Nepal Engineering Council · Civil Engineering · Chapter 3
Basic Water Resources Engineering
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190 questions in 6 syllabus topics.
3.1 Fluids and their properties
32 questions · ACiE0301
1. Which of the following is a Newtonian fluid?
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Think of a fluid with constant viscosity at a given temperature.
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Answer: D. Water
In a Newtonian fluid shear stress is linearly proportional to velocity gradient; water, air and light oils are Newtonian, whereas pastes, slurries and inks are non-Newtonian.
2. An ideal fluid is one that is
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Two properties are set to zero or constant.
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Answer: C. incompressible and non-viscous
An ideal (perfect) fluid is a hypothetical fluid with zero viscosity and constant density, so it cannot sustain shear stress.
3. A fluid is best defined as a substance that
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Think of what happens under shear, not under pressure.
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Answer: B. deforms continuously under the action of a shear stress, however small
The defining property of a fluid is continuous deformation under even a tiny shear stress; a solid deforms by a finite amount.
4. A fluid that requires a minimum yield shear stress before it starts to flow, and then shows a linear stress-rate relation, is called a
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Look for the fluid with a threshold stress.
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Answer: C. Bingham plastic
A Bingham plastic behaves as a rigid body until the yield stress is exceeded, after which τ = τy + μp(du/dy).
5. The SI unit of dynamic viscosity is
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Divide shear stress by velocity gradient.
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Answer: C. N·s/m²
μ = τ/(du/dy) = (N/m²)/(1/s) = N·s/m², i.e. Pa·s.
6. The dimensions of kinematic viscosity are
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It is dynamic viscosity divided by density.
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Answer: B. L²T⁻¹
ν = μ/ρ = (ML⁻¹T⁻¹)/(ML⁻³) = L²T⁻¹; its SI unit is m²/s.
7. With a rise in temperature, the viscosity of a liquid ___ and that of a gas ___.
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Different physical origins in liquids and gases.
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Answer: D. decreases; increases
In liquids viscosity arises from cohesion, which weakens as temperature rises; in gases it arises from molecular momentum exchange, which grows with temperature.
8. The approximate dynamic viscosity of water at 20 °C is
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One centipoise.
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Answer: A. 1.0 × 10⁻³ Pa·s
Water at 20 °C has μ ≈ 1.0 mPa·s (1 centipoise); 1.8 × 10⁻⁵ Pa·s is the value for air and 10⁻⁶ m²/s is its kinematic viscosity.
9. The bulk modulus of elasticity of a fluid is defined as the ratio of
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It has the form stress ÷ strain.
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Answer: C. volumetric stress (pressure change) to volumetric strain
K = −dp/(dV/V); its reciprocal is the compressibility of the fluid.
10. For an ideal gas undergoing an isothermal process, the bulk modulus of elasticity equals
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Differentiate pV = constant.
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Answer: C. the absolute pressure p
From pV = const, dp/dV = −p/V so K = p. For an adiabatic process K = k·p.
11. The rise of liquid in a capillary tube is
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Write the expression h = 4σcosθ/(ρgd).
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Answer: B. inversely proportional to the tube diameter
h = 4σcosθ/(ρ g d), so h ∝ 1/d.
12. Mercury in a glass tube shows capillary depression because
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Compare cohesion and adhesion.
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Answer: A. cohesion between mercury molecules exceeds adhesion to glass
When cohesion exceeds adhesion the contact angle is greater than 90° and the meniscus is convex, so the level falls.
13. The dimensions of surface tension are
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Force per unit length.
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Answer: D. MT⁻²
Surface tension is force per unit length: N/m = (MLT⁻²)/L = MT⁻².
14. The excess pressure inside a soap bubble compared with that inside a liquid droplet of the same diameter and surface tension is
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Count the number of liquid-air interfaces.
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Answer: A. twice as much
A bubble has two surfaces (Δp = 8σ/d) while a droplet has one (Δp = 4σ/d).
15. Cavitation in a flowing liquid begins when the local absolute pressure falls to
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Think of boiling at low pressure.
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Answer: C. the vapour pressure of the liquid at that temperature
When local pressure drops to the vapour pressure, vapour bubbles form and later collapse violently in higher-pressure zones.
16. The vapour pressure of a liquid
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Why does water boil at lower pressure when hot?
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Answer: C. increases with increase in temperature
Higher temperature increases molecular escape, so vapour pressure rises (water: 2.34 kPa at 20 °C, 101.3 kPa at 100 °C).
17. In a Venturi meter, cavitation is most likely to occur at the
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Where is velocity highest?
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Answer: B. throat
Velocity is greatest and pressure lowest at the throat, so it is the first place pressure may drop to vapour pressure.
18. Specific gravity of a liquid is
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It has no unit.
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Answer: A. a dimensionless ratio of its density to that of water at 4 °C
S = ρ/ρwater, a pure number; mercury has S = 13.6.
19. The specific volume of a fluid is the
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Volume per unit mass.
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Answer: D. reciprocal of mass density
v = V/m = 1/ρ, with units m³/kg.
20. Surface tension of a pure liquid such as water, as temperature rises,
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Cohesion weakens when heated.
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Answer: D. decreases
Cohesive forces weaken with temperature, so σ falls (about 0.0728 N/m at 20 °C to 0.0589 N/m at 100 °C).
21. The dimensions of specific weight are
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γ = ρg.
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Answer: D. ML⁻²T⁻²
γ = ρg = (ML⁻³)(LT⁻²) = ML⁻²T⁻²; unit N/m³.
22. The specific weight of an oil of specific gravity 0.85 is (g = 9.81 m/s²)
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γ = S × 9.81 kN/m³.
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Answer: B. 8.34 kN/m³
γ = S·γw = 0.85 × 9.81 = 8.34 kN/m³.
23. A liquid of mass 5 kg occupies 0.004 m³. Its specific gravity is
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Find density first.
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Answer: B. 1.25
ρ = 5/0.004 = 1250 kg/m³, so S = 1250/1000 = 1.25.
24. The specific volume of a liquid of mass density 800 kg/m³ is
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Reciprocal of density.
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Answer: C. 0.00125 m³/kg
v = 1/ρ = 1/800 = 1.25 × 10⁻³ m³/kg.
25. A flat plate moves at 2 m/s over a fixed plate, the 5 mm gap being filled with oil of μ = 0.1 Pa·s. Assuming a linear velocity profile, the shear stress is
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τ = μ × (velocity gradient).
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Answer: B. 40 Pa
τ = μ U/h = 0.1 × 2/0.005 = 40 Pa.
26. An oil has μ = 0.08 Pa·s and specific gravity 0.8. Its kinematic viscosity is
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ν = μ/ρ with ρ in kg/m³.
