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Question 1
ప్రశ్న 1
For a conventional activated sludge aeration tank, which expression correctly defines the food-to-microorganism ratio when the microorganism inventory is specified on an MLVSS basis?
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Explanation:
Food represents the biodegradable organic loading applied to the aeration process, commonly expressed as BOD5 mass per day. Microorganism inventory is represented here by the specified MLVSS mass in the aeration tank. The F/M ratio is the applied BOD5 load divided by this MLVSS inventory. Its units are kg BOD5 per kg MLVSS per day. MLSS-based and MLVSS-based values should not be interchanged without identifying the chosen basis.
వివరణ:
Question 2
ప్రశ్న 2
Which statement best distinguishes hydraulic retention time (HRT) from solids retention time (SRT) in a conventional activated sludge process?
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Explanation:
HRT measures a nominal period of wastewater residence based on the stated tank volume and flow convention. SRT represents the average time that sludge solids remain within the defined biological system. Recycling settled sludge can retain biomass much longer than the hydraulic residence of water. SRT calculations must account for solids wasting and loss in the final effluent. Therefore the two times describe different design and operating characteristics.
వివరణ:
Question 3
ప్రశ్న 3
A tank has MLSS of 3,500 mg/L and MLVSS equal to 70% of MLSS. Which interpretation is most appropriate for design calculations?
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Explanation:
MLVSS is calculated from the volatile fraction of mixed-liquor suspended solids. Multiplying 3,500 mg/L by 0.70 gives 2,450 mg/L. This fraction includes volatile material that is not necessarily living biomass. An F/M calculation using MLVSS has a different denominator from an MLSS-based calculation. Consequently the chosen concentration basis must be stated consistently in the design.
వివరణ:
Question 4
ప్రశ్న 4
At fixed applied BOD5 load and aeration volume, the measured MLVSS concentration doubles while other design assumptions are unchanged. What happens to the calculated F/M ratio?
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Explanation:
The F/M ratio equals applied BOD5 mass per day divided by the specified biomass inventory. At fixed volume, doubling MLVSS concentration doubles that inventory. The applied food load remains the same under the stated assumption. Therefore the calculated F/M ratio is reduced by a factor of two. Whether that operating change is desirable still depends on process performance and secondary-clarifier capacity.
వివరణ:
Question 5
ప్రశ్న 5
Which approach correctly represents the solids-retention-time calculation for a conventional ASP when secondary-clarifier sludge inventory is significant?
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Explanation:
SRT is based on the mass of solids retained within the chosen process boundary. Significant solids held in the secondary clarifier belong in the numerator when that unit is included in the biological system. Solids deliberately wasted and solids carried out with the final effluent belong in the denominator. A consistent suspended-solids or volatile-solids basis is needed throughout the calculation. Neglecting a substantial clarifier blanket can bias the calculated solids age.
వివరణ:
Question 6
ప్రశ్న 6
For a simplified steady-state activated sludge solids balance with negligible fresh suspended solids, WAS and final-effluent solids, which equation correctly relates fresh flow Q, return flow Qr, aeration MLSS X and return-sludge concentration Xr?
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Explanation:
The clarified liquid is assumed to contain negligible solids under the simplified balance. Solids entering the aeration tank with return sludge equal Qr times Xr. Solids leaving the aeration tank toward clarification equal (Q + Qr) times X. Equating these terms and rearranging gives Qr/Q = X/(Xr - X). The relation requires Xr to exceed X and should not be applied unmodified when important additional solids flows exist.
వివరణ:
Question 7
ప్రశ్న 7
Which statement correctly separates carbonaceous oxygen demand from nitrification demand when sizing activated-sludge aeration equipment?
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Explanation:
Carbonaceous oxygen consumption is associated with substrate oxidation and endogenous metabolism. Part of the organic substrate can also be incorporated into new biomass. Biological nitrification creates additional demand when ammonia nitrogen is oxidized. A common stoichiometric allowance for complete oxidation to nitrate is approximately 4.56 kg oxygen per kg ammonia nitrogen oxidized. Actual equipment sizing must also account for field oxygen-transfer efficiency and operating conditions.
