Please keep at least one language enabled. || కనీసం ఒక భాషను ఎంచుకోండి.
Question 1
ప్రశ్న 1
Which approach correctly establishes the hydraulic design discharge of a separate sanitary sewer serving a developing catchment?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
Sanitary sewer capacity is normally assessed for peak wastewater flows expected over the design horizon. Future contributory population and appropriate per-capita sewage generation must first be estimated. A suitable peak factor or measured flow pattern accounts for temporal variation. Legitimate additional discharges and infiltration allowances are then included without double counting. Storm runoff is not deliberately included in the hydraulic design of a properly separate sanitary sewer.
వివరణ:
Question 2
ప్రశ్న 2
For preliminary design using the CPHEEO contributory-population peak-factor table, a branch sewer serves 18,000 people although the entire municipality has 900,000 residents. Which peak factor applies to the branch's contributory population?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The CPHEEO table associates a preliminary peak factor of 3.00 with a contributory population up to 20,000. The branch sewer serves 18,000 people and therefore falls in that category. The larger population of the whole municipality does not itself determine the flow variation in this local branch. Downstream trunk sewers may have different contributing populations and corresponding peak factors. Local flow measurements and project-specific conditions can justify further analysis rather than blind application of the tabulated factor.
వివరణ:
Question 3
ప్రశ్న 3
A sanitary sewer serves 15,000 people, each generating an average of 120 L/day of sewage. Using a peak factor of 3.0 and neglecting additional inflows, what is the peak design discharge?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
Average sewage generation is 15,000 multiplied by 120, or 1,800,000 L/day. This equals 1.8 MLD of average sewage flow. Applying the specified peak factor of 3.0 gives 5.4 MLD. Dividing 5,400,000 L/day by 86,400 seconds/day gives 62.5 L/s. No infiltration or industrial contribution is added because the question excludes additional inflows.
వివరణ:
Question 4
ప్రశ్న 4
A sewer catchment has 100,000 people generating 100 litres of sewage per capita per day. With the specified peak factor of 2.25 and no other inflows, what is the peak discharge?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The average sewage production is 100,000 multiplied by 100 L/day. This gives 10,000,000 L/day, or 10 MLD. Applying the peak factor of 2.25 gives a peak sewage flow of 22.5 MLD. Conversion to litres per second gives 22,500,000/86,400 = approximately 260.4 L/s. The calculation assumes that the specified population, generation rate and peak factor represent the design condition.
వివరణ:
Question 5
ప్రశ్న 5
The average sanitary sewage flow is 4.0 MLD, its peak factor is 2.5, and a separately estimated steady groundwater infiltration of 0.25 MLD must be added after peaking the sanitary component. What is the design discharge?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The peak sanitary sewage component is 4.0 multiplied by 2.5, giving 10.0 MLD. The stated groundwater infiltration is a separate steady contribution. Because the question expressly requires adding it after the sanitary peaking calculation, the infiltration is not multiplied by 2.5. Total design flow is therefore 10.0 plus 0.25 = 10.25 MLD. This avoids applying a peak factor twice to an independently estimated non-storm contribution.
వివరణ:
Question 6
ప్రశ్న 6
Which combination matches the gravity-sewer design velocities stated in the CPHEEO Manual on Sewerage and Sewage Treatment Systems, Part A, for initial peak flow, ultimate peak flow and maximum velocity, respectively?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The CPHEEO gravity-sewer design-velocity table specifies 0.6 m/s as the minimum at initial peak flow. It specifies 0.8 m/s as the minimum at ultimate peak flow. The table also gives 3.0 m/s as the maximum design velocity. These values refer to distinct flow conditions rather than a single universal operating velocity. Material resistance, grit characteristics and site-specific design requirements also need consideration.
