Question 1

Identify the correct dimension (M, L, T) for dynamic viscosity (absolute viscosity) in SI base dimensions.

Accepted Answer

Introduction / Context:Dimensional analysis is fundamental in fluid mechanics. Dynamic viscosity μ relates shear stress to shear rate in Newton’s law of viscosity, and its correct dimensional form is essential for deriving non-dimensional groups and checking equations for consistency.Given Data / Assumptions:Newtonian constitutive equation: τ = μ * (du/dy).Shear stress τ has dimension of force/area; shear rate has dimension of 1/time.Base dimensions: M (mass), L (length), T (time).Concept / Approach:From τ = μ * (du/dy), rearrange to μ = τ / (du/dy). Determine the dimensions of τ and du/dy, then form μ. This ensures the result is independent of the unit system (SI, CGS) as long as base dimensions are used consistently.Step-by-Step Solution:Shear stress τ → force/area = (M L T^-2) / L^2 = M L^-1 T^-2.Shear rate du/dy → velocity gradient = (L T^-1) / L = T^-1.Dynamic viscosity μ = τ / (du/dy) = (M L^-1 T^-2) / (T^-1) = M L^-1 T^-1.Verification / Alternative check:In SI, μ units are Pa·s = N·s/m^2. Since N = kg·m/s^2, Pa·s = (kg·m/s^2)·s / m^2 = kg/(m·s), which corresponds to M L^-1 T^-1.Why Other Options Are Wrong:L^2 T^-1, L T^-2: Lack mass dimension and do not match μ definitions. M L^-1 T^-2: That is stress, not viscosity.Common Pitfalls:Confusing dynamic viscosity with kinematic viscosity (ν), which has dimensions L^2 T^-1.Final Answer:M L^-1 T^-1

Question 2

Which non-dimensional number primarily characterizes compressibility effects in fluid flow and is used to judge subsonic, transonic, or supersonic regimes?

Accepted Answer

Introduction / Context:Dimensionless numbers classify flow regimes and dominant physics. For compressible flows, identifying the parameter that links local flow speed to acoustic speed is essential for shock prediction, choking, and aerodynamic heating considerations.Given Data / Assumptions:Flow may be subsonic, transonic, or supersonic.Local speed of sound a depends on thermodynamic state.Flow speed V compared to a informs compressibility significance.Concept / Approach:Mach number, M = V/a, is the canonical indicator of compressibility effects. For M ≪ 1, compressibility is negligible; for M ≈ 1, transonic phenomena arise; for M > 1, shocks and expansion waves are prevalent. Other numbers target different physics: Reynolds (inertia/viscous), Weber (inertia/surface tension), Euler (pressure/inertia).Step-by-Step Solution:Define M = V / a.Interpret ranges: M 1 → compressible with shocks.Select the option that matches → Mach number.Verification / Alternative check:Aerodynamics and gas dynamics texts consistently use Mach number as the primary criterion for compressibility-driven behavior and wave phenomena.Why Other Options Are Wrong:Weber: Capillarity vs inertia (multiphase). Euler: Pressure vs inertia, not a direct compressibility gauge. Reynolds: Inertia vs viscosity, governs laminar–turbulent transition.Common Pitfalls:Assuming high Reynolds automatically implies compressibility; confusing high speed with high Reynolds without considering speed of sound.Final Answer:Mach number

Question 3

Pipe schedule number is a conventional index of wall thickness used in process piping. Which expression (using allowable internal working pressure P′ and allowable stress S, both in consistent units) defines the schedule number?

Accepted Answer

Introduction / Context:Pipe “schedule” standardizes wall thickness across nominal sizes. Understanding its definition helps engineers relate design pressure and material allowable stress to expected wall thickness selection and mechanical robustness.Given Data / Assumptions:P′: allowable working pressure (e.g., psi).S: allowable stress of pipe material (same unit basis).Schedule is a non-dimensional index used in ASME/ANSI standards.Concept / Approach:Historically, schedule number ≈ 1000 P′/S (with both P′ and S in psi). Larger schedule indicates thicker wall for a given nominal pipe size. The factor 1000 is conventional for scaling to convenient integers in catalogs (e.g., Sch 40, 80).Step-by-Step Solution:Relate pressure capacity to material strength → thicker wall needed as P′ increases or S decreases.Adopt standard relation → Schedule ≈ 1000 P′/S.Interpret: higher pressure rating or lower allowable stress → higher schedule number.Verification / Alternative check:Pipe tables and design references commonly cite the 1000 P′/S rule-of-thumb for schedule index derivation in legacy sizing frameworks.Why Other Options Are Wrong:100 P′/S or 10000 P′/S: Incorrect scaling. 1000 S/P′: Inverse trend (would imply thinner walls at higher pressure), which is illogical.Common Pitfalls:Assuming schedule directly gives thickness independent of diameter; actual wall thickness also depends on nominal size and schedule per tables.Final Answer:1000 P′ / S

