Question 1

Definition of uniform flow in open channels  Uniform flow in an open channel is said to occur when which of the following is true along the reach considered?

Accepted Answer

Introduction / Context:Uniform flow is a basic idealization in open-channel hydraulics used in Manning and Chezy formulations. Correctly identifying uniform conditions is crucial for reliable design of channels, canals, and sewers. Given Data / Assumptions: Steady flow consideration. One-dimensional gradually varied assumptions. No lateral inflow/outflow for strict uniformity. Concept / Approach:Uniform flow means that hydraulic properties do not change spatially: depth y, area A, velocity V, and discharge Q are all constant along the reach. Consequently, the bed slope S, water surface slope, and energy grade line slope Sf are equal, and specific energy and total head per unit weight remain constant except for loss balanced by bed drop. Step-by-Step Solution: Define uniform flow: dV/dx = 0 and dy/dx = 0 along the reach.Implying constant discharge and depth → constant mean velocity for a prismatic channel.Therefore, the correct statement is that depth and mean velocity remain constant with distance. Verification / Alternative check:Energy balance for uniform flow gives Sf = S0 (bed slope) and water surface slope equal to these; profiles are parallel. Why Other Options Are Wrong: Lateral inflow/outflow makes discharge vary → non-uniform. Sf < S or Sf > S indicates accelerating or decelerating gradually varied flow. Uniform flow can occur in non-prismatic channels over short reaches if properties remain effectively constant; prismatic geometry is typical but not a strict logical requirement of the definition. Common Pitfalls:Confusing steady with uniform; equating “prismatic” with “uniform”; ignoring that uniform flow requires absence of spatial changes. Final Answer:Depth and mean velocity remain constant with distance along the channel.

Question 2

Newtonian fluids – what is shear stress directly proportional to? In a Newtonian fluid under simple shear, identify the quantity to which the shear stress is directly proportional (assume constant temperature and composition).

Accepted Answer

Introduction / Context:In fluid mechanics, correctly relating shear stress to motion is essential for modeling laminar flow between plates, in pipes, and within lubrication films. For Newtonian fluids (water, air, many oils in moderate ranges), a simple linear law connects stress to the rate at which adjacent fluid layers slide past each other. Given Data / Assumptions: Continuum, incompressible Newtonian fluid. Simple shear flow with velocity u varying in the transverse direction y. Temperature and composition are constant so viscosity is constant. Concept / Approach: The Newtonian constitutive relation states that shear stress tau is proportional to the rate of shear strain. In one-dimensional simple shear: tau = mu * (du/dy), where mu is dynamic viscosity and du/dy is the velocity gradient normal to the flow direction. This captures how faster relative sliding of adjacent layers increases internal frictional resistance linearly for Newtonian behavior. Step-by-Step Solution: Identify the motion descriptor: not velocity itself, but how velocity changes with position → velocity gradient du/dy.Write Newton’s law of viscosity: tau = mu * du/dy.Conclude: proportionality is between tau and du/dy (with mu as proportionality constant). Verification / Alternative check: Between parallel plates separated by gap h, moving-plate speed U, the profile is linear, du/dy = U/h, giving tau = mu * U/h. Doubling U doubles tau; halving h doubles tau, as observed experimentally for Newtonian fluids. Why Other Options Are Wrong: (a) Velocity at a point is not the cause; spatial variation of velocity is. (b) Shear strain is a geometric angle; rate of change (per time) matters in fluids. (d) Viscosity sets the proportionality but motion is required; tau is zero if du/dy = 0. (e) Square dependence appears in some turbulent drag laws, not in Newtonian constitutive law. Common Pitfalls: Confusing laminar constitutive law with empirical turbulent wall-shear correlations; mixing up strain (angle) with strain rate (time rate of angle change). Final Answer: the rate of shear strain (velocity gradient du/dy)

Question 3

Classifying a flow that varies in space and time If the velocities of fluid particles change in magnitude and direction from point to point and also from instant to instant, how should the flow be classified?

