Please keep at least one language enabled. || కనీసం ఒక భాషను ఎంచుకోండి.
Question 1
ప్రశ్న 1
What is the principal function of a conventional primary sedimentation tank in a municipal sewage treatment plant?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
A primary sedimentation tank separates a substantial portion of settleable solids from sewage using gravity. It also permits floating substances to accumulate where they can be skimmed. Removing settleable organic solids reduces the load on downstream biological treatment. It does not generally remove most dissolved organic pollutants. Sludge withdrawal and scum removal are integral to proper clarifier operation.
వివరణ:
Question 2
ప్రశ్న 2
Which pairing correctly identifies the settling processes most relevant to grit removal and to dilute organic suspended solids in a primary sewage clarifier?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
Discrete particles settle without appreciable changes in size or shape during their descent. Mineral grit is often approximated using discrete-particle settling concepts. Many raw sewage solids aggregate into flocs as they settle. This makes flocculent settling particularly relevant to primary clarification. Zone and compression settling may occur at higher solids concentrations and in concentrated sludge layers.
వివరణ:
Question 3
ప్రశ్న 3
Which definition correctly states the surface overflow rate (SOR) of a sewage sedimentation tank?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
Surface overflow rate is Q/A, where Q is hydraulic inflow and A is the effective plan area. It is commonly expressed in cubic metres per square metre per day. Outlet weir loading instead divides discharge by effective weir length. Nominal detention time is the operating water volume divided by flow rate. Surface overflow rate is a principal hydraulic sizing parameter for primary sedimentation.
వివరణ:
Question 4
ప్రశ్న 4
According to the CPHEEO Manual on Sewerage and Sewage Treatment Systems, which average-flow and peak-flow surface-overflow-rate ranges apply to a primary clarifier followed by secondary treatment?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
CPHEEO provides distinct clarifier design ranges for different treatment arrangements. For primary settling followed by secondary treatment, the indicated average overflow rate is 35 to 50 m3/m2/day. The corresponding peak-flow range is 80 to 120 m3/m2/day. Primary settling used alone has a different, lower design range. The appropriate row of the guidance must therefore be selected before applying numerical criteria.
వివరణ:
Question 5
ప్రశ్న 5
In an ideal discrete-particle sedimentation model, tank flow and plan area remain fixed while water depth is increased. Which consequence follows?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
Surface overflow rate depends on flow divided by plan area and therefore does not change in this example. Increasing water depth increases operating tank volume. At fixed discharge, the nominal hydraulic detention time accordingly increases. In the ideal discrete-particle model, critical capture is governed by overflow rate rather than tank depth alone. Real flocculent settling, turbulence and sludge handling can respond differently to changes in depth.
వివరణ:
Question 6
ప్రశ్న 6
Which inlet and outlet design provision most directly reduces short-circuiting and uneven hydraulic loading in a primary sedimentation tank?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
A strong inlet jet can transport solids rapidly toward the outlet and shorten effective residence time. Suitable inlet baffling and energy dissipation reduce undesirable turbulence. Good flow distribution allows more of the available settling area to function effectively. Level outlet weirs help prevent localized excessive overflow. These provisions complement, but do not replace, the required area and detention checks.
వివరణ:
Question 7
ప్రశ్న 7
What is the principal reason for providing mechanical sludge collection and regular sludge withdrawal in a primary clarifier?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
Settled sewage solids accumulate on the floor of a primary sedimentation tank. Scrapers or other collection systems move the solids toward a sludge hopper or withdrawal point. Regular withdrawal preserves effective tank volume and reduces prolonged sludge storage. Excessive retention of organic sludge can promote septic conditions and odour. The withdrawn primary sludge is directed to an appropriate sludge-treatment or handling system.
వివరణ:
Question 8
ప్రశ్న 8
Which arrangement is most appropriate for removing floating oil, grease and scum from a primary clarifier?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
Floating grease and other light material can accumulate at the surface of a primary clarifier. Scum baffles help prevent this material from passing directly over the outlet. Surface skimmers direct accumulated matter to a suitable collection system. Bottom sludge collectors handle a different fraction of settled material. Effective scum removal protects downstream equipment and improves effluent appearance.
