Question 1

Venturimeter – Primary Use in Internal Flow Measurement A Venturimeter installed in a pipe is used to determine the volumetric discharge of the liquid flowing through that pipe using pressure differences and Bernoulli’s principle.

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Introduction:A Venturimeter is a differential-pressure flowmeter. By constricting the cross section, it relates measured pressure drop to flow rate using energy and continuity principles.Given Data / Assumptions:Incompressible, steady flow in a full pipe.Known meter geometry (inlet, throat, diffuser).Empirical discharge coefficient accounts for losses.Concept / Approach:Bernoulli between inlet and throat gives a relation between pressure drop and velocity increase. With continuity, one can express Q in terms of measured differential pressure and meter constants. Hence, the Venturimeter measures discharge; velocity can be inferred only after dividing by area.Step-by-Step Solution:1) Measure Δp between inlet and throat using a manometer/transducer.2) Use Bernoulli + continuity to compute velocity at the throat.3) Multiply by throat area, apply discharge coefficient → Q (discharge).Verification / Alternative check:Calibration against gravimetric or volumetric standards confirms accuracy within stated uncertainty when the meter is properly installed.Why Other Options Are Wrong:Velocity alone is not the direct instrument output; pressure is not the primary measured quantity (that is a means to infer Q). Pressure difference is measured but the device is classified by its purpose: discharge measurement. Density is not directly measured.Common Pitfalls:Ignoring installation requirements (straight runs, Reynolds-number effects) and calibration coefficients, which can bias Q.Final Answer:measure the discharge of liquid flowing in a pipe

Question 2

Rectangular Weir with End Contractions – Francis Discharge Formula For a sharp-crested rectangular weir having n end contractions, the practical Francis form in SI units is Q = 1.84 * (L − 0.1 * n * H) * H^(3/2), where L is crest length and H is hea…

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Introduction:The Francis formula is a widely used empirical relation for discharge over sharp-crested rectangular weirs. It corrects the crest length for end contractions and consolidates constants for SI usage.Given Data / Assumptions:Rectangular, sharp-crested weir with free nappe and adequate aeration.Head H measured sufficiently upstream to avoid velocity-of-approach error.n end contractions (typically 2 for a contracted weir).Concept / Approach:The theoretical discharge per unit width from energy considerations scales with H^(3/2). Accounting for contraction and empirical coefficient Cd leads to a compact constant 1.84 in SI so that Q = 1.84 (L − 0.1 n H) H^(3/2).Step-by-Step Solution:1) Effective crest length L_e = L − 0.1 n H.2) Use Q ∝ L_e * H^(3/2) from sharp-crested weir theory.3) Apply consolidated constant → Q = 1.84 L_e H^(3/2).4) Substitute L_e to get the stated Francis form.Verification / Alternative check:Dimensional analysis confirms m^3/s when L and H are in meters. Field calibrations show good agreement within the method’s range.Why Other Options Are Wrong:Options with H^2 or H^(5/2) do not match the sharp-crested exponent. Using n H instead of 0.1 n H overcorrects end contractions. Expressions with Cd 2 g are incomplete without integration and constants consolidation.Common Pitfalls:Measuring H too close to the crest, inadequate aeration of the nappe, or miscounting end contractions.Final Answer:Q = 1.84 (L − 0.1 n H) H^(3/2)

Question 3

Temperature Effect – Dynamic Viscosity of Gases For most common gases, the dynamic viscosity increases as temperature rises due to enhanced molecular momentum transport.

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Introduction:This item contrasts the temperature–viscosity trends of gases versus liquids. Understanding the trend helps in heat-transfer and flow-friction predictions at elevated temperatures.Given Data / Assumptions:Ideal or near-ideal gases over engineering temperature ranges.No phase change or chemical reaction.Pressure variations modest.Concept / Approach:Gas viscosity arises from molecular momentum exchange. As temperature increases, molecular speeds rise, increasing the rate and distance of momentum transport, hence dynamic viscosity mu increases (e.g., Sutherland’s formula for air).Step-by-Step Solution:1) Recognize that in gases, intermolecular collisions transport momentum.2) Higher temperature → higher molecular speed.3) Net effect: increased mu with temperature.Verification / Alternative check:Empirical correlations (Sutherland) and tabulated data show monotonic increase of mu for air from ambient to high temperatures.Why Other Options Are Wrong:Unaffected or decreases contradict gas behavior (decrease applies to liquids). Other options introduce patterns not generally observed.Common Pitfalls:Applying liquid trends to gases; confusing dynamic viscosity with kinematic viscosity (which also depends on density).Final Answer:increases

