Question 1

Nozzle geometry for liquids (general practice): Which overall shape is a nozzle generally made in for subsonic liquid jets and efficient discharge?

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Introduction / Context:Nozzles convert pressure energy into kinetic energy. Their geometry depends on the working fluid and target Mach number. For water and most engineering liquid applications (subsonic), practical nozzles are convergent to accelerate the flow and form a coherent jet for power transmission (e.g., Pelton wheels), cutting, or cleaning.Given Data / Assumptions:Incompressible liquid (water-like), subsonic regime.Applications requiring high exit velocity and compact jet.Negligible compressibility effects compared with gases.Concept / Approach:For incompressible or low-Mach flows, continuity and Bernoulli imply that reducing area increases velocity and reduces static pressure toward the outlet. A convergent profile assists smooth acceleration with minimal separation. Convergent–divergent (de Laval) profiles are used for compressible gases to produce supersonic jets; not typically necessary for liquids.Step-by-Step Solution:Identify fluid: liquid at M ≪ 1 → density nearly constant.Decrease area (convergent) → velocity increases (v ∝ 1/A).Select the geometry minimizing losses and maximizing jet coherence: convergent.Verification / Alternative check:Hydraulic machinery references list convergent nozzles for water jets; C-D nozzles are reserved for gas dynamics when choking and supersonic expansion are required.Why Other Options Are Wrong:Cylindrical: Does not accelerate the flow appreciably; large losses at entry/exit.Divergent: Decelerates subsonic flow; opposite of desired effect.Convergent–divergent: Appropriate for compressible gas flows aiming for M > 1, not typical water jets.Common Pitfalls:Generalizing gas-dynamics nozzle shapes to liquids; overlooking separation risk from abrupt contractions.Final Answer:convergent shape

Question 2

Darcy–Weisbach friction loss in a circular pipe: Select the correct head loss expression (f = Darcy's friction coefficient, l = pipe length, d = diameter, v = mean velocity).

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Introduction / Context:The Darcy–Weisbach equation gives a unified way to compute head loss due to wall friction in pipes, valid across laminar and turbulent regimes when the appropriate friction factor is known. Many textbooks define f as Darcy’s coefficient and include a factor of 4 compared with the Fanning factor. Choosing the correct constant and velocity power is essential.Given Data / Assumptions:Steady, incompressible, fully developed internal flow in a straight circular pipe.f is the Darcy (not Fanning) friction factor.v is mean velocity; elevation changes are negligible for the segment considered.Concept / Approach:The Darcy–Weisbach head loss is h_f = f_D * (l/d) * (v^2/(2g)) when f_D denotes the Darcy factor. Some Indian texts denote “coefficient of friction f” such that the formula appears as h_f = 4 f (l/d) (v^2/(2g)). Here the options map to that convention, making option A correct.Step-by-Step Solution:Start with h_f = f_D * (l/d) * (v^2/(2g)).If f in the options = f/4 (Fanning) then constants differ; but the question states “Darcy’s coefficient”, implying Darcy form.Match to choices: (4 * f * l * v^2)/(2 g d) equals f_D (l/d) (v^2/(2g)) with f_D = 4 f.Verification / Alternative check:Using Moody chart values (Darcy factor), numerical results align with the A-form; converting to Fanning requires dividing by 4.Why Other Options Are Wrong:B: Missing the factor 4 under this notation.C: Uses v to the first power, which is dimensionally inconsistent for energy loss.D: Same as A but doubled overall—incorrect constant.Common Pitfalls:Mixing Darcy and Fanning definitions; using centerline velocity instead of mean; ignoring minor losses (separate from friction loss).Final Answer:h_f = (4 * f * l * v^2) / (2 * g * d)

Question 3

Onset of flow separation (boundary layer concept): “The separation of flow occurs when the hydrodynamic boundary-layer thickness is reduced to zero.” State whether this is correct.

