Question 1

Hydrostatics — Pressure at a Depth in Water Calculate the gauge pressure at a point located 4 m below the free surface of water (assume rho ≈ 1000 kg/m^3 and g ≈ 9.81 m/s^2).

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Introduction:Hydrostatic pressure in a liquid at rest increases linearly with vertical depth below the free surface. This simple but powerful relation is fundamental for manometer readings, dam design, and underwater equipment ratings.Given Data / Assumptions:Depth h = 4 m below a water free surface.Density of water rho ≈ 1000 kg/m^3.Acceleration due to gravity g ≈ 9.81 m/s^2.Gauge reference at the free surface (p = 0 gauge at surface).Concept / Approach:Use the hydrostatic relation p = rho * g * h. This gives pressure directly in pascals when SI units are used. Convert to kilopascals by dividing by 1000.Step-by-Step Solution:Write p = rho * g * h.Substitute: p = 1000 * 9.81 * 4.Compute: p = 39,240 Pa.Convert to kPa: 39,240 Pa / 1000 = 39.24 kPa.Verification / Alternative check:A quick estimate using g ≈ 10 m/s^2 gives p ≈ 40 kPa, which is close and confirms the detailed result 39.24 kPa.Why Other Options Are Wrong:19.24 kPa and 29.24 kPa correspond to shallower depths; 49.24 kPa would imply a depth near 5 m, not 4 m.Common Pitfalls:Forgetting the gauge vs absolute reference; mixing units (using g in cm/s^2 or rho in g/cm^3 without converting); using slanted distance instead of vertical depth.Final Answer:39.24 kPa

Question 2

Orifices — Typical Average Value of Coefficient of Velocity (Cv) For a sharp-edged orifice under ordinary laboratory conditions, what is a commonly accepted average value of the coefficient of velocity Cv?

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Introduction:Three principal coefficients describe orifice discharge: coefficient of contraction Cc, coefficient of velocity Cv, and coefficient of discharge Cd = Cc * Cv. Recognizing typical magnitudes helps in quick estimates and cross-checks of experimental data.Given Data / Assumptions:Sharp-edged thin-plate orifice.Clean approach flow, moderate heads.Water at room temperature.Concept / Approach:Coefficient of velocity Cv accounts for the difference between theoretical and actual jet velocities at the vena contracta. With small energy losses, Cv is close to unity, typically around 0.97–0.99 for sharp-edged orifices.Step-by-Step Solution:Recall Cd ≈ 0.61 for sharp edges and Cc ≈ 0.62.Compute a check: Cv ≈ Cd / Cc ≈ 0.61 / 0.62 ≈ 0.98.Thus a representative average is Cv ≈ 0.97.Verification / Alternative check:Laboratory measurements of jet trajectory (using horizontal range) yield velocities very close to theoretical, confirming Cv near 1.Why Other Options Are Wrong:0.62 is typical of Cc, not Cv; 0.76 and 0.84 are too low for velocity coefficient under normal conditions.Common Pitfalls:Confusing Cv with Cd or Cc; using atypical values from rough-edged or partially blocked orifices.Final Answer:0.97

Question 3

Velocity Distribution in Open Channels — Is It Uniform? State whether the following is true or false: “The velocity of flow is the same at all points across the cross-section of an open channel.”

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Introduction:Understanding velocity distribution across a cross-section is crucial for accurate discharge measurement and sediment transport analysis. Real flows exhibit shear near boundaries, creating non-uniform profiles.Given Data / Assumptions:Steady, fully developed open-channel flow.No slip at the boundaries (bed and banks) and free surface exposed to atmosphere.Prismatic channel for conceptual clarity.Concept / Approach:Viscous and turbulent shear cause velocity to vary from near zero at the bed and banks to a maximum just below the free surface. Hence, only an average (mean) velocity is uniform in the sense of a single representative value; point velocities differ across the section.Step-by-Step Solution:Recall boundary condition: velocity approaches zero at solid boundaries.Recognize free-surface shear is small; thus peak velocity occurs slightly below the surface.Therefore the statement of uniform velocity across the section is false.Verification / Alternative check:Velocity-area method uses a velocity distribution (via current meters or ADCP) precisely because velocities vary spatially; this confirms non-uniformity.Why Other Options Are Wrong:Choosing “True” would neglect shear and turbulence—contrary to basic fluid mechanics.Common Pitfalls:Assuming plug flow because textbooks draw simplified diagrams; confusing time-steady flow with spatially uniform profiles.Final Answer:False

