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Question 1
ప్రశ్న 1
Which statement best distinguishes the biological mechanism of a trickling filter from that of a conventional activated-sludge aeration tank?
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Explanation:
A trickling filter is an attached-growth biological treatment process. Microorganisms colonize the surfaces of stationary support media. Wastewater passes over the biofilm, where biodegradable constituents are metabolized. In conventional activated sludge, the active biomass is principally suspended as mixed-liquor flocs. Both processes require suitable oxygen supply for aerobic carbon oxidation.
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Question 2
ప్రశ్న 2
For a conventional stone-media trickling-filter treatment train, which sequence most appropriately handles incoming solids and detached biological growth?
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Explanation:
Preliminary treatment removes debris and grit that can interfere with downstream equipment. Primary clarification reduces settleable solids before application to the media. The trickling filter then oxidizes biodegradable pollutants through attached biological growth. Some of that growth periodically detaches and enters the filter effluent. A secondary clarifier is consequently required to capture sloughed solids before the final effluent is released or further treated.
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Question 3
ప్రశ్న 3
Which property combination makes a medium suitable for a conventional aerobic trickling filter?
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Explanation:
Trickling-filter media must provide stable surfaces for biofilm attachment. Open voids permit wastewater to pass through the filter without persistent ponding. They also support the airflow needed to maintain aerobic conditions. Structural durability is important because the bed must withstand its own weight and hydraulic loading. Stone and suitably designed synthetic modules can meet these requirements in different configurations.
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Question 4
ప్రశ్న 4
What are the main hydraulic functions of rotary distributor arms and underdrains in a conventional trickling filter?
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Explanation:
The distributor applies wastewater over the plan area of the media bed. Good distribution limits local dry zones and excessive localized hydraulic loading. Underdrains convey treated liquid and detached solids toward downstream collection. Their void space and outlet arrangement can also assist passage of ventilation air. Inadequate distribution or drainage can impair aerobic biological treatment.
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Question 5
ప్రశ్న 5
Compared with a standard-rate stone-media trickling filter, which description most appropriately characterizes a high-rate trickling filter?
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Explanation:
High-rate trickling filters are operated at higher loadings than conventional standard-rate units. Recirculation is commonly used to maintain wetting and promote flushing of the media. Low-rate units can have more opportunity for nitrification under favorable loading and temperature conditions. High-rate operation does not guarantee nitrification or any fixed removal percentage. Both designs require suitable downstream solids separation where biofilm detaches into the effluent.
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Question 6
ప్రశ్న 6
During operation of a trickling filter, what is meant by sloughing and why is a downstream clarifier important?
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Explanation:
Biofilm thickness changes as microorganisms grow and consume substrate. Portions of the attached growth eventually detach because of biological and hydraulic effects. This detachment is commonly called sloughing. The detached material increases suspended solids in the filter effluent. A downstream settling unit removes these biological solids and protects final effluent quality.
వివరణ:
Question 7
ప్రశ్న 7
For a trickling filter receiving fresh wastewater flow Q and recirculated liquid flow Qr, which expression represents the hydraulic loading rate based on the total flow applied to media of plan area A?
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Explanation:
Hydraulic loading describes liquid volume applied per unit filter plan area per unit time. Both fresh influent and recirculated flow pass through the distributor and over the media. Their volumetric flow rates therefore enter the total applied hydraulic load. Dividing Q + Qr by plan area yields the loading rate in m3/(m2 day) when consistent daily units are used. Organic loading is a different parameter based on the oxygen-demanding mass associated with a specified flow basis.
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Question 8
ప్రశ్న 8
A designer explicitly defines external organic loading as the BOD5 mass brought by fresh primary effluent per unit filter-media volume per day. Which formula follows for fresh flow Q, BOD5 concentration S and media volume V?
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Explanation:
The question defines loading specifically on the fresh incoming BOD mass. One MLD at one mg/L represents one kg of BOD mass per day. Thus Q in MLD multiplied by S in mg/L gives the fresh organic load in kg/day. Dividing by filter-media volume gives the stated external loading in kg/(m3 day). If recycled liquid also contains BOD, its contribution to total BOD physically applied to the distributor must be distinguished from this fresh-load definition.
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Question 9
ప్రశ్న 9
Which definition correctly describes a trickling-filter recirculation ratio R?
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Explanation:
The recirculation ratio compares the recycled liquid flow with the fresh incoming wastewater flow. It is conventionally expressed as R = Qr/Q. For example, a ratio of one means the recycled flow equals the fresh flow. Total filter-applied liquid flow is then Q(1 + R). The ratio by itself does not specify the BOD concentration of the recycled stream.
