---
title: "HR trickling filter, BOD 280 mg/l"
subject: "Sanitary Engineering"
chapter: 7
key: "semester_6/sanitary_engineering:trickling-hr-5mld-280"
source: https://www.bctengineer.com/q/semester_6/sanitary_engineering/trickling-hr-5mld-280
site: BCT Engineering
---

# HR trickling filter, BOD 280 mg/l

Sanitary Engineering · Chapter 7: Wastewater Treatment

- Asked in: 2076 Baisakh, 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/trickling-hr-5mld-280
- Whole chapter: https://www.bctengineer.com/md/semester_6/sanitary_engineering/chapter-7

## Question

Raw sewage contain BOD of 280 mg/l with flow of 5 MLD is required to be treated through a single stage high rate trickling filter to convert effluent BOD of 25 mg/l to dispose off in the nearby river. Assuming that the primary sedimentation tank removes 25% of BOD, propose the dimension of filter with recirculation ratio of 1.5.

## Short answer

Single-stage high-rate filter volume 5974.6 m³, depth 2.0 m, diameter 61.7 m; in practice use two filters of 43.6 m diameter each.

## Answer

#### Data and approach

High-rate filter design uses the **NRC (National Research Council) equation**; $W$ is in kg/day of BOD applied to the filter, $V$ in m³:

$$
E = \frac{100}{1+0.4432\sqrt{\dfrac{W}{VF}}},\qquad F=\frac{1+R}{(1+0.1R)^2}
$$

#### Step 1: BOD applied to the filter and required efficiency

$$
\begin{aligned}
Q &= 5000\ \text{m}^3/\text{day}\\
\text{BOD after primary} &= 280\times(1-0.25) = 210.0\ \text{mg/l}\\
W &= Q\times BOD = 5000\times 210.0\times 10^{-3} = 1050.0\ \text{kg/day}\\
E &= \frac{210.0-25}{210.0}\times 100 = 88.10\%
\end{aligned}
$$

#### Step 2: Recirculation factor

$$
F = \frac{1+1.5}{(1+0.1\times 1.5)^2} = 1.890
$$

#### Step 3: Volume

$$
88.10 = \frac{100}{1+0.4432\sqrt{W/(VF)}}\Rightarrow \sqrt{\frac{W}{VF}} = 0.3049 \Rightarrow VF = \frac{1050.0}{0.3049^2} = 11294.1
$$

$$
V = \frac{VF}{F} = \frac{11294.1}{1.890} = 5974.6\ \text{m}^3
$$

#### Step 4: Diameter

Adopt depth $d = 2.0$ m (high-rate filters are 1.8-3 m deep).

$$
A = \frac{V}{d} = 2987.3\ \text{m}^2,\qquad D = \sqrt{\frac{4A}{\pi}} = 61.67\ \text{m (adopt 61.7 m)}
$$

#### Step 5: Checks

$$
\begin{aligned}
\text{Organic loading} &= \frac{W}{V} = 0.18\ \text{kg/m}^3\text{/day}\ (\text{high rate: 0.32-1.0})\\
\text{Hydraulic loading} &= \frac{Q(1+R)}{A} = 4.2\ \text{m}^3/\text{m}^2\text{/day}\ (\text{high rate: 10-40})
\end{aligned}
$$

Both loadings are below the usual high-rate range. This is because a single stage must give the high efficiency of 88%, so the NRC equation requires a large volume. A diameter of 61.7 m is also more than the usual maximum of about 60 m for a rotary distributor. In practice, provide **two filters in parallel**, each of area $2987.3/2 = 1493.7$ m² and diameter $\sqrt{4\times1493.7/\pi} = 43.6$ m (depth 2.0 m), or use a two-stage filter.

**Answer: single-stage high-rate filter of volume about 5974.6 m³, depth 2.0 m, diameter 61.7 m (in practice 2 filters of 43.6 m diameter each).**

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. Raw BOD is 280 mg/l with 25% removed in the PST at 5 MLD. What load W goes onto the filter?
   A. 1400 kg/day
   B. 1050 kg/day
   C. 350 kg/day
   D. 700 kg/day
   Answer: B. Settled BOD = 280 x 0.75 = 210 mg/l; W = 5000 x 210 x 10^-3 = 1050 kg/day. (Key point: BOD load on the filter)

2. What required efficiency is used in the NRC equation?
   A. 91.07%, from (280-25)/280
   B. 75%, the primary removal
   C. 88.10%, from (210-25)/210
   D. 85.7%, from 25/210
   Answer: C. Efficiency is calculated on the BOD entering the filter (210 mg/l), giving 88.10%. (Key point: Required efficiency)

3. For R = 1.5, what is F?
   A. 1.890
   B. 2.5
   C. 1.500
   D. 2.174
   Answer: A. F = (1+1.5)/(1+0.15)^2 = 2.5/1.3225 = 1.890. (Key point: Recirculation factor)

4. The answer finds an organic loading of 0.18 kg/m3/day, against 0.8-1.6 for high rate. What does this show?
   A. The filter will fail because it is overloaded
   B. The depth must be raised to 3 m
   C. The recirculation ratio is too high to be valid
   D. The filter is large (lightly loaded) for the stated BOD target, so the computed size is on the generous side
   Answer: D. A loading below the high-rate range means the volume is big relative to load; the NRC design still meets the required effluent BOD. (Key point: Checking the design)

## Related in Sanitary Engineering

- [Effluent BOD of two-stage trickling filter](https://www.bctengineer.com/question.md?key=semester_6%2Fsanitary_engineering%3Atwo-stage-trickling-effluent-bod) — Sanitary Engineering, chapter 7
- [HR trickling filter, 5 MLD, RR 1.5](https://www.bctengineer.com/question.md?key=semester_6%2Fsanitary_engineering%3Atrickling-hr-5mld-rr1p5) — Sanitary Engineering, chapter 7, asked 2×
- [HR trickling filter, 5 MLD, RR 1.3](https://www.bctengineer.com/question.md?key=semester_6%2Fsanitary_engineering%3Atrickling-single-stage-5mld-rr1p3) — Sanitary Engineering, chapter 7
- [HR trickling filter, 2.5 MLD; change if standard](https://www.bctengineer.com/question.md?key=semester_6%2Fsanitary_engineering%3Atrickling-hr-2p5mld-std-compare) — Sanitary Engineering, chapter 7
- [Two-stage trickling filter diameters, 4 MLD](https://www.bctengineer.com/question.md?key=semester_6%2Fsanitary_engineering%3Atwo-stage-trickling-4mld-200) — Sanitary Engineering, chapter 7, asked 2×
- [Two-stage trickling filter diameters, 5 MLD](https://www.bctengineer.com/question.md?key=semester_6%2Fsanitary_engineering%3Atwo-stage-trickling-5mld-180) — Sanitary Engineering, chapter 7, asked 2×

## Related in other subjects

- [Rapid sand filter design for 25 MLD](https://www.bctengineer.com/question.md?key=semester_5%2Fwater_supply_engineering%3Arsf-b81) — Water Supply Engineering, chapter 6