Question 1

Submerged Body Stability – Relative Positions of Centers A fully submerged body is in stable equilibrium if its center of gravity lies below its center of buoyancy, so a small angular displacement produces a righting moment.

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Introduction:Stability of submerged bodies is governed by how the weight and buoyant force lines of action separate when the body is tilted. Understanding the roles of the center of gravity (G) and the center of buoyancy (B) is fundamental in naval architecture and underwater vehicle design.Given Data / Assumptions:Rigid body completely submerged in a static liquid.Small angular displacements only (initial stability).Liquid is homogeneous and at rest.Concept / Approach:For a fully submerged body of fixed volume, the point about which the buoyant force effectively acts for small tilts is at B (the metacenter coincides with B, unlike surface-piercing bodies). When the body is disturbed, if G is below B, the restoring couple brings the body back (stable). If G is above B, the couple is overturning (unstable). If G coincides with B, the body is neutrally stable.Step-by-Step Solution:1) Weight W acts vertically downward through G.2) Buoyant force equals displaced liquid weight and acts upward through B.3) On tilting, the line of action of buoyancy shifts to remain through B in the rotated configuration.4) If G is below B, the two forces form a restoring couple, giving stable equilibrium.Verification / Alternative check:Consider a dense sphere fully submerged: adding ballast to lower G increases stability because the lever arm for the restoring moment increases when G is beneath B.Why Other Options Are Wrong:Coincides with: Neutral, not stable. Lies above: Produces an overturning moment (unstable). Same vertical level as: Also neutral stability. Outside the fluid: Not applicable for a submerged body.Common Pitfalls:Confusing submerged bodies with floating bodies where the metacenter M differs from B; assuming stability depends on mass alone rather than geometry and G-B separation.Final Answer:lies below

Question 2

Fluid Behavior – Viscosity Independent of Shear Rate A fluid whose viscosity remains constant regardless of the rate of shear is termed a Newtonian fluid. Water, air, and light oils are common examples.

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Introduction:Classifying fluids by how shear stress relates to rate of deformation is crucial for selecting correct constitutive equations. In many engineering calculations, assuming Newtonian behavior simplifies momentum and energy analysis.Given Data / Assumptions:Steady, simple shear in a continuum fluid.Newtonian law of viscosity: tau = mu * (du/dy).Viscosity mu is constant for given temperature and pressure.Concept / Approach:In Newtonian fluids, shear stress is directly proportional to shear rate with a proportionality constant mu that does not depend on the shear rate itself. In contrast, non-Newtonian fluids have mu_eff that varies with shear rate, time, or both (shear thinning, shear thickening, viscoelasticity).Step-by-Step Solution:1) Define Newtonian behavior: tau proportional to du/dy with constant mu.2) Evaluate examples: water, air, and many oils behave approximately Newtonian over wide shear-rate ranges.3) Therefore the correct classification is Newtonian fluid.Verification / Alternative check:Rheometer data for simple liquids show linear tau versus shear-rate plots passing through the origin with slope mu independent of shear rate, confirming Newtonian behavior.Why Other Options Are Wrong:Real fluid: Generic term not tied to shear-rate independence. Ideal fluid: Zero viscosity, not a physical fluid. Non-Newtonian: Viscosity changes with shear rate. Plastic fluid: Requires yield stress before flow.Common Pitfalls:Equating temperature dependence of viscosity with non-Newtonian behavior; confusing apparent viscosity changes due to turbulence with constitutive properties.Final Answer:newtonian fluid

Question 3

Footstep Bearing – Power Absorbed by Viscous Resistance For a flat circular footstep bearing of radius R rotating at angular speed omega with oil film thickness h and viscosity mu, the power absorbed in viscous shear can be computed from the viscous…