వివరణ:
Question 7
ప్రశ్న 7
A proposed gravity sewer is predicted to have 0.55 m/s at initial peak flow and 0.90 m/s at ultimate peak flow. Considering the CPHEEO gravity-sewer minimum-velocity criteria, which judgement is correct?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The stated initial peak-flow velocity is 0.55 m/s. It is below the 0.6 m/s minimum used in the CPHEEO gravity-sewer criterion. The ultimate peak-flow velocity of 0.90 m/s exceeds its 0.8 m/s minimum. Satisfying the future-flow check does not remove the need to examine the initial-flow condition. The proposed gradient or other design measures should therefore be reconsidered for self-cleansing at the earlier stage.
వివరణ:
Question 8
ప్రశ్న 8
When applying a non-scouring velocity criterion to gravity sewer design, which consideration is most defensible?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
Non-scouring velocity relates to the risk of abrasion and erosion of sewer materials. The damage potential depends on the boundary material and the abrasive solids transported by the flow. General gravity-sewer design guidance provides an upper velocity check, but material-specific restrictions may be relevant. Excessive gradients can require engineering measures to control velocity and dissipate energy. Treating one numerical value as universally safe for every material and wastewater is not a sound design practice.
వివరణ:
Question 9
ప్రశ్న 9
Why is minimum self-cleansing velocity not determined solely from the internal diameter of a sewer?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
Self-cleansing design aims to prevent persistent deposition of solids in the sewer. The motion of granular particles depends substantially on particle diameter and relative density. Boundary shear, resistance and the hydraulic characteristics of the flow also matter. Hence a velocity value alone is a simplified design check rather than a complete sediment-transport model. The selected sewer should be evaluated for both capacity and deposition risk under relevant flow conditions.
వివరణ:
Question 10
ప్రశ్న 10
For conventional circular gravity sewer design using the CPHEEO 2013 Part A hydraulic guidance, which pair of design checks is appropriate?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The CPHEEO guidance limits circular gravity-sewer design depth to 0.8 of the diameter at ultimate peak flow. This retains a free surface and provides allowance for ventilation and hydraulic operation. Diameter is selected to convey the ultimate design discharge within the adopted depth limit. Initial and ultimate flow velocities require separate verification against relevant self-cleansing criteria. Adequate hydraulic capacity alone does not establish satisfactory sewer performance throughout the design horizon.
వివరణ:
Question 11
ప్రశ్న 11
For a combined sewer, a catchment of 18 ha has design rainfall intensity 60 mm/h and a runoff coefficient of 0.65. Using Q = CIA/360, where Q is in m^3/s, I in mm/h and A in ha, what is the design storm-runoff rate?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The rational-method expression in the stated units is Q = CIA/360. The coefficient C is 0.65, the intensity I is 60 mm/h, and the catchment A is 18 ha. Multiplying gives 0.65 times 60 times 18 = 702. Dividing by 360 gives a runoff discharge of 1.95 m^3/s. The result assumes the stated rainfall intensity and runoff coefficient are appropriate to the catchment and design event.
వివరణ:
Question 12
ప్రశ్న 12
A combined sewer must convey a coincident storm-runoff peak of 1.95 m^3/s, peak sanitary flow of 0.18 m^3/s and an additional steady non-storm contribution of 0.02 m^3/s. Neglecting storage and attenuation, what is the specified total peak flow?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The question expressly assumes that the stated component peaks are coincident. Under that assumption, the combined sewer design flow is the sum of storm runoff and sanitary contributions. Adding 1.95 and 0.18 gives 2.13 m^3/s. Including the additional 0.02 m^3/s produces 2.15 m^3/s. An actual sewer-design study must also examine rainfall timing, catchment response and applicable routing effects.
వివరణ:
Question 13
ప్రశ్న 13
What is the full-section reference discharge of a 0.60 m internal-diameter circular sewer with Manning n = 0.013 and steady uniform energy slope S = 0.001?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The full-section area is A = pi(0.60)^2/4 = 0.28274 m^2. Its hydraulic radius is R = D/4 = 0.15 m. Manning velocity is R^(2/3)sqrt(S)/n, giving approximately 0.68673 m/s. Multiplying this velocity by area gives approximately 0.19417 m^3/s. This is a full-section hydraulic reference and must be adjusted when an allowable partial-flow design depth is specified.