వివరణ:
Question 9
ప్రశ్న 9
Which comparison between primary and secondary clarifiers in a conventional activated-sludge treatment plant is correct?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
Primary clarification normally follows preliminary sewage treatment and precedes a biological reactor. It removes settleable and floatable material from raw sewage. A secondary clarifier is normally placed after the biological treatment stage. It separates biological flocs from treated mixed liquor or other biologically treated flow. Activated-sludge secondary clarifiers also interact with return and waste sludge management.
వివరణ:
Question 10
ప్రశ్న 10
Which statement best describes primary clarification performance and the effect of excessively long detention?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
Primary sedimentation removes settleable particulate matter more readily than dissolved pollutants. A representative well-operated primary clarifier may remove roughly 60 percent of TSS and about 30 percent of BOD, depending on sewage properties. The removal percentages should not be treated as guaranteed values. Additional detention eventually gives diminishing improvements in separation. Excessive sludge storage and detention can allow septic conditions and odour to develop.
వివరణ:
Question 11
ప్రశ్న 11
A primary sedimentation tank receives 12 MLD at a selected average-flow surface overflow rate of 40 m3/m2/day. What effective horizontal plan area is required?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The daily inflow is 12,000 m3/day. The surface overflow rate is discharge divided by plan area. Rearranging gives area equal to 12,000 divided by 40. The required effective horizontal surface area is 300 m2. Other design requirements such as peak loading, weir length and detention time must also be checked.
వివరణ:
Question 12
ప్రశ్న 12
The total required primary-clarifier plan area is 300 m2, to be shared equally among three identical circular operating tanks. What is the approximate diameter of each tank, neglecting any deductions from effective plan area?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
Three identical tanks share the total 300 m2 area equally. Each circular tank therefore requires an effective plan area of 100 m2. The circular area formula is A = pi D^2/4. Solving gives D = sqrt(4 times 100/pi), approximately 11.28 m. The nearest proposed diameter is 11.3 m, before practical construction adjustments.
వివరణ:
Question 13
ప్రశ్న 13
A sedimentation tank has effective plan area 500 m2 and side-water depth 3.0 m. At an average flow of 18 MLD, what is its nominal hydraulic detention time?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The effective operating volume is plan area multiplied by water depth. This gives 500 times 3.0 = 1500 m3. A flow of 18 MLD is 18,000 m3/day, or 750 m3/h. Nominal detention time equals 1500 divided by 750. The resulting detention time is 2.0 hours.
వివరణ:
Question 14
ప్రశ్న 14
A rectangular primary clarifier receives 12 MLD, has plan area 400 m2, and must provide a specified nominal detention time of 2.4 h at average flow. What effective water depth is needed?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The flow of 12 MLD equals 12,000 m3/day or 500 m3/h. The required active volume is flow rate multiplied by detention time. This gives 500 times 2.4 = 1200 m3. Dividing volume by the specified plan area of 400 m2 gives 3.0 m. Additional depth allowances for sludge collection and freeboard are separate from this calculated water depth.
వివరణ:
Question 15
ప్రశ్న 15
A rectangular settling tank needs 400 m2 of effective plan area with its length four times its width. Which length and width meet these conditions?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
Let the tank width be B metres. Its length is then 4B because the length-to-width ratio is four. The area condition is 4B^2 = 400. Therefore B = 10 m and length = 40 m. These dimensions meet the stated plan area and ratio but do not by themselves establish complete hydraulic performance.
వివరణ:
Question 16
ప్రశ్న 16
A primary clarifier followed by secondary treatment receives 12 MLD average and 30 MLD peak flow, with effective plan area 300 m2. Which assessment of surface overflow rates is correct against the CPHEEO ranges for this arrangement?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The average overflow rate is 12,000/300 = 40 m3/m2/day. The peak overflow rate is 30,000/300 = 100 m3/m2/day. The CPHEEO ranges for primary settling followed by secondary treatment are 35 to 50 at average flow and 80 to 120 at peak flow. Both calculated rates fall within those ranges. This verifies only the stated overflow-rate criteria, not the remaining clarifier design checks.