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Answer: B. 1.0 × 10⁻⁴ m²/s
ν = μ/ρ = 0.08/800 = 1.0 × 10⁻⁴ m²/s.
27. A dynamic viscosity of 2.5 poise is equal to
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1 poise = 0.1 N·s/m².
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Answer: A. 0.25 Pa·s
1 poise = 0.1 Pa·s (1 dyne·s/cm²), so 2.5 poise = 0.25 Pa·s.
28. Water (K = 2.2 GPa) is subjected to a pressure increase of 11 MPa. The percentage reduction in its volume is
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ΔV/V = Δp/K.
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Answer: A. 0.5 %
ΔV/V = Δp/K = 11 × 10⁶/2.2 × 10⁹ = 0.005 = 0.5 %.
29. Water (σ = 0.0728 N/m, ρ = 998 kg/m³, contact angle 0°) rises in a clean glass tube of 2 mm diameter by about
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h = 4σcosθ/(ρgd).
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Answer: A. 14.9 mm
h = 4σ/(ρ g d) = 4 × 0.0728/(998 × 9.81 × 0.002) = 0.0149 m ≈ 14.9 mm.
30. The excess pressure inside a water droplet of diameter 0.5 mm in air (σ = 0.073 N/m) is
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One surface; use Δp = 4σ/d.
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Answer: D. 584 Pa
Δp = 4σ/d = 4 × 0.073/0.0005 = 584 Pa.
31. A soap bubble of diameter 40 mm has surface tension 0.025 N/m. The excess pressure inside it is
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Two interfaces.
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Answer: D. 5.0 Pa
A bubble has two surfaces: Δp = 8σ/d = 8 × 0.025/0.04 = 5 Pa.
32. The speed of sound in water of bulk modulus 2.2 GPa and density 1000 kg/m³ is about
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c = √(K/ρ).
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Answer: A. 1483 m/s
c = √(K/ρ) = √(2.2×10⁹/1000) = 1483 m/s.
3.2 Hydrostatics
32 questions · ACiE0302
33. Pascal's law states that, in a fluid at rest, the pressure at a point
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Think of pressure as isotropic.
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Answer: A. is the same in all directions
In a static fluid there is no shear, so pressure at a point is equal in all directions; an applied pressure is transmitted undiminished through the fluid.
34. In a static liquid of constant density, pressure varies with depth h as
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Integrate dp = ρ g dh.
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Answer: B. p = ρ g h, linearly with depth
dp/dz = ρg gives a linear increase of gauge pressure with depth.
35. The absolute pressure at a point is equal to
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Gauge pressure is measured from atmospheric.
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Answer: B. gauge pressure plus atmospheric pressure
p_abs = p_gauge + p_atm; for a vacuum, p_abs = p_atm − p_vacuum.
36. Standard atmospheric pressure is approximately equivalent to a column of water of height
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Divide 101.3 kPa by γw.
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Answer: C. 10.3 m
101.3 kPa/(9.81 kN/m³) = 10.33 m of water (equal to 760 mm of mercury).
37. A simple piezometer tube is NOT suitable for measuring the pressure of
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Which fluid cannot form a free surface in the tube?
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Answer: C. a gas
A piezometer needs a liquid column open to the atmosphere; it cannot hold a gas pressure, and large pressures need impractically tall tubes.
38. A differential manometer is used to measure
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Two connections are made.
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Answer: B. the pressure difference between two points
Its two limbs are connected to two different points (or pipes), so the manometric reading gives p1 − p2.
39. An inclined-tube manometer is preferred for measuring very small pressures because
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Reading length vs vertical height.
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Answer: C. the scale reading is magnified for a given vertical head
For a fixed vertical rise h, the length read along an inclined tube is h/sinθ, so sensitivity increases.
40. The total hydrostatic force on a plane surface submerged in a liquid equals
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Pressure at the centroid × area.
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Answer: B. ρ g A x̄, where x̄ is the depth of the centroid
F = γ A x̄ = pressure at centroid × area.
41. The centre of pressure of a plane surface submerged in a static liquid
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Pressure increases with depth.
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Answer: A. lies below the centroid, except for a horizontal surface
h_cp = x̄ + I_G sin²θ/(A x̄) ≥ x̄; for a horizontal surface pressure is uniform, so it coincides with the centroid.
42. For a vertical plane surface, as its depth of submergence increases the centre of pressure
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Examine I_G/(A x̄).
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Answer: D. moves closer to the centroid
The offset I_G/(A x̄) decreases as x̄ increases.
43. The horizontal component of the hydrostatic force on a curved surface is equal to
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Project the surface onto a vertical plane.
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Answer: C. the force on the vertical projection of the surface
F_H = ρ g A_v x̄_v acts at the centre of pressure of the vertical projection.
44. The vertical component of the hydrostatic force on a curved surface is equal to
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Think of the liquid column above.
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Answer: A. the weight of the real or imaginary liquid vertically above the surface up to the free surface
F_V = ρ g × (volume between the surface and the free surface); it acts through the centroid of that volume.
45. The resultant hydrostatic force on a vertical dam face of height H acts at what height above the base (full reservoir, triangular pressure diagram)?
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Centroid of a triangle.
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Answer: D. H/3
The triangular pressure diagram has its centroid H/3 above the base, which is where the resultant acts.
46. The pressure at the base of a vessel filled with a liquid to depth h depends on
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Pressure is a function of depth only.
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Answer: D. h and liquid density, and not on the shape of the vessel
p = ρ g h only (hydrostatic paradox); the total force still depends on base area.
47. Archimedes' principle states that a body immersed in a fluid experiences an upthrust equal to
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Compare with the displaced fluid.
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Answer: B. the weight of the fluid displaced by the body
The buoyant force F_B = ρ g V_displaced.
48. The centre of buoyancy of a floating body is located at the
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It depends on the displaced shape, not the body.
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Answer: A. centroid of the displaced volume of fluid
The buoyant force acts through the centre of gravity of the displaced fluid.
49. A floating body is in stable equilibrium if its
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Look at the sign of GM.
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Answer: D. metacentre lies above its centre of gravity
GM > 0 produces a righting moment; GM = 0 is neutral, GM < 0 is unstable.
50. A completely submerged body is in stable equilibrium when its
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A submerged body has no waterplane.
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Answer: C. centre of gravity lies below the centre of buoyancy
For a submerged body, a tilt creates a righting couple only if G is below B; if they coincide the equilibrium is neutral.
51. The gauge pressure at a depth of 10 m in a lake is (γw = 9.81 kN/m³)
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p = γh.
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Answer: A. 98.1 kPa
p = ρ g h = 9.81 × 10 = 98.1 kPa (gauge); adding 101.3 kPa would give the absolute pressure.
