Q: Ideal (inviscid, incompressible) fluid behavior: Which fundamental relationship does an ideal flow strictly obey in continuum mechanics?

Introduction / Context: An ideal fluid is inviscid (μ = 0) and incompressible. Recognizing which governing equations remain applicable helps in simplifying many hydraulics and aerodynamics problems, particularly when using Bernoulli’s theorem and potential flow theory. Given Data / Assumptions: μ = 0 (no viscous stresses), ρ = constant (incompressible). Continuum hypothesis holds; body forces and pressure act normally. Concept / Approach: All real flows respect conservation laws. For an ideal fluid, the viscous constitutive relation (Newton’s law of viscosity) is irrelevant. However, mass conservation — the continuity equation — must be satisfied in any flow. Momentum conservation (Euler’s equations) also apply, but the question asks what an ideal flow “obeys” distinctively compared with real-fluid constitutive behavior. Step-by-Step Solution: Note that τ = μ dv/dy → 0 for μ = 0; hence Newton’s viscosity law is not the operative relation.Continuity: ∂ρ/∂t + ∇·(ρV) = 0 simplifies to ∇·V = 0 for incompressible ideal flow—this is always valid.Euler/Bernoulli may be derived from momentum conservation under inviscid assumptions, but continuity is universally required. Verification / Alternative check: In potential flow (ideal), velocity field derives from a scalar potential satisfying Laplace’s equation with ∇·V = 0 — a direct expression of continuity. Why Other Options Are Wrong: Newton’s law of viscosity characterizes viscous fluids, not ideal ones. F = m a is true but not a constitutive relation specific to ideal flow; it governs dynamics in general. “Dynamic viscosity law” reiterates viscous behavior. Bernoulli without continuity is incomplete; continuity is essential. Common Pitfalls: Assuming Bernoulli alone characterizes ideal flow; mass conservation still constrains velocity fields and discharge. Final Answer: Continuity equation (mass conservation)

Q: Pipes in series (equivalent single pipe): A compound pipeline made of segments with different lengths and diameters carries the same discharge through each segment. If replaced by one equivalent pipe of diameter D and length L_eq with the same head …

Introduction / Context: Pipelines in series often change diameter due to design constraints. To analyze head losses compactly, engineers replace the compound line with an equivalent single pipe carrying the same discharge and experiencing the same total friction head loss. Given Data / Assumptions: Steady, incompressible flow with uniform roughness and friction factor f taken equal in all segments (or appropriately averaged). Darcy–Weisbach equation for friction losses. Pipes are in series → discharge Q is the same in every segment. Concept / Approach: For a segment i, head loss h_fi = f * (L_i / D_i) * (V_i^2 / (2g)). With common discharge Q, velocity V_i = 4Q/(π D_i^2). Therefore, h_fi ∝ (L_i / D_i^5) * Q^2. Equating the sum of segment losses to the single equivalent pipe loss yields the L_eq formula. Step-by-Step Solution: Write h_f,total = Σ h_fi = f * Q^2/(2g) * Σ(L_i / D_i^5) * (constant factor 32/π^2 absorbed).For the equivalent pipe: h_f,eq = f * (L_eq / D^5) * (Q^2 / (2g)) * same constant.Equate h_f,total = h_f,eq ⇒ L_eq / D^5 = Σ(L_i / D_i^5) ⇒ L_eq = D^5 * Σ(L_i / D_i^5). Verification / Alternative check: Dimensional consistency: L_eq has units of length; D^5 multiplies a sum of L_i / D_i^5 (dimensionless inside except L_i), giving length overall. Why Other Options Are Wrong: Σ(L_i * D_i^5)/D^5 inverts the dependency; larger D_i should reduce loss, not increase it. Linear diameter combinations (Σ L_i / D_i) ignore the D^5 sensitivity from velocity-squared and area effects. Heuristic averages (options d, e) lack the correct fifth-power relation for series flow. Common Pitfalls: Mixing Darcy–Weisbach with Hazen–Williams forms (different exponents). Final Answer: L_eq = D^5 * Σ(L_i / D_i^5)

Q: Wall shear distribution in fully developed laminar pipe flow: For a viscous fluid in a circular pipe, where is the shear stress magnitude maximum across the cross-section?

Introduction / Context: Understanding shear stress distribution is crucial for estimating friction losses and wall stresses in internal flows. In steady, fully developed laminar flow through a circular pipe, velocity is parabolic and shear stress varies linearly from the centre to the wall. Given Data / Assumptions: Newtonian fluid; steady, incompressible, fully developed laminar flow. No slip at the wall; axisymmetric conditions. Concept / Approach: From the momentum balance, shear stress τ(r) varies with radius r as τ(r) = −(dp/dx) * r / 2, where dp/dx is the (negative) pressure gradient. Thus, τ increases linearly with r, reaching zero at the centre (r = 0) and maximum at the pipe wall (r = R). Step-by-Step Solution: Write τ(r) = −(dp/dx) * r / 2.At r = 0 (centre), τ = 0 → no shear at the centre.At r = R (wall), τ_w = −(dp/dx) * R / 2 → maximum magnitude. Verification / Alternative check: The parabolic velocity profile v(r) = v_max * (1 − (r/R)^2) has derivative dv/dr proportional to −r, which is largest in magnitude at the wall, consistent with τ = μ dv/dr being maximum there. Why Other Options Are Wrong: Uniform shear contradicts the linear τ(r) dependence on r. Maximum at centreline is incorrect; τ(0) = 0. Zero at wall (no-slip means zero velocity, not zero shear). Common Pitfalls: Confusing velocity (zero at wall) with shear stress (maximum at wall). Final Answer: Maximum at the inner wall (pipe surface)

Q: Viscous shear in fluids: When two adjacent fluid layers separated by dy have a relative velocity dv, the resulting shear stress τ = μ * (dv/dy). The symbol μ is commonly called what?

