Tamil Nadu PSC — Agricultural Engineering (Code 164, degree standard) · India · Tamil Nadu Public Service Commission
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)
How our questions for Tamil Nadu PSC spread across the units of Fluid Mechanics & Open Channel Hydraulics. Difficulty: 26 easy, 16 medium, 2 hard.
| Weirs and notches | 7 | |
| Bernoulli's equation: venturimeter, orifice meter and pitot tube | 4 | |
| Forces on submerged surfaces and buoyancy | 3 | |
| Minor losses | 3 | |
| Orifices, mouthpieces and tank emptying | 3 | |
| Uniform flow in open channels: Manning and Chezy | 3 | |
| Specific energy and critical depth | 3 | |
| Dimensional analysis and similitude | 3 | |
| Properties of fluids | 2 | |
| Hydrostatics and manometry | 2 | |
| Laminar and turbulent flow in pipes | 2 | |
| Darcy-Weisbach equation and Moody diagram | 2 | |
| Pipes in series and parallel | 2 | |
| Most economical channel section | 2 | |
| Continuity equation | 1 | |
| Hazen-Williams equation | 1 | |
| Hydraulic jump | 1 |
Choose an answer, then open the working. Each explanation says why the right answer is right and why the others are not.
A horizontal venturimeter 150 mm × 75 mm carries water. A mercury-water differential manometer (mercury relative density 13.6) connected between inlet and throat reads 200 mm. With a discharge coefficient of 0.97, what is the discharge?
A. 31.1 L/s
h = 0.2 × (13.6 − 1) = 2.52 m of water. a₁ = 0.0177 m², a₂ = 0.00442 m². Q = C_d a₁a₂√(2gh)/√(a₁² − a₂²) = 0.97 × 0.0177 × 0.00442 × √(2 × 9.81 × 2.52)/√(0.0177² − 0.00442²) = 0.0311 m³/s = 31.1 L/s. Using 13.6x for h gives 32.3 L/s; omitting C_d gives 32.1 L/s. 8.77 L/s takes the 0.2 m mercury reading as the water head.
Further reading: Modi and Seth, Hydraulics and Fluid Mechanics
Water issues from a small orifice 5 m below the free surface of a large open tank. Neglecting losses, what is the jet velocity?
D. 9.90 m/s
Torricelli's theorem: V = √(2gH) = √(2 × 9.81 × 5) = 9.90 m/s. 7.00 m/s drops the 2, and 98.1 forgets the square root. 49.1 m/s drops both the 2 and the square root (9.81 × 5).
Further reading: Modi and Seth, Hydraulics and Fluid Mechanics
Oil with kinematic viscosity 1 × 10⁻⁴ m²/s flows at 1.5 m/s in a 100 mm pipe. What is the Reynolds number and the flow regime?
C. 1500, laminar
Re = VD/ν = 1.5 × 0.1/10⁻⁴ = 1500, below about 2000, so laminar. Using water's ν (10⁻⁶ m²/s) gives 150000; using the radius gives 750.0.
Further reading: Modi and Seth, Hydraulics and Fluid Mechanics
An Indian textbook writes the pipe friction loss as h_f = 4fLV²/(2gD) and quotes f = 0.005. What is the equivalent Darcy friction factor used in h_f = fLV²/(2gD)?
B. 0.020
The 4f form uses the Fanning coefficient, which is one quarter of the Darcy factor. 4 × 0.005 = 0.020. Mixing the two forms is a common source of answers four times too large or too small. 0.00125 divides by 4 instead of multiplying; 0.010 doubles the Fanning value.
Further reading: Modi and Seth, Hydraulics and Fluid Mechanics
For pipes connected in series, which quantity is the same in every pipe?
D. The discharge
In series the same flow passes through each pipe in turn and the losses add. In parallel the head loss is the same in each branch and the flows add. Velocity and Re change with diameter.
Further reading: Modi and Seth, Hydraulics and Fluid Mechanics
A 90° V-notch with C_d = 0.6 is used to measure a tubewell's discharge. The head over the vertex is 0.2 m. What is the discharge?
D. 25.4 L/s
Q = (8/15) C_d √(2g) tan(θ/2) H^2.5 = (8/15) × 0.6 × 4.429 × tan 45° × 0.2^2.5 = 1.417 × 0.0179 = 0.0254 m³/s = 25.4 L/s. Using H^1.5 gives 127 L/s, a large overestimate. 56.7 L/s uses H² instead of H^2.5; 31.7 L/s uses the rectangular 2/3 instead of 8/15.
Further reading: USBR Water Measurement Manual; Modi and Seth
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
A wide rectangular channel carries a discharge of 3 m³/s per metre of width. What is the critical depth, and the minimum specific energy?
A. y_c = 0.972 m; E_min = 1.46 m
y_c = (q²/g)^(1/3) = (9/9.81)^(1/3) = 0.972 m. At critical depth the velocity head is y_c/2, so E_min = 1.5 y_c = 1.46 m. Forgetting to square q gives 0.674 m; a square root instead of cube root gives 0.958 m.
Further reading: Subramanya, Flow in Open Channels
For a given discharge in an open channel, what is true of the flow at critical depth?
B. The specific energy is a minimum
On the specific-energy curve for a fixed q, the critical depth is the point of minimum E; equivalently, for a fixed E, discharge is maximum there. In a rectangular channel the velocity head at critical depth is y_c/2, not y_c, and Fr = 1.
Further reading: Subramanya, Flow in Open Channels
A hydraulic jump forms in a horizontal rectangular channel. The depth before the jump is 0.4 m and the upstream Froude number is 5. What is the sequent depth after the jump?
A. 2.64 m
y₂/y₁ = ½[√(1 + 8Fr₁²) − 1] = ½[√201 − 1] = 6.59. y₂ = 0.4 × 6.59 = 2.64 m. Forgetting to square Fr₁ gives 1.08 m. 3.04 m uses + 1 instead of − 1 in the bracket; 2.00 m multiplies y₁ by Fr₁.
Further reading: Subramanya, Flow in Open Channels
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