What is computed first and what is its value?

E1 = 100/(1+0.4432 sqrt(W1/(V1 F))) = 66.50%

What load and equation are used for stage 2?

W2 = W1(1 - E1) = 586.3 kg/day, with the 0.4432/(1 - E1) factor in the formula

W2 = W1 = 1750 kg/day with the same formula as stage 1

W2 = 586.3 kg/day with the same formula as stage 1

How is the final effluent BOD obtained?

BOD from stage 1 (117.3 mg/l) x (1 - E2) = 60 mg/l

2075 Bhadra · 8 marks

Effluent BOD of two-stage trickling filter

Short answer

First-stage efficiency 66.50% leaves BOD 117.3 mg/l; second-stage efficiency 48.87% reduces it further. Final effluent BOD is 60.0 mg/l.

Calculate the effluent BOD of a two stage trickling filter with the following data: i) Sewage flow = 5 MLD; ii) Influent BOD in first trickling filter = 350 mg/l; iii) Volume of first filter = 650 m³; iv) Volume of second filter = 450 m³; v) Recirculation ratio for both filters = 2.0

Equations (NRC, two-stage; $W$ in kg/day, $V$ in m³)

E_1=\frac{100}{1+0.4432\sqrt{\dfrac{W_1}{V_1F}}},\qquad E_2=\frac{100}{1+\dfrac{0.4432}{1-E_1}\sqrt{\dfrac{W_2}{V_2F}}},\qquad F=\frac{1+R}{(1+0.1R)^2}

( $E_1$ is used as a fraction in the second equation.)

Step 1: Loads and recirculation factor

\begin{aligned} W_1 &= 5000\times350\times10^{-3}=1750.0\ \text{kg/day}\\ F &= \frac{1+2}{(1+0.2)^2}=2.0833 \end{aligned}

Step 2: First stage

E_1=\frac{100}{1+0.4432\sqrt{1750.0/(650\times 2.0833)}}=66.50\%

\text{BOD leaving stage 1}=350\,(1-0.6650)=117.3\ \text{mg/l},\qquad W_2=1750.0\times(1-0.6650)=586.3\ \text{kg/day}

Step 3: Second stage

E_2=\frac{100}{1+\dfrac{0.4432}{1-0.6650}\sqrt{586.3/(450\times 2.0833)}}=48.87\%

Step 4: Final effluent BOD

BOD_e=117.3\times(1-0.4887)=60.0\ \text{mg/l}

Overall filter efficiency $=1-60.0/350=82.9\%$ .

Answer: effluent BOD $\approx 60.0$ mg/l.

✓ Answer checked by Claude Opus 5.5 (high effort)

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