Question 4

Newton’s Law of Viscosity – Relationship Between Shear Stress and Shear Rate According to Newton’s law, the shear stress on a fluid layer is directly proportional to the rate of shear strain (velocity gradient), with proportionality constant equal t…

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Introduction:Newton’s law of viscosity provides the linear constitutive relation for Newtonian fluids, forming the basis for many laminar-flow analyses and engineering correlations.Given Data / Assumptions:Simple shear between parallel plates or near a wall.Newtonian fluid behavior.Continuum hypothesis valid.Concept / Approach:The law states tau = mu * (du/dy), where tau is shear stress, mu is dynamic viscosity, and du/dy is the velocity gradient normal to the flow. Thus tau increases linearly with shear rate when mu is constant.Step-by-Step Solution:1) Identify shear flow with gradient du/dy.2) Apply Newton’s law: tau = mu * (du/dy).3) Conclude the dependence is direct proportionality.Verification / Alternative check:Rheometer data for water and many oils produce straight-line tau vs. (du/dy) plots, confirming direct proportionality within certain ranges.Why Other Options Are Wrong:Equal to: Omits the proportionality constant mu. Inversely proportional or independent: Contradict the constitutive model. Logarithmic: Not applicable for Newtonian fluids.Common Pitfalls:Confusing Newtonian with non-Newtonian behavior (e.g., power-law fluids) where proportionality is not linear.Final Answer:directly proportional

Question 5

Applications of Bernoulli’s Equation – Ideal Energy Balance Under appropriate assumptions (steady, incompressible, negligible losses), Bernoulli’s equation underpins the working principles of the Venturimeter, orifice meter, and Pitot tube.

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Introduction:Bernoulli’s equation relates pressure, velocity, and elevation along a streamline. Many classical flow-measurement devices are derived from this ideal energy balance with suitable coefficients to account for losses and non-idealities.Given Data / Assumptions:Steady, incompressible flow.Negligible shaft work and heat transfer.Use of discharge/velocity coefficients for practical devices.Concept / Approach:A Venturimeter and an orifice meter infer discharge from measured pressure differences created by area change. A Pitot tube converts dynamic pressure to a measurable stagnation pressure to infer velocity. All rely on Bernoulli between appropriate points with correction factors.Step-by-Step Solution:1) Write Bernoulli between tap points.2) Combine with continuity to get velocity at a constriction or from stagnation pressure.3) Apply coefficients (Cd, Cc, Cv) as needed to account for losses.Verification / Alternative check:Standard calibrations show close agreement with Bernoulli-based predictions when correction factors are used.Why Other Options Are Wrong:Picking any single instrument ignores that all three are rooted in the same principle.Common Pitfalls:Using Bernoulli without including real-flow coefficients; misplacing pressure taps; neglecting elevation heads.Final Answer:all of these

Question 6

Pipe Flow Transition – Definition of Higher Critical Velocity In internal pipe flow, the higher critical velocity corresponds to the speed (or Reynolds number) above which turbulence is fully established; hence the statement that turbulent flow star…