Question 4

Identify the quantity that is NOT dimensional: which of the following is not dimensional (i.e., is a dimensionless parameter)?

Accepted Answer

Introduction / Context:Dimensionless groups are the backbone of similarity analysis, scale-up, and correlation development in fluid mechanics. Recognizing which listed quantities are dimensionless avoids unit mistakes and aids in proper use of charts and correlations.Given Data / Assumptions:Euler number: pressure forces normalized by inertial forces.Specific gravity: density ratio relative to a reference (usually water/air).Fanning friction factor: wall shear normalization in internal flows.Concept / Approach:Each listed quantity is a ratio of like-dimensioned terms and is therefore dimensionless. Hence none of them carry base dimensions (M, L, T). The correct response is that all are dimensionless, so there is no non-dimensional among them to exclude.Step-by-Step Solution:Euler number Eu = Δp / (ρ V^2) → dimensionless.Specific gravity SG = ρ/ρ_ref → dimensionless.Fanning factor f = τ_w / (0.5 ρ V^2) → dimensionless.Verification / Alternative check:Handbooks list these as classic π-groups used in correlating pressure drop, scale-up, and buoyancy/pressure-driven flows.Why Other Options Are Wrong:Since all three are dimensionless, choosing any one as “not dimensionless” would be incorrect; therefore “None of these” is the only correct choice.Common Pitfalls:Confusing “unitless” with “dimensionless”; some dimensionless groups still have numerical values dependent on unit-consistent inputs but no underlying dimensions.Final Answer:None of these

Question 5

Surface tension represents force acting along a line at a liquid interface. What is the correct dimension of surface tension in terms of force F and length L?

Accepted Answer

Introduction / Context:Surface tension appears in capillarity, bubble formation, and atomization. It is physically the force per unit length acting along a line in the surface, or equivalently energy per unit area. Getting its dimensions right is essential for correct use in correlations like Weber number and capillary rise equations.Given Data / Assumptions:Surface tension symbol γ (or σ) with common units N/m (SI).Force denoted by F; length by L.Alternative energy interpretation: J/m^2 also equals N/m.Concept / Approach:By definition, surface tension = force per unit length. Hence its dimensions are force divided by length. This aligns with many interface mechanics derivations and balances at three-phase contact lines.Step-by-Step Solution:Define γ = force / length.Dimensional form → γ ~ F / L = F L^-1.Check with energy view: energy/area = (F·L)/(L^2) = F/L → consistent.Verification / Alternative check:SI unit check: N/m corresponds exactly to the stated dimension. In base dimensions, γ = M T^-2 (since N = M L T^-2, divided by L → M T^-2), which agrees with F L^-1.Why Other Options Are Wrong:F^-1·L, F·L^-2, F^-2·L: These do not match the fundamental definition and lead to inconsistent units in capillarity balances.Common Pitfalls:Confusing surface tension with pressure (F L^-2) or elastic modulus; mixing up with interfacial energy units without the area relation.Final Answer:F L^-1

Question 6

Cavitation in fluid flow occurs under which condition within pumps, valves, or constricted passages?