Question 4

Chezy’s Formula — Correct Expression for Discharge in Open Channels According to Chezy’s formula, which expression correctly gives the discharge Q through an open channel, where A = area of flow, C = Chezy’s constant, m = hydraulic mean depth (hydra…

Accepted Answer

Introduction:Chezy’s relation is one of the earliest uniform-flow formulas. It connects the mean velocity in an open channel to the hydraulic radius and bed slope through an empirical constant C, enabling designers to estimate discharge for given geometry and slope.Given Data / Assumptions:Uniform, steady open-channel flow.Prismatic channel with constant roughness and slope.Hydraulic mean depth m = A / P (also called hydraulic radius R).Concept / Approach:Chezy’s velocity formula is V = C * sqrt(m * i). Multiplying by area yields the discharge: Q = A * V = A * C * sqrt(m * i). This expression shows that Q scales with the square root of the product of hydraulic radius and slope, not linearly.Step-by-Step Solution:Start with V = C * sqrt(m * i).Compute discharge: Q = A * V.Therefore, Q = A * C * sqrt(m * i).Verification / Alternative check:Manning’s formula V = (1/n) * m^(2/3) * i^(1/2) also implies a square-root dependence on slope, consistent with Chezy’s structure.Why Other Options Are Wrong:Linear forms with m * i (options A, B, D) contradict the square-root relationship and have incorrect dimensions for velocity before multiplying by A.Common Pitfalls:Confusing hydraulic mean depth with flow depth; treating C as constant across roughness/scale without calibration.Final Answer:Q = A * C * sqrt(m * i)

Question 5

Surface Tension Force — Dependence on Line of Contact Is the following statement correct? “Surface tension force equals the product of surface tension per unit length and the cross-sectional area of flow.”

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Introduction:Surface tension arises at interfaces and manifests as a force acting along lines (perimeters) of contact, not over cross-sectional areas like pressure forces. Clarifying this distinction prevents dimensional and conceptual mistakes in capillarity problems.Given Data / Assumptions:Clean liquid–gas or liquid–liquid interface.Contact angle alpha defined at the solid–liquid interface where relevant.Quasi-static conditions.Concept / Approach:The magnitude of surface tension force is F = sigma * L * cos(alpha_component) where sigma is surface tension (N/m) and L is the length of contact (e.g., tube inner perimeter). This already shows that multiplying by an area would give wrong units and an incorrect physical interpretation.Step-by-Step Solution:Identify units: sigma has units N/m.Multiply by a length L (m): sigma * L gives N, a force.Multiplying by area (m^2) would produce N*m, not a force, hence the statement is false.Verification / Alternative check:Capillary rise formula h = (4 * sigma * cos(alpha)) / (w * d) derives from balancing sigma * perimeter against the weight of the column, again confirming perimeter (length), not area.Why Other Options Are Wrong:Choosing “Correct” ignores the fundamental line-force nature of surface tension and leads to dimensional inconsistency.Common Pitfalls:Confusing surface tension (line force) with surface energy (J/m^2) and with pressure forces (which act over area).Final Answer:Incorrect

Question 6

External Mouthpiece — Parameter Affecting Coefficient of Discharge For an external (projecting) mouthpiece attached to a tank orifice, which geometric or flow parameter most significantly influences the coefficient of discharge Cd?

Accepted Answer

Introduction:Mouthpieces are short tubes fitted to orifices that alter jet contraction and energy losses. The discharge coefficient Cd depends on how the jet forms and attaches within the mouthpiece, which is strongly affected by its length and shape.Given Data / Assumptions:External (projecting) mouthpiece, sharp entry from the tank.Steady discharge under a given head.Negligible air entrainment and cavitation.Concept / Approach:For very short lengths, the jet exhibits a vena contracta with significant contraction losses; as length increases moderately, the jet expands to fill the tube (“running full”), reducing contraction losses and changing Cd. Thus, the mouthpiece length relative to diameter governs the regime and resulting discharge coefficient.Step-by-Step Solution:Classify regimes: running free vs running full.Increase length beyond about 2–3 diameters to promote running full.Observe Cd changes accordingly; therefore, length is the key influence.Verification / Alternative check:Laboratory measurements show Cd for external mouthpieces increases as the mouthpiece length permits the jet to adhere and fill the passage.Why Other Options Are Wrong:Velocity and pressure are consequences of head; area changes capacity but does not by itself determine loss mechanisms the way length does.Common Pitfalls:Ignoring entrance shape and edge sharpness; confusing with nozzles where contouring dominates.Final Answer:length of mouthpiece

Question 7

Specific Weight from Weight and Volume — Oil in a Vessel A closed vessel contains V = 4 m^3 of oil with total weight W = 30 kN. Determine the specific weight (unit weight) gamma of the oil in kN/m^3.

