Question 1

Surface tension comparison – mercury vs water at normal temperature At ordinary laboratory conditions, the surface tension of mercury is __________ that of water:

Accepted Answer

Introduction: Surface tension governs phenomena such as droplet formation, capillary rise/depression, and wetting. This item checks qualitative knowledge of how mercury compares with water in terms of surface tension at room temperature. Given Data / Assumptions: Temperature near 20–25 °C. Clean, uncontaminated surfaces (no surfactants). Standard laboratory pressure; gravity effects typical. Concept / Approach: Surface tension is the energy per unit area needed to create new surface. Mercury has strong cohesive metallic bonding and poor wetting on glass, leading to a high surface tension and a convex meniscus. Water, while relatively high among common liquids, still has significantly lower surface tension than mercury. Step-by-Step Solution: 1) Recall qualitative values: water ~ 0.072 N/m at 20 °C; mercury ~ 0.48–0.49 N/m.2) Compare magnitudes: mercury » water.3) Conclude: mercury's surface tension is higher than that of water at normal temperature. Verification / Alternative check: Capillary behavior: water rises in clean glass (concave meniscus), while mercury depresses (convex meniscus) because its surface tension is higher and contact angle exceeds 90°, consistent with reduced wetting. Why Other Options Are Wrong: Same as / lower than: contradict benchmark data and observable capillary effects. Not comparable / time varying: surface tension is well-defined at given T; no periodic time variation exists in a clean system. Common Pitfalls: Confusing density (also higher for mercury) with surface tension; overlooking contamination that can reduce water's surface tension. Final Answer: higher than

Question 2

Dynamics to statics transformation – D’Alembert’s principle in fluids The D’Alembert’s principle is used to convert a dynamic equilibrium problem of a fluid mass into an equivalent static equilibrium problem. Do you agree with this statement?

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Introduction: D’Alembert’s principle introduces an inertia force equal to mass * acceleration but opposite in direction, allowing dynamic systems to be treated as if in static equilibrium. In fluid mechanics, this conceptual tool simplifies analyses like potential flow around bodies. Given Data / Assumptions: Consider a fluid element of mass m undergoing acceleration a. Body forces and surface forces act on the element. Continuum hypothesis holds; properties vary smoothly. Concept / Approach: By adding a fictitious inertia force F_i = −m * a to the real forces, the sum can be set to zero to enforce an equivalent static equilibrium for instantaneous motion states. This is particularly useful in deriving Bernoulli-like relations in non-inertial frames and for potential-flow formulations. Step-by-Step Solution: 1) Identify all real forces on a fluid element (pressure, shear, body forces).2) Compute inertia term m * a from the local/convective accelerations.3) Add F_i = −m * a to the system.4) Impose static equilibrium: ΣF_real + F_i = 0, enabling static-like balance equations. Verification / Alternative check: Potential-flow solutions treat unsteady accelerations via velocity potentials while invoking D’Alembert’s concept to avoid explicit time-dependent force sums in some derivations. Why Other Options Are Wrong: Disagree / restricted cases: the principle is general; material model (compressible/incompressible) or Re does not negate the conceptual transformation. Rigid bodies only: fluids also obey Newtonian mechanics; the inertia concept applies to fluid elements. Common Pitfalls: Interpreting inertia force as real; it is a mathematical device to recast dynamics into static form at each instant. Final Answer: Agree

Question 3

Hydrostatic resultants – liquid pressures on both sides of a vertical wall When a vertical wall is subjected to hydrostatic pressures due to liquids on both sides, the net resultant hydrostatic pressure on the wall equals the __________ of the two i…