వివరణ:
Question 8
ప్రశ్న 8
An ASP designer increases the selected MLSS concentration in order to reduce aeration-tank volume at unchanged BOD loading and F/M basis. Which further checks are especially important?
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Explanation:
At a fixed F/M ratio, greater biomass concentration can reduce the reactor volume needed for a specified food load. However, concentrating more solids in less liquid can place additional demands on aeration and mixing. The secondary clarifier must reliably separate and concentrate those solids. Return-sludge pumping and the response to hydraulic surges also affect feasibility. Reactor volume should not be minimized without checking the associated unit-process constraints.
వివరణ:
Question 9
ప్రశ్న 9
An aeration tank receives 5.0 MLD of primary effluent at 180 mg/L BOD5 and contains 3,000 m3 of mixed liquor with MLVSS = 2,500 mg/L. What is the F/M ratio based on applied BOD5 and MLVSS?
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Explanation:
The incoming BOD5 load is 5.0 times 180 = 900 kg/day. The MLVSS concentration of 2,500 mg/L equals 2.5 kg/m3. Multiplying this concentration by 3,000 m3 gives an MLVSS inventory of 7,500 kg. The F/M ratio is 900/7,500 = 0.12 per day on the stated BOD5 and MLVSS basis. Return-sludge flow need not be added to the fresh external BOD5 load for this defined calculation.
వివరణ:
Question 10
ప్రశ్న 10
A conventional ASP is to treat 4.0 MLD with applied BOD5 concentration 200 mg/L. If the selected F/M is 0.20 kg BOD5/(kg MLVSS day) and design MLVSS is 2,500 mg/L, what aeration-tank volume is required?
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Explanation:
Fresh incoming BOD5 load is 4.0 times 200 = 800 kg/day. The design MLVSS concentration equals 2.5 kg/m3. Required biomass inventory is 800/0.20 = 4,000 kg MLVSS. Dividing this inventory by 2.5 kg/m3 gives a tank volume of 1,600 m3. Hydraulic retention time and other design criteria must subsequently be checked.
వివరణ:
Question 11
ప్రశ్న 11
An ASP receives 3.0 MLD of wastewater with BOD5 = 210 mg/L. Aeration volume is 1,500 m3, MLSS is 3,000 mg/L and MLVSS constitutes 70% of MLSS. What is the applied-BOD5-to-MLVSS F/M ratio?
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Explanation:
The applied BOD5 load is 3.0 times 210 = 630 kg/day. The MLVSS concentration is 0.70 times 3,000 = 2,100 mg/L. This is equivalent to 2.1 kg/m3 and gives an MLVSS inventory of 3,150 kg. The F/M ratio is therefore 630/3,150 = 0.20 per day. Using the larger MLSS inventory instead would yield a different ratio and would not follow the stated basis.
వివరణ:
Question 12
ప్రశ్న 12
Aeration-tank operating volume is 2,400 m3 and the fresh sewage flow is 8.0 MLD. Using nominal HRT = V/Q based on fresh flow, what is the HRT?
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Explanation:
Fresh flow of 8.0 MLD is 8,000 m3/day. The specified nominal HRT equals operating volume divided by fresh daily flow. The ratio is 2,400/8,000 = 0.30 day. Multiplying by 24 gives 7.2 hours. This definition does not treat internal RAS circulation as an additional external wastewater inflow.
వివరణ:
Question 13
ప్రశ్న 13
An aeration tank has operating volume 1,800 m3 and receives 4.0 MLD of primary effluent containing 180 mg/L BOD5. What is the applied volumetric BOD5 loading?
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Explanation:
The incoming organic load is 4.0 times 180 = 720 kg BOD5/day. Volumetric loading is the applied food mass per day divided by the reactor liquid volume. The specified aeration volume is 1,800 m3. Dividing 720 by 1,800 gives 0.40 kg BOD5/(m3 day). This parameter differs from F/M because it does not divide by the biomass inventory.