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Introduction:Transition from laminar to turbulent flow in pipes is commonly described using two threshold velocities (or Reynolds numbers). This question targets recognition of the role of the higher critical value in initiating sustained turbulence.Given Data / Assumptions:Newtonian fluid in a circular pipe.Disturbances and surface roughness within normal ranges.Reynolds number Re = (rho * V * D) / mu.Concept / Approach:Below the lower critical value the flow is stably laminar. Between the lower and higher critical values, the flow is transitional and sensitive to disturbances. Above the higher critical value, turbulence is fully sustained. Therefore saying “turbulent flow starts” at the higher critical velocity is acceptable in practical engineering usage.Step-by-Step Solution:1) Identify the transitional range of Re (roughly 2000–4000 for smooth pipes).2) Associate the upper bound with the onset of sustained turbulence.3) Conclude the statement is true in the conventional sense.Verification / Alternative check:Laboratory observations show intermittent turbulence in the range between the critical values; fully turbulent profiles dominate beyond the higher critical threshold.Why Other Options Are Wrong:No: Conflicts with standard terminology. Other options impose conditions unrelated to the definition.Common Pitfalls:Assuming a single sharp universal value; actual thresholds depend on inlet conditions and roughness.Final Answer:Yes

Question 7

Surface Tension – Ability of a Liquid Surface to Resist Tension Statement: The property of a liquid that enables its free surface to resist a tensile stress (pulling apart) is called surface tension.

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Introduction:Surface tension quantifies the energetic cost of increasing liquid surface area. It manifests as a membrane-like behavior that resists stretching, affecting droplets, bubbles, capillarity, and wetting.Given Data / Assumptions:Clean liquid–gas interface at equilibrium.Isothermal conditions; negligible contamination.No surfactants unless stated.Concept / Approach:Surface tension has units of force per unit length or energy per unit area. Molecules at the surface experience net inward cohesive forces, creating a tendency to minimize area and resist tensile deformation of the interface.Step-by-Step Solution:1) Recognize that pulling on the surface increases area.2) The work required per unit area is the surface energy, equal to the surface tension for isothermal changes.3) Hence the surface behaves as if it resists tensile stress.Verification / Alternative check:Experiments such as ring/plate tensiometry directly measure the force needed to detach or expand a liquid surface.Why Other Options Are Wrong:Disagree: Contradicts the physical definition. Statements about solids, vacuum, or viscosity are irrelevant to the interfacial property being defined.Common Pitfalls:Confusing surface tension with viscosity; assuming it acts in the bulk rather than at the interface; overlooking surfactant effects that reduce surface tension.Final Answer:Agree

Question 8

Most Efficient Rectangular Channel Section – Proportion for Maximum Discharge For a given area and bed slope, a rectangular channel conveys maximum discharge when the breadth is twice the flow depth (b = 2 y), which maximizes hydraulic radius.

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Introduction:The “most economical” (most efficient) channel section minimizes wetted perimeter for a given area, thereby maximizing hydraulic radius and discharge for a given slope and roughness. For rectangles, this yields a simple proportion between breadth and depth.Given Data / Assumptions:Steady, uniform open-channel flow.Rectangular prismatic channel with width b and depth y.Manning or Chezy friction formulation where Q ∝ A * R^(2/3) (or similar).Concept / Approach:For fixed area A = b * y, minimizing wetted perimeter P = b + 2 y maximizes hydraulic radius R = A / P. Using calculus with the constraint b * y = A gives the optimal proportion b = 2 y for a rectangle.Step-by-Step Solution:1) Let A be fixed; write P = b + 2 y with b = A / y.2) Substitute to get P(y) = A / y + 2 y.3) Minimize P(y) w.r.t. y → dP/dy = −A / y^2 + 2 = 0 → y = sqrt(A / 2).4) Then b = A / y = 2 y, proving the proportion.Verification / Alternative check:Using b = 2 y yields R = (b y)/(b + 2 y) = (2 y^2)/(2 y + 2 y) = y/2, the known efficient-section relation for rectangles.Why Other Options Are Wrong:Other ratios increase wetted perimeter for the same area, reducing hydraulic radius and thus discharge capacity under the same slope/roughness.Common Pitfalls:Confusing the efficient rectangle with trapezoidal or triangular optimal conditions; assuming the ratio depends on slope or roughness (it depends on geometry for a given A).Final Answer:its breadth is twice the depth

Question 9

Hemispherical Tank Draining – Time to Lower Level from H1 to H2 A hemispherical tank of radius R drains through a small orifice of area a at its bottom. Including coefficient of discharge Cd, the time to reduce the liquid head from H1 to H2 (measure…