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Introduction:Footstep bearings support axial loads using a thin lubricant film between a stationary base and a rotating circular face. Estimating viscous power loss is essential for heat generation calculations and lubricant selection in machine design.Given Data / Assumptions:Flat circular disk of radius R rotates at angular speed omega over a stationary surface.Uniform lubricant film thickness h and Newtonian lubricant with viscosity mu.No slip at boundaries; velocity varies linearly across the film.Concept / Approach:At radius r, the circumferential velocity is v = omega * r. The shear rate across the film is dv/dy = v / h = (omega * r) / h, giving local shear stress tau = mu * (omega * r) / h. The viscous shear produces a torque dT on ring area at radius r which integrates over the disk.Step-by-Step Solution:1) For an annular ring of radius r and width dr, area dA = 2 * pi * r * dr.2) Shear stress tau = mu * (omega * r) / h.3) Shear force dF = tau * dA = mu * (omega * r) / h * 2 * pi * r * dr.4) Torque element dT = dF * r = 2 * pi * mu * omega / h * r^3 * dr.5) Integrate from r = 0 to R: T = integral(0..R) 2 * pi * mu * omega / h * r^3 dr = (pi * mu * omega * R^4) / (2 * h).6) Power absorbed P = T * omega = (pi * mu * omega^2 * R^4) / (2 * h).Verification / Alternative check:Dimensional check: mu has Ns/m^2, multiply by omega^2 (1/s^2) and R^4/h (m^4/m = m^3) to get Nm/s = W, confirming a power expression.Why Other Options Are Wrong:Option B: Missing one omega factor and wrong power dependence. Option C: Excess power of omega and radius. Option D: Missing geometric constants and radius dependence incorrect. Option E: Omits omega entirely, cannot be power loss for rotating shear.Common Pitfalls:Using mean radius incorrectly; forgetting the linear velocity distribution with r; mixing collar bearing friction models with dry friction instead of viscous shear.Final Answer:P = (pi * mu * omega^2 * R^4) / (2 * h)

Question 4

Hydrostatics – Total Force on an Inclined Plane Surface For a plane area A immersed in a liquid and inclined at angle theta to the free surface, the total hydrostatic pressure equals w * A * x_cg, where w is specific weight and x_cg is the vertical …

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Introduction:Determining the resultant hydrostatic force on plane surfaces is basic to gate, dam, and tank design. The magnitude depends on the area, specific weight, and the centroidal depth, not directly on the surface orientation for the force magnitude itself.Given Data / Assumptions:Liquid at rest with specific weight w.Plane surface area A inclined at angle theta to the free surface.Centroid located at vertical depth x below the free surface.Concept / Approach:Pressure varies linearly with depth: p = w * z. The resultant force equals the integral of pressure over area, which reduces to F = w * A * x_cg, where x_cg is the centroid depth measured vertically. Orientation affects center of pressure location but not the magnitude expression.Step-by-Step Solution:1) Write elemental force dF = p * dA = w * z * dA.2) Integrate over the surface: F = w * integral(z dA).3) By definition of centroid depth x_cg, integral(z dA) = A * x_cg.4) Therefore F = w * A * x_cg = w * A * x.Verification / Alternative check:For a vertical rectangular gate, x is the depth of its centroid; substituting gives the standard textbook expression and matches experimental measurements of total force.Why Other Options Are Wrong:wA: Ignores depth variation of pressure. wx: Misses area dimension. wAx/sin theta: Angle does not enter the force magnitude in this form; it affects the center of pressure. wAcos theta: Not dimensionally consistent for force without x.Common Pitfalls:Measuring x along the plate instead of vertically; confusing the force magnitude with the line of action location (which does depend on theta).Final Answer:wAx

Question 5

Flow Regime – Definition of Non-Uniform Flow A flow in a pipe or channel is called non-uniform when the mean velocity changes from section to section along the direction of flow, even if discharge is steady in time.

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Introduction:Engineers distinguish spatial variation from temporal variation when classifying flows. Recognizing non-uniform flow is key to choosing the right governing equations and approximations for gradually varied or rapidly varied conditions.Given Data / Assumptions:Single-phase liquid in a duct or open channel.Focus on spatial variation along the streamline direction.Time variation may or may not be present.Concept / Approach:Uniform flow means velocity is the same at all cross sections along the flow direction (though it can vary within a section). Non-uniform flow means the sectional mean velocity varies with position. Steady flow concerns variation with time at a fixed location. The two concepts are independent.Step-by-Step Solution:1) Define uniform flow: V(x) constant along length.2) If V changes with x, the flow is non-uniform.3) Therefore the correct statement is that velocities at different sections differ.Verification / Alternative check:A gradually varied backwater curve in an open channel is steady in time but non-uniform in space; measured depth and velocity change along the reach.Why Other Options Are Wrong:Option A: Describes uniform flow. Option C: Constant discharge defines steady flow, not spatial uniformity. Option D: Definite path is a streamline concept, not a uniformity criterion. Option E: Equal pressure at all sections is rarely true and unrelated to uniformity.Common Pitfalls:Equating steady with uniform; ignoring that internal velocity profiles can be non-uniform even when the mean is spatially uniform.Final Answer:the liquid particles at different sections have different velocities

Question 6

Venturi Flume – What It Measures in Open Channels A Venturi flume is an open-channel primary device that accelerates flow through a throat to relate water level to discharge using calibrated curves.

