Question 1

Fluids change volume under external pressure primarily due to which material property? (Assume ordinary liquids and gases in classical fluid mechanics; choose the most appropriate fundamental property.)

Accepted Answer

Introduction / Context:In fluid mechanics and engineering, we often need to know whether a fluid's volume will change when external pressure is applied. This determines how we model flow (compressible vs incompressible), affects wave speed, cavitation, and the behavior of hydraulic systems. Given Data / Assumptions: Ordinary fluids (liquids or gases) under external pressure. No phase change, chemical reaction, or temperature-driven expansion is considered. Small to moderate pressure variations relevant to engineering analysis. Concept / Approach: The property that quantifies volume change with pressure is compressibility (or its inverse, bulk modulus). Viscosity quantifies internal friction and shear stress–strain rate relation; plasticity is a solid-material concept describing permanent deformation after yield; elasticity describes stress–strain but for fluids in bulk it collapses to compressibility for volumetric changes. Step-by-Step Solution: Define compressibility: k = −(1/V) * (dV/dp).If k > 0 and finite, volume decreases as pressure increases.Liquids have small k (nearly incompressible), gases have larger k (noticeably compressible). Verification / Alternative check: Speed of sound c in a fluid is related to compressibility through c^2 = 1/(ρ * k_eff). If volume could not change, c would be infinite, which is not physical except as an approximation for water at low speeds. Why Other Options Are Wrong: Plasticity: pertains to solids and permanent shape change, not bulk volume change in fluids. Viscosity: measures resistance to shear, not compressive volumetric change. “None of these” and “elasticity only (no volume change)” are incorrect because fluids do exhibit finite compressibility. Common Pitfalls: Assuming “incompressible” means zero compressibility; it is an approximation valid when density changes are negligible for the problem scale. Final Answer: Compressibility

Question 2

Momentum correction factor β for fully developed laminar flow in a circular pipe In internal laminar (parabolic) flow of a Newtonian fluid through a round pipe, what is the appropriate value of the momentum correction factor β?

Accepted Answer

Introduction / Context:The momentum correction factor β accounts for the non-uniform velocity profile when converting between average-velocity-based momentum flux and the actual momentum flux. It is essential in accurate force and thrust computations for ducts and nozzles. Given Data / Assumptions: Newtonian fluid, steady, incompressible, fully developed laminar flow in a circular pipe. Parabolic velocity distribution, no slip at the wall. Concept / Approach: Definition: β = (∫_A ρ u^2 dA) / (ρ A u_avg^2). For a parabolic profile u(r) = u_max * (1 − (r/R)^2), the integrals can be evaluated analytically to obtain a constant β for laminar flow in a circular pipe. Step-by-Step Solution: Compute u_avg = (1/A) ∫_A u dA = u_max/2.Compute ∫_A u^2 dA using polar area element 2π r dr and the parabolic u(r).After integration, β = (∫ u^2 dA) / (A u_avg^2) = 4/3 ≈ 1.333. Verification / Alternative check: Known standard results: energy correction factor α = 2 (laminar pipe), momentum correction factor β = 4/3. For turbulent flow, both factors are closer to 1 due to flatter profiles. Why Other Options Are Wrong: 1/2, 2/3, 3/4, 3/2 do not match the parabolic-profile integration result for laminar pipe flow. Common Pitfalls: Confusing α (energy correction) with β (momentum correction), or using turbulent values (near 1) for laminar cases. Final Answer: 4/3

Question 3

Condition for maximum horsepower at a nozzle fed by a long pipeline For maximum power delivered by a nozzle at the end of a pipe, the total head supplied must be equal to how many times the head lost due to pipe friction?

