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Question 1
ప్రశ్న 1
In a conventional sewage treatment plant, which comparison correctly identifies primary sludge and waste activated sludge (WAS)?
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Explanation:
Primary sludge originates from solids settling during primary clarification. Waste activated sludge is removed to control the biomass inventory of the suspended-growth treatment process. Primary sludge often contains readily putrescible raw organics and may thicken more readily by gravity. WAS commonly contains hydrated biological flocs and may be harder to gravity-thicken. Their origins and properties affect the selection of sludge-handling equipment.
వివరణ:
Question 2
ప్రశ్న 2
Which statement most accurately distinguishes sludge thickening from sludge stabilization before final disposal?
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Explanation:
Thickening concentrates sludge solids by separating a portion of its liquid. The resulting smaller sludge volume can reduce downstream pumping and digester volume requirements. Stabilization instead alters degradable organic material to reduce odour and instability. Anaerobic and aerobic digestion are examples of biological stabilization. A higher total-solids percentage alone does not prove that sludge is stabilized.
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Question 3
ప్రశ్న 3
A sludge analysis reports total solids (TS) and volatile solids (VS) on a dry-solids basis. Which interpretation is correct?
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Explanation:
Total solids represent the residue remaining after the prescribed drying procedure. Volatile solids are estimated from the loss in mass during a subsequent ignition procedure. The volatile fraction commonly indicates organic-rich matter but is not an exact assay of living cells. The fixed-solids fraction is the residue that remains after ignition. Both quantities can inform sludge digestion and handling assessments.
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Question 4
ప్రశ్న 4
A plant has readily settling primary sludge and dilute waste activated sludge with poor gravity thickening. Which preliminary thickening strategy is most technically defensible?
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Explanation:
Primary sludge commonly has better gravity-thickening characteristics than waste activated sludge. Waste activated sludge is often light and highly hydrated. Flotation and mechanical thickening can be more appropriate for some WAS streams. Actual performance depends on sludge characteristics and operational conditions. Selection should consider solids capture, supernatant quality, costs and the downstream treatment train.
వివరణ:
Question 5
ప్రశ్న 5
Before feeding a digester with intermittently produced primary sludge and WAS, what is the main benefit of a properly operated sludge blending or equalization tank?
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Explanation:
Primary and secondary sludge may arrive at different times and with different solids concentrations. Controlled blending reduces rapid variations in the digester feed. Equalization can also support steadier pumping and thickener operation. Adequate mixing should prevent uncontrolled solids deposition in the tank. Holding times and odour controls must be managed so that raw sludge does not become excessively septic.
వివరణ:
Question 6
ప్రశ్న 6
Why may polymer or coagulant conditioning be used when preparing sludge for thickening or dewatering?
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Explanation:
Chemical conditioning alters the interaction of sludge particles and water. Suitable polymers can promote floc formation and improve solids capture during separation. The optimum chemical type and dosage depend on sludge characteristics and the chosen equipment. Some additives increase the total dry-solids mass or affect downstream processes. Conditioning is therefore selected when needed and is not inherently a universal pretreatment step before digestion.
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Question 7
ప్రశ్న 7
A sludge pump delivers 45 m3/day of primary sludge at a bulk density of 1,020 kg/m3 and a total-solids content of 3.0% by mass. What dry-solids mass is conveyed per day?
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Explanation:
The daily wet sludge mass equals volume multiplied by bulk density. This is 45 times 1,020 = 45,900 kg/day. Total solids constitute 3.0% of that wet mass. Multiplying 45,900 by 0.03 gives 1,377 kg/day of dry solids. Using a density of 1,000 kg/m3 instead would give a different, unrequested approximation.
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Question 8
ప్రశ్న 8
A batch of wet sewage sludge has a mass of 18 tonnes and contains 96% moisture by mass. What is the mass of dry solids in the batch?
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Explanation:
Moisture and dry solids together account for the entire wet sludge mass. A moisture content of 96% means that dry solids are 4% by mass. The wet mass of 18 tonnes is 18,000 kg. Dry solids therefore equal 18,000 times 0.04 = 720 kg. The remaining 17,280 kg is water under the specified two-component model.
వివరణ:
Question 9
ప్రశ్న 9
Sludge has 5.0% total solids by mass and a measured bulk density of 1,030 kg/m3. What is its total-solids concentration expressed as kg/m3 of wet sludge?
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Explanation:
The mass fraction of solids is 0.05. Each cubic metre of wet sludge has a mass of 1,030 kg at the stated density. Multiplying 1,030 by 0.05 gives 51.5 kg of dry solids. The concentration is therefore 51.5 kg/m3 of wet sludge. This corresponds to 51,500 mg/L and shows why a mass percentage needs density for conversion to a volumetric concentration.
