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Question 1
ప్రశ్న 1
What is the primary purpose of anaerobic digestion of municipal sewage sludge?
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Explanation:
Anaerobic digestion stabilizes biodegradable organic matter in sludge without deliberate oxygen supply. Microorganisms convert a portion of this material into methane-rich biogas and carbon dioxide. This lowers the biodegradable volatile-solids content and can reduce odour potential. Digested sludge still contains water, residual solids and nutrients requiring further management. The process does not guarantee complete pathogen destruction or suitability for unrestricted disposal.
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Question 2
ప్రశ్న 2
Which sequence correctly describes the major biological conversions during anaerobic sludge digestion?
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Explanation:
Hydrolysis converts complex particulate organic compounds into smaller soluble molecules. Acidogenesis converts many hydrolysis products into organic acids and other intermediates. Acetogenesis produces substrates such as acetate, hydrogen and carbon dioxide. Methanogenic archaea convert suitable intermediates into methane-containing gas. These are linked biochemical transformations rather than necessarily separate physical tanks.
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Question 3
ప్రశ్న 3
Why is the methanogenic population especially important when judging the stability of an anaerobic sewage-sludge digester?
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Explanation:
Methane-forming microorganisms are generally slower-growing than many acid-forming organisms. Rapid substrate changes can allow volatile acids to accumulate faster than methanogens consume their precursors. A substantial acid accumulation can reduce pH and inhibit methane production. Temperature shifts and toxic compounds can further disrupt the methanogenic community. Stable loading, mixing and environmental control help maintain balanced digestion.
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Question 4
ప్రశ్న 4
Which comparison best distinguishes conventional low-rate from high-rate anaerobic sludge digestion?
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Explanation:
Conventional low-rate digesters may be fed intermittently and develop scum, supernatant and sludge layers. High-rate digestion emphasizes effective mixing and relatively uniform sludge feeding. Heating and feed thickening may also be used to support higher-rate operation. Effective mixing reduces stagnant zones and improves contact between organisms and substrate. Selected detention time and loading must still be justified for the particular process and temperature.
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Question 5
ప్రశ్న 5
Which operating-temperature comparison is appropriate for conventional anaerobic sewage-sludge digesters?
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Explanation:
Mesophilic digesters are often operated near 35°C with an adapted microbial population. Thermophilic digesters commonly operate around 50 to 55°C with a different microbial community. Higher temperature can support faster digestion but may introduce greater heating needs and process sensitivity. Temperature should be maintained consistently rather than changed abruptly. Design detention times and stabilization targets depend on the selected regime and feed characteristics.
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Question 6
ప్రశ్న 6
For a continuously fed, completely mixed digester with no selective solids retention, which statement best describes nominal HRT and SRT?
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Explanation:
Nominal hydraulic retention time is the working liquid volume divided by the incoming liquid volume rate. Solids retention time instead relates retained solids inventory to the rate of solids leaving the system. In a fully mixed system with no selective solids retention, the two can be approximately equal. Settling, separated supernatant, recycle or biomass retention can cause them to differ. The chosen design must provide sufficient retention for slow-growing methanogens.
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Question 7
ప్రశ్న 7
Which formula defines volumetric volatile-solids loading in a sludge digester when based on feed VS and effective operating volume?
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Explanation:
Volatile-solids loading measures the organic solids mass applied to each unit of effective digester volume per day. It is calculated from the VS mass rate in the feed and the active working volume. Volatile-solids loading differs from total-solids loading because fixed solids are excluded. It also differs from hydraulic detention time, which is based on liquid volume flow. Digester design should check organic loading and biological retention together.
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Question 8
ప్రశ్న 8
Which statement about anaerobic-digester supernatant and downstream sludge handling is technically correct?
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Explanation:
Supernatant is the liquid separated from sludge during or after digestion. It may contain appreciable dissolved nitrogen and residual organic pollutants. Returning it to the liquid-treatment train adds a sidestream load that should be included in plant design. Digested sludge generally still needs dewatering, storage or other final handling. Stabilization does not make separated liquors automatically suitable for uncontrolled discharge.
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Question 9
ప్రశ్న 9
Which gas-management arrangement is most appropriate for a municipal anaerobic sludge digester?
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Explanation:
Anaerobic digestion produces combustible biogas containing methane and carbon dioxide. Hydrogen sulfide, water vapour and trace contaminants can create corrosion or equipment problems. Gas collection and use require appropriate pressure control and flame or ignition safeguards. Air ingress can produce a hazardous mixture with methane. Proper gas cleaning and controlled flaring or energy recovery reduce safety and operational risks.
