Question 1

Hydrostatics: through which point does the resultant hydrostatic force act on a submerged plane surface?  In fluid mechanics and civil engineering hydraulics, identify the point through which the single resultant of distributed hydrostatic pressure …

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Introduction / Context: The fluid pressure on a submerged plane surface increases linearly with depth, producing a distributed load. To replace this distribution with a single equivalent force for design and analysis (e.g., gates, dams, tank covers), we use the concept of the resultant hydrostatic force and its line of action. Given Data / Assumptions: Incompressible, static fluid. Plane surface fully or partially submerged. Atmospheric effects uniform and thus cancel for force location. Concept / Approach: Resultant magnitude on a plane area A is F = rho * g * ȳ * A, where ȳ is depth of area centroid below free surface. However, the line of action is not through the centroid unless pressure is uniform. The correct point is the centre of pressure, located below the centroid because pressure increases with depth. Step-by-Step Solution: Recognize pressure distribution: p = rho * g * y (linear with depth).Equivalent resultant acts at the point where the moment of the distributed pressure about any axis equals the moment of the resultant.That intersection point is defined as the centre of pressure; its depth y_cp = I_G/(ȳ*A) + ȳ for an inclined surface (I_G being second moment of area about centroidal horizontal axis). Verification / Alternative check: Compute moments of pressure diagram; the centroid of a triangular (or trapezoidal) pressure diagram lies below the centroid of the area, consistent with y_cp > ȳ. Why Other Options Are Wrong: Centre of top/bottom edge: generally not the correct line of action. Metacentre: relates to floating body stability, not fixed plane surfaces. None of these: incorrect because “centre of pressure” is the standard answer. Common Pitfalls: Confusing force magnitude point (centroid depth for F) with force location (centre of pressure). Assuming uniform pressure leads to centroid, which is not valid for varying pressure. Final Answer: centre of pressure

Question 2

Open container accelerating vertically upward: effect on pressure at the bottom  An open container completely filled with water is moved vertically upward with a uniform linear acceleration. Compared with the static case at rest, how does the pressu…

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Introduction / Context: Accelerating containers create an apparent body force field in addition to gravity. In vertical accelerations, the hydrostatic pressure distribution is modified by an effective gravity g_eff, impacting bottom pressures and gauge readings. Given Data / Assumptions: Open container, water fully filling the container. Uniform upward linear acceleration a. Neglect sloshing and compressibility; quasi-static distribution. Concept / Approach: In a non-inertial frame accelerating upward with acceleration a, the effective body force per unit mass is (g + a) downward. Hence pressure varies as p = rho * (g + a) * h below the free surface (which stays horizontal for uniform acceleration). Step-by-Step Solution: Define effective gravity: g_eff = g + a (upward acceleration).Hydrostatic relation becomes dp/dz = rho * g_eff.At the bottom, p_bottom(accel) = rho * (g + a) * H, which exceeds rho * g * H at rest. Verification / Alternative check: Energy viewpoint: additional inertial field increases potential gradient, thus higher pressure at the same depth. Why Other Options Are Wrong: Equal/lesser: contradict the increase in effective gravity. None of these: there is a definite, predictable increase. Common Pitfalls: Confusing upward vs downward acceleration; for downward acceleration, effective gravity is (g − a) and pressure reduces (becoming zero at free fall). Final Answer: greater than static pressure

Question 3

Mass density (ρ): equivalent definitions used in hydraulics  Select the correct comprehensive statement(s) for the mass density of a liquid ρ.

