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Question 1
ప్రశ్న 1
What does biochemical oxygen demand (BOD) primarily indicate in wastewater engineering?
Explanation:
BOD measures oxygen consumption associated with microbial oxidation under stated test conditions. Its principal application is assessing biodegradable pollution in water and wastewater. Depending on microbial activity and nitrification control, the measured demand may include nitrogenous as well as carbonaceous oxidation. BOD is different from the dissolved oxygen already present in a sample. It is also not identical to chemical oxygen demand, which is based on chemical oxidation.
వివరణ:
Question 2
ప్రశ్న 2
Which pairing correctly distinguishes the commonly used Indian BOD incubation convention from the international five-day BOD convention?
Explanation:
The Indian oxygen-depletion procedure in IS 3025 (Part 44) uses three days of incubation at 27°C. The widely used conventional BOD5 test uses five days at 20°C. Both results represent oxygen demand over a specified time and temperature. The two reported values should not be assumed numerically identical for every wastewater. A meaningful comparison must state which incubation procedure was used.
వివరణ:
Question 3
ప్రశ్న 3
Why are BOD bottles kept in the dark during incubation?
Explanation:
The BOD test estimates the decrease in dissolved oxygen during microbial activity. Exposure to light may permit photosynthetic organisms to release oxygen. Such production can offset part of the oxygen consumed and distort the test result. Incubation in darkness reduces this interference. Bottles are also carefully stoppered to prevent exchange of oxygen with the atmosphere.
వివరణ:
Question 4
ప్రశ్న 4
For a routine diluted BOD determination using the usual oxygen-depletion acceptability criteria, which bottle result is acceptable for calculation?
Explanation:
Routine diluted BOD measurements generally require a measurable oxygen depletion of at least 2.0 mg/L. A residual dissolved oxygen concentration of at least 1.0 mg/L must also remain. These conditions reduce uncertainty and avoid oxygen-limited incubation. A bottle with 3.0 mg/L depletion and 2.0 mg/L residual meets both requirements. The other options fail at least one of the stated checks.
వివరణ:
Question 5
ప్రశ్న 5
An unseeded BOD bottle contains a diluted sample with a sample-volume fraction of 0.05. Its initial and final DO are 8.6 and 4.1 mg/L after the specified incubation. Assuming negligible blank demand, what is the BOD of the original sample?
Explanation:
The observed DO depletion of the diluted sample is 8.6 minus 4.1, or 4.5 mg/L. The sample-volume fraction P is 0.05. For an unseeded valid bottle with negligible blank demand, BOD is calculated as DO depletion divided by P. Thus BOD equals 4.5/0.05 = 90 mg/L. The depleted oxygen and residual oxygen also satisfy the stated routine acceptance criteria.
వివరణ:
Question 6
ప్రశ్న 6
A seeded BOD dilution uses 10 mL of original sample in a final volume of 300 mL. Diluted-sample DO falls from 8.8 to 3.3 mg/L; seed-control DO depletion is 1.0 mg/L and the seed correction factor is 0.50. What is the seed-corrected BOD of the original sample?
Explanation:
The diluted bottle has a measured DO depletion of 8.8 minus 3.3, or 5.5 mg/L. The seed correction is 1.0 multiplied by 0.50, or 0.5 mg/L. The corrected oxygen uptake attributable to the sample is therefore 5.0 mg/L. The sample fraction is 10/300 = 1/30. Dividing 5.0 by 1/30 gives a seed-corrected BOD of 150 mg/L.
వివరణ:
Question 7
ప్రశ్న 7
A diluted BOD test bottle begins with 8.0 mg/L dissolved oxygen and ends with 0.2 mg/L. What is the most appropriate interpretation under routine minimum-residual-DO requirements?
Explanation:
The measured DO depletion is 8.0 minus 0.2, or 7.8 mg/L. However, the residual DO is below the usual 1.0 mg/L minimum. Oxygen limitation can distort the measured uptake and lead to underestimation of the sample's demand. A more dilute preparation should be analysed to obtain an acceptable residual. A large depletion by itself does not make a BOD result valid.
వివరణ:
Question 8
ప్రశ్న 8
An undiluted, unseeded surface-water sample with sufficient natural microorganisms has initial DO of 7.9 mg/L and final DO of 5.1 mg/L after the specified BOD incubation. What is its measured BOD?