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Question 10
ప్రశ్న 10
A mesophilic digester experiences rising volatile fatty acids, falling alkalinity reserve and declining methane generation after a sudden feed increase. What is the most defensible interpretation?
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Explanation:
Acid-forming organisms can respond rapidly to increased organic substrate. Methanogens may not convert the resulting intermediates at the same rate. Volatile-acid buildup can exhaust buffering capacity and cause a pH decline. Falling methane production is a warning of possible biological imbalance. The appropriate response begins with checking feed loading and process conditions rather than assuming that greater acid production is beneficial.
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Question 11
ప్రశ్న 11
A continuously fed sludge digester receives 45 m3/day of thickened sludge. If a project-specified nominal hydraulic detention time is 18 days and no separate storage allowance is included, what effective liquid volume is required?
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Explanation:
Nominal hydraulic detention time is working liquid volume divided by daily feed flow. Rearranging gives required volume equal to feed flow multiplied by detention time. The specified values are 45 m3/day and 18 days. Multiplying them gives 810 m3 of effective working volume. Freeboard, inactive deposits and any separate storage requirement would be considered in addition to this hydraulic calculation.
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Question 12
ప్రశ్న 12
A completely mixed digester has an effective working volume of 1,260 m3 and receives 70 m3/day of sludge. What is its nominal hydraulic detention time?
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Explanation:
The nominal hydraulic detention time is t = V/Q. The stated effective working volume is 1,260 m3. The sludge feed rate is 70 m3/day. Dividing 1,260 by 70 gives 18 days. This calculation assumes steady flow and does not establish actual biological SRT if solids are selectively retained or lost.
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Question 13
ప్రశ్న 13
A digester receives 1,260 kg/day of volatile solids. If the project-specified maximum volumetric VS loading is 1.40 kg VS/(m3 day), what minimum effective digester volume follows from the loading criterion alone?
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Explanation:
Volumetric volatile-solids loading equals the daily VS mass feed divided by the digester working volume. Rearranging gives required volume equal to the feed mass rate divided by the permissible loading. Substituting 1,260 kg/day and 1.40 kg/(m3 day) gives 900 m3. This volume meets only the specified organic-loading limit. An independent retention-time and operational-capacity check is still necessary.
వివరణ:
Question 14
ప్రశ్న 14
A sludge feed is 30 m3/day at 4.0% total solids by mass, with volatile solids equal to 75% of dry total solids. Assuming wet-sludge density of 1,000 kg/m3, what is the daily VS feed mass?
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Explanation:
The daily wet-sludge mass is 30 times 1,000 = 30,000 kg/day. Total solids constitute 4.0% of this mass. Thus dry total solids are 1,200 kg/day. The volatile fraction is 75% of dry solids, giving 1,200 times 0.75 = 900 kg VS/day. The two percentages have different denominators and must be applied in sequence.
వివరణ:
Question 15
ప్రశ్న 15
A digester receives 60 m3/day of sludge carrying 1,800 kg VS/day. Design requires HRT at least 20 days and VS loading no more than 1.20 kg VS/(m3 day). What minimum effective volume satisfies both constraints?
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Explanation:
The detention-time requirement gives V at least 60 times 20 = 1,200 m3. The volatile-solids loading requirement gives V at least 1,800/1.20 = 1,500 m3. Both constraints must be met simultaneously by the selected working volume. The larger required volume is 1,500 m3 and therefore governs. At this volume, nominal HRT is 25 days and VS loading equals the specified 1.20 kg/(m3 day) cap.
వివరణ:
Question 16
ప్రశ్న 16
The steady influent to a digester contains 1,400 kg VS/day and the digested sludge outlets together contain 840 kg VS/day. With these measured VS mass flows and no significant VS accumulation, what is the VS destruction percentage?
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Explanation:
The volatile-solids mass destroyed equals VS entering minus VS leaving. The stated input is 1,400 kg/day and the output is 840 kg/day. Their difference is 560 kg/day. Percentage destruction is 560/1,400 times 100 = 40%. The calculation relies on a mass balance rather than merely subtracting the feed and digested-sludge VS percentages.
వివరణ:
Question 17
ప్రశ్న 17
A digester receives 2,400 kg/day of dry total solids, 70% of which are volatile. Exactly 50% of the incoming volatile-solids mass is destroyed, fixed solids remain unchanged and no other dry solids are added or lost. What dry-solids mass leaves the digester?