వివరణ:
Question 10
ప్రశ్న 10
A process must handle 1,200 kg/day of dry sludge solids as a slurry containing 4.0% solids by mass. Assume the slurry density is 1,000 kg/m3. What slurry volume is produced each day?
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Explanation:
The dry-solids mass fraction is 0.04. Required wet sludge mass is 1,200 divided by 0.04 = 30,000 kg/day. The assumed density is 1,000 kg/m3. Dividing wet mass by density gives 30 m3/day. The volume would differ if the density or total-solids fraction were different.
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Question 11
ప్రశ్న 11
A thickener receives 120 m3/day of sludge at 1.0% solids by mass and produces 4.0%-solids underflow. Assume a density of 1,000 kg/m3 for both streams and 100% recovery of dry solids in the underflow. What are underflow volume and water-rich overflow volume?
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Explanation:
The incoming dry-solids mass is 120 times 1,000 times 0.01 = 1,200 kg/day. With full solids recovery, the underflow also contains 1,200 kg/day. At 4.0% solids and the stated density, underflow volume is 1,200/(0.04 times 1,000) = 30 m3/day. The overflow is 120 - 30 = 90 m3/day by the assumed volumetric balance. The calculation demonstrates water removal rather than destruction of solids.
వివరణ:
Question 12
ప్రశ్న 12
A thickener receives 100 m3/day at 1.2% solids by mass, captures 92% of influent dry solids in an underflow at 5.0% solids by mass, and returns the rest with the overflow. Both streams have a density of 1,000 kg/m3. What are the underflow volume and overflow dry-solids loss?
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Explanation:
Incoming solids mass equals 100 times 1,000 times 0.012 = 1,200 kg/day. Underflow solids mass is 92% of 1,200, giving 1,104 kg/day. At 5.0% solids by mass, its wet volume is 1,104/(0.05 times 1,000) = 22.08 m3/day. Solids not captured are 8% of 1,200 = 96 kg/day. This illustrates why volume reduction and solids-capture efficiency must be distinguished.
వివరణ:
Question 13
ప్రశ్న 13
Primary sludge at 20 m3/day and 4.0% solids by mass is blended with WAS at 40 m3/day and 0.5% solids by mass. Assuming both densities are 1,000 kg/m3 and volumes are additive, what is the combined solids percentage by mass?
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Explanation:
The primary-sludge solids mass is 20 times 1,000 times 0.04 = 800 kg/day. The WAS stream contains 40 times 1,000 times 0.005 = 200 kg/day. Total dry solids are 1,000 kg/day and total wet mass is 60,000 kg/day. Dividing gives a solids mass fraction of 1,000/60,000 = 0.016667. Hence the combined sludge contains approximately 1.67% solids by mass.
వివరణ:
Question 14
ప్రశ్న 14
A sludge batch contains 500 kg of dry mineral solids with specific gravity 1.50 and 9,500 kg of water. Assume water density 1,000 kg/m3 and additive solid and water volumes. What is the approximate bulk density of the sludge?
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Explanation:
The mineral-solids density is 1.50 times 1,000 = 1,500 kg/m3. Their volume equals 500/1,500 = 0.3333 m3. Water volume equals 9,500/1,000 = 9.5 m3. Total wet mass is 10,000 kg and total volume is approximately 9.8333 m3. The bulk density is 10,000/9.8333, approximately 1,017 kg/m3.
వివరణ:
Question 15
ప్రశ్న 15
A thickener is assigned a project-specific limiting solids loading of 60 kg dry solids/(m2 day). For a sludge dry-solids feed of 1,800 kg/day, what minimum effective thickener surface area is required?
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Explanation:
Limit the solids mass applied to each square metre of thickener area. The specified feed is 1,800 kg dry solids/day. The allowable loading is 60 kg/(m2 day). Required area equals 1,800/60 = 30 m2. The selection must also satisfy hydraulic loading, sludge-thickening performance and other project constraints.
వివరణ:
Question 16
ప్రశ్న 16
A thickener receives 60 m3/day of WAS at 0.8% total solids by mass. It captures 85% of the solids in an underflow containing 4.0% solids by mass. Assume 1,000 kg/m3 density for both streams. What is the underflow volume?
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Explanation:
Incoming solids equal 60 times 1,000 times 0.008 = 480 kg/day. The captured solids are 0.85 times 480 = 408 kg/day. The thickened underflow contains 4% dry solids by mass. At 1,000 kg/m3, its solids concentration is 40 kg/m3. Underflow volume is therefore 408/40 = 10.2 m3/day.
