TGPSC HMWSSB Manager Water Resources & Irrigation previous year questions with answers
Every Water Resources & Irrigation 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 Water Resources & Irrigation onlineHydrology
In a catchment there are six rain gauge stations with average depth of rainfall of 92.8cm and standard deviation of the rainfall is recorded in these rain gauge stations is 30.7cm. For a 10% degree of error in the measurement of mean rainfall, the optimum number of stations required is
- 5 nos
- 6 nos
- 10 nos
- 11 nos
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Answer: (4) 11 nos
Coefficient of variation C_v = 100 × 30.7/92.8 = 33.08%. Optimum number of stations N = (C_v/ε)² = (33.08/10)² = 10.9, which is rounded up to 11 stations.
The area between the two isohyets 45cm and 55cm is 100km² and that between 55cm and 65cm is 150km². What is the average depth of annual precipitation over the basin of 250km²
- 50cm
- 52cm
- 56cm
- 60cm
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Answer: (3) 56cm
By the isohyetal method each strip takes the mean of its bounding isohyets (50 cm and 60 cm), so P = (100 × 50 + 150 × 60)/250 = 14000/250 = 56 cm.
The total observed runoff volume during a 4hr storm with a uniform intensity of 2.8cm/hr is 25.2×10⁶ m³ from a basin of 280 km² area. What is the average infiltration rate for the basin
- 3.6mm/hr
- 4.8mm/hr
- 5.2mm/hr
- 5.5mm/hr
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Answer: (4) 5.5mm/hr
Rainfall = 2.8 × 4 = 11.2 cm; runoff depth = 25.2×10⁶ / (280×10⁶) = 0.09 m = 9 cm; so infiltration = 11.2 − 9 = 2.2 cm in 4 h = 0.55 cm/h = 5.5 mm/hr.
A volume of 3×10⁶ m³ of groundwater was pumped out from an unconfined aquifer, uniformly over an area of 5km². The pumping lowered the water table from initial level of 102m to 99m. The specific yield of the aquifer is
- 0.20
- 0.30
- 0.40
- 0.50
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Answer: (1) 0.20
Specific yield = volume drained / (area × fall of water table) = 3×10⁶ / (5×10⁶ × 3) = 0.20.
A hyetograph is a plot of
- Cumulative rainfall Vs time
- Rainfall intensity Vs time
- Rainfall depth Vs duration
- Discharge Vs time
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Answer: (2) Rainfall intensity Vs time
A hyetograph shows rainfall intensity against time (usually as a bar diagram); cumulative rainfall against time is a mass curve and discharge against time is a hydrograph.
The unit of intrinsic permeability is
- cm/day
- m/day
- Darcy/day
- cm²
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Answer: (4) cm²
Intrinsic permeability k = Kμ/γ depends only on the pore geometry of the medium and has dimensions of L² (cm², or the darcy); cm/day and m/day are units of hydraulic conductivity, which also depends on the fluid.
If S_y = specific yield and S_r = specific retention then
- S_y + S_r = void ratio
- S_y + S_r = porosity
- S_y + S_r = 1.0
- S_y + S_r = permeability
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Answer: (2) S_y + S_r = porosity
Of the total pore volume, the part that drains by gravity is the specific yield and the part held back against gravity is the specific retention, so S_y + S_r = porosity (n), not void ratio.
Irrigation Engineering
If duty (D) is 1428 hectare/cumec and base period (B) is 120 days for an irrigated crop, then delta (Δ) in metres is given by
- 102.8
- 0.73
- 1.38
- 0.01
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Answer: (2) 0.73
Δ = 8.64B/D (Δ in m, B in days, D in ha/cumec) = 8.64 × 120/1428 = 0.726 ≈ 0.73 m.
Which one of the following equations represents the downstream profile of Ogee spillway with vertical upstream face? (X, Y) are the coordinates of the point on the downstream profile with origin at the crest of the spillway and H_d is the design head.
- Y/H_d = -0.5(X/H_d)^1.85
- Y/H_d = -0.5(X/H_d)^(1/185)
- Y/H_d = -2.0(X/H_d)
- Y/H_d = -2.0(X/H_d)^(1/185)
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Answer: (1) Y/H_d = -0.5(X/H_d)^1.85
The WES standard profile for a vertical upstream face is X^1.85 = 2.0 H_d^0.85 Y, i.e. Y/H_d = 0.5(X/H_d)^1.85, with the minus sign because Y is measured downward from the crest.
For no tension to develop in the gravity dam the eccentricity 'e' of the resultant force should be
- Less than b/3
- Less than b/6
- Less than b/4
- Less than b/12
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Answer: (2) Less than b/6
Base stress = (ΣV/b)(1 ± 6e/b), so the heel stays in compression only if e ≤ b/6, i.e. the resultant lies within the middle third of the base.
Lacey's equations can be used for the design of
- Unlined channels only
- Lined channels only
- Both lined and unlined channels
- Neither lined nor unlined channels
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Answer: (1) Unlined channels only
Lacey's regime theory applies to unlined alluvial channels that adjust their own width, depth and slope to the silt load until they reach regime; lined channels have a fixed section and are designed with Manning's formula.
Syphon aqueduct is a cross drainage work provided to carry canal over a natural drain when
- Canal bed is well above the HFL of the natural drain
- Canal bed is at the same level as the bed of the natural drain
- Canal bed is below the HFL of the natural drain
- Canal bed is below bed of the natural drain
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Answer: (3) Canal bed is below the HFL of the natural drain
When the drain's HFL rises above the canal bed, the drain water passes under the canal through barrels flowing under pressure (syphonic action), so the structure is a syphon aqueduct; if the canal bed is well above the HFL it is an ordinary aqueduct.