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Introduction:Venturi flumes are widely used for flow measurement in irrigation canals, wastewater channels, and environmental monitoring. They produce a known head-discharge relationship by constricting the cross section and monitoring upstream and throat depths.Given Data / Assumptions:Free-surface open-channel flow under gravity.Properly installed Venturi flume with standard geometry.Calibrated relationship between measured head(s) and discharge.Concept / Approach:By continuity and energy principles, accelerating flow through a throat changes depth and velocity in a predictable way. Empirical or semi-theoretical calibration provides Q as a function of measured head. Unlike Venturi meters for pressurized pipes, flumes operate with a free surface and measure discharge directly from level readings.Step-by-Step Solution:1) Constrict the channel with a Venturi flume geometry.2) Measure specified head locations (for example, upstream and at the throat).3) Use the device calibration to compute Q from head reading(s).4) Report discharge for the channel.Verification / Alternative check:Device manuals provide rating tables or equations. Field spot checks with current meters or dilution gaging confirm the flume discharge within stated accuracy when installation is correct.Why Other Options Are Wrong:Pressure measurement options relate to pressurized conduits or differential devices, not free-surface flumes. Fluid density is not measured by a Venturi flume.Common Pitfalls:Improper submergence conditions invalidating the free-flow equation; incorrect staff-gage locations; sediment deposition altering geometry.Final Answer:discharge of liquid

Question 7

Pressure Scales – What a Gauge Indicates The pressure read directly on a mechanical or electronic pressure gauge equals gauge pressure, which is pressure above or below the local atmospheric reference.

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Introduction:Many industrial instruments report pressure relative to ambient air. Correctly interpreting gauge, absolute, and vacuum readings avoids errors in design calculations and safety assessments.Given Data / Assumptions:Standard pressure gauge calibrated to read zero when exposed to local atmosphere.Local atmospheric pressure serves as reference.Positive readings indicate above atmospheric; negative readings indicate vacuum relative to atmosphere.Concept / Approach:Absolute pressure p_abs relates to gauge pressure p_g and atmospheric pressure p_atm by p_abs = p_g + p_atm. Gauges therefore do not directly provide absolute values unless specifically designed as absolute transducers with a sealed reference vacuum.Step-by-Step Solution:1) Identify instrument type: standard gauge zeroed at atmosphere.2) Recognize display equals p_g = p_abs - p_atm.3) Conclude the measurement reported is gauge pressure.Verification / Alternative check:When a gauge is disconnected and vented, it reads zero. However, the absolute pressure at that moment equals p_atm, confirming the relationship p_abs = p_g + p_atm.Why Other Options Are Wrong:Atmospheric pressure: This is the reference, not the measurement. Absolute pressure: Requires vacuum reference sensor. Mean pressure and vapor pressure: Not what a general pressure gauge indicates.Common Pitfalls:Using gauge readings in gas law calculations without adding p_atm; misreading compound gauges that can show vacuum in negative units.Final Answer:gauge pressure

Question 8

Transition in Internal Flow – Definition of Critical Velocity The critical velocity is the characteristic mean velocity at which internal flow in a duct shifts from laminar to turbulent, commonly identified using a threshold Reynolds number.

