---
title: "Recirculation ratio, filter volume 510 m³"
subject: "Sanitary Engineering"
chapter: 7
key: "semester_6/sanitary_engineering:trickling-recirculation-ratio-510m3"
source: https://www.bctengineer.com/q/semester_6/sanitary_engineering/trickling-recirculation-ratio-510m3
site: BCT Engineering
---

# Recirculation ratio, filter volume 510 m³

Sanitary Engineering · Chapter 7: Wastewater Treatment

- Asked in: 2074 Bhadra, 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-recirculation-ratio-510m3
- Whole chapter: https://www.bctengineer.com/md/semester_6/sanitary_engineering/chapter-7

## Question

If the effluent BOD is to be equal to or less than 35 mg/l, what will be the recirculation ratio required of a single high rate trickling filter having volume of 510 m³ which receives a flow of 2.8 MLD. The raw sewage has BOD of 210 mg/l. The primary treatment removes 20% BOD.

## Short answer

Recirculation ratio required is about 4.5, determined from NRC equation with required efficiency 79.17% and factor F = 2.616.

## Answer

#### Data and approach

NRC equation for a high-rate filter: $E=\dfrac{100}{1+0.4432\sqrt{W/(VF)}}$, with $F=\dfrac{1+R}{(1+0.1R)^2}$ ($W$ in kg/day, $V$ in m³).

#### Step 1: BOD load and required efficiency

$$
\begin{aligned}
Q &= 2800\ \text{m}^3/\text{day}\\
BOD\ \text{applied} &= 210\times(1-0.2) = 168.0\ \text{mg/l}\\
W &= 2800\times 168.0\times10^{-3} = 470.4\ \text{kg/day}\\
E &= \frac{168.0-35}{168.0}\times100 = 79.17\%
\end{aligned}
$$

#### Step 2: Required recirculation factor

Let $s=\sqrt{W/(VF)}$:

$$
s=\frac{100/E-1}{0.4432}=\frac{100/79.17-1}{0.4432}=0.5938\ \Rightarrow\ F=\frac{W}{V\,s^2}=\frac{470.4}{510\times 0.5938^2}=2.616
$$

#### Step 3: Recirculation ratio

$$
F=\frac{1+R}{(1+0.1R)^2}\ \Rightarrow\ 0.01F\,R^2+(0.2F-1)R+(F-1)=0
$$

$$
0.0262R^2 - 0.4768R + 1.6162 = 0\ \Rightarrow\ R=4.50
$$

(The other root, $R\approx 13.7$, lies beyond $R=8$, where $F$ is at its maximum, and is not used.)

Check: $F(R=4.50)=2.616$ and $E=79.17\%$, equal to the required 79.17%.

**Answer: recirculation ratio required $R \approx 4.50$ (about 4.5).**

## Check yourself

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

1. Here the volume is known and R is wanted. What is the first step after finding W and E?
   A. Assume R and compute V
   B. Rearrange the NRC equation to find F = W/(V s^2), where s = (100/E - 1)/0.4432
   C. Use R = (1-E)/E directly
   D. Find the diameter first
   Answer: B. With E, W and V known, F is the only unknown in the NRC equation, and R is then found from F. (Key point: Strategy)

2. With raw BOD 210 mg/l, 20% PST removal, final BOD 35 mg/l and 2.8 MLD, what are W and E?
   A. W = 470.4 kg/day, E = 79.17%
   B. W = 588 kg/day, E = 83.3%
   C. W = 470.4 kg/day, E = 83.3%
   D. W = 588 kg/day, E = 79.17%
   Answer: A. Settled BOD = 168 mg/l; W = 2800 x 168 x 10^-3 = 470.4 kg/day; E = (168-35)/168 = 79.17%. (Key point: BOD load and efficiency)

3. How is R obtained once F = 2.616 is known?
   A. R = F - 1 directly
   B. R = F/(1+0.1F)
   C. Solve F = (1+R)/(1+0.1R)^2, a quadratic in R: 0.01F R^2 + (0.2F - 1)R + (F - 1) = 0
   D. R = sqrt(F)
   Answer: C. Expanding the F equation gives a quadratic in R; the practical root is R = 4.50. (Key point: Solving for R)

4. What recirculation ratio is needed?
   A. About 1.3
   B. About 2.6
   C. About 0.5
   D. About 4.5
   Answer: D. The quadratic gives R = 4.50, which reproduces F = 2.616 and E = 79.17%. (Key point: Result)

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