వివరణ:
Question 14
ప్రశ్న 14
A 0.50 m internal-diameter sewer has Manning n = 0.013 and uniform energy slope S = 0.001. Which pair is closest to its full-section reference mean velocity and discharge?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
For D = 0.50 m, the full-section hydraulic radius is 0.125 m. Manning equation with n = 0.013 and S = 0.001 gives V approximately 0.60813 m/s. The circular flow area is pi(0.50)^2/4 = 0.19635 m^2. Multiplying area by velocity gives Q approximately 0.11941 m^3/s. The correctly paired velocity and discharge are therefore approximately 0.608 m/s and 0.119 m^3/s.
వివరణ:
Question 15
ప్రశ్న 15
For a 0.40 m circular sewer evaluated at full-section reference conditions, n = 0.013. What minimum uniform energy slope approximately produces a mean velocity of 0.60 m/s?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
At full-section reference conditions, the hydraulic radius is R = D/4 = 0.10 m. Manning equation can be rearranged as S = [Vn/R^(2/3)]^2. Substituting V = 0.60 m/s and n = 0.013 gives S approximately 0.0013108. The equivalent uniform gradient is about 1 in 763. This is a velocity calculation for the specified reference geometry, not a substitute for the actual operating-depth checks.
వివరణ:
Question 16
ప్రశ్న 16
For a 0.50 m circular sewer at full-section reference conditions with n = 0.013, which uniform slope is approximately required to produce V = 0.80 m/s?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The hydraulic radius for a full 0.50 m circular section is 0.125 m. Rearranging Manning equation gives S = [Vn/R^(2/3)]^2. The specified product Vn is 0.80 times 0.013 = 0.0104. Substitution gives S approximately 0.0017306. The equivalent gradient is about 1 in 578 under the uniform-flow assumption.
వివరణ:
Question 17
ప్రశ్న 17
An ultimate peak sanitary flow of 0.25 m^3/s must be conveyed at or below 0.80D. For circular sewers with n = 0.013 and uniform slope S = 0.001, and available internal diameters of 0.50, 0.60, 0.70 and 0.80 m, which is the smallest hydraulically adequate diameter using constant-n Manning analysis?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
At 0.80D, a circular sewer with constant Manning n carries approximately 0.9775 of its full-section reference discharge. At the stated slope, the approximate 0.80D capacities for diameters 0.50, 0.60, 0.70 and 0.80 m are 0.117, 0.190, 0.286 and 0.409 m^3/s. The 0.60 m sewer fails to carry the required 0.25 m^3/s within the depth limit. The 0.70 m sewer is the smallest listed size with adequate discharge capacity. A complete design must also verify its actual peak and initial-flow velocities and other project constraints.
వివరణ:
Question 18
ప్రశ్న 18
A circular sewer has a full-section reference capacity of 0.24 m^3/s at its selected slope and constant Manning n. If it is limited to half-full depth at the design condition, can it carry a specified discharge of 0.15 m^3/s?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
At half-full depth, a circular sewer has half its full-section flow area. Its hydraulic radius is equal to the full-section reference hydraulic radius. Consequently, mean velocity is unchanged for equal slope and constant Manning n. Its discharge is therefore half the full-section reference capacity, or 0.12 m^3/s. The specified 0.15 m^3/s exceeds this capacity and fails the stated depth restriction.
వివరణ:
Question 19
ప్రశ్న 19
A circular sewer has a full-section reference discharge of 0.20 m^3/s under fixed slope and constant Manning n. Which discharge and mean-velocity comparisons are approximately correct at depth 0.80D?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
At a relative depth of 0.80, circular-section geometry yields a flow area less than the full circular area. Under constant Manning n and slope, the mean velocity ratio is approximately 1.1397. The discharge ratio is approximately 0.9775 because discharge also depends on the reduced area. Multiplying 0.20 m^3/s by 0.9775 gives about 0.1955 m^3/s. Therefore the mean velocity is higher than the full-section reference value even though the discharge is slightly lower.
