---
title: "Recirculation ratio, filter volume 350 m³"
subject: "Sanitary Engineering"
chapter: 7
key: "semester_6/sanitary_engineering:trickling-recirculation-ratio-350m3"
source: https://www.bctengineer.com/q/semester_6/sanitary_engineering/trickling-recirculation-ratio-350m3
site: BCT Engineering
---

# Recirculation ratio, filter volume 350 m³

Sanitary Engineering · Chapter 7: Wastewater Treatment

- Asked in: 2073 Magh, 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-350m3
- Whole chapter: https://www.bctengineer.com/md/semester_6/sanitary_engineering/chapter-7

## Question

What will be the recirculation ratio required of a single stage filter having volume of 350 m³. A effluent having maximum BOD concentration of 35 mg/lit, for a flow sewage of 5 MLD having BOD of 180 mg/lit and 33% BOD is removed in PST.

## Short answer

Recirculation ratio required is about 1.84, obtained from NRC equation with required efficiency 70.98% and factor F = 2.024.

## 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

$$
\begin{aligned}
Q &= 5000\ \text{m}^3/\text{day}\\
BOD\ \text{applied} &= 180\times(1-0.33) = 120.6\ \text{mg/l}\\
W &= 5000\times 120.6\times10^{-3} = 603.0\ \text{kg/day}\\
E &= \frac{120.6-35}{120.6}\times100 = 70.98\%
\end{aligned}
$$

#### Step 2: Required recirculation factor

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

$$
s=\frac{100/E-1}{0.4432}=\frac{100/70.98-1}{0.4432}=0.9226\ \Rightarrow\ F=\frac{W}{V\,s^2}=\frac{603.0}{350\times 0.9226^2}=2.024
$$

#### Step 3: Recirculation ratio

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

$$
0.0202R^2 - 0.5952R + 1.0242 = 0\ \Rightarrow\ R=1.84
$$

(The smaller root is taken; the other root, about 27.6, is not practical.)

Check: $F(R=1.84)=2.024$ and $E=70.98\%$, equal to the required 70.98%.

**Answer: recirculation ratio required $R \approx 1.84$ (about 1.8).**

## Check yourself

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

1. Raw BOD 180 mg/l, 33% PST removal, effluent 35 mg/l, 5 MLD. What are W and E?
   A. W = 900 kg/day, E = 80.6%
   B. W = 603 kg/day, E = 70.98%
   C. W = 603 kg/day, E = 80.6%
   D. W = 900 kg/day, E = 70.98%
   Answer: B. Settled BOD = 120.6 mg/l; W = 5000 x 120.6 x 10^-3 = 603 kg/day; E = (120.6-35)/120.6 = 70.98%. (Key point: BOD load and efficiency)

2. How is F obtained for the 350 m3 filter?
   A. From the NRC equation: F = W/(V s^2) with s = (100/E - 1)/0.4432 = 0.9226, giving F = 2.024
   B. F = V/W
   C. F = E/100
   D. F = 1 + R with R assumed 1
   Answer: A. The NRC equation is rearranged for F since V is fixed at 350 m3. (Key point: Finding F)

3. Which equation links F to the recirculation ratio R?
   A. F = (1+R)^2/(1+0.1R)
   B. F = 1/(1+R)
   C. F = (1+R)/(1+0.1R)^2, which becomes a quadratic in R
   D. F = R/(1+0.1R)
   Answer: C. The recirculation factor is (1+R)/(1+0.1R)^2; rearranged it is 0.01F R^2 + (0.2F-1)R + (F-1) = 0. (Key point: Solving for R)

4. What recirculation ratio is required?
   A. About 0.5
   B. About 4.5
   C. About 2.0
   D. About 1.8
   Answer: D. Solving the quadratic gives R = 1.84, which gives back F = 2.024 and E = 70.98%. (Key point: Result)

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