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Introduction:Internal flow regimes are classified using Reynolds number Re = (rho * V * D) / mu. The velocity associated with the onset of turbulence is often referred to as the critical velocity, important for selecting correlations for head loss and heat transfer.Given Data / Assumptions:Newtonian fluid in a circular pipe under steady conditions.Onset of transition defined by empirical threshold of Re (about 2000 for smooth pipes).Mean bulk velocity V relates to Re at given diameter and properties.Concept / Approach:For specified fluid properties and pipe diameter, the velocity at which Re reaches the threshold is calculated and termed critical velocity. Below it, viscous forces dominate (laminar). Above it, inertial instabilities amplify, leading to turbulence.Step-by-Step Solution:1) Define Re = (rho * V * D) / mu.2) Set Re to transitional value (about 2000) for smooth pipes.3) Solve for V to obtain the critical velocity for the given fluid and diameter.Verification / Alternative check:Laboratory pipe-flow experiments show parabolic velocity profiles and Hagen-Poiseuille law validity below the critical value, with breakdown to flatter turbulent profiles above it.Why Other Options Are Wrong:Velocity of approach: Used in weir and orifice corrections, not regime transition. Sub-sonic and super-sonic velocities concern compressible gas dynamics, not laminar-turbulent transition in incompressible internal flow. Drift velocity: Not a standard fluid mechanics term for this context.Common Pitfalls:Treating the critical value as universal and exact; neglecting roughness, inlet disturbances, and measurement uncertainty that shift the transition range.Final Answer:critical velocity

Question 9

Compressible Flow Regimes – Hypersonic Definition by Mach Number A flow is termed hypersonic when the Mach number is significantly greater than supersonic values; in engineering practice hypersonic usually means Mach number greater than 5.

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Introduction:Compressible flow regimes are categorized by the Mach number M = V / a, where V is flow speed and a is local speed of sound. Hypersonic flows exhibit strong compressibility effects, high kinetic temperature rise, and additional phenomena such as real-gas and high-temperature chemistry effects.Given Data / Assumptions:Standard atmosphere and ideal-gas baseline for defining sound speed.Regime boundaries are conventional engineering thresholds.Focus on external aerodynamics parlance.Concept / Approach:Typical regime names: subsonic (M 1), and hypersonic (commonly M > 5). As M increases, shock waves strengthen, boundary layers heat up, and dissociation or ionization may occur at very high speeds.Step-by-Step Solution:1) Identify the hypersonic threshold used in engineering: M > 5.2) Compare choices to this definition.3) Select the option specifying Mach number greater than 5.Verification / Alternative check:Standard aerospace references and design guides for reentry vehicles and scramjet concepts adopt M > 5 as the working definition for hypersonic, distinguishing it from ordinary supersonic flight.Why Other Options Are Wrong:Less than 1: Subsonic. Equal to 1: Sonic. Between 1 and 5: Supersonic but not hypersonic. Between 0.3 and 0.8: Incompressible to subsonic range.Common Pitfalls:Assuming a sharp universal boundary without context; forgetting that local Mach numbers on a body can differ from freestream values due to acceleration over surfaces.Final Answer:Mach number greater than 5

Question 10

Fluid Mechanics — Definition of Reynolds Number In civil and mechanical engineering fluid dynamics, the Reynolds number (Re) is defined as the ratio of the inertia force to which other characteristic force acting within the fluid? Choose the most ap…

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Introduction:The Reynolds number is a fundamental dimensionless quantity used to predict flow regimes—laminar, transitional, or turbulent—by comparing the dominance of inertia effects to viscous effects in a flowing fluid. It guides pipe design, open-channel hydraulics, and external aerodynamics.Given Data / Assumptions:Steady, incompressible flow is assumed for basic interpretation.Characteristic length L and velocity V are identifiable (e.g., pipe diameter and average speed).Fluid properties density rho and dynamic viscosity mu are uniform.Concept / Approach:By definition, Reynolds number Re compares inertia force to viscous force. In engineering form, Re = (rho * V * L) / mu. High Re means inertia dominates and turbulence is more likely; low Re means viscosity dominates and laminar behavior persists.Step-by-Step Solution:Start with dimensional analysis of forces: inertia force ~ rho * V^2 * A, viscous force ~ mu * V * A / L.Form the ratio: (rho * V^2 * A) / (mu * V * A / L) = (rho * V * L) / mu.Thus Re = (rho * V * L) / mu = inertia force / viscous force.Verification / Alternative check:In pipe flow, thresholds around Re ≈ 2000 (laminar) and Re > 4000 (turbulent) are widely used. This practical usage aligns with the inertia/viscosity competition that the definition encodes.Why Other Options Are Wrong:Surface tension and elastic forces are not in the canonical Re definition for internal flow; gravity enters other nondimensional groups (e.g., Froude number), not Reynolds number.Common Pitfalls:Mixing up Re with Froude (inertia/gravity) or Weber (inertia/surface tension); using kinematic viscosity nu incorrectly—remember Re = V * L / nu where nu = mu / rho.Final Answer:viscous force

Question 11

Orifice and Vena Contracta — Coefficient of Contraction A jet of water issues from a 40 mm diameter orifice. At the vena contracta, the jet diameter is measured as 32 mm. Compute the coefficient of contraction (Cc) using area ratio.