Accepted Answer

Introduction / Context:Power transmission by water through pipelines is limited by frictional head losses. There exists an optimal operating condition where the product of discharge and available head at the nozzle is maximized, yielding maximum horsepower (HP) at the outlet. Given Data / Assumptions: Total head at the pipe entrance: H. Head loss due to friction along the pipe: h_f. Head available at the nozzle: H_o = H − h_f. Other losses (minor losses) neglected for the classical result. Concept / Approach: Delivered power P ∝ Q * H_o. For a given pipe, h_f ∝ V^2 and Q ∝ V (V is mean velocity). Thus P ∝ V * (H − k V^2). Maximizing this with respect to V provides the optimum balance between flow and head loss. Step-by-Step Solution: Let P(V) = c * V * (H − k V^2).dP/dV = c * (H − 3 k V^2) = 0 ⇒ H = 3 k V^2.But h_f = k V^2, hence H = 3 h_f ⇒ H/h_f = 3. Verification / Alternative check: At this optimum, the efficiency of transmission is η_max = (H − h_f)/H = (H − H/3)/H = 2/3 (a well-known benchmark value). Why Other Options Are Wrong: Equating H to 1×, 2×, or 4× h_f does not satisfy the optimum condition from calculus; those settings yield less outlet power. Common Pitfalls: Maximizing head alone or flow alone instead of their product; ignoring that friction rise with V^2 shifts the optimum. Final Answer: Thrice the head loss (3×)

Question 4

Identify the parameter that is NOT dimensionless for flowing liquids Which of the following is not a dimensionless parameter used to characterize fluid flow?

Accepted Answer

Introduction / Context:Fluid dynamicists routinely use dimensionless numbers to compare flows across scales and media. Recognizing which quantities are dimensionless prevents errors when applying similarity laws and correlating experimental data. Given Data / Assumptions: Standard SI or consistent unit systems. Typical external or internal flows without phase change. Concept / Approach: Reynolds number, Weber number, pressure coefficient, and Darcy–Weisbach friction factor are all non-dimensional by construction. Kinematic viscosity ν has units of area per time (L^2/T) and is not dimensionless. Step-by-Step Solution: Check dimensions: ν = μ/ρ has units m^2/s, hence dimensional.Re, We, C_p (pressure coefficient), and f are ratios of forces or pressures scaled by dynamic quantities ⇒ dimensionless. Verification / Alternative check: Dimensional analysis (Buckingham π theorem) confirms that Re, We, etc., arise as π-groups; ν is a base property, not a π-group. Why Other Options Are Wrong: They are standard dimensionless measures widely tabulated in fluid handbooks. Common Pitfalls: Confusing ν with the reciprocal of Reynolds number; they are related but not interchangeable. Final Answer: Kinematic viscosity

Question 5

Stability of a floating body: position of the metacentre A floating body will have stable equilibrium when the metacentre (M) is located where, relative to the body’s centre of gravity (G)?

Accepted Answer

Introduction / Context:Ship stability, pontoon design, and floating structures rely on metacentric height GM. The sign and magnitude of GM determine whether small angular disturbances are self-correcting (stable) or amplifying (unstable). Given Data / Assumptions: Small heel angles (initial stability). Homogeneous fluid and standard hydrostatic definitions of G, B (buoyancy center), and M (metacentre). Concept / Approach: Initial stability criterion: GM = BM − BG. If GM > 0, the righting moment restores equilibrium. Geometrically, this is equivalent to metacentre M lying above center of gravity G for small angles. Step-by-Step Solution: At small heel, buoyancy shifts to a new line of action; lines of action intersect at M.If M above G ⇒ righting couple forms ⇒ stable.If M below G ⇒ overturning couple ⇒ unstable. Verification / Alternative check: Righting arm GZ ≈ GM * sin θ for small θ. Positive GM yields positive GZ, giving restoring moment Δ * GZ (Δ is displacement). Why Other Options Are Wrong: M at G (GM = 0) is neutral, not stable. M below G is unstable. “Anywhere” and “at the waterline” do not reflect the hydrostatic stability criterion. Common Pitfalls: Confusing centroid with centre of gravity and with centre of buoyancy; only the position relative to G matters for initial stability. Final Answer: Above the centre of gravity (M above G)

Question 6

Froude number at critical flow In open-channel hydraulics, what is the value of the Froude number at critical depth/critical flow conditions?