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Introduction: Structures like partition walls or lock gates may have liquids on both sides. Determining the net hydrostatic force guides sizing and stability checks. Given Data / Assumptions: Each side has a hydrostatic pressure distribution producing a resultant force F1 and F2 acting normal to the wall and directed from liquid to wall. Liquids act on opposite faces, so their forces are in opposite directions. Wall is planar and vertical; densities and depths may differ on the two sides. Concept / Approach: Since the forces act oppositely, the net resultant magnitude is |F1 − F2|, directed toward the side with the larger hydrostatic resultant. Line of action is determined by centers of pressure, which generally differ for the two sides. Step-by-Step Solution: 1) Compute each side's resultant: F = w * A * x_c for horizontal plates or via integration for vertical plates.2) Determine directions: they oppose each other through the wall.3) Combine vectorially along the wall normal: F_net = |F1 − F2|.4) Locate the net line of action by taking moments of individual forces about a reference. Verification / Alternative check: Special case with equal liquids and equal depths on both sides gives F1 = F2, so net force is zero, matching intuition. Why Other Options Are Wrong: Sum / means: these do not reflect opposing directions; only difference captures cancellation. Common Pitfalls: Forgetting to recompute the line of action after taking the difference; ignoring different densities or depths. Final Answer: difference

Question 4

Jet/orifice coefficients – definition of coefficient of velocity (Cv) The coefficient of velocity is defined as the ratio of the following quantities for flow through an orifice:

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Introduction: Orifice flow is characterized by three standard coefficients: coefficient of velocity (Cv), coefficient of contraction (Cc), and coefficient of discharge (Cd). Correct definitions prevent misuse in calculations. Given Data / Assumptions: Sharp-edged orifice with a vena contracta downstream. Steady incompressible flow. Negligible upstream approach velocity or accounted separately. Concept / Approach: Cv compares the actual jet speed to the ideal Bernoulli speed at the orifice plane. Cc captures area contraction (A_jet/A_orifice). Cd measures discharge shortfall: Cd = Cc * Cv. Step-by-Step Solution: 1) Compute theoretical velocity: V_th = sqrt(2 * g * H).2) Measure actual velocity at vena contracta: V_act (e.g., via jet range or Pitot).3) Evaluate Cv = V_act / V_th.4) Use Cd = Cc * Cv for discharge calculations. Verification / Alternative check: Typical sharp-edge values: Cv ~ 0.95–0.99; Cc ~ 0.61–0.64; Cd ~ 0.58–0.64, confirming the relationships. Why Other Options Are Wrong: Cc definition (option b) and Cd definition (option c) are different coefficients. Head ratio (option d) and squared ratio (option e) are not standard definitions of Cv. Common Pitfalls: Using Cd in place of Cv causes underestimation of jet speed; mislocating the measurement point (must be at vena contracta). Final Answer: actual velocity of jet at vena contracta to the theoretical velocity

Question 5

Compressible flows – definition of sonic flow A flow is termed “sonic” (critically choked) when the Mach number (M) is unity. Do you agree?

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Introduction: Mach number M is the ratio of local flow speed to local speed of sound. Sonic conditions play a central role in nozzle choking, shock formation, and compressible-flow similarity. Given Data / Assumptions: Local speed of sound a depends on thermodynamic state (e.g., a = sqrt(gamma * R * T) for ideal gases). Flow speed V may vary along streamlines. Continuum, compressible regime applies. Concept / Approach: By definition, M = V / a. Sonic flow occurs when V equals a, i.e., M = 1. In ducts and nozzles, this condition marks the critical station that controls mass flow when choking occurs. Step-by-Step Solution: 1) Determine local a from fluid properties.2) Measure or compute V at the point of interest.3) Compute M = V / a.4) If M = 1, the state is sonic; M < 1 is subsonic; M > 1 is supersonic. Verification / Alternative check: In converging–diverging nozzles, the throat reaches M = 1 at choked conditions; further backpressure reduction shifts acceleration to the diverging section with supersonic exit speeds. Why Other Options Are Wrong: Disagree: contradicts the definition of Mach number. Restrictions to ideal gases, nozzles, or sea level are unnecessary; M = 1 is definitional regardless of medium. Common Pitfalls: Confusing “sonic” with “sound speed in still air” rather than the local medium; ignoring temperature dependence of a. Final Answer: Agree

Question 6

Viscosity – property governing resistance to flow The property of a liquid that controls its resistance to deformation and thus its rate of flow is called viscosity. True or false?