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Introduction:When fluid discharges through a sharp-edged orifice, the jet contracts to a minimum cross-section called the vena contracta. The coefficient of contraction (Cc) quantifies the ratio between the jet area at vena contracta and the orifice area, reflecting streamline curvature and energy losses.Given Data / Assumptions:Orifice diameter D = 40 mm.Vena contracta diameter d = 32 mm.Sharp-edged orifice, steady discharge, water at standard conditions.Concept / Approach:The contraction coefficient is Cc = A_vc / A_o. Since areas scale with diameter squared, Cc = (d/D)^2 for circular sections.Step-by-Step Solution:Compute diameter ratio: d/D = 32/40 = 0.8.Compute area ratio: Cc = (0.8)^2 = 0.64.Hence, the coefficient of contraction is 0.64.Verification / Alternative check:Typical Cc for sharp-edged orifices lies around 0.61–0.65. The computed 0.64 is within the expected practical range, supporting the calculation.Why Other Options Are Wrong:0.46 and 0.78 or 0.87 do not match the squared ratio (0.8)^2 and are outside typical sharp-edge values.Common Pitfalls:Using diameter ratio directly instead of squaring; mixing Cc with coefficient of discharge Cd or velocity coefficient Cv. Remember Cd = Cc * Cv.Final Answer:0.64

Question 12

Hydrostatics — Pressure at Depth in a Liquid For a static liquid, the pressure intensity p (in kN/m^2 or kPa) at a depth h below the free surface is given by which expression in terms of specific weight w?

Accepted Answer

Introduction:Hydrostatic pressure increases linearly with depth due to the weight of the overlying fluid. This principle underlies manometers, dams, submarine hulls, and pressure vessels in contact with liquids.Given Data / Assumptions:Liquid at rest (no shear stresses).Constant specific weight w = rho * g (in kN/m^3).Depth measured vertically from the free surface by h (in m).Concept / Approach:The hydrostatic equation is dp/dz = -rho * g. Integrating from the surface (p = 0 gauge) to depth h gives p = rho * g * h = w * h (in consistent units).Step-by-Step Solution:Define w = rho * g.Write pressure-depth relation: p = w * h.Interpretation: pressure increases linearly with h; slope equals w.Verification / Alternative check:Units check: w (kN/m^3) times h (m) yields kN/m^2, which equals kPa. This confirms dimensional consistency and correctness.Why Other Options Are Wrong:Using w alone ignores depth; w/h or h/w have incorrect dimensions and do not represent hydrostatic pressure.Common Pitfalls:Confusing specific weight with specific gravity; forgetting to convert units; using density without multiplying by g; measuring oblique distance rather than vertical depth.Final Answer:w * h

Question 13

Rectangular Notch (Sharp-Crest) — Head–Discharge Relation In a standard sharp-crested rectangular notch, how does discharge Q vary with the head H measured above the crest, ignoring submergence and applying an appropriate coefficient?

Accepted Answer

Introduction:Weirs and notches are primary measuring devices for open-channel flow. For a sharp-crested rectangular notch, the discharge–head relation is a well-known power law that enables simple flow metering with head measurements.Given Data / Assumptions:Free, fully aerated nappe conditions.Velocity of approach correction small or included in Cd.Rectangular, sharp crest; head H measured above crest.Concept / Approach:The theoretical discharge through a strip of width b at depth y is dQ = b * sqrt(2gy) dy. Integrating from 0 to H, Q_theoretical = (2/3) * b * sqrt(2g) * H^(3/2). Real flow uses Q = Cd * (2/3) * b * sqrt(2g) * H^(3/2). Thus Q is proportional to H^(3/2).Step-by-Step Solution:Set up elemental discharge using Torricelli speed sqrt(2gy).Integrate across depth 0 to H to obtain H^(3/2) dependence.Apply discharge coefficient Cd to account for contraction and viscosity.Verification / Alternative check:Empirical calibrations confirm the 3/2 power for sharp-crested rectangular weirs under free overflow. Deviations arise mainly from approach velocity and submergence effects.Why Other Options Are Wrong:Inverses with H are physically incorrect for free overflow; H^(5/2) appears with some combined relations but not the basic rectangular notch formula.Common Pitfalls:Measuring head from the wrong datum; neglecting velocity of approach; using the formula under submergence without correction.Final Answer:directly proportional to H^(3/2)