Explanation:
For an undiluted and otherwise valid unseeded sample, BOD is the measured decrease in DO. The initial reading is 7.9 mg/L and the final reading is 5.1 mg/L. Their difference is 2.8 mg/L. No dilution-factor adjustment is needed because the sample fraction is one. The depletion exceeds 2.0 mg/L and the residual DO exceeds 1.0 mg/L.
వివరణ:
Question 9
ప్రశ్న 9
A wastewater has an expected BOD of 240 mg/L. For a 300 mL unseeded BOD bottle, which sample volume should theoretically yield about 4.0 mg/L DO depletion, assuming proportional response and negligible blank demand?
Explanation:
For the stated simplified calculation, diluted-bottle oxygen depletion equals original BOD multiplied by the sample fraction. The required fraction is 4.0/240 = 1/60. Multiplying 300 mL by 1/60 gives 5 mL of wastewater. The remainder of the 300 mL bottle is made up with suitable dilution water. Practical testing should still use appropriate quality controls and, where needed, several dilutions.
వివరణ:
Question 10
ప్రశ్న 10
A treatment plant receives 2.0 MLD of sewage having a BOD concentration of 200 mg/L. What is the incoming BOD mass load?
Explanation:
BOD mass loading depends on both wastewater concentration and daily flow. In convenient units, load in kg/day equals flow in MLD multiplied by BOD in mg/L. Substitution gives 2.0 times 200 = 400 kg/day. The unit conversion follows because 1 MLD at 1 mg/L carries 1 kg/day. This mass load is useful for assessing biological treatment requirements.
వివరణ:
Question 11
ప్రశ్న 11
A sewage-treatment unit lowers BOD from 250 mg/L to 40 mg/L. Assuming inlet and outlet concentrations are comparable under the same BOD test procedure, what is the percentage BOD concentration reduction?
Explanation:
The reduction in BOD concentration is 250 minus 40 = 210 mg/L. Percentage concentration reduction equals this difference divided by the influent concentration. Thus the percentage is 210/250 multiplied by 100. The result is 84 percent. A true mass-removal percentage also requires flow information if inlet and outlet flows differ.
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Question 12
ప్రశ్న 12
Two domestic wastewater samples are tested using identical BOD procedures without microbial inhibition. Sample X has BOD of 320 mg/L and sample Y has BOD of 80 mg/L. What is the most defensible conclusion?
Explanation:
Both samples were measured using the same specified BOD test. The ratio of their measured demands is 320/80 = 4. Sample X therefore shows four times the test-period oxygen demand per litre. BOD does not directly quantify pathogenic microorganism numbers. Nor does the ratio establish a corresponding ratio of COD or overall toxicity.
వివరణ:
Question 13
ప్రశ్న 13
What is the principal purpose of adding a suitable nitrification inhibitor when measuring carbonaceous BOD (CBOD)?
Explanation:
Carbonaceous BOD is associated with microbial oxidation of biodegradable organic material. Nitrifying microorganisms can also consume oxygen while oxidising reduced nitrogen compounds. An appropriate nitrification inhibitor reduces this nitrogenous contribution during the test. The resulting CBOD value better represents the carbonaceous oxygen demand. The use of an inhibitor and the reported incubation period should be identified with the result.
వివరణ:
Question 14
ప్రశ్న 14
A heavily polluted industrial wastewater gives unexpectedly low BOD although other evidence indicates a high organic load. Which explanation should be investigated first?
Explanation:
BOD is a biological assay requiring active microorganisms. Toxicants and extreme pH can inhibit microbial oxidation during incubation. Under such conditions the measured oxygen depletion may underestimate biodegradable demand. Appropriate dilution, neutralisation, seeding and quality-control checks can help investigate the cause. A low measured BOD alone does not establish that the wastewater is environmentally harmless.
వివరణ:
Question 15
ప్రశ్న 15
For an ideal first-order carbonaceous BOD model using a natural-base deoxygenation constant k in day^-1, which expression gives BOD exerted by time t from ultimate demand L0?
Explanation:
The ideal first-order model assumes that the rate of oxygen-demand exertion is proportional to the demand remaining. The remaining demand is L(t) = L0 exp(-kt). The exerted demand equals the initial ultimate demand minus the remaining demand. Hence BODt = L0[1 - exp(-kt)]. The given k uses natural exponential notation, not a base-10 rate constant.
వివరణ:
Question 16
ప్రశ్న 16
Wastewater has ultimate carbonaceous BOD L0 = 300 mg/L and a natural-base first-order deoxygenation constant k = 0.23 day^-1 at the specified temperature. What is the BOD exerted after five days?
