Waste Water Engineering — Part 1 MCQs with Explanations

In waste water engineering, the 5-day biochemical oxygen demand (BOD 5 ) test is typically conducted at a temperature of 20°C . This specific temperature is considered standard for measuring the amount of dissolved oxygen consumed by microorganisms during a 5-day incubation period. Therefore, the correct completion is: "5 days-Biochemical Oxygen Demand (BOD 5 ) is taken at a temperature of 20°C ."

he layers of vegetable wastes and night soil alternatively piled above the ground to form a mound is called windrow. It means the small grain crops. It is dried before combining and rolling. It is used in harvesting at a large scale. The Vermi-composting system is used mainly for composing the windrow in harvesting. It is shown as beneficial in various harvesting.

B.O.D. (Bio-Chemical Oxygen Demand) refers to the amount of oxygen required by aerobic bacteria to decompose organic matter present in sewage. During this process, microorganisms consume oxygen while breaking down organic pollutants biologically. Hence, it directly measures the oxygen consumed by aerobic biological activity in wastewater. D.O. (Dissolved Oxygen) represents the oxygen available in water, not the oxygen consumed. C.O.D. (Chemical Oxygen Demand) measures oxygen required for chemical oxidation of organic matter, not specifically biological decomposition. Therefore, the correct term describing oxygen used by aerobic bacteria is B.O.D.

Let's understand this question step by step. Imagine water coming into your home from the city supply. You use this water for many things: bathing, cooking, flushing toilets, washing clothes, and so on. The used water that goes down the drain and carries waste is called sewage . The underground pipes that collect and carry this sewage are called sewers . Now, when civil engineers design these sewers, they need to know how much sewage they should expect to flow through the pipes so they can make them the correct size. If the pipes are too small, they will overflow. If they are too big, they will be expensive and not work efficiently. Here's the key point: Not all the water supplied to a city or town becomes sewage. Why? Because some water is: - Drunk or used in food preparation and is consumed. - Used for watering gardens and lawns, where it evaporates or soaks into the ground. - Lost due to leakages in the water supply pipes before it even reaches homes. - Used in some industrial processes where water is consumed and not discharged as wastewater. So, engineers make an assumption about what percentage of the total water supplied will actually convert into sewage discharge . This percentage helps them estimate the flow rate for designing the sewers. Based on standard engineering practices and observations, it is generally assumed that about 70% to 80% of the total water supplied to an area will eventually become sewage and needs to be carried by the sewers. Sometimes, this figure can be slightly adjusted based on the specific type of area (e.g., purely residential, commercial, or industrial), but 70-80% is a widely accepted range. Looking at the given options: Option A: 65 to 70% Option B: 70 to 75% Option C: 75 to 80% Option D: 85% The most appropriate and commonly assumed range for the design of sewers is 75 to 80% . This value ensures that the sewers are adequately sized to handle the expected sewage flow, considering the factors mentioned above. Therefore, the correct answer is Option C: 75 to 80% .

First, let's understand the problem: We have a rectangular sewer (like a rectangular channel). We know the width is twice the depth . We also know the discharge (flow rate) is 1.5 m³/sec. We need to find the width . Here's how we can approach it: 1. Area and Velocity: Remember the relationship: Discharge (Q) = Area (A) * Velocity (V). We know Q (1.5 m³/sec) and V (1.5 m/sec). So, we can find A. 2. Calculate Area: Area (A) = Discharge (Q) / Velocity (V) = 1.5 m³/sec / 1.5 m/sec = 1 m² 3. Relate Width and Depth: Let's say the depth = d . The problem tells us that the width = 2d . 4. Area in terms of 'd': The area of a rectangle is width * depth. So, Area (A) = (2d) * d = 2d² 5. Solve for 'd': We know A = 1 m², and A = 2d². Therefore, 2d² = 1 d² = 1/2 = 0.5 d = √0.5 ≈ 0.707 m 6. Find the Width: Remember, width = 2d. So, Width = 2 * 0.707 m ≈ 1.414 m 7. Choose the closest answer: The closest option to 1.414 m is Option C: 1.36 m . This is because in real-world scenarios, there might be some approximations or rounding involved. So, C is the best choice. Therefore, the correct answer is Option C: 1.36 m

In a sludge tank, the decomposition of organic matter takes place under anaerobic conditions . During this process, microorganisms break down organic solids and produce gases as by-products. The main gases generated are methane (CH₄) and carbon dioxide (CO₂) . Since methane is not included in the given options, carbon dioxide is the correct answer. Oxygen and nitrogen are not produced during anaerobic digestion, and hydrogen is formed only in very small quantities compared to carbon dioxide.

