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Question 1
ప్రశ్న 1
Which feature best distinguishes the conventional activated sludge process (ASP) from a conventional trickling filter?
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Explanation:
Conventional ASP is an aerobic suspended-growth biological treatment process. Its microorganisms are maintained as flocs in the aeration tank. A conventional trickling filter instead supports much of its biological growth on fixed media. Both processes oxidize biodegradable material using microbial activity. Activated sludge normally requires a secondary clarifier and a return-sludge system to retain enough biomass.
వివరణ:
Question 2
ప్రశ్న 2
Which flow arrangement correctly represents a conventional ASP treatment train after preliminary treatment?
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Explanation:
Primary sedimentation commonly reduces settleable solids before conventional ASP aeration. Incoming primary effluent is mixed with return activated sludge at or near the aeration stage. Oxygen transfer and mixing sustain the suspended microbial community. The secondary clarifier separates the resulting biological flocs from treated liquid. Part of the settled sludge is returned and the excess is removed for sludge handling.
వివరణ:
Question 3
ప్రశ్న 3
What happens to biodegradable organic matter during aerobic activated sludge treatment?
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Explanation:
Activated sludge microorganisms use biodegradable compounds as substrates. A portion of this substrate is oxidized to obtain metabolic energy. Another portion is incorporated into new biological cells. The resulting cells contribute to the suspended solids that must later be separated. Consequently, oxygen consumption cannot automatically be equated to the entire incoming BOD mass without considering synthesis and other oxygen demands.
వివరణ:
Question 4
ప్రశ్న 4
In the conventional ASP, what does the term 'mixed liquor' describe?
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Explanation:
Mixed liquor is the suspension found in the activated sludge aeration basin. It contains wastewater constituents together with biological flocs and other suspended solids. Return activated sludge supplies concentrated biomass to this mixture. The mixed liquor proceeds to a secondary clarifier for solids separation. It is therefore distinct from both the clarified final effluent and the concentrated return sludge.
వివరణ:
Question 5
ప్రశ్న 5
Which interpretation of MLSS and MLVSS in an aeration tank is technically sound?
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Explanation:
MLSS denotes the concentration of total mixed-liquor suspended solids. These solids include active microorganisms, dead cells and inert materials. MLVSS is the portion of suspended solids lost upon the prescribed ignition procedure. It is commonly used as a proxy for organic biological solids rather than an exact measurement of active cells. Distinguishing the two avoids overinterpreting an operating biomass measurement.
వివరణ:
Question 6
ప్రశ్న 6
Why is an appropriate aeration and mixing system essential in a conventional aerobic activated sludge tank?
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Explanation:
Aerobic microorganisms consume oxygen while oxidizing biodegradable pollutants. Aerators or diffusers must supply oxygen at an adequate rate. Mixing also keeps the biological solids in suspension and promotes wastewater-biomass contact. Excessive or insufficient aeration can create operational problems. A secondary clarifier is still needed to separate biological solids from the treated liquid.
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Question 7
ప్రశ్న 7
Which description correctly states the function of the secondary clarifier in a conventional ASP?
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Explanation:
The aeration-tank mixed liquor enters the secondary clarifier. Settleable microbial flocs collect in the lower part of the clarifier. The clarified liquid passes to downstream treatment or discharge. Concentrated underflow provides return activated sludge and, in many layouts, waste activated sludge. Poor separation can cause high effluent suspended solids and loss of process biomass.
వివరణ:
Question 8
ప్రశ్న 8
An ASP operator proposes doubling RAS pumping because a higher return rate is assumed always to improve treatment. What is the soundest assessment?
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Explanation:
Return activated sludge helps maintain the desired suspended biomass in the aeration tank. Its required flow depends on sludge concentration, settleability and other plant conditions. Increasing RAS also increases internal liquid circulation and may affect clarifier loading. An excessive or inappropriate return rate cannot guarantee improved effluent quality. The control decision should be based on measured solids and hydraulic conditions.
వివరణ:
Question 9
ప్రశ్న 9
Which statement most accurately distinguishes waste activated sludge (WAS) from return activated sludge (RAS)?
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Explanation:
Both RAS and WAS can be taken from concentrated sludge produced by secondary clarification. RAS is directed back to the aeration system. WAS leaves the biological process for subsequent sludge treatment or handling. Wasting regulates solids inventory and is an important control of sludge age. The two streams are distinguished mainly by their destinations and operating purposes, not by a necessary difference in origin.
వివరణ:
Question 10
ప్రశ్న 10
An ASP develops a rising sludge blanket, very poor settling and high suspended solids in secondary effluent. Which explanation and response are most relevant?