Accepted Answer

Introduction / Context:Cavitation is the formation and subsequent collapse of vapor bubbles in a liquid, a damaging phenomenon in pumps, turbines, propellers, and throttling devices. Recognizing the precise trigger condition helps in design and troubleshooting.Given Data / Assumptions:Liquid with known vapor pressure at the local temperature.Flow domain includes regions of high velocity or sharp pressure drops (e.g., impeller eye, valve vena contracta).Basic Bernoulli and vapor–liquid equilibrium considerations apply.Concept / Approach:As static pressure falls at constant temperature, once it approaches the liquid’s vapor pressure, nucleation of vapor cavities can occur. These cavities convect into higher-pressure regions and collapse violently, producing noise, vibration, efficiency loss, and surface pitting. Hence designs ensure NPSHa > NPSHr and avoid abrupt pressure drops.Step-by-Step Solution:Identify low-pressure locations (e.g., pump suction, vena contracta).Compare local static pressure p_local to vapor pressure p_vap(T).If p_local ≲ p_vap(T), vapor cavities form → cavitation onset.Mitigation: raise suction head, reduce temperature, lower velocity/ losses, redesign inlets.Verification / Alternative check:Bernoulli reasoning: increases in velocity cause static pressure reductions; if the drop is severe, pressure can approach vapor pressure and trigger cavitation. Field indicators include characteristic crackling noise and reduced pump head.Why Other Options Are Wrong:Velocity decreases to zero: Not causative; low velocity tends to raise static pressure. Total energy decreases: Vague and not the defining condition. Both: Combines two incorrect/irrelevant statements.Common Pitfalls:Using gauge instead of absolute pressure near zero; neglecting temperature dependence of vapor pressure; ignoring suction pipe losses in NPSH calculations.Final Answer:When the local flow pressure approaches the liquid’s vapor pressure at the prevailing temperature

Question 7

In centrifugal pump design, the developed head follows the pump affinity laws. Which basic geometric/operating parameters primarily govern the head produced by a centrifugal pump?

Accepted Answer

Introduction / Context:Centrifugal pump performance is commonly estimated using the affinity (similarity) laws. Understanding which parameters control the developed head helps in scale-up, troubleshooting, and variable-speed control in chemical and mechanical engineering applications.Given Data / Assumptions:Pump type: centrifugal, geometrically similar for comparisons.Fluid properties and suction conditions are unchanged.We consider the idealized relationships from the affinity laws.Concept / Approach:For geometrically similar centrifugal pumps operating within similar hydraulic regimes, the affinity laws state: H ∝ N^2 and H ∝ D^2, where H is head, N is rotational speed, and D is impeller diameter. Flow rate scales as Q ∝ N and Q ∝ D^3; power scales as P ∝ N^3 and P ∝ D^5. Therefore, both speed and impeller diameter directly influence the head developed.Step-by-Step Solution:Use the head–speed relation: H1/H2 = (N1/N2)^2 (constant D).Use the head–diameter relation: H1/H2 = (D1/D2)^2 (constant N).Conclude that changing either N or D (or both) changes H in proportion to the square of the change.Verification / Alternative check:Manufacturer pump curves plotted for different speeds show head varying with N^2. Trimmed impeller curves (reduced D) similarly show head reduction approximately with D^2, confirming the law empirically within practical ranges.Why Other Options Are Wrong:Speed only: incomplete; diameter also governs head.Impeller diameter only: incomplete; speed also governs head.Neither: incorrect, as both parameters control head per similarity laws.Common Pitfalls:Ignoring system curve interactions; even if pump head increases with N^2, the operating point depends on the pipeline/system resistance. Also, extreme extrapolation beyond the tested range can introduce inaccuracies.Final Answer:both (a) and (b)

Question 8

Piping standards and nominal size: For a given nominal pipe size (NPS), which dimensional feature remains the same across different schedules (wall thicknesses)?

Accepted Answer

Introduction / Context:Piping selection often uses nominal pipe size (NPS) with various schedules (e.g., 40, 80). Knowing which dimension stays constant simplifies fittings compatibility, insulation, and support design.Given Data / Assumptions:Standard steel piping to ASME/ANSI dimensional practice.Comparisons at the same NPS but different schedules.Room-temperature geometric considerations only.Concept / Approach:By convention, pipes of a given NPS share a standard outside diameter so that flanges, couplings, and fittings match universally. Changing the schedule alters the wall thickness, which in turn changes the inside diameter (ID). Therefore, OD remains fixed, while ID varies with schedule.Step-by-Step Solution:Recognize that NPS controls OD (e.g., NPS 2 has OD ≈ 60.3 mm).Schedule defines wall (e.g., Sch 40 vs Sch 80 → different t).ID = OD − 2t → varies with schedule; OD unchanged.Verification / Alternative check:Dimensional tables show identical OD for a given NPS across multiple schedules; IDs differ according to wall thickness entries.Why Other Options Are Wrong:Inside diameter: varies with schedule.Wall thickness: schedule changes wall thickness; not constant.None of these: incorrect since OD does remain constant.Common Pitfalls:Confusing tube sizing (often by OD) with pipe sizing (nominal). Also, assuming NPS equals actual ID for larger sizes—only approximately true for some ranges.Final Answer:outside diameter

Question 9

Fluid mechanics concept check: At high Reynolds number (Re ≫ critical), which force balance best characterizes the flow regime?