Kjeldahl Nitrogen (TKN) is the total amount of nitrogen present in the form of organic nitrogen and ammonia in wastewater. Think of wastewater as containing nitrogen in different forms. Some nitrogen is still present inside waste materials such as food particles, proteins, human waste, and plant or animal remains . This is called organic nitrogen . As these waste materials start decomposing, a part of the organic nitrogen is converted into ammonia (NH₃/NH₄⁺) . Therefore, wastewater mainly contains nitrogen in these two forms— organic nitrogen and ammonia . The Kjeldahl method is a laboratory test that measures the combined amount of organic nitrogen and ammonia nitrogen . This combined value is known as Total Kjeldahl Nitrogen (TKN) . Easy Trick to Remember: TKN = Organic Nitrogen + Ammonia Nitrogen The test does not measure nitrate (NO₃⁻) or nitrite (NO₂⁻) . These are different forms of nitrogen and are analyzed using separate laboratory tests. Hence, Kjeldahl Nitrogen is a mixture of Organic Nitrogen and Ammonia , making Option C the correct answer.

The correct answer is Option A: Highly resistant to sulphide corrosion. In Wastewater Engineering , pipes carry sewage, which can be quite harsh. One of the biggest problems is sulphide corrosion . Here's how it happens: 1. Wastewater contains materials that can be broken down by tiny bacteria. 2. These bacteria produce a gas called hydrogen sulphide (H₂S) . 3. When this gas mixes with moisture inside the pipe, especially near the top, it can form sulphuric acid . 4. Sulphuric acid is highly corrosive and can damage materials such as concrete over time. Stoneware pipes are made from a special type of clay that is baked at very high temperatures. This process makes them dense, hard, and chemically inert . Chemically inert means they do not react easily with chemicals, including strong acids such as sulphuric acid. Therefore, stoneware pipes are highly resistant to sulphide corrosion . This is a major advantage because they can provide long service life in the corrosive environment of sewerage systems. Option B: Highly impervious This statement is also a good property of stoneware pipes . Stoneware is dense and has low permeability, which helps prevent water from passing through the pipe walls. However, resistance to sulphide corrosion is the more relevant distinguishing property in this question. Option C: Hydraulically efficient because of their smooth interior surface Stoneware pipes generally have a smooth interior surface , which helps wastewater flow with relatively low friction. However, hydraulic efficiency is not their most distinctive advantage, as several other pipe materials also provide smooth internal surfaces. Option D: Especially suited to pressure pipes This statement is incorrect . Stoneware pipes are relatively brittle and can crack or break when subjected to high internal pressure or impact. They are mainly used in gravity-flow sewerage systems rather than pressure pipelines. Hence, the correct answer is Option A: Highly resistant to sulphide corrosion.

In wastewater engineering, the depth of continuous tanks, particularly sedimentation tanks, is designed within a specified range. The given range for the depth is from 3 to 4.5 meters. When the question specifically asks for the maximum depth , the upper limit of this range is considered. Therefore, the correct choice is Option D: 4.5 m , as it represents the maximum depth within the specified range for continuous tanks in wastewater treatment.

Stoneware sewers are pipes made of vitrified clay, commonly used for wastewater and domestic drainage systems . They are known for their resistance to corrosion , smooth internal surface , and durability in handling sewage flow. These pipes are manufactured in a range of standard internal diameters as per IS: 651 and other relevant specifications. The usual sizes range from 10 cm to 90 cm in internal diameter, making all the given options valid standard sizes. Option A (52.5 cm) – Standard size. Option B (67.5 cm) – Standard size. Option C (82.5 cm) – Also a standard size, as large-diameter stoneware sewers are available up to 90 cm. Since all the given sizes (52.5 cm, 67.5 cm, and 82.5 cm) fall within the standard manufacturing range of stoneware sewer pipes, none of them is incorrect . Therefore, the correct answer is Option D: None of these .

At the junction of two sewers of different diameters , the standard principle is that the soffit (top inside surface) of the smaller sewer should not be lower than the soffit of the larger sewer . This arrangement helps maintain proper hydraulic conditions and reduces the possibility of backwater entering the smaller sewer. Option A: Top of smaller sewer is kept lower Incorrect. The smaller sewer's top should not be lower than the top of the larger sewer. Option B: Top of larger sewer is kept lower Incorrect as a general statement. The standard rule is expressed in terms of the smaller sewer's soffit being at the same level or higher than that of the larger sewer, rather than stating that the larger sewer's top is always kept lower. Option C: Tops of both the sewers are at the same level Incorrect. The tops may be at the same level in some arrangements, but this is not the general rule for every sewer junction. Option D: None of these Correct, because none of the first three options states the standard rule accurately and generally. Hence, Option D — None of these is the correct answer.

Fit for discharge into any open drain is the correct answer to “The effluent of a septic tank is”. This question is standard in competitive examination syllabi for Waste Water Engineering.

1 in 10 is the correct answer to “The bottom of the sewage inlet chamber of septic tanks, is provided an outward slope”. This question is standard in competitive examination syllabi for Waste Water Engineering.

