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Question 1
ప్రశ్న 1
In wastewater engineering, which statement best defines chemical oxygen demand (COD) as determined by the standard dichromate method?
Explanation:
COD expresses the oxygen equivalent of substances oxidized by the prescribed chemical procedure. The standard dichromate method oxidizes many organic compounds and can also respond to certain inorganic reducing substances. Its result is reported as milligrams of oxygen per litre. COD is not a direct measurement of the dissolved oxygen already present in a sample. It also does not by itself distinguish readily biodegradable material from biologically resistant material.
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Question 2
ప్రశ్న 2
Which pairing correctly identifies the current BIS COD testing standard and the principal oxidant in its open-reflux method?
Explanation:
BIS identifies IS 3025 (Part 58):2023 as the standard covering chemical oxygen demand. Its open-reflux determination employs potassium dichromate in strongly acidic medium. The dichromate consumed is related to the oxygen equivalent of chemically oxidizable substances. IS 3025 Part 44 concerns biochemical oxygen demand rather than COD. The distinction matters because different chemical and biological tests have different response characteristics.
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Question 3
ప్రశ్న 3
Which comparison of COD and BOD testing is technically most accurate for routine wastewater assessment?
Explanation:
COD commonly uses an oxidizing reagent and heating rather than multiday microbial incubation. Results can generally be obtained within a few hours under the specified COD procedure. BOD instead measures microbial oxygen consumption over a prescribed incubation period. The Indian three-day and conventional five-day incubation conditions are not interchangeable without appropriate evidence. Neither test alone completely characterizes all aspects of wastewater pollution.
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Question 4
ప్రశ్న 4
A wastewater has BOD5/COD = 0.45, measured on comparable samples without apparent toxicity. Which interpretation is most defensible?
Explanation:
The BOD5-to-COD ratio is frequently used as a preliminary indicator of relative biological treatability. A ratio of 0.45 suggests that appreciable oxygen demand may be exerted biologically over the test period. However, BOD5 is time limited and does not represent all ultimate biodegradable demand. COD can include biologically resistant organics and inorganic reducing substances. Therefore the numerical ratio cannot be equated exactly with a biodegradable mass or COD fraction.
వివరణ:
Question 5
ప్రశ్న 5
A treatment plant operator needs a same-day indication of changing influent oxidizable load while retaining biological-process performance monitoring. Which approach is most appropriate?
Explanation:
COD is valuable for monitoring changes in oxidizable wastewater loading over a relatively short analytical period. BOD supplies information about oxygen consumption by microorganisms over the specified incubation. The two measurements therefore provide related but not identical information. Correlations between them may be developed from representative site-specific data. Replacing one with the other by assuming an identical numerical value is unjustified.
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Question 6
ప్రశ్న 6
A high-chloride industrial wastewater is tested for COD using acidic dichromate. Which interference and control are correctly paired?
Explanation:
Chloride may be oxidized during dichromate digestion and produce an artificially high COD result. Standard procedures address this interference through specified controls, including chloride complexation with mercuric sulfate where applicable. Extremely saline samples may still require specialized analysis or corrections. Mercuric sulfate is hazardous and must be handled and disposed of under laboratory safety procedures. Chloride interference should be considered before interpreting unusually high COD in saline wastewater.
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Question 7
ప్రశ్న 7
What is the principal function of silver sulfate included in the sulfuric-acid reagent of a conventional dichromate COD test?
Explanation:
Silver sulfate is used as a catalyst in the acidic dichromate COD determination. Its presence improves oxidation of certain organic compounds that otherwise react less readily. Potassium dichromate remains the principal oxidizing reagent. Sulfuric acid establishes the strongly acidic conditions required by the procedure. The reagent has no role in supplying a biological seed for BOD measurement.
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Question 8
ప్రశ్న 8
In an open-reflux COD test, the blank and digested sample receive identical dichromate doses and the remaining oxidant is titrated with ferrous ammonium sulfate (FAS). For a positive COD result, which titration relationship normally holds?
Explanation:
A reagent blank contains the specified test reagents without the wastewater oxidizable load. The wastewater sample consumes some dichromate during digestion. Consequently less residual dichromate remains to be titrated in the sample. The FAS volume for the sample is therefore ordinarily lower than that for the blank. Their difference is used to calculate the oxygen-equivalent demand.
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Question 9
ప్రశ్న 9
In a titrimetric COD determination, FAS consumption is 18.0 mL for the reagent blank and 10.0 mL for a 40.0 mL sample. If FAS normality is 0.100 N and COD = (A - B)N(8000)/V, what is the COD?