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Question 10
ప్రశ్న 10
A high-rate trickling filter receives a sudden toxic industrial discharge despite normal recirculation. Which response is most defensible?
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Explanation:
Attached-growth microorganisms can be inhibited by toxic compounds. Recirculation can dilute concentrations at the distributor but does not itself destroy the incoming toxic mass. Source control or industrial pretreatment may therefore be necessary. Equalization can reduce short-term fluctuations in wastewater flow and strength. The proposed response must also account for the filter hydraulic capacity and observed process performance.
వివరణ:
Question 11
ప్రశ్న 11
A filter receives 4.0 MLD of fresh wastewater and recirculates 1.5 times the fresh flow. If the media plan area is 2500 m2, what is the total applied hydraulic loading?
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Explanation:
The recycled flow is 1.5 times 4.0 MLD, which equals 6.0 MLD. The total flow applied to the bed is 4.0 + 6.0 = 10.0 MLD. This is 10000 m3/day. Dividing by 2500 m2 gives a hydraulic loading of 4.0 m3/(m2 day). The calculation includes recirculation because the question asks for total applied hydraulic loading.
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Question 12
ప్రశ్న 12
A trickling filter handles 2.4 MLD of fresh effluent with a recirculation ratio of 0.5. Its plan area is 1200 m2. What is the total applied hydraulic loading?
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Explanation:
A recirculation ratio of 0.5 gives recycled flow of 1.2 MLD. The total applied flow is therefore 2.4 + 1.2 = 3.6 MLD. Converting this gives 3600 m3/day. Dividing by the plan area of 1200 m2 gives 3.0 m3/(m2 day). Using fresh flow alone would underestimate the specified hydraulic application rate.
వివరణ:
Question 13
ప్రశ్న 13
Fresh flow to a trickling filter is 3.0 MLD and the recirculation ratio is 1.0. For a project-specified maximum applied hydraulic loading of 6.0 m3/(m2 day), what minimum filter plan area is required?
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Explanation:
A recirculation ratio of one means recycled flow equals fresh flow. The total applied liquid flow is 3.0 + 3.0 = 6.0 MLD. This is equivalent to 6000 m3/day. Dividing by the stated maximum loading of 6.0 m3/(m2 day) yields 1000 m2. The selected rate is an explicit design assumption rather than a universal limit for all filter media.
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Question 14
ప్రశ్న 14
A trickling filter receives fresh primary effluent at 2.0 MLD with BOD5 = 150 mg/L. If its project-specified external organic loading is 0.30 kg BOD5/(m3 day), what minimum media volume is needed?
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Explanation:
The fresh BOD5 load is 2.0 multiplied by 150, or 300 kg/day. External volumetric organic loading equals fresh BOD mass per day divided by filter-media volume. Rearranging gives volume equal to 300/0.30. The required media volume is 1000 m3. This result uses the stated fresh-load basis and must be accompanied by a separate hydraulic loading check.
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Question 15
ప్రశ్న 15
A proposed trickling filter has required media volume 1500 m3 and a selected plan area of 600 m2. What effective media depth is necessary if the bed has a uniform rectangular or cylindrical plan cross-section?
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Explanation:
Media-bed volume equals horizontal plan area multiplied by effective media depth for a uniform cross-section. The specified required media volume is 1500 m3. The chosen plan area is 600 m2. Dividing 1500 by 600 gives an effective depth of 2.5 m. Practical media selection and ventilation must also be checked for the proposed depth.
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Question 16
ప్రశ్న 16
A single circular trickling filter receives 4.0 MLD of fresh flow and recycles an equal flow. If the selected applied hydraulic loading is 8.0 m3/(m2 day), what is its required approximate diameter?
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Explanation:
Total applied flow is 4.0 + 4.0 = 8.0 MLD. At 8.0 m3/(m2 day), required area is 8000/8.0 = 1000 m2. For a circular plan, A = pi D squared divided by four. Therefore D = sqrt(4 times 1000 divided by pi), approximately 35.7 m. The calculation treats the entire stated area as effective media-plan area.
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Question 17
ప్రశ్న 17
A filter with media volume 2000 m3 receives 3.0 MLD of fresh primary effluent at BOD5 = 200 mg/L. What is its external volumetric organic loading?
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Explanation:
The fresh daily BOD5 load is 3.0 times 200 = 600 kg/day. The volume of the trickling-filter media is 2000 m3. Organic loading is defined in this question as fresh BOD5 mass divided by media volume. Hence the result is 600/2000 = 0.30 kg BOD5/(m3 day). Recycled BOD mass is not included in this specifically defined external loading.