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Introduction:Unsteady draining of curved tanks requires accounting for how the cross-sectional area varies with head. For a hemisphere, the area changes quadratically with head, leading to an integral that produces powers 3/2 and 5/2 in the time expression.Given Data / Assumptions:Hemispherical tank of radius R, orifice of area a at the bottom center.Head H measured vertically above the orifice (0 ≤ H ≤ R).Torricelli efflux with coefficient Cd; inviscid elsewhere; free surface velocity negligible.Concept / Approach:The free-surface area at head H is A(H) = pi * (2 * R * H − H^2). Continuity gives A(H) dH/dt = −Cd * a * sqrt(2 * g * H). Separating variables yields an integral of the form ∫ (2 R H − H^2) / sqrt(H) dH, producing the stated closed form for time.Step-by-Step Solution:1) Write A(H) = pi (2 R H − H^2).2) Use A(H) dH/dt = −Cd a sqrt(2 g H).3) Rearrange and integrate from H1 to H2: t = (pi / (Cd a sqrt(2 g))) ∫_{H2}^{H1} (2 R H − H^2) / sqrt(H) dH.4) Evaluate: ∫(2 R H^{1/2} − H^{3/2}) dH = (4 R / 3) H^{3/2} − (2 / 5) H^{5/2}.5) Apply limits to obtain the final expression shown in Option A.Verification / Alternative check:Setting H2 = 0 yields the emptying time, matching known textbook forms for hemispherical tanks. Units reduce to seconds.Why Other Options Are Wrong:Option B is for a prismatic tank of constant area A. Option C assumes linear head dependence. Option E is for laminar tube draining or viscous head-loss models. Hence Option A is the correct hemispherical result; Option D is false.Common Pitfalls:Using the constant-area formula; misdefining H from the free surface; neglecting Cd which shortens predicted times if omitted.Final Answer:t = (pi / (Cd * a * sqrt(2 * g))) * [ (4 * R / 3) * (H1^(3/2) − H2^(3/2)) − (2 / 5) * (H1^(5/2) − H2^(5/2)) ]

Question 10

Venturi Flume — Typical Range of Coefficient of Discharge (Cd) For a properly calibrated Venturi flume operating under free-flow (submergence not excessive), the overall coefficient generally lies in which range?

Accepted Answer

Introduction:Venturi flumes (including Parshall and other throat-control flumes) are open-channel primary devices that accelerate flow to near-critical conditions at a throat. Their rating curves are usually expressed as Q = C * H^n, where C embeds geometry and a discharge coefficient. Because losses are small and flow passes at atmospheric pressure, the overall coefficient often approaches unity, higher than for sharp-crested weirs or orifices.Given Data / Assumptions:Properly constructed and installed Venturi/Parshall-type flume.Free-flow (not excessively submerged); head measured at the prescribed gage point.Reynolds number sufficiently high; approach is tranquil and well-conditioned with minimal sediment/debris.Concept / Approach:In a flume, energy loss between approach and throat is modest because there is no nappe contraction or enclosed jet as in weirs/orifices. The calibration therefore yields coefficients close to 1.00. Most practical tables show Cd clustered from about 0.97 to 1.00 depending on size and approach conditions.Step-by-Step Solution:Recognize that flumes operate under atmospheric pressure with streamlined contractions/expansions.Small head losses imply Cd near unity in rating equation Q = Cd * f(geometry, H).Hence the representative range is 0.95–1.00.Verification / Alternative check:Standard manuals list calibration constants that effectively correspond to Cd ≈ 1.0 for free flow. Submergence corrections reduce effective discharge, but under recommended submergence ratios the coefficient remains high.Why Other Options Are Wrong:0.30–0.45 and 0.50–0.75 are typical of highly lossy set-ups or sharp contractions, not Venturi flumes. 0.75–0.95 is closer but still low for well-designed free-flow flumes.Common Pitfalls:Applying free-flow calibration when submergence is significant; ignoring sediment build-up that alters throat geometry; measuring head at the wrong gage point.Final Answer:0.95 to 1.00

Question 11

Internal Mouthpiece — Running Full Criterion An internal (re-entrant) mouthpiece is said to be running full if the length of the mouthpiece exceeds which condition relative to the orifice diameter?