Accepted Answer

Introduction / Context:Reynolds number (Re = ρVD/μ) compares inertial to viscous forces. Recognizing which forces dominate at different Re values is foundational to predicting laminar, transitional, or turbulent behavior in pipes and around bodies.Given Data / Assumptions:High Re implies large characteristic velocity/length or small viscosity.Flow is fully developed in a representative geometry.No special roughness or compressibility effects dominate.Concept / Approach:Re is proportional to inertial/viscous force ratio. At large Re, inertial effects dominate; viscous stresses are relatively small except within thin boundary layers. This leads to turbulence, eddy formation, and energy cascades, while viscous forces mainly dissipate energy at small scales.Step-by-Step Solution:Start from definition: Re = ρVD/μ.Large Re ⇒ ρVD ≫ μ ⇒ inertial terms dominate the Navier–Stokes momentum balance.Therefore, the qualitative statement is: inertial forces control; viscous forces are less important globally.Verification / Alternative check:Classic pipe-flow behavior: for Re ≫ 4000, turbulent regime with friction factors depending weakly on Re and roughness; viscous sublayer becomes thin relative to the pipe radius.Why Other Options Are Wrong:Options (b) and (c) describe low Re (creeping/laminar) conditions.None of these: incorrect because option (a) correctly states the high Re regime.Common Pitfalls:Assuming viscous forces vanish entirely; they still act within boundary layers and in energy dissipation at small eddies.Final Answer:inertial forces control and viscous forces are unimportant

Question 10

Pipeline transients: What is the primary cause of water hammer in pipelines during rapid operational changes (e.g., sudden valve closures or pump trips)?

Accepted Answer

Introduction / Context:Water hammer (hydraulic shock) is a transient phenomenon causing pressure surges in piping systems. Understanding its cause is essential for designing surge protection, valve sequencing, and safe operating procedures in water and process systems.Given Data / Assumptions:Incompressible liquid approximation with finite wave speed.Events such as sudden valve closure, pump start/stop, or rapid load change.Elastic pipe walls and fluid compressibility allow pressure waves to propagate.Concept / Approach:Water hammer arises when the flow rate changes quickly, imparting a momentum change to the fluid column. The resulting pressure wave travels at the acoustic wave speed, reflecting at boundaries and superimposing to create high transient pressures. The fundamental cause is the rapid change in flow (velocity), not the structural failure itself.Step-by-Step Solution:Apply unsteady momentum: Δp = ρaΔV, where a is wave speed and ΔV is velocity change.Sudden closure/opening → large ΔV over short time → large Δp.These pressure surges can induce fatigue or rupture if unmitigated.Verification / Alternative check:Joukowsky equation: Δp = ρaΔV confirms proportionality of surge to velocity change and wave speed, matching field observations.Why Other Options Are Wrong:Option (a): Describes a consequence (bursting) rather than the cause.Option (c): Too specific; while valve closure is a cause, the general principle is rapid change in flow rate in any form.None of these: Incorrect because (b) correctly states the primary cause.Common Pitfalls:Confusing steady-state overpressure with transient surge; water hammer is dynamic and depends on rate of change, pipe length, and wave reflections.Final Answer:rapid pressure change due to a rapid change in the rate of flow

Question 11

Particle settling and micro-scale motion: In which particle size range is Brownian motion most prominent, significantly affecting sedimentation behavior in fluids?

