Question 1

Centrifugal pumps — effect of throttling discharge valve When the discharge of a centrifugal pump is throttled on the delivery side, what happens to the suction lift (all else unchanged)?

Accepted Answer

Introduction / Context: Operators often adjust flow by throttling the discharge valve of a centrifugal pump. Understanding which variables change (flow, head, power) and which remain essentially unaffected (suction lift/NPSH available) is a key field skill. Given Data / Assumptions: Pump speed and suction piping conditions remain constant. Discharge valve is partially closed to reduce flow. Suction source level and vapor pressure do not change. Concept / Approach: Throttling a centrifugal pump reduces flow and moves the operating point along the pump curve to higher head and typically lower power. Suction lift (static suction head/height) depends on liquid level and piping, not on the discharge valve setting. NPSH required usually decreases as flow decreases; NPSH available remains determined by system suction conditions. Step-by-Step Solution: Close discharge valve slightly → flow decreases.Operating head increases modestly, pump power often drops.Suction side geometry and levels unchanged → suction lift remains the same. Verification / Alternative check: Consult typical pump performance curves: throttling shifts along the same speed curve; suction static head is independent of discharge throttling. Why Other Options Are Wrong: Increases/decreases: would require change in suction level, friction, or vapor conditions—not caused by discharge throttling alone. Data insufficient: standard practice shows negligible effect on suction lift. Common Pitfalls: Confusing suction lift with NPSH required; the latter can vary with flow, but the static lift does not. Final Answer: remains unchanged

Question 2

Pipeline engineering — common joints for underground lines In water and utility distribution networks, which joint type is most commonly used for underground pipelines to accommodate minor misalignment and facilitate assembly?

Accepted Answer

Introduction / Context: Underground pipelines require joints that are easy to assemble, tolerant of small alignment errors, and capable of sealing under burial conditions. The choice affects installation speed, leak tightness, and maintenance. Given Data / Assumptions: Underground service (e.g., water mains). Need for simple assembly and slight flexibility. Standard civil/municipal practice is implied. Concept / Approach: Sleeve (sometimes called “Dresser”) joints utilize a sleeve and gaskets to connect pipe ends, allowing slight angular and axial tolerance. They are widely used for buried pipelines where flanged joints (rigid, bolt-intensive) are impractical except at appurtenances. Expansion joints are specialized and uncommon for general buried service. “Coupling” is a generic term but sleeve joint specifically denotes the widely used flexible coupling assembly. Step-by-Step Solution: Identify the installation constraints: buried, long runs, limited access.Match to joint offering flexibility and ease: sleeve joint.Select “sleeve joint.” Verification / Alternative check: Civil standards and catalogs show sleeve-type flexible couplings extensively used for underground lines and repairs. Why Other Options Are Wrong: Flange: rigid, labor-intensive; mostly for connections to valves/equipment. Expansion joint: not common underground; movement is handled by soil/pipe flexibility and restrained joints. “Coupling” generic term lacks specificity; sleeve joint is the established type. Common Pitfalls: Equating plant piping practice (flanges) with buried utility practice. Final Answer: sleeve joint

Question 3

Orifice and jet coefficients — Cv vs. Cd For an orifice or nozzle, how does the coefficient of velocity (Cv) compare to the coefficient of discharge (Cd)?

Accepted Answer

Introduction / Context: Three key coefficients describe jet behavior at orifices: coefficient of contraction (Cc), coefficient of velocity (Cv), and coefficient of discharge (Cd). Their relationships are fundamental in hydraulics and flow measurement. Given Data / Assumptions: Cd = actual discharge / theoretical discharge. Cd = Cc * Cv is the standard relationship. Cc is typically less than 1. Concept / Approach: Because Cc < 1 for a contracting jet, multiplying Cv by Cc yields Cd < Cv. Thus, Cv is greater than Cd for real orifices/nozzles. Step-by-Step Solution: Start from Cd = Cc * Cv.Recognize Cc < 1 (vena contracta phenomenon).Therefore, Cv > Cd. Verification / Alternative check: Typical values: Cc ≈ 0.62–0.64, Cv ≈ 0.97–0.99, Cd ≈ 0.61–0.62 (for sharp-edged orifice), confirming Cv > Cd. Why Other Options Are Wrong: Less than/equal to/not related: contradict empirical relations and definitions. Common Pitfalls: Mixing up the three coefficients or forgetting the contraction effect. Final Answer: more than

Question 4

Open-channel hydraulics — circular channel condition for maxima For a circular open channel, when the hydraulic mean depth Dm equals 0.30 D (D = diameter), this condition corresponds to the maximum of which quantity?

