---
title: "Standard rate trickling filter, 4 MLD, NRC"
subject: "Sanitary Engineering"
chapter: 7
key: "semester_6/sanitary_engineering:trickling-standard-4mld-180-nrc"
source: https://www.bctengineer.com/q/semester_6/sanitary_engineering/trickling-standard-4mld-180-nrc
site: BCT Engineering
---

# Standard rate trickling filter, 4 MLD, NRC

Sanitary Engineering · Chapter 7: Wastewater Treatment

- Asked in: 2072 Asoj, 8 marks
- Video: [numericals on determination of dimension of Standard Rate Trickling Filter](https://www.youtube.com/watch?v=ck9WKAEaRxE) — Geo & Structure Academy
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/sanitary_engineering/trickling-standard-4mld-180-nrc
- Whole chapter: https://www.bctengineer.com/md/semester_6/sanitary_engineering/chapter-7

## Question

The effluent from PST is applied to a standard rate trickling filter at the rate of 4 MLD having a settled sewage BOD of 180 mg/l. Determine the depth and volume of the filter considering (hydraulic) surface loading of 2000 liter/m² day and organic loading of 150 gram/m³ day. Also determine the efficiency of the filter using NRC equation.

## Short answer

Volume = 4800 m³, depth = 2.40 m, area = 2000 m²; NRC efficiency = 85.35%.

## Answer

Data: $Q=4$ MLD $=4000$ m³/day, settled sewage BOD $=180$ mg/l, hydraulic (surface) loading $=2000$ l/m²/day, organic loading $=150$ g/m³/day.

#### Step 1: BOD load

$$
W=Q\times BOD=4000\times 180\times10^{-3}=720.0\ \text{kg/day}
$$

#### Step 2: Volume (organic loading 150 g/m³/day)

$$
V=\frac{W}{\text{organic loading}}=\frac{720.0}{0.150}=4800.0\ \text{m}^3
$$

#### Step 3: Area (hydraulic loading 2000 l/m²/day) and depth

$$
A=\frac{Q}{\text{surface loading}}=\frac{4000\times1000}{2000}=2000.0\ \text{m}^2,\qquad \text{Depth}=\frac{V}{A}=\frac{4800.0}{2000.0}=2.40\ \text{m}
$$
Diameter of the circular filter: $D=\sqrt{4A/\pi}=50.5$ m.

#### Step 4: Efficiency by the NRC equation (no recirculation, $F=1$)

$$
E=\frac{100}{1+0.4432\sqrt{W/V}}=\frac{100}{1+0.4432\sqrt{720.0/4800.0}}=85.35\%
$$

Effluent BOD $=180\,(1-85.35/100)=26.4$ mg/l.

**Answer: volume $=4800.0$ m³; depth $=2.40$ m; area $=2000.0$ m²; NRC efficiency $=85.35\%$.**

## Check yourself

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

1. With W = 720 kg/day and organic loading 150 g/m3/day, what is the volume?
   A. 108 m3
   B. 4800 m3
   C. 1200 m3
   D. 2000 m3
   Answer: B. V = W / organic loading = 720 / 0.150 = 4800 m3. (Key point: Volume)

2. What fixes the plan area?
   A. Organic loading: A = W/OL
   B. Depth assumed at 3 m
   C. Hydraulic loading: A = Q / 2000 l/m2/day = 2000 m2
   D. Volume divided by 2 m
   Answer: C. Area comes from flow divided by surface loading: 4 000 000 / 2000 = 2000 m2. (Key point: Area)

3. What depth results?
   A. 2.4 m (V/A = 4800/2000)
   B. 1.2 m
   C. 4.0 m
   D. 2.0 m
   Answer: A. Depth = V/A = 4800/2000 = 2.4 m, within the 1.8-3 m range for a standard-rate filter. (Key point: Depth)

4. How is the efficiency found and what is it?
   A. E = (BOD in - BOD out)/BOD in with an assumed effluent of 30 mg/l
   B. E = 100/(1+0.4432 sqrt(W/V)) with F = 2, giving 90%
   C. E = organic loading x 100, giving 15%
   D. E = 100/(1+0.4432 sqrt(W/V)) with F = 1, giving 85.35%
   Answer: D. With no recirculation, W/V = 0.15 and E = 100/(1+0.4432 x 0.3873) = 85.35%. (Key point: NRC efficiency)

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