Question 14

Capillarity in a Narrow Tube — Rise/Fall Height Formula A clean glass tube of small inside diameter d is dipped vertically into a liquid. What is the expression for the capillary rise (or fall) height h in terms of specific weight w, angle of contac…

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Introduction:Capillarity causes liquids to rise or fall in narrow tubes depending on wetting characteristics. This phenomenon is central to soil suction, ink flow in pens, medical diagnostics, and porous media flow.Given Data / Assumptions:Tube is narrow, clean, and circular with internal diameter d.Liquid has surface tension sigma and angle of contact alpha with glass.Specific weight w = rho * g is constant; temperature effects neglected.Concept / Approach:At equilibrium, upward surface tension component along the contact line balances the weight of the liquid column displaced relative to the free surface. For a circular tube, the contact line length equals the tube circumference, and geometry gives the standard expression for h.Step-by-Step Solution:Upward force due to surface tension: F_up = sigma * (perimeter) * cos(alpha) = sigma * (pi * d) * cos(alpha).Weight of liquid column of height h: W = (area) * h * w = (pi * d^2 / 4) * h * w.Set equilibrium F_up = W to obtain: sigma * pi * d * cos(alpha) = (pi * d^2 / 4) * h * w.Solve for h: h = (4 * sigma * cos(alpha)) / (w * d).Verification / Alternative check:For water–glass, alpha ≈ 0, so cos(alpha) ≈ 1 and rise is positive; for mercury–glass, alpha > 90°, cos(alpha) negative, predicting a depression, which matches observations.Why Other Options Are Wrong:The factor 4 is essential from geometry; dividing by d^2 is incorrect because perimeter/area ratio yields 1/d; the reciprocal form in option D inverts the physics.Common Pitfalls:Using surface tension units inconsistently; forgetting that contamination changes alpha; applying the formula to oversized tubes where meniscus curvature is negligible.Final Answer:h = (4 * sigma * cos(alpha)) / (w * d)

Question 15

Pressure Measurement — Can Vacuum Pressure Be Read with a Piezometer? Consider a simple piezometer tube (open to atmosphere at the top). Can it be used to measure vacuum (below-atmospheric) pressure at a tapping point in a pipeline?

Accepted Answer

Introduction:A piezometer is the simplest pressure-measuring device: a vertical transparent tube connected to the point of interest and open to atmosphere at the top. It is suitable for measuring positive gauge pressure head but has limitations for negative gauge (vacuum) pressures.Given Data / Assumptions:Tube is open to atmosphere at the top.Working fluid is incompressible and compatible with the tube.No sealing or auxiliary vacuum equipment is used.Concept / Approach:In a piezometer, the liquid level rises above the tapping if the pressure is above atmospheric (positive gauge). For vacuum at the tapping, the column would need to fall below the connection level, which an open-topped tube cannot display reliably; air would simply enter, and cavitation or separation can occur. Hence piezometers cannot measure vacuum; manometers with a low-vapor-pressure fluid (e.g., mercury) or vacuum gauges are used instead.Step-by-Step Solution:For p > p_atm, height h = (p - p_atm) / (rho * g) is observed as a rise.For p Therefore, vacuum pressure measurement needs alternative arrangements (e.g., U-tube manometer or Bourdon-type vacuum gauge).Verification / Alternative check:Standard textbooks explicitly restrict piezometers to positive gauge measurements and recommend U-tube or inverted manometers for vacuum readings.Why Other Options Are Wrong:Sealing with a pump changes the instrument type; arbitrary small limits (Common Pitfalls:Connecting piezometers where vapor pressure may be approached; assuming any negative reading is meaningful; ignoring capillary effects in very small tubes.Final Answer:False

Question 16

Weir Calibration — Bazin’s Observation on Coefficient of Discharge According to Bazin's experiments on sharp-crested weirs, does the coefficient of discharge (Cd) vary with the height of water (head) over the sill/crest?