Introduction / Context: Newton’s law of viscosity defines the proportionality between shear stress and velocity gradient in a Newtonian fluid. Different textbooks use synonymous terms for μ, which can confuse learners during exams and practice. Given Data / Assumptions: Newtonian fluid behavior: τ = μ * (dv/dy). Small deformations and steady shear. Concept / Approach: The parameter μ has SI units Pa·s (N·s/m^2). It is referred to by several equivalent names: dynamic viscosity, absolute viscosity, coefficient of viscosity, or simply viscosity. The term “kinematic viscosity” (ν = μ/ρ) is different and has units m^2/s. Step-by-Step Solution: Identify the constitutive law: τ = μ * (dv/dy).Recognize μ naming conventions: coefficient of viscosity = absolute viscosity = dynamic viscosity.Distinguish from ν (kinematic viscosity), which is not asked here. Verification / Alternative check: Standards (e.g., SI) give μ in Pa·s; handbooks equate “absolute” and “dynamic” to avoid confusion with kinematic viscosity. Why Other Options Are Wrong: Individually, each of A–D is acceptable; the comprehensive correct choice is “All of the above”. Common Pitfalls: Confusing dynamic viscosity μ with kinematic viscosity ν. Final Answer: All of the above

Q: Water hammer basics: The minimum valve-closing time to qualify as “gradual closure” (so that the full Joukowsky rise does not occur) is expressed in terms of pipe length L and wave speed C as what?

Introduction / Context: Water hammer is a transient pressure surge caused by rapid changes in flow. The characteristic time for a pressure wave to travel the pipe length and return (a round trip) governs whether a closure is considered “sudden” or “gradual” for surge calculations. Given Data / Assumptions: Rigid-pipe approximation with wave speed C. One-dimensional transient behavior; reflection at reservoir/valve ends. Valve located at one end of a line of length L. Concept / Approach: If the valve-closing time t_c is less than 2L/C, the closure is “sudden”, and the Joukowsky equation Δp = ρ C ΔV applies for the maximum surge. If t_c ≥ 2L/C, the closure is “gradual”; the maximum head rise reduces and scales with t_c. Thus, 2L/C is the dividing line between sudden and gradual closures. Step-by-Step Solution: Compute the wave round-trip time: t_round = 2L/C.Define criterion: t_c ≥ t_round ⇒ gradual closure.Therefore, the minimum time for “gradual” classification is t_g = 2L/C. Verification / Alternative check: Method of characteristics solutions and classic water-hammer charts corroborate the 2L/C threshold for end-valve systems. Why Other Options Are Wrong: L/C (one-way travel) underestimates the interaction time. Other multiples (4L/C, L/2C) do not reflect the physics of wave reflection timing. Common Pitfalls: Using local pipe segment length instead of the full length to the reflecting boundary. Final Answer: t_g = 2L / C

Q: Nozzle optimization at end of pipeline: A pipe of diameter D and length L feeds a nozzle of diameter d. Neglecting minor losses, what d maximizes the power carried by the jet (friction factor f by Darcy–Weisbach)?

Introduction / Context: Nozzles at the end of pipelines convert pressure head into kinetic energy. For a given supply head H and line length L with friction factor f, there exists an optimal nozzle size that maximizes jet power delivered to a turbine or impact plate downstream. Given Data / Assumptions: Pipe diameter D, length L; nozzle diameter d at the end. Single straight run; Darcy–Weisbach friction with constant f in the pipe (minor losses neglected). Steady incompressible flow; nozzle discharges to atmosphere at the same elevation. Concept / Approach: Energy: H = hf + v^2/(2g), where v is jet velocity, hf = f * (L/D) * V^2/(2g), and V is pipe velocity. Continuity gives V = v * (d^2 / D^2). Jet power P = ρ g Q * v^2/(2g) = 0.5 * ρ * A_n * v^3 (A_n = π d^2/4). Maximizing P with respect to d (or y = d^2) yields the optimum. Step-by-Step Solution: Express H in terms of v and d: H = v^2/(2g) * [ 1 + (f L / D) * (d^4 / D^4) ].For fixed H, v = sqrt( 2gH / (1 + K d^4) ), where K = f L / D^5.Power P ∝ d^2 * v^3 ∝ d^2 * (1 + K d^4)^(-3/2).Differentiate w.r.t. y = d^2 and set derivative to zero ⇒ 1 − 2 K y^2 = 0.Hence y^2 = 1/(2K) ⇒ d^4 = D^5 / (2 f L). Verification / Alternative check: The equivalent form d/D = ( D / (2 f L) )^(1/4) follows by dividing both sides by D^4. Both representations are standard results. Why Other Options Are Wrong: Forms with inverted ratios (2 f L / D)^(1/4) or d^4 = 2 f L / D^5 reverse the dependence and contradict optimization. Exponents like 1/5 lack derivational basis. Common Pitfalls: Maximizing discharge instead of jet power; these are not identical under frictional losses. Final Answer: d^4 = D^5 / (2 f L)

Q: For fully developed laminar (viscous) flow in a circular pipe, the momentum correction factor β equals:

Introduction / Context: The momentum correction factor β adjusts between the idealized “plug” velocity assumption and the actual velocity distribution when computing momentum flux. In laminar flow through a circular pipe, the parabolic velocity profile makes the momentum flux larger than if the mean velocity acted uniformly; β corrects for this discrepancy. Given Data / Assumptions: Steady, incompressible, laminar flow in a straight circular pipe. Parabolic velocity profile: u(r) = umax * (1 − (r/R)^2). Mean velocity Um = Q / A. Concept / Approach: By definition, β = (∫_A rho * u^2 dA) / (rho * Um^2 * A). For the parabolic profile, integration over the circular cross-section yields β = 4/3 = 1.333… . This exceeds 1 because the squared velocity weights the faster core more heavily than the mean. Step-by-Step Solution: Write u/umax = 1 − (r/R)^2; compute Um = umax/2.Compute ∫ u^2 dA over the circle using polar coordinates.Form β = (∫ u^2 dA) / (Um^2 * A) = ( (πR^2) * (3/8) * u_max^2 ) / ( (u_max/2)^2 * πR^2 ) = 4/3.Conclude β = 1.33 for laminar pipe flow. Verification / Alternative check (if short method exists): Known results: α (energy correction) = 2 and β (momentum correction) = 4/3 for laminar circular pipe; for turbulent, both approach ≈1 due to flatter profiles. Why Other Options Are Wrong: 1.25, 1.50, 1.66, and 2.00 do not match the exact integral result for laminar flow (though α = 2.00 is the energy correction factor, not β). Common Pitfalls (misconceptions, mistakes): Mixing up α and β; assuming β = 1 (valid only for uniform or very flat profiles typical in high-Re turbulent flow). Final Answer: 1.33

Q: Hydraulic power from a nozzle: if the total head at a pipe nozzle is 37.5 m and the discharge is 1 m^3/s, estimate the power output (in horsepower).