Accepted Answer

Introduction / Context: Design of circular open channels (partially full pipes, sewers) often uses optimal depth conditions for maximum discharge or maximum velocity. Hydraulic mean depth Dm = A/P (area over wetted perimeter) is a convenient parameter. This question targets the specific condition Dm = 0.30 D. Given Data / Assumptions: Circular channel partially full (open-channel flow). D = full pipe diameter; Dm = A/P. Known hydraulic optima exist for max discharge and max velocity. Concept / Approach: For a circular channel: maximum discharge (flow rate) occurs at a depth close to 0.94–0.96 D, for which hydraulic mean depth is about 0.29–0.30 D. Maximum mean velocity occurs at a shallower depth (roughly 0.80–0.82 D), corresponding to Dm about 0.24–0.25 D. Therefore, Dm = 0.30 D signals the maximum discharge condition, not the maximum velocity condition. Step-by-Step Solution: Recall optimums: Dm ≈ 0.30 D → max discharge; Dm ≈ 0.25 D → max velocity.Match the given Dm to the appropriate maximum.Select “flow rate.” Verification / Alternative check: Derivations from Manning or Chezy formulations with geometric relations for circular segments confirm the distinct depths for maxima. Why Other Options Are Wrong: Mean velocity: associated with Dm ≈ 0.25 D, not 0.30 D. Both/neither: contradict standard geometric-hydraulic results. Common Pitfalls: Assuming maxima for velocity and discharge occur at the same depth—they do not. Final Answer: flow rate

Question 5

Micro-hydrodynamics — drag coefficient at very low Reynolds number A bacterium of size 1 µm swims in water at 1 mm/s (kinematic viscosity ν = 1×10^-6 m^2/s). What is the order of magnitude of its drag coefficient?

Accepted Answer

Introduction / Context: At microscopic scales, flows are dominated by viscosity (low Reynolds number). The drag coefficient for small particles or organisms scales inversely with Reynolds number in the creeping-flow regime, illustrating why swimming at micro-scales is fundamentally different from macro-scales. Given Data / Assumptions: Characteristic length L ≈ 1 µm = 1×10^-6 m. Speed V = 1 mm/s = 1×10^-3 m/s. Kinematic viscosity ν = 1×10^-6 m^2/s. Reynolds number Re = V * L / ν. Concept / Approach: For very small Re (creeping flow), an approximate relation for bluff-body drag coefficient is Cd ≈ 24/Re (Stokes regime). We compute Re and then estimate Cd order of magnitude. Step-by-Step Solution: Compute Re: Re = (1×10^-3 m/s) * (1×10^-6 m) / (1×10^-6 m^2/s) = 1×10^-3.Estimate Cd: Cd ≈ 24 / Re = 24 / (1×10^-3) = 24,000.Round to order of magnitude as required → 24,000. Verification / Alternative check: The creeping-flow assumption holds since Re ≪ 1; empirical corrections (Oseen) are minor for this order-of-magnitude question. Why Other Options Are Wrong: 24: corresponds to Re ≈ 1. 0.44 and 0.24: typical of high-Re streamlined bodies, not micro-swimmers. Common Pitfalls: Using high-Re intuition; micro-swimmers operate where inertia is negligible. Final Answer: 24000

Question 6

Internal flows — shape of velocity profile in turbulence For fully developed turbulent flow through a closed conduit (e.g., a smooth pipe), which functional form best describes the mean velocity profile across the radius?