వివరణ:
Question 17
ప్రశ్న 17
Two parallel primary clarifiers receive a total average flow of 10 MLD. Each has an effective outlet weir length of 40 m, and the flow divides equally. What is the combined average weir overflow loading?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
Average weir loading is the combined daily discharge divided by total effective outlet weir length. The total wastewater flow is 10,000 m3/day. The two tanks provide 40 + 40 = 80 m of effective weir. Dividing 10,000 by 80 gives 125 m3 per metre per day. The calculation assumes uniform distribution and fully effective overflow weirs.
వివరణ:
Question 18
ప్రశ్న 18
A primary clarifier must pass an average flow of 15 MLD. If a project-specified maximum average weir loading is 125 m3/(m day), what is the minimum total effective overflow-weir length required?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The average daily wastewater flow is 15,000 m3/day. Weir overflow loading equals flow divided by effective outlet weir length. Rearranging gives required length equal to 15,000 divided by 125. The minimum effective length is therefore 120 m. Weir layout must also maintain even withdrawal and avoid excessive local hydraulic loading.
వివరణ:
Question 19
ప్రశ్న 19
Two circular clarifiers each have a 20 m diameter and a fully effective single peripheral overflow weir. With 12 MLD divided equally between the tanks, what is their combined average weir loading? Use pi = 3.1416.
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The effective weir length of each circular clarifier is pi D. With D = 20 m, each tank provides approximately 62.832 m of weir. Two tanks together provide about 125.664 m of effective weir. Dividing the total daily flow of 12,000 m3 by this length gives 95.5 m3/(m day). The result assumes the whole stated circumference acts as an effective overflow weir.
వివరణ:
Question 20
ప్రశ్న 20
A primary clarifier has plan area 300 m2 and effective water depth 3.0 m. Flow rises from 12 MLD to 18 MLD without any change in operating tank geometry. What are its new surface overflow rate and nominal detention time?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The new daily flow is 18,000 m3/day. Dividing this by 300 m2 gives surface overflow rate of 60 m3/m2/day. The operating volume is 300 times 3.0 = 900 m3. At 18 MLD, flow is 750 m3/h and the resulting detention time is 900/750 = 1.2 h. The increase in flow raises overflow rate and reduces residence time.
వివరణ:
Question 21
ప్రశ్న 21
An operating primary sedimentation tank has 600 m2 of effective plan area. At a project-selected maximum average surface overflow rate of 40 m3/m2/day, what is its corresponding maximum average-flow capacity?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
Hydraulic capacity at a specified surface overflow rate equals the rate multiplied by effective plan area. The selected rate is 40 m3/m2/day and the area is 600 m2. Their product is 24,000 m3/day. Dividing by 1000 converts this value to 24 MLD. Other operating constraints may make the actual permissible flow lower.
వివరణ:
Question 22
ప్రశ్న 22
A primary clarifier receives 8 MLD of sewage containing 240 mg/L total suspended solids (TSS) and removes 60% of influent TSS. Assuming equal influent and effluent water flows, what are daily dry TSS mass removed and effluent TSS concentration?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The incoming TSS mass load is 8 times 240 = 1920 kg/day. A 60 percent removal fraction gives 1920 times 0.60 = 1152 kg/day of dry suspended solids removed. The remaining TSS concentration is 240 times 0.40 = 96 mg/L under the equal-flow assumption. This is a solids mass balance rather than a direct calculation of wet sludge volume. Wet sludge volume additionally depends on underflow solids concentration and density.
వివరణ:
Question 23
ప్రశ్న 23
A primary sedimentation unit treats 12 MLD and reduces measured BOD from 200 mg/L to 140 mg/L, with equal influent and effluent flow. What daily BOD load is removed?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The reduction in BOD concentration is 200 - 140 = 60 mg/L. At constant flow, the daily removed load equals flow in MLD multiplied by the concentration reduction in mg/L. Thus removed BOD load is 12 times 60 = 720 kg/day. The corresponding concentration removal is 60/200 = 30 percent. This represents BOD-equivalent load removed, not the wet mass of collected sludge.
వివరణ:
Question 24
ప్రశ్న 24
A clarifier produces 900 kg/day of dry sludge solids. The withdrawn slurry contains 4% solids by mass and has an assumed density of 1000 kg/m3. What volume of wet sludge is withdrawn daily?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
Dry solids make up 4 percent of the total wet sludge mass. The wet sludge mass is therefore 900/0.04 = 22,500 kg/day. The problem specifies a slurry density of 1000 kg/m3. Dividing the wet mass by density gives 22.5 m3/day. A different solids concentration or density would change the required sludge pumping volume.