వివరణ:
Question 17
ప్రశ్న 17
A sludge storage tank must be emptied at a total withdrawal rate of 48 m3/day, but the duty pump operates only 4 hours per day. What average pumping capacity is needed during the operating hours?
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Explanation:
Average operating pump capacity equals required daily volume divided by actual pumping hours. Dividing 48 m3/day by 4 h/day gives 12 m3/h. Converting cubic metres per hour to litres per second requires multiplying by 1,000 and dividing by 3,600. This gives approximately 3.33 L/s. Practical pump selection must additionally consider sludge rheology, head losses, reliability and standby needs.
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Question 18
ప్రశ్న 18
A thickening unit treats 80 m3/day of sludge containing 1.5% solids by mass. Its overall dry-solids capture is 90%, and the overflow is returned to the wastewater treatment plant. Assuming a feed density of 1,000 kg/m3, how many kilograms of solids per day return in the overflow?
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Explanation:
Incoming dry solids are 80 times 1,000 times 0.015 = 1,200 kg/day. A 90% capture rate means that 10% of these solids do not enter the thickened underflow. The uncaptured solids flow is 0.10 times 1,200. Therefore 120 kg/day of dry solids return with the overflow. This return load should be recognized in the overall plant solids balance.
వివరణ:
Question 19
ప్రశ్న 19
Primary sludge and WAS arrive at steady rates of 16 and 32 m3/day, respectively. What working volume is needed to provide a nominal 6-hour combined-sludge equalization period, excluding freeboard and standby storage?
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Explanation:
Combined sludge flow is 16 + 32 = 48 m3/day. A six-hour period is one-quarter of a day. The nominal equalization volume is the daily flow multiplied by 6/24. This gives 48 times 0.25 = 12 m3. Actual tank sizing should also consider mixing, peak sludge flows, operating cycles and avoidance of septic storage.
వివరణ:
Question 20
ప్రశ్న 20
A conditioning trial selects an active-polymer dose of 5 kg per tonne of incoming dry sludge solids. If the daily dry-solids feed is 800 kg and chemical solution strength is not relevant to the question, what active-polymer mass is needed per day?
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Explanation:
The incoming dry solids are 800 kg/day or 0.8 tonne/day. The trial dosage is 5 kg active polymer per tonne of dry solids. Multiplying 0.8 by 5 gives 4.0 kg/day of active polymer. This is not automatically the required mass of a diluted or commercial polymer solution. Solution concentration would be needed to convert active-polymer mass to solution consumption.
వివరణ:
Question 21
ప్రశ్న 21
A sludge line handles 1,800 kg/day of total dry solids, of which 65% is volatile solids on the dry-mass basis. What volatile-solids and fixed-solids loads does it carry?
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Explanation:
Volatile solids account for 65% of the total dry solids. Their mass flow is 0.65 times 1,800 = 1,170 kg/day. The fixed-solids fraction is 35% of total dry solids. Its mass flow is 0.35 times 1,800 = 630 kg/day. The calculation treats VS and fixed solids as complementary fractions under the specified analytical procedure.
వివరణ:
Question 22
ప్రశ్న 22
A sludge stream initially occupies 100 m3 at 95% moisture by mass. It is thickened to 90% moisture by mass with no loss of dry solids and with wet-sludge density of 1,000 kg/m3 before and after thickening. What is the final sludge volume?
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Explanation:
Initial dry-solids percentage is 100 - 95 = 5% by mass. The initial wet sludge mass is 100 times 1,000 = 100,000 kg. Therefore dry solids equal 5,000 kg. At final moisture of 90%, solids represent 10% of wet mass, which requires 50,000 kg of thickened sludge. Dividing by 1,000 kg/m3 gives a final volume of 50 m3.
వివరణ:
Question 23
ప్రశ్న 23
A primary-sludge holding tank begins producing pronounced odour after long periods of poorly mixed storage. Which operating response is most appropriate before sludge digestion?
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Explanation:
Primary sewage sludge can contain readily degradable organic material. Prolonged storage in poorly controlled conditions can lead to septic decomposition and nuisance odours. Good sludge management reduces unnecessary retention and prevents deposition in holding tanks. Mixing and appropriate covered ventilation or odour-control arrangements may be needed. These measures support safe handling but are not a substitute for the designed stabilization process.
వివరణ:
Question 24
ప్రశ్న 24
Which statement is the most appropriate design judgement when selecting a thickening or conditioning step before sludge digestion?