Accepted Answer

Introduction / Context: Velocity profiles reveal the physics of flow regimes. Laminar pipe flow exhibits a parabolic profile, whereas turbulent flow shows a much flatter core and a characteristic near-wall behavior described by empirical laws. Given Data / Assumptions: Flow is fully developed and turbulent in a circular conduit. No strong roughness effects altering the classic wall law. Mean velocity profile is considered (time-averaged). Concept / Approach: The law of the wall for turbulent boundary layers and internal flows gives a logarithmic relation between mean velocity and distance from the wall in inner variables. Hence, the overall shape is commonly termed “logarithmic,” not parabolic (laminar) or linear. Step-by-Step Solution: Recall laminar Hagen–Poiseuille profile → parabolic.Recall turbulent wall law → logarithmic mean velocity in the inner region and flattened core.Select “logarithmic.” Verification / Alternative check: Experimental data and turbulence models (e.g., log-law with von Kármán constant) support the logarithmic near-wall form. Why Other Options Are Wrong: Parabolic: laminar regime only. Linear: typical only in special Couette flows. Hyperbolic: not a standard descriptor of turbulent pipe profiles. Common Pitfalls: Overgeneralizing parabolic profiles to all internal flows. Final Answer: logarithmic

Question 7

Instrumentation in fluid mechanics: A Pitot–static tube directly senses which types of pressure, and which type does it not measure (being obtained only by difference)?

Accepted Answer

Introduction / Context: The Pitot–static tube is a standard device in fluid mechanics and aerodynamics for measuring flow speed in pipes and around bodies. It uses pressure measurements to infer velocity via Bernoulli’s equation. Understanding which pressures are measured directly versus inferred is a core skill for instrumentation questions. Given Data / Assumptions: Pitot–static tube has two ports: an impact (Pitot) opening facing the flow and static side ports. Fluid is steady, incompressible, and along a streamline; elevation changes are negligible. Bernoulli relation between free stream and stagnation point applies without major losses. Concept / Approach: Static pressure acts normal to surfaces and is sampled from side holes aligned to avoid dynamic effects. Total (stagnation) pressure is the pressure when the fluid is brought to rest isentropically at the Pitot opening. Dynamic pressure is not tapped; it is computed as the difference between total and static pressures. Step-by-Step Solution: Let p_s = static pressure measured at the lateral ports.Let p_0 = total (stagnation) pressure measured at the forward-facing Pitot opening.Dynamic pressure q = p_0 - p_s.Velocity magnitude V follows V = sqrt(2 * q / rho) where rho is density. Verification / Alternative check: A differential manometer connected between the Pitot and static ports reads a head proportional to p_0 - p_s, confirming that the device infers dynamic pressure by subtraction rather than direct measurement. Why Other Options Are Wrong: Static pressure: directly sampled by side ports; hence measured. Total pressure: directly sampled by the facing Pitot port; hence measured. Both with a single tap: requires two separate ports; a single tap cannot give both simultaneously. Vapor pressure: unrelated to Pitot–static operation and not measured. Common Pitfalls: Confusing “reading on a differential gauge” with “direct measurement.” The gauge shows the difference, but the tube itself provides static and total; dynamic is derived. Final Answer: Dynamic pressure (total minus static)

Question 8

Hot-wire anemometry: The resistance wire element used to sense air velocity is commonly made from which material?

Accepted Answer

Introduction / Context: Hot-wire anemometers measure fluid velocity by relating convective heat loss from a heated, fine wire to flow speed. The choice of wire material is crucial because the sensor operates at elevated temperature, requires mechanical robustness at microscopic diameters, and must have stable electrical properties. Given Data / Assumptions: The sensor is a very fine wire (often 2–10 micrometers in diameter). It must sustain high temperatures and repeated thermal cycling. Stable resistance–temperature characteristics are desirable. Concept / Approach: Tungsten is widely used for hot-wire probes due to its very high melting point, good strength at temperature, and ability to be drawn into extremely fine, robust wires. While platinum or platinum–iridium wires are also used (especially for constant-temperature anemometry and calibration stability), tungsten is the classic choice for many laboratory and industrial probes. Step-by-Step Solution: Identify required properties: high melting point, fine drawability, mechanical toughness.Compare materials: copper and aluminium oxidize/soften and lack high-temperature strength; chromium alone is not used for such wires.Conclude tungsten best satisfies hot-wire demands. Verification / Alternative check: Instrument catalogs and experimental fluid mechanics texts list tungsten and platinum-family wires as standard, with tungsten being common for constant-current or constant-temperature anemometers in air. Why Other Options Are Wrong: Copper/aluminium: low melting points, oxidation, poor high-temperature strength. Chromium: not commonly used as a standalone filament in hot-wire probes. Platinum–iridium alloy: used in some designs, but the most common textbook answer is tungsten. Common Pitfalls: Assuming general electrical conductivity determines suitability; in reality, high-temperature mechanical stability and survivability at micro-scale are decisive. Final Answer: Tungsten

Question 9

Venturimeter design practice: What is the typical included angle of the diffuser (exit cone) in a venturimeter to minimize separation and losses?

