Q: Short-tube mouthpiece behavior — a short tube mouthpiece will NOT run full at its outlet under which head condition?

Introduction / Context: A short-tube (Borda-type) mouthpiece may discharge with a separated jet that does not fill the tube (“running free”) or with the tube running completely full. Whether it runs full depends on the available head and the ability of ambient pressure to sustain the separated core. Given Data / Assumptions: Sharp-edged short tube attached to a tank/orifice. Atmospheric pressure at the outlet; negligible losses aside from separation. Head measured above the centerline of the mouthpiece. Concept / Approach: For heads below a critical value (~12.2 m of water), the separated jet persists without reattaching, and the mouthpiece does not run full. For heads exceeding this, pressure recovery can cause the jet to fill the tube (“running full”). Thus, “will not run full” corresponds to heads less than about 12.2 m of water. Step-by-Step Solution: Compare operating head H to the critical value ~12.2 m.If H 12.2 m → tube tends to run full at outlet. Verification / Alternative check (if short method exists): Empirical data for Borda mouthpieces report Cc ≈ 0.5 (free) and Cc ≈ 1 (full) with the transition near 12.2 m head. Why Other Options Are Wrong: “More than” reverses the condition; “equal” is a threshold case not aligned with the “will not” phrasing; “none” and “1 m” do not reflect the known criterion. Common Pitfalls (misconceptions, mistakes): Confusing Borda mouthpiece behavior with long pipes; ignoring the role of separation and reattachment. Final Answer: less than 12.2 m of the water

Q: Laminar loss in a circular pipe: For a viscous liquid flowing at mean velocity v through a circular pipe of length L and diameter d, the head loss due to viscous effects is approximately what?

Introduction / Context: In fully developed laminar flow inside a circular pipe, the Hagen–Poiseuille law governs the pressure drop and head loss. Recognizing the correct dependence on viscosity μ, length L, diameter d, and mean velocity v is important for microfluidics, lubrication problems, and low-Reynolds-number flows. Given Data / Assumptions: Steady, incompressible, Newtonian fluid. Fully developed laminar flow (Re < 2100 typically). Gravity head g included to convert pressure loss to head loss. Concept / Approach: Start from Darcy–Weisbach form h_f = f (L/d) * v^2/(2g) with laminar friction factor f = 64/Re, where Re = ρ v d / μ. Substituting f and simplifying yields the laminar head-loss expression directly proportional to μ, v, and L, and inversely proportional to d^2. Step-by-Step Solution: Take f = 64 / Re = 64 μ / (ρ v d).Insert into h_f: h_f = (64 μ / (ρ v d)) * (L/d) * v^2/(2g).Simplify: h_f = (64 μ L v) / (ρ 2 g d^2) = 32 μ v L / (ρ g d^2). Verification / Alternative check: Using the Hagen–Poiseuille equation Δp = 32 μ v L / d^2 also leads to the same head loss since h_f = Δp/(ρ g). Why Other Options Are Wrong: Option A is the general Darcy–Weisbach form but not a closed form for laminar flow unless f is replaced appropriately. Option C has incorrect diameter dependence. Option D is dimensionally inconsistent. Common Pitfalls: Confusing mean velocity v with centerline velocity (which is 2v in laminar flow). Final Answer: h_f = 32 μ v L / (ρ g d^2)

Q: Velocity profile in a pipe: value at the centreline In steady, fully developed internal flow through a pipe, the velocity of the fluid particle at the centre of the circular section is:

Introduction / Context: Understanding velocity distributions in pipe flow is foundational for computing discharge, shear stress, and head loss. Whether the regime is laminar or turbulent, the no-slip condition dictates zero velocity at the wall and a peak at the centreline for steady, fully developed flows in a round pipe. Given Data / Assumptions: Flow is steady and fully developed. Pipe is circular and straight. No swirl; single-phase Newtonian fluid. Concept / Approach: No-slip condition enforces u = 0 at the wall. Momentum balance and symmetry about the pipe axis create a monotonic profile rising from the wall to the axis. In laminar flow the profile is parabolic with u_max = 2 * u_avg at the centreline. In turbulent flow the profile is fuller (flatter) but still peaks at the centreline due to symmetry and wall shear effects. Step-by-Step Solution: Apply no-slip: u(r = R) = 0.By symmetry, du/dr = 0 at r = 0 (centreline), indicating an extremum.Energy and momentum considerations show the extremum is a maximum (not minimum) since shear stresses decelerate fluid near the wall. Verification / Alternative check: For laminar Poiseuille flow: u(r) = u_max * (1 − (r/R)^2), clearly maximum at r = 0. Turbulent empirical profiles (e.g., 1/nth-power laws) are also maximal at the centreline. Why Other Options Are Wrong: Minimum occurs at the wall, not at the centre; uniform velocity (option c) would violate no-slip; zero at the centre contradicts both theory and experiments. Common Pitfalls: Confusing average velocity with centreline velocity; assuming plug flow in all cases. Final Answer: Maximum

Question 1

Power transmission by pipeline — the maximum efficiency attainable for transmitting power through a pipeline occurs at what efficiency value?

