--- title: "Critical DO deficit, location, sag curve" subject: "Sanitary Engineering" chapter: 6 key: "semester_6/sanitary_engineering:sag-curve-effluent-1p5-stream-7p5" source: https://www.bctengineer.com/q/semester_6/sanitary_engineering/sag-curve-effluent-1p5-stream-7p5 site: BCT Engineering --- # Critical DO deficit, location, sag curve Sanitary Engineering · Chapter 6: Wastewater Disposal - Asked in: 2076 Baisakh, 8 marks - Video: [Oxygen Sag Curve Solved Examples | Waste Water Engineering](https://www.youtube.com/watch?v=6SYuc_3dAtg) — APSEd - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_6/sanitary_engineering/sag-curve-effluent-1p5-stream-7p5 - Whole chapter: https://www.bctengineer.com/md/semester_6/sanitary_engineering/chapter-6 ## Question From the following data find: a. Critical DO deficit. b. Location of critical DO deficit. c. BOD5 at 20°C of the sample taken at critical point. d. Draw a sag curve. | Item | Effluent | Stream | |---|---|---| | Discharge (m³/s) | 1.5 | 7.5 | | BOD5 20°C (mg/l) | 280 | 1.2 | | DO (mg/l) | 0.2 | 90% of saturated | | Temperature (°C) | 30 | 18 | Assume K = 0.1 day⁻¹, R = 0.5 day⁻¹ at 20°C. Saturation DO at 20°C and 18°C is 9.17 mg/l and 9.54 mg/l respectively. Velocity of river = 0.8 m/s. ## Short answer Mixing at 20°C gives initial deficit 1.98 mg/L, ultimate BOD 69.71 mg/L; K=0.1/d, R=0.5/d (base 10). Critical deficit = 9.61 mg/L at 111.72 km downstream (1.62 days). BOD5 at critical point = 32.85 mg/L; minimum DO becomes negative, indicating severe pollution. ## Answer Stream DO is 90% of saturation at 18°C: $DO_r = 0.9\times 9.54 = 8.586$ mg/l. The mixture temperature is $(1.5\times 30+7.5\times 18)/9 = 20^\circ$C, so the given $K=0.1$ and $R=0.5$ (base 10) and $DO_{sat}=9.17$ mg/l apply without correction. #### Step 1: Conditions just after mixing $$ \begin{aligned} Q_{mix} &= Q_s+Q_r = 1.5 + 7.5 = 9\ m^3/s\\ BOD_5^{mix} &= \frac{Q_sB_s+Q_rB_r}{Q_s+Q_r} = \frac{1.5\times 280 + 7.5\times 1.2}{9} = 47.67\ \text{mg/l}\\ DO_{mix} &= \frac{Q_s DO_s+Q_r DO_r}{Q_s+Q_r} = \frac{1.5\times 0.2 + 7.5\times 8.586}{9} = 7.19\ \text{mg/l}\\ D_0 &= DO_{sat}-DO_{mix} = 9.17 - 7.19 = 1.98\ \text{mg/l} \end{aligned} $$ #### Step 2: Ultimate BOD and self-purification constant $$ \begin{aligned} L_0 &= \frac{BOD_5}{1-10^{-5K}} = \frac{47.67}{1-10^{-5\times 0.1}} = 69.71\ \text{mg/l}\\ f &= \frac{R}{K} = \frac{0.5}{0.1} = 5.00 \end{aligned} $$ #### Step 3: Critical time and critical deficit $$ \begin{aligned} t_c &= \frac{1}{K(f-1)}\log_{10}\left[f\left\{1-(f-1)\frac{D_0}{L_0}\right\}\right]\\ &= \frac{1}{0.1(5.00-1)}\log_{10}\left[5.00\left\{1-(5.00-1)\frac{1.98}{69.71}\right\}\right] = 1.616\ \text{days}\\ D_c &= \frac{L_0}{f}\,10^{-K t_c} = \frac{69.71}{5.00}\times 10^{-0.1\times 1.616} = 9.61\ \text{mg/l} \end{aligned} $$ #### (a) Critical