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Explanation:
The incoming volatile-solids mass is 0.70 times 2,400 = 1,680 kg/day. Incoming fixed solids are the other 30%, or 720 kg/day. Half the volatile solids are destroyed, leaving 840 kg/day of volatile solids. Adding unchanged fixed solids gives 840 plus 720 = 1,560 kg/day of outgoing dry solids. The calculation explicitly excludes other solids losses or additions.
వివరణ:
Question 18
ప్రశ్న 18
Influent sludge contains 1,000 kg VS/day and 400 kg fixed solids/day. Digestion destroys 450 kg VS/day, fixed solids are conserved and no other dry-solids streams exist. What percentage of the digested dry solids is volatile?
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Explanation:
The remaining volatile solids are 1,000 minus 450 = 550 kg/day. Fixed solids remain 400 kg/day under the stated assumption. Total digested dry solids are therefore 950 kg/day. The volatile fraction is 550/950 times 100, approximately 57.9%. This percentage differs from the 45% reduction of influent volatile-solids mass.
వివరణ:
Question 19
ప్రశ్న 19
A digester destroys 700 kg VS/day and has an explicitly assumed biogas yield of 0.80 m3 per kg VS destroyed, with both quantities on the same stated gas-volume basis. What daily biogas volume is predicted?
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Explanation:
Biogas yield in this question is specified per kilogram of volatile solids destroyed. The daily destroyed mass is 700 kg/day. Multiplying by the given yield of 0.80 m3/kg gives 560 m3/day. The result refers to total biogas at the stated volume basis, not methane alone. Real gas production depends on sludge composition, stabilization and measurement conditions.
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Question 20
ప్రశ్న 20
A gas holder receives 900 m3/day of digester biogas containing 60% methane by volume. Assuming both gas components are measured at the same conditions, what methane volume is present per day?
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Explanation:
The methane volume fraction is given as 0.60. Total biogas production is 900 m3/day. Multiplying the volume by the methane fraction gives 900 times 0.60 = 540 m3/day. The balance of the gas includes carbon dioxide and possible trace gases. The calculation is a compositional split and does not assume complete conversion of all feed solids to methane.
వివరణ:
Question 21
ప్రశ్న 21
A plant produces 600 m3/day of biogas with a stipulated heating value of 20 MJ/m3. If 30% of its chemical energy is converted into electricity and generation is continuous, what is the approximate average electrical power?
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Explanation:
The daily chemical energy in biogas is 600 times 20 = 12,000 MJ/day. At 30% electrical conversion, useful electrical energy is 3,600 MJ/day. One kilowatt corresponds to one kilojoule per second. Therefore average electrical power is 3,600,000 kJ/day divided by 86,400 s/day. The result is approximately 41.7 kW, neglecting any additional plant parasitic load.
వివరణ:
Question 22
ప్రశ్న 22
A sludge stream contains 1,500 kg/day of dry solids and is thickened from 3% to 6% total solids by mass without losing solids. Assume density 1,000 kg/m3 in both cases and specify a digester HRT of 20 days. What effective working-volume requirements correspond to before and after thickening?
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Explanation:
At 3% solids and the assumed density, sludge volume is 1,500/(0.03 times 1,000) = 50 m3/day. At 6% solids it becomes 1,500/(0.06 times 1,000) = 25 m3/day. Using a 20-day nominal HRT gives 50 times 20 = 1,000 m3 before thickening. After thickening, the corresponding required working volume is 25 times 20 = 500 m3. The calculation isolates the hydraulic-volume effect and does not eliminate independent VS-loading and mixing checks.
వివరణ:
Question 23
ప్రశ్న 23
A vertical cylindrical digester has internal diameter 12 m and effective liquid depth 6 m. With sludge feed at 36 m3/day, what are its effective working volume and nominal HRT? Use pi = 3.1416.
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Explanation:
Effective cylindrical volume is pi D squared h divided by four. Substituting D = 12 m and h = 6 m gives approximately 678.6 m3. Nominal hydraulic retention time is volume divided by the daily feed rate. Dividing 678.6 by 36 gives approximately 18.85 days. Freeboard and any inactive storage are not included in the stated effective depth.
వివరణ:
Question 24
ప్రశ్న 24
Two identical digesters are operated in parallel and receive a total sludge feed of 80 m3/day equally. If each must provide 15 days nominal HRT, what effective working volume is required in each digester?
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Explanation:
Each of the two parallel units receives half the total sludge flow. The sludge flow to one unit is therefore 80/2 = 40 m3/day. Working volume for a 15-day HRT equals flow multiplied by the retention time. Thus each digester requires 40 times 15 = 600 m3. The combined effective working volume is 1,200 m3, before separate operational allowances.
