Why does the answer state assumptions like Y = 0.5, kd = 0.06/day and theta_c = 10 days?

The question did not give them, so typical design values are assumed

They are derived from the BOD data

They are fixed by law for all plants

They are required only for the recycle ratio

Which expression gives the volume for a completely mixed reactor?

V = theta_c Y Q (S0 - S) / (X (1 + kd theta_c)) = 2083 m3

How is the recycle ratio obtained?

From a biomass balance X(Q + Qr) = Xr Qr, so r = X/(Xr - X) = 3000/7000 = 0.43

2072 Asoj · 8 marks

Completely mixed ASP reactor, 10,000 m³/day

Short answer

Reactor volume 2083 m³ using Lawrence-McCarty equation with θc=10 days, Y=0.5, kd=0.06; detention time 5 h; recycle ratio 0.43 assuming return sludge concentration 10,000 mg/L.

A activated sludge system is to be used for secondary treatment of 10,000 m³/day of municipal wastewater. After primary clarification, the BOD is 250 mg/l and is desired to have not more than 50 mg/l of soluble BOD in the effluent. A completely mixed reactor is to be used. MLSS concentration is of 3000 mg/l. Determine: i) The volume of reactor; ii) Detention time; iii) The recycle ratio.

Assumptions (not given, typical values)

Yield $Y=0.5$ kg cells/kg BOD removed; endogenous decay $k_d=0.06$ /day; mean cell residence time (sludge age) $\theta_c=10$ days; effluent suspended solids neglected; recycled sludge concentration $X_R=10000$ mg/l (SVI about 100 ml/g).

(i) Volume of the reactor

For a completely mixed reactor (Lawrence-McCarty):

V=\frac{\theta_c\,Y\,Q\,(S_0-S)}{X\,(1+k_d\theta_c)}=\frac{10\times0.5\times10000\times(250-50)}{3000\times(1+0.06\times10)}=2083\ \text{m}^3

(Check: $F/M=\dfrac{QS_0}{VX}=\dfrac{10000\times250}{2083\times3000}=0.40$ /day, within the usual 0.2-0.6.)

(ii) Detention time

t=\frac{V}{Q}=\frac{2083}{10000}=0.208\ \text{day}=5.0\ \text{h}

(iii) Recycle ratio

A balance on the biomass in the reactor (no solids in the effluent) gives, for the aeration tank, $X(Q+Q_R)=X_RQ_R$ :

r=\frac{Q_R}{Q}=\frac{X}{X_R-X}=\frac{3000}{10000-3000}=0.43

So the return sludge flow is $Q_R=4286$ m³/day. (Waste sludge: $Q_w=\dfrac{VX}{\theta_cX_R}=62.5$ m³/day.)

Answer: (i) $V\approx 2083$ m³; (ii) detention time $\approx 5.0$ h; (iii) recycle ratio $\approx 0.43$ .

✓ Answer checked by Claude Opus 5.5 (high effort)

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Completely mixed ASP reactor, 10,000 m³/day

Assumptions (not given, typical values)

(i) Volume of the reactor

(ii) Detention time

(iii) Recycle ratio

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