Accepted Answer

Introduction / Context: Venturimeters measure flow rate using a converging section, throat, and a diffuser (exit cone). The diffuser recovers pressure by decelerating the flow. Its angle is critical: too steep causes boundary-layer separation and loss; too shallow makes the device unnecessarily long. Given Data / Assumptions: Subsonic, incompressible flow context typical of industrial metering. Well-designed venturimeter uses a modest diffuser angle to balance pressure recovery and length. Surface finish is adequate; Reynolds number sufficiently high for diffuser design rules of thumb. Concept / Approach: Diffuser (exit cone) angles around 7 to 15 degrees (included) are widely adopted to avoid separation while achieving reasonable pressure recovery. Values below ~7 degrees increase length and cost; above ~15 degrees raise the risk of separation and energy loss. Step-by-Step Solution: Goal: maximize static pressure recovery with minimal length.Constraint: avoid boundary-layer separation in the diffuser.Empirical practice sets included angle near 7–15 degrees, hence Option B. Verification / Alternative check: Handbooks on flow measurement and standards for venturi meters specify diffuser angles in this common range, confirming the choice. Why Other Options Are Wrong: 2–5 degrees: unnecessarily long device for marginal extra recovery. 15–25 or >25 degrees: prone to separation and higher losses. 45–60 degrees: would definitely separate; unsuitable for a venturi diffuser. Common Pitfalls: Confusing half-angle with included angle; if quoting half-angles, the typical values are roughly 3.5–7.5 degrees. Final Answer: 7 to 15 degrees

Question 10

Centrifugal pumps and suction: If the maximum theoretical suction lift for water at 15°C is 34 ft, what would you expect for hot water at 90°C (considering much higher vapor pressure)?

Accepted Answer

Introduction / Context: Suction lift for pumps is limited by the available Net Positive Suction Head (NPSH), which depends on atmospheric pressure, static lift, friction losses, and—critically—the liquid’s vapor pressure. As temperature rises, water’s vapor pressure increases sharply, reducing allowable suction lift to avoid cavitation. Given Data / Assumptions: Theoretical maximum at 15°C is 34 ft (based on atmospheric head minus negligible vapor pressure and losses). At 90°C, water vapor pressure is high, dramatically reducing NPSH available. Neglect line losses and assume sea-level atmospheric pressure for comparison. Concept / Approach: Maximum theoretical suction lift ≈ (p_atm/γ) - (p_v/γ). At 15°C, p_v is small; at 90°C, p_v is large, so the difference shrinks. Practically, hot water service demands flooded suction or very short suction lifts to prevent cavitation. Step-by-Step Solution: At 15°C: p_v is low ⇒ theoretical lift ~34 ft.At 90°C: p_v is much higher ⇒ the allowable lift drops steeply.Rule-of-thumb and charts place hot water suction lift at only a few feet; 8 ft is a representative upper bound in idealized theory.Therefore, select ~8 ft, reflecting the reduction due to high vapor pressure. Verification / Alternative check: Steam tables show water’s vapor pressure near 90°C is substantial (tens of kPa), leaving little headroom under 1 atm for suction. Practical pump manuals recommend minimal or zero lift for hot liquids. Why Other Options Are Wrong: 40 ft/37 ft/34 ft: ignore the significant increase in vapor pressure at 90°C. 20 ft: still too high; cavitation risk would be unacceptable. Common Pitfalls: Assuming suction lift depends only on atmospheric pressure; neglecting the strong temperature dependence via vapor pressure. Final Answer: About 8 ft (significantly reduced)

Question 11

Fluid–fluid devices: A device that uses the momentum of one fluid stream to entrain and transport another fluid is called a/an