వివరణ:
Question 18
ప్రశ్న 18
A high-rate trickling filter receives 2.5 MLD of fresh wastewater and has recirculation ratio R = 1.2. What are the recycled flow and total applied flow, respectively?
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Explanation:
Recirculation ratio is Qr/Q = 1.2. Thus recycled flow equals 1.2 times 2.5 = 3.0 MLD. The trickling-filter distributor receives both fresh and recycled liquid. Total applied flow is 2.5 + 3.0 = 5.5 MLD. This flow must be used when evaluating hydraulic loading over the media surface.
వివరణ:
Question 19
ప్రశ్న 19
A trickling-filter distributor blends 2.0 MLD of fresh primary effluent at BOD5 = 180 mg/L with 2.0 MLD of recirculated clarified effluent at BOD5 = 30 mg/L. Assuming immediate mixing without reaction, what is the resulting BOD5 concentration?
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Explanation:
The fresh-stream BOD load is 2.0 times 180 = 360 kg/day. The recycled stream contributes 2.0 times 30 = 60 kg/day. The total applied BOD load at the distributor is 420 kg/day. Dividing by the combined flow of 4.0 MLD gives 105 mg/L. Recirculation lowers the mixed concentration here but does not eliminate the fresh influent BOD mass.
వివరణ:
Question 20
ప్రశ్న 20
A trickling filter blends 1.8 MLD of fresh wastewater at BOD5 = 220 mg/L with 1.8 MLD of recycled clarified effluent at BOD5 = 40 mg/L. Neglecting reaction during mixing, which pair of mixed concentration and fresh BOD5 load is correct?
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Explanation:
The fresh BOD5 load is 1.8 times 220 = 396 kg/day. The recycled stream contains 1.8 times 40 = 72 kg/day of BOD5. The total flow applied to the filter is 3.6 MLD and total applied BOD5 is 468 kg/day. The mixed concentration is therefore 468/3.6 = 130 mg/L. The fresh sewage contributes only 396 kg/day, illustrating the distinction between external load and mixed distributor load.
వివరణ:
Question 21
ప్రశ్న 21
Primary effluent enters a trickling-filter process at BOD5 = 160 mg/L, and the clarified final effluent has BOD5 = 40 mg/L at the same net plant flow. What is the overall BOD5 concentration-removal efficiency?
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Explanation:
The BOD5 concentration reduction is 160 minus 40 = 120 mg/L. Percentage removal equals the reduction divided by influent concentration. Dividing 120 by 160 and multiplying by 100 gives 75 percent. Equal net influent and effluent flows allow the same value to represent net BOD load removal. The calculation refers to the stated process including the final clarification step, not exclusively to oxidation on the media.
వివరణ:
Question 22
ప్రశ్న 22
Two trickling-filter treatment stages are arranged in series without bypass, dilution or recycle effects. If each stage removes 60% of the BOD5 entering that stage, what is the overall BOD5 removal fraction?
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Explanation:
After the first stage, 40 percent of the incoming BOD5 remains. The second stage removes 60 percent of that remainder, leaving 40 percent of 40 percent. Thus the final remaining fraction is 0.4 multiplied by 0.4 = 0.16. Overall removal is 1 - 0.16 = 0.84, or 84 percent. Percentages for successive stages must not be added directly.
వివరణ:
Question 23
ప్రశ్న 23
A filter receives 2.4 MLD of fresh primary effluent at 200 mg/L BOD5 and returns an equal clarified flow. Its effective media area is 600 m2 and media depth is 3.0 m. What are total applied hydraulic loading and external volumetric organic loading, respectively?
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Explanation:
The recycle flow equals the fresh 2.4 MLD, so total applied liquid is 4.8 MLD. Dividing 4800 m3/day by 600 m2 gives 8.0 m3/(m2 day). The fresh BOD5 load is 2.4 times 200 = 480 kg/day. Media volume is 600 times 3.0 = 1800 m3. The external organic loading is therefore 480/1800 = approximately 0.267 kg BOD5/(m3 day).
వివరణ:
Question 24
ప్రశ్న 24
A plant sends 4.0 MLD of fresh wastewater through two identical duty trickling filters, with a total recirculation ratio of 0.5 relative to fresh flow. Each filter has 500 m2 effective plan area and the liquid splits equally. What is the applied hydraulic loading on each filter?