Explanation:
The titrant-volume difference is 18.0 minus 10.0, or 8.0 mL. The given titrimetric formula is COD = (A - B)N(8000)/V. Substitution gives 8.0 times 0.100 times 8000 divided by 40.0. This evaluates to 160 mg/L as oxygen. The calculation assumes the method blank and interference corrections have been appropriately addressed.
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Question 10
ప్రశ్న 10
The reagent blank needs 25.0 mL of 0.100 N FAS, while a digested 20.0 mL wastewater aliquot needs 13.0 mL. Using COD = (A - B)N(8000)/V, determine COD of the aliquot before any further dilution correction.
Explanation:
The blank-to-sample FAS difference is 25.0 minus 13.0 = 12.0 mL. COD is computed as (A - B)N(8000)/V. Inserting the stated figures yields 12.0 times 0.100 times 8000 divided by 20.0. The resulting concentration is 480 mg/L as oxygen. No additional dilution factor is applied because the question asks for the concentration of the tested aliquot.
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Question 11
ప్రశ్న 11
An industrial effluent is diluted by making 1 volume of original sample up to 5 total volumes before COD digestion. The diluted sample gives COD = 130 mg/L with negligible dilution-water COD. What is the original effluent COD?
Explanation:
The original wastewater constitutes one-fifth of the final diluted volume. The dilution factor is therefore 5. The COD measured on the diluted sample is 130 mg/L. Multiplication by the dilution factor gives 130 times 5 = 650 mg/L. This assumes the dilution water has negligible COD and the chemical determination responds proportionally.
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Question 12
ప్రశ్న 12
A wastewater sample has BOD5 = 180 mg/L and COD = 500 mg/L under comparable testing conditions. What is its BOD5/COD ratio?
Explanation:
The requested ratio is BOD5 divided by COD. Substituting the values gives 180/500. The result is 0.36, or 36% when expressed as a percentage ratio. This ratio is a screening indicator of biological oxygen demand relative to chemical oxygen demand. It must not be described as an exact fraction of total organic matter that is biodegradable.
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Question 13
ప్రశ్న 13
Two comparable wastewaters have the following results: X, BOD5 = 210 mg/L and COD = 420 mg/L; Y, BOD5 = 80 mg/L and COD = 400 mg/L. Which conclusion is most defensible?
Explanation:
The ratio for X is 210/420 = 0.50. The ratio for Y is 80/400 = 0.20. The greater ratio for X suggests a larger share of measurable short-term biological oxygen demand relative to its COD. This may indicate more favourable biological treatability under comparable test conditions. It does not guarantee complete biodegradation or establish a universal treatment efficiency.
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Question 14
ప్రశ్న 14
A wastewater treatment plant receives 3.5 MLD at an influent COD concentration of 400 mg/L. What is the COD mass loading, assuming the stated average flow and concentration represent the same period?
Explanation:
COD mass loading equals flow multiplied by COD concentration with consistent units. One MLD carrying 1 mg/L corresponds to 1 kg/day. The loading is therefore 3.5 multiplied by 400 kg/day. This gives 1400 kg/day of oxygen-equivalent COD load. The calculation does not imply that the plant must physically inject that mass of oxygen into every treatment process.
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Question 15
ప్రశ్న 15
Two streams mix without reaction or loss: stream 1 is 2 MLD at COD 300 mg/L, and stream 2 is 3 MLD at COD 500 mg/L. What is the mixed COD concentration?
Explanation:
The COD load in stream 1 is 2 times 300 = 600 kg/day. Stream 2 contributes 3 times 500 = 1500 kg/day. Total COD load is therefore 2100 kg/day. The total wastewater flow is 2 plus 3 = 5 MLD. The mixed concentration is 2100/5 = 420 mg/L.
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Question 16
ప్రశ్న 16
A treatment unit reduces COD concentration from 720 mg/L to 180 mg/L. Assuming equal influent and effluent flow rates, what percentage of COD mass is removed?
Explanation:
The reduction in concentration is 720 minus 180 = 540 mg/L. Under the stated equal-flow assumption, concentration reduction equals mass removal percentage. Divide 540 by the inlet concentration of 720. Multiplying by 100 gives 75%. If inlet and outlet flows differed appreciably, a mass-balance calculation would be needed instead.
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Question 17
ప్రశ్న 17
A receiving-water sample has measured DO of 1.5 mg/L and COD of 400 mg/L. Which interpretation correctly distinguishes the two results?