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Introduction / Context:Flow separation is a key phenomenon in external aerodynamics and internal flows with adverse pressure gradients. It changes drag, lift, and pressure recovery dramatically. A clear understanding of the boundary-layer behavior at separation avoids common misconceptions stated in the prompt.Given Data / Assumptions:Viscous flow over a wall with possible adverse pressure gradient.Boundary layer defined by significant velocity gradient near the wall.No special assumption on laminar/turbulent beyond applicability.Concept / Approach:Separation is characterized by the wall shear stress falling to zero and then becoming negative as the near-wall flow reverses. The boundary layer does not shrink to zero thickness; rather, it thickens markedly approaching separation. The velocity profile develops an inflection and near-wall backflow forms; the locus of zero wall shear marks the separation point.Step-by-Step Solution:Apply boundary-layer momentum integral: wall shear τ_w → 0 at separation.Adverse pressure gradient causes deceleration within the layer.Near-wall u decreases, reversal can occur, and δ (thickness) grows, not vanishes.Therefore, the statement that δ → 0 at separation is false.Verification / Alternative check:Classical measurements on airfoils and diffusers show rising shape factor H and growing δ before separation. Surface oil-flow and tuft visualizations confirm flow reversal at the wall where τ_w = 0.Why Other Options Are Wrong:Agree: Contradicts boundary-layer theory and experiments.Only true at high Re / only for laminar: Separation criterion is τ_w = 0, not δ = 0, in both laminar and turbulent contexts.Common Pitfalls:Confusing “thin boundary layer” with “no boundary layer”; believing separation is a thinning event rather than a detachment triggered by adverse pressure gradient.Final Answer:Disagree

Question 4

Gradually varied flow (GVF) in open channels: How is the flow classified when depth varies slowly along the channel but does not change with time?

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Introduction / Context:Open-channel flows are classified by temporal variation (steady vs unsteady) and spatial variation (uniform vs non-uniform). Gradually varied flow (GVF) is ubiquitous—occurring upstream of dams, along mild slopes, and near control structures. Correct classification underpins the use of the GVF differential equation and profile analysis (M1, M2, S1, etc.).Given Data / Assumptions:Depth and velocity change with distance but slowly (gradually), so hydrostatic pressure distribution holds.No change with time at a fixed section (steady).Prismatic channel with slowly varying boundary conditions.Concept / Approach:“Steady” means ∂()/∂t = 0; “uniform” means properties are constant along the channel (d()/dx = 0). GVF has d()/dx ≠ 0 but small, and ∂()/∂t = 0. Hence it is steady and non-uniform. The gradual assumption allows using hydrostatic pressure and the energy (or momentum) approach with small-slope approximations.Step-by-Step Solution:Identify temporal behavior: depth does not change with time → steady.Identify spatial behavior: depth varies with x → non-uniform.Therefore classification = steady non-uniform.Verification / Alternative check:GVF equation dy/dx = (S_0 − S_f)/(1 − Fr^2) is a spatial differential relation, appropriate only when the flow is steady and depth varies slowly with x.Why Other Options Are Wrong:Steady uniform: Would require constant depth along x (normal depth everywhere).Unsteady uniform / unsteady non-uniform: These imply time variation, not given here.Common Pitfalls:Confusing GVF with rapidly varied flow (RVF) such as hydraulic jumps; assuming “gradually varied” means time-varying rather than space-varying.Final Answer:steady non-uniform flow

Question 5

Relative discharge of mouthpieces: The discharge through a convergent mouthpiece is ________ the discharge through an internal (re-entrant/Borda's) mouthpiece of the same diameter and head.