Accepted Answer

Introduction:A mouthpiece is a short tube attached to an orifice. For internal or re-entrant mouthpieces, the flow pattern depends on the length-to-diameter ratio. When the jet fills the passage and issues without vena contracta inside the tube, the mouthpiece is said to run full, which affects the applicable coefficients used in discharge formulas.Given Data / Assumptions:Internal mouthpiece attached to a thin-plate orifice.Steady, incompressible liquid; gravitational discharge.Criterion stated in terms of length vs. orifice diameter.Concept / Approach:Empirically, an internal mouthpiece runs full when its length L is sufficiently large compared to the orifice diameter d. The classical rule used in textbooks is: if L > 3d, the mouthpiece runs full. In that state, the flow behaves more like through a short tube than through an orifice with vena contracta, changing Cd and head-loss characteristics.Step-by-Step Solution:Identify the type: internal/re-entrant mouthpiece.Apply the criterion: L > 3d implies running full.Therefore the descriptive term to select is ‘‘full’’.Verification / Alternative check:Hydraulics lab observations show that as L increases beyond ~3d, the contraction diminishes and the jet adheres to the walls, filling the mouthpiece.Why Other Options Are Wrong:‘‘Free’’ refers to a jet discharging freely with a vena contracta; ‘‘partially’’ is non-standard terminology for the classification.Common Pitfalls:Confusing external (Borda) mouthpieces with internal; mixing the ‘‘running free’’ vs ‘‘running full’’ regimes.Final Answer:full

Question 12

Terminology — Short Pipe Fitted to an Orifice A pipe whose length is greater than about twice the orifice diameter, fitted externally or internally to the orifice, is termed as:

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Introduction:Distinguishing between notches, weirs, orifices, mouthpieces, and nozzles is foundational. Each device has a characteristic geometry and flow regime leading to different discharge equations and coefficients.Given Data / Assumptions:Short pipe attached to an orifice.Length exceeds roughly 2 times the orifice diameter.Liquid discharge under hydrostatic head.Concept / Approach:Mouthpieces are short tubes that modify the jet formation and reduce contraction relative to a thin-plate orifice. This changes coefficients (Cc, Cv, Cd) and energy losses. Notches/weirs are open-channel overflow structures; a nozzle is a shaped duct usually to convert pressure to velocity efficiently but is much longer and shaped than a simple short tube.Step-by-Step Solution:Check definition: short tube L ≳ 2d attached to an orifice = mouthpiece.Other terms do not match the described geometry.Therefore, select ‘‘mouthpiece’’.Verification / Alternative check:Textbook definitions consistently identify the L/d criterion to distinguish a mouthpiece from a mere orifice opening.Why Other Options Are Wrong:Notch/weir are overflow devices; nozzle implies a longer, shaped passage; they are not simply short pipes at orifices.Common Pitfalls:Calling any short attachment a nozzle; forgetting that mouthpieces may be internal or external and may run free or full.Final Answer:mouthpiece

Question 13

Venturi Flume — Pressure Regime During Flow In a Venturi flume operating correctly in an open channel, the water through the throat flows at which pressure condition?

Accepted Answer

Introduction:A Venturi/Parshall flume accelerates open-channel flow through a contraction to control discharge without creating enclosed pressurization. Knowing the pressure regime clarifies why head measurements are taken relative to the free surface and why coefficients are close to unity.Given Data / Assumptions:Open channel, free surface present.Flume properly aerated and not submerged beyond recommended ratio.Gages referenced to the free surface elevations.Concept / Approach:Open-channel flows communicate with the atmosphere; thus the static pressure at the free surface is atmospheric. In a Venturi flume, even at the throat where velocity is highest, the flow is still at atmospheric pressure because the water surface remains open and aerated. The energy change manifests as velocity head variation and a depression of water depth, not as a sealed internal pressure drop.Step-by-Step Solution:Identify the boundary condition: free surface ⇒ p = p_atm.At the throat, depth reduces and velocity increases, but the free surface still enforces p = p_atm.Therefore the correct statement is ‘‘atmospheric pressure’’.Verification / Alternative check:Calibration equations use water depth (head) rather than enclosed pressure; submerged flow corrections account for tailwater effects but still reference atmospheric conditions.Why Other Options Are Wrong:‘‘Gauge’’ or ‘‘absolute’’ here are distractions; the operative fact is that static pressure equals atmospheric in open-channel sections.Common Pitfalls:Confusing closed-conduit Venturi meters (pressurized) with open-channel Venturi flumes.Final Answer:atmospheric pressure