Explanation:
Dissolved oxygen is the concentration of molecular oxygen already present in water. COD is the oxygen equivalent of oxidizable substances determined chemically under prescribed conditions. The two quantities therefore describe different aspects of water quality. A high COD can indicate substantial pollution loading while a low DO can indicate oxygen stress in the receiving water. Neither result can be calculated by directly subtracting the other without a separate model and relevant data.
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Question 18
ప్రశ్న 18
An unfiltered wastewater sample has total COD of 560 mg/L, while a representative filtered aliquot has COD of 210 mg/L. Assuming comparable test performance and a defined filtration cutoff, what is the estimated particle-associated COD?
Explanation:
Total COD is measured on the specified unfiltered sample. The filtered-sample result represents COD that passes the defined filtration step. An operational particle-associated COD estimate is obtained by subtracting the filtered result from the total. The difference is 560 minus 210 = 350 mg/L. The estimate depends on sample representativeness, the filtration cutoff and consistent analytical measurements.
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Question 19
ప్రశ్న 19
An industrial wastewater has high dichromate COD but an unexpectedly low BOD5 result. Which investigation is technically justified before declaring the wastewater non-biodegradable?
Explanation:
BOD depends on active microorganisms and suitable incubation conditions. Toxic constituents or unsuitable pH may inhibit microbial oxygen consumption. Inadequate seeding can also reduce the response for some industrial wastewaters. Certain compounds can be chemically oxidizable yet biodegrade only slowly over the BOD incubation period. Treatability assessment and appropriate quality controls are therefore needed before assigning a definitive biodegradability classification.
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Question 20
ప్రశ్న 20
Why can a comparison of total BOD5 and COD be misleading when nitrification is significant during BOD incubation?
Explanation:
BOD incubation can include oxygen consumed by nitrifying organisms when nitrification occurs. This nitrogenous demand is distinct from the carbonaceous oxygen demand of biodegradable organics. Dichromate COD is a chemical oxidation assay rather than an incubation of nitrifying microorganisms. Therefore differences between total BOD5 and COD cannot always be interpreted solely as differences in biodegradable organic carbon. Reporting carbonaceous BOD with suitable nitrification control can make comparisons clearer.
వివరణ:
Question 21
ప్రశ్న 21
Which statement about COD, total suspended solids (TSS) and dissolved solids in wastewater is correct?
Explanation:
COD quantifies the chemical oxygen equivalent of oxidizable constituents. TSS is measured as retained solids under an operational filtration and drying procedure. Dissolved solids are assessed through a different residue-based measurement. Both dissolved and particulate wastewater constituents can contribute to COD. Therefore solids concentrations are not interchangeable with COD without appropriate characterization.
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Question 22
ప్రశ్న 22
A sewer carries 2.0 MLD with COD = 600 mg/L and BOD5 = 240 mg/L. What are the COD and BOD5 loads, respectively?
Explanation:
Mass loading in kg/day equals concentration in mg/L multiplied by flow in MLD. COD load is 2.0 times 600 = 1200 kg/day. BOD5 load is 2.0 times 240 = 480 kg/day. The two loads represent oxygen-equivalent measurements made by different procedures. Their numerical difference is not necessarily equal to the mass of nonbiodegradable organic compounds.
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Question 23
ప్రశ్న 23
In a titrimetric COD test, blank FAS consumption is 18.0 mL, sample aliquot is 20.0 mL, FAS normality is 0.100 N, and the sample COD is 320 mg/L. What sample FAS consumption is consistent with COD = (A - B)N(8000)/V?
Explanation:
The COD formula can be rearranged to A - B = COD times V divided by N times 8000. Substituting the specified values gives (320 times 20) divided by (0.100 times 8000). The titrant-volume difference is therefore 8.0 mL. Since the blank requires 18.0 mL, the sample needs 18.0 minus 8.0 = 10.0 mL. This is consistent with the sample having consumed some dichromate during digestion.
వివరణ:
Question 24
ప్రశ్న 24
A plant's influent has BOD5 = 200 mg/L and COD = 500 mg/L, while its effluent has BOD5 = 30 mg/L and COD = 150 mg/L. Which ratio comparison is correct?
Explanation:
Influent BOD5/COD is 200/500 = 0.40. Effluent BOD5/COD is 30/150 = 0.20. The measured ratio therefore decreases across the process. This is consistent with removal of readily biodegradable constituents leaving a relatively less biodegradable residual, under comparable test conditions. It is not an exact measurement of a nonbiodegradable mass fraction, and other sample-specific factors can affect the ratio.