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Introduction: Viscosity quantifies internal friction in fluids. It influences velocity profiles, pressure drops, and energy losses in pipelines and open channels. Given Data / Assumptions: Continuum fluid behavior; temperature affects viscosity but definition remains. Flow may be laminar or turbulent; viscosity is relevant in both regimes (e.g., laminar friction factor depends solely on Re). Newtonian behavior assumed for simple constitutive relation tau = mu * (du/dy). Concept / Approach: Dynamic viscosity mu links shear stress tau to velocity gradient. Higher mu means greater resistance to shear, slowing the rate at which layers slide past each other and thus controlling how readily the liquid flows under a given driving head. Step-by-Step Solution: 1) Recognize that for laminar pipe flow, Δp ∝ mu * L * Q / (R^4), showing direct proportionality to mu.2) For a given pressure drop, increasing mu reduces Q (rate of flow).3) Therefore viscosity governs the ease/difficulty of flow. Verification / Alternative check: Compare water (low mu) with glycerin or oils (high mu): under the same head, water flows much more readily, consistent with the definition. Why Other Options Are Wrong: False or restricted cases: viscosity remains the controlling property, though non-Newtonian liquids need generalized relations. Density primarily affects inertia and hydrostatics, not shear resistance at a given gradient. Common Pitfalls: Confusing dynamic (mu) with kinematic viscosity (nu = mu / rho); overlooking strong temperature dependence. Final Answer: True

Question 7

Weirs/notches – error propagation for a triangular (V) notch If the measured head over a triangular notch crest (H) has a +1% error, the resulting percentage error in computed discharge is approximately:

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Introduction: Discharge over notches and weirs is a power function of head. Understanding error propagation helps prioritize measurement accuracy for key variables. Given Data / Assumptions: Triangular (V) notch with standard calibration. Empirical relation: Q ∝ H^(5/2) (ignoring minor corrections). Small error approximation for differential changes. Concept / Approach: For Q ∝ H^n, the relative error relation is dQ/Q = n * dH/H for small perturbations. For a V-notch, n = 5/2 = 2.5, which amplifies head measurement errors by a factor of 2.5. Step-by-Step Solution: 1) Write Q = K * H^(5/2).2) Take logarithms and differentiate: dQ/Q = (5/2) * dH/H.3) With dH/H = 0.01, compute dQ/Q = 2.5 * 0.01 = 0.025 = 2.5%.4) Hence, a 1% head error gives about 2.5% discharge error. Verification / Alternative check: A rectangular notch has Q ∝ H^(3/2), so a 1% head error would yield ~1.5% Q error—smaller than the triangular notch case, consistent with the higher exponent for V-notches. Why Other Options Are Wrong: 1%, 1.5%, 2%: underestimate the amplification factor n = 2.5. 5%: doubles the correct amplification; not supported by Q–H relation. Common Pitfalls: Applying rectangular-notch exponents to V-notches; ignoring that error propagation assumes small percentage errors. Final Answer: 2.5%

Question 8

Hydrostatics – pressure distribution on a vertical wall (single liquid) A vertical wall retains a single, homogeneous liquid on one side only. Which statement best describes the hydrostatic pressure along the wall from the free surface to the bottom?

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Introduction: Hydrostatic pressure increases with depth due to the weight of the overlying fluid. Understanding the distribution is essential for computing forces on retaining walls, tanks, and gates. Given Data / Assumptions: Liquid is at rest; density is constant. Free surface is exposed to atmosphere. Wall is vertical; gravity is uniform. Concept / Approach: The hydrostatic relation is p = p_atm + w * h, where h is depth below the free surface. Gauge pressure p_g = p − p_atm = w * h increases linearly with h, from zero at the free surface to maximum at the deepest point. Step-by-Step Solution: 1) At h = 0 (free surface), p_g = 0 → pressure equals atmospheric.2) At the bottom (h = H), p_g = w * H → maximum pressure.3) The wall therefore experiences a triangular pressure distribution.4) The resultant hydrostatic force acts at H/3 above the bottom (center of pressure for a vertical plane). Verification / Alternative check: Tank gauge readings and force integrations on test panels match the linear rise in pressure with depth. Why Other Options Are Wrong: Minimum but not zero: at the free surface, gauge pressure is precisely zero. Bottom zero or uniform pressure: contradicts hydrostatic law. Common Pitfalls: Confusing absolute and gauge pressure; forgetting free-surface reference. Final Answer: The pressure at the liquid level is zero, and the pressure at the bottom is maximum.