Accepted Answer

Introduction / Context: Many process plants employ momentum-transfer devices where a high-velocity motive fluid entrains a secondary fluid. These devices have no moving parts and are widely used for pumping, mixing, vacuum generation, and gas–liquid contacting. Given Data / Assumptions: A high-pressure motive fluid accelerates through a nozzle. Low-pressure suction fluid is entrained and mixed. Diffuser section converts kinetic energy back to pressure for discharge. Concept / Approach: The generic names are jet pump, ejector, or eductor. They rely on Bernoulli acceleration in the nozzle, momentum mixing in the throat, and pressure recovery in the diffuser. Typical motive/suction combinations include steam–gas, liquid–liquid, or liquid–gas. Step-by-Step Solution: Identify device principle: momentum transfer from motive to suction stream.Match to terminology: jet pump/ejector/eductor.Select the option naming this device. Verification / Alternative check: Standards and vendor catalogs consistently use “jet pump,” “ejector,” and “eductor” for this principle. Blowers and compressors are mechanical rotating equipment; an “acid egg” is a pressure vessel for liquid transfer, not a momentum device. Why Other Options Are Wrong: Blower/rotary compressor: add energy via rotating elements, not via motive jet entrainment. Acid egg: uses pressure differential in a vessel; not a jet ejector. Vent cap: unrelated to pumping. Common Pitfalls: Confusing ejectors (no moving parts) with compressors (mechanical). Also, assuming “pump” always implies mechanical impellers. Final Answer: Jet pump (ejector/eductor)

Question 12

Particle settling by Stokes’ law: If a 10 μm sphere settles with velocity v in a Newtonian oil (SG 0.9, μ = 10 poise), what is the settling velocity of a 20 μm sphere in the same oil (all else equal)?

Accepted Answer

Introduction / Context: In the creeping-flow regime (Reynolds number much less than 1), Stokes’ law gives the terminal settling velocity of small spheres in a viscous fluid. Understanding how velocity scales with particle diameter is essential in sedimentation and centrifugation design. Given Data / Assumptions: Spherical particles with diameters d1 = 10 μm and d2 = 20 μm. Same oil, so identical density and viscosity. Stokes regime applies (laminar drag, Re << 1). Concept / Approach: Under Stokes’ law, terminal velocity v_t is proportional to d^2 for a given density difference and viscosity. Thus, doubling the diameter multiplies v_t by 2^2 = 4. Step-by-Step Solution: Let v1 ∝ d1^2 and v2 ∝ d2^2.Given d2 = 2 * d1, then v2 / v1 = (d2/d1)^2 = 2^2 = 4.Therefore, v2 = 4 * v1. Verification / Alternative check: Dimensionally and from the Stokes expression v_t = (g * (ρ_p - ρ_f) * d^2) / (18 * μ), the quadratic dependence on diameter is explicit, confirming the 4x scaling. Why Other Options Are Wrong: Same or one fourth: contradict the d^2 proportionality. Twice: would correspond to linear scaling, not Stokes regime. Eight times: would imply cubic scaling, incorrect for Stokes settling. Common Pitfalls: Applying turbulent or intermediate drag correlations instead of Stokes law; forgetting that shape and agglomeration can alter effective drag. Final Answer: Four times that of the 10 μm particle

Question 13

Fluid statics at a point: The normal stress is equal in all directions at a point in a fluid under which condition?

Accepted Answer

Introduction / Context: Pascal’s law states that in a fluid at rest, the pressure at a point acts equally in all directions. This equality of normal stresses underpins hydrostatics and the definition of scalar pressure. In moving fluids, shear stresses can be present, complicating the stress state. Given Data / Assumptions: We are considering a point in a fluid continuum. We distinguish between hydrostatic and general motion conditions. “No relative motion” implies no velocity gradients and thus no viscous shear. Concept / Approach: For a fluid at rest (or a fluid moving as a rigid body with no shear), the only stresses are normal stresses, and these are equal in all directions, defining pressure as a scalar. Viscosity, compressibility, or absolute pressure do not alone guarantee isotropy of normal stress if shear exists. Step-by-Step Solution: Identify condition for hydrostatic state: zero shear stress.Zero shear occurs when there is no relative motion between fluid layers.Therefore, normal stress (pressure) is equal in all directions under that condition. Verification / Alternative check: From the Cauchy stress tensor, in the absence of shear components, the tensor reduces to −p I (isotropic), confirming equal normal stresses. Why Other Options Are Wrong: Non-viscous or incompressible alone: a fluid can be inviscid or incompressible yet still have shear if velocity gradients exist. Both together: still not sufficient without the “no shear” condition. High pressure: magnitude of pressure is irrelevant to isotropy. Common Pitfalls: Assuming “inviscid” implies no shear; shear stress can be zero in the inviscid model, but real fluids with motion have shear unless velocity is uniform. Final Answer: When there is no relative motion between adjacent fluid layers (no shear)

Question 14

Compression limits: Industrial compressors can deliver gases up to approximately what maximum pressure (order of magnitude)?