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Explanation:
Sludge treatment units should be designed on a plant-wide solids balance. Thickening reduces liquid volume and can improve the use of digester capacity. Conditioning may improve separation but is not inherently mandatory for every feed stream. Separated liquor can contain solids and soluble loads that affect wastewater treatment when recycled. A suitable scheme considers capture, stability, pumping, handling, safety and treatment performance together.
వివరణ:
Question 25
ప్రశ్న 25
A blended sludge feed contains 600 kg/day of primary dry solids and 300 kg/day of WAS dry solids. A thickener captures 90% of the combined dry-solids feed and discharges underflow at 3.0% solids by mass, with underflow density 1,000 kg/m3. What is the underflow volume?
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Explanation:
The total incoming dry-solids load is 600 + 300 = 900 kg/day. Underflow recovery of 90% gives 810 kg/day of captured dry solids. At 3.0% solids by mass, wet underflow mass is 810/0.03 = 27,000 kg/day. The assumed underflow density is 1,000 kg/m3. Dividing wet mass by density gives 27 m3/day.
వివరణ:
Answer Key సమాధానాల పట్టిక
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Question 1 ప్రశ్న 1Answer: B. Primary sludge is withdrawn from primary sedimentation and contains settleable raw-sewage solids, while WAS is excess biological solids removed from the activated-sludge system. సమాధానం: B.
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Question 2 ప్రశ్న 2Answer: D. Thickening increases the solids concentration mainly by removing water, whereas stabilization reduces the sludge’s tendency to putrefy by biological or chemical transformation. సమాధానం: D.
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Question 3 ప్రశ్న 3Answer: D. VS represents the fraction of dry solids lost during a prescribed ignition test and is often used as an approximate indicator of organic matter, not an exact viable-biomass measurement. సమాధానం: D.
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Question 4 ప్రశ్న 4Answer: B. Evaluate gravity thickening for primary sludge and a suitable flotation or mechanical method for WAS, based on tests and plant constraints. సమాధానం: B.
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Question 5 ప్రశ్న 5Answer: A. It helps provide a more uniform combined feed in concentration and flow while allowing controlled transfer to downstream units. సమాధానం: A.
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Question 6 ప్రశ్న 6Answer: C. It can promote particle aggregation and improve liquid–solids separation, but suitability and dosage require sludge-specific evaluation. సమాధానం: C.
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Question 7 ప్రశ్న 7Answer: A. 1,377 kg/day సమాధానం: A.
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Question 8 ప్రశ్న 8Answer: B. 720 kg సమాధానం: B.
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Question 9 ప్రశ్న 9Answer: C. 51.5 kg/m3 సమాధానం: C.
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Question 10 ప్రశ్న 10Answer: B. 30 m3/day సమాధానం: B.
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Question 11 ప్రశ్న 11Answer: B. 30 m3/day and 90 m3/day సమాధానం: B.
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Question 12 ప్రశ్న 12Answer: A. 22.08 m3/day and 96 kg/day సమాధానం: A.
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Question 13 ప్రశ్న 13Answer: A. Approximately 1.67% సమాధానం: A.
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Question 14 ప్రశ్న 14Answer: C. 1,017 kg/m3 సమాధానం: C.
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Question 15 ప్రశ్న 15Answer: D. 30 m2 సమాధానం: D.
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Question 16 ప్రశ్న 16Answer: C. 10.2 m3/day సమాధానం: C.
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Question 17 ప్రశ్న 17Answer: C. 12 m3/h, approximately 3.33 L/s సమాధానం: C.
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Question 18 ప్రశ్న 18Answer: D. 120 kg/day సమాధానం: D.
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Question 19 ప్రశ్న 19Answer: B. 12 m3 సమాధానం: B.
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Question 20 ప్రశ్న 20Answer: A. 4.0 kg/day సమాధానం: A.
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Question 21 ప్రశ్న 21Answer: D. 1,170 kg VS/day and 630 kg fixed solids/day సమాధానం: D.
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Question 22 ప్రశ్న 22Answer: C. 50 m3 సమాధానం: C.
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Question 23 ప్రశ్న 23Answer: A. Limit avoidable holding time, provide appropriate mixing and ventilation or odour controls, and maintain planned sludge transfer to treatment. సమాధానం: A.
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Question 24 ప్రశ్న 24Answer: D. Select thickening based on feed-solids concentration, solids capture, return-liquid quality and downstream digester needs; add conditioning only when testing and process requirements justify it. సమాధానం: D.
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Question 25 ప్రశ్న 25Answer: A. 27 m3/day సమాధానం: A.