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Introduction / Context: Mass density is a primary thermophysical property appearing in continuity, momentum, and energy equations. Engineers routinely move between mass-based and weight-based formulations, so recognizing equivalent forms is essential. Given Data / Assumptions: Standard gravity symbol g denotes acceleration due to gravity. Specific weight γ = W / V = ρ * g. m is mass; W is weight; V is volume. Concept / Approach: By definition, density ρ is mass per unit volume. Since weight W = m * g, the ratio W/(gV) reduces to (mg)/(gV) = m/V = ρ. Similarly, γ = ρ * g, so ρ = γ / g. These interchangeable expressions are used depending on whether measurements are mass- or weight-based. Step-by-Step Solution: Start with ρ = m/V.Use W = mg ⇒ ρ = (W/g)/V = W/(gV).Use γ = ρg ⇒ ρ = γ/g. Verification / Alternative check: Dimensional analysis: [ρ] = M L^-3; W/(g V) has units (M L T^-2)/(L T^-2 * L^3) = M L^-3, and γ/g likewise yields M L^-3. Why Other Options Are Wrong: Each of (a), (b), (c) by itself is correct. Therefore “all the above” is the complete correct option. Common Pitfalls: Confusing ρ with specific weight γ; forgetting to divide by g when converting weight-based values to mass density. Final Answer: all the above

Question 4

Definition of streamline (laminar) flow  The flow in which each fluid particle follows a definite path and the paths of adjacent particles do not intersect is called:

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Introduction / Context: Fluid motions are categorized to analyze stability, mixing, and energy losses. A key distinction is between orderly, layered motion and random, fluctuating motion. The former is described by streamlines that never cross. Given Data / Assumptions: Inviscid or viscous fluids may both exhibit streamline (laminar) behavior at sufficiently low Reynolds numbers. Paths are conceptual curves tangent to the velocity vector at every point. Concept / Approach: Streamline (laminar) flow is characterized by smooth, parallel layers; each particle follows a fixed path, and neighboring paths do not intersect. This differs from “steady flow” (no time variation at a point) and “uniform flow” (no spatial variation along a reach). Turbulent flow has random fluctuations and eddies. Step-by-Step Solution: Identify the defining phrase “definite path” and “paths do not cross”.These are the hallmark features of streamline/laminar flow.Thus, choose “stream line flow”. Verification / Alternative check: In dye-streak experiments at low Re, a thin colored filament stays coherent—indicating streamlines that do not intersect. Why Other Options Are Wrong: Uniform flow: property constancy along space, not about particle paths crossing. Steady flow: time invariance at a fixed point; streamlines can still be complex. Turbulent flow: inherently involves fluctuating, intermingling paths. Common Pitfalls: Equating “steady” with “laminar”—they are independent attributes; a steady flow can be turbulent. Final Answer: stream line flow

Question 5

Venturimeter orientation: effect of inclination on indicated differential head  A venturimeter is installed in an inclined position instead of perfectly horizontal. For the same discharge, how does its indicated differential head (and hence computed…

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Introduction / Context: Venturimeters measure discharge by relating the pressure difference between the inlet and the throat to the velocity change. Practitioners often wonder whether pipe inclination alters the indicated head and thus the computed flow rate. Given Data / Assumptions: Proper pressure tappings at the inlet and throat. Same volumetric flow rate and fluid properties. Differential manometer (or transmitter) measures pressure difference between the two tapping points. Concept / Approach: Bernoulli between the two sections includes elevation terms. However, the instrument measures piezometric head difference that already accounts for elevation. For a given discharge, the velocity distribution at inlet and throat is the same regardless of the pipe’s overall inclination, so the measured differential head corresponding to velocity difference remains effectively unchanged (neglecting minor secondary effects). Step-by-Step Solution: For fixed Q, V1 and V2 (and hence dynamic head change) are fixed.The manometer measures p1/γ − p2/γ + (z1 − z2). The z-difference is inherent in what is measured, so the meter’s differential reflects the same flow condition.Therefore, indicated differential head and computed Q remain the same. Verification / Alternative check: Manufacturers specify that a venturimeter can be mounted in any orientation; calibration coefficients remain valid. Why Other Options Are Wrong: More/less reading: would imply orientation changes the inherent pressure-velocity relation, which it does not for the same Q. None of these: incorrect since “same reading” is standard guidance. Common Pitfalls: Confusing static head changes along an inclined run with the differential reading between two fixed stations; ignoring proper datum handling. Final Answer: same reading

Question 6

Weber number (We): identify the force ratio it represents  In dimensionless analysis of fluid flows with free surfaces, the Weber number expresses the ratio of which two effects?