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Question 25
ప్రశ్న 25
An industrial plant's discharge changes from 3.0 MLD at COD 450 mg/L to 2.0 MLD at COD 700 mg/L. What does a COD mass-load comparison show?
Explanation:
The initial COD load is 3.0 times 450 = 1350 kg/day. The new load is 2.0 times 700 = 1400 kg/day. Thus the discharge flow has decreased while the COD concentration has increased. The resulting mass load nevertheless rises by 50 kg/day. This illustrates why pollution loading cannot be inferred from concentration or flow in isolation.
వివరణ:
Answer Key సమాధానాల పట్టిక
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Question 1 ప్రశ్న 1Answer: C. The oxygen equivalent of substances chemically oxidized under specified test conditions, including some inorganic reducing substances. సమాధానం: C.
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Question 2 ప్రశ్న 2Answer: A. IS 3025 (Part 58):2023 and potassium dichromate in acidic medium. సమాధానం: A.
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Question 3 ప్రశ్న 3Answer: D. COD is determined chemically in hours, whereas the conventional BOD test requires a stated multiday incubation such as 3 days at 27°C or 5 days at 20°C. సమాధానం: D.
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Question 4 ప్రశ్న 4Answer: B. The ratio suggests appreciable biological treatability, but it is not an exact measurement of the biodegradable COD fraction. సమాధానం: B.
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Question 5 ప్రశ్న 5Answer: C. Use COD for relatively rapid chemical oxygen-demand tracking and maintain BOD or respirometric checks for biological interpretation. సమాధానం: C.
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Question 6 ప్రశ్న 6Answer: A. Chloride can cause a positive COD interference; mercuric sulfate is used in specified methods to complex chloride. సమాధానం: A.
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Question 7 ప్రశ్న 7Answer: D. It acts as a catalyst that promotes oxidation of certain organic compounds. సమాధానం: D.
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Question 8 ప్రశ్న 8Answer: B. The reagent blank requires more FAS than the sample because the sample has consumed part of the dichromate. సమాధానం: B.
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Question 9 ప్రశ్న 9Answer: C. 160 mg/L సమాధానం: C.
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Question 10 ప్రశ్న 10Answer: A. 480 mg/L సమాధానం: A.
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Question 11 ప్రశ్న 11Answer: D. 650 mg/L సమాధానం: D.
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Question 12 ప్రశ్న 12Answer: B. 0.36 సమాధానం: B.
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Question 13 ప్రశ్న 13Answer: C. X has the higher BOD5/COD ratio and may be more amenable to biological treatment, subject to treatability checks. సమాధానం: C.
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Question 14 ప్రశ్న 14Answer: A. 1400 kg/day సమాధానం: A.
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Question 15 ప్రశ్న 15Answer: D. 420 mg/L సమాధానం: D.
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Question 16 ప్రశ్న 16Answer: B. 75% సమాధానం: B.
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Question 17 ప్రశ్న 17Answer: C. DO is the oxygen currently dissolved in the water, whereas COD represents potential chemical oxygen-equivalent demand under the test. సమాధానం: C.
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Question 18 ప్రశ్న 18Answer: A. 350 mg/L సమాధానం: A.
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Question 19 ప్రశ్న 19Answer: D. Check for toxicity, pH effects, microbial seeding problems and slowly biodegradable compounds that may suppress short-term BOD. సమాధానం: D.
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Question 20 ప్రశ్న 20Answer: B. Total BOD5 may include nitrogenous microbial oxygen demand, whereas dichromate COD does not measure that biological nitrification demand in the same way. సమాధానం: B.
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Question 21 ప్రశ్న 21Answer: C. COD indicates chemical oxygen-equivalent demand, whereas TSS and dissolved-solids tests classify material by operational physical separation and residue methods. సమాధానం: C.
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Question 22 ప్రశ్న 22Answer: A. 1200 kg/day and 480 kg/day సమాధానం: A.
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Question 23 ప్రశ్న 23Answer: D. 10.0 mL సమాధానం: D.
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Question 24 ప్రశ్న 24Answer: B. BOD5/COD decreases from 0.40 to 0.20, suggesting that the residual demand is relatively less exerted biologically over five days. సమాధానం: B.
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Question 25 ప్రశ్న 25Answer: C. The COD load rises slightly from 1350 kg/day to 1400 kg/day despite the reduced flow. సమాధానం: C.