Accepted Answer

Introduction / Context:The Froude number Fr compares inertial forces to gravitational wave effects in open-channel flow. Identifying whether flow is subcritical, critical, or supercritical is crucial for control structures, hydraulic jumps, and energy dissipation design. Given Data / Assumptions: Definition: Fr = V / sqrt(g * D), where D is hydraulic depth (A/T) or depth y for wide channels. Steady, prismatic channel conditions. Concept / Approach: Critical flow is the transition point between subcritical (Fr < 1, gravity-dominated) and supercritical (Fr > 1, inertia-dominated). At the transition, Fr equals unity; disturbances neither propagate upstream nor downstream preferentially. Step-by-Step Solution: Set Fr = 1 at critical: V_c = sqrt(g * D_c).Relate to specific energy minimum: dE/dy = 0 occurs at Fr = 1.Design gates and weirs to achieve or avoid Fr ≈ 1 depending on objectives. Verification / Alternative check: Empirical observations of standing waves and control sections align with the condition Fr = 1 at critical depth. Why Other Options Are Wrong: Values other than 1 indicate subcritical or supercritical flow regimes, not the transition point. Common Pitfalls: Using depth y instead of hydraulic depth D for non-wide channels; misclassifying flow regime when estimating V or D. Final Answer: 1.0

Question 7

Critical velocity in terms of critical depth in open-channel flow If h_c is the critical depth, what is the expression for the critical velocity V_c in a wide channel (hydraulic depth ≈ depth)?

Accepted Answer

Introduction / Context:Critical flow concepts are used to size control sections and predict transitions such as hydraulic jumps. The relation between critical depth and critical velocity provides a quick check in design and analysis for wide, prismatic channels. Given Data / Assumptions: Wide channel so hydraulic depth D ≈ flow depth y. Critical state satisfies Fr = 1. Acceleration due to gravity = g. Concept / Approach: Froude number Fr = V / sqrt(g * D). At critical flow, Fr = 1, hence V_c = sqrt(g * D_c). For wide channels, D_c ≈ h_c, yielding the familiar square-root relation. Step-by-Step Solution: Write Fr = V / sqrt(g * D) = 1 at critical.Replace D by h_c in wide-channel approximation.Obtain V_c = sqrt(g * h_c). Verification / Alternative check: The result is consistent with specific energy theory where minimum specific energy occurs at Fr = 1 and V_c relates to depth through the same relation. Why Other Options Are Wrong: g * h_c or h_c / sqrt(g) have incorrect dimensions; sqrt(g / h_c) is dimensionally inconsistent for velocity. Common Pitfalls: Using hydraulic radius or area/width inconsistently; the wide-channel simplification avoids those complications. Final Answer: V_c = sqrt(g * h_c)

Question 8

When can a flow net be drawn? Select the essential flow characteristics under which a valid flow net (equipotential lines and streamlines orthogonal) can be constructed for seepage or potential flow analysis.

Accepted Answer

Introduction / Context:Flow nets (families of orthogonal streamlines and equipotential lines) are powerful graphical tools for groundwater seepage, sheet pile design, and potential flow around structures. They rely on the mathematics of Laplace’s equation. Given Data / Assumptions: Steady conditions (time-invariant). Incompressible fluid (constant density). Two-dimensional domain (planar section). Irrotational flow (zero vorticity). Concept / Approach: Under these assumptions, the velocity potential φ and stream function ψ exist and satisfy Laplace’s equation. Their level sets are orthogonal, enabling the construction of a conformal mesh known as the flow net. Step-by-Step Solution: Confirm incompressibility ⇒ continuity reduces to ∇·V = 0.Confirm irrotationality ⇒ ∇×V = 0 ⇒ existence of potential φ.In 2D steady flow, φ and ψ satisfy ∇^2φ = 0, ∇^2ψ = 0 ⇒ orthogonal families are drawable. Verification / Alternative check: Classic seepage under dams and around sheet piles is modeled with flow nets because porous media flow is well-approximated by 2D potential flow in many practical cases. Why Other Options Are Wrong: Turbulent, rotational, compressible with shocks, or massively unsteady separated flows violate the Laplacian assumptions; flow nets cease to be valid. Common Pitfalls: Forgetting to enforce boundary conditions on impervious and constant-head boundaries; drawing non-square curvilinear cells leads to inaccurate head loss estimates. Final Answer: Two-dimensional, incompressible, irrotational (potential) flow

Question 9

Head–discharge proportionality for notches Choose the correct statement(s) relating discharge Q to head H over a weir/notch under free-flow conditions.

