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Question 1
ప్రశ్న 1
What is the principal treatment objective of a grit chamber in a municipal sewage treatment plant?
Explanation:
Grit chambers are preliminary treatment units intended primarily to remove dense inorganic particles. Sand, gravel and similar grit can abrade pumps and mechanical equipment. Grit deposited in channels or digesters can also reduce working capacity. The unit is operated to avoid excessive removal of lighter putrescible organic matter. Dissolved organic removal and disinfection require other treatment processes.
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Question 2
ప్రశ్న 2
Which conventional sequence most appropriately places a grit chamber within a sewage treatment plant?
Explanation:
Screens remove coarse debris such as rags that can obstruct grit-handling equipment. Grit removal normally follows the screening stage. Primary sedimentation then separates settleable solids including more organic material. Biological treatment follows to reduce biodegradable pollution. This order protects downstream pumps, tanks and other treatment equipment.
వివరణ:
Question 3
ప్రశ్న 3
For a conventional grit-chamber design reference, which target particle characteristics are commonly assumed, while recognizing that actual requirements can differ?
Explanation:
A nominal particle diameter of about 0.20 mm is commonly used in preliminary grit design. A specific gravity near 2.65 is a standard reference for mineral grit. Smaller or lighter particles may require more demanding removal arrangements. Actual wastewater can contain grit with a wide range of sizes and densities. The reference values must not be interpreted as guaranteed characteristics of every influent.
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Question 4
ప్రశ్న 4
A velocity-controlled horizontal-flow grit chamber is operated near a longitudinal mean velocity of 0.30 m/s. What is the principal reason for selecting a moderate, controlled velocity?
Explanation:
A controlled horizontal velocity allows relatively dense grit particles to settle. The velocity should not be so low that excessive organic material deposits. It should not be so high that target grit is swept away or resuspended. A value near 0.30 m/s is a widely used horizontal-flow design reference, not a universal requirement for every technology. Flow-control provisions are important because influent discharge varies during the day.
వివరణ:
Question 5
ప్రశ్న 5
Which outcome is most likely when horizontal flow velocity deviates substantially from its suitable operating range in a conventional grit channel?
Explanation:
At excessively low velocity, more slowly settling organic solids may be deposited with the grit. Such deposited organics increase grit washing and disposal difficulties. At excessively high velocity, target mineral particles may not settle in the available time. High bottom velocities may also re-entrain previously deposited grit. Maintaining an appropriate velocity range is thus central to horizontal-flow grit chamber operation.
వివరణ:
Question 6
ప్రశ్న 6
Which pairing correctly describes aerated and vortex-type grit chambers?
Explanation:
An aerated grit chamber introduces air along one side to generate a spiral or helical liquid motion. Heavier grit moves toward the bottom while suitable aeration helps transport lighter organic particles onward. A vortex unit typically introduces wastewater tangentially to create rotational flow. Grit is concentrated into a collection hopper for removal. Selection between these arrangements depends on hydraulics, available land, grit characteristics and operation needs.
వివరణ:
Question 7
ప్రశ్న 7
A rectangular grit channel carries 0.18 m^3/s through a 0.90 m water width and a 0.80 m flowing depth. What is the average horizontal flow velocity?
Explanation:
The flowing area of the rectangular channel is width multiplied by depth. Here the area is 0.90 times 0.80 = 0.72 m^2. Continuity gives horizontal mean velocity as Q/A. Substituting 0.18/0.72 gives 0.25 m/s. This value should be compared with the selected velocity criterion at the stated flow condition.
వివరణ:
Question 8
ప్రశ్న 8
A horizontal-flow grit chamber must convey Q = 0.20 m^3/s at a specified mean velocity of 0.25 m/s. What flowing cross-sectional area is required?
Explanation:
Continuity for the flowing water gives Q = AV. The required area is therefore A = Q/V. Dividing 0.20 m^3/s by 0.25 m/s yields 0.80 m^2. The area is the wetted flow cross-section, not the plan surface area of the tank. Width and flow depth can be selected subsequently to satisfy this area and other constraints.
వివరణ:
Question 9
ప్రశ్న 9
A single rectangular grit channel carries 0.24 m^3/s at a design horizontal mean velocity of 0.30 m/s and a flowing water depth of 1.00 m. What flowing width is required?
