Critical oxygen deficit and location (table 2)

The BOD is the 5-day BOD at 20°C. Rates are base $e$, corrected to the mixture temperature by $K_T=K_{20}(1.047)^{T-20}$ and $R_T=R_{20}(1.016)^{T-20}$. The saturation DO is read from the table at the mixture temperature.

Step 1: Mixture properties

$$\begin{aligned} Q &= 0.35+0.60 = 0.95\ \text{m}^3/\text{s}\\ T_{mix} &= \frac{0.35\times 29+0.60\times 22}{0.95} = 24.58^\circ\text{C}\\ DO_{mix} &= \frac{0.35\times 2+0.60\times 7}{0.95} = 5.16\ \text{mg/l}\\ BOD_5^{mix} &= \frac{0.35\times 155+0.60\times 2}{0.95} = 58.37\ \text{mg/l}\\ DO_{sat} &= 8.440\ \text{mg/l (interpolated between 24 and 26}^\circ\text{C)}\\ D_0 &= 8.440-5.16 = 3.28\ \text{mg/l} \end{aligned}$$

Step 2: Ultimate BOD and corrected constants

$$\begin{aligned} L_0 &= \frac{58.37}{1-e^{-5\times 0.2}} = 92.34\ \text{mg/l}\\ k'_T &= 0.2(1.047)^{4.58} = 0.2468\ \text{/day}, \qquad r'_T = 0.35(1.016)^{4.58} = 0.3764\ \text{/day} \end{aligned}$$

Step 3: Critical time, deficit and location

$$\begin{aligned} t_c &= \frac{1}{r'-k'}\ln\left[\frac{r'}{k'}\left\{1-\frac{D_0(r'-k')}{k'L_0}\right\}\right] = 3.111\ \text{days}\\ D_c &= \frac{k'L_0}{r'}e^{-k't_c} = 28.09\ \text{mg/l}\\ x_c &= 0.25\times 86400\times 3.111/1000 = 67.21\ \text{km} \end{aligned}$$

Answer: critical oxygen deficit $D_c = 28.09$ mg/l at about $67.21$ km downstream of point A ($t_c=3.111$ days). The computed deficit is far larger than the saturation DO (8.44 mg/l), so the formula would give a minimum DO of $8.440-28.09=-19.65$ mg/l, which is physically impossible: the DO actually falls to zero well before the critical point, the stream turns anaerobic, and the effluent must be treated before discharge.