---
title: "Standard rate trickling filter, 1.2 MLD, NRC"
subject: "Sanitary Engineering"
chapter: 7
key: "semester_6/sanitary_engineering:trickling-standard-1p2mld-nrc"
source: https://www.bctengineer.com/q/semester_6/sanitary_engineering/trickling-standard-1p2mld-nrc
site: BCT Engineering
---

# Standard rate trickling filter, 1.2 MLD, NRC

Sanitary Engineering · Chapter 7: Wastewater Treatment

- Asked in: 2070 Bhadra, 8 marks; 2070 Magh, 5+3 marks (asked 2 times)
- Video: [Design of standard rate trickling filter, volume of tank, depth of tank, BOD removal, organic loadin](https://www.youtube.com/watch?v=WO60Ro82fdg) — Civil Engineering Exam
- 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-1p2mld-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 1.2 million liters/day having BOD5 of 200 mg/l. Determine the depth and volume of filter considering surface loading of 1200 liters/m².day and organic loading of 250 gm/m³.day. Also, calculate the efficiency of filter using NRC equation.

## Short answer

From organic loading 250 g/m³·d and surface loading 1200 L/m²·d, volume is 960 m³ and area 1000 m², giving depth 0.96 m. NRC efficiency (no recirculation) is 81.86%, yielding an effluent BOD of 36.3 mg/l.

## Answer

#### Data

Flow $Q=1.2\times 10^6$ l/day $=1200$ m³/day; influent BOD (after PST) $=200$ mg/l; surface (hydraulic) loading $=1200$ l/m²/day; organic loading $=250$ g/m³/day.

#### Step 1: BOD load

$$
W = Q\times BOD = 1200\times 200\times 10^{-3} = 240.0\ \text{kg/day}
$$

#### Step 2: Volume (from organic loading)

$$
V = \frac{W}{\text{organic loading}} = \frac{240.0}{0.25} = 960.0\ \text{m}^3
$$

#### Step 3: Surface area (from hydraulic loading)

$$
A = \frac{Q}{\text{surface loading}} = \frac{1.2\times 10^6}{1200} = 1000.0\ \text{m}^2
$$

#### Step 4: Depth

$$
\text{Depth} = \frac{V}{A} = \frac{960.0}{1000.0} = 0.96\ \text{m}
$$

(Diameter of the circular filter $D=\sqrt{4A/\pi}=35.7$ m. The calculated depth is smaller than the usual 1.8-3 m of a standard-rate filter; if a depth of 1.8 m is wanted, the area is reduced to 533 m² and the hydraulic loading rises accordingly.)

#### Step 5: Efficiency by the NRC equation (standard rate, no recirculation, $F=1$)

$$
E=\frac{100}{1+0.4432\sqrt{W/V}} = \frac{100}{1+0.4432\sqrt{240.0/960.0}} = 81.86\%
$$

Effluent BOD $=200\,(1-81.86/100) = 36.3$ mg/l.

**Answer: volume $=960.0$ m³, depth $=0.96$ m (area 1000.0 m²); NRC efficiency $=81.86\%$.**

## Check yourself

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

1. Which loading gives the filter volume?
   A. Surface loading
   B. Organic loading: V = W / 0.25 kg/m3/d = 960 m3
   C. Velocity gradient
   D. Detention time
   Answer: B. BOD load 240 kg/d over 0.25 kg/m3/d gives 960 m3. (Key point: Volume)

2. Which loading gives the surface area?
   A. Recirculation ratio
   B. BOD removal
   C. Organic loading
   D. Surface (hydraulic) loading: A = 1.2x10^6/1200 = 1000 m2
   Answer: D. Area = Q / surface loading. (Key point: Area)

3. How is depth obtained?
   A. Depth = A/V
   B. Depth = Q/V
   C. Depth = W/A
   D. Depth = V/A = 960/1000 = 0.96 m
   Answer: D. Depth is volume over area. (Key point: Depth)

4. What is special about applying the NRC equation to a standard-rate filter?
   A. No recirculation, so F = 1 and the formula uses W/V
   B. Efficiency is fixed at 50%
   C. W is replaced with Q
   D. F is taken as 10
   Answer: A. With no recirculation F = 1, giving E = 81.86%. (Key point: NRC for standard rate)

## Close variants set in other exams

- [Standard rate trickling filter, 3 MLD, 300 mg/l](https://www.bctengineer.com/question.md?key=semester_6%2Fsanitary_engineering%3Atrickling-standard-3mld-300) — Sanitary Engineering, chapter 7, asked 3×

## Related in Sanitary Engineering

- [Standard rate trickling filter, 4 MLD, NRC](https://www.bctengineer.com/question.md?key=semester_6%2Fsanitary_engineering%3Atrickling-standard-4mld-180-nrc) — Sanitary Engineering, chapter 7
- [Standard rate filter, 3 MLD, 175 mg/l, NRC](https://www.bctengineer.com/question.md?key=semester_6%2Fsanitary_engineering%3Atrickling-standard-3mld-175-nrc) — Sanitary Engineering, chapter 7
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