Accepted Answer

Introduction / Context: Dimensionless numbers classify flow regimes and similarity. The Weber number is critical in atomization, droplet breakup, wave formation, and jet impingement where surface tension competes with inertia. Given Data / Assumptions: Characteristic length L, velocity V, density ρ, and surface tension σ. Concept / Approach: Weber number is defined as We = ρ * V^2 * L / σ. It therefore measures the relative importance of inertial forces to surface-tension forces. Large We implies inertia dominates and interfaces deform or break; small We means surface tension stabilizes the interface. Step-by-Step Solution: Start from dimensional forces: inertia ~ ρ * V^2; surface tension ~ σ / L.Form the ratio We = (ρ * V^2) / (σ/L) = ρ * V^2 * L / σ.Interpretation: increasing V or L increases We, reducing stability of drops/films. Verification / Alternative check: Related groups: Capillary number Ca = μ V / σ (viscous vs surface). Froude number Fr = V / √(gL) (inertia vs gravity). Reynolds number Re = ρ V L / μ (inertia vs viscosity). Why Other Options Are Wrong: They correspond to Fr (gravity), Re (viscosity), or compressibility effects (elastic), not to Weber. Common Pitfalls: Confusing We with Fr or Ca; forgetting that σ appears in the denominator. Final Answer: inertia force to surface-tension force

Question 7

Continuity equation in fluid mechanics: foundational conservation principle  The continuity equation for a control volume or along a streamline is fundamentally based on the conservation of which quantity?

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Introduction / Context: The continuity equation is one of the three cornerstone equations in fluid mechanics (with momentum and energy). It enforces that matter is neither created nor destroyed within a system. Given Data / Assumptions: Continuum hypothesis holds (density is well-defined locally). Single-phase flow without phase change (for simple forms). Concept / Approach: The general, integral continuity equation states: rate of increase of mass within a control volume + net mass outflow across its control surface = 0. For steady, incompressible flow, it reduces to A1 V1 = A2 V2 in a single streamtube. Step-by-Step Solution: Start with conservation of mass: d/dt ∫_CV ρ dV + ∮_CS ρ (V · n) dA = 0.For steady incompressible ρ = constant: ∮_CS (V · n) dA = 0 → A1 V1 = A2 V2 for a duct.Hence, continuity is fundamentally a mass balance, not momentum or force balance. Verification / Alternative check: Differential form for compressible flows: ∂ρ/∂t + ∇·(ρ V) = 0; still a statement of mass conservation. Why Other Options Are Wrong: Momentum and force relate to Newton’s second law and lead to the Navier–Stokes equations, not continuity. “None” is incorrect because mass conservation is explicit. Common Pitfalls: Using A V = constant where density varies significantly; in compressible flows, ρ A V = constant is the correct form. Final Answer: mass

Question 8

Velocity of approach for a tank–orifice system  A large tank (area a1) drains through a sharp-edged orifice of area a2 located on its side/bottom. If the head on the orifice is h (measured to the centre of the orifice), the mean velocity in the tank…

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Introduction / Context: When a tank empties through an orifice, the free surface drops. The finite, though often small, downward velocity of the free surface is termed the “velocity of approach” and slightly increases the actual discharge beyond the simple Torricelli estimate if accounted for. Given Data / Assumptions: a1 = plan area of tank free surface (assumed constant). a2 = area of the orifice. h = head on the orifice, measured to its centre. Neglect coefficient effects for the relation between Va and h. Concept / Approach: Continuity requires the volumetric flow leaving through the orifice to equal the rate of decrease of volume in the tank. If v is the jet velocity at the orifice, and Va is the downward velocity of the free surface, then Q = a2 * v = a1 * Va. For ideal conditions, v ≈ √(2 g h). Therefore, Va = (a2 / a1) * √(2 g h). Step-by-Step Solution: Write Q_orifice = a2 * √(2 g h).Set Q_tank = a1 * Va = Q_orifice.Solve for Va: Va = (a2 / a1) * √(2 g h). Verification / Alternative check: If a1 ≫ a2, Va is very small, justifying its neglect in many basic problems while recognizing its existence for accuracy. Why Other Options Are Wrong: √(2 g h) is the orifice jet velocity, not the free-surface approach velocity. Inverted or squared area ratios give wrong dimensions or magnitudes. Common Pitfalls: Confusing jet velocity with surface velocity; ignoring that Va scales with a2/a1. Final Answer: Va = (a2 / a1) * √(2 g h)