Accepted Answer

Introduction / Context: Compressors are used across industry to raise gas pressure. The achievable delivery pressure depends on compressor type (centrifugal, screw, reciprocating), staging, inter-cooling, materials, and application. Knowing realistic pressure ceilings helps in selecting equipment. Given Data / Assumptions: Multistage reciprocating compressors provide the highest delivery pressures among standard industrial equipment. We seek an order-of-magnitude upper bound commonly cited for industrial practice. Concept / Approach: Centrifugal and rotary compressors excel at high flow, moderate pressures; reciprocating compressors in multiple stages with intercooling can achieve very high pressures. Typical upper limits cited in chemical engineering references are on the order of hundreds to about a thousand atmospheres for specialized service. Step-by-Step Solution: Identify high-pressure class: multistage reciprocating units.Compare options: 10 atm (too low), 100 atm (common), 250 atm (high), 1000 atm (upper industrial limit).Select 1000 atm as the representative maximum in the list. Verification / Alternative check: Manufacturers’ data and design texts document reciprocating compressors used to several hundred bar (approaching 1000 atm in specialized cases). Why Other Options Are Wrong: 10–250 atm: achievable but not the upper extreme. 5000 atm: far beyond standard compressor design and enters research regimes. Common Pitfalls: Assuming all compressors have the same limits; overlooking staging and intercooling, which enable high discharge pressures. Final Answer: 1000 atmospheres

Question 15

Pumps for viscous masses: Which pump is most suitable for transporting a viscous, homogeneous molten soap mass in process plants?

Accepted Answer

Introduction / Context: Molten soap is a viscous liquid that benefits from smooth, continuous pumping with minimal pulsation and good volumetric efficiency. Selecting the proper pump type improves product quality and energy efficiency while reducing shear-related issues. Given Data / Assumptions: Fluid: viscous molten soap, relatively homogeneous and lubricating. Need: steady flow, good suction characteristics, self-priming if possible. Solids content minimal; moderate temperatures. Concept / Approach: Gear pumps (external or internal gear) are classic positive displacement rotary pumps for viscous liquids such as syrups, oils, and molten soaps. They provide near-constant flow proportional to speed, handle high viscosities, and maintain good efficiency where centrifugal pumps lose head. Step-by-Step Solution: Match fluid to pump class: viscous → positive displacement rotary.Shortlist options: gear, lobe, screw; among given choices, gear pump fits best.Select gear pump for molten soap transfer. Verification / Alternative check: Pump selection guides consistently recommend gear or screw pumps for viscous, clean liquids due to favorable performance curves at high viscosity. Why Other Options Are Wrong: Diaphragm/reciprocating: better for slurries or metering; can be pulsatile and less ideal for hot viscous melts. Centrifugal/axial: efficiency drops with viscosity; poor choice for thick, slow-moving liquids. Common Pitfalls: Choosing centrifugal pumps for high-viscosity service and then encountering inadequate head/flow and overheating. Final Answer: Gear pump (positive displacement, rotary)

Question 16

Handling thick slurries: For transferring a solution containing a thick slurry (abrasive, solids-laden), which pump type is generally preferred?