52. A gauge pressure of 200 kPa in an oil of specific gravity 0.8 corresponds to a head of
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h = p/γ.
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Answer: A. 25.5 m of oil
h = p/(ρ g) = 200/(0.8 × 9.81) = 25.5 m of oil.
53. Two points in a horizontal water pipe are connected to a U-tube differential manometer containing mercury (S = 13.6). The mercury levels differ by 200 mm. The pressure difference is
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Use the difference of specific weights.
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Answer: C. 24.7 kPa
Δp = (S_m − S_w) γw h = (13.6 − 1) × 9.81 × 0.2 = 24.7 kPa.
54. A vertical rectangular gate 2 m wide and 3 m high has its top edge at the water surface. The hydrostatic force on it is
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Centroid depth = 1.5 m.
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Answer: D. 88.3 kN
F = γ x̄ A = 9.81 × 1.5 × (2 × 3) = 88.3 kN.
55. For a vertical rectangular gate 3 m high whose top edge is at the free water surface, the centre of pressure is at a depth of
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Two-thirds of the height.
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Answer: B. 2.0 m
h_cp = 2H/3 = 2 m below the free surface (x̄ + I_G/(A x̄) = 1.5 + 0.5 = 2.0 m).
56. A vertical circular plate of diameter 2 m has its centre 4 m below the water surface. The hydrostatic force on it is
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A = πd²/4.
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Answer: A. 123.3 kN
F = γ x̄ A = 9.81 × 4 × (π × 1²) = 123.3 kN.
57. In a hydraulic press the ram area is 50 times the plunger area. A force of 100 N on the plunger can lift a load of (neglecting friction)
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Equal pressure on both pistons.
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Answer: D. 5000 N
Pressure is transmitted equally, so F2 = F1 × (A2/A1) = 100 × 50 = 5000 N.
58. A flat wooden raft 0.5 m thick and of specific gravity 0.6 floats in fresh water. Its depth of immersion is
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Fraction submerged = specific gravity.
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Answer: D. 0.30 m
Weight = buoyancy gives S × a = draft, so draft = 0.6 × 0.5 = 0.30 m.
59. Water stands to the top of a quadrant gate of radius 2 m and width 3 m. The horizontal component of water force on the gate is
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Project onto the vertical plane.
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Answer: C. 58.9 kN
The vertical projection is 2 m × 3 m with centroid 1 m deep: F_H = 9.81 × 1 × 6 = 58.9 kN.
60. The total water force per metre width on a vertical dam face holding 12 m of water is
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Area of the triangular pressure diagram.
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Answer: B. 706 kN
F = ½ γ H² = 0.5 × 9.81 × 144 = 706 kN per metre.
61. A mercury barometer reads 750 mm. The atmospheric pressure is (S = 13.6)
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p = Sγwh.
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Answer: D. 100.1 kPa
p = ρ g h = 13.6 × 9.81 × 0.75 = 100.1 kPa.
62. An iceberg of specific gravity 0.92 floats in sea water of specific gravity 1.025. The fraction of its volume below the water surface is
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Equate weight and buoyancy.
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Answer: C. 89.8 %
Fraction submerged = ρ_body/ρ_fluid = 0.92/1.025 = 0.898 = 89.8 %.
63. A rectangular barge 6 m wide and 20 m long has a draft of 2 m, and its centre of gravity is 2 m above the bottom. Its metacentric height is
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GM = KB + BM − KG.
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Answer: B. 0.50 m
KB = 1 m, BM = b²/(12d) = 36/24 = 1.5 m, so GM = KB + BM − KG = 1 + 1.5 − 2 = 0.5 m (stable).
64. A layer of oil (S = 0.8, 2 m deep) floats on 3 m of water. The gauge pressure at the bottom is
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Add the pressures of each layer.
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Answer: A. 45.1 kPa
p = γw (0.8 × 2 + 3) = 9.81 × 4.6 = 45.1 kPa.
3.3 Hydro-kinematics and hydro-dynamics
30 questions · ACiE0303
65. In a steady flow, the fluid properties at a fixed point
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Time versus space.
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Answer: B. do not change with time
Steady flow means ∂(any property)/∂t = 0 at a point; uniform flow means no change with position.
66. Flow through a gradually converging pipe at a constant discharge is
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Check time-dependence and space-dependence separately.
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Answer: C. steady and non-uniform
Discharge is constant (steady), but velocity changes along the pipe (non-uniform).
67. Flow in a circular pipe is generally laminar when the Reynolds number is less than about
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Critical Reynolds number for pipes.
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Answer: D. 2000
Re = vd/ν < 2000 gives laminar flow; between 2000 and 4000 is transitional; above about 4000 turbulent.
68. A streamline is a line such that
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Instantaneous direction of velocity.
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Answer: D. the tangent at any point gives the direction of velocity at that point
Streamlines are instantaneous curves tangent to the velocity vector; there is no flow across them. They coincide with path lines only in steady flow.
69. Irrotational flow is one in which
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Rotation of the particle itself.
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Answer: A. fluid particles do not rotate about their own axes
Irrotational flow has zero vorticity (ωz = ½(∂v/∂x − ∂u/∂y) = 0); a free vortex is irrotational despite circular streamlines.
70. The continuity equation for steady, incompressible flow in a pipe is
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Mass in = mass out.
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Answer: D. A₁v₁ = A₂v₂
Conservation of mass with constant density gives constant discharge Q = A v.
71. Bernoulli's equation along a streamline is valid for flow that is
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Assumptions of Euler's equation.
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Answer: C. steady, incompressible and frictionless
The derivation integrates Euler's equation along a streamline, assuming steady, inviscid, incompressible flow without energy added or removed.
72. In Bernoulli's equation p/γ + v²/2g + z = constant, each term has the dimension of
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Energy per unit weight.
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Answer: B. length
All three are heads (energy per unit weight), measured in metres of fluid.
73. The energy line (total energy line) lies above the hydraulic grade line by
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What does the HGL leave out?
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Answer: C. v²/2g
TEL = p/γ + z + v²/2g, HGL = p/γ + z, so the gap is the velocity head.
74. According to the momentum equation, the net force on a control volume of fluid in a given direction equals
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Newton's second law for a control volume.
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Answer: B. the rate of change of momentum flux in that direction
ΣF = ρQ(v₂ − v₁) for steady flow with one inlet and one outlet.
75. In a Venturi meter, the discharge is computed from the pressure difference between
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The narrowest section.
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Answer: B. the inlet section and the throat
The manometer measures the pressure drop between the upstream section and the throat where velocity is higher.