Accepted Answer

Introduction:Specific weight (unit weight) gamma is the weight per unit volume of a fluid. It is central to hydrostatics because pressure head p/w translates pressure into an equivalent height of fluid.Given Data / Assumptions:Total weight W = 30 kN.Volume V = 4 m^3.Fluid is homogeneous; gravity standard.Concept / Approach:The definition is gamma = W / V. Units: kN divided by m^3 produces kN/m^3, which is the desired unit.Step-by-Step Solution:Write gamma = W / V.Substitute: gamma = 30 kN / 4 m^3.Compute: gamma = 7.5 kN/m^3.Verification / Alternative check:Compare with water's specific weight ≈ 9.81 kN/m^3; the obtained value 7.5 kN/m^3 is lower, which is reasonable for many oils that are lighter than water.Why Other Options Are Wrong:4.5 and 6 kN/m^3 would correspond to lower W or higher V; 10 kN/m^3 would require heavier fluid or smaller volume than given.Common Pitfalls:Confusing specific weight with specific gravity; using mass density instead of weight density; unit mistakes between N and kN.Final Answer:7.5 kN/m^3

Question 8

Flow Description — Pathlines That Never Cross A flow in which each liquid particle follows a definite path and the paths of individual particles do not intersect is called:

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Introduction:Describing flow patterns helps choose appropriate models. One useful idealization is streamline (laminar) flow, where orderly layers slide over one another and particle paths are well defined.Given Data / Assumptions:Incompressible liquid with moderate Reynolds number.No mixing across streamlines; negligible instabilities.Observation over a period where the pattern remains representative.Concept / Approach:In streamline (laminar) flow, neighboring particles follow smooth, non-intersecting paths. Streamlines show the instantaneous direction of the velocity field and cannot cross because velocity at a point has a unique direction.Step-by-Step Solution:Define streamline: line everywhere tangent to the instantaneous velocity vector.Argue non-intersection: two directions at one point would be required to cross, which is impossible for a single-valued velocity field.Therefore the correct term is streamline flow.Verification / Alternative check:Low-Re flows in narrow tubes or viscous oils exhibit dye streaks that remain layered and do not mix rapidly, matching the streamline behavior.Why Other Options Are Wrong:Steady refers to time invariance; uniform refers to spatial constancy; turbulent features random fluctuations and enhanced mixing, where pathlines may get highly convoluted.Common Pitfalls:Equating steady with laminar; assuming streamlines represent time-averaged motion in turbulence (they do not).Final Answer:streamline flow

Question 9

Footstep (Pivot) Bearing — Effect of Shaft Radius on Viscous Torque For a vertical shaft running in a footstep bearing with a uniform oil film of thickness t and viscosity mu, if the shaft radius R is doubled (same speed), the torque required to ove…

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Introduction:Power loss in bearings often arises from viscous shear in thin lubricant films. In a flat footstep (pivot) bearing supporting a vertical shaft, torque due to viscosity grows rapidly with shaft radius, which is critical for sizing and thermal analysis.Given Data / Assumptions:Flat circular contact area with oil film thickness t (constant).Dynamic viscosity mu and angular speed omega are constant.No slip at boundaries; laminar shear within the film.Concept / Approach:The local shear stress is tau = mu * (du/dy) = mu * (omega * r / t). The shear force on an annulus of radius r and width dr is dF = tau * dA = mu * (omega * r / t) * (2 * pi * r * dr). The elemental torque is dT = r * dF.Step-by-Step Solution:Write dT = r * dF = mu * omega / t * 2 * pi * r^3 * dr.Integrate from r = 0 to R: T = mu * omega / t * 2 * pi * (R^4 / 4).Simplify: T = (pi * mu * omega * R^4) / 2.Thus T ∝ R^4. Doubling R gives T_new / T_old = (2R)^4 / R^4 = 16.Verification / Alternative check:Dimensional check shows torque scales with viscosity, speed, area, and the fourth power of radius in this configuration—consistent with classical tribology results.Why Other Options Are Wrong:Linear, quadratic, or cubic scaling (2, 4, 8 times) underestimate the strong dependence of viscous torque on radius for a flat film.Common Pitfalls:Using average radius approximations without integration; forgetting that film thickness variations or step bearings change exponents.Final Answer:sixteen times