Question 9

Internal mouthpieces – condition for running free vs running full An internal mouthpiece is said to be running free if the length of the mouthpiece is __________ the diameter of the orifice:

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Introduction: Flow through internal (Borda's) mouthpieces can occur in two regimes: running free (jet springs clear of the walls after vena contracta) or running full (the mouthpiece runs full like a short pipe). The regime depends largely on the mouthpiece length-to-diameter ratio. Given Data / Assumptions: Sharp-edged entrance forming a vena contracta. Mouthpiece aligned with the orifice axis. Steady incompressible flow; minor losses present. Concept / Approach: When the length is relatively short, the contracted jet does not expand enough to contact the walls; the flow is said to run free, exhibiting coefficients characteristic of orifice discharge. As the length increases beyond a few diameters, wall attachment occurs and the passage runs full, behaving closer to short-pipe flow with different coefficients. Step-by-Step Solution: 1) Define L/D as the length-to-diameter ratio.2) Empirical criterion: running free for L/D less than about 3; running full when L/D exceeds about 3.3) Therefore the correct comparative phrase is “less than three times.” Verification / Alternative check: Observed discharge coefficients shift as L/D crosses ~3, confirming regime transition in laboratory studies. Why Other Options Are Wrong: Less than twice: too restrictive; many designs run free up to near 3D. More than twice / more than three times / exactly four times: these tend toward running full, not free. Common Pitfalls: Ignoring entrance shape; a rounded entrance alters contraction and regime boundaries. Final Answer: less than three times

Question 10

Parallel Pipes – Governing Conditions for Head Loss and Discharge In a pipeline system where multiple branches run in parallel between the same two junctions, the head loss (energy drop) across each branch is identical, and the total discharge equal…

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Introduction: Parallel pipe networks are common in water distribution and process plants. Correctly identifying how head loss and discharge partition among branches is essential for sizing lines, selecting pumps, and predicting system behavior. Given Data / Assumptions: Pipes connect the same upstream and downstream nodes (parallel configuration). Steady, incompressible flow; elevation difference between nodes is fixed. Losses are represented by standard head-loss formulas (e.g., Darcy–Weisbach). Concept / Approach: Between two common nodes, energy grade lines must meet the same boundary conditions. Therefore, each branch experiences the same head loss. Continuity at the junctions requires that the total inflow equals total outflow, so total discharge splits among branches and sums to the supply/exit flow. Step-by-Step Solution: 1) Apply energy between common nodes for each branch: H_in − H_out = h_f,branch. The left and right nodes are the same for all branches, so each branch has identical head loss.2) Apply continuity at a node: Sum of branch discharges = system discharge.3) Conclude: Statements (A) and (B) are both true; statement (C) describes series systems, not parallel. Verification / Alternative check: Solving a two-branch example using Darcy–Weisbach with different diameters shows different velocities and individual Q values, yet identical head losses; adding the Q values gives the total system discharge. Why Other Options Are Wrong: Total head loss is the sum...: This applies to series, not parallel. None of the above: Incorrect because (A) and (B) are established principles. Common Pitfalls: Confusing series vs parallel rules; forgetting that minor losses per branch must also match the same overall drop between nodes. Final Answer: Both (A) and (B)

Question 11

Compressible Flow – Definition of Stagnation Point In a compressible flow field, any point where the local fluid velocity reduces to zero relative to the body (for example, at the nose of a blunt object) is known as the stagnation point.

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Introduction: Flow nomenclature distinguishes special points that control pressure and temperature fields. The stagnation point is central in aerodynamics and turbomachinery because it ties measured stagnation (total) properties to local flow behavior. Given Data / Assumptions: Steady compressible flow past a body. Reference frame attached to the body; local speed may fall to zero. Negligible viscous heating in the free stream. Concept / Approach: At a stagnation point, velocity V = 0, so static pressure attains a local maximum on the surface (stagnation pressure for inviscid assumptions). Total enthalpy relates stagnation temperature and static temperature via standard compressible relationships along streamlines absent losses. Step-by-Step Solution: 1) Identify point with V → 0 on the body surface or symmetry line.2) Recognize that, by Bernoulli-type relations for compressible flow (with appropriate assumptions), total pressure converts to static pressure at V = 0.3) Conclude the correct term is “stagnation point”. Verification / Alternative check: Pressure taps at a sphere’s front face register the highest static pressure where the local velocity is zero, confirming the stagnation-point definition. Why Other Options Are Wrong: Critical point: Refers to Mach 1 condition or mathematical criticality, not necessarily V = 0. Vena contracta: Minimum jet section downstream of an orifice. None of these: Incorrect since the standard term exists. Sonic point: Where M = 1, not V = 0. Common Pitfalls: Confusing stagnation with sonic conditions; ignoring viscous effects near walls that modify but do not change the V = 0 definition at the exact stagnation point. Final Answer: stagnation point