Accepted Answer

Introduction / Context: When power is transmitted hydraulically through a long pipeline to a turbine/nozzle, friction losses increase with velocity. There exists an operating condition that maximizes delivered power for a given supply head, which implies a specific efficiency at the optimum. Given Data / Assumptions: Total supply head H (fixed). Head loss due to friction h_f increases with velocity squared. Delivered head at outlet H_d = H − h_f. Discharge Q ∝ velocity; delivered power P ∝ Q * H_d. Concept / Approach: Write P ∝ V * (H − k * V^2), where k aggregates friction factors. Differentiate dP/dV = 0 to find the optimum. The result shows h_f = H/3 at maximum power. Therefore efficiency η = H_d / H = (H − H/3) / H = 2/3 = 66.67%. Step-by-Step Solution: Let h_f = k * V^2; then P ∝ V * (H − k * V^2).Set derivative zero: dP/dV ∝ H − 3 k V^2 = 0 → k V^2 = H/3 → h_f = H/3.Compute η = (H − h_f)/H = (H − H/3)/H = 2/3 = 0.6667.Convert to percentage: 66.67%. Verification / Alternative check (if short method exists): Substitute back into P(V) to confirm a maximum (second derivative negative). Classic textbooks report the same optimum: loss = one-third of head, efficiency = two-thirds. Why Other Options Are Wrong: 25%, 33.3%, and 50% correspond to non-optimal operating points; 75% would require h_f < H/4, which does not satisfy the maximum-power condition. Common Pitfalls (misconceptions, mistakes): Maximizing efficiency instead of power; forgetting friction grows with V^2, not V. Final Answer: 66.67%

Question 2

Short-tube mouthpiece behavior — a short tube mouthpiece will NOT run full at its outlet under which head condition?

Accepted Answer

Introduction / Context: A short-tube (Borda-type) mouthpiece may discharge with a separated jet that does not fill the tube (“running free”) or with the tube running completely full. Whether it runs full depends on the available head and the ability of ambient pressure to sustain the separated core. Given Data / Assumptions: Sharp-edged short tube attached to a tank/orifice. Atmospheric pressure at the outlet; negligible losses aside from separation. Head measured above the centerline of the mouthpiece. Concept / Approach: For heads below a critical value (~12.2 m of water), the separated jet persists without reattaching, and the mouthpiece does not run full. For heads exceeding this, pressure recovery can cause the jet to fill the tube (“running full”). Thus, “will not run full” corresponds to heads less than about 12.2 m of water. Step-by-Step Solution: Compare operating head H to the critical value ~12.2 m.If H < 12.2 m → jet runs free (tube not full).If H > 12.2 m → tube tends to run full at outlet. Verification / Alternative check (if short method exists): Empirical data for Borda mouthpieces report Cc ≈ 0.5 (free) and Cc ≈ 1 (full) with the transition near 12.2 m head. Why Other Options Are Wrong: “More than” reverses the condition; “equal” is a threshold case not aligned with the “will not” phrasing; “none” and “1 m” do not reflect the known criterion. Common Pitfalls (misconceptions, mistakes): Confusing Borda mouthpiece behavior with long pipes; ignoring the role of separation and reattachment. Final Answer: less than 12.2 m of the water

Question 3

Select the most familiar Bernoulli energy equation form used along a streamline for steady, incompressible, inviscid flow.

Accepted Answer

Introduction / Context: Bernoulli’s equation is the energy conservation statement for steady, inviscid, incompressible flow along a streamline. It relates pressure head, velocity head, and elevation head, and is ubiquitous in hydraulics and aerodynamics. Given Data / Assumptions: Steady flow. Incompressible, homogeneous fluid. Negligible viscous dissipation (inviscid). Evaluation along a streamline. Concept / Approach: The “familiar” head form sums three terms: pressure head p/γ (where γ = ρ * g), velocity head v^2/(2g), and elevation head z. Their sum equals a constant along a streamline in the absence of pumps, turbines, or losses. Extensions add pump/turbine heads and losses as needed. Step-by-Step Solution: Start from Euler’s equation along a streamline: dp/ρ + g * dz + v * dv = 0.Integrate to obtain: p/ρ + g * z + v^2/2 = constant (per unit mass).Divide by g to convert to head form: p/γ + z + v^2/(2g) = constant. Verification / Alternative check (if short method exists): Dimensional analysis confirms each term has dimensions of length (head). Laboratory ventures (Venturi, Pitot) rely on this relation. Why Other Options Are Wrong: Option (b) is the differential Euler form; (c) mixes per-unit-mass constants with raw terms; (d) and (e) omit key terms, valid only under very restrictive conditions. Common Pitfalls (misconceptions, mistakes): Applying Bernoulli across devices with energy addition/removal without correction; using it across widely separated streamlines in rotational flows. Final Answer: p/γ + v^2/(2g) + z = constant

Question 4

Siphon applications — a siphon is primarily used for which of the following hydraulic purposes?

