ICAR AICE-JRF/SRF (PhD) — Agricultural Engineering subjects 38–43 · India · National Testing Agency for ICAR
Original practice questions written to the published syllabus · not past papersFluids at rest and in motion, pipe flow and open channels: the base of irrigation, drainage, pumping and conservation structures.
In the exams: GATE AG Section 4 (hydraulics asked in 6 of 8 papers 2019–26); ICAR AIEEA-PG Unit III; ICAR AICE subject 43 Unit 2; NCEES FE 12–18 questions (largest FE area)
ICAR AICE-JRF/SRF (PhD) — subject 43 Irrigation and Drainage Engineering, Unit 2
PG-depth fluid mechanics and open-channel hydraulics: derivations and limits of Darcy-Weisbach and Hazen-Williams, Moody zones, Hardy Cross, specific energy and transitions, critical flow in non-rectangular sections, jump classification, gradually varied flow ideas and model laws.
How our questions for ICAR AICE-JRF/SRF spread across the units of Fluid Mechanics & Open Channel Hydraulics. Difficulty: 5 easy, 38 medium, 24 hard.
| Hydraulic jump | 7 | |
| Pipes in series and parallel | 6 | |
| Specific energy and critical depth | 6 | |
| Dimensional analysis and similitude | 6 | |
| Momentum equation | 5 | |
| Darcy-Weisbach equation and Moody diagram | 4 | |
| Orifices, mouthpieces and tank emptying | 4 | |
| Most economical channel section | 4 | |
| Properties of fluids | 3 | |
| Forces on submerged surfaces and buoyancy | 3 | |
| Laminar and turbulent flow in pipes | 3 | |
| Weirs and notches | 3 | |
| Uniform flow in open channels: Manning and Chezy | 3 | |
| Hydrostatics and manometry | 2 | |
| Continuity equation | 2 | |
| Bernoulli's equation: venturimeter, orifice meter and pitot tube | 2 | |
| Hazen-Williams equation | 2 | |
| Minor losses | 2 |
Choose an answer, then open the working. Each explanation says why the right answer is right and why the others are not.
A soap bubble 40 mm in diameter is made from a solution with surface tension 0.04 N/m. What is the excess of pressure inside the bubble over the outside?
D. 8.00 Pa
A soap bubble has two surfaces, so Δp = 8σ/d = 8 × 0.04/0.04 = 8.00 Pa. 4σ/d = 4.00 Pa is the result for a liquid droplet (one surface); 2σ/d is for a cylindrical jet; using the radius in 8σ/d gives 16.0 Pa.
Further reading: White, Fluid Mechanics
An inverted U-tube manometer with oil of relative density 0.8 in its upper bend is connected between two points on a water pipeline. The oil-water interfaces differ in level by 200 mm. What is the difference in piezometric head between the two points, in metres of water?
C. 0.0400 m
For a lighter manometric liquid, h = x(1 − S_light/S) = 0.2 × (1 − 0.8) = 0.0400 m of water. Using x × S_light gives 0.160 m; the inverted manometer is chosen precisely because it gives a large deflection for a small head difference. 0.360 m uses (1 + 0.8) instead of (1 − 0.8).
Further reading: Modi and Seth, Hydraulics and Fluid Mechanics
A vertical rectangular gate 1.5 m wide and 2 m high has its top edge 1 m below the water surface. At what depth below the water surface does the resultant hydrostatic force act?
A. 2.17 m
h_cp = h̄ + I_G/(A h̄). h̄ = 2 m, A = 3 m², I_G = bd³/12 = 1.5 × 8/12 = 1.00 m⁴. h_cp = 2 + 1/(3 × 2) = 2.17 m. The centre of pressure is always below the centroid (2.00 m); forgetting to divide by h̄ gives 2.33 m. 1.83 m subtracts I_G/(A h̄) and so puts the centre of pressure above the centroid.