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Explanation:
The total recirculated flow is 0.5 times 4.0 = 2.0 MLD. Fresh and recycled flows sum to 6.0 MLD over the two operating filters. Each identical filter receives 3.0 MLD under the equal-split assumption. Dividing 3000 m3/day by 500 m2 gives 6.0 m3/(m2 day). The duty-filter count is used here without adding any standby area.
వివరణ:
Question 25
ప్రశ్న 25
A trickling-filter scheme receives 2.0 MLD at primary-effluent BOD5 = 150 mg/L, with recycle ratio R = 1.0 and media depth fixed at 2.0 m. If design caps are 0.50 kg BOD5/(m3 day) on fresh load and 6.0 m3/(m2 day) on total hydraulic load, which minimum plan area governs?
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Explanation:
The fresh BOD5 load is 2.0 times 150 = 300 kg/day. At 0.50 kg/(m3 day), the required media volume is 600 m3, or 300 m2 at 2.0 m depth. Including equal recycle, the applied flow is 4.0 MLD. The hydraulic cap requires area of at least 4000/6.0 = 666.7 m2. Therefore approximately 667 m2 governs, subject to practical rounding and other design checks.
వివరణ:
Answer Key సమాధానాల పట్టిక
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Question 1 ప్రశ్న 1Answer: B. A trickling filter uses attached biofilm on fixed media, whereas a conventional activated-sludge system maintains biological flocs in suspension. సమాధానం: B.
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Question 2 ప్రశ్న 2Answer: D. Screening and grit removal, primary clarification, trickling filter, secondary clarification. సమాధానం: D.
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Question 3 ప్రశ్న 3Answer: A. Adequate attachment surface, suitable void space, structural durability and air passage through the bed. సమాధానం: A.
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Question 4 ప్రశ్న 4Answer: C. The distributor spreads wastewater across the media surface, while underdrains collect drainage and assist ventilation. సమాధానం: C.
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Question 5 ప్రశ్న 5Answer: B. It generally operates at higher hydraulic and organic loading, commonly using recirculation to improve wetting and flushing. సమాధానం: B.
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Question 6 ప్రశ్న 6Answer: D. The detachment of excess biofilm from the media, producing solids that must be separated from the treated liquid. సమాధానం: D.
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Question 7 ప్రశ్న 7Answer: A. (Q + Qr)/A, expressed for example in m3/(m2 day). సమాధానం: A.
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Question 8 ప్రశ్న 8Answer: C. QS/V, with consistent units such that Q in MLD and S in mg/L give QS in kg/day. సమాధానం: C.
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Question 9 ప్రశ్న 9Answer: B. The recycled liquid flow rate Qr divided by the fresh influent flow rate Q. సమాధానం: B.
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Question 10 ప్రశ్న 10Answer: D. Investigate source control, pretreatment and equalization while assessing inhibition and hydraulic loading. సమాధానం: D.
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Question 11 ప్రశ్న 11Answer: A. 4.0 m3/(m2 day). సమాధానం: A.
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Question 12 ప్రశ్న 12Answer: C. 3.0 m3/(m2 day). సమాధానం: C.
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Question 13 ప్రశ్న 13Answer: B. 1000 m2. సమాధానం: B.
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Question 14 ప్రశ్న 14Answer: D. 1000 m3. సమాధానం: D.
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Question 15 ప్రశ్న 15Answer: A. 2.5 m. సమాధానం: A.
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Question 16 ప్రశ్న 16Answer: C. 35.7 m. సమాధానం: C.
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Question 17 ప్రశ్న 17Answer: B. 0.30 kg BOD5/(m3 day). సమాధానం: B.
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Question 18 ప్రశ్న 18Answer: D. 3.0 MLD and 5.5 MLD. సమాధానం: D.
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Question 19 ప్రశ్న 19Answer: A. 105 mg/L. సమాధానం: A.
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Question 20 ప్రశ్న 20Answer: C. 130 mg/L and 396 kg/day. సమాధానం: C.
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Question 21 ప్రశ్న 21Answer: B. 75%. సమాధానం: B.
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Question 22 ప్రశ్న 22Answer: A. 84%. సమాధానం: A.
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Question 23 ప్రశ్న 23Answer: D. 8.0 m3/(m2 day) and 0.267 kg BOD5/(m3 day). సమాధానం: D.
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Question 24 ప్రశ్న 24Answer: C. 6.0 m3/(m2 day). సమాధానం: C.
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Question 25 ప్రశ్న 25Answer: A. About 667 m2, governed by the applied hydraulic-loading limit. సమాధానం: A.