Question 14

Nozzle at Pipe End — Nature of Exit Jet A nozzle fitted at the end of a water pipeline primarily delivers water at which condition at the outlet?

Accepted Answer

Introduction:Nozzles are area-reducing devices that convert pressure energy into kinetic energy. Recognizing what changes predominantly at the exit—pressure or velocity—helps in selecting and sizing nozzles for jets, firefighting, turbine inlets, and cleaning applications.Given Data / Assumptions:Incompressible liquid water; negligible compressibility effects.Nozzle discharges to atmosphere.Steady operating condition; upstream has available pressure head.Concept / Approach:Bernoulli's principle for a streamline through the nozzle shows conversion of pressure head into velocity head. As area decreases, velocity rises while static pressure at the exit approaches atmospheric (for a free jet). Thus the signature trait of a nozzle is a high-velocity jet leaving at near-atmospheric pressure.Step-by-Step Solution:Apply energy equation: p/gamma + V^2/(2g) + z = constant (minus losses).Across the nozzle, p term decreases; V term increases.At the exit to atmosphere, p ≈ p_atm, and V is maximized ⇒ ‘‘high velocity’’.Verification / Alternative check:Measured jet speeds from hose nozzles confirm large velocities with only atmospheric static pressure at the exit.Why Other Options Are Wrong:‘‘High pressure’’ is upstream; ‘‘low velocity’’ contradicts nozzle function; ‘‘low pressure’’ alone is incomplete and misleading because the defining feature is high kinetic energy.Common Pitfalls:Assuming pressure remains high at the outlet; ignoring head losses that slightly limit achievable velocity.Final Answer:high velocity

Question 15

Venturimeter Design — Throat/Pipe Diameter Ratio to Avoid Separation To avoid flow separation at the throat in a Venturimeter, a commonly recommended ratio of throat diameter (d_t) to pipe diameter (D) is approximately:

Accepted Answer

Introduction:A Venturimeter measures discharge in closed conduits using a smooth converging section, a short throat, and a long diffuser. The throat/pipe diameter ratio (beta = d_t/D) influences pressure recovery, head loss, and the risk of separation. Picking a suitable beta ensures accurate differential readings while avoiding excessive losses or cavitation.Given Data / Assumptions:Incompressible liquid; Reynolds number high enough for fully turbulent core.Smooth, gradual contraction and diffuser angles within standard practice.No cavitation; adequate upstream straight length.Concept / Approach:Too small a throat (beta very low) yields very high velocities and large pressure drops, risking cavitation and unnecessary losses; too large a throat (beta too high) reduces measurable differential and can exacerbate sensitivity to installation effects. A widely adopted compromise is beta ≈ 0.5, which balances measurement differential and recovery while keeping diffuser angles gentle enough to prevent separation.Step-by-Step Solution:Define beta = d_t / D.Choose beta that ensures moderate acceleration and manageable diffuser expansion.Select the commonly recommended value beta ≈ 0.50 to avoid separation at the throat and allow good pressure recovery.Verification / Alternative check:Design handbooks list typical Venturimeter betas in the 0.4–0.6 range, with 0.5 a frequent default for water services to balance accuracy and losses.Why Other Options Are Wrong:0.25 and 0.33 produce very high velocities and risk cavitation; 0.80 yields too small a differential and can reduce accuracy and stability.Common Pitfalls:Ignoring diffuser angle limits (~5–7° half-angle typical) which are critical for preventing separation; neglecting upstream disturbances.Final Answer:0.50

Question 16

Open-Channel Regime — Depth Compared to Critical Depth If the flow depth y in an open channel is less than the critical depth y_c, how is the flow classified?