Accepted Answer

Introduction / Context:Engineers distinguish flows by how properties vary with time (steady vs unsteady) and space (uniform vs non-uniform). This classification underpins when Bernoulli can be applied directly and how to set up control-volume balances or CFD boundary conditions. Given Data / Assumptions: Velocity changes with location → spatial variation (non-uniform). Velocity changes with time at fixed points → temporal variation (unsteady). No assumptions on viscosity or Reynolds number are explicitly given. Concept / Approach: Uniform means no change along the direction considered; steady means no time change at a fixed point. The statement explicitly says the velocity varies with position and with time, so the flow is both non-uniform and unsteady. Many such flows are turbulent, but turbulence is not strictly required by the wording; still, the description often matches turbulent behavior in practice. Step-by-Step Solution: If ∂V/∂t ≠ 0 at a fixed point → unsteady.If ∂V/∂x, ∂V/∂y, or ∂V/∂z ≠ 0 along the flow → non-uniform.Therefore classification: unsteady and non-uniform. Verification / Alternative check: Examples: gusty wind in a street (changes with both location and time); pump start-up transients in a pipe with varying diameter. Both require time-dependent and spatially varying analyses. Why Other Options Are Wrong: (a) Steady & uniform contradicts the description. (b) Uniformity fails (flow varies in space). (c) Steadiness fails (flow varies in time). (e) Laminar refers to orderly layer motion at low Reynolds numbers; the question concerns variability, not laminarity. Common Pitfalls: Equating “unsteady” with “turbulent.” Turbulence implies irregular fluctuations, but a laminar flow can still be unsteady (e.g., oscillatory Poiseuille flow). Final Answer: unsteady and non-uniform flow (often turbulent)

Question 4

Most economical trapezoidal channel section (maximum discharge) For the most economical (maximum discharge) trapezoidal channel section, which of the following conditions hold true simultaneously?

Accepted Answer

Introduction / Context:The “most economical” open-channel section maximizes discharge (or minimizes wetted perimeter) for a given flow area, reducing friction losses and construction cost. For trapezoidal channels (widely used in irrigation and drainage), elegant geometric conditions characterize this optimum. Given Data / Assumptions: Prismatic trapezoidal channel: bed width b, side slope z (horizontal:vertical), depth y. Uniform, steady, subcritical flow; resistance described by Manning or Chezy. Objective: minimize wetted perimeter P for given area A to maximize A/P (hence hydraulic radius R). Concept / Approach: For a given A, discharge Q ∝ R^(2/3) A^(5/3) (Manning). Maximizing Q reduces to maximizing R = A/P at fixed A, i.e., minimizing P. Using calculus of variations or standard derivations yields relations among b, y, and z for the optimum trapezoid. Step-by-Step Solution: Area: A = b y + z y^2; Wetted perimeter: P = b + 2 y √(1 + z^2).Set ∂P/∂b subject to constant A: leads to b = 2 y (√(1 + z^2) − z).Geometric consequences: (i) half of top width T/2 equals the sloping side length y √(1 + z^2); (ii) hydraulic mean depth R = A/P = y/2; (iii) perpendiculars from the top-width midpoint to the sloping sides are equal by symmetry. Verification / Alternative check: For the most economical triangle (b = 0), the side slope satisfies z = 1/√2; for a nonzero b the above relationships collapse consistently and reproduce classic textbook results such as R = y/2. Why Other Options Are Wrong: Each of (a), (b), and (c) is individually correct, so any single choice alone is incomplete. The comprehensive answer is that all stated geometric conditions are satisfied at the optimum. Common Pitfalls: Confusing hydraulic depth (A/T) with hydraulic radius (A/P); forgetting that “most economical” refers to geometry, not necessarily cost-only factors like lining roughness or freeboard. Final Answer: all the above

Question 5

Energy grade line (TEL) and hydraulic grade line (HGL) Which statement is correct regarding the graphical interpretation of head lines in pipe flow?

Accepted Answer

Introduction / Context:Plotting energy lines helps visualize head distribution along pipes, nozzles, and fittings. The total energy line (TEL) and hydraulic grade line (HGL) separate dynamic and static contributions and reveal where pumps add head or friction removes it. Given Data / Assumptions: Total head H = z + p/γ + V^2/(2g). Hydraulic grade line is z + p/γ (piezometric head). Steady, incompressible flow in a conduit with losses. Concept / Approach: The HGL expresses static energy (elevation plus pressure head). The TEL adds velocity head on top. Their vertical separation is precisely the kinetic term V^2/(2g). Friction and minor losses make both lines slope downward in the flow direction unless a pump elevates the TEL (and HGL accordingly). Step-by-Step Solution: Define TEL: H = z + p/γ + V^2/(2g).Define HGL: h = z + p/γ.Vertical gap between lines: H − h = V^2/(2g) → velocity head. Verification / Alternative check: At a reservoir (V ≈ 0), TEL and HGL coincide. At a nozzle exit to atmosphere, HGL meets the exit elevation line (p/γ = 0), while TEL remains above by the exit velocity head. Why Other Options Are Wrong: (a) TEL includes elevation, pressure, and velocity heads. (c) The separation is kinetic, not static pressure. (d) HGL is always at or below TEL; it cannot lie above. (e) With friction, TEL drops; zero difference implies no losses or a frictionless, idealization-only case. Common Pitfalls: Confusing pressure head with piezometric head; forgetting to include minor-loss-induced TEL drops at fittings and valves. Final Answer: The vertical distance between the total energy line (TEL) and hydraulic grade line (HGL) equals the velocity head

