Question 5 of 5: Dissolved Oxygen Sag Below a Meat-Processing Discharge

Question 5: Dissolved Oxygen Sag Below a Meat-Processing Discharge (25 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given.

Find. Whether the minimum downstream dissolved oxygen falls below the 6.0 mg/L regulatory limit.

Approach. Mix the effluent and river flows to get the blended BOD₅, DO and temperature at the discharge point, correct the rate constants to that stream temperature, convert BOD₅ to ultimate BOD using the standard 20 °C lab-test rate constant, then apply the Streeter-Phelps critical-deficit solution to find the minimum downstream DO and compare it to the 6.0 mg/L limit.

1. Mixed (blended) conditions. $Q_{mix}=10+50=60\ \text{m}^3/\text{h}$. Flow-weighted mixing gives $$BOD_{mix}=\frac{10(50)+50(3)}{60}=\boxed{10.8\ \text{mg/L}},\quad DO_{mix}=\frac{10(2)+50(7)}{60}=\boxed{6.17\ \text{mg/L}},\quad T_{mix}=\frac{10(15)+50(10)}{60}=10.8\ ^\circ\text{C}$$
2. Temperature-corrected rate constants. Correcting $k_1,k_2$ from 20 °C to the mixed stream temperature via $k_T = k_{20}\,\theta^{T-20}$: $$k_{1,T}=0.15(1.135)^{10.8-20}=\boxed{0.047\ \text{d}^{-1}},\qquad k_{2,T}=0.28(1.056)^{10.8-20}=\boxed{0.170\ \text{d}^{-1}}$$
3. Ultimate BOD (L₀). The BOD₅-to-ultimate-BOD conversion uses $k_1$ at the standard 20 °C BOD-bottle test condition (not the temperature-corrected river rate, which governs decay IN the river, not the lab test itself): $$L_0 = \frac{BOD_{mix}}{1-10^{-k_{1,20}(5)}} = \frac{10.8}{1-10^{-0.75}} = \boxed{13.2\ \text{mg/L}}$$
4. Initial deficit and critical point. $D_0 = DO_{sat}-DO_{mix}=9.2-6.17=3.03\ \text{mg/L}$. Since $k_1,k_2$ are explicitly base-10 constants here, the deficit curve is $D(t)=A(10^{-k_{1,T}t}-10^{-k_{2,T}t})+D_0\,10^{-k_{2,T}t}$ with $A=k_{1,T}L_0/(k_{2,T}-k_{1,T})$, and setting $dD/dt=0$ (using $\log_{10}$, consistent with the base-10 exponent — NOT the natural-log form used for base-e rate constants) gives the time to the critical (minimum-DO) point: $$t_c=\frac{1}{k_{2,T}-k_{1,T}}\log_{10}\!\left[\frac{(A-D_0)\,k_{2,T}}{A\,k_{1,T}}\right] = \boxed{1.28\ \text{d}}$$ and the critical deficit is $$D_c = \frac{k_{1,T}}{k_{2,T}}L_0\,10^{-k_{1,T}t_c} = \boxed{3.17\ \text{mg/L}}$$
5. Minimum downstream DO and compliance check. $$DO_{min} = DO_{sat}-D_c = 9.2-3.17 = \boxed{6.03\ \text{mg/L}}$$ 6.03 mg/L stays above the 6.0 mg/L regulatory limit, but only just — the margin is a mere 0.03 mg/L, occurring about 1.3 days (equivalently, at a river velocity of, say, 0.3 m/s, roughly 33 km) downstream of the discharge. The dissolved oxygen does NOT fall below the regulatory limit, but the result is essentially right at the boundary: within the precision of the given rate constants and temperature-correction factors, this discharge should be treated as marginal rather than comfortably compliant, and any small increase in effluent BOD or drop in reaeration capacity would push it into violation.