Accepted Answer

Introduction:Open-channel flow regimes are often described using the Froude number Fr = V / sqrt(g * y). The relationship between actual depth and critical depth determines whether the flow is tranquil (subcritical), critical, or torrential (supercritical), with major implications for transitions, control structures, and energy dissipation.Given Data / Assumptions:Prismatic channel with steady, uniform approximation locally.Depth y compared to critical depth y_c for the given discharge and geometry.Incompressible liquid; gravity-dominated flow.Concept / Approach:Critical flow corresponds to Fr = 1. For y 1 (supercritical). Traditional hydraulics terms this regime ‘‘torrential’’ flow. For y > y_c, Fr Step-by-Step Solution:Recall Fr = V / sqrt(g * y).If y decreases below y_c at fixed discharge, V increases and Fr > 1.Thus the correct classification is ‘‘torrential flow’’ (supercritical).Verification / Alternative check:Hydraulic jumps occur when torrential flow transitions abruptly to tranquil flow, a standard example demonstrating Fr > 1 upstream and Fr Why Other Options Are Wrong:‘‘Tranquil’’ corresponds to y > y_c; ‘‘critical’’ is y = y_c; ‘‘turbulent’’ refers to laminar vs turbulent (Reynolds-based), not Froude regime.Common Pitfalls:Confusing turbulence (a Re effect) with regime (a Fr effect); assuming depth alone determines discharge without considering velocity.Final Answer:torrential flow

Question 17

Fluid Classification — Definition of Non-Newtonian Fluid State whether the following statement is correct: ‘‘A fluid whose viscosity changes with the rate of deformation (shear rate) is called a non-Newtonian fluid.’’

Accepted Answer

Introduction:Rheology categorizes fluids by how shear stress relates to shear rate. Newtonian fluids have a linear relation with constant dynamic viscosity mu. Non-Newtonian fluids deviate from this linear law; their apparent viscosity depends on shear rate and sometimes time, enabling behaviors like shear-thinning, shear-thickening, or yielding.Given Data / Assumptions:Isothermal conditions; no phase change.Homogeneous fluid without large particles (unless considered as a suspension).Shear rate may vary over a practical range.Concept / Approach:Newtonian: tau = mu * (du/dy), mu constant. Non-Newtonian: functional relation tau = f(du/dy) with effective viscosity mu_app = tau / (du/dy) varying with shear rate. Examples include polymer solutions (shear-thinning), corn-starch mixtures (shear-thickening), and Bingham plastics (yield stress).Step-by-Step Solution:Examine the definition given.Compare with rheological laws: if viscosity depends on shear rate, it is non-Newtonian.Therefore the statement is correct.Verification / Alternative check:Viscometer data for paints or blood show viscosity decreasing with increasing shear rate (shear-thinning), validating the definition.Why Other Options Are Wrong:‘‘False’’ would imply shear-rate independence, which is the Newtonian case, contradicting the statement.Common Pitfalls:Confusing temperature dependence (all fluids have it) with shear-rate dependence; assuming time-dependent thixotropy is required—many non-Newtonian fluids are time-independent but shear-rate dependent.Final Answer:True

Question 18

Rheology — Fluid Obeying Newton’s Law of Viscosity A fluid that obeys Newton’s law of viscosity (shear stress proportional to shear rate with constant viscosity) is termed a:

Accepted Answer

Introduction:Newton's law of viscosity underpins many engineering calculations: shear stress tau is proportional to velocity gradient du/dy with proportionality mu (dynamic viscosity). Fluids following this linear relation with mu approximately constant are classified as Newtonian.Given Data / Assumptions:Continuum assumption valid; single-phase fluid.Shear rates within a range where mu does not depend on shear rate.Temperature uniform (to keep mu constant).Concept / Approach:Newtonian: tau = mu * (du/dy). Non-Newtonian fluids violate this linearity or constancy of mu. ‘‘Ideal’’ fluid is inviscid (mu = 0) — a mathematical abstraction. ‘‘Real’’ includes all actual fluids, Newtonian or not, so it is not specific enough.Step-by-Step Solution:Match the definition given to the rheological class.Linear stress–strain-rate with constant mu ⇒ Newtonian fluid.Therefore choose ‘‘newtonian fluid’’.Verification / Alternative check:Water and air behave Newtonian across wide shear-rate ranges, confirming the classification.Why Other Options Are Wrong:‘‘Real’’ is too broad; ‘‘ideal’’ ignores viscosity; ‘‘non-Newtonian’’ contradicts the constant-mu law.Common Pitfalls:Assuming any temperature dependence disqualifies Newtonian behavior—Newtonian refers to shear-rate independence, not temperature independence.Final Answer:newtonian fluid