Question 9

Most economical rectangular channel (minimum perimeter for given area)  For a rectangular open channel section that is hydraulically most economical (minimum wetted perimeter for a given area), the width b must be:

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Introduction / Context: “Most economical” sections minimize the wetted perimeter for a given flow area, maximizing hydraulic radius and reducing friction losses. For rectangular channels, a simple geometric proportion yields the optimal shape for efficient conveyance. Given Data / Assumptions: Rectangular cross-section of width b and depth y. Target: minimize wetted perimeter P for a given area A = b * y. Steady, uniform design context (Manning/Chezy). Concept / Approach: For a rectangle: P = b + 2y. With A fixed, use calculus (or Lagrange multipliers) to minimize P. The result is b = 2y, which maximizes hydraulic radius R = A / P and thus reduces head loss for given discharge. Step-by-Step Solution: Let A be fixed; express b = A / y.Write P(y) = (A / y) + 2y.Set dP/dy = −A / y^2 + 2 = 0 ⇒ y^2 = A/2 ⇒ b = A / y = 2y. Verification / Alternative check: At b = 2y, hydraulic radius R = A / P simplifies to y/2, which is maximal for the rectangular family under fixed A. Why Other Options Are Wrong: b = y or b = y/2 do not minimize P for fixed area; they lead to larger wetted perimeter and lower R. “None of these” is incorrect because a known optimum exists. Common Pitfalls: Confusing the result with the triangular or trapezoidal optima; forgetting that this condition pertains to rectangular sections only. Final Answer: twice the depth of flow

Question 10

Water hammer in pressurized pipelines In a closed water-conveyance pipe system, the transient pressure rise (surge) produced when a flowing stream is decelerated by valve closure depends primarily on which factors?

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Introduction / Context: Water hammer is a classic unsteady-flow phenomenon in hydraulics. When a valve is closed or a pump trips, the flowing water column is suddenly decelerated, generating pressure waves that can far exceed normal steady-state pressures. Understanding what controls the surge magnitude is essential for safe pipeline design, surge-tank sizing, and valve-closure specifications. Given Data / Assumptions: Rigid or elastic pipes carrying water under pressure. Valve operation causes a transient from an initial steady velocity. Waves propagate at c (celerity) determined by water and pipe elasticity and geometry. Concept / Approach: The Joukowsky relation gives the ideal instantaneous surge: Δp = ρ c ΔV. The wave speed c depends on bulk modulus of water, pipe wall modulus, diameter, and wall thickness. The resulting maximum pressure rise also depends on valve-closure time relative to the pipe period (2L/c) and on reflections governed by pipeline length and end conditions. Step-by-Step Solution: Wave speed: c = 1 / √(ρ(1/K + D/(E t))).Instantaneous stop: Δp = ρ c V (larger V → larger surge).Closure time effect: if t_close < 2L/c, near-Joukowsky maximum; if longer, surge is smaller.System effect: pipe length L and boundaries set reflection timing and superposition. Verification / Alternative check: Compare quick vs slow closure using method of characteristics: shorter closure yields higher initial surge; more elastic pipes reduce c, reducing Δp for the same ΔV. Why Other Options Are Wrong: Each single factor (a, b, c, d) affects the surge, but not exclusively. The correct comprehensive choice is that all listed factors contribute. Common Pitfalls: Ignoring pipe-wall elasticity; assuming only valve speed matters; neglecting reflections that can compound pressures at tees or dead ends. Final Answer: All of the above

Question 11

Center of pressure for a vertical plane surface fully submerged Where is the center of pressure located relative to the area for a vertical plane surface immersed in a liquid (with its centroid below the free surface)?