Question 6

Where is Manning’s formula applicable? Identify the correct application area for Manning’s equation relating velocity/discharge to roughness, hydraulic radius, and slope.

Accepted Answer

Introduction / Context:Manning’s equation is a cornerstone of open-channel hydraulics, used to compute normal depth, discharge, and backwater profiles in rivers, canals, storm sewers running partially full, and lined channels. Given Data / Assumptions: Steady, uniform (or gradually varied) flow in a prismatic open channel. Empirical roughness coefficient n represents boundary resistance. Hydraulic radius R = A/P; bed slope approximates energy slope for uniform flow. Concept / Approach: Manning’s formula: V = (1/n) * R^(2/3) * S^(1/2), Q = A * V. It is intended for free-surface flows where the hydraulic radius and slope characterize resistance. While sometimes adapted to pipes running partially full (treated as open channels), it is not the standard for full pressurized pipe friction (Darcy–Weisbach is preferred). Step-by-Step Solution: Identify regime: free-surface flow → open channel.Apply V = (1/n) R^(2/3) S^(1/2) to compute velocity, then Q = A V.Iterate on normal depth or design dimensions as required. Verification / Alternative check: Compare results with Chezy or Darcy–Weisbach adapted for open channels; consistent results validate the use of Manning in its empirical range. Why Other Options Are Wrong: (b) Manning gives velocity/discharge, not directly “head loss only.” (c) Pressurized full-pipe friction should use Darcy–Weisbach/Hazen–Williams, not Manning. (d) Compressible gas flow requires different models. (e) Unsteady surge needs waterhammer equations, not Manning alone. Common Pitfalls: Using an inappropriate n; assuming S equals bed slope in rapidly varied flow; applying Manning outside its empirical calibration range. Final Answer: flow in open channels

Question 7

Center of pressure on an inclined plane surface For a plane surface fully submerged and inclined, the center of pressure is located where relative to the centroid of the area?

Accepted Answer

Introduction / Context:Designing gates, tanks, and dam faces requires knowing where the resultant hydrostatic force acts. For plane surfaces, the resultant acts at the center of pressure, generally not at the centroid due to the pressure variation with depth. Given Data / Assumptions: Incompressible fluid at rest; hydrostatic pressure increases with depth. Plane surface fully submerged; surface is inclined with its centroid below the free surface. Pressure distribution is linear with depth. Concept / Approach: The resultant hydrostatic force equals pressure at the centroid times the area, but its line of action passes below the centroid. The depth of the center of pressure y_cp (measured vertically) is y_cp = y_c + I_G / (y_c A), where y_c is centroid depth, I_G is the second moment of area about the free-surface-parallel axis through the centroid, and A is area. Since I_G, y_c, A are positive, y_cp > y_c, i.e., deeper than the centroid. Step-by-Step Solution: Compute pressure at depth y: p = γ y (linear).Resultant force: F = γ A y_c (acts normal to surface).Center of pressure depth: y_cp = y_c + I_G/(y_c A) → strictly greater than y_c. Verification / Alternative check: For a rectangular plate of height h with centroid depth y_c, the added term simplifies to h^2/(12 y_c) for the axis choice, which is positive, confirming the center of pressure is below the centroid for any finite y_c. Why Other Options Are Wrong: (a) Only true if pressure were uniform, which it is not in a fluid at rest with depth. (b) Above the centroid is impossible for the linear hydrostatic distribution. (d) Position depends on geometry and submergence. (e) Free surface is not a feasible location for a submerged plane’s resultant. Common Pitfalls: Confusing line of action with magnitude; forgetting the inclination angle does not change the conclusion about being deeper than the centroid. Final Answer: below the centroid (deeper than the centroid)

Question 8

Non-dimensional number governing jet breakup/spray formation The formation of liquid sprays from an orifice or nozzle is primarily characterized by which dimensionless number?