వివరణ:
Question 20
ప్రశ్న 20
A 0.50 m diameter gravity sewer has n = 0.013 and uniform slope S = 0.001. At initial peak flow the depth is 0.50D, while at ultimate peak flow it is 0.80D. Under constant-n Manning analysis, which CPHEEO minimum-velocity checks are satisfied?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The full-section reference velocity is approximately 0.6081 m/s. At half-full depth, the mean velocity equals that reference value, so initial peak velocity exceeds 0.6 m/s. At 0.80D, the velocity is approximately 1.1397 times 0.6081, or 0.693 m/s. This is below the CPHEEO ultimate peak-flow minimum of 0.8 m/s. The proposed sewer therefore satisfies the stated initial criterion but not the ultimate criterion.
వివరణ:
Question 21
ప్రశ్న 21
A 0.60 m sewer has n = 0.013 and uniform slope S = 0.002. Its ultimate peak flow is 0.272 m^3/s and the design depth may not exceed 0.80D. What is the correct capacity assessment under constant-n Manning analysis?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
Manning equation gives a full-section reference velocity of approximately 0.9712 m/s at slope 0.002. Multiplying by the full circular area gives a reference discharge of approximately 0.2746 m^3/s. The constant-n discharge ratio at 0.80D is approximately 0.9775. Thus the maximum permitted-depth discharge is approximately 0.2684 m^3/s. The specified demand of 0.272 m^3/s is greater than this allowable-depth capacity, so the proposal fails the stated check.
వివరణ:
Question 22
ప్రశ్న 22
A gravity sewer must convey 0.12 m^3/s at a target mean flow velocity of 0.80 m/s. What wetted flow area follows from continuity, and what further step is necessary to select circular sewer diameter?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The continuity equation for mean flow is Q = AV. Rearranging gives A = Q/V = 0.12/0.80 = 0.15 m^2. This is the required wetted cross-sectional flow area at the selected velocity. The internal diameter of a partly full circular sewer cannot be determined from this area alone without a design flow depth. A valid final design also checks gradient, roughness and velocity criteria at relevant operating flows.
వివరణ:
Question 23
ప్రశ్న 23
For a preliminary full-circular-area estimate only, a discharge of 0.24 m^3/s is to be conveyed at mean velocity 1.20 m/s. What is the corresponding equivalent circular diameter, before applying partial-flow design checks?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
Continuity requires area A = Q/V = 0.24/1.20 = 0.20 m^2. For a full circular area, A = pi D^2/4. Rearranging gives D = sqrt(4A/pi). Substitution gives D approximately 0.5046 m. This is only an equivalent full-area estimate and does not replace sizing for permissible partial-flow depth and selected gradient.
వివరణ:
Question 24
ప్రశ్న 24
A gravity sewer runs 300 m along a uniform gradient of 1 in 600. Ignoring local drops and changes of alignment, what is the vertical fall of its invert over this length?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
A gradient of 1 in 600 means one unit of vertical fall for 600 units of horizontal distance. The sewer run is 300 m. The total fall is therefore 300/600 m. This evaluates to 0.50 m. The calculation presumes a constant invert gradient and excludes any additional drops introduced at structures.
వివరణ:
Question 25
ప్రశ్న 25
A 0.40 m diameter circular sewer flows exactly half full with n = 0.013 and steady uniform slope S = 0.05. Its Manning mean velocity is approximately 3.71 m/s. Which design response is appropriate when applying the CPHEEO general gravity-sewer maximum of 3.0 m/s?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
At half-full depth, the hydraulic radius of a circular pipe is D/4 = 0.10 m. Manning equation with n = 0.013 and S = 0.05 gives mean velocity approximately 3.71 m/s. This exceeds the cited general maximum gravity-sewer design velocity of 3.0 m/s. Abrasive grit and the pipe material may make an excessively steep sewer vulnerable to damage. The engineer should investigate feasible grade control, energy dissipation or protective measures rather than accept the velocity without examination.