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Introduction / Context: Hydrostatic forces on submerged gates and plates act at a point called the center of pressure, which is not generally the centroid because pressure increases linearly with depth. Correctly locating this point is essential for computing moment and designing supports. Given Data / Assumptions: Plane, vertical surface fully submerged in a static liquid. Centroid of the area is at depth y_c below the free surface. Liquid is incompressible; hydrostatic pressure varies as p = γ y. Concept / Approach: The resultant hydrostatic force on a plane surface equals F = γ A y_c. The center of pressure depth for a vertical surface is y_cp = y_c + I_G / (y_c A), where I_G is the second moment of area about a horizontal axis through the centroid. Because I_G, y_c, and A are all positive, y_cp > y_c; i.e., the center of pressure lies below the centroid but not at any special edge point. Step-by-Step Solution: Compute resultant: F = γ A y_c.Locate line of action: y_cp = y_c + I_G/(y_c A).Interpretation: position is below the centroid; it is not tied to upper or lower edges in a simple way. Verification / Alternative check: For a rectangle of height h, I_G = b h^3/12. With y_c known, y_cp evaluates to a value below y_c but above the lower edge unless very shallow submergence. Why Other Options Are Wrong: (a) Centroid is not the center of pressure under linear pressure variation. (b) and (c) tie the point to edges, which is incorrect. (e) A fixed fraction independent of geometry is not valid. Common Pitfalls: Confusing magnitude (F) with location (y_cp); overlooking the effect of shape via I_G. Final Answer: none of these

Question 12

Pressure difference across an inclined pipe – contributing heads In a steady flow through an inclined pipe with entrance and exit, the pressure difference between the two ends arises due to which components?

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Introduction / Context: Applying the energy equation between two pipe sections reveals that several terms contribute to head differences. Recognizing all contributors (major losses, minor losses, and elevation) is essential for accurate sizing and pump selection. Given Data / Assumptions: Steady, incompressible flow. Uniform pipe diameters at sections, with defined inlet and outlet. Significant friction and minor (local) losses at fittings and ends. Concept / Approach: Bernoulli with head-loss terms: z1 + p1/γ + V1^2/(2g) = z2 + p2/γ + V2^2/(2g) + h_f + Σh_m. For equal diameters, velocities may cancel, leaving elevation head (z1 − z2), frictional loss h_f, and local losses (entrance/exit). Thus the pressure difference is due to all three contributions listed. Step-by-Step Solution: Write energy equation with losses between ends.Recognize elevation change adds/subtracts static head.Include frictional (Darcy–Weisbach) and minor-loss coefficients. Verification / Alternative check: Compute p1 − p2 = γ[(z2 − z1) + h_f + Σh_m] when velocities are equal; each term contributes to the total pressure difference. Why Other Options Are Wrong: Choosing only one source underestimates the actual pressure difference; real systems include all listed effects. Common Pitfalls: Ignoring exit loss to atmosphere or assuming friction is the only contributor on long lines. Final Answer: all the above

Question 13

Laminar flow criterion in open channels For open-channel flow, the motion is considered laminar when the Reynolds number (based on hydraulic radius) is below what threshold?

Accepted Answer

Introduction / Context: Classifying open-channel flow as laminar or turbulent guides the choice of resistance equations and roughness parameters. While pipes commonly use a 2300 threshold, open channels use a different criterion due to the free surface and characteristic length choice (hydraulic radius). Given Data / Assumptions: Reynolds number for channels: Re = V R_h / ν, where R_h = A/P. Newtonian fluid (water), steady uniform flow conceptually. Free surface effects present. Concept / Approach: Empirical evidence places laminar-channel behavior at Re below approximately 500. For 500 < Re < 2000, transitional effects may occur, and above roughly 2000, fully turbulent regimes dominate for most practical channels. Step-by-Step Solution: Compute Re = V R_h / ν.If Re < 500 → laminar; Re ≫ 500 → turbulent/transition.Select appropriate resistance relation (e.g., Hagen–Poiseuille analog only for laminar). Verification / Alternative check: Compare observed velocity profiles: laminar exhibits parabolic profile; turbulent shows fuller profile and eddy viscosity behavior. Why Other Options Are Wrong: (b) and (c) do not represent the typical laminar threshold in channels. (d) is incorrect as a practical criterion exists. (e) is overly conservative for channels. Common Pitfalls: Using the pipe threshold (2300) directly for open channels; choosing wrong characteristic length (use hydraulic radius, not depth alone). Final Answer: less than 500