వివరణ:
Question 14
ప్రశ్న 14
A conventional ASP treats 6.0 MLD of fresh sewage and operates at Qr/Q = 0.40. Neglecting WAS and other withdrawals upstream of the secondary clarifier, what are return-sludge flow and aeration-tank outlet flow?
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Explanation:
Return-sludge flow equals the RAS ratio multiplied by fresh flow. Here Qr = 0.40 times 6.0 = 2.4 MLD. The aeration tank receives fresh flow and return sludge together. Their sum is 6.0 + 2.4 = 8.4 MLD. Because RAS is an internal recycle, the net plant effluent should not be inferred to equal this larger mixed flow.
వివరణ:
Question 15
ప్రశ్న 15
At steady state, 8.0 MLD fresh wastewater with negligible incoming suspended solids is mixed with RAS. Aeration-tank MLSS is 3,000 mg/L and RAS solids are 12,000 mg/L. Neglect WAS and final-effluent solids for this balance. What RAS flow is needed?
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Explanation:
The simplified sludge return ratio is X/(Xr - X). Substituting X = 3,000 and Xr = 12,000 mg/L gives 3,000/9,000 = 1/3. Multiplying fresh flow of 8.0 MLD by this ratio gives Qr = 2.667 MLD. The nearest answer is approximately 2.67 MLD. This calculation depends on the specified simplifications of the solids balance.
వివరణ:
Question 16
ప్రశ్న 16
For an instantaneous mixing calculation, 4.0 MLD of wastewater at 200 mg/L suspended solids is blended with 1.0 MLD of RAS at 8,200 mg/L. Neglect growth and solids removal during mixing. What mixed suspended-solids concentration results?
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Explanation:
The fresh-flow solids contribution is 4.0 times 200 = 800 in concentration-flow units. The RAS contribution is 1.0 times 8,200 = 8,200 in the same units. Total suspended-solids contribution is 9,000 and total mixed flow is 5.0 MLD. Dividing 9,000 by 5.0 gives 1,800 mg/L. The problem is a mixing balance and does not model subsequent biological growth or decay.
వివరణ:
Question 17
ప్రశ్న 17
An ASP has 6,000 kg of relevant suspended-solids inventory and requires SRT = 10 days. Solids loss through the final effluent is 100 kg/day. If WAS contains 10,000 mg/L suspended solids, what waste-sludge flow is required, assuming these are the only solids outlets?
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Explanation:
For the stated process boundary, target daily solids loss equals inventory divided by SRT. Thus total loss must be 6,000/10 = 600 kg/day. The effluent already carries 100 kg/day, leaving 500 kg/day for WAS. A WAS concentration of 10,000 mg/L equals 10 kg/m3. Required WAS flow is 500/10 = 50 m3/day.
వివరణ:
Question 18
ప్రశ్న 18
An ASP retains 4,500 kg of solids in the aeration tanks and 500 kg in the secondary clarifier. WAS removes 600 kg/day and final effluent carries 100 kg/day on the same solids basis. What is the overall SRT?
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Explanation:
Total relevant solids inventory is 4,500 + 500 = 5,000 kg. Total daily solids leaving the process is 600 + 100 = 700 kg/day. SRT equals inventory divided by daily solids loss. The result is 5,000/700 = approximately 7.14 days. Including the clarifier inventory is appropriate because the question explicitly defines it within the system boundary.
వివరణ:
Question 19
ప్రశ్న 19
An aeration basin initially contains 1,600 m3 of mixed liquor at MLSS = 3,500 mg/L. A controlled operation is intended to lower inventory to the equivalent of 3,000 mg/L at the same 1,600 m3 operating volume. If wasted sludge remains at 8,000 mg/L and growth and other solids flows are neglected, what total WAS volume removes the required solids mass?