Accepted Answer

Introduction / Context:Brownian motion is the random thermal movement of small particles suspended in a fluid. It becomes important when designing colloid separation processes, ultrafiltration, and nanoparticle characterization, where gravity settling is minimal.Given Data / Assumptions:Isothermal conditions; fluid at rest on average.Particles are rigid spheres for conceptual discussion.Continuum assumptions begin to break down at extremely small scales.Concept / Approach:As particle diameter decreases, the gravitational settling velocity (Stokes regime) decreases with d^2, while thermal agitation remains significant. In the submicron range (tens of nanometers to about 0.1 micron), Brownian diffusion dominates, countering sedimentation and leading to stable suspensions absent aggregation.Step-by-Step Solution:Use Stokes settling: Vs ∝ d^2; for very small d, Vs → very small.Brownian diffusivity from Stokes–Einstein: D ∝ 1/d; smaller particles diffuse more rapidly.Compare effects: At 0.01–0.10 μm, Brownian diffusion overwhelms gravity settling.Verification / Alternative check:Colloid science texts define colloids roughly 1 nm to 1 μm, with strongest Brownian effects below ~0.1 μm, consistent with the given range.Why Other Options Are Wrong:2–3 μm and larger ranges: Gravity settling predominates; Brownian motion is negligible for most practical fluids.100–1000 μm and 200–300 μm: Granular settling fully dominates; Brownian motion is irrelevant.Common Pitfalls:Assuming turbidity stability at micrometer scales without considering flocculation; interparticle forces and surfactants also affect suspension stability.Final Answer:0.01 to 0.10 microns

Question 12

Aerodynamics of bluff bodies: Pressure (form) drag predominantly depends on flow separation and body shape. Which of the following has the least influence on pressure drag?

Accepted Answer

Introduction / Context:Total drag on a body in a fluid consists of skin-friction (viscous) drag and pressure (form) drag. Separating which parameters mainly affect pressure drag is essential for designing streamlined bodies and reducing energy consumption in flow systems.Given Data / Assumptions:External flow over a body at moderate to high Reynolds number.Comparison between effects of shape/length and surface roughness.Focus on pressure drag, not skin-friction drag.Concept / Approach:Pressure drag results from separation and the large-scale wake; it depends strongly on body shape, bluffness, and length in the flow direction (which affects separation and pressure recovery). Surface roughness primarily influences skin-friction drag. While roughness can sometimes alter separation for spheres/cylinders, its first-order effect relates to viscous drag, not pressure drag, for many engineering shapes; thus, it is the least influential of the listed factors on pressure drag.Step-by-Step Solution:Identify governing factor: shape and streamlining control separation → pressure drag magnitude.Body length in flow direction improves pressure recovery → reduces form drag.Surface roughness mainly affects wall shear stress → skin friction, hence least direct effect on pressure drag.Verification / Alternative check:Wind-tunnel data for streamlined vs bluff bodies show significant pressure-drag reduction with elongation; roughness changes mainly shift friction coefficients.Why Other Options Are Wrong:Length in flow direction: Directly alters separation and pressure recovery.Pressure of main flow only: By itself does not define drag without geometry, but compared to roughness, shape-related parameters govern form drag more strongly.All of these: Not correct since one factor (roughness) is least relevant to pressure drag.Common Pitfalls:Conflating skin friction with form drag; they respond to different geometric and surface factors. Also, special cases exist where controlled roughness triggers earlier transition and changes separation, but the general rule remains.Final Answer:roughness of surface of the body

Question 13

Centrifugal pump characteristics: As the flow rate through a centrifugal pump increases from low to high values, how does the hydraulic efficiency typically vary?

Accepted Answer

Introduction / Context:Hydraulic efficiency reflects how effectively a pump converts input energy into fluid energy, excluding mechanical and volumetric losses. Its variation with flow rate guides selection of the best efficiency point (BEP) for reliable operation.Given Data / Assumptions:Single-stage centrifugal pump with typical radial-flow characteristics.Clear liquid and steady operation.Standard impeller and casing geometry.Concept / Approach:Hydraulic efficiency is low at very small flows due to recirculation and incidence losses, rises to a maximum near the BEP where flow aligns with blade angles, then falls at high flows due to increased shock, separation, and hydraulic losses at the impeller exit and casing tongue.Step-by-Step Solution:At low Q: misalignment and internal recirculation → low efficiency.As Q increases toward BEP: flow becomes well-guided → efficiency increases.Beyond BEP: exit losses and incidence losses rise → efficiency decreases.Verification / Alternative check:Manufacturer pump curves show efficiency contours forming a ridge at BEP, confirming the increase-then-decrease trend with Q.Why Other Options Are Wrong:Monotonic decrease/increase or constant: inconsistent with standard performance curves.Decrease then increase: opposite to observed behavior; low-flow efficiency increases toward BEP before dropping off.Common Pitfalls:Operating far from BEP increases vibration, noise, and bearing loads; users may misinterpret overall efficiency changes with motor and drive losses included.Final Answer:increases and then decreases

Question 14

Wettability and interfacial forces: Which fluid property explains why mercury forms a convex meniscus and does not wet glass?