Question 12

Collar Bearing – Torque to Overcome Viscous Shear (Annular Film) For a collar (annular) bearing with outer radius R1 and inner radius R2, lubricant viscosity mu, angular speed omega, and film thickness h, the viscous torque resisting rotation is T =…

Accepted Answer

Introduction: Viscous losses in collar bearings influence heating and power requirements. The torque expression is derived by summing shear contributions from concentric annular rings across the lubricated area. Given Data / Assumptions: Annular contact, inner radius R2, outer radius R1. Newtonian lubricant of viscosity mu, uniform film thickness h. No slip at boundaries; rigid-body rotation at angular speed omega. Concept / Approach: Local tangential velocity at radius r is v = omega * r. Shear rate across film is dv/dy = v / h = (omega * r) / h; shear stress tau = mu * (omega * r) / h. Differential torque is shear force times radius integrated over the annulus. Step-by-Step Solution: 1) Ring area dA = 2 * pi * r * dr.2) Shear stress tau = mu * (omega * r) / h.3) Shear force dF = tau * dA = 2 * pi * mu * omega / h * r^2 * dr.4) Elemental torque dT = dF * r = 2 * pi * mu * omega / h * r^3 * dr.5) Integrate r from R2 to R1: T = (pi * mu * omega / (2 * h)) * (R1^4 − R2^4).6) Use P = T * omega for viscous power loss if required. Verification / Alternative check: Dimensional analysis: [mu] Ns/m^2, multiply by omega/h (1/s / m) and R^4 (m^4) gives Nm, confirming torque units. Why Other Options Are Wrong: Option B uses area moment (R^2) rather than the correct r^3 integrand. Option C adds an extra omega factor (power, not torque). Option D uses a sum instead of the difference in fourth powers. Option E has wrong radius exponent and missing constants. Common Pitfalls: Assuming uniform shear independent of r; forgetting that torque weighting introduces r^3 in the integral; mixing footstep (solid disk) and collar (annulus) formulas. Final Answer: T = (pi * mu * omega / (2 * h)) * (R1^4 − R2^4)

Question 13

Instrumentation – What a Barometer Measures A barometer is an instrument that measures the atmospheric pressure (the pressure exerted by the weight of the air column above the measurement point).

Accepted Answer

Introduction: Understanding pressure measurement devices prevents misapplication. Barometers quantify ambient atmospheric pressure, which serves as a reference for gauge instruments and weather analysis. Given Data / Assumptions: Standard mercury or aneroid barometer exposed to the atmosphere. Static conditions (no dynamic head components). Calibration to units such as Pa, kPa, or mm Hg. Concept / Approach: Atmospheric pressure is the force per unit area due to the air column. Barometers either balance this pressure with a mercury column (mercury barometer) or translate it via elastic elements (aneroid). Pipe pressure or differential pressure in internal flows require other devices like manometers or gauges. Step-by-Step Solution: 1) Identify the measured quantity: ambient air pressure.2) Recognize barometer types and outputs.3) Conclude: a barometer measures atmospheric pressure. Verification / Alternative check: A mercury barometer reading plus local gravity corrections gives standard atmospheric pressure values comparable to weather-station data. Why Other Options Are Wrong: Velocity and pipe/channel pressure require pitot tubes or pressure gauges. Differential pressure between two points uses a manometer or differential transmitter. Absolute vacuum is a limiting concept, not directly read by a standard barometer. Common Pitfalls: Confusing atmospheric (absolute) pressure with gauge pressure; ignoring altitude and temperature corrections for precise measurements. Final Answer: atmospheric pressure

Question 14

Jet Power from a Nozzle – Optimal Pipe Friction Loss For a nozzle fed by a long pipe from a supply head, the power delivered by the exit jet is maximum when the head lost due to pipe friction equals one-third of the total supply head.