76. Compared with a Venturi meter of the same size, an orifice meter has
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Gradual vs abrupt geometry.
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Answer: A. a larger permanent head loss
The orifice plate causes abrupt contraction and expansion with no gradual recovery, so more head is lost; it is however cheaper and compact.
77. A Pitot tube measures the velocity at a point from
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Stagnation vs static.
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Answer: C. the difference between stagnation and static pressure
v = √(2Δp/ρ) = √(2gh), where Δp = p₀ − p is the dynamic pressure.
78. The relation between the coefficients of an orifice is
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Actual Q = Cc a × Cv √(2gh).
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Answer: A. Cd = Cc × Cv
Discharge coefficient = contraction coefficient × velocity coefficient.
79. The vena contracta of a jet from a sharp-edged orifice is the section
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Contraction of the jet.
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Answer: C. where the jet area is minimum
Streamlines continue to converge outside the orifice, giving a minimum area at about d/2 downstream (Cc ≈ 0.61–0.64).
80. The discharge over a rectangular weir and over a triangular (V-notch) weir varies with the head H as, respectively,
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The V-notch width varies with H.
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Answer: A. H^1.5 and H^2.5
Q ∝ L H^(3/2) for a rectangular weir and Q ∝ tan(θ/2) H^(5/2) for a V-notch.
81. A triangular notch is preferred to a rectangular weir for measuring small discharges because
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Sensitivity of head to discharge.
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Answer: B. the head changes appreciably for a small change in discharge
Since Q ∝ H^2.5 and the notch is narrow at the bottom, a given Q gives a larger H, so reading error is smaller.
82. At a stagnation point in a flow, the
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Bernoulli at the nose of a body.
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Answer: C. velocity is zero and pressure is maximum
The kinetic energy is fully converted to pressure, giving p₀ = p + ½ρv².
83. Water flows in a 300 mm diameter pipe at 1.2 m/s and enters a 150 mm diameter pipe. The velocity in the smaller pipe is
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Area ratio is the square of the diameter ratio.
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Answer: B. 4.8 m/s
v₂ = v₁(d₁/d₂)² = 1.2 × 4 = 4.8 m/s.
84. The discharge in a 300 mm diameter pipe flowing at a mean velocity of 2 m/s is
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Q = A v.
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Answer: D. 141.4 L/s
Q = A v = (π/4)(0.3)² × 2 = 0.1414 m³/s = 141.4 L/s.
85. In a horizontal pipe, water of pressure 200 kPa flows at 2 m/s and then enters a section where its velocity is 4 m/s. Neglecting losses, the pressure in the second section is
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Velocity head rises, pressure head falls.
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Answer: A. 194 kPa
p₂ = p₁ + ½ρ(v₁² − v₂²) = 200 − 0.5 × 1000 × (16 − 4)/1000 = 194 kPa.
86. The theoretical velocity of efflux from a small orifice under a constant head of 5 m is (g = 9.81 m/s²)
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Torricelli's theorem.
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Answer: A. 9.90 m/s
v = √(2gH) = √(2 × 9.81 × 5) = 9.90 m/s (Torricelli).
87. A 50 mm diameter sharp-edged orifice (Cd = 0.6) discharges under a head of 4 m. The discharge is
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Q = Cd a √(2gH).
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Answer: B. 10.44 L/s
Q = Cd a √(2gH) = 0.6 × 0.001963 × 8.859 = 0.01044 m³/s = 10.44 L/s.
88. A Pitot tube in a water pipe shows a difference of 0.20 m of water between stagnation and static heads. The velocity is
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v = √(2gh).
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Answer: D. 1.98 m/s
v = √(2gh) = √(2 × 9.81 × 0.2) = 1.98 m/s.
89. A Venturi meter has inlet 300 mm and throat 150 mm diameter. Taking Cd = 1, the discharge for a differential head of 0.50 m of water is
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Use the Venturi formula with both areas.
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Answer: D. 57.2 L/s
Q = a₁a₂√(2gh)/√(a₁² − a₂²) = 0.0572 m³/s = 57.2 L/s.
90. Using Q = 1.84 L H^1.5, the discharge over a rectangular weir of crest length 2 m under a head of 0.30 m is
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Raise H to the power 3/2.
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Answer: A. 0.605 m³/s
Q = 1.84 × 2 × 0.3^1.5 = 1.84 × 2 × 0.1643 = 0.605 m³/s.
91. A right-angled V-notch (Cd = 0.6) works under a head of 0.20 m. Using Q = (8/15) Cd √(2g) tan(θ/2) H^2.5, the discharge is
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tan(90°/2) = 1.
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Answer: D. 25.4 L/s
tan 45° = 1: Q = 0.5333 × 0.6 × 4.429 × 0.2^2.5 = 0.0254 m³/s = 25.4 L/s.
92. A jet of water 50 mm in diameter with velocity 20 m/s strikes a fixed flat plate normally. The force on the plate is
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Rate of loss of momentum of the jet.
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Answer: A. 785 N
F = ρ a v² = 1000 × 0.001963 × 400 = 785 N.
93. Water flows at 0.5 m/s in a 50 mm pipe (ν = 1.0 × 10⁻⁶ m²/s). The Reynolds number and flow type are
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Re = vd/ν.
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Answer: B. 25 000, turbulent
Re = vd/ν = 0.5 × 0.05/10⁻⁶ = 25 000 > 4000, so turbulent.
94. A tank of plan area 2 m² has a 0.002 m² orifice (Cd = 0.6) at its bottom. The time to fall from 2 m to 0 head is
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Integrate A dh = −Cd a √(2gh) dt.
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Answer: C. 1064 s
t = 2A(√H₁ − √H₂)/(Cd a √(2g)) = 2 × 2 × 1.414/(0.6 × 0.002 × 4.429) = 1064 s.
3.4 Pipe flow
32 questions · ACiE0304
95. The Darcy–Weisbach equation for head loss due to friction in a pipe is
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Head loss is proportional to velocity head.
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Answer: D. h_f = f L v²/(2 g d)
h_f = f (L/d)(v²/2g), where f is the Darcy friction factor.
96. For laminar flow in a circular pipe, the Darcy friction factor is
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Valid for Re < 2000.
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Answer: D. 64/Re
From Hagen–Poiseuille, f = 64/Re (the Fanning factor is 16/Re).
97. In laminar flow through a pipe, the head loss due to friction varies with mean velocity as
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Examine h_f = 32μLv/(γd²).
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Answer: B. v¹
Hagen–Poiseuille: h_f = 32 μ L v/(γ d²), linear in velocity.
98. In the fully rough (complete turbulence) zone of the Moody diagram, the friction factor depends on
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Look at the flat portions of the Moody curves.
