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ASRB NET Watershed Hydrology: practice questions

ASRB NET — Farm Machinery & Power / Land & Water Management Engineering / Agricultural Structures & Process Engineering · India · Agricultural Scientists Recruitment Board (ICAR)

Original practice questions written to the published syllabus · not past papers
6worked here, free
24written to this exam in this subject
12syllabus units

What this subject covers

Rain to runoff: measuring precipitation, subtracting losses, estimating runoff and routing floods.

In the exams: GATE AG Section 4 Hydrology (SWCE is the heaviest GATE section, ~16 marks a year 2019–26); ICAR AIEEA-PG Unit VII; ICAR AICE subjects 40 and 43; NCEES PE hydrology; state PSCs

What ASRB NET asks from it

ASRB NET — Land & Water Management Engineering

Hydrology and soil and water conservation at PG depth: similar to the ICAR PhD papers, with emphasis on concepts and standard formulas.

Where the 24 questions fall

How our questions for ASRB NET spread across the units of Watershed Hydrology. Difficulty: 1 easy, 6 medium, 17 hard.

S-curve, IUH and synthetic unit hydrographs4
Drought classification4
Infiltration: Horton, Philip and the phi-index3
Frequency analysis, return period and IDF curves2
Runoff estimation: SCS curve number2
Streamflow measurement2
Flood routing: Muskingum and reservoir (Puls) methods2
The hydrologic cycle1
Precipitation measurement and rain-gauge networks1
Missing rainfall data and double-mass analysis1
Interception and evaporation (pan)1
Hydrographs and the unit hydrograph1

6 practice questions, worked

Choose an answer, then open the working. Each explanation says why the right answer is right and why the others are not.

1 · The hydrologic cycle · hard

In a humid, forested headwater catchment with deep permeable soils, storm runoff comes mainly from narrow saturated zones beside the stream that expand during the storm and shrink afterwards. This behaviour is best described as:

  1. Hortonian infiltration-excess runoff over the whole catchment
  2. baseflow from a deep confined aquifer
  3. saturation-excess runoff from a variable source area
  4. runoff from channel precipitation only
Answer and working

C. saturation-excess runoff from a variable source area

Where infiltration capacity is high, overland flow arises only where the soil has saturated from below, first near streams; the contributing area varies with wetness (the variable source area concept). Hortonian flow over the whole catchment needs intensity above infiltration capacity everywhere, which deep forest soils rarely allow.

Further reading: Chow, Maidment and Mays, Applied Hydrology

2 · Interception and evaporation (pan) · hard

Using Meyer's formula E = C(ew − ea)(1 + u₉/16) with C = 0.36 for a large deep reservoir, estimate daily evaporation when ew = 23.76 mm Hg, ea = 11.88 mm Hg and u₉ = 16 km/h.

  1. 11.88 mm/day
  2. 8.55 mm/day
  3. 4.28 mm/day
  4. 17.11 mm/day
Answer and working

B. 8.55 mm/day

E = 0.36 × (23.76 − 11.88) × (1 + 16/16) = 0.36 × 11.88 × 2 = 8.55 mm/day. C = 0.50 is for small shallow waters; omitting the wind term gives 4.28; using ew instead of the difference gives 17.11.

Further reading: Subramanya, Engineering Hydrology

3 · Infiltration: Horton, Philip and the phi-index · hard

A 5-hour storm has hourly rainfall of 8, 18, 30, 12 and 5 mm. The direct runoff depth is 32 mm. The φ-index is:

  1. 8.20 mm/h
  2. 9.33 mm/h
  3. 9.00 mm/h
  4. 10.25 mm/h
Answer and working

B. 9.33 mm/h

Assume the 18, 30 and 12 mm hours exceed φ: 60 − 3φ = 32, so φ = 9.33 mm/h. Check: 8 and 5 are below it and 12 is above, so the assumption holds. 8.20 is total loss divided by duration, which ignores that low-intensity hours lose less than φ; 9.00 assumes four hours contribute, but then 8 < 9.0 fails the check. 10.25 mm/h divides the total loss (73 − 32 = 41 mm) by four hours.

Further reading: Subramanya, Engineering Hydrology

4 · Flood routing: Muskingum and reservoir (Puls) methods · hard

Using Muskingum coefficients C₀ = 0.0323, C₁ = 0.6129 and C₂ = 0.3548, with inflow rising from I₁ = 40 to I₂ = 85 m³/s and outflow O₁ = 38 m³/s, the outflow at the end of the step is:

  1. 66.9 m³/s
  2. 40.7 m³/s
  3. 38.0 m³/s
  4. 62.5 m³/s
Answer and working

B. 40.7 m³/s

O₂ = C₀I₂ + C₁I₁ + C₂O₁ = 0.0323×85 + 0.6129×40 + 0.3548×38 = 40.7 m³/s. Swapping I₁ and I₂ gives 66.9; dropping the C₀I₂ term gives 38.0. 62.5 m³/s simply averages the two inflows.

Further reading: Subramanya, Engineering Hydrology

5 · Flood routing: Muskingum and reservoir (Puls) methods · hard

For Muskingum routing with K = 10 h and x = 0.2, the range of routing interval Δt that keeps all three coefficients non-negative is:

  1. 2 h ≤ Δt ≤ 10 h
  2. 4 h ≤ Δt ≤ 16 h
  3. 4 h ≤ Δt ≤ 10 h
  4. 0 < Δt ≤ 20 h
Answer and working

B. 4 h ≤ Δt ≤ 16 h

C₀ ≥ 0 requires Δt ≥ 2Kx = 4 h; C₂ ≥ 0 requires Δt ≤ 2K(1 − x) = 16 h. The upper limit is not K itself, a common simplification.

Further reading: Subramanya, Engineering Hydrology

6 · Drought classification · hard

For a district and week, current NDVI is 0.42, and the historical minimum and maximum NDVI for that week are 0.30 and 0.70. The Vegetation Condition Index is:

  1. 60%
  2. 42%
  3. 70%
  4. 30%
Answer and working

D. 30%

VCI = (NDVI − NDVImin)/(NDVImax − NDVImin) × 100 = (0.42 − 0.30)/(0.40) × 100 = 30%. 60% divides by the maximum only; 70% measures distance from the maximum instead. 42% simply reads the current NDVI as a percentage.

Further reading: WMO Handbook of Drought Indicators and Indices

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Read the subject

Watershed HydrologyThe full study manual, free ASRB NETFormat, sections and how it has changed

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Watershed Hydrology for other exams

ICAR AIEEA-PG · IndiaGATE · IndiaRajasthan PSC · IndiaTamil Nadu PSC · IndiaICAR AICE-JRF/SRF · India

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