Question 10

Open-channel flow in a circular section: The mean velocity through a circular channel (partially full) becomes maximum at a particular depth of flow. The depth of water for maximum velocity is approximately what fraction of the channel diameter?

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Introduction / Context:For a given circular channel carrying open-channel flow under gravity, the mean velocity v depends on the hydraulic radius R = A/P through resistance formulas (Chezy or Manning). Because area A and wetted perimeter P both change with depth y, there exists a particular y/D that maximizes R and hence the achievable mean velocity for a given slope and roughness. This question asks for that optimum fractional depth relative to the diameter D.Given Data / Assumptions:Circular channel of diameter D, partially full (free surface present).Steady, uniform open-channel flow with fixed bed slope and roughness.Velocity proportional to hydraulic radius: v ∝ R^m (m = 1/2 for Chezy, 2/3 for Manning in discharge), so maximizing R leads to maximum v for fixed slope and roughness.Concept / Approach:For a circular arc of central angle 2θ (in radians) filled with liquid, A = (D^2/8) * (2θ − sin 2θ) and P = (D/2) * 2θ = Dθ. The hydraulic radius is R(θ) = A/P. The depth-to-diameter ratio is y/D = (1 − cos θ)/2. Differentiating R with respect to θ and setting dR/dθ = 0 gives an optimum at θ ≈ 2.639 rad (≈151.2°), which corresponds to y/D ≈ 0.81. This classical result is widely used for design checks in sewers and storm drains.Step-by-Step Solution:Relate geometry: y/D = (1 − cos θ)/2.Area: A = (D^2/8) * (2θ − sin 2θ); Wetted perimeter: P = Dθ.Hydraulic radius: R(θ) = A/P = (D/8) * (2θ − sin 2θ) / θ.Optimize: set dR/dθ = 0 → numerical solution gives θ ≈ 2.639 rad.Compute y/D = (1 − cos 2.639)/2 ≈ 0.81 → maximum mean velocity depth.Verification / Alternative check:If one computes R across several y/D values (e.g., 0.67, 0.75, 0.81, 0.90), R peaks near 0.81, confirming the analytic optimum. Practical charts in handbooks reproduce this peak at approximately 0.81D for velocity and around 0.94D for maximum discharge, emphasizing the distinction between criteria (velocity vs discharge).Why Other Options Are Wrong:0.34 times: Far too shallow; wetted perimeter is small but area is also too small.0.67 times: Below the optimum; R has not yet peaked.0.95 times: Closer to the value for maximum discharge, not maximum velocity.Common Pitfalls:Confusing the depth for maximum discharge with that for maximum mean velocity; forgetting that both A and P change with depth in a circular section, so maximizing A alone is not sufficient.Final Answer:0.81 times

Question 11

Capillarity in a narrow glass tube dipped in water: The water column rises with which type of free-surface curvature at the meniscus?

Accepted Answer

Introduction / Context:Capillary action in narrow tubes depends on the balance of adhesive forces (liquid–solid) and cohesive forces (liquid–liquid), producing a meniscus that determines whether the liquid rises or depresses. Water in clean glass typically wets the surface, creating a concave meniscus and a capillary rise. This phenomenon is central in soil mechanics, porous media flow, and microfluidics.Given Data / Assumptions:Clean glass tube of small internal diameter.Liquid is water at ordinary conditions.No contamination altering contact angle.Concept / Approach:When adhesion (water–glass) exceeds cohesion (water–water), the contact angle is less than 90 degrees, the meniscus curves upward at the walls (concave), and the vertical component of surface tension supports the column against gravity. The height of rise h satisfies h = (4 * σ * cos θ) / (ρ * g * d) for a circular tube, where σ is surface tension, θ is contact angle (θ Step-by-Step Solution:Identify wetting condition: θ Meniscus shape: concave (curves upward near the wall).Consequence: upward capillary rise until hydrostatic head balances surface tension component.Verification / Alternative check:In contrast, mercury in glass has θ > 90°, forms a convex meniscus, and experiences capillary depression. Observing both cases side-by-side in a lab demonstrates the role of contact angle and surface tension direction.Why Other Options Are Wrong:Convex: Corresponds to non-wetting liquids like mercury on glass.Planar / irregular: Do not describe the stable equilibrium shape in a narrow, clean tube for water.Common Pitfalls:Assuming all liquids rise in capillary tubes; some depress. Also, small contamination or surfactants can alter θ and the observed curvature.Final Answer:concave