Introduction / Context: Power available from a jet/nozzle is the rate at which potential head is converted to kinetic energy. For water systems, hydraulic power can be computed directly from discharge and head, then converted to horsepower (HP) for conventional rating comparisons. Given Data / Assumptions: Total head H = 37.5 m of water at the nozzle. Discharge Q = 1 m^3/s (1 cumec). Water density rho ≈ 1000 kg/m^3; g ≈ 9.81 m/s^2. Neglect losses at the nozzle (ideal conversion). Use 1 HP ≈ 735.5 W (metric horsepower) for comparison in many civil/mech contexts. Concept / Approach: Hydraulic power P equals rho * g * Q * H. This is the mechanical power theoretically obtainable from the head and discharge. Step-by-Step Solution: Compute P_watts = rho * g * Q * H = 1000 * 9.81 * 1 * 37.5 = 367,875 W.Convert to HP (metric): HP = 367,875 / 735.5 ≈ 500.5 HP.Round sensibly to the nearest option: ≈ 500 H.P. Verification / Alternative check (if short method exists): Using 1 HP (imperial) = 746 W gives 367,875/746 ≈ 493 HP, still closest to 500 H.P. among the given choices. Why Other Options Are Wrong: 400 and 450 H.P. underestimate the computed power; 550 H.P. overestimates. 500 H.P. is the nearest correct estimate with standard conversions. Common Pitfalls (misconceptions, mistakes): Mixing metric and imperial horsepower without noting small differences; omitting density or g; forgetting that losses reduce usable power (here neglected by assumption). Final Answer: 500 H.P.

Q: Ship rolling: a floating ship has radius of gyration about the waterline k = 4 m and metacentric height GM = 72.5 cm. Its small-angle rolling period is closest to which value?

Introduction / Context: For small-amplitude rolling of a floating body, the restoring moment is proportional to the metacentric height GM, yielding a simple harmonic oscillator. The period depends on the radius of gyration about the waterline and the hydrostatic stiffness m * g * GM. Given Data / Assumptions: Radius of gyration k = 4 m. Metacentric height GM = 72.5 cm = 0.725 m. Small-angle (linear) oscillations; g ≈ 9.81 m/s^2. Ship mass cancels from the period expression. Concept / Approach: The roll period T is T = 2 * π * sqrt( I / (m * g * GM) ), where I = m * k^2. Cancelling m gives T = 2 * π * sqrt( k^2 / (g * GM) ). Step-by-Step Solution: Compute denominator g * GM = 9.81 * 0.725 = 7.1145.Compute k^2 = 4^2 = 16.Form the ratio: 16 / 7.1145 ≈ 2.249.Take square root: sqrt(2.249) ≈ 1.5.Compute T = 2 * π * 1.5 = 3 * π. Verification / Alternative check (if short method exists): Order-of-magnitude check: larger GM shortens period; here GM ~ 0.7 m reasonable for a mid-sized vessel, giving a few π seconds—consistent with 3π. Why Other Options Are Wrong: π and 2π are too short (would require much larger GM); 4π is too long. π/2 is unphysical for a ship of this size. Common Pitfalls (misconceptions, mistakes): Forgetting to convert GM to metres; using diameter instead of radius of gyration; omitting the 2π factor typical of SHM periods. Final Answer: 3π

Question 1

Sharp-crested weirs: To ensure a clean, fully aerated nappe for accurate discharge measurement, the crest (plate) thickness is generally kept less than what fraction of the head over the sill?

Accepted Answer

Introduction / Context: Sharp-crested (thin-plate) weirs require a thin knife-edge crest so that the nappe springs clear and aerates, matching standard discharge calibrations. Excessive crest thickness distorts the flow contraction and invalidates empirical coefficients. Given Data / Assumptions: Sharp upstream edge maintained; crest not rounded. Free overfall with ventilated nappe below the sheet. Head over crest within typical laboratory calibration range. Concept / Approach: Guidelines recommend the crest thickness be small compared to the head on the sill (H). A widely accepted rule of thumb is t < H/3, which preserves the thin-plate behavior and minimizes interference with the contracted section at the vena contracta. Step-by-Step Solution: Identify the governing principle: maintain a true sharp edge with negligible thickness effect.Apply the guideline: crest thickness less than one-third of the head H.Ensure aeration of the nappe to avoid submergence corrections. Verification / Alternative check: Standard references (laboratory practice manuals) specify similar limits; many also recommend that upstream face be vertical and downstream edge beveled to sustain a clean separation. Why Other Options Are Wrong: One-half or two-thirds of H are too thick and disrupt sharp-crested behavior. One-fourth of H is conservative but not the commonly cited threshold (question asks the general limit). “None of these” is incorrect because the H/3 guideline is standard. Common Pitfalls: Allowing burrs or rounding at the crest, which alters discharge coefficient more than thickness alone. Final Answer: One-third of the head on the sill

Question 2

Ideal (inviscid, incompressible) fluid behavior: Which fundamental relationship does an ideal flow strictly obey in continuum mechanics?

Accepted Answer

Introduction / Context: An ideal fluid is inviscid (μ = 0) and incompressible. Recognizing which governing equations remain applicable helps in simplifying many hydraulics and aerodynamics problems, particularly when using Bernoulli’s theorem and potential flow theory. Given Data / Assumptions: μ = 0 (no viscous stresses), ρ = constant (incompressible). Continuum hypothesis holds; body forces and pressure act normally. Concept / Approach: All real flows respect conservation laws. For an ideal fluid, the viscous constitutive relation (Newton’s law of viscosity) is irrelevant. However, mass conservation — the continuity equation — must be satisfied in any flow. Momentum conservation (Euler’s equations) also apply, but the question asks what an ideal flow “obeys” distinctively compared with real-fluid constitutive behavior. Step-by-Step Solution: Note that τ = μ dv/dy → 0 for μ = 0; hence Newton’s viscosity law is not the operative relation.Continuity: ∂ρ/∂t + ∇·(ρV) = 0 simplifies to ∇·V = 0 for incompressible ideal flow—this is always valid.Euler/Bernoulli may be derived from momentum conservation under inviscid assumptions, but continuity is universally required. Verification / Alternative check: In potential flow (ideal), velocity field derives from a scalar potential satisfying Laplace’s equation with ∇·V = 0 — a direct expression of continuity. Why Other Options Are Wrong: Newton’s law of viscosity characterizes viscous fluids, not ideal ones. F = m a is true but not a constitutive relation specific to ideal flow; it governs dynamics in general. “Dynamic viscosity law” reiterates viscous behavior. Bernoulli without continuity is incomplete; continuity is essential. Common Pitfalls: Assuming Bernoulli alone characterizes ideal flow; mass conservation still constrains velocity fields and discharge. Final Answer: Continuity equation (mass conservation)

