1. **State the problem:** We need to find the ultimate carbonaceous BOD ($L_0$) and the ultimate nitrogenous BOD ($NBOD_0$) of the river water immediately after mixing with the wastewater effluent. 2. **Given data:** - Effluent ultimate BOD, $L_0^{effluent} = 25$ mg/L - Effluent NH$_3$-N concentration = 15 mg/L - River upstream flow rate, $Q_r = 1.5$ m$^3$/s - River upstream ultimate BOD, $L_0^{river} = 2.0$ mg/L - Effluent flow rate, $Q_e = 0.5$ m$^3$/s 3. **Formulas and important rules:** - The total flow after mixing is $Q_{total} = Q_r + Q_e$ - The ultimate carbonaceous BOD after mixing is the flow-weighted average of the carbonaceous BODs: $$L_0^{mix} = \frac{Q_r L_0^{river} + Q_e L_0^{effluent}}{Q_r + Q_e}$$ - The ultimate nitrogenous BOD ($NBOD_0$) is related to the nitrogen concentration. The nitrogenous BOD is typically calculated as: $$NBOD_0 = \text{NH}_3\text{-N concentration} \times \frac{16}{14}$$ where 16 is the oxygen equivalent of nitrogen and 14 is the atomic weight of nitrogen. - The nitrogenous BOD after mixing is also a flow-weighted average: $$NBOD_0^{mix} = \frac{Q_r \times 0 + Q_e \times NBOD_0^{effluent}}{Q_r + Q_e}$$ assuming the river upstream has negligible NH$_3$-N concentration. 4. **Calculate total flow:** $$Q_{total} = 1.5 + 0.5 = 2.0 \text{ m}^3/\text{s}$$ 5. **Calculate ultimate carbonaceous BOD after mixing:** $$L_0^{mix} = \frac{1.5 \times 2.0 + 0.5 \times 25}{2.0} = \frac{3.0 + 12.5}{2.0} = \frac{15.5}{2.0} = 7.75 \text{ mg/L}$$ 6. **Calculate ultimate nitrogenous BOD of effluent:** $$NBOD_0^{effluent} = 15 \times \frac{16}{14} = 15 \times 1.1429 = 17.14 \text{ mg/L}$$ 7. **Calculate ultimate nitrogenous BOD after mixing:** $$NBOD_0^{mix} = \frac{1.5 \times 0 + 0.5 \times 17.14}{2.0} = \frac{0 + 8.57}{2.0} = 4.285 \text{ mg/L}$$ **Final answers:** - Ultimate carbonaceous BOD after mixing, $L_0^{mix} = 7.75$ mg/L - Ultimate nitrogenous BOD after mixing, $NBOD_0^{mix} = 4.29$ mg/L (rounded to two decimals)