Accepted Answer

Introduction:Weir coefficients account for non-idealities such as viscosity, surface tension, and approach velocity. Historical calibrations by researchers like Bazin demonstrated how coefficients depend on head and geometry for sharp-crested weirs.Given Data / Assumptions:Sharp-crested weir under free overflow (fully aerated nappe).Measured head H above crest.Standard laboratory conditions and careful head measurement.Concept / Approach:The discharge relation Q = Cd * (2/3) * b * sqrt(2*g) * H^(3/2) shows Cd as an empirical factor. Experiments indicate Cd is not strictly constant; it varies with H, crest condition, approach velocity, and Reynolds effects. Bazin reported such dependence with head over sill.Step-by-Step Solution:Recognize Q depends on both H and Cd.Empirical data sets show Cd changes with H, especially at small heads where surface tension and approach influence are significant.Thus Bazin’s statement that Cd varies with head is correct.Verification / Alternative check:Modern standards provide tables/curves of Cd vs. H for different crest conditions and end contractions, confirming non-constancy.Why Other Options Are Wrong:Restricting variation only to submerged flow or triangular weirs ignores evidence for rectangular sharp-crested weirs; calling it incorrect contradicts experimental results.Common Pitfalls:Using a single Cd across a wide H range; neglecting velocity-of-approach corrections; ignoring submergence thresholds where different formulas apply.Final Answer:Correct

Question 17

Kinematic Viscosity from Poise and Specific Gravity An oil has specific gravity S = 0.95 and dynamic viscosity mu = 0.011 poise. What is its kinematic viscosity nu in stokes (1 stoke = 1 cm^2/s)?

Accepted Answer

Introduction:In cgs units, dynamic viscosity is measured in poise and kinematic viscosity in stokes. They are related through fluid density. Quick conversions are common in lubrication problems and Reynolds number estimates.Given Data / Assumptions:Specific gravity S = 0.95 ⇒ density rho = 0.95 g/cm^3.Dynamic viscosity mu = 0.011 poise.Temperature effects ignored; single-phase liquid.Concept / Approach:Kinematic viscosity is nu = mu / rho. In cgs, mu in poise and rho in g/cm^3 give nu directly in stokes (cm^2/s).Step-by-Step Solution:Compute rho from specific gravity: rho = S * 1 g/cm^3 = 0.95 g/cm^3.Apply nu = mu / rho = 0.011 / 0.95.Calculate value: 0.0115789… stokes ≈ 0.0116 stoke (rounded to 4 significant figures).Verification / Alternative check:Dimensional check: poise/(g/cm^3) = (g/(cm*s))/(g/cm^3) = cm^2/s = stoke. Magnitude is reasonable for light oils.Why Other Options Are Wrong:0.116 and 0.611 are an order of magnitude larger; 0.0611 is too high for the given numbers; only 0.0116 matches mu/rho.Common Pitfalls:Confusing SI and cgs; dividing by specific gravity incorrectly; rounding too early leading to noticeable error.Final Answer:0.0116 stoke

Question 18

Rheological Classification — Water is Best Described As: In terms of how shear stress relates to rate of shear strain (rheology), water is best classified as which type of fluid?

Accepted Answer

Introduction:Fluids can be classified by rheology (stress–strain-rate behavior) and by idealization versus reality. The question asks specifically for rheological classification, focusing on the relation between shear stress and shear rate.Given Data / Assumptions:Single-phase, pure water at ordinary temperature and pressure.No particulate load or polymer additives.Continuum mechanics assumptions apply.Concept / Approach:A Newtonian fluid obeys tau = mu * (du/dy), where tau is shear stress and du/dy is shear rate, with constant dynamic viscosity mu independent of shear rate. Water closely follows this linear law across wide shear rates, making it Newtonian. An ideal (inviscid) fluid is a mathematical abstraction and not physically realizable for water; non-Newtonian fluids exhibit shear-dependent viscosity or yield stress, which water does not in normal conditions.Step-by-Step Solution:Identify the requested basis: rheology, not idealization.Recall the Newtonian constitutive equation tau = mu * (du/dy) with mu approximately constant for water.Conclude that water is Newtonian.Verification / Alternative check:Experimental viscosity measurements show mu for water varies primarily with temperature, not shear rate, within engineering ranges—consistent with Newtonian behavior.Why Other Options Are Wrong:Real: true but not a rheological category by itself; the question seeks the stress–strain relation.Ideal: inviscid model; water is viscous.Non-Newtonian: would require shear-thinning/thickening or yield behavior absent in pure water.Common Pitfalls:Choosing “real” because it sounds correct; mixing modeling idealizations (ideal vs real) with rheological classes (Newtonian vs non-Newtonian).Final Answer:newtonian