Question 14

Equalising levels between two identical tanks via a small orifice Two identical cylindrical vessels (cross-sectional area A each) are interconnected by an orifice of area a and coefficient of discharge C_d. If the initial difference in liquid levels…

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Introduction / Context: Interconnected-tank transients are common in laboratory hydraulics and process engineering. When two identical tanks are connected by a small orifice, the level difference decays in time, and the equalisation time can be derived from unsteady continuity combined with the orifice equation. Given Data / Assumptions: Two tanks, each of area A, connected by an orifice of area a with discharge coefficient C_d. Head loss confined to the orifice; inviscid elsewhere; free discharge between tanks. Initial level difference Δh(0) = H; final equalised difference Δh(T) = 0. Concept / Approach: Flow through the orifice is Q = C_d a √(2 g Δh). Because one tank empties while the other fills, the rate of change of difference is doubled: d(Δh)/dt = −(2Q)/A. This yields a first-order, separable differential equation in √Δh, leading to a closed-form time to reach Δh = 0 from H. Step-by-Step Solution: Q = C_d a √(2 g Δh).A d(h_1)/dt = −Q, A d(h_2)/dt = +Q ⇒ d(Δh)/dt = −(2 Q)/A.d(Δh)/dt = −(2 C_d a/A) √(2 g) √(Δh).Integrate: ∫ dΔh/√(Δh) = −(2 C_d a/A) √(2 g) ∫ dt.2 √(Δh) = −(2 C_d a/A) √(2 g) t + 2 √H.Set Δh = 0 at t = T ⇒ T = (A / (C_d a √(2 g))) √H. Verification / Alternative check: Dimensional analysis: [T] = [A]/([a] √(g) √[H]) → seconds, consistent. Why Other Options Are Wrong: (b) and (c) have incorrect factors of 2; (d) is dimensionally inconsistent; (e) omits √2 in the orifice velocity. Common Pitfalls: Forgetting the factor of 2 because both tanks change level; misapplying Torricelli’s law without the discharge coefficient. Final Answer: T = (A / (C_d a √(2 g))) * √H

Question 15

Hydraulic radius comparison – full circular pipe vs full square conduit If a circular pipe runs full and a square conduit runs full with its side equal to the pipe diameter D, what is the ratio R_pipe / R_square of their hydraulic radii?

Accepted Answer

Introduction / Context: The hydraulic radius R = A/P (flow area divided by wetted perimeter) is a key geometric parameter in resistance equations such as Manning or Chezy. Comparing R across shapes running full helps benchmark frictional performance per unit area. Given Data / Assumptions: Circular pipe of diameter D, running full. Square conduit with side equal to D, running full. Wetted perimeter equals the entire inner boundary for both, because they are full. Concept / Approach: Compute R for each shape: R = A/P. For the circle, A_c = π D^2/4 and P_c = π D ⇒ R_c = D/4. For the square, A_s = D^2 and P_s = 4D ⇒ R_s = D/4. The ratio is therefore 1. Step-by-Step Solution: Circle: A = π D^2/4; P = π D ⇒ R = D/4.Square: A = D^2; P = 4D ⇒ R = D/4.Ratio R_pipe / R_square = (D/4)/(D/4) = 1. Verification / Alternative check: Dimensional consistency holds; both yield the same characteristic depth scale under full flow. Why Other Options Are Wrong: All other ratios differ from unity and contradict the direct A/P computation. Common Pitfalls: Accidentally using hydraulic depth A/T (open-channel concept) instead of hydraulic radius A/P for full-closed conduits. Final Answer: 1

Question 16

Most economical circular channel for maximum velocity (open-channel flow) For a circular channel carrying water (not running full), the section that gives maximum mean velocity satisfies which of the following conditions?