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Explanation:
Bulking or other poor-settling behaviour can prevent adequate separation of mixed-liquor solids. Excess sludge may then escape into the clarified effluent. Settling tests and sludge blanket measurements help characterize the problem. Oxygen supply, influent loading and microbial characteristics should also be examined. The evidence does not justify a single universal adjustment without diagnosing the cause.
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Question 11
ప్రశ్న 11
An aeration tank receives a sustained oxygen supply below the biological requirement. Which consequence is most plausible?
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Explanation:
Aerobic treatment depends on enough dissolved oxygen for the active microbial population. Sustained oxygen limitation can slow oxidation of biodegradable substrates. Nitrifying organisms, when present, may also be inhibited by inadequate oxygen. Poor tank conditions can contribute to changes in floc structure and solids settleability. The plant response must consider both oxygen demand and actual mixing performance.
వివరణ:
Question 12
ప్రశ్న 12
Fresh wastewater flow to an ASP is 4.0 MLD and the return activated sludge ratio Qr/Q is 0.50. Neglecting minor withdrawals, what are the RAS flow and the total liquid flow entering the aeration tank?
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Explanation:
The RAS ratio is defined as recycled sludge flow divided by fresh wastewater flow. Therefore Qr = 0.50 times 4.0 = 2.0 MLD. The aeration tank receives both the fresh and recycled liquid. Their total is 4.0 + 2.0 = 6.0 MLD. The calculation does not imply that 6.0 MLD leaves the entire treatment plant as net final effluent.
వివరణ:
Question 13
ప్రశ్న 13
An aeration tank holds 1500 m3 of well-mixed liquor at MLSS = 3200 mg/L. What is the approximate total mass of suspended solids in the aeration tank?
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Explanation:
A concentration of 3200 mg/L is equivalent to 3.2 kg/m3. The aeration tank liquid volume is 1500 m3. Total suspended-solids inventory equals concentration multiplied by volume. Thus the inventory is 3.2 times 1500 = 4800 kg. This is an MLSS inventory, not necessarily the mass of living microorganisms alone.
వివరణ:
Question 14
ప్రశ్న 14
Mixed liquor contains 3200 mg/L MLSS, and MLVSS is 75% of MLSS by mass. For an aeration volume of 1200 m3, what are MLVSS concentration and MLVSS inventory, respectively?
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Explanation:
The MLVSS fraction is 0.75 of the given MLSS concentration. Thus MLVSS concentration is 0.75 times 3200 = 2400 mg/L. This equals 2.4 kg/m3 in mass-per-volume units. Multiplying 2.4 kg/m3 by 1200 m3 gives an MLVSS inventory of 2880 kg. MLVSS is a useful organic-solids proxy but cannot be identified entirely with viable active biomass.
వివరణ:
Question 15
ప్రశ్న 15
An ASP treatment train receives 3.0 MLD of primary effluent at BOD5 = 200 mg/L and produces clarified effluent at BOD5 = 50 mg/L with equal net influent and effluent water flows. What is the net BOD5 load reduction?
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Explanation:
The incoming BOD5 mass load is 3.0 times 200 = 600 kg/day. The outgoing mass load is 3.0 times 50 = 150 kg/day. Net removal equals incoming load minus outgoing load. Hence the removal is 600 - 150 = 450 kg/day. This process-level mass balance includes the resulting clarified effluent quality and does not identify how much oxygen was supplied.
వివరణ:
Question 16
ప్రశ్న 16
Five MLD of fresh wastewater with negligible suspended solids is mixed instantaneously with 2.5 MLD of RAS at 9000 mg/L suspended solids. Assuming no losses or growth during mixing, what is the resulting mixed-liquor suspended-solids concentration?
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Explanation:
The returned sludge contributes 2.5 times 9000 in concentration-flow units. The fresh flow contributes negligible suspended solids under the stated assumption. Total mixed flow equals 5.0 + 2.5 = 7.5 MLD. The mixed solids concentration is 22500/7.5 = 3000 mg/L. This represents instantaneous mixing rather than a full biological steady-state prediction.
వివరణ:
Question 17
ప్రశ్న 17
For a simplified steady secondary-clarifier solids balance, ignore incoming fresh solids, effluent solids and sludge wasting. If aeration-tank MLSS is 2500 mg/L and RAS concentration is 10000 mg/L at fresh flow 9 MLD, what RAS flow follows?