Accepted Answer

Introduction / Context: A siphon is a closed conduit shaped to rise above the free surfaces of the reservoirs it connects, allowing flow from a higher reservoir to a lower one even when a ridge (hill) lies between them, provided the summit pressure does not drop below vapor pressure significantly. Given Data / Assumptions: Two reservoirs at different elevations with an intervening high point. The siphon is initially primed (filled with liquid). Atmospheric pressure sustains the liquid column at the summit without cavitation. Concept / Approach: Flow occurs due to the net head difference between the reservoirs. The siphon’s summit experiences sub-atmospheric pressure; if the elevation exceeds permissible limits, vaporization can break the siphon. Thus, the classic application is bypassing an obstacle like a hill between reservoirs. Step-by-Step Solution: Prime the siphon to remove air.Open the outlet to create flow from the higher to the lower reservoir.Ensure summit elevation and losses do not reduce pressure below vapor pressure. Verification / Alternative check (if short method exists): Bernoulli analysis between free surfaces demonstrates flow direction and confirms sub-atmospheric summit pressure. Why Other Options Are Wrong: Filling a higher tank from a lower by siphon violates energy considerations; ordinary town supply does not require crossing a summit; “none” is incorrect; laminarization is unrelated. Common Pitfalls (misconceptions, mistakes): Ignoring priming; placing the summit too high; neglecting air leaks that break siphon action. Final Answer: to connect water reservoirs at different levels intervened by a hill

Question 5

Head loss due to sudden enlargement — a pipe expands from area 0.1 m² to 0.3 m². If Q = 0.3 m³/s, the head loss across the expansion is:

Accepted Answer

Introduction / Context: Sudden expansions cause flow separation and mixing, producing a dissipative head loss. The Borda–Carnot relation gives this loss in terms of the velocity change across the area enlargement. Given Data / Assumptions: Upstream area A1 = 0.1 m²; downstream area A2 = 0.3 m². Discharge Q = 0.3 m³/s (steady, incompressible water). Loss formula for sudden expansion: h_L = (v1 − v2)^2 / (2 * g). Concept / Approach: Compute velocities from continuity: v = Q / A. Substitute into the Borda–Carnot expression to obtain head loss in metres of water. Step-by-Step Solution: Compute upstream velocity v1 = Q / A1 = 0.3 / 0.1 = 3 m/s.Compute downstream velocity v2 = Q / A2 = 0.3 / 0.3 = 1 m/s.Apply loss formula: h_L = (v1 − v2)^2 / (2 * g) = (3 − 1)^2 / (2 * g) = 4 / (2 * g) = 2 / g m.Thus the head loss equals 2/g metres of water. Verification / Alternative check (if short method exists): Energy-grade line drop computed by momentum balance across a control volume yields the same value for sudden expansion loss. Why Other Options Are Wrong: g/2 or g m are dimensionally inconsistent in this context; “1g” is not meaningful; zero loss contradicts separation realities. Common Pitfalls (misconceptions, mistakes): Using (v1^2 − v2^2)/(2g) instead of (v1 − v2)^2/(2g); forgetting to recompute v after the area change. Final Answer: 2/g m of water

Question 6

Instrumentation — differential manometers are primarily used to measure which of the following?

Accepted Answer

Introduction / Context: Manometers are simple yet precise pressure-measuring devices based on hydrostatics. Differential manometers use a U-tube (or similar arrangement) filled with manometric fluid to compare pressures at two locations directly. Given Data / Assumptions: Incompressible manometric fluid (e.g., mercury or colored water). Two tap points connected to each limb. Quasi-static readings; negligible dynamic effects. Concept / Approach: The height difference between the columns reflects the pressure difference according to p1 − p2 = rho_m * g * (Δh) corrected for the densities of the working fluids. This makes differential manometers ideal for measuring pressure drops across fittings, Venturi meters, orifices, and filters. Step-by-Step Solution: Connect points to each limb of the manometer.Record column height difference Δh.Compute Δp using Δp = rho_m * g * Δh (with density corrections as needed).Interpret result as the pressure difference between the two points. Verification / Alternative check (if short method exists): Cross-check with calibrated differential pressure transmitters; for steady laminar pipe flow, compare to Hagen–Poiseuille prediction to validate. Why Other Options Are Wrong: Atmospheric pressure is measured by barometers; very low absolute pressures by McLeod or Pirani gauges; “pressure in water channels…” is generic and not specific to differential measurement. Common Pitfalls (misconceptions, mistakes): Using an inappropriate manometric fluid causing capillary or vapor issues; ignoring temperature effects on density. Final Answer: difference in pressure at two points

Question 7

Stream function properties in 2D incompressible flow — which statements about the stream function ψ are correct?