Question 19

Dam Spillways — Weir Profile Commonly Used Which type of weir profile is most commonly adopted for dam spillways to match the underside of the free overfall (design nappe) and pass large discharges efficiently?

Accepted Answer

Introduction:Spillways must pass flood flows safely while minimizing cavitation and structural loads. The crest shape that conforms to the underside of the ideal nappe for a sharp-crested overflow is called the ogee profile and is widely used in dam engineering.Given Data / Assumptions:High-head spillway on a dam.Free overflow with adequate aeration.Crest designed for the project design head.Concept / Approach:An ogee weir has an S-shaped crest approximating the lower nappe of a sharp-crested weir at the design head. This alignment reduces flow separation, minimizes negative pressures, and improves discharge capacity compared to rectangular or broad-crested forms for the same head.Step-by-Step Solution:Identify the functional need: efficient, safe spillway discharge.Match to crest profile that follows the design nappe: ogee.Conclude that ogee weirs are the standard for dam spillways.Verification / Alternative check:Dam design manuals and case studies show ogee crests as the prevalent choice for gravity and arch dams under free overflow conditions.Why Other Options Are Wrong:Narrow or broad-crested weirs are used in canals and measurement; submerged weirs operate under tailwater control and are not typical spillway crests.Common Pitfalls:Using an ogee crest without considering aeration or cavitation protection; failing to adjust the crest for different design heads.Final Answer:Ogee weir

Question 20

Critical flow in open channels: At the critical depth (y_c), what happens to the discharge for a given specific energy or channel control condition?

Accepted Answer

Introduction / Context:Critical flow is a cornerstone idea in open-channel hydraulics. It occurs at the critical depth y_c where the Froude number equals 1. For a given specific energy or at a control (such as a sluice or section where energy is fixed), the discharge reaches an extremum at y_c. Understanding this relationship helps in designing flumes, measuring structures, and recognizing flow regime transitions.Given Data / Assumptions:Prismatic channel, typically analyzed as rectangular for unit-width reasoning.Steady, incompressible flow with negligible losses at the control section.Specific energy E is considered given (e.g., by an upstream control).Concept / Approach:For a rectangular channel of unit width, discharge per unit width is q = v * y. Specific energy is E = y + v^2/(2g) = y + (q^2)/(2g y^2). For fixed E, differentiate q with respect to y and set derivative to zero to locate the extremum. The condition for extremum yields v^2 = g y, i.e., critical state (Fr = 1). Evaluating the second-derivative or using standard results shows that the extremum is a maximum, not a minimum.Step-by-Step Solution:Write E(y) = y + q^2/(2g y^2) for unit width.For given E, solve dE/dy = 0 → v^2 = g y (critical).At this y_c, the achievable q is extremized; analysis shows it is the maximum discharge for the given E.Hence, “discharge corresponding to critical depth” (at a control with fixed E) is maximum.Verification / Alternative check:Dimensionally, the well-known formula for a rectangular channel is y_c = (q^2/g)^{1/3}. Rearranged, q = (g y_c^3)^{1/2}. For any other y at the same E, q is smaller, confirming maximality at y_c.Why Other Options Are Wrong:Zero: Critical flow is a finite, nonzero discharge state.Minimum: Minimum specific energy occurs at y_c, not minimum discharge for given E.Indeterminate: It is determinate and known to be a maximum.Common Pitfalls:Confusing “minimum specific energy” at y_c with “minimum discharge”; mixing up conditions “for given E” versus “for given q”.Final Answer:maximum

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