Explanation:
The first-order exerted-demand equation is BODt = L0[1 - exp(-kt)]. Here L0 = 300 mg/L, k = 0.23 day^-1 and t = 5 days. The exponent is -1.15, giving exp(-1.15) approximately 0.31664. Thus BOD5 = 300(1 - 0.31664), approximately 205.0 mg/L. The remaining ultimate demand is approximately 95.0 mg/L and is not the quantity requested.
వివరణ:
Question 17
ప్రశ్న 17
For a wastewater with ultimate carbonaceous BOD of 200 mg/L and natural-base k = 0.20 day^-1, how much oxygen demand remains unexerted after three days?
Explanation:
Remaining first-order BOD is L(t) = L0 exp(-kt). The product kt equals 0.20 times 3 = 0.60. Therefore L(3) = 200 exp(-0.60). Its value is approximately 109.8 mg/L. The oxygen demand already exerted is approximately 90.2 mg/L, which is a different quantity.
వివరణ:
Question 18
ప్రశ్న 18
In an ideal first-order BOD process, what fraction of ultimate BOD has been exerted when the elapsed time equals the half-life ln(2)/k?
Explanation:
For first-order decay, the remaining-demand fraction is exp(-kt). At t = ln(2)/k, this fraction becomes exp[-ln(2)] = 1/2. Hence half the initial ultimate BOD remains unexerted. The exerted fraction is one minus one-half, which also equals one-half. The result is independent of the particular positive value of k.
వివరణ:
Question 19
ప్రశ్న 19
The measured five-day carbonaceous BOD is 180 mg/L at a specified temperature. If the natural-base first-order constant is k = 0.23 day^-1, what is the corresponding estimated ultimate carbonaceous BOD?
Explanation:
The first-order relation gives BOD5 = L0[1 - exp(-5k)]. Rearranging gives L0 = BOD5/[1 - exp(-5k)]. At k = 0.23 day^-1, the denominator is approximately 0.68336. Dividing 180 mg/L by this fraction yields approximately 263.4 mg/L. The estimate assumes that the specified first-order model describes the carbonaceous oxygen demand.
వివరణ:
Question 20
ప్రశ్న 20
For ideal first-order BOD exertion with natural-base k = 0.20 day^-1, approximately how much time is required to exert 75% of the ultimate BOD?
Explanation:
Exerting 75 percent means that 25 percent of ultimate BOD remains. The first-order remaining fraction is exp(-kt) = 0.25. Taking natural logarithms gives t = -ln(0.25)/k = ln(4)/k. With k = 0.20 day^-1, this equals approximately 6.93 days. This is a model result and not a fixed incubation duration for a laboratory standard.
వివరణ:
Question 21
ప్రశ్న 21
At one fixed temperature, a wastewater follows first-order carbonaceous BOD kinetics with L0 = 250 mg/L and natural-base k = 0.20 day^-1. Which pair gives the approximate BOD exerted after 3 days and 5 days, respectively?
Explanation:
The first-order expression is BODt = 250[1 - exp(-0.20t)] mg/L. At three days, the exponent is -0.60 and the exerted demand is approximately 112.8 mg/L. At five days, the exponent is -1.00 and the exerted demand is approximately 158.0 mg/L. Exerted BOD increases with time while demand remains available. Both calculated values refer to the same stated temperature, so no temperature conversion is involved.
వివరణ:
Question 22
ప్రశ్న 22
In a first-order BOD model with natural-base k = 0.25 day^-1, the oxygen demand remaining at a particular instant is 120 mg/L. What is the instantaneous rate of BOD exertion at that instant?
Explanation:
The first-order rate of exertion equals k multiplied by the demand remaining. The remaining demand is stated as 120 mg/L. Multiplication by k = 0.25 day^-1 gives 30 mg/L per day. The unit is concentration per unit time rather than merely concentration. This rate progressively decreases as the remaining demand declines under the model.
వివరణ:
Question 23
ప్రశ్న 23
Assume the empirical temperature relation kT = k20(1.047)^(T-20) for a natural-base BOD deoxygenation constant. If k20 = 0.20 day^-1, what is k30 approximately?
Explanation:
The question specifies an empirical temperature factor of 1.047 per degree Celsius. Increasing temperature from 20°C to 30°C makes T - 20 equal to 10. Therefore k30 = 0.20(1.047)^10 day^-1. The calculated value is approximately 0.3166 day^-1. This relationship is a stipulated model approximation rather than a universal temperature law for all wastewaters.
