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Recirculation ratio, filter volume 510 m³

Short answer

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

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.

Data and approach

NRC equation for a high-rate filter: E=1001+0.4432W/(VF)E=\dfrac{100}{1+0.4432\sqrt{W/(VF)}}, with F=1+R(1+0.1R)2F=\dfrac{1+R}{(1+0.1R)^2} (WW in kg/day, VV in m³).

Step 1: BOD load and required efficiency

Q=2800 m3/dayBOD applied=210×(1−0.2)=168.0 mg/lW=2800×168.0×10−3=470.4 kg/dayE=168.0−35168.0×100=79.17%\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=W/(VF)s=\sqrt{W/(VF)}:

s=100/E−10.4432=100/79.17−10.4432=0.5938 ⇒ F=WV s2=470.4510×0.59382=2.616s=\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=1+R(1+0.1R)2 ⇒ 0.01F R2+(0.2F−1)R+(F−1)=0F=\frac{1+R}{(1+0.1R)^2}\ \Rightarrow\ 0.01F\,R^2+(0.2F-1)R+(F-1)=0 0.0262R2−0.4768R+1.6162=0 ⇒ R=4.500.0262R^2 - 0.4768R + 1.6162 = 0\ \Rightarrow\ R=4.50

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

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

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

✓ Answer checked by Claude Opus 5.5 (high effort)

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