Question 4 of 5

Question 4 (25 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

(a) Oxygen demand and air requirement for 95% BOD removal (10 marks)

Given.

Find. The net oxygen demand (kg O₂/d) and the corresponding air requirement (m³ air/d).

Approach. Convert the BOD₅ removed to an oxygen demand using the ultimate-BOD conversion factor, credit back the oxygen equivalent of the biomass that leaves as waste sludge, then convert the net oxygen demand to an air volume via air's oxygen content and an assumed field oxygen-transfer efficiency.

1. BOD₅ removed load. Effluent $S_e = S_0(1-0.95) = 150(0.05) = 7.5$ mg/L, so $$\Delta BOD_5 = \frac{Q(S_0-S_e)}{1000} = \frac{15{,}000\times(150-7.5)}{1000} = \boxed{2137.5\ \text{kg BOD}_5/\text{d}}$$
2. Biomass (waste sludge) produced. Using the given observed yield, $$P_x = Y_{obs}\times\Delta BOD_5 = 0.8\times2137.5 = \boxed{1710\ \text{kg TSS/d}}$$
3. Net oxygen demand. Converting BOD₅ to ultimate oxygen demand with $f=BOD_L/BOD_5\approx1.47$ (from the standard $BOD_5/BOD_L=0.68$ relation) and crediting 1.42 kg O₂ per kg of biomass carbon that leaves as waste sludge rather than being oxidized (Metcalf & Eddy AOTR relation): $$AOTR = f\cdot\Delta BOD_5 - 1.42\,P_x = 1.47(2137.5)-1.42(1710) = 3142.1-2428.2 = \boxed{713.9\ \text{kg O}_2/\text{d}}$$
4. Air requirement. Air at standard conditions has density ≈1.2 kg/m³ and an oxygen mass fraction of 23.2%, so 1 m³ of air carries $1.2\times0.232=0.2784$ kg O₂. Assuming a field oxygen-transfer efficiency $OTE=8\%$ (typical for diffused/mechanical aeration, per check note below): $$\text{Air} = \frac{AOTR}{1.2\times0.232\times OTE} = \frac{713.9}{1.2\times0.232\times0.08} = \frac{713.9}{0.02227} = \boxed{32{,}055\ \text{m}^3\ \text{air/d}}$$

(b) Aeration tank volume (8 marks)

Given. $SRT = 4$ d, $MLSS = 2000$ mg/L, and the biomass production $P_x=1710$ kg TSS/d from part (a).

Find. Required aeration tank volume $V$.

Approach. At steady state, the mass of biomass in the reactor ($MLSS\times V$) turns over once every SRT (biomass wasted per day, $P_x$, times SRT), so $SRT = (MLSS\cdot V)/P_x$ rearranges directly for $V$.

1. Aeration volume. With $MLSS=2000\ \text{mg/L}=2.0\ \text{kg/m}^3$: $$V = \frac{P_x\cdot SRT}{MLSS} = \frac{1710\times4}{2.0} = \boxed{3420\ \text{m}^3}$$
2. Hydraulic retention time check. $$HRT = \frac{V}{Q}\times24 = \frac{3420}{15{,}000}\times24 = \boxed{5.47\ \text{h}}$$ This falls within the typical 4–8 h HRT range for conventional activated sludge, confirming the tank size is physically reasonable.

(c) Secondary clarifier area (7 marks)

Given. Peak hourly flow factor $= 3.0$; maximum allowable peak hourly surface overflow rate $SOR_{max}=35\ \text{m}^3/\text{m}^2\text{-d}$.

Find. Minimum required clarifier surface area.

1. Peak hourly flow. $$Q_{peak} = Q\times PF = 15{,}000\times3.0 = \boxed{45{,}000\ \text{m}^3/\text{d}}$$
2. Clarifier area. $$A = \frac{Q_{peak}}{SOR_{max}} = \frac{45{,}000}{35} = \boxed{1285.7\ \text{m}^2}$$