TGPSC HMWSSB Manager Environmental Engineering previous year questions with answers
Every Environmental Engineering question asked in the TGPSC HMWSSB Manager (Engineering) papers on this site, grouped by topic. Answers come from TSPSC's final key; each has a short explanation.
Practise Environmental Engineering onlineWater Supply
A sedimentation tank 6m wide, 15m long and 3m water depth is treating 2MLD of water. The surface overflow rate in lt/hr/m² is
- 858
- 926
- 1028
- 748
Show answer
Answer: (2) 926
Surface overflow rate = Q/(plan area) = 2×10⁶ litres/day ÷ (6 × 15 = 90 m²) = 22,222 lt/day/m² ÷ 24 ≈ 926 lt/hr/m²; the depth does not enter the calculation.
The efficiency of a sediment removal in a continuous sedimentation tank does not depend upon the
- Discharge through the tank
- Width of the tank
- Length of the tank
- Depth of the tank
Show answer
Answer: (4) Depth of the tank
For an ideal continuous-flow settling tank, a particle is removed if its settling velocity is at least the overflow rate Q/(B × L), so efficiency depends on discharge and plan area (length and width) but not on depth.
The disinfection efficiency of chlorine in water treatment
- Is not dependant on pH value
- Is increased by increase in pH value
- Remains constant at all pH values
- Is reduced by increase in pH value
Show answer
Answer: (4) Is reduced by increase in pH value
Chlorine forms hypochlorous acid (HOCl), which dissociates into the much weaker hypochlorite ion (OCl⁻) as pH rises; so disinfection is best at low pH (below about 7) and falls as pH increases.
If total hardness of water is less than its total alkalinity the non-carbonate hardness will be equal to
- Total alkalinity
- Total hardness
- Total alkalinity - total hardness
- Zero
Show answer
Answer: (4) Zero
When total hardness ≤ total alkalinity, all the hardness is carbonate hardness (carbonate hardness = total hardness), so non-carbonate hardness is zero; the extra alkalinity is due to sodium/potassium salts.
If for diluting 25ml of water sample 175ml of taste free water is required to be added to make the water sample to just lose its taste then the Flavour Threshold Number (FTN) will be
- 6
- 7
- 8
- 9
Show answer
Answer: (3) 8
FTN = (A + B)/A, where A is the sample volume and B the taste-free dilution water, so FTN = (25 + 175)/25 = 8.
In network of pipes
- The algebraic sum of discharges around each circuit is zero
- The algebraic sum of head losses around each circuit is zero
- The elevation of hydraulic grade line is assumed for each junction point
- Elementary circuits are replaced by equivalent pipes
Show answer
Answer: (2) The algebraic sum of head losses around each circuit is zero
In a pipe network (Hardy Cross method), the algebraic sum of head losses around every closed circuit (loop) must be zero, because the pressure head at a junction has a single value; the algebraic sum of discharges is zero at each junction (continuity), not around a circuit.
Sewerage & Wastewater Treatment
A 2% solution of a sewage sample is kept at an incubation temperature of 20°C. If initial DO and final DO values after 5 days of incubation period are 8.5 mg/lit and 5.5 mg/lit respectively, then the BOD will be
- 50mg/lit
- 150mg/lit
- 250mg/lit
- 350mg/lit
Show answer
Answer: (2) 150mg/lit
BOD₅ = (DO_initial − DO_final)/dilution fraction = (8.5 − 5.5)/0.02 = 150 mg/lit.
If a sewer carrying a discharge of 3 cumecs outfalls into a river having a discharge of 10 cumecs and DO equal to 9.1mg/lt, the resultant DO of the mix will be equal to
- 5mg/lt
- 6mg/lt
- 7mg/lt
- 8mg/lt
Show answer
Answer: (3) 7mg/lt
Taking the DO of raw sewage as zero, the mixed DO = (10 × 9.1 + 3 × 0)/(10 + 3) = 91/13 = 7 mg/lt.
The natural process under which the flowing river gets cleaned is known as
- Oxidation
- Photosynthesis
- Reduction
- Self-purification
Show answer
Answer: (4) Self-purification
Self-purification is the combined natural action of dilution, sedimentation, oxidation by bacteria, reaeration, sunlight and photosynthesis by which a river cleans itself; oxidation and photosynthesis are only parts of it.
Recirculation in activated sludge process is done to
- Dilute the incoming sewage
- Dampen the effect of the flow variation
- Operate the plant continuously
- Supply seed materials to the aeration tank
Show answer
Answer: (4) Supply seed materials to the aeration tank
Part of the settled sludge from the secondary clarifier is returned to the aeration tank so that the incoming sewage is seeded with an active population of acclimatised micro-organisms, keeping the MLSS at the required level.
Lower Food to Micro-organism (F/M) ratio in a conventional activated treatment plant will mean
- Lower BOD removal
- Higher BOD removal
- No effect on BOD removal
- Medium BOD removal
Show answer
Answer: (2) Higher BOD removal
At a low F/M ratio the micro-organisms are starved of food relative to their number, so they consume the organic matter more completely, giving higher BOD removal (as in extended aeration), though with a larger tank.
The relative stability of a sewage sample whose dissolved oxygen levels equals the total oxygen required to satisfy its BOD is
- Zero
- 1%
- 100%
- Infinity
Show answer
Answer: (3) 100%
Relative stability is the oxygen available in the sample expressed as a percentage of the oxygen needed to satisfy its BOD, so when the two are equal the relative stability is 100%.
Solid Waste & Pollution
The dispersion of pollutants in atmosphere is maximum when
- Environmental lapse rate is greater than adiabatic lapse rate
- Environmental lapse rate is less than adiabatic lapse rate
- Environmental lapse rate is equal to adiabatic lapse rate
- Maximum mixing depth is equal to zero
Show answer
Answer: (1) Environmental lapse rate is greater than adiabatic lapse rate
When the environmental lapse rate exceeds the adiabatic lapse rate (super-adiabatic condition), the atmosphere is unstable, rising air keeps rising and vertical mixing is strongest; a lower environmental lapse rate gives stable air and poor dispersion.
Which one of the following pollutant (or) pairs of pollutants is formed due to photochemical reactions
- CO alone
- O₃ and PAN
- PAN and NH₃
- NH₃ and CO
Show answer
Answer: (2) O₃ and PAN
Ozone and peroxyacetyl nitrate (PAN) are secondary pollutants formed when sunlight acts on nitrogen oxides and hydrocarbons (photochemical smog); CO and NH₃ are emitted directly as primary pollutants.