Accepted Answer

Introduction / Context:Wetting behavior in capillaries and containers depends on the balance between cohesive forces within a liquid and adhesive forces between the liquid and a solid surface. Mercury–glass interaction provides a classic counterexample to water, which wets glass.Given Data / Assumptions:Room temperature behavior in clean glass and pure mercury.No contamination or oxide layers altering surface chemistry.Qualitative comparison of forces.Concept / Approach:When cohesion (liquid–liquid attraction) exceeds adhesion (liquid–solid attraction), the liquid minimizes contact area with the solid, producing a convex meniscus and non-wetting behavior. Mercury’s strong metallic bonding leads to high cohesion relative to adhesion to silica glass; thus, it does not wet the glass surface.Step-by-Step Solution:Identify forces: cohesion (within liquid) vs adhesion (liquid–solid).For mercury on glass: cohesion > adhesion → convex meniscus; contact angle > 90 degrees.Conclude non-wetting behavior explained by dominant cohesion.Verification / Alternative check:Capillary action reverses: mercury is depressed in capillary tubes (negative capillarity), consistent with non-wetting and high contact angle observations.Why Other Options Are Wrong:Surface tension: a related property, but the key explanation is the balance of cohesion vs adhesion; surface tension alone does not specify wetting without considering adhesion.Viscosity: flow resistance, not directly responsible for wetting.Adhesion: If adhesion dominated, mercury would wet glass (concave meniscus), which is not observed.Common Pitfalls:Equating high surface tension with non-wetting universally; water has high surface tension yet wets glass because adhesion exceeds cohesion at that interface.Final Answer:cohesion

Question 15

Hydraulic roughness criteria: Two pipes are said to have the same hydraulic roughness when which dimensionless parameter is equal for the two pipes?

Accepted Answer

Introduction / Context:In turbulent pipe flow, the friction factor depends on Reynolds number and roughness effects. Properly characterizing hydraulic roughness is key to using the Moody chart and Colebrook–White equation for pressure drop calculations.Given Data / Assumptions:Commercial pipes with uniformly distributed roughness height k.Same flow regime comparison (turbulent).Focus on geometric similarity of roughness effect.Concept / Approach:Hydraulic roughness impact on friction scales with the nondimensional parameter k/D, known as relative roughness. Two pipes with the same k/D plot along the same family of curves on the Moody chart, yielding comparable friction behavior at the same Reynolds number. Absolute roughness alone is insufficient because diameter D also influences the effect of roughness on flow.Step-by-Step Solution:Define relative roughness: ε = k/D.For pipes A and B, if εA = εB, they exhibit the same hydraulic roughness effect.Thus, same ε implies similar friction trends versus Re.Verification / Alternative check:On the Moody chart, lines of constant k/D overlay friction factor behavior; identical k/D aligns pipes to identical roughness curves.Why Other Options Are Wrong:Absolute roughness: Different diameters would yield different ε and different friction behavior.Friction coefficient equality: This depends on both Re and k/D; equality at one Re does not define hydraulic roughness generically.All (a), (b) & (c): Overbroad; only relative roughness is the defining equality.Common Pitfalls:Using absolute roughness tables without accounting for pipe diameter; misreading Moody chart scales or regimes (hydraulically smooth vs fully rough).Final Answer:relative roughness

Question 16

Pick the wrong statements about fluids and viscosity: Which option correctly identifies the incorrect assertions regarding equilibrium shear stress and non-Newtonian viscosity?