Question 3

Pipes in series (equivalent single pipe): A compound pipeline made of segments with different lengths and diameters carries the same discharge through each segment. If replaced by one equivalent pipe of diameter D and length L_eq with the same head …

Accepted Answer

Introduction / Context: Pipelines in series often change diameter due to design constraints. To analyze head losses compactly, engineers replace the compound line with an equivalent single pipe carrying the same discharge and experiencing the same total friction head loss. Given Data / Assumptions: Steady, incompressible flow with uniform roughness and friction factor f taken equal in all segments (or appropriately averaged). Darcy–Weisbach equation for friction losses. Pipes are in series → discharge Q is the same in every segment. Concept / Approach: For a segment i, head loss h_fi = f * (L_i / D_i) * (V_i^2 / (2g)). With common discharge Q, velocity V_i = 4Q/(π D_i^2). Therefore, h_fi ∝ (L_i / D_i^5) * Q^2. Equating the sum of segment losses to the single equivalent pipe loss yields the L_eq formula. Step-by-Step Solution: Write h_f,total = Σ h_fi = f * Q^2/(2g) * Σ(L_i / D_i^5) * (constant factor 32/π^2 absorbed).For the equivalent pipe: h_f,eq = f * (L_eq / D^5) * (Q^2 / (2g)) * same constant.Equate h_f,total = h_f,eq ⇒ L_eq / D^5 = Σ(L_i / D_i^5) ⇒ L_eq = D^5 * Σ(L_i / D_i^5). Verification / Alternative check: Dimensional consistency: L_eq has units of length; D^5 multiplies a sum of L_i / D_i^5 (dimensionless inside except L_i), giving length overall. Why Other Options Are Wrong: Σ(L_i * D_i^5)/D^5 inverts the dependency; larger D_i should reduce loss, not increase it. Linear diameter combinations (Σ L_i / D_i) ignore the D^5 sensitivity from velocity-squared and area effects. Heuristic averages (options d, e) lack the correct fifth-power relation for series flow. Common Pitfalls: Mixing Darcy–Weisbach with Hazen–Williams forms (different exponents). Final Answer: L_eq = D^5 * Σ(L_i / D_i^5)

Question 4

Wall shear distribution in fully developed laminar pipe flow: For a viscous fluid in a circular pipe, where is the shear stress magnitude maximum across the cross-section?

Accepted Answer

Introduction / Context: Understanding shear stress distribution is crucial for estimating friction losses and wall stresses in internal flows. In steady, fully developed laminar flow through a circular pipe, velocity is parabolic and shear stress varies linearly from the centre to the wall. Given Data / Assumptions: Newtonian fluid; steady, incompressible, fully developed laminar flow. No slip at the wall; axisymmetric conditions. Concept / Approach: From the momentum balance, shear stress τ(r) varies with radius r as τ(r) = −(dp/dx) * r / 2, where dp/dx is the (negative) pressure gradient. Thus, τ increases linearly with r, reaching zero at the centre (r = 0) and maximum at the pipe wall (r = R). Step-by-Step Solution: Write τ(r) = −(dp/dx) * r / 2.At r = 0 (centre), τ = 0 → no shear at the centre.At r = R (wall), τ_w = −(dp/dx) * R / 2 → maximum magnitude. Verification / Alternative check: The parabolic velocity profile v(r) = v_max * (1 − (r/R)^2) has derivative dv/dr proportional to −r, which is largest in magnitude at the wall, consistent with τ = μ dv/dr being maximum there. Why Other Options Are Wrong: Uniform shear contradicts the linear τ(r) dependence on r. Maximum at centreline is incorrect; τ(0) = 0. Zero at wall (no-slip means zero velocity, not zero shear). Common Pitfalls: Confusing velocity (zero at wall) with shear stress (maximum at wall). Final Answer: Maximum at the inner wall (pipe surface)

Question 5

Viscous shear in fluids: When two adjacent fluid layers separated by dy have a relative velocity dv, the resulting shear stress τ = μ * (dv/dy). The symbol μ is commonly called what?

Accepted Answer

Introduction / Context: Newton’s law of viscosity defines the proportionality between shear stress and velocity gradient in a Newtonian fluid. Different textbooks use synonymous terms for μ, which can confuse learners during exams and practice. Given Data / Assumptions: Newtonian fluid behavior: τ = μ * (dv/dy). Small deformations and steady shear. Concept / Approach: The parameter μ has SI units Pa·s (N·s/m^2). It is referred to by several equivalent names: dynamic viscosity, absolute viscosity, coefficient of viscosity, or simply viscosity. The term “kinematic viscosity” (ν = μ/ρ) is different and has units m^2/s. Step-by-Step Solution: Identify the constitutive law: τ = μ * (dv/dy).Recognize μ naming conventions: coefficient of viscosity = absolute viscosity = dynamic viscosity.Distinguish from ν (kinematic viscosity), which is not asked here. Verification / Alternative check: Standards (e.g., SI) give μ in Pa·s; handbooks equate “absolute” and “dynamic” to avoid confusion with kinematic viscosity. Why Other Options Are Wrong: Individually, each of A–D is acceptable; the comprehensive correct choice is “All of the above”. Common Pitfalls: Confusing dynamic viscosity μ with kinematic viscosity ν. Final Answer: All of the above

Question 6

Water hammer basics: The minimum valve-closing time to qualify as “gradual closure” (so that the full Joukowsky rise does not occur) is expressed in terms of pipe length L and wave speed C as what?

