For R = 1.5 what is F, and what does it do to the required volume?

F = 1.890; the volume from the NRC equation is VF/F, so recirculation reduces the volume

F = 2.5; volume increases by 2.5 times

How is the standard-rate filter treated for the same efficiency?

Take R = 0 so F = 1; the volume then equals VF = 2578.1 m3

Keep R = 1.5 but halve the depth

Use the same volume 1363.8 m3 with a deeper tank

Since D is proportional to sqrt(V) at equal depth, by what percentage must the diameter change for a standard-rate filter?

Increase by about 37.5% (sqrt(1.890) - 1)

2076 Bhadra · 8 marks

HR trickling filter, 2.5 MLD; change if standard

Short answer

High-rate filter volume 1363.8 m³, depth 2.0 m, diameter 29.5 m. If designed as standard-rate, diameter must be increased by 37.5% (to 40.5 m).

Determine the dimensions of a high rate trickling filter for the following data: (i) Sewage flow = 2.5 MLD; (ii) Recirculation ratio = 1.5; (iii) BOD of raw sewage = 300 mg/l; (iv) BOD removal in primary settling tank = 30%; (v) Final effluent BOD desired = 35 mg/l. By what % the diameter of the filter will have to be modified if it is to be designed as a standard rate trickling filter for above requirements?

Data and approach

High-rate filter design uses the NRC (National Research Council) equation; $W$ is in kg/day of BOD applied to the filter, $V$ in m³:

E = \frac{100}{1+0.4432\sqrt{\dfrac{W}{VF}}},\qquad F=\frac{1+R}{(1+0.1R)^2}

Step 1: BOD applied to the filter and required efficiency

\begin{aligned} Q &= 2500\ \text{m}^3/\text{day}\\ \text{BOD after primary} &= 300\times(1-0.3) = 210.0\ \text{mg/l}\\ W &= Q\times BOD = 2500\times 210.0\times 10^{-3} = 525.0\ \text{kg/day}\\ E &= \frac{210.0-35}{210.0}\times 100 = 83.33\% \end{aligned}

Step 2: Recirculation factor

F = \frac{1+1.5}{(1+0.1\times 1.5)^2} = 1.890

Step 3: Volume

83.33 = \frac{100}{1+0.4432\sqrt{W/(VF)}}\Rightarrow s=\sqrt{\frac{W}{VF}} = 0.4513 \Rightarrow VF = \frac{525.0}{0.4513^2} = 2578.1

V = \frac{VF}{F} = \frac{2578.1}{1.890} = 1363.8\ \text{m}^3

Step 4: Diameter

Adopt depth $d = 2.0$ m (high-rate filters are 1.8-3 m deep).

A = \frac{V}{d} = 681.9\ \text{m}^2,\qquad D = \sqrt{\frac{4A}{\pi}} = 29.47\ \text{m (adopt 29.5 m)}

Step 5: Checks

\begin{aligned} \text{Organic loading} &= \frac{W}{V} = 0.38\ \text{kg/m}^3\text{/day}\ (\text{high rate: 0.32-1.0, OK})\\ \text{Hydraulic loading} &= \frac{Q(1+R)}{A} = 9.2\ \text{m}^3/\text{m}^2\text{/day}\ (\text{high rate: 10-40; just below, acceptable}) \end{aligned}

Answer: single-stage high-rate filter of volume about 1363.8 m³, depth 2.0 m, diameter 29.5 m.

Change in diameter if designed as a standard-rate filter

A standard-rate filter has no recirculation ( $R=0$ , $F=1$ ). The same efficiency $E=83.33\%$ (same $s=0.4513$ ) then needs

V_{std}=\frac{W}{s^2}=\frac{525.0}{0.2036}=2578.1\ \text{m}^3

With the same depth of 2.0 m, $D_{std}=\sqrt{4V_{std}/(\pi\times2)}=40.5$ m, compared with $D=29.5$ m for the high-rate filter.

\text{Change in diameter}=\frac{40.5-29.5}{29.5}\times100=37.5\%\ (\text{since } D\propto\sqrt{V}\propto\sqrt{F}=\sqrt{1.890})

The diameter must be increased by about 37.5% if the filter is built as a standard-rate unit.

✓ Answer checked by Claude Opus 5.5 (high effort)

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