---
title: "Standard rate trickling filter, 3 MLD, 300 mg/l"
subject: "Sanitary Engineering"
chapter: 7
key: "semester_6/sanitary_engineering:trickling-standard-3mld-300"
source: https://www.bctengineer.com/q/semester_6/sanitary_engineering/trickling-standard-3mld-300
site: BCT Engineering
---

# Standard rate trickling filter, 3 MLD, 300 mg/l

Sanitary Engineering · Chapter 7: Wastewater Treatment

- Asked in: 2078 Chaitra, 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-standard-3mld-300
- 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 3 million liters/day with BOD5 300 mg/l. Calculate the depth and volume of filter considering the surface loading of 3000 liters/m²day and organic loading of 300 gm/m³day.

## Short answer

Volume = 3000 m³, depth = 3.00 m, area = 1000 m²; NRC efficiency = 80.47%.

## Answer

Data: $Q=3\times10^6$ l/day $=3000$ m³/day, settled-sewage BOD$_5$ $=300$ mg/l, surface loading $=3000$ l/m²/day, organic loading $=300$ g/m³/day. (The NRC efficiency is also given below for completeness.)

#### Step 1: BOD load

$$
W=Q\times BOD=3000\times 300\times10^{-3}=900.0\ \text{kg/day}
$$

#### Step 2: Volume (organic loading 300 g/m³/day)

$$
V=\frac{W}{\text{organic loading}}=\frac{900.0}{0.300}=3000.0\ \text{m}^3
$$

#### Step 3: Area (hydraulic loading 3000 l/m²/day) and depth

$$
A=\frac{Q}{\text{surface loading}}=\frac{3000\times1000}{3000}=1000.0\ \text{m}^2,\qquad \text{Depth}=\frac{V}{A}=\frac{3000.0}{1000.0}=3.00\ \text{m}
$$
Diameter of the circular filter: $D=\sqrt{4A/\pi}=35.7$ m.

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

$$
E=\frac{100}{1+0.4432\sqrt{W/V}}=\frac{100}{1+0.4432\sqrt{900.0/3000.0}}=80.47\%
$$

Effluent BOD $=300\,(1-80.47/100)=58.6$ mg/l.

**Answer: volume $=3000.0$ m³; depth $=3.00$ m; area $=1000.0$ m²; NRC efficiency $=80.47\%$.**

## Check yourself

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

1. In sizing a standard-rate filter from two loadings, what does each give?
   A. Organic loading gives area; hydraulic loading gives volume
   B. Organic loading gives the volume (V = W/OL); hydraulic (surface) loading gives the area (A = Q/SL)
   C. Both give volume and the larger is taken
   D. Organic loading gives depth directly
   Answer: B. BOD load / organic loading = volume; flow / surface loading = plan area; depth = V/A. (Key point: Role of each loading)

2. For 3000 m3/day at 300 mg/l BOD and organic loading 300 g/m3/day, what volume is needed?
   A. 900 m3
   B. 10 000 m3
   C. 3000 m3
   D. 1000 m3
   Answer: C. W = 3000 x 300 x 10^-3 = 900 kg/day; V = 900/0.300 = 3000 m3. (Key point: Volume)

3. Surface loading is 3000 l/m2/day. What depth results?
   A. 3.0 m, since A = 1000 m2 and depth = V/A
   B. 1.0 m, since depth = A/V
   C. 3.0 m, since it equals the loading in m
   D. 0.3 m
   Answer: A. A = 3 000 000 / 3000 = 1000 m2; depth = 3000/1000 = 3.0 m. (Key point: Depth)

4. What F value is used in the NRC equation for a standard-rate filter without recirculation?
   A. F = 0, giving E = 100%
   B. F = 2, giving E = 90%
   C. F = 1.3, giving E = 75%
   D. F = 1, giving E = 80.47%
   Answer: D. No recirculation means R = 0 and F = 1; E = 100/(1+0.4432 sqrt(900/3000)) = 80.47%. (Key point: NRC efficiency of a standard filter)

## Close variants set in other exams

- [Standard rate trickling filter, 1.2 MLD, NRC](https://www.bctengineer.com/question.md?key=semester_6%2Fsanitary_engineering%3Atrickling-standard-1p2mld-nrc) — Sanitary Engineering, chapter 7, asked 3×

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