Accepted Answer

Introduction / Context:When a liquid jet emerges from a nozzle, it may remain intact, form ligaments, or break into droplets. Predicting atomization is vital for fuel injectors, sprinklers, and spray dryers. Surface tension competes with inertia to resist breakup, and their ratio is captured by a specific dimensionless group. Given Data / Assumptions: Largely incompressible liquid; ambient gas resistance present. Characteristic length typically the jet diameter; characteristic velocity the exit speed. Gravity may be secondary compared to capillarity and inertia during initial breakup. Concept / Approach: The Weber number We = ρ V^2 L / σ compares inertial forces to surface-tension forces. Higher We means inertia overwhelms surface tension, promoting ligament formation and droplet breakup. Reynolds number governs viscous effects, while Froude assesses gravity influence; Mach concerns compressibility of gases and is irrelevant at typical liquid-jet speeds. Step-by-Step Solution: Identify competing mechanisms: inertia (destabilizing) vs surface tension (stabilizing).Form We = ρ V^2 L / σ; check magnitude for the application.If We exceeds a critical value (geometry dependent), atomization occurs. Verification / Alternative check: Empirical spray maps classify regimes (dripping, jetting, atomization) primarily by We and Re; breakup boundaries shift with nozzle design and ambient gas properties. Why Other Options Are Wrong: (a) Gravity is less dominant at nozzle exit for small L, high V. (c) Viscosity influences internal losses and ligament thinning but does not set the primary breakup threshold. (d) Liquids are effectively incompressible; Mach applies to high-speed gases. (e) Euler is not the standard criterion for jet breakup. Common Pitfalls: Using Re alone to predict atomization; neglecting σ changes with temperature or surfactants, which alter We and thus breakup behavior. Final Answer: Weber number (inertia vs surface tension)

Question 9

Hydraulic coefficients of an orifice – correct relation Given the standard hydraulic coefficients of an orifice (C_v: coefficient of velocity, C_c: coefficient of contraction, C_d: coefficient of discharge, C_r: coefficient of resistance), which rel…

Accepted Answer

Introduction / Context:Discharge through a sharp-edged orifice differs from the ideal due to jet contraction and frictional/viscous effects. Engineers summarize these effects with three widely used coefficients that multiply to the net discharge coefficient. Given Data / Assumptions: C_v accounts for the actual jet speed at the vena contracta relative to ideal velocity. C_c accounts for the reduced jet area at the vena contracta relative to the orifice area. C_d accounts for the overall discharge reduction: C_d = Q_actual / Q_ideal. Concept / Approach: The actual discharge Q_actual = A_c * V_actual, where A_c = C_c * A_orifice and V_actual = C_v * V_ideal. Therefore, Q_actual = (C_c * A_orifice) * (C_v * V_ideal) = (C_c * C_v) * (A_orifice * V_ideal). Hence, C_d = Q_actual / Q_ideal = C_c * C_v. Step-by-Step Solution: Define Q_ideal = A * V_ideal (from Torricelli/Bernoulli).Define A_c = C_c A; V_actual = C_v V_ideal.Compute Q_actual = A_c V_actual = C_c C_v A V_ideal.Therefore C_d = Q_actual/Q_ideal = C_c C_v. Verification / Alternative check: Typical sharp-edged values: C_c ≈ 0.62, C_v ≈ 0.97, giving C_d ≈ 0.60, consistent with laboratory measurements. Why Other Options Are Wrong: (b) C_r is not defined by that formula; it is sometimes related to head-loss but not as shown. (c) and (e) are dimensionally inconsistent with definitions. (d) Inverts the correct product relation. Common Pitfalls: Confusing coefficients at the vena contracta with discharge coefficient at the orifice plane; mixing coefficients from different orifice geometries. Final Answer: C_d = C_c * C_v

Question 10

Water density of approximately 1000 kg/m^3 – at what condition? The statement “the density of water is about 1000 kg/m^3” is most accurate under which condition?

