---
title: "HR trickling filter, 5 MLD, RR 1.5"
subject: "Sanitary Engineering"
chapter: 7
key: "semester_6/sanitary_engineering:trickling-hr-5mld-rr1p5"
source: https://www.bctengineer.com/q/semester_6/sanitary_engineering/trickling-hr-5mld-rr1p5
site: BCT Engineering
---

# HR trickling filter, 5 MLD, RR 1.5

Sanitary Engineering · Chapter 7: Wastewater Treatment

- Asked in: 2075 Baisakh, 10 marks; 2069 Bhadra, 8 marks (asked 2 times)
- 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-rr1p5
- Whole chapter: https://www.bctengineer.com/md/semester_6/sanitary_engineering/chapter-7

## Question

Determine the size of a high-rate single stage trickling filter for the following data: Sewage flow = 5 MLD; Recirculation ratio = 1.5; BOD of raw sewage = 250 mg/l; BOD removal in primary clarifier = 30%; Final effluent BOD desired = 30 mg/l.

## Short answer

Using the NRC equation with BOD load after primary 875 kg/d, required efficiency 82.86%, and recirculation factor F=1.89, the required filter volume is 2124 m³. Depth 2.0 m gives a diameter of 36.8 m.

## 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

$$
\begin{aligned}
Q &= 5000\ \text{m}^3/\text{day}\\
\text{BOD after primary} &= 250\times(1-0.3) = 175.0\ \text{mg/l}\\
W &= Q\times BOD = 5000\times 175.0\times 10^{-3} = 875.0\ \text{kg/day}\\
E &= \frac{175.0-30}{175.0}\times 100 = 82.86\%
\end{aligned}
$$

#### Step 2: Recirculation factor

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

#### Step 3: Volume

$$
82.86 = \frac{100}{1+0.4432\sqrt{W/(VF)}}\Rightarrow \sqrt{\frac{W}{VF}} = 0.4668 \Rightarrow VF = \frac{875.0}{0.4668^2} = 4015.1
$$

$$
V = \frac{VF}{F} = \frac{4015.1}{1.890} = 2124.0\ \text{m}^3
$$

#### Step 4: Diameter

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

$$
A = \frac{V}{d} = 1062.0\ \text{m}^2,\qquad D = \sqrt{\frac{4A}{\pi}} = 36.77\ \text{m (adopt 36.8 m)}
$$

#### Step 5: Checks

$$
\begin{aligned}
\text{Organic loading} &= \frac{W}{V} = 0.41\ \text{kg/m}^3\text{/day}\ (\text{high rate: 0.8-1.6})\\
\text{Hydraulic loading} &= \frac{Q(1+R)}{A} = 11.8\ \text{m}^3/\text{m}^2\text{/day}\ (\text{high rate: 10-40})
\end{aligned}
$$

The organic loading comes out below the usual high-rate range because a single stage has to give a high efficiency (about 83%); the hydraulic loading is within the high-rate range.

**Answer: single-stage high-rate filter of volume about 2124.0 m³, depth 2.0 m, diameter 36.8 m.**

## Check yourself

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

1. Which equation is used for the high-rate filter?
   A. Streeter-Phelps equation
   B. Hazen formula
   C. NRC equation E = 100/(1+0.4432 sqrt(W/VF))
   D. Camp-Shields equation
   Answer: C. High-rate filters are designed with the NRC equation. (Key point: Equation)

2. What BOD load W is applied to the filter?
   A. 5000 m3/d x 175 mg/l = 875 kg/d (after 30% primary removal)
   B. 875 g/d
   C. 5000 x 250 = 1250 kg/d
   D. 5000 x 30 = 150 kg/d
   Answer: A. Primary removal lowers 250 to 175 mg/l, giving W = 875 kg/d. (Key point: BOD to filter)

3. What does F = (1+R)/(1+0.1R)^2 = 1.89 represent?
   A. The hydraulic loading
   B. The efficiency
   C. The filter depth
   D. The effective number of passes of the organic matter through the filter due to recirculation
   Answer: D. F accounts for recirculation, which improves treatment as if the filter were larger. (Key point: Recirculation factor)

4. How is V found after F and E are known?
   A. Solve E for W/(VF) to get VF, then V = VF/F = 2124 m3
   B. V = W x F
   C. V = Q / E
   D. V = A x F
   Answer: A. VF = 875/0.4668^2 = 4015; dividing by 1.89 gives 2124 m3. (Key point: Volume)

## Related in Sanitary Engineering

- [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, BOD 280 mg/l](https://www.bctengineer.com/question.md?key=semester_6%2Fsanitary_engineering%3Atrickling-hr-5mld-280) — Sanitary Engineering, chapter 7
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- [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
- [Recirculation ratio, filter volume 510 m³](https://www.bctengineer.com/question.md?key=semester_6%2Fsanitary_engineering%3Atrickling-recirculation-ratio-510m3) — 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×