DO deficit $D_c = 9.61$ mg/l (minimum DO $=9.17-9.61=-0.44$ mg/l). #### (b) Location of the critical deficit $$ x_c = v\,t_c = 0.8\times 86400\times 1.616/1000 = 111.72\ \text{km downstream} $$ #### (c) BOD5 of a sample taken at the critical point BOD remaining at time $t_c$ is $L_t = L_0\,10^{-Kt_c} = 69.71\times 10^{-0.1\times 1.616} = 48.05$ mg/l (ultimate). The 5-day BOD of that sample is $$ BOD_5 = L_t\left(1-10^{-5K}\right) = 48.05\times 0.6838 = 32.85\ \text{mg/l} $$ #### (d) Sag curve Deficit at time $t$: $D_t=\dfrac{K L_0}{R-K}\left(10^{-Kt}-10^{-Rt}\right)+D_0\,10^{-Rt}$, and $DO_t = 9.17-D_t$. | t (days) | Deficit $D_t$ (mg/l) | DO (mg/l) | |---|---|---| | 0 | 1.98 | 7.19 | | 0.5 | 6.85 | 2.32 | | 1 | 8.96 | 0.21 | | 1.5 | 9.59 | -0.42 | | 2 | 9.45 | -0.28 | | 2.5 | 8.93 | 0.24 | | 3 | 8.25 | 0.92 | | 4 | 6.78 | 2.39 | | 5 | 5.46 | 3.71 | | 6 | 4.36 | 4.81 | | 8 | 2.76 | 6.41 | | 10 | 1.74 | 7.43 | ```text 7.43 |* * * * * 6.45 | * * * 5.47 | * * 4.48 | * * 3.50 | * 2.52 | * * * 1.54 | * 0.56 | * * * -0.42 | * +-------------------------------------------- 0 10 t, days ``` The DO falls from 7.19 mg/l to a minimum of -0.44 mg/l at $t_c=1.616$ days (the critical point), then recovers towards 9.17 mg/l as reaeration exceeds deoxygenation. Negative DO values only mean that the oxygen is exhausted (the stream turns anaerobic in that stretch); this effluent needs treatment. **Answer: $D_c=9.61$ mg/l; $x_c=111.72$ km; BOD5 at critical point $=32.85$ mg/l.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Why can K = 0.1 and R = 0.5 be used without correction? A. They are base e values B. The mixture temperature works out to 20 C C. The stream is at 20 C D. Temperature does not affect them Answer: B. T = (1.5 x 30 + 7.5 x 18)/9 = 20 C, so the 20 C constants and DO_sat = 9.17 apply. (Key point: Mixture temperature) 2. How is the 5-day BOD of a sample at the critical point obtained? A. BOD5 = L0 B. BOD5 = Dc C. BOD5 = Lt (1 - 10^(-5K)) where Lt = L0 10^(-K tc) D. BOD5 = Lt x 5 Answer: C. The remaining ultimate BOD is 48.05 mg/L and only 68.38% of it is exerted in 5 days, giving 32.85 mg/L. (Key point: BOD at critical point) 3. What does the sag curve show after the critical point? A. DO continues to fall B. DO stays constant C. DO becomes zero permanently D. DO rises toward saturation as reaeration exceeds deoxygenation Answer: D. After tc the deficit decreases, so DO recovers toward 9.17 mg/L. (Key point: Sag curve shape) 4. What does a computed DO of -0.44 mg/L mean? A. Oxygen is exhausted and the stretch turns anaerobic B. DO is above saturation C. The data are in error by sign D. Reaeration is infinite Answer: A. DO cannot be negative in reality; it means oxygen is used up and the stream is anaerobic in that stretch. 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