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Introduction / Context: For a given circular conduit running partially full as an open channel, there is a specific depth that maximizes the mean velocity (equivalently minimizes hydraulic losses per unit area). This is often called the most economical circular section for velocity in open-channel hydraulics. Given Data / Assumptions: Circular conduit flowing with a free surface (not pressurized). Resistance described by Manning or Chezy; objective is to maximize V ∝ R^(2/3) for given slope and roughness. Geometry relates depth, area, and wetted perimeter via the central angle of the wetted segment. Concept / Approach: By differentiating V with respect to depth (or equivalently maximizing R = A/P), the optimal depth occurs at y ≈ 0.81 D. The corresponding geometric relations yield R ≈ 0.304 D and wetted perimeter P ≈ 2.245 D. Step-by-Step Solution: Set up A(θ) and P(θ) for a circular segment; express R(θ) = A/P.Differentiate R with respect to θ and set derivative to zero to find optimal θ.Back-substitute to find y/D ≈ 0.81, R/D ≈ 0.304, and P/D ≈ 2.245. Verification / Alternative check: Standard hydraulics tables list the same ratios for the most economical circular section for maximum velocity, confirming the results. Why Other Options Are Wrong: Each item (a), (b), and (c) is individually correct; therefore, any single selection is incomplete. The comprehensive correct choice is “all the above.” Common Pitfalls: Confusing “maximum velocity” with “maximum discharge” depth (which has a different optimal ratio); applying full-pipe relations to free-surface flow. Final Answer: all the above

Question 17

Backwater curve length from specific-energy and slope data If E1 is the specific energy at the start of a backwater curve and E2 at the maximum water-surface rise, and i_b and i_e are bed slope and energy-grade-line slope respectively, the backwater…

Accepted Answer

Introduction / Context: Gradually varied flow (GVF) profiles like backwater curves are evaluated using the specific-energy form of the GVF equation. Over a reach, changes in specific energy relate directly to the difference between bed slope and energy grade line slope, allowing estimation of the water-surface profile length. Given Data / Assumptions: Steady, gradually varied open-channel flow. Specific energies E1 and E2 at the start and at the section of maximum rise. Bed slope i_b and energy-grade-line slope i_e approximately uniform over the reach. Concept / Approach: The differential form is dE/dx = i_b − i_e. Integrating from section 1 to section 2 gives E2 − E1 = ∫(i_b − i_e) dx ≈ (i_b − i_e) L if slopes are roughly constant. Hence L = (E2 − E1)/(i_b − i_e). Step-by-Step Solution: Start from GVF: dE/dx = i_b − i_e.Assume i_b and i_e constant over the small reach.Integrate: E2 − E1 = (i_b − i_e) L ⇒ L = (E2 − E1)/(i_b − i_e). Verification / Alternative check: Backwater curves have i_b > i_e; thus denominator is positive, giving a positive L for E2 > E1, consistent with a rise in the water surface along the flow. Why Other Options Are Wrong: (b) Changes sign incorrectly; (c) ignores bed slope; (e) is dimensionally inconsistent. (d) is unnecessary since (a) is correct. Common Pitfalls: Confusing specific energy with total head; overlooking that i_e includes both friction and minor losses where relevant. Final Answer: L = (E2 − E1) / (i_b − i_e)

Question 18

Sensitivity of rectangular-notch discharge to head-measurement error For a rectangular weir/notch where Q ∝ H^(3/2), the ratio of percentage error in discharge to percentage error in head measurement is

Accepted Answer

Introduction / Context: Discharge over a rectangular notch is commonly expressed as Q = C_d b √(2 g) H^(3/2). Because Q depends on H to the 3/2 power, small errors in head gauging can translate into larger fractional errors in discharge. Quantifying this sensitivity is essential in flow measurement practice. Given Data / Assumptions: Rectangular notch, free nappe conditions. Discharge varies as H^(3/2), other factors held constant. Small measurement errors (use differentials/relative errors). Concept / Approach: If Q ∝ H^n, then dQ/Q = n (dH/H). Here n = 3/2. Thus the ratio of fractional (percentage) errors is 3/2. Step-by-Step Solution: Q = K H^(3/2) for constant K.Take logarithmic differential: dQ/Q = (3/2) dH/H.Therefore %ΔQ / %ΔH = 3/2. Verification / Alternative check: A 1% head error yields approximately a 1.5% discharge error, matching field experience that accurate head measurement is critical. Why Other Options Are Wrong: (a), (b), and (d) understate the sensitivity; (e) overstates it for a rectangular notch. Common Pitfalls: Confusing rectangular notch (exponent 3/2) with triangular notch where exponent differs (5/2), changing sensitivity. Final Answer: 3/2