Question 12

Specific gravity versus density (SI units): “The density of a liquid in kg/m^3 is numerically equal to its specific gravity.” State whether this is correct.

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Introduction / Context:Specific gravity (SG) is a dimensionless ratio comparing a substance’s density to a reference density, typically water at about 4°C in many engineering contexts. Confusion frequently arises when switching between cgs units (g/cm^3) and SI units (kg/m^3). This question clarifies that numerical equality does not hold in SI unless a conversion is applied.Given Data / Assumptions:Density ρ expressed in SI units kg/m^3.Reference density of water ρ_w ≈ 1000 kg/m^3 (near room temperature).SG defined as SG = ρ / ρ_w.Concept / Approach:Because SG is ρ normalized by 1000 kg/m^3, the numerical relation in SI is ρ (kg/m^3) = SG * 1000. Therefore, ρ and SG are not numerically equal unless ρ is expressed in g/cm^3 (where water’s density is approximately 1 g/cm^3). In cgs units, numerical equality happens because ρ_w ≈ 1 g/cm^3, so SG equals ρ in g/cm^3; this does not carry over to SI without multiplying by 1000.Step-by-Step Solution:Define SG = ρ / ρ_w.Take ρ_w ≈ 1000 kg/m^3.Thus ρ (kg/m^3) = SG * 1000 → not numerically equal unless SG is multiplied by 1000.Verification / Alternative check:Example: A light oil with SG = 0.8 has density ρ = 0.8 * 1000 = 800 kg/m^3 in SI, clearly not equal to 0.8 numerically when using kg/m^3. In cgs, ρ = 0.8 g/cm^3 matches SG numerically because the reference is 1 g/cm^3.Why Other Options Are Wrong:True: Incorrect in SI kg/m^3.True only when density is in g/cm^3: While numerically true in cgs, the prompt explicitly states kg/m^3, so still false.True only at 4°C: Temperature affects ρ_w slightly, but the factor of 1000 in SI remains; equality still does not hold numerically.Common Pitfalls:Dropping units and comparing bare numbers; mixing SG with specific weight or relative density for gases (which may use air as reference).Final Answer:False

Question 13

Bulk modulus (compressibility of a fluid): Select the correct definition of bulk modulus K from the choices below.

Accepted Answer

Introduction / Context:The bulk modulus K quantifies a fluid’s resistance to uniform compression. It is essential in acoustics (speed of sound), water hammer calculations, and compressible-flow corrections even for liquids that are nearly incompressible. Knowing its definition helps distinguish between different elastic and viscous properties in fluids and solids.Given Data / Assumptions:Small, uniform changes in pressure and volume.Volumetric strain defined as ΔV/V (dimensionless).Isothermal or adiabatic context determines the numerical value but not the definition.Concept / Approach:By definition, bulk modulus K = Δp / (ΔV/V) = −V * dp/dV (the negative sign indicates that volume decreases as pressure increases). The larger the K, the less compressible the fluid. Water has a high K compared to gases, thereby supporting the approximation of incompressibility in many hydraulic problems.Step-by-Step Solution:Define volumetric strain ε_v = ΔV/V.Relate pressure and volume: K = Δp / ε_v.Interpretation: big K → small volume change for a given pressure increase.Verification / Alternative check:Speed of sound a in a fluid satisfies a = sqrt(K/ρ). Higher K yields higher a, as seen by comparing liquids (fast) to gases (slower), confirming consistency with physical intuition about compressibility.Why Other Options Are Wrong:Shear stress to shear strain: That is shear modulus (solids), not bulk modulus.Increase in volume to viscosity: Mixes incompatible quantities.Critical velocity to viscosity: Unrelated ratio with mismatched dimensions.Common Pitfalls:Confusing bulk modulus with modulus of rigidity; forgetting that fluids cannot sustain shear in static conditions, so K concerns volumetric changes only.Final Answer:increase in pressure to the volumetric strain

Question 14

Froude number (open-channel and free-surface flows): Froude’s number compares inertia force to which other characteristic force?