Accepted Answer

Introduction / Context: The 2D stream function ψ(x, y) is a scalar potential whose contours represent streamlines in incompressible flow. It enables automatic satisfaction of continuity and simplifies visualization and analysis of planar flows. Given Data / Assumptions: Two-dimensional, incompressible flow. ψ is single-valued and differentiable in the flow domain. Velocity components are related to ψ via partial derivatives. Concept / Approach: In 2D, velocity components can be obtained from ψ by u = ∂ψ/∂y and v = −∂ψ/∂x (sign convention may vary). Thus, gradients of ψ determine velocity components normal to directions, and constant-ψ lines are streamlines. Circulation around a closed path composed of streamlines is zero for steady, continuous flow without singularities, consistent with the definition. Step-by-Step Solution: State streamline property: along a streamline, ψ is constant.Relate velocity: u = ∂ψ/∂y, v = −∂ψ/∂x → velocity obtained by differentiating ψ.Directional change: dψ/dn = |∇ψ| along the normal is proportional to normal velocity component, linking spacing of streamlines to speed.For continuous 2D incompressible flow, the net flow across a closed path composed of streamlines is zero, aligning with the stated property. Verification / Alternative check (if short method exists): Check continuity: ∂u/∂x + ∂v/∂y = ∂²ψ/∂x∂y − ∂²ψ/∂y∂x = 0, always satisfied, confirming incompressibility. Why Other Options Are Wrong: Individually they are each correct; hence the combined correct choice is “all the above.” Common Pitfalls (misconceptions, mistakes): Mixing stream function with velocity potential (φ); forgetting sign conventions; assuming ψ exists in 3D the same way (it does not). Final Answer: all the above

Question 8

Dimensional analysis in fluid mechanics: What is the product of the Reynolds number (Re), the Froude number (Fr), and the Mach number (Ma), and how is it best described?

Accepted Answer

Introduction / Context: This question tests conceptual understanding of the principal dimensionless numbers used in fluid mechanics—Reynolds (Re), Froude (Fr), and Mach (Ma). These arise from balancing inertia against viscous, gravitational, and elastic (compressibility) effects, respectively. Students often memorize definitions, but deeper mastery includes recognizing what combinations do or do not correspond to widely adopted named groups. Given Data / Assumptions: Re = (rho * v * L) / mu (inertial/viscous). Fr = v / sqrt(g * L) (inertial/gravity). Ma = v / a (inertial/elastic; a is the speed of sound). All three are dimensionless for any consistent units. Concept / Approach: Each number isolates one physical effect versus inertia. Named numbers like Weber (We = rho * v^2 * L / sigma) or Euler (Eu = Δp / (rho * v^2)) have direct ties to surface tension and pressure forces. By contrast, the product Re * Fr * Ma multiplies three independent ratios. The product remains dimensionless, but there is no ubiquitous textbook name or standard use for it because it conflates three distinct balances into one value that lacks a clear single-physics interpretation. Step-by-Step Solution: Write Re * Fr * Ma = [(rho * v * L) / mu] * [v / sqrt(g * L)] * [v / a].This simplifies to (rho * v^3 * sqrt(L)) / (mu * a * sqrt(g)), which is a valid dimensionless grouping but not a standard named number.Compare with known groups: none of We, Eu, Ca (= Ma^2), or Ga match this product. Verification / Alternative check: Checking standard references shows We involves surface tension sigma, Eu involves pressure differences, Ca involves viscosity versus surface tension (not to be confused with Cauchy). Cauchy number in compressible flow equals Ma^2, which is not the given product. Why Other Options Are Wrong: Weber number requires sigma and depends on v^2, not v^3. Euler number depends on pressure ratio, absent here. Cauchy number equals Ma^2, not Re * Fr * Ma. Galileo number (also called Gravitational number) involves nu and g differently. Common Pitfalls: Assuming any product of familiar numbers must itself be a named parameter. Confusing similarly named “Ca” (Capillary number) with Cauchy number. Final Answer: It forms a dimensionless group without a standard, widely used name

Question 9

Open-channel flow transition: The phenomenon in an open channel where a high-velocity (supercritical) stream abruptly converts to a low-velocity (subcritical) stream with a sudden rise in water surface is called what?

Accepted Answer

Introduction / Context: A hydraulic jump is a classic rapidly varied open-channel flow transition used to dissipate energy downstream of spillways, sluices, and culverts. It occurs when supercritical flow (high velocity, low depth, Fr > 1) is forced to transition to subcritical flow (lower velocity, greater depth, Fr < 1), producing turbulence and a visible rise in the water surface. Given Data / Assumptions: Open-channel flow under gravity. Upstream flow is supercritical; downstream control enforces subcritical flow. Idealized horizontal rectangular channel considered for simple explanation (general ideas still hold otherwise). Concept / Approach: The jump conserves mass and (approximately) momentum across the roller; energy is not conserved because substantial energy is dissipated by turbulence and air entrainment. Conjugate (sequent) depths y1 and y2 are related by momentum principles; the energy loss ΔE is a function of Froude number Fr1 of the incoming stream. Step-by-Step Solution: Identify the regime change: Fr1 > 1 (supercritical) to Fr2 < 1 (subcritical).Apply the momentum equation between sections just before and after the roller to obtain y2 in terms of y1 and Fr1.Compute energy loss as E1 − E2; note the increase in depth (surface rise) visible as the jump. Verification / Alternative check: Observation of roller, intense turbulence, and surface boil confirms a hydraulic jump. In contrast, “water hammer” is a pressure transient in closed conduits due to rapid valve action, not an open-channel phenomenon. Why Other Options Are Wrong: Critical discharge refers to the condition at Fr = 1, not an abrupt surface rise with strong turbulence. Water hammer happens in pressurized pipes, not open channels. None of these is incorrect because “hydraulic jump” is the established term. Common Pitfalls: Confusing sequent depth relationship (momentum-based) with gradually varied flow profiles (energy-based). Final Answer: Hydraulic jump

Question 10

Laminar loss in a circular pipe: For a viscous liquid flowing at mean velocity v through a circular pipe of length L and diameter d, the head loss due to viscous effects is approximately what?