Accepted Answer

Introduction / Context:Discharge measurement by notches and weirs is fundamental in hydraulic laboratories and field channels. The head–discharge exponent depends on the crest shape and flow contraction behavior. Given Data / Assumptions: Free, fully aerated overflow without submergence. Standard coefficients near calibrated values; end-contraction corrections omitted for proportionality. Sharp-crested geometry for the rectangular case; V-notch with fixed angle (e.g., 90°) for the triangular case. Concept / Approach: Classical results: rectangular sharp-crested weir Q ∝ H^(3/2); triangular (V-notch) Q ∝ H^(5/2). The larger exponent for the V-notch reflects the narrowing flow area with decreasing head, increasing sensitivity to H. Step-by-Step Solution: Rectangular: Q = C_d * b * √(2 g) * H^(3/2).Triangular: Q = (8/15) * C_d * √(2 g) * tan(θ/2) * H^(5/2).Hence proportionalities in (a) and (b) are both correct. Verification / Alternative check: Calibration charts and standard texts confirm these exponents under non-drowned, sharp-crested conditions. Why Other Options Are Wrong: Options denying either exponent contradict well-established discharge formulas. Common Pitfalls: Using rectangular exponents for V-notches; ignoring submergence or approach velocity corrections when high accuracy is needed. Final Answer: Both (a) and (b)

Question 10

Unit of dynamic viscosity in the CGS system In the centimetre–gram–second (CGS) system, what is the unit of dynamic viscosity?

Accepted Answer

Introduction / Context:Viscosity units vary with unit systems. Converting and recognizing units correctly is essential when reading material property tables or correlating experimental data across metric and CGS literature. Given Data / Assumptions: Dynamic viscosity μ is being asked (not kinematic viscosity ν). CGS (centimetre–gram–second) unit system. Concept / Approach: In CGS, the unit of dynamic viscosity is the poise (P). One poise equals 1 g/(cm·s), and 1 P = 0.1 Pa·s in SI. Kinematic viscosity in CGS uses stokes (St), where 1 St = 1 cm^2/s. Step-by-Step Solution: Recall: shear stress τ = μ * (du/dy). In CGS, τ in dyn/cm^2, gradient in s^−1 ⇒ μ unit dyn·s/cm^2 = g/(cm·s) = 1 poise.Relation to SI: 1 Pa·s = 10 P ⇒ 1 P = 0.1 Pa·s. Verification / Alternative check: Typical dynamic viscosity of water at 20°C is about 1 cP (centipoise) = 0.001 Pa·s, matching the conversion above. Why Other Options Are Wrong: Dyne and Newton are force units; joule is energy; “none” is unnecessary because poise is the standard CGS unit. Common Pitfalls: Confusing centipoise (cP) with poise (P); many data sheets list viscosity in cP, where 1 cP = 0.01 P. Final Answer: Poise

Question 11

In jet outflow devices, compare jet contraction between two types of mouthpieces. Even though the angle of deviation of the liquid is larger for an internal (re-entrant/Borda) mouthpiece, the contraction of the issuing jet is:

Accepted Answer

Introduction / Context:Mouthpieces are short tubes fitted to an orifice. Two common types are the internal (re-entrant, often called Borda mouthpiece) and the external (short cylindrical) mouthpiece. A key behavioural difference is the extent of jet contraction (how much the actual jet area reduces relative to the physical opening). Understanding this helps in predicting discharge and energy losses. Given Data / Assumptions: Internal (re-entrant) mouthpiece causes the flow to turn sharply and separate internally. External mouthpiece allows the jet to form outside and may run free or full depending on pressure conditions. Coefficient of contraction Cc qualitatively indicates the degree of contraction (smaller jet area means stronger contraction). Concept / Approach:In a re-entrant mouthpiece, the jet forms a pronounced vena contracta immediately at the plane of the orifice and the effective jet area is significantly reduced compared with the opening. In contrast, the short external mouthpiece tends to have a fuller jet and less contraction when running full due to wall-guided flow. Step-by-Step Solution:Identify flow deflection: internal mouthpiece requires a larger turn → stronger separation.Separation intensifies contraction → smaller effective jet area.Hence, contraction is greater for the internal mouthpiece than for the external one. Verification / Alternative check:Typical experimental data show lower discharge coefficient for the Borda (internal) mouthpiece compared to a short external mouthpiece running full, consistent with a stronger contraction in the internal type. Why Other Options Are Wrong:Less in the internal mouthpiece: contradicts separation-driven contraction.Equal contraction: ignores geometric and flow-path differences.None of these: unnecessary; the correct comparative statement is available. Common Pitfalls: Confusing discharge coefficient with contraction alone; while related, Cc, Cv, and Cd differ. Assuming all short tubes behave like sharp-edged orifices. Final Answer:more in the internal mouthpiece

Question 12

A liquid rotates as a rigid body with constant angular velocity about a vertical axis. How does pressure vary with radial distance from the axis?

Accepted Answer

Introduction / Context:In forced vortex motion (rigid-body rotation), every fluid particle rotates with the same angular velocity. The free surface becomes parabolic and pressure increases radially outward. Recognizing the correct pressure–radius relation is essential for problems involving rotating tanks and centrifuges. Given Data / Assumptions: Angular velocity = ω (constant). Axis of rotation is vertical; gravity acts downward. Negligible viscous shear in the bulk (idealized rigid-body rotation). Concept / Approach:Radial equilibrium requires the pressure gradient to balance centripetal acceleration: (1/ρ) * dp/dr = ω^2 * r. Integrating from the axis (r = 0) gives p(r) = p(0) + 0.5 * ρ * ω^2 * r^2, showing a quadratic dependence on r. Step-by-Step Solution:Set radial equilibrium: dp/dr = ρ * ω^2 * r.Integrate: p(r) − p(0) = ∫0^r ρ ω^2 r dr = 0.5 ρ ω^2 r^2.Therefore, pressure increases proportional to r^2. Verification / Alternative check:The parabolic free surface z = (ω^2 r^2)/(2g) has the same r^2 dependence, consistent with the pressure field in the rotating liquid. Why Other Options Are Wrong:Linear increase: would correspond to a constant radial body force, not centripetal acceleration proportional to r.Inverse altitude statement is unrelated to radial pressure variation.Decrease with r^2 contradicts equilibrium. Common Pitfalls: Mixing up free-vortex (irrotational) and forced-vortex (rigid-body) behaviours. Forgetting that p varies both with r and z; here the question isolates radial dependence. Final Answer:Increases as the square of the radial distance

Question 13

During sudden valve closure in a water-filled pipe with flow velocity v, the surge pressure (water hammer) intensity is directly proportional to which property?

Accepted Answer

Introduction / Context:Water hammer is a transient pressure rise that occurs when flow is rapidly changed. For a rigid column approximation with elastic fluid/pipe, the Joukowsky equation estimates the pressure rise: Δp = ρ * a * Δv, where a is the wave speed. Understanding how Δp scales with material properties is key for surge protection. Given Data / Assumptions: Sudden closure so Δv ≈ v. Elastic fluid with bulk modulus K; pipe-wall elasticity may also matter, but the core dependence of a on K remains via a ∝ sqrt(K/ρ) (modified for wall effects). Density ρ approximately constant. Concept / Approach:The wave speed a in a fluid–structure system is approximately a = sqrt( K / ρ / (1 + (KD)/(Ee)) ). With pipe rigidity large, a ≈ sqrt(K/ρ). Therefore, Δp ∝ a ∝ sqrt(K) for fixed ρ and fixed Δv. Hence, the direct proportionality to the square root of bulk modulus is the most accurate among the given choices. Step-by-Step Solution:Start with Δp = ρ * a * v.Note a ≈ sqrt(K/ρ) (rigid pipe limit).Therefore, Δp ∝ ρ * sqrt(K/ρ) * v = v * sqrt(K * ρ) → proportional to sqrt(K) for fixed ρ and v. Verification / Alternative check:Including wall elasticity changes only the prefactor in a; the dependence on sqrt(K) persists, confirming the chosen proportionality trend. Why Other Options Are Wrong:Linear in K: contradicts wave-speed theory.Linear in specific weight: Δp depends on ρ via sqrt(ρ), not γ directly; γ also introduces g which is not part of the Joukowsky relation.None of these: incorrect because a √K dependence exists. Common Pitfalls: Assuming incompressible (K → ∞) behaviour which would imply infinite a and unrealistic Δp. Ignoring the effect of valve closure time relative to 2L/a. Final Answer:Square root of the bulk modulus of elasticity of water