Accepted Answer

Introduction / Context:Dimensionless numbers encapsulate the competition between forces in fluid flow. Froude number Fr is central to free-surface flows such as rivers, spillways, and ship hydrodynamics. Recognizing the force pairing behind each number helps decide dynamic similarity and interpret scale models correctly.Given Data / Assumptions:Characteristic velocity V and length L (e.g., depth).Gravitational acceleration g acts on the free surface.Inertia force represents ρ * L^2 * V^2 scaling.Concept / Approach:Froude number is defined as Fr = V / sqrt(g * L). Interpreting via force ratios, Fr^2 ∝ (inertia forces) / (gravity forces). When Fr 1 indicates supercritical regimes with distinct wave propagation behavior. This controls hydraulic jump formation, specific energy relationships, and surface wave patterns.Step-by-Step Solution:Recall definition: Fr = V / sqrt(gL).Hence Fr^2 = V^2 / (gL) ∝ inertia/gravity.Therefore, Froude's number compares inertia to gravity forces.Verification / Alternative check:In spillway design, matching Fr between model and prototype ensures similar wave and jump behavior since gravity dominates surface dynamics; other numbers like Reynolds may be unmatched without breaking similarity critically for free-surface patterns.Why Other Options Are Wrong:Pressure force: Leads to Euler number, not Froude.Elastic force: Associated with Mach number.Surface tension force: Associated with Weber number.Common Pitfalls:Confusing Froude with Reynolds (inertia/viscous) or Weber (inertia/surface tension); forgetting that appropriate characteristic length for Fr is usually depth in open-channel flow.Final Answer:gravity force

Question 15

Classification of orifices by head–size ratio: An orifice is termed “large” when which condition on head and orifice height (or diameter) is satisfied?

Accepted Answer

Introduction / Context:Small and large orifices are distinguished not merely by physical size but by the ratio of available head to the characteristic dimension of the opening. This classification determines whether the velocity distribution changes appreciably between the top and bottom of the orifice and whether simple small-orifice formulas remain valid.Given Data / Assumptions:Head H measured from the upstream free surface to the orifice center (or top).Characteristic orifice height or diameter a.Steady flow under gravity.Concept / Approach:A “small” orifice assumes nearly uniform head across the opening, typically valid when H ≥ 5a (some texts use 5 times the orifice depth). When H is not sufficiently large (H Step-by-Step Solution:Check ratio: If H/a ≥ 5 → small orifice.If H/a Therefore, condition for large orifice is H less than 5 times the orifice height.Verification / Alternative check:Deriving discharge by integrating over the orifice depth shows velocity v(y) = sqrt(2g(H − y)) varies from top to bottom; this variation is negligible only when H ≫ a, underpinning the 5-times guideline.Why Other Options Are Wrong:Physical “size is large” or “velocity is large”: Vague and not the accepted criterion.Head more than 5 times: That defines a small orifice, not a large one.Common Pitfalls:Using small-orifice coefficients for large-orifice conditions; defining a using radius instead of height/diameter inconsistently; forgetting elevation head variation across the opening.Final Answer:the available head of liquid is less than 5 times the height of orifice

Question 16

Terminology in open-channel structures: A structure that dams up a stream or river and over which water flows is called what?

Accepted Answer

Introduction / Context:Open-channel measurement and control structures include notches, weirs, flumes, and dams. The terms are sometimes used loosely, but in hydraulics a weir has a specific function and geometry: a barrier or crest across a channel over which water flows, enabling head–discharge relationships and level regulation.Given Data / Assumptions:Flow passes over a crest with a defined head upstream.Free overfall or submerged (drowned) conditions may occur.Permanent structure across the full channel width (usually).Concept / Approach:A weir is a low dam or barrier designed so that flow passes over its crest. It is used to measure discharge (via stage–discharge calibration) and to control upstream water levels. Notches are smaller openings used in laboratory tanks or small channels. An “orifice” is a submerged opening through a wall; a “dam” generally impounds water but is not primarily defined by overflow measurement functions.Step-by-Step Solution:Identify structure type: barrier with overflow across a crest → weir.Match to choices: “weir” best fits the description.Verification / Alternative check:Standard weir types include sharp-crested, broad-crested, ogee, and Cipolletti (trapezoidal). Each has known head–discharge formulas, confirming the measurement role that distinguishes weirs from general dams.Why Other Options Are Wrong:Orifice: Submerged opening through which water issues.Notch: Laboratory-scale opening in a tank wall.Dam: Generic impounding structure; may not be designed for calibrated overflow.Common Pitfalls:Using “dam” and “weir” interchangeably; overlooking that notches are essentially small weirs used for measurement at smaller scales.Final Answer:weir