Accepted Answer

Introduction / Context: In fully developed laminar flow inside a circular pipe, the Hagen–Poiseuille law governs the pressure drop and head loss. Recognizing the correct dependence on viscosity μ, length L, diameter d, and mean velocity v is important for microfluidics, lubrication problems, and low-Reynolds-number flows. Given Data / Assumptions: Steady, incompressible, Newtonian fluid. Fully developed laminar flow (Re < 2100 typically). Gravity head g included to convert pressure loss to head loss. Concept / Approach: Start from Darcy–Weisbach form h_f = f (L/d) * v^2/(2g) with laminar friction factor f = 64/Re, where Re = ρ v d / μ. Substituting f and simplifying yields the laminar head-loss expression directly proportional to μ, v, and L, and inversely proportional to d^2. Step-by-Step Solution: Take f = 64 / Re = 64 μ / (ρ v d).Insert into h_f: h_f = (64 μ / (ρ v d)) * (L/d) * v^2/(2g).Simplify: h_f = (64 μ L v) / (ρ 2 g d^2) = 32 μ v L / (ρ g d^2). Verification / Alternative check: Using the Hagen–Poiseuille equation Δp = 32 μ v L / d^2 also leads to the same head loss since h_f = Δp/(ρ g). Why Other Options Are Wrong: Option A is the general Darcy–Weisbach form but not a closed form for laminar flow unless f is replaced appropriately. Option C has incorrect diameter dependence. Option D is dimensionally inconsistent. Common Pitfalls: Confusing mean velocity v with centerline velocity (which is 2v in laminar flow). Final Answer: h_f = 32 μ v L / (ρ g d^2)

Question 11

Velocity profile in a pipe: value at the centreline In steady, fully developed internal flow through a pipe, the velocity of the fluid particle at the centre of the circular section is:

Accepted Answer

Introduction / Context: Understanding velocity distributions in pipe flow is foundational for computing discharge, shear stress, and head loss. Whether the regime is laminar or turbulent, the no-slip condition dictates zero velocity at the wall and a peak at the centreline for steady, fully developed flows in a round pipe. Given Data / Assumptions: Flow is steady and fully developed. Pipe is circular and straight. No swirl; single-phase Newtonian fluid. Concept / Approach: No-slip condition enforces u = 0 at the wall. Momentum balance and symmetry about the pipe axis create a monotonic profile rising from the wall to the axis. In laminar flow the profile is parabolic with u_max = 2 * u_avg at the centreline. In turbulent flow the profile is fuller (flatter) but still peaks at the centreline due to symmetry and wall shear effects. Step-by-Step Solution: Apply no-slip: u(r = R) = 0.By symmetry, du/dr = 0 at r = 0 (centreline), indicating an extremum.Energy and momentum considerations show the extremum is a maximum (not minimum) since shear stresses decelerate fluid near the wall. Verification / Alternative check: For laminar Poiseuille flow: u(r) = u_max * (1 − (r/R)^2), clearly maximum at r = 0. Turbulent empirical profiles (e.g., 1/nth-power laws) are also maximal at the centreline. Why Other Options Are Wrong: Minimum occurs at the wall, not at the centre; uniform velocity (option c) would violate no-slip; zero at the centre contradicts both theory and experiments. Common Pitfalls: Confusing average velocity with centreline velocity; assuming plug flow in all cases. Final Answer: Maximum

Question 12

Effect of droplet size on internal pressure due to surface tension As the diameter of a liquid droplet decreases, how does the excess pressure inside the droplet change?

Accepted Answer

Introduction / Context: Surface tension produces a pressure jump across curved interfaces. For small droplets or bubbles, this capillary pressure becomes significant. Knowing how it scales with size is essential in atomization, emulsions, spray drying, and cloud microphysics. Given Data / Assumptions: Liquid droplet (single interface), spherical. Surface tension T is constant (isothermal, clean interface). Static equilibrium (no acceleration/flow inside). Concept / Approach: For a spherical interface, the excess internal pressure over the outside pressure is Δp = 2 * T / r for a liquid droplet (one surface). In terms of diameter d, Δp = 4 * T / d. Hence, as size decreases (smaller r or d), Δp increases inversely with size. Step-by-Step Solution: Formula: Δp_droplet = 2 * T / r = 4 * T / d.Let d decrease while T is constant ⇒ Δp increases as 1/d.Therefore, smaller droplets sustain higher internal pressure above ambient. Verification / Alternative check: For a soap bubble (two interfaces), Δp_bubble = 4 * T / r = 8 * T / d, even larger for the same diameter. The inverse proportionality to size is consistent across spherical capillarity relations. Why Other Options Are Wrong: Decrease or no change contradicts Δp ∝ 1/r; zero or negative is physically incorrect for a clean droplet under positive T. Common Pitfalls: Confusing coefficients for droplet (single interface) and bubble (double interface); mixing radius and diameter forms. Final Answer: Increases

Question 13

Core assumptions behind Bernoulli’s equation (inviscid, steady, along a streamline) Which set of statements best reflects the assumptions under which Bernoulli’s equation applies?