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Answer: D. relative roughness only
In fully rough flow the viscous sublayer is thinner than the roughness projections, so f becomes independent of Re.
99. The head loss due to a sudden enlargement of a pipe from velocity v₁ to v₂ is given by
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It is based on the velocity difference.
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Answer: C. (v₁ − v₂)²/2g
By momentum and energy balance (Borda–Carnot), h_L = (v₁ − v₂)²/2g.
100. The head loss at the entrance of a pipe from a large reservoir with a sharp-cornered entrance is approximately
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Contraction of flow at entry.
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Answer: C. 0.5 v²/2g
The entrance loss coefficient K ≈ 0.5 for a square-edged entrance; for exit into a reservoir K = 1.0.
101. A minor loss is called 'minor' because
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Compare with friction loss in long pipes.
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Answer: A. it is usually small in comparison with friction loss in a long pipe
In long pipelines losses at bends, valves and transitions are small relative to friction, but in short pipes they can dominate.
102. The Hazen–Williams coefficient C is
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Smoother pipe, smaller loss.
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Answer: A. higher for smoother pipes
C is about 140 for new smooth pipes and falls to 100 or less for old rough cast iron; head loss decreases as C increases.
103. Two pipes are said to be equivalent when, for the same head loss,
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Same discharge, same head loss.
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Answer: B. they carry the same discharge
An equivalent pipe replaces a compound system with a single pipe carrying the same Q for the same total head loss.
104. For pipes connected in series, the
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Think of a single flow path.
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Answer: A. discharge is the same and head losses add up
Continuity gives constant Q in series; total head loss is the sum of the individual losses.
105. For pipes connected in parallel between two junctions, the
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Same two end points.
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Answer: D. head loss is the same in each pipe and the discharges add up
Both branches span the same two nodes, hence the same head difference; flows sum to the total.
106. The Hardy Cross method for pipe networks is based on satisfying
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Two conditions: nodes and loops.
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Answer: A. continuity at each node and zero net head loss around each loop
Initial flows satisfy continuity; they are iteratively corrected by ΔQ = −Σh_f/Σ(2h_f/Q) until Σh_f ≈ 0 around each loop.
107. Water hammer in a pipe is caused by
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Sudden closure of a valve.
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Answer: B. sudden change in flow velocity, such as rapid valve closure
Sudden stopping of flow converts kinetic energy to pressure energy, generating pressure waves that travel with celerity c.
108. The maximum pressure rise in a pipe due to instantaneous closure of a valve (Joukowsky) is
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Product of density, wave speed and velocity change.
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Answer: B. Δp = ρ c Δv
The head rise is Δh = cΔv/g, or Δp = ρcΔv, where c is the wave celerity.
109. Valve closure in a pipe of length L is considered 'rapid' (water hammer is at its maximum) if the time of closure is
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Round trip of the pressure wave.
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Answer: B. less than 2L/c
2L/c is the time for the pressure wave to return from the reservoir; closing within it gives the full Joukowsky rise.
110. A surge tank in a hydropower scheme is provided to
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It acts as a nearby free surface.
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Answer: D. reduce water hammer pressure in the penstock by absorbing flow variations
It gives a free water surface near the powerhouse, damping pressure waves and supplying water during load acceptance.
111. Which of the following is a relief device used against water hammer in pressurised pipelines?
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It opens at high pressure.
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Answer: A. A pressure relief (surge relief) valve
Relief valves open when pressure exceeds a set value, discharging water to limit the pressure surge; air vessels and surge tanks are other devices.
112. If the hydraulic grade line falls below the pipe centre line at a point, the pressure there is
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HGL = p/γ + z.
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Answer: A. below atmospheric (negative gauge)
HGL height above the pipe = gauge pressure head; if HGL is lower than the pipe, p/γ is negative.
113. For maximum power transmission through a pipeline of given total head H, the head lost to friction should be
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Differentiate P with respect to Q.
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Answer: C. H/3
P ∝ (H − h_f)Q with h_f ∝ Q²; maximising gives h_f = H/3 and a transmission efficiency of 66.7 %.
114. Water flows at 2 m/s through a 200 mm diameter, 500 m long pipe with f = 0.02. The friction head loss is
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Darcy–Weisbach.
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Answer: D. 10.19 m
h_f = f L v²/(2 g d) = 0.02 × 500 × 4/(2 × 9.81 × 0.2) = 10.19 m.
115. Flow in a pipe has Reynolds number 1600. The Darcy friction factor is
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f = 64/Re.
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Answer: C. 0.04
Laminar flow: f = 64/Re = 64/1600 = 0.04.
116. Oil (μ = 0.1 Pa·s, ρ = 900 kg/m³) flows laminarly at 0.5 m/s in a 50 mm pipe 100 m long. The head loss is
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Hagen–Poiseuille.
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Answer: B. 7.25 m
h_f = 32 μ L v/(ρ g d²) = 32 × 0.1 × 100 × 0.5/(900 × 9.81 × 0.0025) = 7.25 m.
117. Water flowing at 3 m/s through a pipe suddenly enters a larger pipe where velocity becomes 1 m/s. The head loss at the enlargement is
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Use the velocity difference squared.
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Answer: B. 0.204 m
h_L = (v₁ − v₂)²/2g = (3 − 1)²/19.62 = 0.204 m.
118. A valve with loss coefficient K = 10 is in a pipe carrying water at 2 m/s. The head loss across the valve is
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h_L = K v²/2g.
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Answer: C. 2.04 m
h_L = K v²/2g = 10 × 4/19.62 = 2.04 m.
119. Using the Hazen–Williams equation h_f = 10.67 L Q^1.852/(C^1.852 d^4.87), the head loss in a 1000 m long, 200 mm pipe with C = 120 carrying 50 L/s is about
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SI units: Q in m³/s, d in m.
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Answer: D. 14.9 m
h_f = 10.67 × 1000 × 0.05^1.852/(120^1.852 × 0.2^4.87) = 14.9 m.
120. Two pipes of diameters 300 mm and 150 mm have the same length and friction factor and are connected in parallel. The ratio of their discharges (large to small) is about
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Q ∝ d^(5/2) for equal head loss, L and f.
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Answer: A. 5.66
Same h_f gives Q ∝ d^2.5, so ratio = 2^2.5 = 5.66.
121. Two pipes in series, (L = 300 m, d = 0.2 m) and (L = 200 m, d = 0.1 m), both with f = 0.02, carry 0.02 m³/s. Neglecting minor losses, the total friction head loss is
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Compute each velocity from Q = Av.
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Answer: B. 13.8 m
v₁ = 0.637 m/s, v₂ = 2.546 m/s; h₁ = 0.62 m, h₂ = 13.22 m; total = 13.8 m.