Accepted Answer

Introduction / Context:Texts often quote water’s density as 1000 kg/m^3, but this is temperature dependent. Recognizing the correct reference condition avoids small but important errors in buoyancy, hydrometry, and calibration of volumetric glassware. Given Data / Assumptions: Fresh water (not saline), near 1 atm pressure. Density varies primarily with temperature over typical laboratory ranges. Compressibility effects due to pressure are minor at near-atmospheric conditions. Concept / Approach: Pure water reaches its maximum density near 4°C, about 999.97 kg/m^3, commonly rounded to 1000 kg/m^3. At higher temperatures (e.g., 20°C), density is lower (~998 kg/m^3), and at 0°C it is also lower (~999.8 kg/m^3), with small pressure dependencies around atmospheric pressure. Step-by-Step Solution: Recognize that “1000 kg/m^3” is an approximation.Identify the temperature of maximum density for pure water: ~4°C.Conclude that the quoted value is most accurate near 4°C at approximately 1 atm. Verification / Alternative check: Density tables used for hydrometers and volumetric calibration confirm the maximum near 4°C, with deviations of about 0.2% across typical room temperatures. Why Other Options Are Wrong: (a) Pressure specification alone is insufficient; temperature matters. (c) “Any temperature” is incorrect because density changes with temperature. (d) Not all conditions are equally accurate. (e) Vacuum conditions are irrelevant to liquid water’s density at standard states. Common Pitfalls: Confusing density with specific weight (≈9.81 kN/m^3); assuming salinity has no effect (seawater is denser than pure water). Final Answer: at approximately 4°C (fresh water, near standard pressure)

Question 11

In fluid mechanics instrumentation, what does a piezometer specifically measure when connected to a point in a water channel or pipe (assuming the fluid is a liquid and the pressure is not below atmospheric)?

Accepted Answer

Introduction / Context:A piezometer is one of the simplest pressure-measuring devices in hydraulics. It is widely used in civil engineering laboratories and field work to read the pressure head at a point in a liquid system such as a pipe, channel, or tank. Given Data / Assumptions: The fluid is a liquid (not a gas) and is exposed to atmospheric pressure at the piezometer tube top. Pressures are not below atmospheric (no cavitation or vacuum conditions). Steady conditions and negligible capillary effects (tube sufficiently large). Concept / Approach: A piezometer is simply a vertical, transparent tube tapped into the flow at the point of interest. The static liquid column rises to an elevation where the hydrostatic head balances the local gauge pressure. The piezometric reading directly gives pressure head p/(rhog) above atmospheric at that tap location. Step-by-Step Solution: Connect the small vertical tube to the pipe or channel at the point of interest.Allow the liquid column in the tube to rise to a stable level open to atmosphere.Measure the height h of the liquid column above the tap centreline; this equals pressure head p/(rhog).Convert to pressure if required by p = rho * g * h. Verification / Alternative check: Manometers with two limbs can measure pressure differences; a single-limb piezometer reads only the local pressure head and cannot measure vacuum (sub-atmospheric) pressure correctly because the column would break. Why Other Options Are Wrong: (b) A piezometer does not provide differential pressure; that is the role of differential manometers. (c) Atmospheric pressure is measured by a barometer. (d) Very low (vacuum) pressure cannot be handled by a simple open piezometer. (e) Not applicable. Common Pitfalls: Using a piezometer for gases or for negative gauge pressures; ignoring elevation of tap point when converting head to pressure. Final Answer: pressure in water channels and pipes (pressure head at a point)

Question 12

Free-surface shape in a rotating cylinder: how does the rise of liquid at the wall compare to the depression at the rotation axis?

Accepted Answer

Introduction / Context:When a cylindrical vessel of liquid rotates steadily about its vertical axis, the free surface forms a paraboloid of revolution. Engineers use this fact to predict head distribution in rotating equipment and mixing tanks. Given Data / Assumptions: Rigid-body rotation (no sloshing), steady angular speed. Incompressible liquid, negligible evaporation and capillarity. Container is completely filled to leave a free surface exposed to atmosphere. Concept / Approach: The free surface adjusts until pressure is hydrostatic in a rotating frame: p/ρ + g z − (ω^2 r^2)/2 = constant. This gives a parabolic surface z = (ω^2 r^2)/(2g) + constant. Conservation of volume requires that the volume rising near the wall equals the volume depressed near the centreline, so the rise at the wall equals the depression at the axis measured from the original horizontal level. Step-by-Step Solution: Start from rotating hydrostatic relation: p/ρ + g z − (ω^2 r^2)/2 = constant.Free surface has p = atmospheric ⇒ z = (ω^2 r^2)/(2g) + C.Compute average z over the cross-section: volume conservation keeps mean level unchanged.Hence, rise at wall equals depression at axis from the initial level. Verification / Alternative check: Integrating z over the radius yields zero net change in volume relative to the original plane surface, confirming symmetry of rise and depression magnitudes. Why Other Options Are Wrong: (a) and (b) violate volume conservation in steady rotation; (d) viscosity affects transient approach, not the final static shape; (e) the relationship is deterministic, not random. Common Pitfalls: Confusing transient waves with steady rigid-body rotation; forgetting the reference is the original mean level. Final Answer: equal to the depression at the axis

Question 13

Weirs and nappes terminology: when no air is entrained below the nappe and the jet adheres to the downstream face, what is the nappe called?