Accepted Answer

Introduction / Context: Bernoulli’s equation is derived from the mechanical energy balance for steady, incompressible, inviscid flow along a streamline with no pumps/turbines and only gravity as a body force. Recognizing its assumptions prevents misuse in diffusers, across valves, or in highly viscous flows. Given Data / Assumptions: Steady flow (no local acceleration). Incompressible fluid of constant density. Negligible viscosity (inviscid), hence no dissipation between sections. Body force is gravity only; no shaft work added or removed. Uniform section-wise velocity is an engineering approximation used when applying average velocity with a kinetic energy correction factor near 1. Concept / Approach: Bernoulli states p/γ + v^2/(2 g) + z = constant along a streamline for the above conditions. In practice, a kinetic energy correction factor may be used if velocity is non-uniform; the “uniform velocity” statement reflects that simplified use without correction factors. Step-by-Step Solution: Identify inviscid assumption ⇒ no head loss.Restrict body forces to gravity ⇒ potential head term z.Steady, incompressible ⇒ no density/time variations.Uniform velocity (or correction factor near unity) allows use of average velocity in v^2/(2 g). Verification / Alternative check: Compare with the full energy equation including head loss and pump/turbine head; Bernoulli is the special case with h_L = 0 and h_pump = h_turbine = 0. Why Other Options Are Wrong: Options (a), (b), and (c) individually are partial statements; the correct comprehensive assumption set is captured by “All of the above”. Common Pitfalls: Applying Bernoulli across components with significant losses or work exchange; ignoring the need for correction factors when velocity is highly non-uniform. Final Answer: All of the above

Question 14

Terminal settling velocity of a small sphere (Stokes’ law form) A sphere of radius r and specific weight γ_s falls slowly through a viscous liquid of viscosity μ and specific weight γ. For creeping flow (very low Reynolds number), the terminal veloc…

Accepted Answer

Introduction / Context: Stokes’ law gives terminal settling velocity for tiny particles where viscous forces dominate inertial forces (Re << 1). This is fundamental in sedimentation, centrifugation scaling, and particle sizing in process engineering and environmental engineering. Given Data / Assumptions: Rigid sphere of radius r. Very low Reynolds number (creeping flow, no wake). Fluid is Newtonian with dynamic viscosity μ. Specific weights: particle γ_s, fluid γ; gravity acts downward. Concept / Approach: At terminal velocity, net force is zero: buoyant-reduced weight equals viscous drag. Balance: (γ_s − γ) * (4/3) * pi * r^3 = 6 * pi * μ * r * v, leading to the Stokes formula. Using γ = ρ g automatically incorporates g. Step-by-Step Solution: Reduced weight W_r = (γ_s − γ) * (4/3) * pi * r^3.Stokes drag D = 6 * pi * μ * r * v.Set W_r = D ⇒ v = [(γ_s − γ) * (4/3) * pi * r^3] / [6 * pi * μ * r] = (2/9) * r^2 * (γ_s − γ) / μ. Verification / Alternative check: Dimensional check: [γ] = N/m^3, so r^2 * γ / μ gives m/s as required. The dependence v ∝ r^2 is characteristic of Stokes settling. Why Other Options Are Wrong: Options with r, r^3, or sum (γ_s + γ) do not follow from the drag–weight balance; (9/2) factor is inverted. Common Pitfalls: Using density difference without g when specific weight is already used; applying Stokes’ law at higher Reynolds numbers where it no longer holds. Final Answer: v = (2/9) * r^2 * (γ_s − γ) / μ

Question 15

Criterion for “sudden” valve closure in pipelines (water hammer) A valve closure in a pipeline is termed “sudden” when the actual closure time t_c satisfies which condition (L = pipe length, a = wave speed)?

Accepted Answer

Introduction / Context: Water hammer analysis classifies valve closures as “sudden” or “gradual” depending on how the closure time compares to the characteristic time for a pressure wave to travel to the upstream reservoir and reflect back. This classification determines the peak transient pressure predicted by the Joukowsky equation. Given Data / Assumptions: One-dimensional elastic water hammer model. Wave (celerity) speed a is constant. Uniform pipe of length L, liquid initially in steady motion. Concept / Approach: The critical time is 2 L / a, the round-trip travel time of a pressure wave in the pipe. If the valve closes in less time than this, the flow cannot adjust gradually; the transient is a “sudden” closure yielding the maximum pressure rise Δp = ρ a Δv (Joukowsky). For longer times, the event is “gradual,” and the peak pressure is smaller, depending on the closure law. Step-by-Step Solution: Characteristic time t* = 2 L / a.Define sudden: t_c < t* ⇒ “sudden” closure.Define gradual: t_c ≥ t* ⇒ “gradual” closure. Verification / Alternative check: Method of characteristics solutions show that if the boundary condition (valve opening) changes fully before the first reflected wave returns, the maximum Joukowsky rise occurs, consistent with the t_c < 2 L / a criterion. Why Other Options Are Wrong: Equal or greater than 2 L / a are not “sudden”; L / a and 4 L / a are not the accepted threshold in standard water hammer theory. Common Pitfalls: Confusing single-trip time L / a with the round-trip 2 L / a; overlooking that wave speed depends on both fluid compressibility and pipe wall elasticity. Final Answer: t_c < 2 L / a