122. In a Hardy Cross iteration for a loop, Σ(kQ|Q|) = 4.2 m and Σ(2k|Q|) = 56 s/m². The flow correction ΔQ is
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ΔQ = −(Σh)/(Σ2h/Q).
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Answer: D. -0.075 m³/s
ΔQ = −Σ(kQ|Q|)/Σ(2k|Q|) = −4.2/56 = −0.075 m³/s.
123. A pressure wave of celerity 1000 m/s travels in a pipe in which flow of 2 m/s is instantly stopped. The pressure rise (water, ρ = 1000 kg/m³) is
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Joukowsky.
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Answer: C. 2.0 MPa
Δp = ρcΔv = 1000 × 1000 × 2 = 2 × 10⁶ Pa = 2 MPa.
124. A penstock 1500 m long has a pressure-wave celerity of 1000 m/s. The critical period of valve closure 2L/c is
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Round-trip time of the wave.
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Answer: A. 3 s
T = 2L/c = 2 × 1500/1000 = 3 s.
125. A pipeline of total available head 90 m is to transmit maximum power. The friction head loss in the pipe should be
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One-third of the supply head.
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Answer: C. 30 m
For maximum power, h_f = H/3 = 30 m.
126. A pipe 100 m long and 100 mm diameter (f = 0.02) discharges freely from a reservoir whose water level is 20 m above the outlet. Including entrance (0.5) and exit (1.0) losses, the velocity in the pipe is
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Add all losses in series.
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Answer: C. 4.27 m/s
H = (0.5 + 1.0 + fL/d) v²/2g = (21.5) v²/2g, so v = √(2 × 9.81 × 20/21.5) = 4.27 m/s.
3.5 Open channel flow
32 questions · ACiE0305
127. The hydraulic radius of an open channel section is
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Wetted perimeter, not top width.
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Answer: C. flow area divided by wetted perimeter
R = A/P; the hydraulic depth, by contrast, is D = A/T.
128. The hydraulic depth of a channel section (used in the Froude number) is
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Top width, not wetted perimeter.
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Answer: A. flow area divided by top width
D = A/T; for a rectangular channel D equals the flow depth y.
129. The Froude number in an open channel is defined as
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The speed of a small surface wave is √(gD).
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Answer: C. v/√(g D)
Fr = v/√(gA/T) is the ratio of inertial to gravity forces; Fr < 1 is subcritical, Fr = 1 critical, Fr > 1 supercritical.
130. Uniform flow in an open channel requires that
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Depth does not change with distance.
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Answer: C. the bed slope, water-surface slope and energy slope are all equal
In uniform flow depth, velocity and area are constant along the channel, hence S₀ = S_w = S_f.
131. Manning's formula for velocity of uniform flow is
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Hydraulic radius carries the 2/3 power.
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Answer: C. v = (1/n) R^(2/3) S^(1/2)
Manning (SI): v = (1/n)R^(2/3)S^(1/2), related to Chezy's C by C = R^(1/6)/n.
132. The most efficient (best hydraulic) rectangular channel section has
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Minimum wetted perimeter.
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Answer: B. depth equal to half the width
For a given area, minimum perimeter occurs at b = 2y, giving R = y/2.
133. For the most efficient trapezoidal section, the hydraulic radius equals
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Same as for rectangle and triangle.
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Answer: D. half the flow depth
For any best hydraulic section R = y/2 (half hexagon for the trapezoid).
134. Specific energy in an open channel is the energy per unit weight of water measured with respect to
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The datum moves with the bed.
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Answer: B. the channel bed
E = y + αv²/2g, taken with the channel bed as datum.
135. At critical depth, for a given discharge, the specific energy is
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Differentiate E = y + Q²/2gA².
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Answer: C. a minimum
dE/dy = 0 gives Fr = 1; the specific energy curve has its minimum at yc.
136. For a rectangular channel at critical flow, the velocity head is equal to
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Use Fr = 1.
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Answer: B. half of the depth
v²/2g = yc/2 since v² = g yc; hence E_min = 1.5 yc.
137. For a given specific energy greater than the minimum, flow in an open channel can occur at
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See the E–y diagram.
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Answer: D. two alternate depths, one subcritical and one supercritical
The E–y curve has two branches for E > Emin: the upper (subcritical) and the lower (supercritical).
138. The specific force (momentum function) at two sections is equal for the
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Which principle is conserved across a jump?
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Answer: A. conjugate (sequent) depths of a hydraulic jump
Momentum is conserved across a jump (losses in energy), so M₁ = M₂ for y₁ and y₂.
139. A backwater curve upstream of a dam in a channel of mild slope is a
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Zone 1, mild slope.
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Answer: D. M1 profile
Depth exceeds both normal and critical depth (y > yn > yc) and increases downstream toward the dam: M1.
140. On a mild slope, a draw-down curve approaching a free overfall has the profile
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Depth falls toward the brink.
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Answer: D. M2
Depth lies between yn and yc and decreases downstream (zone 2); this is the M2 curve.
141. A gradually varied flow profile is classified as 'mild' when the
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Compare yn and yc.
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Answer: D. normal depth exceeds the critical depth
Mild: yn > yc (Fr < 1 in uniform flow); steep: yn < yc; critical: yn = yc.
142. A hydraulic jump forms when the flow changes from
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The upstream flow is fast.
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Answer: A. supercritical to subcritical
Rapidly varied flow with large energy dissipation occurs when Fr₁ > 1 changes to Fr₂ < 1.
143. A hydraulic jump with an upstream Froude number between 4.5 and 9 is classified as a
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Best performing for stilling basins.
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Answer: C. steady jump
Undular 1–1.7, weak 1.7–2.5, oscillating 2.5–4.5, steady 4.5–9, strong > 9; the steady jump gives about 45–70 % energy dissipation.
144. The Shields parameter is the dimensionless ratio
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Bed shear over submerged weight per unit area.
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Answer: D. τ₀/[(γs − γ) d]
θ = τ₀/[(γs − γ)d] compares bed shear to the submerged weight of grains; it is plotted against the boundary Reynolds number.
145. According to the Shields diagram, sediment motion is considered to begin when
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Think of bed shear stress.
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Answer: D. the bed shear stress exceeds the critical value for the grain size
Incipient motion occurs when τ₀ > τc, i.e. the Shields parameter exceeds its critical value (about 0.05 for coarse grains).
146. In a uniform flow, the average bed shear stress on a wide channel is
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Gravity component balances boundary shear.
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Answer: A. τ₀ = γ R S₀
A force balance on a control volume of uniform flow gives τ₀P = γ A S₀ so τ₀ = γRS₀.