వివరణ:
Question 25
ప్రశ్న 25
A digester receives 20 m3/day of primary sludge at 5% total solids with VS/TS = 70%, and 15 m3/day of WAS at 4% total solids with VS/TS = 80%. Both have density 1,000 kg/m3. If effective digester volume is 1,000 m3, what are the total VS feed and volumetric VS loading?
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Explanation:
Primary sludge adds 20 times 1,000 times 0.05 times 0.70 = 700 kg VS/day. WAS adds 15 times 1,000 times 0.04 times 0.80 = 480 kg VS/day. Total volatile-solids feed is therefore 700 + 480 = 1,180 kg/day. Dividing by 1,000 m3 working volume gives 1.18 kg VS/(m3 day). Both the solids fraction and volatile fraction are applied separately for each sludge stream.
వివరణ:
Answer Key సమాధానాల పట్టిక
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Question 1 ప్రశ్న 1Answer: B. To biologically stabilize putrescible sludge, reduce part of its volatile solids and recover methane-rich gas. సమాధానం: B.
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Question 2 ప్రశ్న 2Answer: D. Hydrolysis of complex organics, acidogenesis, acetogenesis and methanogenesis. సమాధానం: D.
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Question 3 ప్రశ్న 3Answer: A. Methanogens generally grow relatively slowly and are sensitive to rapid changes in pH, temperature and inhibitory substances. సమాధానం: A.
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Question 4 ప్రశ్న 4Answer: C. High-rate digestion uses effective mixing and relatively uniform feeding to improve active-volume use, while low-rate digesters may stratify and be fed intermittently. సమాధానం: C.
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Question 5 ప్రశ్న 5Answer: B. Mesophilic operation is commonly near 35°C, while thermophilic operation is commonly near 50–55°C. సమాధానం: B.
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Question 6 ప్రశ్న 6Answer: A. HRT equals working liquid volume divided by daily feed flow, and SRT may be close to HRT if solids leave with the mixed discharge without selective retention. సమాధానం: A.
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Question 7 ప్రశ్న 7Answer: D. Daily volatile-solids mass fed divided by effective digester volume, usually in kg VS/(m3 day). సమాధానం: D.
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Question 8 ప్రశ్న 8Answer: C. Separated supernatant can contain ammonia and soluble organic loads and requires appropriate management when returned to the wastewater plant. సమాధానం: C.
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Question 9 ప్రశ్న 9Answer: B. Collect methane-rich biogas using gas-tight equipment with suitable safety controls, and address hydrogen sulfide and other contaminants before sensitive energy uses. సమాధానం: B.
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Question 10 ప్రశ్న 10Answer: D. Acid production may be outpacing methanogenesis, so loading, pH, alkalinity and inhibitory conditions should be investigated. సమాధానం: D.
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Question 11 ప్రశ్న 11Answer: A. 810 m3. సమాధానం: A.
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Question 12 ప్రశ్న 12Answer: C. 18 days. సమాధానం: C.
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Question 13 ప్రశ్న 13Answer: B. 900 m3. సమాధానం: B.
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Question 14 ప్రశ్న 14Answer: D. 900 kg VS/day. సమాధానం: D.
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Question 15 ప్రశ్న 15Answer: A. 1,500 m3. సమాధానం: A.
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Question 16 ప్రశ్న 16Answer: C. 40%. సమాధానం: C.
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Question 17 ప్రశ్న 17Answer: B. 1,560 kg/day. సమాధానం: B.
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Question 18 ప్రశ్న 18Answer: D. Approximately 57.9%. సమాధానం: D.
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Question 19 ప్రశ్న 19Answer: A. 560 m3/day. సమాధానం: A.
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Question 20 ప్రశ్న 20Answer: C. 540 m3/day. సమాధానం: C.
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Question 21 ప్రశ్న 21Answer: B. 41.7 kW. సమాధానం: B.
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Question 22 ప్రశ్న 22Answer: D. 1,000 m3 before and 500 m3 after. సమాధానం: D.
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Question 23 ప్రశ్న 23Answer: A. Approximately 678.6 m3 and 18.85 days. సమాధానం: A.
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Question 24 ప్రశ్న 24Answer: C. 600 m3. సమాధానం: C.
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Question 25 ప్రశ్న 25Answer: D. 1,180 kg/day and 1.18 kg/(m3 day). సమాధానం: D.