Explanation:
The channel flow area is Q/V = 0.24/0.30 = 0.80 m^2. For a rectangular channel, area equals flowing width multiplied by water depth. The specified water depth is 1.00 m. Hence required width is 0.80/1.00 = 0.80 m. Practical construction dimensions and grit-removal clearances may require further adjustment.
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Question 10
ప్రశ్న 10
A grit channel has effective flow-path length 15 m and a uniform mean horizontal velocity of 0.25 m/s. What is the nominal travel or detention time along that path?
Explanation:
Nominal travel time equals effective flow-path length divided by mean horizontal velocity. Here the path length is 15 m. Dividing 15 by 0.25 gives 60 seconds. The same result follows from volume divided by discharge for uniform rectangular flow. Actual particle residence times can vary because of inlet effects and short-circuiting.
వివరణ:
Question 11
ప్రశ్న 11
For a horizontal-flow grit channel designed at a mean velocity of 0.25 m/s and a specified nominal travel time of 60 s, what effective channel length is required?
Explanation:
Effective travel length is calculated from L = Vh t. The specified velocity is 0.25 m/s and the time is 60 seconds. Their product is 15 m. This is the hydraulic flow-path length under ideal uniform-flow conditions. Allowances for inlet and outlet disturbances should be evaluated separately in a real design.
వివరణ:
Question 12
ప్రశ్న 12
At a design discharge of 0.12 m^3/s, a grit-removal unit is assigned a nominal hydraulic detention time of 50 s. What operating water volume is required?
Explanation:
The hydraulic detention time is t = Volume/Q. Rearranging gives Volume = Q t. Multiplying 0.12 m^3/s by 50 s yields 6.0 m^3. The result is the active hydraulic volume at the specified design condition. Grit storage volume and other geometrical allowances must be considered separately.
వివరణ:
Question 13
ప్రశ్న 13
A target grit particle settles at 0.020 m/s in an ideal quiescent-water approximation. How long does it take to settle vertically through a 1.20 m water depth?
Explanation:
Ideal settling time is water depth divided by particle settling velocity. The stated settling distance is 1.20 m. Dividing 1.20 by 0.020 gives 60 seconds. The calculation assumes the particle maintains the specified settling velocity throughout its descent. Turbulence and nonideal flow can require a more conservative design.
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Question 14
ప్రశ్న 14
An ideal horizontal grit channel is 1.00 m deep; the target grit has settling velocity 0.025 m/s and horizontal mean water velocity is 0.30 m/s. Neglecting inlet and outlet allowances, what minimum effective settling length is required for a particle entering at the free surface?
Explanation:
The particle must settle through the 1.00 m depth before it reaches the outlet. Its ideal settling time is 1.00/0.025 = 40 s. During that time the water travels horizontally at 0.30 m/s. Required length is therefore 0.30 times 40 = 12.0 m. Real chamber design should allow for turbulence, flow distribution and collection details.
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Question 15
ప్రశ్న 15
For an ideal horizontal-flow grit chamber, the design discharge is 0.15 m^3/s and the target particle settling velocity is 0.025 m/s. Ignoring nonideal-flow allowances, what minimum plan surface area corresponds to Q/vs?
Explanation:
For ideal discrete-particle settling, the relevant surface loading is Q divided by plan area. An ideal capture threshold occurs when surface loading is no greater than the specified settling velocity. Therefore required plan area is Q/vs. Substituting 0.15/0.025 gives 6.0 m^2. Nonideal flow and the desired removal reliability may justify a larger area in practice.
వివరణ:
Question 16
ప్రశ్న 16
A rectangular horizontal grit channel has effective length 13.5 m, flowing depth 0.90 m and mean horizontal velocity 0.30 m/s. What minimum grit settling velocity is needed for a particle entering at the water surface to settle by the outlet in an ideal trajectory model?
Explanation:
The water travel time is 13.5/0.30 = 45 seconds. A particle entering at the surface must descend 0.90 m within this time. Its minimum ideal settling velocity is therefore 0.90/45. The result is 0.020 m/s. Actual capture requires allowance for turbulence and variations in particle trajectories.
వివరణ:
Question 17
ప్రశ్న 17
A total sewage flow of 0.18 m^3/s is distributed equally to two identical rectangular grit channels, each 0.60 m wide with a water depth of 0.75 m. What is the mean flow-through velocity in each channel?
