DO deficit 50 km downstream, 1.9 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

$$\begin{aligned} Q_{mix}&=1.9+0.6=2.5\ \text{m}^3/\text{s}\\ DO_{mix}&=\frac{1.9(9.17)+0.6(1.8)}{2.5}=7.401\ \text{mg/L}\\ D_0&=9.17-7.401=1.769\ \text{mg/L}\\ L_0&=\frac{1.9(2.5)+0.6(210)}{2.5}=52.30\ \text{mg/L} \end{aligned}$$

Step 2: Reaeration constant and time of flow

$$\begin{aligned} K_{2(e)}&=\frac{3.9\times0.15^{0.5}}{1.4^{1.5}}=0.912\ \text{day}^{-1}\Rightarrow K_2=0.396\ \text{day}^{-1}\ (\text{base }10)\\ t&=\frac{x}{v}=\frac{50000\ \text{m}}{0.15\ \text{m/s}}=333333\ \text{s}=3.858\ \text{days} \end{aligned}$$

Step 3: Deficit at the section

$$D_t=\frac{K_1L_0}{K_2-K_1}\left(10^{-K_1t}-10^{-K_2t}\right)+D_010^{-K_2t}$$ $$\begin{aligned} D_t&=\frac{0.1\times52.30}{0.396-0.1}\left(10^{-0.3858}-10^{-1.5278}\right)+1.769\times10^{-1.5278}\\ &=6.80\ \text{mg/L} \end{aligned}$$

Answer: DO deficit 50 km downstream = 6.80 mg/L (DO = 2.37 mg/L).

Note: if the given BODs were 5-day values, $L_0$ would be $76.5$ mg/L, and the deficit would rise further (to about 9.9 mg/L, which is more than the saturation value, so the stream would turn septic).