Further reading: Modi and Seth, Hydraulics and Fluid Mechanics
Oil (density 900 kg/m³, dynamic viscosity 0.2 Pa·s) flows at a mean velocity of 1 m/s through 100 m of 100 mm pipe. What is the friction head loss?
D. 7.25 m of oil
Re = ρVD/μ = 900 × 1 × 0.1/0.2 = 450, laminar. Hagen-Poiseuille: h_f = 32μVL/(ρgD²) = 32 × 0.2 × 1 × 100/(900 × 9.81 × 0.01) = 7.25 m. The same comes from f = 64/Re = 0.142 in Darcy-Weisbach. Assuming a turbulent f = 0.02 gives 1.02 m, far too low. 3.62 m uses 16 instead of 32 in Hagen-Poiseuille; 14.5 m uses 64.
Further reading: Modi and Seth, Hydraulics and Fluid Mechanics
Use the Swamee-Jain equation to estimate the Darcy friction factor for turbulent flow at Re = 2 × 10⁵ in a pipe of relative roughness ε/D = 0.001.
A. 0.0212
f = 0.25/[log₁₀(ε/3.7D + 5.74/Re^0.9)]² = 0.25/[log₁₀(0.000270 + 0.0000973)]² = 0.0212. The Blasius smooth-pipe value, 0.316/Re^0.25 = 0.0149, ignores roughness; 64/Re applies only to laminar flow. 0.000320 is 64/Re, valid only for laminar flow; 0.00530 is the Fanning value (one quarter of Darcy).
Further reading: White, Fluid Mechanics
A single 300 mm pipe is to be replaced by two identical pipes laid in parallel, of the same length, carrying the same total discharge with the same head loss and friction factor. What diameter should each new pipe be?
D. 227 mm
Each new pipe carries Q/2 at the same head loss: (Q/2)²/d⁵ = Q²/D⁵, so d⁵ = D⁵/4 and d = D/4^0.2 = D/2^0.4 = 0.758D = 227 mm. Keeping the same total area (D/√2 = 212 mm) is not enough, because smaller pipes lose more head per unit flow. 261 mm takes the fifth root of 2 instead of 4; 150 mm simply halves the diameter.
Further reading: Modi and Seth, Hydraulics and Fluid Mechanics
The head over a triangular notch is measured with an error of 2 per cent. What is the resulting percentage error in the computed discharge?
D. 5 %
Q ∝ H^2.5, so dQ/Q = 2.5 dH/H = 2.5 × 2 = 5 %. For a rectangular weir (Q ∝ H^1.5) it would be 3 %, which is why head must be measured more precisely on a V-notch. 4 % uses an exponent of 2.
Further reading: Modi and Seth, Hydraulics and Fluid Mechanics
If both the bed width and the side slope of a trapezoidal channel are free to be chosen, what shape is the most economical section?
B. Half a regular hexagon, with sides at 60° to the horizontal
Optimising side slope as well as width gives z = 1/√3 (sides at 60°), with the bed and the sloping sides of equal length: half a regular hexagon. The b = 2y rectangle is the best only among rectangles.
Further reading: Chow, Open-Channel Hydraulics
A hydraulic jump in a horizontal rectangular channel has depths of 0.5 m before and 2.0 m after the jump. What is the head loss in the jump?
B. 0.844 m
ΔE = (y₂ − y₁)³/(4y₁y₂) = 1.5³/(4 × 0.5 × 2.0) = 3.375/4 = 0.844 m. Omitting the 4 gives 3.38 m; squaring instead of cubing gives 0.562 m; 1.50 m is the jump height, not the loss.
Further reading: Subramanya, Flow in Open Channels
In a 1:36 Froude model of a canal fall, a flood wave takes 1 minute to pass through the model. How long does it take in the prototype?
D. 6 minutes
Under Froude similarity T_r = L_r^0.5 = √36 = 6, so 1 min in the model is 6 min in the prototype. 36 would be the length ratio and 216 = 36^1.5 has no meaning here.
Further reading: Modi and Seth, Hydraulics and Fluid Mechanics
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