147. A trapezoidal channel with bottom width 5 m, side slope 1.5 H : 1 V and flow depth 2 m has hydraulic radius
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R = A/P with slant length y√(1+z²).
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Answer: A. 1.31 m
A = (5 + 1.5 × 2) × 2 = 16 m²; P = 5 + 2 × 2 × √(1 + 1.5²) = 12.21 m; R = 1.31 m.
148. A rectangular channel 3 m wide carries uniform flow at 1.2 m depth, with n = 0.015 and S₀ = 0.001. The discharge is
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Compute R, then v, then Q = Av.
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Answer: B. 5.79 m³/s
R = 3.6/5.4 = 0.667 m; v = (1/0.015)(0.667)^(2/3)(0.001)^(1/2) = 1.609 m/s; Q = 5.79 m³/s.
149. Water flows at 3.5 m/s with depth 1.0 m in a rectangular channel. The Froude number and flow type are
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Fr = v/√(gy).
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Answer: C. 1.12, supercritical
Fr = v/√(gy) = 3.5/√9.81 = 1.12 > 1, so supercritical.
150. The critical depth in a wide rectangular channel for a discharge of 3 m³/s per metre width is
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yc = (q²/g)^(1/3).
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Answer: A. 0.97 m
yc = (q²/g)^(1/3) = (9/9.81)^(1/3) = 0.97 m.
151. The specific energy of a flow of depth 1.2 m and mean velocity 2 m/s is
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E = y + v²/2g.
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Answer: B. 1.40 m
E = y + v²/2g = 1.2 + 4/19.62 = 1.40 m.
152. The minimum specific energy for a discharge of 4 m³/s per metre width in a rectangular channel is
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E_min = 1.5 yc.
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Answer: B. 1.77 m
yc = (16/9.81)^(1/3) = 1.177 m; E_min = 1.5 yc = 1.77 m.
153. A hydraulic jump forms in a horizontal rectangular channel with upstream depth 0.4 m and velocity 6 m/s. The sequent depth is
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Belanger's equation.
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Answer: B. 1.53 m
Fr₁ = 6/√(9.81 × 0.4) = 3.03; y₂ = (y₁/2)(√(1+8Fr₁²) − 1) = 1.53 m.
154. In a horizontal rectangular channel, the sequent depth ratio y₂/y₁ for an upstream Froude number of 3 is
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Belanger's equation with Fr₁ = 3.
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Answer: C. 3.77
y₂/y₁ = ½(√(1 + 8 × 9) − 1) = ½(√73 − 1) = 3.77.
155. Uniform flow occurs in a wide channel with hydraulic radius 1.0 m and bed slope 0.001. The average bed shear stress is
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τ₀ = γRS.
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Answer: D. 9.81 Pa
τ₀ = γ R S₀ = 9810 × 1.0 × 0.001 = 9.81 Pa.
156. The critical shear stress for sand of diameter 1 mm (ρs = 2650 kg/m³) at a critical Shields parameter of 0.045 is
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Use the submerged density (ρs − ρ).
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Answer: A. 0.73 Pa
τc = θc(ρs − ρ) g d = 0.045 × 1650 × 9.81 × 0.001 = 0.73 Pa.
157. A rectangular channel is to carry 18 m² of flow area as the best hydraulic section. Its depth is
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b = 2y.
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Answer: A. 3.0 m
A = b y = 2y² = 18 gives y = 3 m and b = 6 m.
158. Using Lacey's regime formula v = (Q f²/140)^(1/6), the regime velocity for Q = 64 m³/s and silt factor f = 1 is
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Take the sixth root.
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Answer: B. 0.88 m/s
v = (64 × 1/140)^(1/6) = 0.457^(0.1667) = 0.88 m/s.
3.6 Hydrology
32 questions · ACiE0306
159. In the hydrologic cycle, the process by which water vapour changes directly from ice to vapour, bypassing the liquid phase, is called
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Solid to gas.
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Answer: A. sublimation
Sublimation is solid-to-vapour change; transpiration is vapour loss from plants, interception is retention on vegetation.
160. Precipitation that occurs when moist air is forced to rise over a mountain barrier is called
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Mountains lift the air.
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Answer: C. orographic precipitation
Orographic lifting causes cooling and condensation on windward slopes, which is a key feature of the Himalayan monsoon rainfall.
161. In the Thiessen polygon method of estimating mean areal rainfall, the weights assigned to the stations are
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Weights come from the area closest to each gauge.
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Answer: D. the areas of the polygons around the stations
P̄ = Σ(Pi Ai)/ΣAi, assuming each point is best represented by its nearest gauge.
162. In hilly and mountainous areas the most accurate method of computing average rainfall over a basin is generally the
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Contours of equal rainfall.
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Answer: D. isohyetal method
The isohyetal method accounts for orographic variation of rainfall by drawing contours of equal rainfall.
163. According to Horton's equation, the infiltration capacity of a soil during a prolonged storm
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Soil gets wetter and fills pores.
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Answer: A. decreases exponentially with time to a constant final rate
f = fc + (f0 − fc)e^(−kt); the initially dry soil absorbs water fast and then approaches a constant rate.
164. The φ-index is the
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Constant loss rate.
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Answer: A. constant rainfall intensity above which the rainfall volume equals the runoff volume
Rainfall in excess of φ is runoff (effective rainfall); the portion below φ is treated as infiltration and other losses.
165. The Soil Conservation Service (SCS) curve number method gives, for the same rainfall,
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CN = 100 means an impervious surface.
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Answer: C. more runoff for higher curve numbers
S = 25400/CN − 254 (mm); a higher CN means a smaller storage S and so a larger runoff.
166. The time of concentration of a catchment is the
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Travel time of the remotest drop.
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Answer: C. time for runoff from the farthest point of the basin to reach the outlet
It is the travel time for a drop from the hydraulically most remote point to the outlet; storm durations equal to tc give peak discharge in the rational method.
167. The rational formula for peak discharge is considered applicable to
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Assumes uniform rain over the basin.
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Answer: D. small catchments, because rainfall is assumed uniform over the area
The rational method (Q = CiA/360) assumes uniform intensity over tc, which only holds for small areas (typically below a few km²).
168. The stage–discharge (rating) curve of a river gauging section is generally expressed as
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Power-law form of rating curve.
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Answer: B. Q = C (G − a)ⁿ
G is the gauge height, a the gauge reading at zero flow and C, n are constants found from measurements (n ≈ 1.5–2).
169. In the two-point method of velocity measurement with a current meter, the mean velocity in a vertical is taken as the average of the velocities at depths of
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Symmetrical about mid-depth, but not at the ends.