వివరణ:
Question 24
ప్రశ్న 24
A report gives BOD3 measured at 27°C for one sample and BOD5 measured at 20°C for another. Which interpretation is technically valid?
Explanation:
BOD exertion depends on both incubation duration and microbial reaction rate. Temperature affects the rate, while a longer duration permits additional oxygen demand to be exerted. The Indian three-day procedure and conventional five-day procedure consequently do not produce universally interchangeable values. An assumed kinetic model can estimate a conversion only if its parameters and applicability are justified. Reports should identify the testing convention before values are compared or used for design.
వివరణ:
Question 25
ప్రశ్న 25
Two wastewater streams mix without reaction or loss: 2 MLD at BOD 200 mg/L and 3 MLD at BOD 100 mg/L. Assume comparable BOD tests and additive oxygen-demand loads. What is the mixed BOD concentration?
Explanation:
The first stream contributes 2 times 200 = 400 kg/day of BOD load. The second contributes 3 times 100 = 300 kg/day. The combined BOD load is 700 kg/day and the total flow is 5 MLD. The mixed concentration is therefore 700/5 = 140 mg/L. This calculation assumes compatible BOD measurements and no reaction or inhibitory interaction during mixing.
వివరణ:
Answer Key సమాధానాల పట్టిక
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Question 1 ప్రశ్న 1Answer: C. The oxygen consumed by microorganisms in stabilizing biologically oxidizable substances under specified test conditions. సమాధానం: C.
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Question 2 ప్రశ్న 2Answer: A. Indian BOD3: 3 days at 27°C; BOD5: 5 days at 20°C. సమాధానం: A.
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Question 3 ప్రశ్న 3Answer: B. To avoid photosynthetic oxygen production that can distort dissolved-oxygen depletion. సమాధానం: B.
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Question 4 ప్రశ్న 4Answer: D. DO depletion of 3.0 mg/L and final DO of 2.0 mg/L. సమాధానం: D.
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Question 5 ప్రశ్న 5Answer: C. 90 mg/L సమాధానం: C.
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Question 6 ప్రశ్న 6Answer: A. 150 mg/L సమాధానం: A.
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Question 7 ప్రశ్న 7Answer: B. The bottle fails the residual-oxygen criterion and a more suitable dilution should be evaluated. సమాధానం: B.
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Question 8 ప్రశ్న 8Answer: D. 2.8 mg/L సమాధానం: D.
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Question 9 ప్రశ్న 9Answer: C. 5 mL సమాధానం: C.
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Question 10 ప్రశ్న 10Answer: A. 400 kg/day సమాధానం: A.
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Question 11 ప్రశ్న 11Answer: B. 84% సమాధానం: B.
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Question 12 ప్రశ్న 12Answer: D. Sample X exhibits four times the measured biochemical oxygen demand per unit volume under the stated test. సమాధానం: D.
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Question 13 ప్రశ్న 13Answer: C. To suppress nitrogenous oxygen demand so that the measured uptake primarily reflects carbonaceous oxidation. సమాధానం: C.
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Question 14 ప్రశ్న 14Answer: A. Toxic substances or unsuitable pH may have inhibited the microorganisms used in the BOD bioassay. సమాధానం: A.
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Question 15 ప్రశ్న 15Answer: B. BODt = L0[1 - exp(-kt)] సమాధానం: B.
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Question 16 ప్రశ్న 16Answer: D. 205.0 mg/L సమాధానం: D.
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Question 17 ప్రశ్న 17Answer: C. 109.8 mg/L సమాధానం: C.
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Question 18 ప్రశ్న 18Answer: A. 50% సమాధానం: A.
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Question 19 ప్రశ్న 19Answer: B. 263.4 mg/L సమాధానం: B.
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Question 20 ప్రశ్న 20Answer: D. 6.93 days సమాధానం: D.
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Question 21 ప్రశ్న 21Answer: C. 112.8 mg/L and 158.0 mg/L సమాధానం: C.
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Question 22 ప్రశ్న 22Answer: A. 30 mg/L per day సమాధానం: A.
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Question 23 ప్రశ్న 23Answer: B. 0.317 day^-1 సమాధానం: B.
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Question 24 ప్రశ్న 24Answer: D. The methods have different time-temperature conditions, so conversion or comparison requires appropriate sample-specific assumptions or evidence. సమాధానం: D.
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Question 25 ప్రశ్న 25Answer: C. 140 mg/L సమాధానం: C.