Accepted Answer

Introduction / Context:Distinguishing correct from incorrect statements about viscous behavior is essential in rheology and fluid statics. This item targets understanding of shear stress in equilibrium states and how viscosity depends on shear for non-Newtonian fluids.Given Data / Assumptions:Equilibrium refers to fluid at rest with no velocity gradients.Non-Newtonian fluids show apparent viscosity that varies with shear rate and sometimes time.Temperature dependence applies to all fluids but is not the only dependency for non-Newtonian types.Concept / Approach:In a static fluid (equilibrium), shear stress is zero; only normal (pressure) stresses act. For non-Newtonian fluids, apparent viscosity depends on shear rate (and possibly time), not just temperature. Momentum transfer by viscosity indeed arises from velocity gradients (Newton’s law of viscosity) in laminar flow.Step-by-Step Solution:Evaluate (a): True — shear stress τ = μ du/dy in laminar flow.Evaluate (b): False — at equilibrium (no motion), τ = 0; fluids cannot sustain shear at rest.Evaluate (c): False — non-Newtonian viscosity depends on shear rate (and sometimes time), not solely on temperature.Thus, both (b) and (c) are wrong statements.Verification / Alternative check:Rheograms of shear-thinning/thickening fluids (e.g., polymer melts, paints) show viscosity as a function of shear rate; static fluids exhibit hydrostatic pressure with zero shear.Why Other Options Are Wrong:(a) is correct and should not be selected as wrong.Any single selection misses that two statements are wrong here.Common Pitfalls:Confusing non-Newtonian behavior with temperature-only effects; also, assuming fluids at rest can support shear like solids—incorrect by definition.Final Answer:both (b) and (c)

Question 17

Particle settling in water at terminal velocity: which force balance condition is correct for a rigid sphere? (Assume steady fall in a quiescent liquid with constant properties.)

Accepted Answer

Introduction:This question tests core fluid–particle mechanics: how forces balance on a particle falling through a liquid at its terminal (steady) settling velocity. At terminal velocity the acceleration is zero, so the net force must be zero. Understanding which forces are present and how they sum is foundational for designing thickeners, classifiers, and sedimentation tanks.Given Data / Assumptions:Spherical rigid particle falling in water.Steady fall at terminal velocity (no acceleration).Quiescent, homogeneous fluid; constant properties.Forces considered: weight (W), buoyancy (B), and drag (D).Concept / Approach:At terminal velocity, Newton’s second law gives the algebraic sum of forces equal to zero. Taking downward as positive, weight acts downward, buoyancy upward, and drag opposite to motion (upward for a falling particle). The correct balance is W = B + D. Equivalently, D = W − B (drag equals the apparent weight).Step-by-Step Solution:Identify forces: W downward; B upward; D upward.Apply equilibrium at terminal state: sum of vertical forces = 0.Write balance: W − B − D = 0.Rearrange: W = B + D.Verification / Alternative check:In Stokes regime (very small Reynolds number), the analytical solution gives D = 3πμ d v_t. At terminal velocity D precisely equals W − B, confirming the general balance.Why Other Options Are Wrong:Buoyancy = weight + drag: would imply buoyancy exceeds weight, contradicting settled motion.Drag = buoyancy + weight: would require drag to exceed weight, unphysical for downward fall.Drag = weight: ignores buoyancy; only true if fluid density were negligible.Apparent weight = 0: at terminal velocity the apparent weight equals drag, not zero.Common Pitfalls:Omitting buoyancy in liquids, or confusing instantaneous force equality with time-averaged relations. Also, students sometimes assume terminal velocity requires drag = weight, which is only valid in gases of negligible density.Final Answer:Weight = buoyancy + drag

Question 18

Head loss in fully rough turbulent pipe flow: how does the frictional head loss (for a given length) vary with average fluid velocity?