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Explanation:
The initial suspended-solids mass is 1,600 times 3.5 = 5,600 kg. The target inventory is 1,600 times 3.0 = 4,800 kg. Thus the mass to be removed is 800 kg. At a stipulated constant WAS concentration of 8 kg/m3, the required withdrawal volume is 800/8 = 100 m3. This simplified inventory adjustment assumes that the basin volume is maintained and other solids gains or losses are negligible.
వివరణ:
Question 20
ప్రశ్న 20
For a carbonaceous-only ASP oxygen estimate, 800 kg/day of BOD5 is removed, the specified ratio f = BOD5/ultimate BOD is 0.80, and net biological sludge production is 250 kg VSS/day. Using O2 = BOD5 removed/f - 1.42 times net VSS production, what oxygen requirement results?
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Explanation:
The specified ratio converts 800 kg/day of removed BOD5 to 800/0.80 = 1,000 kg/day ultimate oxygen-equivalent demand. The oxygen-equivalent allowance for retained net biological solids is 1.42 times 250 = 355 kg/day. The stated carbonaceous formula subtracts this amount from ultimate demand. The calculated carbonaceous oxygen requirement is 1,000 - 355 = 645 kg O2/day. Additional nitrification requirements and field transfer adjustments have been excluded explicitly.
వివరణ:
Question 21
ప్రశ్న 21
If an ASP is designed to oxidize 60 kg/day of ammonia nitrogen completely to nitrate, what additional theoretical oxygen demand follows using 4.56 kg O2 per kg NH3-N oxidized?
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Explanation:
The question supplies a stoichiometric oxygen requirement for nitrification. Multiply 60 kg ammonia nitrogen per day by 4.56 kg oxygen per kg nitrogen. This gives 273.6 kg oxygen per day. The value is additional to the separately calculated carbonaceous oxygen requirement. Actual aerator capacity must also consider transfer conditions, treatment targets and operational safety factors.
వివరణ:
Question 22
ప్రశ్న 22
Aeration equipment must deliver 960 kg O2/day to mixed liquor under field conditions. If its verified field oxygen-transfer performance is 1.6 kg O2 per kWh and it operates 24 hours daily, what average power is required solely for this transfer duty?
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Explanation:
The electrical energy needed per day equals oxygen-transfer demand divided by field transfer performance. Dividing 960 by 1.6 gives 600 kWh/day. The equipment operates continuously over 24 hours. Average power is therefore 600/24 = 25 kW. This calculation does not add standby capacity or other energy demands such as sludge pumps.
వివరణ:
Question 23
ప్రశ్న 23
A conventional ASP needs nominal HRT of 8 hours at fresh flow of 6.0 MLD. If the selected effective liquid depth is 4.0 m and a single rectangular basin is 8.0 m wide, what effective volume and length are required?
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Explanation:
The fresh flow is 6,000 m3/day. Eight hours equals 8/24 = one-third day. Required operating volume is 6,000 times one-third = 2,000 m3. The specified depth and width provide 4.0 times 8.0 = 32 m2 of cross-section. Dividing volume by cross-section gives a basin length of 62.5 m.
వివరణ:
Question 24
ప్రశ్న 24
Two identical rectangular aeration basins share a required operating volume of 3,000 m3 equally. Each has 5.0 m effective depth and a plan length-to-width ratio of 3:1. What length and width are required for each basin?
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Explanation:
Each basin has operating volume 3,000/2 = 1,500 m3. At 5.0 m depth, each plan area is 1,500/5 = 300 m2. Writing length as three times width gives 3B squared = 300. Thus width is 10 m and length is 30 m. These dimensions satisfy the specified plan area, depth and length-to-width ratio.
వివరణ:
Question 25
ప్రశ్న 25
An ASP treats 6.0 MLD at applied BOD5 = 200 mg/L with design MLVSS = 2,500 mg/L. The specified limits are F/M no greater than 0.20 kg BOD5/(kg MLVSS day) and nominal HRT not less than 8 hours. Which minimum volume governs both constraints?