Question 16

Dimensionless group representing inertia-to-gravity force ratio The dimensionless number that compares inertial forces to gravitational forces in a flow (neglecting other forces) is the:

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Introduction / Context: Dimensionless numbers classify dynamic similarity in fluid flows. Choosing the right group helps scale models and predict dominant forces. When gravity plays a major role (open-channel flows, ship hydrodynamics), the inertia-to-gravity ratio is key. Given Data / Assumptions: We compare inertial and gravitational forces only. Characteristic velocity V and length L are defined. Fluid density is constant for scaling purposes. Concept / Approach: The Froude number Fr is defined as Fr = V / √(g L). It can also be viewed from force scaling as Fr^2 ∼ inertia/gravity. It governs phenomena with free surfaces and wave-making, such as spillways, ship waves, and open-channel transitions. Step-by-Step Solution: Identify target forces: inertia vs gravity ⇒ use Froude number.Recall: Fr = V / √(g L) and Fr^2 ∝ F_inertia / F_gravity.Therefore, the required dimensionless group is the Froude number. Verification / Alternative check: Other groups: Reynolds compares inertia to viscous (ρ V L / μ); Weber compares inertia to surface tension (ρ V^2 L / σ); Euler links pressure to inertia; Mach compares inertia to compressibility (V / a_sound). None are inertia/gravity except Froude. Why Other Options Are Wrong: Euler, Reynolds, Weber, and Mach address different force balances and are not suited for gravity-dominated similarity. Common Pitfalls: Confusing Froude with Reynolds; overlooking that Fr controls wave behavior and critical/supercritical flows in open channels. Final Answer: Froude number

Question 17

Chezy application to a full rectangular channel (b = 6 m, y = 3 m) A rectangular channel is 6 m wide and 3 m deep, running full on a bed slope S = 1/2000. If Chezy’s constant C = 54.8, identify the correct specifications (hydraulic mean depth, veloc…

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Introduction / Context: Chezy’s formula is widely used for uniform open-channel flow to relate mean velocity to hydraulic radius and slope. This problem checks competence in computing section properties and applying the velocity formula correctly. Given Data / Assumptions: Rectangular section: width b = 6 m; depth y = 3 m (running full). Chezy constant C = 54.8; bed slope S = 1/2000. Uniform, steady flow; prismatic channel; water at ordinary conditions. Concept / Approach: Hydraulic mean depth (hydraulic radius) R = A / P for open channels. Chezy’s velocity: V = C * √(R * S). Discharge: Q = A * V. Step-by-Step Solution: Area A = b * y = 6 * 3 = 18 m^2.Wetted perimeter P = b + 2 y = 6 + 6 = 12 m.Hydraulic radius R = A / P = 18 / 12 = 1.5 m.Mean velocity V = C * √(R * S) = 54.8 * √(1.5 * 1/2000).Compute inside root: 1.5/2000 = 0.00075; √(0.00075) ≈ 0.027386; V ≈ 54.8 * 0.027386 ≈ 1.5 m/s.Discharge Q = A * V = 18 * 1.5 = 27 m^3/s. Verification / Alternative check: Dimensions: R in meters, S dimensionless; V in m/s; Q in m^3/s. Values are internally consistent and typical for a channel of this size and slope. Why Other Options Are Wrong: Since (a), (b), and (c) are all correct, any option omitting one of them is incomplete. Common Pitfalls: Using hydraulic depth A/b instead of hydraulic radius A/P; arithmetic slips with slope; forgetting the square root in Chezy’s relation. Final Answer: All of the above

Question 18

In open-channel hydraulics, a “most economical” rectangular channel section minimizes wetted perimeter for a given area (or maximizes discharge for a given perimeter). For such a rectangular channel, what proportional relationship between depth and …

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Introduction / Context: In civil engineering water resources and irrigation design, the “most economical” (or hydraulically efficient) channel shape minimizes the wetted perimeter for a given flow area, reducing friction losses. For a rectangular channel, there is a well-known proportionality between depth and width that achieves this condition. Given Data / Assumptions: Prismatic open channel with rectangular cross-section. Objective: minimize wetted perimeter P for a given area A. Steady, uniform flow; roughness constant; gravity drainage. Concept / Approach: The hydraulic radius R is A / P. For fixed A, maximizing R (equivalently minimizing P) yields the most economical section. For a rectangle of width b and depth y: A = b * y, P = b + 2y. Using calculus, differentiate P with respect to b (or y) under a constant A constraint to find the optimum proportion. Step-by-Step Solution: Let A = b * y = constant.Express b = A / y.Wetted perimeter P = b + 2y = (A / y) + 2y.Minimize P: dP/dy = d/dy[(A / y) + 2y] = (-A / y^2) + 2.Set derivative to zero: (-A / y^2) + 2 = 0.Solve for y: A / y^2 = 2 → A = 2y^2 → by = 2y^2 → b = 2y.Therefore, depth y = b / 2. Verification / Alternative Check: Compute hydraulic radius at optimum: R = A / P = (by) / (b + 2y) = (2y*y) / (2y + 2y) = (2y^2) / (4y) = y / 2, confirming standard textbook result for rectangular channels at economical section. Why Other Options Are Wrong: One-fourth of the width: y = b/4 gives larger P for same A, not minimum.Three times hydraulic radius: y = 3R is not the defining condition here; for the economical rectangle R = y/2.Hydraulic mean depth: this is another name for hydraulic radius in some contexts, not a direct proportion b–y.None of these: incorrect because a correct option exists. Common Pitfalls: Confusing triangular or trapezoidal optimal proportions with the rectangular case. Maximizing area instead of minimizing wetted perimeter for a fixed area. Mixing up hydraulic radius R with depth y; for the economical rectangle R = y/2. Final Answer: depth equals half the width