Question 17

Pipes in series replaced by an equivalent pipe: Which of the following statements about an equivalent pipe for a compound (series) pipe system is incorrect?

Accepted Answer

Introduction / Context:Multiple pipes in series with different diameters and lengths can be replaced by one “equivalent” pipe that yields the same discharge under the same head loss. This simplification is frequently used in network calculations and preliminary sizing. Understanding what stays the same and what changes in the equivalent representation is key to correct modeling.Given Data / Assumptions:All segments are in series, carrying the same discharge Q.Total head loss across the series is known or specified.Friction losses dominate; minor losses may be included by adjusting length.Concept / Approach:For a compound pipe in series, the total head loss h_f,total is the sum of losses over each segment. An equivalent single pipe is defined such that it carries the same Q with the same total h_f across the same overall length (often taken as the geometric length sum or an adjusted “equivalent length” including fittings). The required equivalent diameter is then computed from Darcy–Weisbach or Hazen–Williams; it generally differs from any individual segment diameter and from a simple average.Step-by-Step Solution:Set Q_equiv = Q_series (same discharge).Set h_f,eq = h_f,total (same head loss across endpoints).Solve for D_eq from the friction formula; D_eq seldom equals any component diameter.Verification / Alternative check:Using Darcy–Weisbach with f assumed constant for a first pass: h_f ∝ (L/D^5) * Q^2 for laminar or with appropriate friction relations in turbulent range. Aggregating L/D^5 terms shows D_eq depends on a weighted combination, not equality to compound-pipe diameters.Why Other Options Are Wrong:Equivalent pipe used as ordinary pipe: Correct for calculations.Length equal to compound pipe: Typically taken as the sum (or adjusted equivalent length), acceptable statement.Discharge equal: By definition, equivalence preserves Q.Common Pitfalls:Assuming D_eq equals arithmetic or geometric mean of component diameters; neglecting minor-loss contributions to equivalent length; mixing Darcy and Hazen–Williams forms without consistent parameters.Final Answer:The diameter of an equivalent pipe is equal to that of a compound pipe

Question 18

Velocity profile in pipe flow: When a liquid flows through a circular pipe, how does the local velocity vary from the wall to the centreline (typical laminar or turbulent cases)?

Accepted Answer

Introduction / Context:The no-slip condition at a solid boundary requires fluid velocity relative to the wall to vanish at the wall. Away from the wall, velocity increases toward the pipe centreline, reaching a maximum. This holds for both laminar (parabolic profile) and turbulent (flatter, but still with zero at the wall) flows, making the general qualitative statement robust.Given Data / Assumptions:Circular pipe, steady internal flow.No-slip boundary condition applies.Incompressible fluid; fully developed region considered.Concept / Approach:For laminar flow, the exact solution of the Navier–Stokes equations yields a parabola: u(r) = umax * (1 − (r/R)^2), where u = 0 at r = R (wall) and u = umax at r = 0 (centre). For turbulent flow, the mean profile is blunter (log-law in the inner/outer layers), yet still satisfies u = 0 at the wall and u = umax at the centreline. Therefore, the most precise of the given choices is “maximum at the centre and zero near the walls.”Step-by-Step Solution:Apply no-slip: u = 0 at pipe wall.Recognize viscous or turbulent transport sets the shape, but the centreline has the highest speed.Select option that explicitly states zero at wall and maximum at centre.Verification / Alternative check:Pitot traverses show stagnation reading rising from wall toward centre; in laminar theory, umax = 2 * v_mean; in turbulent, umax ≈ 1.2–1.3 * v_mean depending on Reynolds number, corroborating the qualitative pattern.Why Other Options Are Wrong:Maximum near walls: Violates no-slip.Zero at centre: Physically impossible for pressure-driven pipe flow.“Minimum near walls” is true but imprecise; the exact boundary value is zero, making option D superior.Common Pitfalls:Assuming plug flow for turbulent cases; ignoring viscous sublayer presence where velocity ramps from 0 at the wall.Final Answer:maximum at the centre and zero near the walls

Question 19

Water hammer basics: How does the peak pressure rise due to a sudden valve closure vary with the fluid density (all else equal)?