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Introduction / Context:Mouthpieces are short tubes fitted to orifices. Their discharge differs with geometry because coefficients change. A re-entrant (Borda’s) mouthpiece has a much smaller coefficient of discharge than a well-designed convergent mouthpiece, so, for the same head and diameter, the convergent type delivers substantially more flow.Given Data / Assumptions:Same diameter and same head of water.Well-formed convergent mouthpiece running full.Internal re-entrant mouthpiece representing Borda’s type.Concept / Approach:Discharge Q = C_d * A * sqrt(2 * g * H). For re-entrant mouthpiece, C_d is about 0.5. For a convergent mouthpiece, C_d commonly lies around 0.85–0.95 (depending on details). The ratio of discharges is therefore roughly 0.9/0.5 ≈ 1.8, often summarized in basic MCQs as “about twice.”Step-by-Step Solution:Q_conv / Q_reentrant = (C_d,conv / C_d,reentrant) * (A/A) * (same head factor)≈ 0.9 / 0.5 = 1.8 → “approximately double”.Therefore, among the discrete options, “double” is the best match.Verification / Alternative check:Handbook tables consistently show re-entrant coefficients near 0.5 and convergent near 0.9 for water at moderate heads, supporting a near-two ratio.Why Other Options Are Wrong:Equal to: Ignores large coefficient differences.One-half: Reverses the comparison.Three fourth: Too low for typical data; underestimates improvement.Common Pitfalls:Confusing orifice coefficients (≈0.62) with mouthpiece coefficients; comparing to a short cylindrical mouthpiece running free instead of re-entrant.Final Answer:double

Question 6

Stagnation concepts in fluid flow: “The pressure at a stagnation point is always low.” State whether this statement is correct.

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Introduction / Context:A stagnation point is where the local fluid velocity relative to the body becomes zero. Bernoulli’s principle (for incompressible, inviscid regions) shows that as kinetic energy per unit weight converts to pressure head, the static pressure rises. Thus, stagnation pressure is the highest pressure felt by a probe aligned with the flow (ignoring viscous losses within the probe).Given Data / Assumptions:Subsonic, incompressible flow near a body.Negligible height change locally.Use of Bernoulli along a streamline into the stagnation point.Concept / Approach:Bernoulli: p/ρg + v^2/(2g) ≈ constant along a streamline (inviscid). At a stagnation point, v → 0, so p_stag = p_static + ρ v^2/2. Therefore, stagnation pressure exceeds the local static pressure, not lower. In compressible flows, total pressure includes compressibility effects but still represents a maximum for isentropic deceleration.Step-by-Step Solution:Take a streamline from free stream (p∞, v∞) to the stagnation point.Apply Bernoulli: p_stag = p∞ + ρ v∞^2 / 2.Since ρ v∞^2/2 > 0, p_stag > p∞; it is not low.Verification / Alternative check:Pitot tubes measure stagnation pressure, then subtract static pressure to yield dynamic pressure; the reading is positive, reinforcing that stagnation pressure is higher.Why Other Options Are Wrong:Agree: Contradicts Bernoulli and standard measurements.Only true in turbulent / high Mach: Neither turbulence nor Mach alone reverses the pressure rise on deceleration to zero speed.Common Pitfalls:Confusing base pressure in separated wakes (often low) with stagnation pressure at the nose; mixing static and total (stagnation) pressure terminology.Final Answer:Disagree

Question 7

Venturimeter design detail: Why is the divergent (diffuser) portion made longer than the convergent portion?

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Introduction / Context:A venturimeter incorporates a diffuser to recover pressure after the throat. Proper diffuser design is essential to convert kinetic energy back to static pressure while keeping losses small. The length and angle of the diffuser are selected to prevent boundary-layer separation and to minimize loss coefficients.Given Data / Assumptions:Incompressible, subsonic internal flow.Adverse pressure gradient present in the diffuser.Goal: high pressure-recovery coefficient with low loss.Concept / Approach:Diffusers with too-steep divergence angles cause boundary-layer separation (“breaking away of the stream”), which dramatically increases form drag and losses. A longer, gentler diffuser maintains attached flow, improving pressure recovery and reducing friction/form losses. Hence both preventing separation and minimizing losses motivate a longer diffuser than the convergent section.Step-by-Step Solution:Increase area gradually to reduce deceleration rate.Keep wall divergence angle modest to avoid separation.Result: better pressure recovery and lower loss coefficient.Verification / Alternative check:Empirical design charts recommend small half-angles (≈3–7 degrees) for diffusers to balance length and losses, confirming the need for a longer diffuser.Why Other Options Are Wrong:(a) alone or (b) alone: Both reasons are simultaneously valid.None of these: Contradicted by diffuser design practice.Common Pitfalls:Assuming symmetry between convergent and divergent sections; overlooking how adverse pressure gradients trigger separation if expansion is too rapid.Final Answer:both (a) and (b)