Question 19

A hydraulic ram (hydram) is a classic water-supply device used where a small fall of water can lift a portion of that water to a higher elevation. Which statement best describes what a hydraulic ram is designed to do in practical field applications?

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Introduction / Context: A hydraulic ram (hydram) is a water-powered pump widely used in rural water-supply engineering. It exploits the momentum of a relatively large flow at low head to lift a smaller quantity of water to a much greater head, requiring no electricity or fuel. Given Data / Assumptions: Continuous source of water with a modest supply head (fall). Desire to lift a fraction of that water to a higher elevation. No electrical or combustion power source available onsite. Concept / Approach: The hydram uses the water hammer effect. When a waste valve suddenly closes, the momentum of the moving water generates a pressure rise that opens a delivery check valve, forcing part of the water into an air-charged delivery chamber and up the delivery pipe to a higher level. Step-by-Step (Principle of Operation): Water flows down the drive pipe under supply head (Hs), building velocity.Waste valve initially open discharges water to atmosphere; flow accelerates.Waste valve slams shut due to increased dynamic pressure, creating a high transient pressure (water hammer).The transient opens the delivery check valve; water enters the air chamber and delivery pipe against a higher static head (Hd).Pressure decays; delivery valve closes; waste valve reopens; the cycle repeats automatically. Verification / Alternative Check: Energy balance shows that (flow lifted * Hd) is less than (supply flow * Hs); overall efficiency often 50%–70% for well-designed systems. No external power is required beyond the water source itself. Why Other Options Are Wrong: Accelerate water for dissipation: not the purpose; acceleration is a means to create a pressure pulse, not an end.Lift heavy loads: that describes hydraulic jacks, not hydrams.None of these / vacuum priming: incorrect; the defining function is power-free water lifting. Common Pitfalls: Assuming constant flow; hydram operation is cyclic. Confusing hydram with conventional pumps that need external power. Ignoring the need for a proper drive pipe length and rigid support to harness water hammer safely. Final Answer: lift water to a higher elevation without using an electric motor or external power

Question 20

When water flows over a rectangular suppressed weir, a nappe forms and air may be entrained beneath it. Considering the pressure beneath the freely ventilated nappe, how does any negative (sub-atmospheric) pressure under the nappe influence the meas…

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Introduction / Context: Free overfall from a sharp-crested, rectangular suppressed weir produces a nappe with atmospheric or sub-atmospheric conditions beneath it depending on aeration. The pressure regime under the nappe impacts the effective head and therefore the discharge for a given upstream head measurement point. Given Data / Assumptions: Sharp-crested, rectangular suppressed weir. Head measured at a standard upstream location where velocity of approach is either negligible or corrected. Nappe ventilation may be present or partially restricted, leading to negative pressure beneath nappe. Concept / Approach: Bernoulli’s principle with the appropriate discharge coefficient shows that reduced pressure under the nappe increases the pressure differential across the crest. This raises the effective head driving flow, hence the actual discharge rises for the same measured upstream head. Step-by-Step Reasoning: Weir formula: Q = C_d * b * (2g)^(1/2) * H^(3/2), where H is the head over crest.If sub-atmospheric pressure develops under the nappe, the pressure at the crest control section is reduced below atmospheric.Effective head becomes H_eff > H_measured, because downstream/under-nappe pressure is lower.Therefore Q_actual = C_d′ * b * (2g)^(1/2) * H_eff^(3/2) increases compared with the fully ventilated case. Verification / Alternative Check: Field practice recommends ventilating the nappe to maintain near-atmospheric pressure beneath it; otherwise discharge coefficients must be adjusted upward to account for increased flow at the same indicated head. Why Other Options Are Wrong: Decreases discharge: opposite of the pressure-head effect.No appreciable effect: contradicted by experiments; effect can be significant.Flow reversal: physically unrealistic for ordinary heads.Only C_v changes: net impact is increased Q unless compensated in coefficients; statement is incomplete. Common Pitfalls: Ignoring ventilation; unventilated nappes can produce measurement errors. Applying standard coefficients without checking nappe aeration. Neglecting approach velocity correction, compounding error. Final Answer: It increases the discharge for the same measured head

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