వివరణ:
Answer Key సమాధానాల పట్టిక
-
Question 1 ప్రశ్న 1Answer: C. Estimate ultimate peak sanitary discharge and add applicable non-storm contributions such as design infiltration without double counting. సమాధానం: C.
-
Question 2 ప్రశ్న 2Answer: A. 3.00 సమాధానం: A.
-
Question 3 ప్రశ్న 3Answer: D. 62.5 L/s సమాధానం: D.
-
Question 4 ప్రశ్న 4Answer: B. 260.4 L/s సమాధానం: B.
-
Question 5 ప్రశ్న 5Answer: C. 10.25 MLD సమాధానం: C.
-
Question 6 ప్రశ్న 6Answer: A. 0.6 m/s minimum, 0.8 m/s minimum, and 3.0 m/s maximum సమాధానం: A.
-
Question 7 ప్రశ్న 7Answer: B. The initial peak-flow criterion fails even though the ultimate peak-flow criterion passes. సమాధానం: B.
-
Question 8 ప్రశ్న 8Answer: D. The design must consider pipe and lining resistance, transported grit and local requirements, even when checking a general maximum-velocity guideline. సమాధానం: D.
-
Question 9 ప్రశ్న 9Answer: C. Transport of deposited particles depends on their size and relative density, boundary resistance and hydraulic conditions. సమాధానం: C.
-
Question 10 ప్రశ్న 10Answer: A. Limit ultimate peak design depth to 0.8D and independently verify adequate velocity at initial and ultimate peak conditions. సమాధానం: A.
-
Question 11 ప్రశ్న 11Answer: D. 1.95 m^3/s సమాధానం: D.
-
Question 12 ప్రశ్న 12Answer: B. 2.15 m^3/s సమాధానం: B.
-
Question 13 ప్రశ్న 13Answer: A. Approximately 0.194 m^3/s సమాధానం: A.
-
Question 14 ప్రశ్న 14Answer: C. 0.608 m/s and 0.119 m^3/s సమాధానం: C.
-
Question 15 ప్రశ్న 15Answer: B. 0.00131 సమాధానం: B.
-
Question 16 ప్రశ్న 16Answer: D. 0.00173 సమాధానం: D.
-
Question 17 ప్రశ్న 17Answer: A. 0.70 m సమాధానం: A.
-
Question 18 ప్రశ్న 18Answer: C. No, because half-full capacity is 0.12 m^3/s, below the specified discharge. సమాధానం: C.
-
Question 19 ప్రశ్న 19Answer: B. Discharge 0.1955 m^3/s and mean velocity 1.14 times full-section velocity. సమాధానం: B.
-
Question 20 ప్రశ్న 20Answer: D. The initial minimum is satisfied, but the ultimate minimum of 0.8 m/s is not satisfied. సమాధానం: D.
-
Question 21 ప్రశ్న 21Answer: C. It fails because its 0.80D capacity is approximately 0.268 m^3/s, despite a slightly higher full-section reference capacity. సమాధానం: C.
-
Question 22 ప్రశ్న 22Answer: A. Area = 0.15 m^2; use the permitted relative depth and circular geometry before finalising diameter. సమాధానం: A.
-
Question 23 ప్రశ్న 23Answer: B. 0.505 m సమాధానం: B.
-
Question 24 ప్రశ్న 24Answer: D. 0.50 m సమాధానం: D.
-
Question 25 ప్రశ్న 25Answer: C. Recognise excessive velocity and investigate a suitable gradient-control, drop or energy-dissipation arrangement and material protection. సమాధానం: C.