Accepted Answer

Introduction / Context:Water hammer (pressure surge) occurs when flow is rapidly decelerated, launching a compression wave that propagates along the pipe. The classic Joukowsky relation captures the instantaneous pressure rise for a sudden change in velocity. Understanding how density affects this rise is critical for surge protection, valve operation strategies, and pipe material selection.Given Data / Assumptions:Elastic pipe–fluid system characterized by wave speed c.Sudden valve closure causing velocity change ΔV.Fluid is slightly compressible; pipe is slightly elastic.Concept / Approach:The Joukowsky equation states Δp = ρ * c * ΔV for a rapid closure without significant reflection effects during the initial instant. For fixed ΔV and fixed c, the pressure rise is directly proportional to fluid density ρ. Even though c itself depends on fluid bulk modulus and pipe elasticity, the primary proportionality to ρ remains explicit in the formula.Step-by-Step Solution:Write Joukowsky: Δp = ρ * c * ΔV.Hold c and ΔV constant (same system and maneuver).Conclude: Δp ∝ ρ → greater density yields greater surge pressure.Verification / Alternative check:Wave speed c ≈ sqrt(K/ρ) adjusted for pipe elasticity. If ρ increases modestly while K and pipe properties remain fixed, c decreases slightly, but the product ρ * c still grows with ρ for common ranges, aligning with the direct proportionality in the Joukowsky relation when c is treated as system-given.Why Other Options Are Wrong:Inverse or square-root relations: Do not match the fundamental Δp = ρ * c * ΔV dependence.Common Pitfalls:Ignoring the role of pipe elasticity; assuming incompressible rigid-pipe conditions where c → ∞ (nonphysical); confusing mass density with specific weight when comparing fluids.Final Answer:directly proportional to density of fluid

Question 20

Discharge coefficients in orifice flow – relationship among Cd, Cc, and Cv In classical hydraulics for a sharp-edged orifice, the overall coefficient of discharge Cd is related to the coefficient of contraction Cc and the coefficient of velocity Cv …

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Introduction:The coefficient of discharge Cd for an orifice accounts for both jet contraction and departure from ideal velocity. This question checks whether you remember the fundamental decomposition linking Cd to the coefficient of contraction Cc and the coefficient of velocity Cv, a staple identity in fluid mechanics labs and exams.Given Data / Assumptions:Sharp-edged orifice issuing into the atmosphere (free jet) or similar condition where standard coefficients apply.Steady flow; head measured appropriately so Torricelli’s law is usable with coefficients.Neglect compressibility; liquid is Newtonian.Concept / Approach:Ideal jet velocity is V_ideal = sqrt(2 * g * H). Real jets contract to a smaller effective area A_e = Cc * A and leave with a velocity V = Cv * V_ideal. The actual discharge is Q = A_e * V = (Cc * A) * (Cv * V_ideal). Comparing with the ideal discharge Q_ideal = A * V_ideal identifies Cd = Q / Q_ideal = Cc * Cv.Step-by-Step Solution:Write Q = A_e * V = (Cc * A) * (Cv * V_ideal).Compute the ratio: Cd = Q / (A * V_ideal) = Cc * Cv.Interpretation: Cc captures geometric contraction; Cv captures velocity shortfall; their product captures overall discharge reduction.Verification / Alternative check:Typical values for sharp-edged orifices: Cc ≈ 0.62, Cv ≈ 0.97, giving Cd ≈ 0.60. This matches widely reported data, supporting Cd = Cc * Cv.Why Other Options Are Wrong:Cc x Cr or Cv x Cr: Coefficient of resistance Cr is not multiplied this way to obtain Cd.Cc/Cr or Cc + Cv: Do not reflect the physical composition of area and velocity effects.Common Pitfalls:Confusing Cv with Cd, or forgetting that contraction and velocity effects are independent phenomena whose influences multiply in discharge.Final Answer:Cc x Cv

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