Accepted Answer

Introduction / Context: Thick slurries present challenges: high solids content, abrasiveness, and potential for clogging. The pump must tolerate solids, provide gentle handling, and often be capable of dry running or at least robust against suction issues. Given Data / Assumptions: Fluid: thick slurry (suspension of solids in liquid). Abrasive particles likely present; viscosity elevated. Need: good solids-handling, non-shearing, clog-resistant flow path. Concept / Approach: Diaphragm pumps (air-operated double-diaphragm or mechanically driven) are widely used for slurries: they are positive displacement, have check valves that pass solids, can run at low speeds, and isolate the drive from the process fluid. They handle abrasives better than gear pumps and resist clogging more than tight-clearance rotary machines. Step-by-Step Solution: Match requirements to pump characteristics: solids tolerance and gentle action → diaphragm pumps.Eliminate options unsuitable for abrasives or high viscosity (e.g., gear pumps with tight clearances).Select diaphragm pump as the best general choice for thick slurries. Verification / Alternative check: Industrial practice favors diaphragm or peristaltic hose pumps for abrasive slurries; among listed options, diaphragm is the most standard answer for chemical/process plants. Why Other Options Are Wrong: Reciprocating plunger: can handle pressure but suffers from valve wear and pulsation; less tolerant of large solids. Gear: tight internal clearances; poor for abrasive solids. Centrifugal: workable for some slurries with special designs, but thick slurries often require PD pumps. Peristaltic: also suitable but not provided as the main textbook choice in many question banks; diaphragm is more common. Common Pitfalls: Choosing gear pumps for slurries and damaging internals; oversizing centrifugal pumps leading to severe wear and low efficiency. Final Answer: Diaphragm pump (air-operated or mechanical)

Question 17

In pump engineering, the net positive suction head (NPSH) of a centrifugal pump is defined as the available suction-side head (static pressure head plus velocity head) referred to the pump datum, after subtracting the liquid’s vapor pressure head at…

Accepted Answer

Introduction / Context: NPSH (Net Positive Suction Head) is a critical concept in turbomachinery and fluid systems. It quantifies how far the local absolute pressure at a pump’s suction is above the liquid’s vapor pressure. Adequate NPSH prevents cavitation, performance loss, and impeller damage. Given Data / Assumptions: Centrifugal pump operating with a liquid at a known suction temperature. Static pressure head and velocity head are evaluated at the pump suction flange (datum). Vapor pressure head corresponds to the same temperature as at suction. Concept / Approach: NPSH is defined on the suction side. The available head is the sum of the static pressure head and velocity head (and elevation if used consistently), minus the vapor pressure head. This expresses the margin against vaporization at the inlet. Engineers compare NPSHa (available) to NPSHr (required by pump) to ensure cavitation-free operation (NPSHa > NPSHr). Step-by-Step Solution: Write head terms at suction: H_suction = p_suction/(rhog) + V_suction^2/(2g).Compute vapor pressure head: H_vap = p_vap/(rho*g) at suction temperature.Define NPSH: NPSH = H_suction − H_vap (with consistent elevation datum).Interpretation: Positive margin above vapor pressure suppresses bubble formation at the eye of the impeller. Verification / Alternative check: Manufacturer curves specify NPSHr; field calculations of NPSHa use suction pressure readings corrected to absolute, velocity head, and the appropriate vapor pressure of the fluid at temperature. This agrees with standard pump handbooks and industry practice. Why Other Options Are Wrong: At discharge: NPSH is a suction-side criterion, not a discharge metric. At suction (without vapor pressure): Missing the essential subtraction of vapor pressure head. At discharge minus vapor pressure at discharge temperature: Location and temperature are incorrect for NPSH definition. Common Pitfalls: Forgetting to use absolute pressure; using gauge pressure without atmospheric correction; using vapor pressure at the wrong temperature; ignoring suction-line losses and resulting velocity head changes. Final Answer: At suction minus the vapor pressure head of the liquid at the suction temperature.

Question 18

Identify the pump type: which class of pumps is designed for primarily axial flow through the impeller passage?

Accepted Answer

Introduction / Context: Pumps are categorized by predominant flow direction: radial, mixed, or axial. Recognizing axial-flow pumps is essential for low head–high flow applications such as circulation, flood control, and condenser water systems. Given Data / Assumptions: Axial-flow geometry implies flow essentially parallel to the pump shaft. Applications demand high volumetric throughput at relatively low heads. Terminology aligns with common industry usage (propeller, axial-flow, tubular pumps). Concept / Approach: Propeller pumps are classic axial-flow pumps. The impeller resembles a boat propeller; energy addition primarily increases the liquid’s axial momentum. By contrast, turbine (radial) pumps develop head via centrifugal action, and diffuser is a stationary component used to recover pressure, not a standalone pump type. Step-by-Step Solution: Define axial-flow pump → velocity vector mainly along shaft axis.Match with propeller-style impeller blades → propeller pump.Exclude radial (turbine) and component-only (diffuser) options. Verification / Alternative check: Manufacturer catalogs and pump handbooks list “propeller pumps” under axial-flow family, often supplied as vertical mixed/axial variants for cooling water and irrigation services. Why Other Options Are Wrong: Turbine pump: Primarily radial flow. Diffuser pump: Diffuser is a pressure-recovery component; not the axial-flow category name. None of these: Incorrect because propeller pump is the standard axial-flow type. Common Pitfalls: Confusing mixed-flow with axial-flow; assuming “turbine” implies axial (it does not in pump context). Final Answer: Propeller pump