Accepted Answer

Introduction / Context: Water hammer is a transient pressure surge caused by rapid changes in flow. The characteristic time for a pressure wave to travel the pipe length and return (a round trip) governs whether a closure is considered “sudden” or “gradual” for surge calculations. Given Data / Assumptions: Rigid-pipe approximation with wave speed C. One-dimensional transient behavior; reflection at reservoir/valve ends. Valve located at one end of a line of length L. Concept / Approach: If the valve-closing time t_c is less than 2L/C, the closure is “sudden”, and the Joukowsky equation Δp = ρ C ΔV applies for the maximum surge. If t_c ≥ 2L/C, the closure is “gradual”; the maximum head rise reduces and scales with t_c. Thus, 2L/C is the dividing line between sudden and gradual closures. Step-by-Step Solution: Compute the wave round-trip time: t_round = 2L/C.Define criterion: t_c ≥ t_round ⇒ gradual closure.Therefore, the minimum time for “gradual” classification is t_g = 2L/C. Verification / Alternative check: Method of characteristics solutions and classic water-hammer charts corroborate the 2L/C threshold for end-valve systems. Why Other Options Are Wrong: L/C (one-way travel) underestimates the interaction time. Other multiples (4L/C, L/2C) do not reflect the physics of wave reflection timing. Common Pitfalls: Using local pipe segment length instead of the full length to the reflecting boundary. Final Answer: t_g = 2L / C

Question 7

Nozzle optimization at end of pipeline: A pipe of diameter D and length L feeds a nozzle of diameter d. Neglecting minor losses, what d maximizes the power carried by the jet (friction factor f by Darcy–Weisbach)?

Accepted Answer

Introduction / Context: Nozzles at the end of pipelines convert pressure head into kinetic energy. For a given supply head H and line length L with friction factor f, there exists an optimal nozzle size that maximizes jet power delivered to a turbine or impact plate downstream. Given Data / Assumptions: Pipe diameter D, length L; nozzle diameter d at the end. Single straight run; Darcy–Weisbach friction with constant f in the pipe (minor losses neglected). Steady incompressible flow; nozzle discharges to atmosphere at the same elevation. Concept / Approach: Energy: H = hf + v^2/(2g), where v is jet velocity, hf = f * (L/D) * V^2/(2g), and V is pipe velocity. Continuity gives V = v * (d^2 / D^2). Jet power P = ρ g Q * v^2/(2g) = 0.5 * ρ * A_n * v^3 (A_n = π d^2/4). Maximizing P with respect to d (or y = d^2) yields the optimum. Step-by-Step Solution: Express H in terms of v and d: H = v^2/(2g) * [ 1 + (f L / D) * (d^4 / D^4) ].For fixed H, v = sqrt( 2gH / (1 + K d^4) ), where K = f L / D^5.Power P ∝ d^2 * v^3 ∝ d^2 * (1 + K d^4)^(-3/2).Differentiate w.r.t. y = d^2 and set derivative to zero ⇒ 1 − 2 K y^2 = 0.Hence y^2 = 1/(2K) ⇒ d^4 = D^5 / (2 f L). Verification / Alternative check: The equivalent form d/D = ( D / (2 f L) )^(1/4) follows by dividing both sides by D^4. Both representations are standard results. Why Other Options Are Wrong: Forms with inverted ratios (2 f L / D)^(1/4) or d^4 = 2 f L / D^5 reverse the dependence and contradict optimization. Exponents like 1/5 lack derivational basis. Common Pitfalls: Maximizing discharge instead of jet power; these are not identical under frictional losses. Final Answer: d^4 = D^5 / (2 f L)

Question 8

In fluid mechanics, the Reynolds number (Re) is defined as the ratio of inertial force to which competing effect?

Accepted Answer

Introduction / Context: Reynolds number (Re) is a cornerstone dimensionless parameter in fluid mechanics. It compares the tendency of a flow's inertia to keep it moving in a straight, mixed state versus the tendency of viscous effects to diffuse momentum and smooth out velocity gradients. Recognizing the correct physical ratio behind Re is essential for predicting laminar–turbulent transitions and for similarity analysis in model testing. Given Data / Assumptions: Characteristic length L and characteristic velocity V are representative of the flow. Fluid density rho and dynamic viscosity mu are constants (for the considered conditions). We are identifying which competing effect appears in the denominator of Re. Concept / Approach: Inertial force scale is proportional to rho * V^2 * (area), whereas viscous force scale is proportional to mu * V * (area / length). Taking their ratio eliminates geometric areas and yields Re = (rho * V * L) / mu. Thus, Reynolds number is the ratio of inertial to viscous effects. It is not a ratio involving gravity, surface tension, elasticity, or compressibility—those appear in other dimensionless groups (e.g., Froude, Weber, Mach numbers). Step-by-Step Solution: Write inertial scale ~ rho * V^2.Write viscous scale ~ mu * V / L.Form ratio: (rho * V^2) / (mu * V / L) = (rho * V * L) / mu.Conclude: denominator corresponds to viscous effects (viscosity). Verification / Alternative check (if short method exists): Dimensional check: (rho * V * L) / mu is dimensionless. Empirically, low Re flows are laminar (viscous dominated), high Re are turbulent (inertial dominated), matching the interpretation. Why Other Options Are Wrong: Surface tension relates to Weber number, gravity to Froude number, elasticity/compressibility to Mach or elastic numbers—not to Reynolds number. Common Pitfalls (misconceptions, mistakes): Confusing Reynolds number with Froude or Weber numbers, or thinking Re compares pressure to viscosity instead of inertia to viscosity. Final Answer: viscosity

Question 9

For fully developed laminar (viscous) flow in a circular pipe, the momentum correction factor β equals:

Accepted Answer

Introduction / Context: The momentum correction factor β adjusts between the idealized “plug” velocity assumption and the actual velocity distribution when computing momentum flux. In laminar flow through a circular pipe, the parabolic velocity profile makes the momentum flux larger than if the mean velocity acted uniformly; β corrects for this discrepancy. Given Data / Assumptions: Steady, incompressible, laminar flow in a straight circular pipe. Parabolic velocity profile: u(r) = umax * (1 − (r/R)^2). Mean velocity Um = Q / A. Concept / Approach: By definition, β = (∫_A rho * u^2 dA) / (rho * Um^2 * A). For the parabolic profile, integration over the circular cross-section yields β = 4/3 = 1.333… . This exceeds 1 because the squared velocity weights the faster core more heavily than the mean. Step-by-Step Solution: Write u/umax = 1 − (r/R)^2; compute Um = umax/2.Compute ∫ u^2 dA over the circle using polar coordinates.Form β = (∫ u^2 dA) / (Um^2 * A) = ( (πR^2) * (3/8) * u_max^2 ) / ( (u_max/2)^2 * πR^2 ) = 4/3.Conclude β = 1.33 for laminar pipe flow. Verification / Alternative check (if short method exists): Known results: α (energy correction) = 2 and β (momentum correction) = 4/3 for laminar circular pipe; for turbulent, both approach ≈1 due to flatter profiles. Why Other Options Are Wrong: 1.25, 1.50, 1.66, and 2.00 do not match the exact integral result for laminar flow (though α = 2.00 is the energy correction factor, not β). Common Pitfalls (misconceptions, mistakes): Mixing up α and β; assuming β = 1 (valid only for uniform or very flat profiles typical in high-Re turbulent flow). Final Answer: 1.33

Question 10

Hydraulic power from a nozzle: if the total head at a pipe nozzle is 37.5 m and the discharge is 1 m^3/s, estimate the power output (in horsepower).

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Introduction / Context: Power available from a jet/nozzle is the rate at which potential head is converted to kinetic energy. For water systems, hydraulic power can be computed directly from discharge and head, then converted to horsepower (HP) for conventional rating comparisons. Given Data / Assumptions: Total head H = 37.5 m of water at the nozzle. Discharge Q = 1 m^3/s (1 cumec). Water density rho ≈ 1000 kg/m^3; g ≈ 9.81 m/s^2. Neglect losses at the nozzle (ideal conversion). Use 1 HP ≈ 735.5 W (metric horsepower) for comparison in many civil/mech contexts. Concept / Approach: Hydraulic power P equals rho * g * Q * H. This is the mechanical power theoretically obtainable from the head and discharge. Step-by-Step Solution: Compute P_watts = rho * g * Q * H = 1000 * 9.81 * 1 * 37.5 = 367,875 W.Convert to HP (metric): HP = 367,875 / 735.5 ≈ 500.5 HP.Round sensibly to the nearest option: ≈ 500 H.P. Verification / Alternative check (if short method exists): Using 1 HP (imperial) = 746 W gives 367,875/746 ≈ 493 HP, still closest to 500 H.P. among the given choices. Why Other Options Are Wrong: 400 and 450 H.P. underestimate the computed power; 550 H.P. overestimates. 500 H.P. is the nearest correct estimate with standard conversions. Common Pitfalls (misconceptions, mistakes): Mixing metric and imperial horsepower without noting small differences; omitting density or g; forgetting that losses reduce usable power (here neglected by assumption). Final Answer: 500 H.P.

Question 11

Ship rolling: a floating ship has radius of gyration about the waterline k = 4 m and metacentric height GM = 72.5 cm. Its small-angle rolling period is closest to which value?

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Introduction / Context: For small-amplitude rolling of a floating body, the restoring moment is proportional to the metacentric height GM, yielding a simple harmonic oscillator. The period depends on the radius of gyration about the waterline and the hydrostatic stiffness m * g * GM. Given Data / Assumptions: Radius of gyration k = 4 m. Metacentric height GM = 72.5 cm = 0.725 m. Small-angle (linear) oscillations; g ≈ 9.81 m/s^2. Ship mass cancels from the period expression. Concept / Approach: The roll period T is T = 2 * π * sqrt( I / (m * g * GM) ), where I = m * k^2. Cancelling m gives T = 2 * π * sqrt( k^2 / (g * GM) ). Step-by-Step Solution: Compute denominator g * GM = 9.81 * 0.725 = 7.1145.Compute k^2 = 4^2 = 16.Form the ratio: 16 / 7.1145 ≈ 2.249.Take square root: sqrt(2.249) ≈ 1.5.Compute T = 2 * π * 1.5 = 3 * π. Verification / Alternative check (if short method exists): Order-of-magnitude check: larger GM shortens period; here GM ~ 0.7 m reasonable for a mid-sized vessel, giving a few π seconds—consistent with 3π. Why Other Options Are Wrong: π and 2π are too short (would require much larger GM); 4π is too long. π/2 is unphysical for a ship of this size. Common Pitfalls (misconceptions, mistakes): Forgetting to convert GM to metres; using diameter instead of radius of gyration; omitting the 2π factor typical of SHM periods. Final Answer: 3π

Question 12

Power transmission with an end nozzle: for a pipe of diameter D, length L, and Darcy–Weisbach friction factor f, the optimum nozzle diameter d (for maximum power at the jet) is:

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Introduction / Context: When a long pipeline ends in a nozzle driving a jet (e.g., impulse turbines, firefighting), the nozzle size controls both discharge and head lost to pipe friction. There exists an optimum nozzle diameter that maximizes power delivered by the jet. Given Data / Assumptions: Total available head at pipe entrance is H (symbolic). Pipe: length L, diameter D, friction factor f (Darcy–Weisbach). Nozzle of diameter d discharging to atmosphere; single, smooth, frictionless nozzle. Concept / Approach: Head loss in pipe: h_f = 4 * f * (L / D) * V^2 / (2 * g), where V is mean pipe velocity. Jet velocity v_j = sqrt( 2 * g * (H − h_f) ). Continuity gives V = v_j * (d^2 / D^2). Jet power P = (rho / 2) * A_n * v_j^3. Maximizing P with respect to d yields the classical condition (4 * f * L / D) * (d^4 / D^4) = 1/2, which rearranges to the formula below. Step-by-Step Solution: Use V = v_j * (d^2 / D^2) to write h_f in terms of v_j and d.Express P in terms of d via v_j and differentiate dP/dd = 0.Obtain optimality: (4 f L / D) * (d^4 / D^4) = 1/2.Solve for d: d = D * ( D / (8 f L) )^(1/4). Verification / Alternative check (if short method exists): Dimensional check confirms d has dimension of length; the ratio inside the fourth-root is dimensionless. Why Other Options Are Wrong: Options (b)-(e) contradict the fourth-root dependence or invert the friction/length effect. Larger f or L should reduce optimum d, which (a) captures. Common Pitfalls (misconceptions, mistakes): Using the Hazen–Williams form directly in optimization; forgetting that nozzle losses are neglected in this ideal result. Final Answer: d = D * ( D / (8 * f * L) )^(1/4)

Question 13

Which form of mechanical energy is not possessed by an independent (unconfined) mass of fluid?