Question 14

For a triangular (V-notch) weir, which notch angle gives the maximum discharge for a given head over the crest?

Accepted Answer

Introduction / Context:The discharge over a V-notch weir is governed by the notch angle θ and the upstream head H. For a given head, the notch angle affects the effective flow area and thus the discharge. Designers select the angle to achieve measurable heads and desired ranges of flow. Given Data / Assumptions: Sharp-crested, fully aerated triangular notch. Coefficient of discharge Cd approximately constant over the operating range. Comparison is for the same head H. Concept / Approach:The theoretical discharge relation for a V-notch is Q = (8/15) * Cd * √(2g) * tan(θ/2) * H^(5/2). For fixed H and Cd, Q is proportional to tan(θ/2), which monotonically increases with θ for 0° < θ < 180°. Practical practice favours θ = 90° as the standard for large, stable discharge with good sensitivity. Step-by-Step Solution:Write Q ∝ tan(θ/2) * H^(5/2).For fixed H, maximize tan(θ/2): larger θ → larger Q.The commonly adopted maximum practical angle is 90°, providing high discharge without geometric issues. Verification / Alternative check:Hydrometry standards frequently use 90° notches as reference because of accuracy and manageable nappe behaviour across a wide range of heads. Why Other Options Are Wrong:30° and 60° produce smaller tan(θ/2), hence smaller discharge.120° increases tan(θ/2) further but is generally not used for standard maximum discharge calibration; 90° is the recognized practical maximum choice in typical exam contexts. Common Pitfalls: Comparing discharges at different heads; the head must be the same for fair comparison. Ignoring aeration; a drowned nappe alters coefficients. Final Answer:90°

Question 15

Identify the condition(s) that imply non-uniform flow in an open channel. Choose the most comprehensive statement.

Accepted Answer

Introduction / Context:Open-channel flow classification depends on how flow variables change in space and time. Non-uniform (spatially varied) flow occurs when depth and velocity change along the channel length due to bed slope, geometry, or added/removed energy. Given Data / Assumptions: Steady flow unless otherwise noted (time-invariant at a fixed section). Channel may have prismatic or non-prismatic geometry. Energy grade line and bed slope may differ. Concept / Approach:Uniform flow requires constant depth and velocity along the channel, implying water surface parallel to the bed and energy slope equal to bed slope. Any departure from these conditions (accelerating or decelerating flow) implies non-uniform flow. Step-by-Step Solution:Option (a): If the free surface is not parallel to the bed, depth varies with x → non-uniform.Option (b): If head to overcome friction is less than bed drop, excess head accelerates the flow → depth/velocity vary → non-uniform.Option (c): If head to overcome friction exceeds bed drop, additional losses or backwater effects decelerate the flow → again non-uniform. Verification / Alternative check:Gradually varied flow equations (dY/dx) predict profiles whenever So ≠ Sf or geometry changes. Both accelerating and decelerating cases are encompassed by non-uniform flow. Why Other Options Are Wrong:Choosing only one of (a), (b), or (c) ignores other valid mechanisms for non-uniformity; the complete answer is the collective condition. Common Pitfalls: Confusing non-uniform with unsteady; non-uniform concerns spatial variation, not time variation. Assuming prismatic channels always yield uniform flow—this is only true when slopes match and flow is fully developed. Final Answer:All of the above

Question 16

In compressible-flow approximations, gases may be treated as incompressible when the Mach number (M) is approximately:

