Environmental Studies · Chemistry

Water and Wastewater Treatment

1,345 Questions

This topic deals with water purification, wastewater treatment methods, and industrial effluent management. It includes questions on filtration, disinfection, and water hardness removal. These concepts are tested in environmental science and engineering exams.

Water purificationWastewater treatmentBioremediation strategiesWater hardness removalIndustrial effluent management

Water and Wastewater Treatment Questions

Multiple choice

Match Group I (Terminology) with Group II (Definition / Brief Description) for wastewater treatment systems.

 
Group I Group II
P. Primary treatment 1. Contaminant removal by physical forces
Q. Secondary treatment 2. Involving biological and / or chemical reaction
R. Unit operation 3. Conversion of soluble organic matter to business
S. Unit process 4. Removal of solid materials from incoming wastewater

  1. P - 4, Q - 3, R - 1, S - 2

  2. P - 4, Q - 3, R - 2, S - 1

  3. P - 3, Q - 4, R - 2, S - 1

  4. P - 1, Q - 2, R - 3, S - 4
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

 

Multiple choice

An analysis for determination of solids in the return sludge of activated sludge process was done as follows: (1) A crucible was dried to a constant mass of 62.485 g. (2) 75 ml of a well mixed sample was taken in the crucible. (3) The crucible with the sample was dried to a constant mass of 65.020 g in a drying oven at 104o C. The crucible with the dried sample was placed in a muffle furnace at 600o C for an hour. After cooling, the mass of the crucible with residues was 63:145 g.

The concentration of organic fraction of solids present in the return sludge sample is

  1. 8800 mg/l

  2. 25000 mg/l

  3. 33800 mg/l

  4. 42600 mg/l
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The volatile suspended solids, representing the organic fraction, are determined by the loss on ignition between 104 degrees C and 600 degrees C. The initial dry mass of residue at 104 C is 65.020 g - 62.485 g = 2.535 g (or 2535 mg). After ignition at 600 C, the fixed residue is 63.145 g - 62.485 g = 0.660 g (or 660 mg). The volatile solids are 2535 mg - 660 mg = 1875 mg in 75 ml, which translates to 1875 / 0.075 = 25000 mg/l.

Multiple choice

A portion of wastewater sample was subjected to standard BOD test (5 days, 20 oC), yielding a value of 180 mg/l. The reaction rate constant (to the base ‘e’) at 20 oC was taken as 0.18 per day. The reaction rate constant at other temperature may be estimated by kr = k20 (1.047)T-20. The temperature at which the other portion of the sample should be tested, to exert the same BOD in 2.5 days, is

  1. 4.9 oC

  2. 24.9 oC

  3. 31.7 oC

  4. 35.0 oC
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Using the BOD equation BOD_t = L(1 - exp(-kt)), we set the BOD equal for two conditions. Since the BOD is the same, k1*t1 = k2*t2. Solving for k2 at 2.5 days gives k2 = 0.36. Using the temperature correction formula 0.36 = 0.18 * (1.047)^(T-20) leads to T = 35 degrees Celsius.

Multiple choice

The following data are given for a channel-type grit chamber of length 7.5 m.

  1. Flow-through velocity = 0.3 m/s
  2. The depth of wastewater at peak flow in the channel = 0.9 m
  3. Specific gravity of inorganic particles = 2.5
  4. g = 9.80 m/s2, m = 1.002 x 10-3 N-s/m2 at 20 oC, $\rho w$ = 1000 kg/m3 Assuming that the Stokes is valid, the largest diameter particle that would be removed with 100 percent efficiency is

  1. 0.04 mm

  2. 0.21 mm

  3. 1.92 mm

  4. 6.64 mm
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Using Stokes' law, settling velocity vs = g(G-1)d^2 / (18*nu). The removal efficiency is 100% if vs >= depth / (length / velocity). Solving for d gives approximately 0.21 mm.

Multiple choice

The results of analysis of a raw water sample are given below:

 
Turbidity 5 mg/l
pH 7.4
Fluorides 2.5 mg/l
Total Hardness 300 mg/l
Iron 3.0 mg/l
MPN 50 per 100 ml

From the data given above, it can be inferred that water needs removal of

  1. turbidity followed by disinfection

  2. fluorides and hardness

  3. iron, followed by disinfection

  4. both (2) and (3)
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

 Fluoride should be less than 1.5 mg/L, hardness between 75 - 115 mg/L as $CaCO_3$ iron 0.3 mg/L and MPN 1 per 100 mL (presence of pathogenic bacteria). Hence, water needs removal of fluoride, hardness iron and followed by disinfection to reduce MPN to 1 per 100 mL.