Accepted Answer

Introduction: Maximizing jet power at a nozzle connected to a reservoir through a pipe involves trading off discharge (which increases with frictional drop) against delivery head (which decreases as losses rise). The classical result identifies the best split for maximum power output. Given Data / Assumptions: Total supply head H available at pipe inlet. Head loss due to friction h_f along the pipe. Jet exit head H_d = H − h_f; jet power proportional to Q * H_d. Concept / Approach: Under typical assumptions, Q increases with h_f (since more head is spent overcoming friction, velocity and Q adjust). Optimizing P = rho * g * Q * (H − h_f) with respect to h_f yields the one-third rule for maximum power transfer to the jet. Step-by-Step Solution: 1) Express P ∝ Q * (H − h_f).2) For a given line, Q ∝ h_f^(1/2) (from head loss proportional to V^2).3) Write P ∝ h_f^(1/2) * (H − h_f) and set derivative to zero.4) Solve to get optimal h_f = H / 3, so H_d = 2H / 3. Verification / Alternative check: Substitute h_f = H/3 and check that small deviations reduce P on both sides (second-derivative negative), confirming a maximum. Why Other Options Are Wrong: One-half, two-third, one-fourth, three-fourths split do not satisfy the optimum condition and yield lower power. Common Pitfalls: Confusing efficiency with maximum power; ignoring nozzle losses or minor losses, which shift numerical details but not the qualitative one-third insight. Final Answer: one-third

Question 15

Rotating Liquid – Pressure Rise at Drum Rim For a drum of radius r completely filled with a liquid of density rho and rotating at angular speed omega (rad/s), the increase in pressure between the center and the outer edge equals Δp = (1/2) * rho * o…

Accepted Answer

Introduction: Liquids in rigid-body rotation develop a radial pressure gradient that increases outward. Knowing the exact pressure rise is vital for centrifuges, rotating tanks, and turbomachinery cavities. Given Data / Assumptions: Rigid-body rotation at constant omega. Incompressible Newtonian liquid; no free surface curvature effects on pressure difference along a radius. Neglect gravity along the radial balance. Concept / Approach: Radial equilibrium in the rotating frame gives dp/dr = rho * omega^2 * r. Integrating from r = 0 to r = R yields Δp = (1/2) * rho * omega^2 * R^2. This is independent of axial position for a completely filled drum (no free surface pressure variation along a radius). Step-by-Step Solution: 1) Write equilibrium: dp/dr = rho * omega^2 * r.2) Integrate: p(R) − p(0) = rho * omega^2 * ∫ r dr = (1/2) * rho * omega^2 * R^2.3) Therefore the pressure rise to the rim is (1/2) * rho * omega^2 * r^2. Verification / Alternative check: Experiments with manometer taps at different radii confirm quadratic pressure variation with radius in rigid-body rotation. Why Other Options Are Wrong: rho * omega^2 * r^2 and 2 * rho * omega^2 * r^2 overestimate the integral. Terms linear in omega or missing squared dependence are dimensionally inconsistent for pressure. Common Pitfalls: Mixing this result with the free-surface parabolic shape relation z = (omega^2 * r^2) / (2g); both share the 1/2 factor but represent different phenomena. Final Answer: (1/2) * rho * omega^2 * r^2

Question 16

Unsteady Tank Draining – Time to Lower Level from H1 to H2 A tank of uniform cross-sectional area A with a small orifice of area a at the bottom drains under gravity. The time required to drop the liquid level from H1 to H2 (including discharge coef…