Accepted Answer

Introduction / Context:Compressibility effects in gases become significant as flow speed approaches the speed of sound. For many engineering calculations, a threshold Mach number is used to justify the incompressible assumption and the use of simpler formulas. Given Data / Assumptions: Mach number M = V/a, where a is the local speed of sound. Small density change criterion for incompressible modelling. Low subsonic regime. Concept / Approach:Empirical and theoretical analyses show that when M < 0.3, compressibility effects are minor for many applications; a more conservative and commonly taught rule-of-thumb is M < 0.2, under which density change is negligible and incompressible flow theory applies with good accuracy. Step-by-Step Solution:Check regime: M < 0.2 → negligible compressibility.Therefore, continuity and momentum equations can be used with constant density. Verification / Alternative check:Isentropic relations show density change scales with M^2 for small M; at M = 0.2, changes are within a few percent, acceptable for many designs. Why Other Options Are Wrong:M ≥ 0.5: compressibility influential; shocks not present but density change is non-negligible.M ≈ 1 or 1.5: transonic/supersonic—strongly compressible. Common Pitfalls: Applying incompressible Bernoulli at high Mach numbers. Ignoring temperature variations tied to compressibility. Final Answer:Less than 0.2

Question 17

A vertical retaining wall holds back water to a depth H (free surface at the top, vertical plane surface). The height of the centre of pressure above the bottom of the wall is:

Accepted Answer

Introduction / Context:The centre of pressure locates the line of action of the resultant hydrostatic force on a submerged plane surface. For design of retaining walls and gates, its position determines overturning moments and required stability. Given Data / Assumptions: Vertical plane surface with its top at the free surface, depth H. Hydrostatic pressure distribution increases linearly with depth. Fluid is at rest; density constant. Concept / Approach:For a vertical rectangle with top at the free surface, the resultant force acts at depth 2H/3 below the free surface. Therefore, the distance above the bottom (which is at depth H) is H − 2H/3 = H/3. Step-by-Step Solution:Pressure at depth y: p = ρ g y.Resultant force acts at y_cp = (∫ y p dA)/(∫ p dA) = 2H/3 below free surface for a vertical rectangle starting at the surface.Height above bottom = H − 2H/3 = H/3. Verification / Alternative check:Using the second-moment formula: y_cp = ȳ + I_G/(ȳ A), for this case ȳ = H/2 and I_G = b H^3/12; substituting yields y_cp = 2H/3, consistent with the simple result. Why Other Options Are Wrong:H/2 and 3H/4 contradict the linear pressure centroid location.2H/3 is the depth below the free surface, not height above bottom. Common Pitfalls: Reporting the depth below the surface when the question asks distance above the bottom. Confusing centroid of area with centre of pressure (they differ due to pressure variation). Final Answer:H/3

Question 18

Equivalent diameter for multiple identical parallel pipes: To replace one pipe of diameter D by n parallel pipes each of diameter d (same length and headloss), the relationship is best given by:

Accepted Answer

Introduction / Context:Designers often split a large pipeline into several smaller parallel lines to reduce cost or to ease installation. An equivalent diameter relation helps preserve total discharge under the same headloss and length when using the Darcy–Weisbach equation in the turbulent regime with roughly constant friction factor. Given Data / Assumptions: Same length L for all pipes. Same headloss h_f for the single pipe and the set of n parallel pipes. Turbulent flow with approximately constant friction factor f (fully rough assumption) so f does not vary strongly with Reynolds number. Concept / Approach:For Darcy–Weisbach, h_f ∝ (L/D) * V^2. For a given h_f and L, V ∝ √D. Discharge Q = (π D^2/4) V ∝ D^2 * √D = D^(5/2). For n identical parallel pipes, total Q is the sum of n equal branch discharges, each proportional to d^(5/2). Hence D^(5/2) = n d^(5/2). Step-by-Step Solution:Single pipe capacity: Q_single ∝ D^(5/2).Each branch: q_branch ∝ d^(5/2).Total with n branches: Q_total = n q_branch ∝ n d^(5/2).Equate capacities: D^(5/2) = n d^(5/2). Verification / Alternative check:If laminar flow applied, Q ∝ D^4, yielding D^4 = n d^4 (option (c)); but for most water-distribution designs (turbulent), the 5/2 power rule is used—validating option (b). Why Other Options Are Wrong:D^2 and D^3 relations do not follow from Darcy–Weisbach with constant f.D^4 applies to laminar Hagen–Poiseuille flow, not typical design turbulence. Common Pitfalls: Using laminar equivalence in turbulent contexts. Ignoring unequal distribution if branch minor losses differ (not assumed here). Final Answer:D^(5/2) = n d^(5/2)