Explanation:
Equal distribution sends 0.18/2 = 0.09 m^3/s to each channel. The flowing cross-sectional area of one channel is 0.60 times 0.75 = 0.45 m^2. Mean velocity is discharge divided by area. Dividing 0.09 by 0.45 gives 0.20 m/s. The calculation assumes both duty channels receive equal flow and operate at the stated depth.
వివరణ:
Question 18
ప్రశ్న 18
A peak sewage flow of 0.48 m^3/s is to be divided among identical duty grit channels, each carrying water 0.80 m wide and 1.00 m deep at a limiting mean flow velocity of 0.30 m/s. What is the minimum number of such duty channels needed, excluding standby provision?
Explanation:
Each channel has a wetted cross-section of 0.80 times 1.00 = 0.80 m^2. At 0.30 m/s, its nominal capacity is 0.80 times 0.30 = 0.24 m^3/s. The total peak discharge is 0.48 m^3/s. Dividing 0.48 by 0.24 gives two operating channels. Any standby or isolation provision must be considered in addition to this hydraulic minimum.
వివరణ:
Question 19
ప్రశ్న 19
An ideal horizontal grit basin handles 0.16 m^3/s with plan surface area 8.0 m^2. Target grit has settling velocity 0.025 m/s. Which assessment follows from ideal overflow-rate theory?
Explanation:
The surface overflow rate is Q/Aplan. Here 0.16/8.0 gives 0.020 m/s. The target settling velocity is 0.025 m/s, which exceeds the calculated overflow rate. Under an ideal discrete-particle model, a particle entering near the surface can settle before leaving. Actual efficiency may be lower because of turbulence, short-circuiting or resuspension.
వివరణ:
Question 20
ప్రశ్న 20
A plant has average sewage flow 12 MLD and an explicitly assumed peak factor of 2.4. At that peak flow, a velocity-controlled grit channel is sized for 0.30 m/s. What total flowing cross-sectional area is required?
Explanation:
Peak discharge is 12 times 2.4 = 28.8 MLD. Converting gives 28,800 m^3/day divided by 86,400 s/day = 0.3333 m^3/s. The required cross-sectional area is Q/Vh. Dividing 0.3333 by 0.30 gives approximately 1.11 m^2. This hydraulic area must still be translated into practicable channel widths and operating depths.
వివరణ:
Question 21
ప్రశ్న 21
Which statement correctly compares typical retention and motion in conventional horizontal-flow and aerated grit chambers?
Explanation:
Horizontal-flow grit channels are normally designed for short detention while maintaining controlled longitudinal velocity. An order-of-one-minute detention is a frequently cited reference for such channels. Aerated grit chambers instead use induced spiral flow to separate heavy grit from lighter organic material. They commonly employ several minutes of detention at the stated design-flow condition. The detention requirement is technology-specific and must not be transferred indiscriminately between chamber types.
వివరణ:
Question 22
ప్రశ్న 22
After removing mineral grit from raw sewage, the collected material still contains significant putrescible organic matter. What is the most appropriate downstream grit-handling step?
Explanation:
Grit recovered from sewage often includes attached or entrained organic matter. Washing and classification can reduce this material and improve the handling properties of the mineral fraction. Grit may then be dewatered and conveyed to an appropriate disposal location. Proper handling reduces odour, putrefaction and operational nuisance. The choice of equipment depends on the grit volume, composition and plant arrangement.
వివరణ:
Question 23
ప్రశ్న 23
A sewage treatment plant processes 24 ML/day and measured grit generation is 0.075 m^3 per ML of sewage. What daily volume of collected grit should be used for a simple storage estimate?
Explanation:
The specified grit yield is 0.075 cubic metre for each million litres of sewage. The plant receives 24 million litres per day. Multiplying the yield by the daily volume gives 0.075 times 24. The expected daily grit volume is 1.8 m^3. A real collection and storage design should also account for peak grit arrivals and removal schedules.
వివరణ:
Question 24
ప్రశ్న 24
The mean longitudinal velocity of an existing horizontal-flow grit chamber rises from 0.25 m/s to 0.45 m/s after a peak-flow increase, with chamber geometry unchanged. Which operational concern becomes more likely?