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Explanation:
The simplified balance is (Q + Qr)X = Qr Xr. Rearrangement gives Qr/Q = X/(Xr - X). The ratio is 2500/(10000 - 2500) = 1/3. With fresh flow Q = 9 MLD, the RAS flow is 3.0 MLD. The result is conditional on the stated neglect of other solids contributions and losses.
వివరణ:
Question 18
ప్రశ్న 18
An aeration tank has an operating volume of 1500 m3 and receives 6000 m3/day of fresh sewage. Using the conventional nominal hydraulic retention-time definition based on fresh flow Q, what is HRT?
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Explanation:
The specified nominal HRT is aeration volume divided by fresh influent flow. Thus the residence-time ratio is 1500/6000 = 0.25 day. Multiplying 0.25 by 24 hours/day gives 6.0 hours. Return sludge may influence actual hydraulic travel and contact patterns through the tank. The question explicitly uses the conventional fresh-flow HRT definition to avoid confusion with total mixed flow.
వివరణ:
Question 19
ప్రశ్న 19
An aeration tank receives 4.5 MLD of primary effluent containing 200 mg/L BOD5. Tank volume is 1500 m3 and MLSS is 3000 mg/L. What is F/M based on fresh applied BOD5 and MLSS inventory?
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Explanation:
The daily fresh BOD5 loading equals 4.5 times 200 = 900 kg/day. MLSS concentration of 3000 mg/L equals 3.0 kg/m3. The tank contains 1500 times 3.0 = 4500 kg MLSS. Dividing 900 by 4500 gives F/M = 0.20 kg BOD5/(kg MLSS day). The answer uses an MLSS basis as explicitly specified, rather than an MLVSS basis.
వివరణ:
Question 20
ప్రశ్న 20
Waste activated sludge is pumped at 80 m3/day with suspended-solids concentration 7500 mg/L. What dry suspended-solids mass is removed daily through this WAS line?
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Explanation:
A sludge concentration of 7500 mg/L is equivalent to 7.5 kg/m3. The wasting flow rate is 80 m3/day. Dry suspended-solids removal is flow multiplied by solids concentration. Hence 80 times 7.5 gives 600 kg/day. The calculated solids mass is not the same as the total wet-sludge mass or volume.
వివరణ:
Question 21
ప్రశ్న 21
For a simplified ASP solids-age estimate, the aeration-tank MLSS inventory is 3600 kg, WAS removes 500 kg solids/day, and final effluent carries 100 kg suspended solids/day. Neglect clarifier solids inventory and other losses. What is the solids retention time?
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Explanation:
Under the stated simplified definition, SRT equals retained solids mass divided by total daily solids leaving the system. The aeration-tank inventory is 3600 kg. Daily solids removal is 500 + 100 = 600 kg/day. Therefore SRT = 3600/600 = 6.0 days. Including a significant clarifier solids inventory would require expanding the numerator of the solids-age calculation.
వివరణ:
Question 22
ప్రశ్న 22
An activated sludge aeration tank receives 6.0 MLD fresh flow and 3.0 MLD RAS. Its secondary clarifier withdraws 3.0 MLD as RAS and 0.10 MLD as WAS. Neglect evaporation and other losses. What are the clarifier inlet flow and net clarified effluent flow?
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Explanation:
Aeration-tank flow entering the secondary clarifier is 6.0 + 3.0 = 9.0 MLD. Clarifier underflow is split into 3.0 MLD RAS and 0.10 MLD WAS. By continuity, clarified effluent is 9.0 - 3.0 - 0.10 = 5.90 MLD. The recycled RAS is an internal circulation rather than an additional net external discharge. The difference between fresh influent and net effluent equals the stated WAS liquid withdrawal under these assumptions.
వివరణ:
Question 23
ప్రశ్న 23
A 30-minute settleability test gives a settled sludge volume of 300 mL per litre of mixed liquor. If MLSS is 3000 mg/L, what is the sludge volume index (SVI)?
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Explanation:
SVI represents the settled volume occupied per gram of mixed-liquor suspended solids after a standard settling period. The MLSS concentration is 3000 mg/L, equivalent to 3.0 g/L. The settled sludge volume is 300 mL/L. Dividing 300 by 3.0 gives SVI = 100 mL/g. This is a settleability index and does not directly measure soluble BOD removal.
వివరణ:
Question 24
ప్రశ్న 24
An ASP shows satisfactory reduction of soluble BOD in the aeration tank, but final effluent TSS rises sharply and the secondary clarifier sludge blanket is increasing. Which operational investigation is most directly justified?
