DO deficit 60 km downstream, 2 m³/s stream

Assumptions

Streeter-Phelps equation with rate constants to base 10; the BOD values given are taken as the ultimate (first-stage) BODs, so $L_0$ is the BOD of the mixture; $K_1=0.1$ day⁻¹; the reaeration constant is found from the O'Connor-Dobbins formula (base $e$, 20°C):

$$K_{2(e)}=\frac{3.9\,v^{0.5}}{H^{1.5}}\ \text{day}^{-1}\quad(v\text{ in m/s},\ H\text{ in m}),\qquad K_2(\text{base }10)=\frac{K_{2(e)}}{2.303}$$

Step 1: Mixing at the outfall

$$\begin{aligned} Q_{mix}&=2+0.5=2.5\ \text{m}^3/\text{s}\\ DO_{mix}&=\frac{2(9.17)+0.5(2)}{2.5}=7.736\ \text{mg/L}\\ D_0&=9.17-7.736=1.434\ \text{mg/L}\\ L_0&=\frac{2(3)+0.5(200)}{2.5}=42.40\ \text{mg/L} \end{aligned}$$

Step 2: Reaeration constant and time of flow

$$\begin{aligned} K_{2(e)}&=\frac{3.9\times0.2^{0.5}}{1.2^{1.5}}=1.327\ \text{day}^{-1}\Rightarrow K_2=0.576\ \text{day}^{-1}\ (\text{base }10)\\ t&=\frac{x}{v}=\frac{60000\ \text{m}}{0.2\ \text{m/s}}=300000\ \text{s}=3.472\ \text{days} \end{aligned}$$

Step 3: Deficit at the section

$$D_t=\frac{K_1L_0}{K_2-K_1}\left(10^{-K_1t}-10^{-K_2t}\right)+D_010^{-K_2t}$$ $$\begin{aligned} D_t&=\frac{0.1\times42.40}{0.576-0.1}\left(10^{-0.3472}-10^{-2.0008}\right)+1.434\times10^{-2.0008}\\ &=3.93\ \text{mg/L} \end{aligned}$$

Answer: DO deficit 60 km downstream = 3.93 mg/L (DO = 5.24 mg/L).

Note: if the given BODs were 5-day values, $L_0$ would be $62.0$ mg/L, and the deficit would rise further (to about 5.7 mg/L).