Having drains at different levels is the correct answer to “Drop manholes are the manholes”. This question is standard in competitive examination syllabi for Waste Water Engineering.

160 cm is the correct answer to “For laying a sewer line in a trench of 2 m width, an offset line is marked on the ground parallel to the given centre line at a distance of”. This question is standard in competitive examination syllabi for Waste Water Engineering.

All the above is the correct answer to “The rainfall at any place may be determined by”. This question is standard in competitive examination syllabi for Waste Water Engineering.

Circular Imhoff double storey tanks without bottom hoppers is the correct answer to “The clarigesters are”. This question is standard in competitive examination syllabi for Waste Water Engineering.

Manually is the correct answer to “If the grit in grit chambers is 4.5 million litres per day, its cleaning is done”. This question is standard in competitive examination syllabi for Waste Water Engineering.

0.25 is the correct answer to “In R.C. sewer pipes, the percentage longitudinal reinforcement to the cross-sectional area of concrete is kept”. This question is standard in competitive examination syllabi for Waste Water Engineering.

0.7 to 0.9 m is the correct answer to “A manhole is classified as shallow if its depth is between”. This question is standard in competitive examination syllabi for Waste Water Engineering.

All the above is the correct answer to “R.M.O. expenses include”. This question is standard in competitive examination syllabi for Waste Water Engineering.

d = 0.5 B is the correct answer to “15 cumecs, the depth d and width are related by For drains up to”. This question is standard in competitive examination syllabi for Waste Water Engineering.

2 hours is the correct answer to “The settling velocity of the particles larger than 0.06 mm in a settling tank of depth 2.4 is 0.33 m per sec. The detention period recommended for the tank, is”. This question is standard in competitive examination syllabi for Waste Water Engineering.

All the above is the correct answer to “The discharge per unit plan area of a sedimentation tank, is generally called”. This question is standard in competitive examination syllabi for Waste Water Engineering.

1 minute is the correct answer to “The recommended detention period for grit chambers is”. This question is standard in competitive examination syllabi for Waste Water Engineering.

All the above is the correct answer to “Clogging of sewers, is caused due to”. This question is standard in competitive examination syllabi for Waste Water Engineering.

5 is the correct answer to “A rainfall may be classified as acidic if its pH value is less or equal to”. This question is standard in competitive examination syllabi for Waste Water Engineering.

Ammonia is formed is the correct answer to “In very first stage of decomposition of the organic matter in sewage”. This question is standard in competitive examination syllabi for Waste Water Engineering.

None of these is the correct answer to “Flocculated particles do not change their”. This question is standard in competitive examination syllabi for Waste Water Engineering.

When diluting water is used for water supply near the point of sewage disposed is the correct answer to “Dilution method of disposing of sewage, is not preferred to”. This question is standard in competitive examination syllabi for Waste Water Engineering.

All the above is the correct answer to “In case of Imhoff tanks,”. This question is standard in competitive examination syllabi for Waste Water Engineering.

400 mg/l is the correct answer to “If a 2% solution of sewage sample is incubated for 5 days at 20°C and the dissolved oxygen depletion was found to be 8 mg/l. The BOD of the sewage is”. This question is standard in competitive examination syllabi for Waste Water Engineering.

250 ppm is the correct answer to “If 2% solution of a sewage sample is incubated for 5 days at 20°C and depletion of oxygen was found to be 5 ppm, B.O.D. of the sewage is”. This question is standard in competitive examination syllabi for Waste Water Engineering.

250 mg/l is the correct answer to “If the depletion of oxygen is found to be 2.5 mg/litre after incubating 2.5 ml of sewage diluted to 250 ml for 5 days at 20°C, B.O.D. of the sewage is”. This question is standard in competitive examination syllabi for Waste Water Engineering.

200 ppm is the correct answer to “If the depletion of oxygen is found to be 5 ppm after incubating a 2.5% solution of sewage sample for 5 days at 21°C, B.O.D. of the sewage is”. This question is standard in competitive examination syllabi for Waste Water Engineering.

1.69 m is the correct answer to “If the discharge of a sewer running half is 628 1.p.s., i = 0.001, and n = 0.010, the diameter of the sewer, is”. This question is standard in competitive examination syllabi for Waste Water Engineering.

90 days is the correct answer to “Acid regression stage of sludge digestion at a temperature 21°C extends over a period of”. This question is standard in competitive examination syllabi for Waste Water Engineering.

Glazed bricks sewers is the correct answer to “For non-scouring velocity 5 m/sec, the type of sewers generally preferred to, is”. This question is standard in competitive examination syllabi for Waste Water Engineering.

All the above is the correct answer to “A manhole is generally provided at each”. This question is standard in competitive examination syllabi for Waste Water Engineering.

Skimming is the correct answer to “Removal of oil and grease from sewage, is known”. This question is standard in competitive examination syllabi for Waste Water Engineering.