Accepted Answer

Introduction: Predicting drain-down time is a classic unsteady-flow problem. The solution uses Bernoulli’s principle with Torricelli’s efflux velocity and the continuity equation for a falling head in a reservoir-orifice configuration. Given Data / Assumptions: Tank cross-section A; small orifice area a at bottom. Coefficient of discharge Cd accounts for contraction and velocity effects. Inviscid energy assumptions aside from Cd; negligible velocity of the tank free surface. Concept / Approach: Torricelli’s result gives exit speed V = Cd * sqrt(2 * g * H), so discharge Q = a * V = Cd * a * sqrt(2 * g * H). The falling head yields A * dH/dt = −Q. Separating variables produces an integral in 1/sqrt(H). Step-by-Step Solution: 1) Write A * dH/dt = −Cd * a * sqrt(2 * g * H).2) Rearrange: dH / sqrt(H) = −(Cd * a * sqrt(2 * g) / A) * dt.3) Integrate H from H1 to H2 and t from 0 to t.4) Use ∫ dH / sqrt(H) = 2 (sqrt(H2) − sqrt(H1)).5) Solve for t to get t = (2 * A / (Cd * a * sqrt(2 * g))) * (sqrt(H1) − sqrt(H2)). Verification / Alternative check: Setting H2 = 0 gives complete emptying time, matching standard textbook expressions. Dimensional check confirms seconds for t. Why Other Options Are Wrong: Linear dependence on H (Option B, D) contradicts the square-root head relation. Logarithmic form (Option C) suits viscous laminar draining in tubes, not orifices. Option E has incorrect dependence on H and g. Common Pitfalls: Forgetting Cd; using A ≈ constant assumption where it is invalid (e.g., conical tanks); neglecting approach velocity corrections for large a/A. Final Answer: t = (2 * A / (Cd * a * sqrt(2 * g))) * (sqrt(H1) − sqrt(H2))

Question 17

Forces on an Inclined Plate in a Stream – Naming the Components When a flat plate is immersed at an angle to a liquid stream, the resultant pressure on the plate can be decomposed into components. The component along the direction of flow is called …

Accepted Answer

Introduction: External flows over bodies generate forces that engineers decompose into lift (normal to flow) and drag (along flow). Correct terminology is crucial in aerodynamics, hydraulic structures, and particulate transport. Given Data / Assumptions: Uniform approach flow over a thin plate at angle of attack. Reynolds number sufficiently high to permit boundary-layer formation. Pressure and viscous shear both contribute to resultant force. Concept / Approach: The resultant hydrodynamic force is the vector sum of pressure (form) drag and shear (skin-friction) forces. Its decomposition into streamwise (drag) and normal (lift) components defines performance metrics like drag coefficient and lift coefficient. Step-by-Step Solution: 1) Compute distributed pressure and shear over the plate.2) Integrate to obtain resultant force vector.3) Resolve along the flow direction: this component is drag; resolve normal: this is lift. Verification / Alternative check: Wind-tunnel balances directly report drag and lift components relative to the free-stream direction, validating the definitions. Why Other Options Are Wrong: Lift: By definition the normal component, not along the flow. Stagnation pressure: A local pressure at zero relative velocity point, not a net force component. Bulk modulus: Material property unrelated to external flow force decomposition. Skin friction only: Drag includes both pressure and shear contributions. Common Pitfalls: Assuming drag arises only from viscosity (skin-friction) and ignoring pressure (form) drag; misaligning axes so components are mislabeled. Final Answer: drag

Question 18

Temperature Effect – Dynamic Viscosity of Liquids For most common liquids, the dynamic viscosity decreases as temperature rises, because increased molecular activity reduces resistance to shear.

Accepted Answer

Introduction: Designers must account for temperature-dependent viscosity when sizing pumps, estimating losses, and selecting lubricants. Liquids and gases behave oppositely with temperature: liquids thin out, gases thicken slightly. Given Data / Assumptions: Newtonian liquids (water, oils) over normal engineering temperature ranges. No phase change or chemical reaction. Pressure effects modest compared to temperature effects. Concept / Approach: Viscosity reflects internal friction between fluid layers. As temperature increases, liquid molecules gain energy, weakening intermolecular cohesion that impedes sliding, so dynamic viscosity mu drops. Empirical correlations (e.g., Andrade equation) capture this strong inverse relation. Step-by-Step Solution: 1) Recognize liquid microstructure: cohesive forces dominate.2) Increasing temperature disrupts cohesive networks.3) Result: lower shear stress for the same shear rate → decreased mu. Verification / Alternative check: Viscosity-temperature charts for oils show order-of-magnitude drops from ambient to elevated temperatures; water viscosity roughly halves between 20°C and 50°C. Why Other Options Are Wrong: Unaffected: Contradicts data. Increases: Applies to gases, not liquids. Other options mix unrelated effects or oversimplify complex dependencies. Common Pitfalls: Applying gas trends to liquids; neglecting that kinematic viscosity also changes because density varies mildly with temperature. Final Answer: decreases