Question 19

A Bingham plastic (plastic viscosity μ = 10 Pa·s, yield stress τ0 = 10 kPa) is sheared between parallel plates separated by 1×10^-3 m. The top plate moves at 1 m/s. What shear stress acts on the moving plate?

Accepted Answer

Introduction / Context: Bingham plastic fluids (e.g., some slurries, paints) exhibit a yield stress plus a linear viscous term once yielded. Computing shear stress in simple shear between plates tests understanding of constitutive laws and basic kinematics of shear flow. Given Data / Assumptions: Plastic viscosity μ = 10 Pa·s. Yield stress τ0 = 10 kPa = 10,000 Pa. Gap h = 1×10^-3 m; top plate speed U = 1 m/s; bottom plate stationary. Fully yielded uniform shear assumed (no plug zone with these numbers). Concept / Approach: Bingham law: τ = τ0 + μ * (du/dy) for |τ| ≥ τ0. For simple Couette flow, shear rate du/dy ≈ U/h when fully yielded. First verify τ exceeds τ0 to ensure yielding; then compute τ directly from the linear relation. Step-by-Step Solution: Compute shear rate: du/dy = U/h = 1 / (1×10^-3) = 1000 s^-1.Viscous term: μ * (du/dy) = 10 Pa·s * 1000 s^-1 = 10,000 Pa.Total shear stress: τ = τ0 + μ * (du/dy) = 10,000 Pa + 10,000 Pa = 20,000 Pa.Convert to kPa: 20,000 Pa = 20 kPa. Verification / Alternative check: Check yield: τ (20 kPa) > τ0 (10 kPa), so the fluid is indeed yielded across the gap. Hence the Couette assumption is consistent and the calculated τ is valid. Why Other Options Are Wrong: 10 kPa: Counts only yield, ignores viscous term. 30 kPa or 40 kPa: Overestimates by adding extra multiples of the viscous term not supported by the data. Common Pitfalls: Using Newtonian formula τ = μ*(du/dy) and forgetting τ0; mis-converting kPa to Pa; using the wrong gap or speed units. Final Answer: 20 kPa

Question 20

In pump piping practice, the nominal size of the discharge (delivery) pipe is usually selected to be how, relative to the nominal size of the suction pipe?

Accepted Answer

Introduction / Context: Proper sizing of suction and discharge lines is essential for reliable pump operation. Suction lines are often sized more generously to minimize friction losses and avoid excessive velocity, which helps maintain NPSHa and reduce cavitation risk. Given Data / Assumptions: Conventional centrifugal pump with separate suction and discharge piping. Goal: minimize suction-side losses and maintain adequate NPSHa. Typical design velocities: lower on suction, higher on discharge. Concept / Approach: Suction piping is commonly one size larger (or otherwise lower-velocity) than discharge. Lower suction velocity reduces head losses and entrance losses, preserving NPSH. Discharge piping can accept higher velocities because the pump has already added head and cavitation risk is negligible downstream of the impeller. Step-by-Step Solution: Define objective: protect suction conditions → low velocity → larger pipe.Relate to practice: discharge diameter ≤ suction diameter in many designs.Select statement matching practice → discharge nominal size is smaller than suction size. Verification / Alternative check: Plant standards and pump installation guides recommend larger suction lines, straight runs into the pump, and minimized fittings to reduce NPSH losses; discharge lines are often sized nearer to economic velocity limits. Why Other Options Are Wrong: Larger or same as suction: Contradicts common practice for NPSH protection. Exactly twice: Arbitrary and not an industry rule. Common Pitfalls: Using identical sizes without checking velocities; adding many suction-side elbows/strainers that increase losses and reduce NPSHa. Final Answer: Smaller than the suction size

Fluid Mechanics Questions