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Explanation:
The incoming BOD5 load is 6.0 times 200 = 1,200 kg/day. The F/M limit requires volume at least 1,200/(0.20 times 2.5) = 2,400 m3. The eight-hour HRT requirement gives volume at least 6,000 times 8/24 = 2,000 m3. Both inequalities are satisfied by choosing the larger requirement, 2,400 m3. Its nominal HRT is 2,400/6,000 times 24 = 9.6 hours.
వివరణ:
Answer Key సమాధానాల పట్టిక
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Question 1 ప్రశ్న 1Answer: C. F/M = daily applied BOD5 mass divided by MLVSS mass maintained in the aeration tank. సమాధానం: C.
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Question 2 ప్రశ్న 2Answer: A. HRT is a liquid-volume-to-flow ratio, whereas SRT relates retained biological solids inventory to the rate at which solids leave the process. సమాధానం: A.
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Question 3 ప్రశ్న 3Answer: D. MLVSS is 2,450 mg/L and provides an organic-solids proxy rather than a direct count of viable microorganisms. సమాధానం: D.
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Question 4 ప్రశ్న 4Answer: B. It becomes one-half of its previous value because the MLVSS inventory doubles. సమాధానం: B.
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Question 5 ప్రశ్న 5Answer: B. Divide total relevant biological-system solids inventory by the daily solids lost in WAS and final effluent, using a consistent solids basis. సమాధానం: B.
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Question 6 ప్రశ్న 6Answer: C. Qr/Q = X/(Xr - X). సమాధానం: C.
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Question 7 ప్రశ్న 7Answer: A. Carbonaceous oxygen demand includes oxidation of organic substrate and biomass effects; oxidizing ammonia nitrogen adds a separate oxygen demand if nitrification is required. సమాధానం: A.
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Question 8 ప్రశ్న 8Answer: D. Oxygen-transfer and mixing capacity, return-sludge handling, secondary-clarifier solids loading and hydraulic stability must still be examined. సమాధానం: D.
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Question 9 ప్రశ్న 9Answer: D. 0.12 kg BOD5/(kg MLVSS day). సమాధానం: D.
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Question 10 ప్రశ్న 10Answer: B. 1,600 m3. సమాధానం: B.
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Question 11 ప్రశ్న 11Answer: A. 0.20 per day. సమాధానం: A.
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Question 12 ప్రశ్న 12Answer: C. 7.2 hours. సమాధానం: C.
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Question 13 ప్రశ్న 13Answer: D. 0.40 kg BOD5/(m3 day). సమాధానం: D.
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Question 14 ప్రశ్న 14Answer: B. Qr = 2.4 MLD and outlet = 8.4 MLD. సమాధానం: B.
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Question 15 ప్రశ్న 15Answer: A. Approximately 2.67 MLD. సమాధానం: A.
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Question 16 ప్రశ్న 16Answer: C. 1,800 mg/L. సమాధానం: C.
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Question 17 ప్రశ్న 17Answer: B. 50 m3/day. సమాధానం: B.
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Question 18 ప్రశ్న 18Answer: D. Approximately 7.14 days. సమాధానం: D.
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Question 19 ప్రశ్న 19Answer: C. 100 m3. సమాధానం: C.
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Question 20 ప్రశ్న 20Answer: A. 645 kg O2/day. సమాధానం: A.
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Question 21 ప్రశ్న 21Answer: B. 273.6 kg O2/day. సమాధానం: B.
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Question 22 ప్రశ్న 22Answer: D. 25 kW. సమాధానం: D.
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Question 23 ప్రశ్న 23Answer: A. 2,000 m3 and 62.5 m. సమాధానం: A.
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Question 24 ప్రశ్న 24Answer: C. 30 m and 10 m. సమాధానం: C.
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Question 25 ప్రశ్న 25Answer: A. 2,400 m3, governed by F/M, with resulting HRT of 9.6 hours. సమాధానం: A.