--- title: "Two-stage trickling filter, 7665 m³/day" subject: "Sanitary Engineering" chapter: 7 key: "semester_6/sanitary_engineering:two-stage-trickling-7665" source: https://www.bctengineer.com/q/semester_6/sanitary_engineering/two-stage-trickling-7665 site: BCT Engineering --- # Two-stage trickling filter, 7665 m³/day Sanitary Engineering · Chapter 7: Wastewater Treatment - Asked in: 2071 Bhadra, 8 marks - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_6/sanitary_engineering/two-stage-trickling-7665 - Whole chapter: https://www.bctengineer.com/md/semester_6/sanitary_engineering/chapter-7 ## Question A municipal wastewater having a BOD5 of 190 mg/l is to be treated by a two stage trickling filter. The desired BOD5, 20°C of the final effluent is to be 25 mg/l. If both the filter's depth is to be 1.85 m and recirculation ratio for both filters is 0.5, determine the required filter diameters. Assume the wastewater flow rate of 7665 m³/day, and 35% BOD is removed in primary sedimentation tank. ## Short answer After 35% primary removal, applied BOD is 123.5 mg/l, overall filter efficiency 79.76%. Assuming equal stage efficiencies of 55.01%, recirculation factor 1.3605, volumes 204.3 m³ and 454.1 m³ yield diameters 11.9 m and 17.7 m. ## Answer Data: $Q=7665$ m³/day; raw BOD$_5=190$ mg/l; 35% removed in the primary sedimentation tank, so BOD to the filters $=190\times0.65=123.5$ mg/l; final BOD $=25$ mg/l; depth $=1.85$ m; $R=0.5$ for both filters. #### Data and approach Two-stage filter (NRC equations, $W$ in kg/day, $V$ in m³): $$ E_1=\frac{100}{1+0.4432\sqrt{\dfrac{W_1}{V_1F}}},\qquad E_2=\frac{100}{1+\dfrac{0.4432}{1-E_1}\sqrt{\dfrac{W_2}{V_2F}}},\qquad F=\frac{1+R}{(1+0.1R)^2} $$ where $W_2=W_1(1-E_1)$ is the BOD load leaving the first stage ($E_1$ as a fraction in the second equation). Flow $Q=7665$ m³/day. #### Step 1: Loads and efficiencies $$ \begin{aligned} BOD\ \text{applied} &= 123.5\ \text{mg/l},\quad W_1=7665\times 123.5\times10^{-3}=946.6\ \text{kg/day}\\ \text{Overall filter efficiency } E &= \frac{123.5-25}{123.5}=0.7976 \end{aligned} $$ No ratio is given, so both stages are taken to have equal efficiency $E_1=E_2=E$: $1-(1-E)^2=0.7976\Rightarrow E=1-\sqrt{1-0.7976}=0.5501$, i.e. $E_1=E_2=55.01\%$. #### Step 2: Recirculation factor (same for both stages, $R=0.5$) $$ F=\frac{1+0.5}{(1+0.1\times 0.5)^2}=1.3605 $$ #### Step 3: First stage $$ \sqrt{\frac{W_1}{V_1F}}=\frac{100/E_1-1}{0.4432}=1.8455\ \Rightarrow\ V_1=\frac{946.6}{1.3605\times 1.8455^2}=204.3\ \text{m}^3 $$ #### Step 4: Second stage $$ \begin{aligned} W_2 &= W_1(1-E_1)=946.6\times(1-0.5501)=425.9\ \text{kg/day}\\ \sqrt{\frac{W_2}{V_2F}} &= \frac{(100/E_2-1)(1-E_1)}{0.4432}=0.8303\ \Rightarrow\ V_2=\frac{425.9}{1.3605\times 0.8303^2}=454.1\ \text{m}^3 \end{aligned} $$ #### Step 5: Diameters (depth $1.85$ m) $$ D_1=\sqrt{\frac{4V_1}{\pi\,d}}=\sqrt{\frac{4\times 204.3}{\pi\times 1.85}}=11.9\ \text{m},\qquad D_2=\sqrt{\frac{4\times 454.1}{\pi\times 1.85}}=17.7\ \text{m} $$ **Answer: first-stage filter $D_1\approx 11.9$ m (volume 204.3 m³); second-stage filter $D_2\approx 17.7$ m (volume 454.1 m³).** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. The raw BOD is 190 mg/l with 35% removed in the PST. What BOD is used as the filter influent? A. 190 mg/l B. 66.5 mg/l C. 123.5 mg/l D. 150 mg/l Answer: C. 190 x 0.65 = 123.5 mg/l; the filters see settled sewage. (Key point: BOD to the filters) 2. No ratio between stage efficiencies is given. What does the answer assume? A. Equal efficiencies E1 = E2, from 1-(1-E)^2 = 0.7976, so E = 55.01% B. E1 = 2E2 C. E1 = 0 and the entire load goes to stage 2 D. E1 = E2 = 79.76% Answer: A. With no ratio, equal stage efficiencies are taken: E = 1 - sqrt(1 - 0.7976) = 0.5501. (Key point: Efficiency split) 3. Even though E1 = E2, V2 (454 m3) is larger than V1 (204 m3). Why? A. Stage 2 has a higher BOD load B. Stage 2 carries the 1/(1 - E1) factor in its NRC equation, so the same efficiency needs a larger volume at the lower BOD C. Stage 2 uses a lower recirculation factor D. It is an error; both should be equal Answer: B. The second-stage equation includes 0.4432/(1-E1), which makes the second stage harder to treat and so needs more volume. (Key point: Stage sizes) 4. Given depth 1.85 m, how is each diameter found? A. D = V/d, giving 110 m and 245 m B. D = sqrt(V/d), giving 10.5 m and 15.7 m C. D = 2V/d, giving 220 m and 490 m D. D = sqrt(4V/(pi d)), giving D1 = 11.9 m and D2 = 17.7 m Answer: D. Area = V/d, then D = sqrt(4A/pi). 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