Critical DO deficit, location, sag curve

Stream DO is 90% of saturation at 18°C: $DO_r = 0.9\times 9.54 = 8.586$ mg/l. The mixture temperature is $(1.5\times 30+7.5\times 18)/9 = 20^\circ$C, so the given $K=0.1$ and $R=0.5$ (base 10) and $DO_{sat}=9.17$ mg/l apply without correction.

Step 1: Conditions just after mixing

$$\begin{aligned} Q_{mix} &= Q_s+Q_r = 1.5 + 7.5 = 9\ m^3/s\\ BOD_5^{mix} &= \frac{Q_sB_s+Q_rB_r}{Q_s+Q_r} = \frac{1.5\times 280 + 7.5\times 1.2}{9} = 47.67\ \text{mg/l}\\ DO_{mix} &= \frac{Q_s DO_s+Q_r DO_r}{Q_s+Q_r} = \frac{1.5\times 0.2 + 7.5\times 8.586}{9} = 7.19\ \text{mg/l}\\ D_0 &= DO_{sat}-DO_{mix} = 9.17 - 7.19 = 1.98\ \text{mg/l} \end{aligned}$$

Step 2: Ultimate BOD and self-purification constant

$$\begin{aligned} L_0 &= \frac{BOD_5}{1-10^{-5K}} = \frac{47.67}{1-10^{-5\times 0.1}} = 69.71\ \text{mg/l}\\ f &= \frac{R}{K} = \frac{0.5}{0.1} = 5.00 \end{aligned}$$

Step 3: Critical time and critical deficit

$$\begin{aligned} t_c &= \frac{1}{K(f-1)}\log_{10}\left[f\left\{1-(f-1)\frac{D_0}{L_0}\right\}\right]\\ &= \frac{1}{0.1(5.00-1)}\log_{10}\left[5.00\left\{1-(5.00-1)\frac{1.98}{69.71}\right\}\right] = 1.616\ \text{days}\\ D_c &= \frac{L_0}{f}\,10^{-K t_c} = \frac{69.71}{5.00}\times 10^{-0.1\times 1.616} = 9.61\ \text{mg/l} \end{aligned}$$

(a) Critical DO deficit

$D_c = 9.61$ mg/l (minimum DO $=9.17-9.61=-0.44$ mg/l).

(b) Location of the critical deficit

$$x_c = v\,t_c = 0.8\times 86400\times 1.616/1000 = 111.72\ \text{km downstream}$$

(c) BOD5 of a sample taken at the critical point

BOD remaining at time $t_c$ is $L_t = L_0\,10^{-Kt_c} = 69.71\times 10^{-0.1\times 1.616} = 48.05$ mg/l (ultimate). The 5-day BOD of that sample is

$$BOD_5 = L_t\left(1-10^{-5K}\right) = 48.05\times 0.6838 = 32.85\ \text{mg/l}$$

(d) Sag curve

Deficit at time $t$: $D_t=\dfrac{K L_0}{R-K}\left(10^{-Kt}-10^{-Rt}\right)+D_0\,10^{-Rt}$, and $DO_t = 9.17-D_t$.

 7.43 |*                                   * * *  *
 6.45 |                              * * *         
 5.47 |                         * *                
 4.48 |                     * *                    
 3.50 |                   *                        
 2.52 |  *            * *                          
 1.54 |            *                               
 0.56 |    *   * *                                 
-0.42 |      *                                     
      +--------------------------------------------
       0                                10  t, days

The DO falls from 7.19 mg/l to a minimum of -0.44 mg/l at $t_c=1.616$ days (the critical point), then recovers towards 9.17 mg/l as reaeration exceeds deoxygenation. Negative DO values only mean that the oxygen is exhausted (the stream turns anaerobic in that stretch); this effluent needs treatment.

Answer: $D_c=9.61$ mg/l; $x_c=111.72$ km; BOD5 at critical point $=32.85$ mg/l.