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Explanation:
Good soluble-substrate oxidation does not guarantee successful secondary solids separation. An increasing sludge blanket suggests accumulating solids in the clarifier. Poor settling or inadequate solids withdrawal can release biological flocs with the effluent. Clarifier loading and return and waste sludge operating rates should be examined together. The proper corrective action depends on diagnosis rather than automatically stopping sludge removal.
వివరణ:
Question 25
ప్రశ్న 25
In a steady conventional ASP, influent organic load remains approximately constant but WAS removal is stopped for several days while RAS continues and biomass has positive net growth. What is the most likely initial process effect?
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Explanation:
With positive net biomass production, stopping deliberate sludge wasting reduces an important route for solids loss. Suspended-solids inventory can therefore increase until other losses or process feedbacks change the balance. The average solids retention time also tends to rise initially. Higher solids loading can challenge secondary clarification and effluent solids quality. The magnitude of these changes depends on growth, effluent solids losses, return flow and operational conditions.
వివరణ:
Answer Key సమాధానాల పట్టిక
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Question 1 ప్రశ్న 1Answer: C. ASP maintains microbial flocs suspended in aerated mixed liquor, whereas a trickling filter uses biofilm attached to media. సమాధానం: C.
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Question 2 ప్రశ్న 2Answer: A. Primary sedimentation, aeration with return activated sludge, secondary clarification, clarified effluent outflow, and return or wasting of settled secondary sludge. సమాధానం: A.
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Question 3 ప్రశ్న 3Answer: D. Microorganisms oxidize part of the organics for energy and synthesize another part into new biomass. సమాధానం: D.
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Question 4 ప్రశ్న 4Answer: B. The liquid in the aeration system containing wastewater mixed with suspended microbial flocs, including biomass supplied by returned sludge. సమాధానం: B.
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Question 5 ప్రశ్న 5Answer: C. MLSS includes suspended organic and inorganic solids, whereas MLVSS estimates the volatile fraction and is not an exact count of living biomass. సమాధానం: C.
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Question 6 ప్రశ్న 6Answer: A. It transfers oxygen for biological reactions and keeps microbial flocs in effective contact with wastewater. సమాధానం: A.
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Question 7 ప్రశ్న 7Answer: D. It settles mixed-liquor flocs to produce clarified effluent and concentrated sludge for return or wasting. సమాధానం: D.
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Question 8 ప్రశ్న 8Answer: B. RAS rate must be evaluated with MLSS, sludge settleability and clarifier hydraulic or solids loading; more return is not automatically better. సమాధానం: B.
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Question 9 ప్రశ్న 9Answer: C. RAS is recycled to retain biomass, while WAS removes surplus biomass and helps control solids retention time. సమాధానం: C.
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Question 10 ప్రశ్న 10Answer: A. Sludge bulking or another solids-separation problem should be investigated using settling tests, microscopy, DO and loading information. సమాధానం: A.
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Question 11 ప్రశ్న 11Answer: D. Oxygen-limited treatment may reduce aerobic oxidation and nitrification and can adversely affect sludge characteristics. సమాధానం: D.
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Question 12 ప్రశ్న 12Answer: B. 2.0 MLD and 6.0 MLD, respectively. సమాధానం: B.
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Question 13 ప్రశ్న 13Answer: A. 4800 kg. సమాధానం: A.
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Question 14 ప్రశ్న 14Answer: C. 2400 mg/L and 2880 kg. సమాధానం: C.
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Question 15 ప్రశ్న 15Answer: D. 450 kg/day. సమాధానం: D.
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Question 16 ప్రశ్న 16Answer: B. 3000 mg/L. సమాధానం: B.
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Question 17 ప్రశ్న 17Answer: C. 3.0 MLD. సమాధానం: C.
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Question 18 ప్రశ్న 18Answer: A. 6.0 hours. సమాధానం: A.
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Question 19 ప్రశ్న 19Answer: D. 0.20 kg BOD5/(kg MLSS day). సమాధానం: D.
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Question 20 ప్రశ్న 20Answer: B. 600 kg/day. సమాధానం: B.
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Question 21 ప్రశ్న 21Answer: A. 6.0 days. సమాధానం: A.
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Question 22 ప్రశ్న 22Answer: C. 9.0 MLD and 5.90 MLD, respectively. సమాధానం: C.
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Question 23 ప్రశ్న 23Answer: B. 100 mL/g. సమాధానం: B.
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Question 24 ప్రశ్న 24Answer: D. Check clarifier solids loading, settling characteristics, sludge removal and RAS/WAS management. సమాధానం: D.
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Question 25 ప్రశ్న 25Answer: A. System solids inventory and sludge age tend to rise, potentially increasing secondary-clarifier solids loading if growth continues. సమాధానం: A.