Multiple choice

Setting test on a sample drawn from Aeration Tank liquor of ASP (MLSS = 2800 mg/I) was carried out with 1 litre sample. The test yielded a settled volume of 200 ml. The value of Sludge Volume Index shall be

  1. 14.0

  2. 34.2

  3. 71.4

  4. 271
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Sludge Volume Index (SVI) is defined as the volume in milliliters occupied by 1 gram of activated sludge after 30 minutes of settling. SVI = (Settled sludge volume in ml/l * 1000) / MLSS in mg/l. Plugging in the values: (200 * 1000) / 2800 = 71.4 ml/g.

Multiple choice

Match the following:

Group I (Type of water impurity) Group II (Method of treatment)
P. Hardness 1. Reverse Osmosis
Q. Brackish water from sea 2. Chlorination
R. Residual MPN from filters 3. Zeolite Treatment
S. Turbidity 4. Coagulation and Flocculation
5. Coagulation, Flocculation and Filtration

  1. P - 1, Q - 2, R - 4, S - 2

  2. P - 3, Q - 2, R - 2, S - 4

  3. P - 2, Q - 1, R - 3, S - 5

  4. P - 3, Q - 1, R - 2, S - 5
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Hardness is treated by Zeolite (3), Brackish water by Reverse Osmosis (1), Residual MPN by Chlorination (2), and Turbidity by Coagulation, Flocculation, and Filtration (5). This matches option D.

Multiple choice

A water treatment plant treating 10 mld of water requires 20 mg/I of filter Alum, Al2 (SO4)3.18 H2O. The water has 6 mg/l of alkalinity as CaCo3 (Al = 26.97, S = 32, O = 16, H = 1, Ca = 40, and C = 12).

Total alkalinity requirement (106 mg per day as CaCO3) matching filter Alum, shall be

  1. 180

  2. 120

  3. 90

  4. 60
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The alkalinity requirement is based on the stoichiometry of the reaction between alum and alkalinity. 1 mg of alum reacts with 0.45 mg of alkalinity as CaCO3. For 20 mg/l of alum and 10 mld, the calculation yields 90 mg/l equivalent.

Multiple choice

A water treatment plant treating 10 mld of water requires 20 mg/I of filter Alum, Al2 (SO4)3. 18 H2O. The water has 6 mg/l of alkalinity as CaCo3 (Al = 26.97, S = 32, O = 16, H = 1, Ca = 40, and C = 12).

Quantity of Quick Lime required (106 mg per year as CaO) shall be

  1. 2132

  2. 3000

  3. 4132

  4. 6132
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Quick lime (CaO) is added to provide the necessary alkalinity. Using the stoichiometric relationship between alum and lime, the required amount is calculated based on the daily dosage and converted to an annual requirement.

Multiple choice

A conventional Activated Sludge Plant treating 1000 m3/d of municipal waste water disposes of its anaerobically digested sludge on relatively impervious farmland. Use the following data:

1. Raw Sewage SS = 225 mg/I (70% volatile)

BOD = 190 mg/I (Excess activated sludge returned to primary) | | 2. Primary Setting| SS-50% removal BOD-30% removal| | 3. Excess Activated Sludge applied (80% Volatile of total)| 0.4 g VSS produced per g BOD| | 4. Anaerobic Digester| VSS reduced 50% Digested Sludge Concentration – 60% Sludge Specific Gravity-1| | 5. Application on farmland| 2 m3 /ha.d|

Total volatile suspended solids to be anaerobically digested (kg/d, VSS) shall be

  1. 133

  2. 168

  3. 233

  4. 245
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Total VSS is the sum of VSS from primary sludge and excess activated sludge. Primary sludge VSS is calculated from SS removal and volatile fraction. Excess activated sludge is 0.4 * BOD removed. Summing these gives 133 kg/d.

Multiple choice

Zero hardness of water is achieved by

  1. lime soda process

  2. excess lime treatment

  3. iron exchange treatment

  4. excess alum and lime treatment
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Zero hardness, other than ion exchange or base exchange or zeolite method, leaves behind a substantial amount of hardness or alkalinity. Only by means of zeolite process can zero hardness of water be achieved.

Multiple choice

The aquariums are disinfected by

  1. potassium peroxymonosulphate

  2. potassium permanganate

  3. peracetic acid

  4. ozone
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

It is utilised to disinfect aquariums and is also mainly employed in it. It is used to disinfect aquariums and is also widely used in public swimming pools to disinfect people's feet before using the pool.

Multiple choice

What is the reason for using alum for purifying the water?

  1. It forms complex with clay particles.

  2. It coagulates the mud particles.

  3. It exchanges Ca2+ and Mg2+ ions present in hard water.

  4. Its sulphate ion is a water purifier.

  5. None of these
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Mud is colloidal charged solution which on treatment with Alum (Al3+) gets neutralised or coagulated.