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Answer: D. 0.2 and 0.8 times the depth from the surface
v̄ = (v₀.₂ + v₀.₈)/2; the single-point method uses v at 0.6 depth.
170. A unit hydrograph is the direct runoff hydrograph resulting from
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Effective, not total, rainfall.
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Answer: C. 1 cm of effective rainfall of a specified duration spread uniformly over the basin
A D-hour unit hydrograph is for 1 cm (or 1 mm) of excess rainfall over duration D; ordinates are in m³/s per cm.
171. The unit hydrograph theory is based on the assumptions of
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Principles of proportionality and superposition.
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Answer: C. linearity (superposition) and time invariance of the catchment response
Ordinates scale with rainfall excess and the responses to successive pulses are added; the same rainfall gives the same hydrograph at any time.
172. An S-curve hydrograph is mainly used to
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Think of changing the duration D.
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Answer: C. derive a unit hydrograph of a different duration
The S-curve is the response to continuous rain at constant intensity; subtracting a lagged S-curve gives a UH of another duration.
173. Which of the following is a synthetic unit hydrograph method?
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For ungauged catchments.
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Answer: C. Snyder's method
Snyder, SCS (Mockus) and Clark methods derive UH from catchment characteristics when no rain-runoff data exists; Thiessen is for areal rainfall and Horton for infiltration.
174. The return period T of a flood and its annual exceedance probability p are related by
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Reciprocals.
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Answer: D. T = 1/p
A flood with 1 % annual exceedance probability has a return period of 100 years (on average, not exactly every 100 years).
175. Gumbel's distribution is used in flood hydrology to model
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Extreme value distribution.
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Answer: D. extreme (maximum) values such as annual peak flows
Gumbel is the Type I extreme-value distribution, with Q_T = Q̄ + Kσ and frequency factor K a function of T.
176. A confined aquifer is one in which
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Look at the overlying layer.
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Answer: D. the water is overlain by an impermeable layer and is under pressure greater than atmospheric
In a confined (artesian) aquifer the piezometric surface lies above the top of the aquifer; an unconfined aquifer has a free water table.
177. The specific yield of an unconfined aquifer is always
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Some water cannot drain.
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Answer: B. less than its porosity
Specific yield = porosity − specific retention, so it is less than porosity because some water is held by capillary forces.
178. In a catchment, annual rainfall is 1200 mm, runoff is 400 mm and evaporation plus transpiration is 600 mm. The change in storage is
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Water balance: inputs − outputs.
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Answer: A. 200 mm
ΔS = P − R − ET = 1200 − 400 − 600 = 200 mm (increase).
179. Three raingauges have Thiessen polygon areas of 40, 60 and 100 km² and rainfall of 80, 60 and 50 mm respectively. The average rainfall is
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Area-weighted mean.
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Answer: D. 59 mm
P̄ = (80×40 + 60×60 + 50×100)/200 = 11 800/200 = 59 mm.
180. Using the rational formula Q = CiA/360 (i in mm/h, A in ha, Q in m³/s), the peak flow for C = 0.4, i = 60 mm/h and A = 25 ha is
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Multiply C, i and A, then divide by 360.
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Answer: B. 1.67 m³/s
Q = 0.4 × 60 × 25/360 = 1.67 m³/s.
181. A storm gives hourly rainfall of 10, 20, 15 and 5 mm/h in four successive hours and produces 18 mm of runoff. The φ-index is
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Excess rainfall equals the runoff depth.
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Answer: A. 9 mm/h
Assuming φ between 5 and 10: (10 − φ) + (20 − φ) + (15 − φ) = 18 gives φ = 9 mm/h (5 mm/h is below φ, so it is all loss).
182. Using the SCS formula Q = (P − 0.2S)²/(P + 0.8S) with S = 25400/CN − 254 (mm), the runoff from P = 100 mm on a catchment with CN = 80 is
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First compute potential retention S.
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Answer: A. 50.5 mm
S = 63.5 mm; Q = (100 − 12.7)²/(100 + 50.8) = 50.5 mm.
183. A flood has a return period of 50 years. The probability that it will be equalled or exceeded at least once in the next 10 years is
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Use 1 − (1 − 1/T)ⁿ.
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Answer: B. 0.183
P = 1 − (1 − 1/T)ⁿ = 1 − 0.98¹⁰ = 0.183.
184. For a 100-year flood, the Gumbel frequency factor for a very long record is K ≈ 3.14. For a mean annual peak of 500 m³/s and standard deviation 150 m³/s, Q₁₀₀ is
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Q_T = mean + K × standard deviation.
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Answer: C. 971 m³/s
Q_T = Q̄ + Kσ = 500 + 3.14 × 150 = 971 m³/s.
185. The ordinates of a 1-cm unit hydrograph, read at 1 h intervals, sum to 200 m³/s. The catchment area is
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Volume of 1 cm over the area.
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Answer: B. 72 km²
Volume = ΣQΔt = 200 × 3600 = 7.2 × 10⁵ m³ = 0.01 m × A, so A = 7.2 × 10⁷ m² = 72 km².
186. Groundwater flows through a sand layer of cross-section 10 m², hydraulic conductivity 1 × 10⁻⁴ m/s and hydraulic gradient 0.01. The discharge is
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Q = KiA.
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Answer: A. 1 × 10⁻⁵ m³/s
Q = K i A = 10⁻⁴ × 0.01 × 10 = 10⁻⁵ m³/s.
187. A well in a confined aquifer (T = 500 m²/day) has drawdowns of 3.0 m and 1.5 m at observation wells 10 m and 100 m away. The discharge by the Thiem equation is
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Q = 2πT Δs / ln(r2/r1).
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Answer: B. 2047 m³/day
Q = 2πT(s₁ − s₂)/ln(r₂/r₁) = 2π × 500 × 1.5/ln 10 = 2047 m³/day.
188. The water table in an unconfined aquifer of 5 km² area falls by 2 m. With specific yield 0.15, the volume of water released is
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V = Sy A Δh.
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Answer: A. 1.5 × 10⁶ m³
V = Sy × A × Δh = 0.15 × 5 × 10⁶ × 2 = 1.5 × 10⁶ m³.
189. A Class A pan evaporates 8 mm/day with a pan coefficient of 0.7. The estimated lake evaporation is
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Multiply by the pan coefficient.
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Answer: B. 5.6 mm/day
E_lake = Cp × E_pan = 0.7 × 8 = 5.6 mm/day.
190. A rating curve is Q = 20 (G − 0.5)^1.8 (Q in m³/s, G in m). The discharge at a gauge height of 2.5 m is
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Subtract the zero-flow gauge reading first.
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Answer: B. 69.6 m³/s
Q = 20 × 2^1.8 = 20 × 3.482 = 69.6 m³/s.