--- title: "Critical oxygen deficit and location (table 2)" subject: "Sanitary Engineering" chapter: 6 key: "semester_6/sanitary_engineering:critical-do-effluent-stream-table-71" source: https://www.bctengineer.com/q/semester_6/sanitary_engineering/critical-do-effluent-stream-table-71 site: BCT Engineering --- # Critical oxygen deficit and location (table 2) Sanitary Engineering · Chapter 6: Wastewater Disposal - Asked in: 2071 Bhadra, 8 marks - Video: [oxygen deficit problem | DO deficit solved problem | oxygen deficit in downstream side of river](https://www.youtube.com/watch?v=4wst5Ot1YIg) — Civil Engineering Exam - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_6/sanitary_engineering/critical-do-effluent-stream-table-71 - Whole chapter: https://www.bctengineer.com/md/semester_6/sanitary_engineering/chapter-6 ## Question A wastewater treatment plant disposes off its effluents into a stream at a point A. Characteristics of the stream at a location upstream of point A and of the effluent are as follows: | Item | Effluent | Stream | |---|---|---| | Flow Rate, m³/sec | 0.35 | 0.60 | | Dissolved Oxygen, mg/l | 2 | 7 | | Temperature, °C | 29 | 22 | | BOD5 at 20°C, mg/l | 155 | 2 | Assume that the deoxygenation constant at 20°C (base e) = 0.2 per day and the reaeration constant at 20°C (base e) = 0.35 per day. For the mixture, equilibrium concentration of dissolved oxygen for the freshwater is as follows: | Temperature, °C | 21 | 22 | 23 | 24 | 26 | 28 | |---|---|---|---|---|---|---| | DO, mg/l | 8.99 | 8.83 | 8.68 | 8.53 | 8.22 | 7.92 | The velocity of stream downstream of the point A is 0.25 m/sec. Determine the critical oxygen deficit and its location. ## Short answer Mixing yields initial deficit 3.28 mg/L, ultimate BOD 92.34 mg/L; constants corrected to 24.58°C (k'=0.2468/day, r'=0.3764/day). Critical deficit 28.09 mg/L at 67.21 km (3.11 days). Deficit far exceeds saturation DO (8.44 mg/L), so the stream turns anaerobic and treatment is required. ## Answer The BOD is the 5-day BOD at 20°C. Rates are base $e$, corrected to the mixture temperature by $K_T=K_{20}(1.047)^{T-20}$ and $R_T=R_{20}(1.016)^{T-20}$. The saturation DO is read from the table at the mixture temperature. #### Step 1: Mixture properties $$ \begin{aligned} Q &= 0.35+0.60 = 0.95\ \text{m}^3/\text{s}\\ T_{mix} &= \frac{0.35\times 29+0.60\times 22}{0.95} = 24.58^\circ\text{C}\\ DO_{mix} &= \frac{0.35\times 2+0.60\times 7}{0.95} = 5.16\ \text{mg/l}\\ BOD_5^{mix} &= \frac{0.35\times 155+0.60\times 2}{0.95} = 58.37\ \text{mg/l}\\ DO_{sat} &= 8.440\ \text{mg/l (interpolated between 24 and 26}^\circ\text{C)}\\ D_0 &= 8.440-5.16 = 3.28\ \text{mg/l} \end{aligned} $$ #### Step 2: Ultimate BOD and corrected constants $$ \begin{aligned} L_0 &= \frac{58.37}{1-e^{-5\times 0.2}} = 92.34\ \text{mg/l}\\ k'_T &= 0.2(1.047)^{4.58} = 0.2468\ \text{/day}, \qquad r'_T = 0.35(1.016)^{4.58} = 0.3764\ \text{/day} \end{aligned} $$ #### Step 3: Critical time, deficit and location $$ \begin{aligned} t_c &= \frac{1}{r'-k'}\ln\left[\frac{r'}{k'}\left\{1-\frac{D_0(r'-k')}{k'L_0}\right\}\right] = 3.111\ \text{days}\\ D_c &= \frac{k'L_0}{r'}e^{-k't_c} = 28.09\ \text{mg/l}\\ x_c &= 0.25\times 86400\times 3.111/1000 = 67.21\ \text{km} \end{aligned} $$ **Answer: critical oxygen deficit $D_c = 28.09$ mg/l at about $67.21$ km downstream of point A ($t_c=3.111$ days).** The computed deficit is far larger than the saturation DO (8.44 mg/l), so the formula would give a minimum DO of $8.440-28.09=-19.65$ mg/l, which is physically impossible: the DO actually falls to zero well before the critical point, the stream turns anaerobic, and the effluent must be treated before discharge. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What is the DO of the mixture before oxygen is used? A. Average of 2 and 7 = 4.5 B. Flow-weighted: (0.35 x 2 + 0.60 x 7)/0.95 = 5.16 mg/L C. Stream DO of 7 mg/L D. Effluent DO of 2 mg/L Answer: B. Mixing is weighted by flow, giving 5.16 mg/L. (Key point: Mixture DO) 2. What is D0 if DO_sat at the mixture temperature is 8.44 mg/L? A. 8.44 + 5.16 B. 5.16 - 8.44 C. 8.44 - 5.16 = 3.28 mg/L D. 8.44 x 5.16 Answer: C. D0 = DO_sat - DO_mix = 3.28 mg/L. (Key point: Initial deficit) 3. Why are the constants at 24.58 C larger than the 20 C values? A. Because the stream velocity is higher B. Because BOD is higher C. They are not corrected D. Biological and reaeration rates increase with temperature Answer: D. Both constants are raised using 1.047^(T-20) and 1.016^(T-20), giving k' = 0.2468 and r' = 0.3764. (Key point: Corrected constants) 4. What does the very large Dc = 28.09 mg/L (more than the saturation value) indicate? A. The stream goes anaerobic: oxygen is exhausted and the waste needs treatment B. The stream is very clean C. The data were mismatched to temperature D. Reaeration is very fast Answer: A. A deficit greater than DO_sat gives a negative DO, meaning oxygen is exhausted and the stream turns anaerobic. 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