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Introduction / Context: Mechanical energy of a fluid is commonly partitioned into pressure energy, kinetic energy, and potential (elevation) energy. Correctly identifying when each form exists prevents misuse of Bernoulli’s equation and clarifies where work interactions originate. Given Data / Assumptions: Consider a finite “packet” of fluid moving freely in open space (no container walls, no piston). External fields like gravity may be present; velocity may be nonzero. We distinguish between intrinsic energy stored due to compression at pressure p and the ability to do flow work on boundaries. Concept / Approach: Pressure energy (or flow work per unit weight p/γ) is defined via the work needed to push a fluid element into or out of a control volume against a surrounding pressure field—work that occurs at boundaries. An isolated “independent” mass of fluid, with no surrounding confining pressure and no boundary on which to do flow work, does not possess pressure energy in the Bernoulli sense. However, the fluid can still have kinetic energy (1/2 * v^2 per unit mass) if it is moving, and potential energy (g * z per unit mass) in a gravitational field. Step-by-Step Solution: Identify energy forms: pressure, kinetic, and elevation (potential).Note that pressure energy requires a boundary to perform flow work.An independent, unconfined fluid mass lacks such a boundary → no pressure energy.Kinetic and potential energies are still well-defined for free-flight motion and gravity fields. Verification / Alternative check (if short method exists): Bernoulli along a streamline arises from Euler’s equation and includes p/ρ as “flow work.” Remove boundaries and p/ρ loses mechanical meaning, leaving only v^2/2 and g z terms. Why Other Options Are Wrong: Kinetic and potential energies are properties of the state and position/motion; they do not require confinement. “None” and “all” contradict the definitions above. Common Pitfalls (misconceptions, mistakes): Assuming “pressure energy” exists even without a boundary; confusing thermodynamic internal energy with mechanical pressure energy used in flow work. Final Answer: pressure energy

Question 14

Extended Bernoulli application: when work is done on the fluid between two sections (e.g., a pump adds head), this term should be...

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Introduction / Context: Real systems often include devices that add or remove energy (pumps, turbines). The extended Bernoulli equation accounts for these by adding a pump head h_p (work done on the fluid) or subtracting a turbine head h_t (work extracted from the fluid) between two sections. Given Data / Assumptions: Steady, incompressible flow. Sections 1 (upstream) and 2 (downstream) connected through a system with possible head losses h_f. Sign convention: head added to the fluid is positive h_p; head extracted is h_t. Concept / Approach: A common, consistent form is: p1/γ + v1^2/(2g) + z1 + h_p − h_t − h_f = p2/γ + v2^2/(2g) + z2. Thus, “work done on the flow system” (pump head) is added to the left-hand side (together with the upstream energy terms) to balance the downstream head. Equivalent rearrangements are possible, but the standard teaching convention places h_p on the upstream (left) side as an addition. Step-by-Step Solution: Write the ideal Bernoulli between 1 and 2.Insert head terms: +h_p for a pump, −h_t for a turbine, and −h_f for losses.Group them with the upstream side to yield the canonical textbook form.Solve design/analysis problems accordingly. Verification / Alternative check (if short method exists): Moving h_p to the right-hand side yields identical results; however, the pedagogical standard is the LHS addition convention used above. Why Other Options Are Wrong: (b) Uses a different convention than the one requested; (c) ignores essential energy transfer; (d) is nonsensical for balancing; (e) is unnecessary. Common Pitfalls (misconceptions, mistakes): Forgetting sign conventions; mixing pump head with turbine head; neglecting head losses in long pipelines. Final Answer: added on the left side of the energy equation with the upstream section

Question 15

Critical depth in open-channel flow (rectangular channel): which expression gives the critical depth h_c in terms of discharge per unit width q?

Accepted Answer

Introduction / Context: Critical depth is the depth at which the specific energy is minimum for a given discharge in an open channel. For a rectangular channel, it provides a convenient criterion for transitions between subcritical (tranquil) and supercritical (torrential) flow and is widely used in the design of flumes and spillways. Given Data / Assumptions: Rectangular channel of unit width; discharge per unit width q = Q / b. Gravitational acceleration g constant. Frictional losses neglected in the local specific energy analysis. Concept / Approach: Specific energy E for a rectangular channel is E = h + q^2/(2 * g * h^2). The critical depth h_c occurs when dE/dh = 0, giving the condition Fr = 1 (q^2 = g * h_c^3), from which h_c is obtained. Step-by-Step Solution: Write E(h) = h + q^2 / (2 g h^2).Differentiate and set dE/dh = 0 → 1 − q^2 / (g h^3) = 0.Solve for h: g * h^3 = q^2.Hence h_c = ( q^2 / g )^(1/3). Verification / Alternative check (if short method exists): Use Froude number for rectangular section: Fr = v / sqrt(g * h) with v = q / h. Setting Fr = 1 → (q / h) = sqrt(g * h) → q^2 = g * h^3, identical result. Why Other Options Are Wrong: They invert or misapply powers of q and g, or confuse dimensional consistency (depth must scale as (q^2/g)^(1/3)). Common Pitfalls (misconceptions, mistakes): Forgetting that q is discharge per unit width, not total discharge; misusing Froude number definition. Final Answer: h_c = ( q^2 / g )^(1/3)

Question 16

Rectangular weir discharge equation (neglecting end contractions): choose the correct expression for Q as a function of crest length L and head H.