Question 19

Velocity from stream function ψ = 2 x y (two-dimensional incompressible flow) Given the stream function ψ(x,y) = 2 x y, find the fluid speed at the point (x, y) = (3, 4).

Accepted Answer

Introduction / Context: The stream function ψ(x, y) conveniently describes two-dimensional incompressible flows. Velocity components are obtained by differentiating ψ, and the magnitude (speed) follows from the Euclidean norm of the components. This exercise checks fluency with stream-function relations. Given Data / Assumptions: ψ(x, y) = 2 x y. Two-dimensional, incompressible flow in Cartesian coordinates. Point of interest: (x, y) = (3, 4). Concept / Approach: For 2D incompressible flow in (x, y): u = ∂ψ/∂y and v = −∂ψ/∂x. Compute u and v, then speed V = √(u^2 + v^2). Step-by-Step Solution: u = ∂ψ/∂y = 2 x ⇒ at (3,4): u = 23 = 6 m/s.v = −∂ψ/∂x = −2 y ⇒ at (3,4): v = −24 = −8 m/s.Speed V = √(u^2 + v^2) = √(6^2 + 8^2) = √(36 + 64) = √100 = 10 m/s. Verification / Alternative check: Level curves ψ = constant are orthogonal to velocity potential lines (if defined), and the computed components satisfy the Cauchy–Riemann-type relations for incompressible flow. Why Other Options Are Wrong: 6 and 8 m/s are component magnitudes, not the resultant; 12 and 15 m/s exceed the correct Pythagorean result. Common Pitfalls: Mixing the signs (u, v) definitions or using v = ∂ψ/∂x instead of v = −∂ψ/∂x. Final Answer: 10 m/s

Question 20

Triangular (V-notch) weir: identify the incorrect statement about its use, accuracy, and ventilation requirements in open-channel flow measurement.

Accepted Answer

Introduction / Context: Triangular (V-notch) weirs are widely used in hydraulic laboratories and field gauging to measure small to moderate discharges accurately. Because the discharge varies sensitively with head, V-notches are preferred at low flows where rectangular notches may lose resolution. Given Data / Assumptions: Sharp-crested, well-formed V-notch installed with proper approach conditions. Free, ventilated nappe unless otherwise specified. Standard head measurement taken vertically above the crest apex at a prescribed distance upstream. Concept / Approach: The classical discharge relation for a V-notch is Q = C_d * (8/15) * √(2g) * tan(θ/2) * H^(5/2), where H is the head over the crest. Because Q ∝ H^(5/2), small changes in H produce measurable changes in Q, improving sensitivity at low flows. Only one head reading is needed once the notch angle and coefficient are known. Step-by-Step Solution: Recognize the V-notch is designed for low discharges: the power of 5/2 on H increases sensitivity.Note that Q can be computed from a single measured head H, given θ and C_d.Understand ventilation: air beneath the nappe maintains atmospheric pressure under the sheet, preventing adherence to the downstream face. Without ventilation, sub-atmospheric pressure can form and alter C_d, causing under-reading. Verification / Alternative check: Compare with a rectangular notch (Q ∝ H^(3/2)). For identical measurement resolution in H, the V-notch yields a larger proportional change in Q at small H, confirming its advantage at low flow rates. Why Other Options Are Wrong: (a) Correct: V-notches are indeed more accurate at low flows. (b) Correct: only head H is required. (c) Correct: ventilation is necessary to preserve calibration. (e) Incorrect claim; V-notches are specifically recommended for low flows. Common Pitfalls: Not ventilating the nappe; measuring H too close to the crest; ignoring approach velocity head corrections for higher velocities. Final Answer: None of these.

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