Accepted Answer

Introduction / Context:Over weirs, the sheet of water that flows over the crest is called the nappe. Its behaviour depends on ventilation and pressure beneath the nappe, which influences discharge coefficients and downstream conditions. Given Data / Assumptions: Sharp-crested or broad-crested weir with visible nappe. No air space under the nappe; the jet adheres to the downstream face. Atmospheric conditions elsewhere, steady flow. Concept / Approach: If the space beneath the nappe is freely ventilated, the pressure under the nappe is atmospheric and the nappe separates cleanly (free nappe). When ventilation is insufficient, sub-atmospheric pressure forms, drawing the nappe toward the downstream face; the jet then adheres, producing a clinging (or depressed) nappe depending on the exact profile. The required technical term for adherent behaviour is clinging nappe. Step-by-Step Solution: Check ventilation: no air beneath nappe ⇒ sub-atmospheric pressure develops.Observe adherence to downstream face ⇒ clinging behaviour.Name the condition: clinging nappe is the correct term. Verification / Alternative check: Textbook definitions classify nappes as free (ventilated) and clinging/depressed (unventilated). Adherence distinguishes clinging from a separated free jet. Why Other Options Are Wrong: (a) Free nappe requires ventilation and detachment. (b) Depressed nappe describes lower-than-atmospheric pressure causing contraction but without full adherence; the prompt explicitly says it adheres, so clinging is precise. (d) Ventilated nappe contradicts the condition. (e) Submerged nappe is when tailwater submerges the crest. Common Pitfalls: Using “free” whenever the jet is visible; ignoring the role of ventilation under the nappe. Final Answer: clinging nappe

Question 14

Pipe head loss terminology: what is the ratio between the Darcy–Weisbach friction factor and the Fanning (coefficient of) friction used in head-loss relations?

Accepted Answer

Introduction / Context:Two friction parameters are common in hydraulics: the Darcy–Weisbach friction factor f (used in head loss h_f = f * (L/D) * V^2/(2g)) and the Fanning friction factor f_F (often called the coefficient of friction in some texts). Consistency between them avoids errors in design calculations. Given Data / Assumptions: Steady internal flow in a pipe of diameter D and length L. Mean velocity V; gravitational acceleration g. Standard definitions of Darcy and Fanning friction factors. Concept / Approach: The relationship is f (Darcy) = 4 * f_F (Fanning). This arises because the Darcy form lumps wall shear on all sides, whereas the Fanning form is based directly on wall shear stress relative to dynamic pressure. Using the wrong conversion multiplies head loss or pressure drop by a factor of four in error. Step-by-Step Solution: Write Darcy head loss: h_f = f * (L/D) * V^2/(2g).Write Fanning pressure drop: Δp = 4 f_F * (L/D) * (ρ V^2/2).Match terms: f = 4 f_F.Therefore, the ratio f / f_F = 4. Verification / Alternative check: Dimensional analysis and standard Moody chart annotations confirm the 4:1 relationship. Why Other Options Are Wrong: (a), (b), (c), and (e) contradict the established equivalence; using them would mispredict losses. Common Pitfalls: Reading a Moody chart labeled with the Darcy factor but plugging into a Fanning-based formula (or vice versa) without conversion. Final Answer: 4

Question 15

Hydraulically most economical triangular channel section: which triangular geometry maximizes discharge for given area, slope, and roughness?