Question 19

For a short cylindrical external mouthpiece running free, the vena-contracta forms at a distance from the outlet approximately equal to:

Accepted Answer

Introduction / Context:When fluid issues through a short external mouthpiece, the jet contracts to a minimum cross-section called the vena-contracta before expanding downstream. The distance of this section from the outlet helps estimate coefficients and practical layout clearances. Given Data / Assumptions: Short external cylindrical mouthpiece on a large reservoir wall. Jet runs free (adequate aeration, atmospheric pressure outside). Negligible downstream submergence effects. Concept / Approach:Empirical observations and classical hydraulics indicate the vena-contracta for a short external mouthpiece occurs at about one-third of the orifice diameter downstream of the outlet plane, with a contraction coefficient close to unity when the mouthpiece runs full beyond the contraction. Step-by-Step Solution:Identify the regime: free jet from short external tube.Recall standard distance: approximately D/3 from the outlet face.Therefore the correct selection is one-third of the diameter. Verification / Alternative check:Textbook tabulations of mouthpiece characteristics list the vena-contracta location for short external tubes at roughly 0.33 D from the exit plane, confirming the rule of thumb. Why Other Options Are Wrong:One diameter or two-thirds diameter overestimates the distance.One-fourth diameter underestimates the location for this configuration. Common Pitfalls: Confusing locations for sharp-edged orifices versus short tubes. Ignoring changes if the mouthpiece is running submerged or with suction effects. Final Answer:One-third of the orifice diameter

Question 20

For maximum power transmission through a nozzle at the end of a pipeline (with pipeline friction head loss h_f and available head H at the pipe entrance), the condition for maximum power is:

Accepted Answer

Introduction / Context:When a nozzle is supplied by a long pipeline from a reservoir, the jet power at the nozzle depends on the jet velocity (hence residual head) and on losses in the pipe. There exists an optimal balance between velocity gain and friction loss that maximizes power delivered by the jet. Given Data / Assumptions: Total available head at pipe entrance = H. Head loss in pipeline due to friction = h_f (minor losses neglected for simplicity). Nozzle converts remaining head (H − h_f) to velocity head at the exit. Concept / Approach:Jet power P ∝ ρ g Q (H − h_f). For a given pipe, discharge Q increases with head difference but also increases h_f (typically h_f ∝ Q^2 for turbulent flow with constant f). Maximizing P with respect to Q (or h_f) yields the classical result h_f = H/3. Step-by-Step Solution:Let h_f = K Q^2 (turbulent, Darcy–Weisbach with f constant → K lumps geometry and properties).Power at nozzle: P = ρ g Q (H − K Q^2).Differentiate: dP/dQ = ρ g (H − 3K Q^2) = 0 → K Q^2 = H/3 → h_f = K Q^2 = H/3.Thus maximum power occurs when pipe friction loss equals one-third of total head. Verification / Alternative check:Substituting h_f = H/3 gives nozzle head = 2H/3 and shows that one-third of H is spent in friction while two-thirds appears as jet head, a well-known design proportion for penstocks and fire nozzles (neglecting minors). Why Other Options Are Wrong:H/2 or H/4 do not satisfy the optimum condition from differentiation.2H/3 assigns excessive friction, leaving insufficient energy for the jet. Common Pitfalls: Including minor losses without adjusting the optimum; more complete models change the constant slightly but keep the one-third rule as a strong guideline. Maximizing velocity rather than power; the objectives differ. Final Answer:h_f = H/3

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