg/l)
1	10.33	-1.23
3	20.52	-11.42

Answer: the critical deficit occurs about 99.7 km downstream ( $t_c=5.770$ days). The computed deficits (10.33 and 20.52 mg/l) exceed the saturation value 9.1 mg/l, so the DO after 1 day and 3 days is effectively zero (the formula gives -1.23 and -11.42 mg/l, which is physically not possible): the river would turn anaerobic, and the effluent needs treatment.

✓ Answer checked by Claude Opus 5.5 (high effort)

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Critical DO location, industry 1.8 m³/s

Step 1: Mixing

Step 2: Critical time and location

Step 3: DO after 1 and 3 days

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