Accepted Answer

Introduction / Context:The “most economical” channel section minimizes wetted perimeter P for a given flow area A, thereby maximizing hydraulic radius R = A/P and mean velocity for the same slope and roughness. For a triangular section, this leads to a specific apex (vertex) angle condition. Given Data / Assumptions: Open-channel, steady uniform flow. Triangular cross-section with two equal sloping sides (isosceles). Optimization under Chezy/Manning framework. Concept / Approach: For a triangular section with side slopes symmetric, calculus shows that the wetted perimeter is minimized when each side is inclined at 45° to the channel bed, giving an included vertex angle of 90°. This is commonly expressed as the most economical triangular section. The “right-angled triangle with equal sides” describes precisely this geometry in cross-section. Step-by-Step Solution: Express A and P in terms of depth y and side slope.Maximize R = A/P with respect to the side slope angle.Optimization yields side inclination of 45° to the bed ⇒ vertex angle = 90°.Hence the cross-section is an isosceles right triangle (equal sides meeting at 90°). Verification / Alternative check: Textbook results: for triangles, most economical section has half-apex angle = 45°, equivalent to vertex angle = 90°. Why Other Options Are Wrong: (a) Equilateral gives vertex angle 60°, not optimal. (b) “Right-angled triangle” is too general and could include unequal sides. (c) 45° vertex angle is not correct; the correct vertex angle is 90°. (e) Not applicable. Common Pitfalls: Confusing the condition for rectangles (b = 2y) with the triangular case; misreading “45°” as vertex instead of half-apex angle. Final Answer: right-angled triangle with equal sides (vertex angle 90°)

Question 16

Dimensional analysis: what are the base dimensions of dynamic viscosity μ in M–L–T terms?

Accepted Answer

Introduction / Context:Dynamic viscosity μ quantifies a fluid’s resistance to shear. Correct base dimensions are essential for checking equations, converting units, and building dimensionless groups like Reynolds number. Given Data / Assumptions: Shear stress τ = μ * (du/dy). Shear stress has dimensions of force/area. Velocity gradient du/dy has dimensions of T^-1. Concept / Approach: Start by writing τ in dimensions. Force is M L T^-2; area is L^2; hence τ has dimensions M L^-1 T^-2. Since τ = μ * (du/dy) and du/dy has dimensions T^-1, solving for μ yields μ = τ / (du/dy) with dimensions M L^-1 T^-1. Step-by-Step Solution: Write τ = F/A ⇒ [τ] = (M L T^-2) / L^2 = M L^-1 T^-2.Note [du/dy] = (L T^-1)/L = T^-1.Compute [μ] = [τ]/[du/dy] = (M L^-1 T^-2) / (T^-1) = M L^-1 T^-1.Therefore, μ has dimensions M L^-1 T^-1. Verification / Alternative check: In SI, μ units are Pa·s = (N/m^2)·s = (kg·m^-1·s^-2)·s = kg·m^-1·s^-1, matching M L^-1 T^-1. Why Other Options Are Wrong: (a) corresponds to impulse per area; (b) and (c) are dimensionally inconsistent for μ. (e) not needed. Common Pitfalls: Confusing dynamic viscosity with kinematic viscosity ν whose dimensions are L^2 T^-1. Final Answer: M L^-1 T^-1

Question 17

Chezy’s uniform flow relation — identifying terms: In the classical form V = C √(m i) for open-channel flow, what do the symbols represent?

Accepted Answer

Introduction / Context:Chezy’s formula is a foundational empirical relation for steady, uniform open-channel flow. It links the mean velocity to channel characteristics and slope through the Chezy coefficient C. Given Data / Assumptions: Uniform flow: depth and velocity do not vary along the channel reach. Steady conditions; prismatic channel. Hydraulic mean depth m = A/P, where A = area and P = wetted perimeter. Concept / Approach: The relation is V = C √(m i). Here V is the section-averaged velocity, m is the hydraulic radius (often denoted R = A/P, also called hydraulic mean depth in many texts), and i is the energy gradient (loss of head per unit length), equal to the bed slope in uniform flow. Step-by-Step Solution: State Chezy equation: V = C √(m i).Identify m = A/P, the hydraulic radius.Recognize i as head loss per unit length (S_f), equal to slope for uniform flow.Hence, options (a), (b), and (c) are all correct, so choose (d). Verification / Alternative check: Manning’s equation V = (1/n) m^(2/3) i^(1/2) is consistent and can be rearranged to a Chezy form with C related to n and m. Why Other Options Are Wrong: (e) contradicts standard definitions used across hydraulics literature. Common Pitfalls: Confusing m with depth y; mixing up slope of bed with water surface—only equal in uniform flow. Final Answer: All of the above

Question 18

Power transmission through a pipeline: at what fraction of the supplied head should the friction head loss be for maximum power at the outlet?