వివరణ:
Question 25
ప్రశ్న 25
A primary clarifier design has 12 MLD average flow and 30 MLD peak flow. Project limits are 40 m3/m2/day at average flow and 80 m3/m2/day at peak flow. What minimum effective plan area satisfies both hydraulic overflow-rate limits?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
Average-flow area must be at least 12,000/40 = 300 m2. Peak-flow area must be at least 30,000/80 = 375 m2. A tank system must satisfy both conditions simultaneously. The governing requirement is the larger of these areas, equal to 375 m2. Detention, depth, weir loading and sludge collection remain independent design checks.
వివరణ:
Answer Key సమాధానాల పట్టిక
-
Question 1 ప్రశ్న 1Answer: B. To remove settleable suspended solids and floating matter by gravity-based separation before subsequent treatment. సమాధానం: B.
-
Question 2 ప్రశ్న 2Answer: D. Grit is commonly represented by discrete settling, whereas organic solids in primary clarifiers frequently undergo flocculent settling. సమాధానం: D.
-
Question 3 ప్రశ్న 3Answer: A. SOR is the influent flow divided by the horizontal plan surface area of the operating tank. సమాధానం: A.
-
Question 4 ప్రశ్న 4Answer: C. Average 35–50 and peak 80–120 m3/m2/day. సమాధానం: C.
-
Question 5 ప్రశ్న 5Answer: A. Surface overflow rate remains unchanged, while tank volume and nominal detention time increase. సమాధానం: A.
-
Question 6 ప్రశ్న 6Answer: B. Using suitable inlet energy dissipation, flow-distribution baffles and level, adequately distributed outlet weirs. సమాధానం: B.
-
Question 7 ప్రశ్న 7Answer: C. To transport deposited sludge to a withdrawal point and prevent excessive accumulation or septic deterioration. సమాధానం: C.
-
Question 8 ప్రశ్న 8Answer: D. Using surface skimming devices and scum baffles to intercept and collect floating material. సమాధానం: D.
-
Question 9 ప్రశ్న 9Answer: B. Primary clarifiers settle raw-sewage solids before biological treatment, while secondary clarifiers separate biological flocs after aeration. సమాధానం: B.
-
Question 10 ప్రశ్న 10Answer: A. Proper primary settling can remove a substantial fraction of TSS and a smaller fraction of BOD, but prolonged sludge retention may cause septic conditions. సమాధానం: A.
-
Question 11 ప్రశ్న 11Answer: C. 300 m2. సమాధానం: C.
-
Question 12 ప్రశ్న 12Answer: D. 11.3 m. సమాధానం: D.
-
Question 13 ప్రశ్న 13Answer: A. 2.0 h. సమాధానం: A.
-
Question 14 ప్రశ్న 14Answer: B. 3.0 m. సమాధానం: B.
-
Question 15 ప్రశ్న 15Answer: C. 40 m length and 10 m width. సమాధానం: C.
-
Question 16 ప్రశ్న 16Answer: D. Average rate 40 and peak rate 100 m3/m2/day; both fall within their respective published ranges. సమాధానం: D.
-
Question 17 ప్రశ్న 17Answer: A. 125 m3/(m day). సమాధానం: A.
-
Question 18 ప్రశ్న 18Answer: B. 120 m. సమాధానం: B.
-
Question 19 ప్రశ్న 19Answer: C. 95.5 m3/(m day). సమాధానం: C.
-
Question 20 ప్రశ్న 20Answer: D. 60 m3/m2/day and 1.2 h. సమాధానం: D.
-
Question 21 ప్రశ్న 21Answer: A. 24 MLD. సమాధానం: A.
-
Question 22 ప్రశ్న 22Answer: B. 1152 kg/day and 96 mg/L. సమాధానం: B.
-
Question 23 ప్రశ్న 23Answer: C. 720 kg/day. సమాధానం: C.
-
Question 24 ప్రశ్న 24Answer: D. 22.5 m3/day. సమాధానం: D.
-
Question 25 ప్రశ్న 25Answer: A. 375 m2. సమాధానం: A.