Question 19

Droplet Shape — Why Falling Water Drops Tend to Spheres Falling drops of water tend to assume a spherical shape predominantly due to which physical property of water?

Accepted Answer

Introduction:The equilibrium shape of small liquid droplets results from a competition between interfacial forces and body forces. Understanding which property governs droplet sphericity is important in sprays, inkjet printing, and meteorology.Given Data / Assumptions:Drops are small enough that surface effects dominate over gravity.Ambient air offers negligible viscous resistance to shape at equilibrium.No electric or magnetic fields acting on the droplet.Concept / Approach:Surface tension acts to minimize surface area for a given volume. Among all shapes with the same volume, the sphere has the minimum surface area. Therefore, in the absence of strong external forces, droplets tend toward spherical shapes due to surface tension.Step-by-Step Solution:Recognize that energy of the interface is E = sigma * A, where sigma is surface tension and A is area.For fixed volume, minimizing A minimizes E.A sphere provides the smallest area, hence is energetically favored.Verification / Alternative check:As droplet size increases, gravity distorts the shape, making it oblate; small droplets (e.g., fog, mist) remain nearly spherical—consistent with surface-tension dominance.Why Other Options Are Wrong:Compressibility has negligible effect on shape at normal conditions; capillarity refers to interaction with walls, not free droplets; viscosity resists flow but does not set the equilibrium shape.Common Pitfalls:Assuming large raindrops are spherical—larger drops flatten due to aerodynamic forces; confusing capillarity (wetting) with free-surface tension effects.Final Answer:surface tension of water

Question 20

Statement on coefficient of discharge (Cd) and actual discharge through an orifice: If the value of Cd increases, the discharge through the orifice decreases. State whether this statement is true or false.

Accepted Answer

Introduction / Context:Discharge through a sharp-edged orifice (or nozzle) under a constant head is often estimated using the empirical relation Q = Cd * A * sqrt(2 * g * H). This question checks conceptual understanding of how the coefficient of discharge Cd influences the actual discharge Q when all other quantities remain the same.Given Data / Assumptions:Steady flow through an orifice under head H.Orifice area A is constant.Acceleration due to gravity g is constant.Coefficient of discharge Cd lumps together contraction and velocity effects (Cd = Cc * Cv).Concept / Approach:The discharge formula for an orifice is Q = Cd * A * sqrt(2 * g * H). For fixed A, g, and H, Q is directly proportional to Cd. Therefore, if Cd increases, Q must increase proportionally. The given statement says the opposite and is therefore incorrect.Step-by-Step Solution:Start with Q = Cd * A * sqrt(2 * g * H).Hold A, g, and H constant (same orifice and same head).If Cd increases → Q increases linearly because Q ∝ Cd.Hence, the statement that “discharge decreases” when Cd increases is false.Verification / Alternative check:Consider two orifices of the same size under the same head, one with Cd1 = 0.60 and another with Cd2 = 0.65. The second orifice yields Q2/Q1 = Cd2/Cd1 = 0.65/0.60 ≈ 1.083, that is, roughly 8.3% more discharge, confirming the direct proportionality.Why Other Options Are Wrong:True: Contradicts Q ∝ Cd.Depends on head only: Discharge depends on both head and Cd (and area).Cannot be determined: It can be determined from the standard formula.Common Pitfalls:Confusing Cd with velocity coefficient only; ignoring the role of contraction; assuming Q depends only on H; forgetting that Cd accounts for real flow losses and thus scales the ideal discharge to the actual value.Final Answer:False

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