Question 19

Open-Channel Regime at Critical Depth – True or False? At the critical depth, the flow is termed critical (Froude number exactly 1). Torrential or supercritical flow corresponds to depths less than critical (Froude number greater than 1).

Accepted Answer

Introduction: Terminology in open-channel hydraulics distinguishes critical, subcritical (tranquil), and supercritical (torrential) regimes by the Froude number Fr = V / sqrt(g * y_h). Mislabeling regimes leads to wrong control assumptions in channel design. Given Data / Assumptions: Prismatic channel; hydraulic depth y_h applicable. Uniform gravitational field; negligible compressibility. Standard regime thresholds. Concept / Approach: Critical flow occurs at Fr = 1 and defines the critical depth for a given discharge and section. Supercritical (torrential) flow occurs when Fr > 1 (shallower, faster), while subcritical (tranquil) occurs when Fr < 1 (deeper, slower). Therefore the statement “flow at critical depth is torrential” is false. Step-by-Step Solution: 1) Define Fr = V / sqrt(g * y_h).2) Identify regimes: Fr < 1 subcritical; Fr = 1 critical; Fr > 1 supercritical (torrential).3) At critical depth, by definition, Fr = 1 → critical, not torrential. Verification / Alternative check: Specific-energy curves show minimum specific energy at critical depth; slopes of gradually varied profiles change character at this point, reinforcing that it is a distinct regime. Why Other Options Are Wrong: Yes: Contradicts the definition. Conditions on slope or Reynolds number do not redefine regime names. Depth and discharge jointly set Fr; regime names depend on Fr, not discharge alone. Common Pitfalls: Confusing “torrential” with any high-energy state; mixing Reynolds-number criteria (internal laminar/turbulent) with Froude-number criteria (open-channel regime). Final Answer: No

Question 20

Hydrostatics — Converting Pressure to Water Column Height What is the height of a static water column (in metres) that is equivalent to a pressure of 0.15 MPa? Use water's specific weight w ≈ 9.81 kN/m^3 and the relation p = w * h.

Accepted Answer

Introduction: Converting a given pressure into an equivalent head of liquid is a standard hydrostatics task used in civil engineering (e.g., water supply, reservoir levels, manometer calibration). The key relation is p = w * h, where p is pressure, w is the liquid's specific weight, and h is the vertical depth (or head). Given Data / Assumptions: Pressure p = 0.15 MPa = 150 kPa = 150 kN/m^2. Specific weight of water w ≈ 9.81 kN/m^3. Hydrostatic condition (fluid at rest), uniform w, gravity standard. Concept / Approach: The hydrostatic relation p = w * h directly connects pressure to fluid column height. Rearranging gives h = p / w. With consistent units (kN/m^2 divided by kN/m^3), the result is in metres. This linear relationship is foundational for pressure head, piezometric head, and energy grade line concepts. Step-by-Step Solution: Convert pressure: 0.15 MPa = 150 kN/m^2.Use h = p / w = 150 / 9.81 (m).Compute: 150 / 9.81 ≈ 15.29 m, typically rounded to 15.3 m. Verification / Alternative check: Sanity check using 10 kN/m^3 as an easy round value for w gives 150/10 = 15 m. The precise value with 9.81 kN/m^3 is slightly larger (≈15.29 m), which matches the detailed calculation. Why Other Options Are Wrong: 25.3 m, 35.3 m, and 45.3 m would require much larger pressures (≈0.25–0.45 MPa) for water; they do not satisfy p = w * h for 0.15 MPa. Common Pitfalls: Mixing density (kg/m^3) with specific weight (kN/m^3); forgetting unit conversions between MPa, kPa, and kN/m^2; using oblique depth instead of vertical depth. Final Answer: 15.3 m

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