Explanation:
At fixed chamber length, higher longitudinal flow velocity reduces travel time. Target grit then has less time to reach the bottom. Increased flow velocity can also raise the tendency to resuspend deposits. Both effects can produce additional grit carryover to downstream units. The plant should examine peak-flow distribution and velocity-control performance.
వివరణ:
Question 25
ప్రశ్న 25
A horizontal grit channel shows adequate grit settling at peak flow but excessive organic deposition at low flow. Which operational modification directly addresses the probable hydraulic cause?
Explanation:
At low sewage flow, an uncontrolled channel may experience a sharp decline in velocity. This can allow more putrescible organic particles to settle with the grit. A properly designed downstream hydraulic control can help preserve a suitable velocity over changing flow conditions. Proportional weirs and appropriate flume arrangements are established approaches for velocity-controlled grit channels. The control must be hydraulically checked to avoid excessive head loss and bottom scour.
వివరణ:
Answer Key సమాధానాల పట్టిక
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Question 1 ప్రశ్న 1Answer: A. To separate relatively heavy inorganic particles such as sand and grit while limiting deposition of lighter organic solids. సమాధానం: A.
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Question 2 ప్రశ్న 2Answer: B. Screening, grit removal, primary sedimentation, biological treatment. సమాధానం: B.
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Question 3 ప్రశ్న 3Answer: A. Nominal grit size around 0.20 mm and particle specific gravity around 2.65. సమాధానం: A.
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Question 4 ప్రశ్న 4Answer: D. To encourage dense grit settling while carrying much of the lighter organic material through the chamber. సమాధానం: D.
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Question 5 ప్రశ్న 5Answer: C. Too low a velocity may increase organic deposition, while too high a velocity may increase grit carryover or scour. సమాధానం: C.
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Question 6 ప్రశ్న 6Answer: B. Aerated chambers use helical flow induced by side air injection; vortex chambers commonly use tangential inlet flow and a grit hopper. సమాధానం: B.
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Question 7 ప్రశ్న 7Answer: C. 0.25 m/s సమాధానం: C.
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Question 8 ప్రశ్న 8Answer: C. 0.80 m^2 సమాధానం: C.
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Question 9 ప్రశ్న 9Answer: D. 0.80 m సమాధానం: D.
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Question 10 ప్రశ్న 10Answer: D. 60 s సమాధానం: D.
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Question 11 ప్రశ్న 11Answer: A. 15 m సమాధానం: A.
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Question 12 ప్రశ్న 12Answer: C. 6.0 m^3 సమాధానం: C.
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Question 13 ప్రశ్న 13Answer: B. 60 s సమాధానం: B.
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Question 14 ప్రశ్న 14Answer: D. 12.0 m సమాధానం: D.
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Question 15 ప్రశ్న 15Answer: A. 6.0 m^2 సమాధానం: A.
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Question 16 ప్రశ్న 16Answer: C. 0.020 m/s సమాధానం: C.
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Question 17 ప్రశ్న 17Answer: B. 0.20 m/s సమాధానం: B.
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Question 18 ప్రశ్న 18Answer: D. 2 సమాధానం: D.
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Question 19 ప్రశ్న 19Answer: A. Surface loading is 0.020 m/s, below the target settling velocity; ideal full-depth removal is hydraulically feasible. సమాధానం: A.
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Question 20 ప్రశ్న 20Answer: C. 1.11 m^2 సమాధానం: C.
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Question 21 ప్రశ్న 21Answer: B. Aerated grit chambers commonly use a few minutes of detention with helical mixing, whereas velocity-controlled horizontal channels commonly use roughly one minute as a reference. సమాధానం: B.
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Question 22 ప్రశ్న 22Answer: D. Wash and classify the grit to reduce adhering organics before suitable dewatering and disposal. సమాధానం: D.
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Question 23 ప్రశ్న 23Answer: A. 1.8 m^3/day సమాధానం: A.
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Question 24 ప్రశ్న 24Answer: C. Carryover or resuspension of grit because higher horizontal velocities leave less settling time and can cause scour. సమాధానం: C.
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Question 25 ప్రశ్న 25Answer: B. Use a suitably designed proportional control weir or Parshall-type flow-control arrangement to help maintain the intended flow-through velocity across variable flows. సమాధానం: B.