Accepted Answer

Introduction / Context:When fluid power is delivered through a long pipe to a turbine or nozzle, frictional head loss reduces the net head at the outlet. The power delivered P is maximized at a particular balance between losses and available head. Given Data / Assumptions: Total supply head = H. Friction head loss in pipe = h_f; outlet head = H − h_f. Flow rate Q depends on head and losses (for a given pipe, Q ∝ √(H − h_f) when losses scale with Q^2). Concept / Approach: Power at outlet is P = ρ g Q (H − h_f). For turbulent pipe flow with Darcy–Weisbach losses, h_f ∝ Q^2. Writing h_f = k Q^2, we obtain P = ρ g (H Q − k Q^3). Differentiating with respect to Q and setting dP/dQ = 0 gives H − 3 k Q^2 = 0 ⇒ h_f = k Q^2 = H/3. Thus, maximum power occurs when friction loss is one-third of the total head. Step-by-Step Solution: Model losses: h_f = k Q^2.Write power: P = ρ g Q (H − h_f).Differentiate: dP/dQ = ρ g (H − 3 k Q^2) = 0 ⇒ h_f = H/3.Hence, friction loss should be one-third of supply head for maximum P. Verification / Alternative check: At optimum, outlet head is 2H/3, a known result quoted in standard hydraulics texts. Why Other Options Are Wrong: (a) and (c) do not satisfy the optimization; (d) leaves zero usable head; (e) is impossible in real pipes. Common Pitfalls: Maximizing efficiency rather than absolute power; overlooking that maximizing P allows significant but optimal loss. Final Answer: one-third of the total head supplied

Question 19

Specific weight (unit weight) of a liquid: does it vary with location on Earth?

Accepted Answer

Introduction / Context:Specific weight γ (also called unit weight) is defined as γ = ρ g. Unlike density ρ, which depends mainly on temperature and pressure, γ also depends on local gravitational acceleration g, which varies over Earth’s surface. Given Data / Assumptions: Liquid of given density ρ at a fixed temperature. Local gravity g differs with latitude and elevation. No phase change or strong compressibility effects. Concept / Approach: Because γ = ρ g, variation in g (approximately 9.78–9.83 m/s^2 on Earth) translates to small but real differences in γ by location. Therefore, the specific weight is not strictly constant from place to place on Earth, even if ρ were held constant. Step-by-Step Solution: Define specific weight: γ = ρ g.Recognize that g varies with latitude, elevation, and local geology.Conclude γ must vary geographically for the same liquid state.Hence, it varies from place to place on Earth. Verification / Alternative check: Gravity models (e.g., WGS) publish spatial variations in g, confirming the conclusion. Why Other Options Are Wrong: (a) Incorrect because g is not constant. (b) true in spirit but vague compared to (c). (d) Speculative and irrelevant; γ would vary with gravity on other planets as well. (e) Redundant when (c) states the fact directly. Common Pitfalls: Confusing density (material property) with specific weight (depends on gravity as well). Final Answer: varies from place to place on the earth

Question 20

Cipolletti (often misspelled “Cappoletti”) weir identification: which description best characterizes this standard weir type used to compensate end contractions?

Accepted Answer

Introduction / Context:The Cipolletti weir is a standardized trapezoidal weir shape intended to offset end-contraction effects so that its discharge formula resembles that of a fully contracted rectangular weir. Many exam banks misspell it as “Cappoletti.” Given Data / Assumptions: Sharp-crested weir operated free (ventilated nappe). Standard approach and measurement practices. Fixed trapezoidal side slopes chosen to compensate contractions. Concept / Approach: Cipolletti weirs are trapezoidal: the central portion acts like a rectangular section while the sloping sides provide additional area proportional to head, approximating the discharge of an equivalent contracted rectangular weir. Pedagogically, this is described as a combination of rectangular and triangular (V-notch-like) effects, hence “combination weir.” Step-by-Step Solution: Identify geometry: trapezoid with standard side slope (commonly taken as about 1H:4V in many texts).Recognize purpose: compensate for end contractions so Q–H relation is simplified.Interpretation: effectively blends rectangular core with triangular side components — a combination description.Therefore, option (d) captures the essence in general terms. Verification / Alternative check: Handbooks present Cipolletti as a trapezoidal weir; the “combination” phrasing is widely used to distinguish it from pure rectangular or pure triangular notches. Why Other Options Are Wrong: (a) Arbitrary proportion; (b) is a pure V-notch; (c) states “1H:2V” in some banks, which is not the classical Cipolletti slope; (e) refers to a different weir class. Common Pitfalls: Memorizing an incorrect side slope or confusing Cipolletti with sharp 90° V-notches. Final Answer: a combination weir whose geometry blends rectangular and triangular features

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