Accepted Answer

Introduction:Pump sizing and pipeline design rely on knowing how head loss changes with flow rate. This question probes the classic Darcy–Weisbach relation for turbulent flow in pipes and the dominant velocity-squared dependence of frictional head loss.Given Data / Assumptions:Internal pipe flow, single-phase, steady.Turbulent regime.Length and diameter fixed for the comparison.Concept / Approach:The Darcy–Weisbach equation is h_f = f * (L/D) * (v^2/(2g)), where f is the friction factor. Over many practical turbulent conditions, head loss varies approximately with v^2; f depends weakly on Reynolds number and relative roughness, but the dominant scaling with velocity is quadratic.Step-by-Step Solution:Start with h_f = f (L/D) v^2/(2g).For a given pipe (fixed L and D), the term (L/D)/(2g) is constant.Although f can vary with Re and roughness, for a velocity change the leading dependence is with v^2.Therefore, head loss ∝ v^2.Verification / Alternative check:Moody chart and the Blasius correlation (for smooth pipes, f ≈ 0.316 Re^−0.25) both leave the v^2 dependence intact in h_f, since f varies sublinearly with v, while the kinetic head term scales as v^2.Why Other Options Are Wrong:Linear with v: underestimates losses in turbulent flow.Inverse with diameter squared: dimensional form is incorrect; for fixed v, h_f ∝ 1/D (not 1/D^2).Inverse with velocity or independent: contradicts Darcy–Weisbach.Common Pitfalls:Confusing laminar (Δp ∝ v) with turbulent (h_f ∝ v^2) behavior or misapplying Hazen–Williams empiricism beyond its range.Final Answer:Proportional to (velocity)^2

Question 19

Differential pressure flowmeters: compared with a venturimeter, the permanent power/pressure loss across an orificemeter is generally

Accepted Answer

Introduction:Venturimeters and orificemeters are classic differential-pressure flowmeters. While both create a pressure drop to infer flow rate, their hydraulic recoveries differ markedly, impacting pumping cost and energy efficiency.Given Data / Assumptions:Incompressible, steady flow.Meters correctly installed and sized.Focus is on permanent (unrecovered) pressure loss.Concept / Approach:A venturimeter accelerates then smoothly decelerates the fluid through a converging–throat–diverging nozzle, enabling high pressure recovery. An orificemeter has a sharp-edged plate causing flow separation and a vena contracta downstream, with large mixing losses and poor pressure recovery, resulting in higher permanent loss.Step-by-Step Solution:Both devices create a differential pressure ∆p for metering.Venturi: streamlined diffuser recovers much of the kinetic energy → small permanent loss.Orifice: separated flow and turbulence → low recovery → large permanent loss.Therefore, orificemeter loss > venturimeter loss.Verification / Alternative check:Manufacturer data and handbooks show typical loss coefficients: orifices may lose 50–70% (or more) of ∆p, while venturis often recover most, leaving a small residual loss.Why Other Options Are Wrong:Less than/same as: contradict well-known recovery differences.Unpredictable: qualitative ranking is well established.Zero: no real meter is lossless.Common Pitfalls:Confusing differential pressure used for measurement with permanent loss. The two are related but not identical; geometry controls recovery.Final Answer:More than that in a venturimeter

Question 20

Stokes settling of a small sphere: the terminal velocity in a viscous fluid varies primarily with which parameter (holding others constant)?

Accepted Answer

Introduction:Very small particles settling slowly in a viscous fluid obey Stokes’ law, an analytical solution valid at low Reynolds numbers. Recognizing how terminal velocity depends on material and fluid properties is essential for clarifier design and fine particle separations.Given Data / Assumptions:Spherical particle, creeping flow (Re Uniform fluid, isothermal, quiescent.Terminal (steady) settling velocity v_t.Concept / Approach:Stokes’ law gives v_t = g (ρ_p − ρ_f) d^2 / (18 μ). Thus v_t is directly proportional to d^2 and to the density difference, and inversely proportional to dynamic viscosity μ. Among the options provided, the correct single-statement dependence highlighted is the inverse proportionality to μ.Step-by-Step Solution:Write Stokes formula: v_t = [g (ρ_p − ρ_f) d^2] / (18 μ).Identify dependences: v_t ∝ d^2; v_t ∝ (ρ_p − ρ_f); v_t ∝ 1/μ.Match with options: only the viscosity inverse dependence is stated correctly.Verification / Alternative check:Dimensional analysis of drag in creeping flow (D = 3 π μ d v) plus force balance D = W − B leads directly to v_t inversely proportional to μ.Why Other Options Are Wrong:First power of diameter: should be second power (d^2) under Stokes law.Inverse square of diameter: incorrect trend and dimensionally inconsistent.Square of density difference: dependence is first power, not squared.Fourth power of diameter: not applicable in creeping flow.Common Pitfalls:Confusing Stokes (v_t ∝ d^2) with Newton’s law regime (v_t ∝ d^0.5) at high Reynolds numbers. Always check regime validity first.Final Answer:Inverse of fluid viscosity

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