Accepted Answer

Introduction / Context: Sharp-crested (rectangular) weirs are standard flow-measurement devices in open channels. The discharge depends on crest length L and head H above the crest, with a coefficient of discharge C_d accounting for contraction and viscosity effects. Given Data / Assumptions: Sharp-crested rectangular weir, fully contracted effects embedded in C_d. Neglect velocity of approach or include it within an adjusted head if necessary (here neglected). Hydrostatic pressure distribution at the crest; inviscid, steady flow assumption near crest. Concept / Approach: Integrate the elementary discharge over the depth from y = 0 to y = H: dQ = C_d * L * v(y) * dy with v(y) = sqrt(2 * g * (H − y)). The integral yields the standard result. Step-by-Step Solution: Write dQ = C_d * L * sqrt(2 g (H − y)) dy.Integrate from 0 to H to get Q = (2/3) * C_d * L * sqrt(2 g) * H^(3/2).Identify the exponent 3/2 on H as characteristic of sharp crested weirs. Verification / Alternative check (if short method exists): Dimensional analysis: L * (g)^(1/2) * H^(3/2) gives L^3/T, the correct dimension of discharge; C_d is dimensionless. Why Other Options Are Wrong: (b) and (e) have incorrect exponents on H; (c) uses 1/2 instead of 3/2; (d) corresponds to a different profile (e.g., triangular/V-notch formula has 5/2 with tan terms), not a rectangular weir. Common Pitfalls (misconceptions, mistakes): Forgetting velocity of approach corrections; confusing rectangular and V-notch (triangular) weir formulas. Final Answer: Q = (2/3) * C_d * L * sqrt(2 * g) * H^(3/2)

Question 17

Identify the highest dam in India from the given options (by structural height).

Accepted Answer

Introduction / Context: India has several large dams serving irrigation, hydropower, and flood control. Among widely cited structures, Bhakra (Bhakra–Nangal) is notable for its great height, while Hirakud is famous for its enormous length and reservoir capacity. Given Data / Assumptions: Compare structural heights among the listed dams. Standard indicative heights: Bhakra ~226 m; Idukki ~169 m; Nagarjuna Sagar ~124 m; Hirakud ~60 m (approximate structural height; it is the longest). Minor variations by source do not change the ranking among these options. Concept / Approach: Determine which dam has the maximum height from riverbed/foundation to crest among the given choices. Despite Hirakud’s length, its height is lower than Bhakra’s. Idukki and Nagarjuna Sagar are also lower than Bhakra by several tens of metres. Step-by-Step Solution: List the typical heights for each option.Order them descending: Bhakra (highest) > Idukki > Nagarjuna Sagar > Hirakud.Select Bhakra dam as the highest among the options. Verification / Alternative check (if short method exists): Cross-reference standard engineering handbooks or government sources; rankings remain consistent. Why Other Options Are Wrong: Hirakud’s fame is for length, not height; Idukki and Nagarjuna Sagar are tall but not as tall as Bhakra among the provided choices. Common Pitfalls (misconceptions, mistakes): Confusing “largest” (by volume/length/capacity) with “highest”; overlooking that Tehri (taller) is not among the options. Final Answer: Bhakra dam

Question 18

Archimedes’ principle in fluid mechanics — when a body is fully or partially immersed in a fluid, the buoyant (upthrust) force acting on it is equal to which quantity?

Accepted Answer

Introduction / Context: Archimedes’ principle is a foundational concept in fluid mechanics and naval architecture. It explains why objects float or sink in liquids and gases and quantifies the upward buoyant force that a fluid exerts on an immersed body. Understanding this principle is essential for designing ships, hydrometers, and many process equipment items such as separators and level instruments. Given Data / Assumptions: The body may be totally or partially immersed in a static fluid. The fluid is continuous and at rest (hydrostatics). Pressure increases with depth according to p = p0 + rho * g * h. Gravity acts uniformly with acceleration g. Concept / Approach: The buoyant force arises from the hydrostatic pressure distribution on the body’s surface. Pressure is higher at deeper points, producing a net upward resultant. Mathematically, integrating the pressure over the wetted surface is equivalent to the weight of the displaced fluid volume. Therefore, the magnitude of upthrust depends only on the displaced fluid, not directly on the body’s own weight. Step-by-Step Solution: Identify displaced volume V_disp equal to the submerged volume of the body.Compute displaced fluid weight: W_fluid = rho_fluid * g * V_disp.Recognize that the net hydrostatic resultant (buoyancy) equals W_fluid.Compare W_body with W_fluid to infer floating (W_body <= W_fluid) or sinking (W_body > W_fluid). Verification / Alternative check (if short method exists): A force balance on a floating body at equilibrium gives upthrust = weight. Measuring displaced fluid (e.g., via overflow) directly confirms buoyant force equals the displaced fluid’s weight. Why Other Options Are Wrong: “Weight of the body” confuses buoyancy with gravity; “weight of the body and fluid displaced” double counts; “difference of weights …” is not the definition of buoyancy (though net force uses a difference); “none” contradicts Archimedes’ principle. Common Pitfalls (misconceptions, mistakes): Assuming buoyant force depends on the body’s density directly; forgetting that compressible fluids (gases) also exert buoyancy; neglecting that partial immersion displaces less fluid and thus reduces upthrust. Final Answer: weight of the fluid displaced by the body

Question 19

Unsteady drainage — time required to empty a hemispherical vessel through a bottom orifice (question text lacks complete parameters and is therefore incomplete).

Accepted Answer

None of these

Question 20

Hydrostatics on curved surfaces — the horizontal component of hydrostatic force on a curved surface equals which equivalent force?

Accepted Answer

Introduction / Context: Curved submerged surfaces (e.g., gates, spillway profiles, elbows) experience hydrostatic forces with distinct horizontal and vertical components. Engineers resolve these components to design supports and anchors correctly. Given Data / Assumptions: Static fluid; density rho and gravity g are constant. Curved surface submerged with a well-defined vertical projection area A_v. Hydrostatic pressure varies linearly with depth. Concept / Approach: The horizontal component on a curved surface equals the hydrostatic force on its vertical projection, acting through the center of pressure of that projection. The vertical component equals the weight of the imaginary fluid above the surface up to the free surface (upthrust concept). Step-by-Step Solution: Project the curved area onto a vertical plane → area A_v.Compute F_h as for a plane surface: F_h = rho * g * y_c * A_v, where y_c is centroid depth of the vertical projection.Locate the line of action at the center of pressure of the projection.Combine with vertical component (if needed) via vector addition to get the resultant. Verification / Alternative check (if short method exists): Integrating pressure components over the curved surface and comparing with the vertical projection yields identical F_h; this is a classic hydrostatics result. Why Other Options Are Wrong: “Weight of liquid vertically below …” pertains to vertical component, not horizontal; “product of pressure at centroid and area” ignores depth variation; “weight of liquid retained …” again is vertical component logic; “none” is incorrect. Common Pitfalls (misconceptions, mistakes): Confusing horizontal and vertical component formulae; using pressure at centroid instead of integrating for plane/curved differences. Final Answer: force on a vertical projection of the curved surface

Hydraulics Questions