---
title: "Standard rate filter, 3 MLD, 175 mg/l, NRC"
subject: "Sanitary Engineering"
chapter: 7
key: "semester_6/sanitary_engineering:trickling-standard-3mld-175-nrc"
source: https://www.bctengineer.com/q/semester_6/sanitary_engineering/trickling-standard-3mld-175-nrc
site: BCT Engineering
---

# Standard rate filter, 3 MLD, 175 mg/l, NRC

Sanitary Engineering · Chapter 7: Wastewater Treatment

- Asked in: 2068 Bhadra (old course), 10 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-175-nrc
- Whole chapter: https://www.bctengineer.com/md/semester_6/sanitary_engineering/chapter-7

## Question

The effluent from a primary sedimentation tank is applied to a standard rate filter at the rate of 3 million liters per day, having a BOD5 of 175 mg/l. Determine the depth and volume of filter, adopting a surface loading of 150 gm/m³/day. Also determine the efficiency of such filter unit, using NRC formula. Assume recirculation ratio = 1:2.

## Short answer

Volume = 3500 m³, depth 2.00 m, area = 1750 m²; using recirculation ratio 0.5, NRC efficiency = 87.17%.

## Answer

Data: $Q=3$ MLD $=3000$ m³/day, BOD$_5$ $=175$ mg/l, organic loading $=150$ g/m³/day (the "surface loading" given in g/m³/day is an organic loading). The recirculation ratio 1 : 2 is read as $R=Q_R/Q=0.5$. The hydraulic loading is not given, so a depth is assumed.

#### Step 1: BOD load

$$
W=Q\times BOD=3000\times 175\times10^{-3}=525.0\ \text{kg/day}
$$

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

$$
V=\frac{W}{\text{organic loading}}=\frac{525.0}{0.150}=3500.0\ \text{m}^3
$$

#### Step 3: Depth and area

No hydraulic loading is given, so adopt a depth of 2.00 m (standard-rate filters are 1.8-3 m deep):

$$
A=\frac{V}{2.00}=1750.0\ \text{m}^2,\qquad D=\sqrt{\frac{4A}{\pi}}=47.2\ \text{m},\qquad \text{hydraulic loading}=\frac{3000}{1750.0}=1.71\ \text{m}^3/\text{m}^2\text{/day}
$$

#### Step 4: Efficiency by the NRC equation

With the given recirculation ratio $R=0.5$: $F=\dfrac{1+R}{(1+0.1R)^2}=1.361$

$$
E=\frac{100}{1+0.4432\sqrt{W/(VF)}}=\frac{100}{1+0.4432\sqrt{525.0/(3500.0\times 1.361)}}=87.17\%
$$

Effluent BOD $=175\,(1-87.17/100)=22.4$ mg/l.

(If the ratio is read as $R=2$, then $F=2.083$ and $E\approx 89.4\%$.)

**Answer: volume $=3500.0$ m³; depth $=2.00$ m; area $=1750.0$ m²; NRC efficiency $=87.17\%$.**

## Check yourself

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

1. How is the recirculation ratio '1 : 2' interpreted in this answer?
   A. R = 2
   B. R = Q_R/Q = 0.5
   C. R = 1
   D. R = 0, since it is a standard-rate filter
   Answer: B. Recirculated flow to raw flow 1:2 means R = 0.5. (Key point: Meaning of the 1:2 ratio)

2. Organic loading is 150 g/m3/day and W = 525 kg/day. What volume?
   A. 78.75 m3
   B. 1750 m3
   C. 3500 m3
   D. 7000 m3
   Answer: C. V = 525/0.150 = 3500 m3. (Key point: Volume)

3. Why does the answer simply adopt a 2 m depth?
   A. No hydraulic loading is given, so area cannot come from Q/SL; a depth in the usual 1.8-3 m range is adopted
   B. Depth is always 2 m for any filter
   C. Depth is found from the BOD load
   D. The NRC equation gives depth
   Answer: A. Without a surface loading, the area is found by choosing a depth: A = 3500/2 = 1750 m2. (Key point: Why a depth is assumed)

4. What is F for R = 0.5 and what is the resulting efficiency?
   A. F = 1.5, E = 80%
   B. F = 1.0, E = 85%
   C. F = 1.361, E = 70%
   D. F = 1.361, E = 87.17%
   Answer: D. F = 1.5/(1.05)^2 = 1.361; E = 100/(1+0.4432 sqrt(525/(3500 x 1.361))) = 87.17%. (Key point: Efficiency with recirculation)

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