Question 8

Definition and unit of discharge: “The quantity of liquid flowing per second through a cross-section is called discharge and is measured in m^3/s (cumecs).” State whether this statement is correct.

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Introduction / Context:Discharge (or volumetric flow rate) is the most common measure of flow quantity in hydraulics and many process industries. It directly connects to continuity, pump selection, and measurement with devices like orifices, weirs, and flowmeters.Given Data / Assumptions:Steady or unsteady flow through a defined cross-section.Volumetric measure emphasized.SI units in use.Concept / Approach:By definition, discharge Q is volume per unit time crossing an area: Q = dV/dt. In SI, the unit is cubic meters per second (m^3/s), colloquially “cumec.” The concept applies to liquids and gases; the definition does not depend on regime (laminar/turbulent) or compressibility (though density variation matters for mass flow rate).Step-by-Step Solution:Define Q = ∫_A v_n dA, where v_n is normal velocity component.For uniform velocity: Q = v * A; for nonuniform: use area integral.Units: [Q] = m^3/s (cumecs) in SI.Verification / Alternative check:Other common units include L/s and m^3/h; conversions preserve the meaning. Mass flow rate is ṁ = ρ Q, distinct from Q but related.Why Other Options Are Wrong:False: Conflicts with standard definition and SI unit.Only true for gases / laminar: Discharge definition is universal.Common Pitfalls:Mixing volumetric and mass flow rates; calling m^3/h “cumecs” (the term “cumec” specifically implies per second).Final Answer:True

Question 9

Total head of a moving liquid particle (Bernoulli form): The total head equals the sum of which components?

Accepted Answer

Introduction / Context:Bernoulli’s equation expresses energy conservation along a streamline for steady, incompressible, inviscid flow with negligible shaft work and heat exchange. In “head” form (energy per unit weight), the total head is the sum of elevation, pressure, and velocity heads. Recognizing these components is fundamental for pump sizing, flow measurement, and piping calculations.Given Data / Assumptions:Steady, incompressible flow.No shaft work or heat transfer across the streamline section considered.Negligible viscous losses between the two points (or accounted separately as head loss).Concept / Approach:Total head H_t is H_t = z + p/(ρ g) + v^2/(2 g). Here, z is potential (elevation) head, p/(ρ g) is pressure head, and v^2/(2 g) is kinetic (velocity) head. In real systems, an additional loss term h_f is subtracted when moving from one section to another to account for friction and fittings.Step-by-Step Solution:Write Bernoulli: z_1 + p_1/(ρg) + v_1^2/(2g) = z_2 + p_2/(ρg) + v_2^2/(2g) + h_f − h_p + h_t.Identify “total head” at a point as H_t = z + p/(ρg) + v^2/(2g).Therefore, the correct composition is pressure head + kinetic head + potential head.Verification / Alternative check:Piezometric head is z + p/(ρg); adding kinetic head gives total head. Pitot measurements recover total (stagnation) pressure related to total head via p_t/(ρg) = z_ref difference + v^2/(2g) when referenced appropriately.Why Other Options Are Wrong:Subtractive forms contradict energy conservation statement in the absence of loss/work terms.Common Pitfalls:Confusing piezometric vs total head; forgetting to include elevation in sloped piping; misapplying head loss signs.Final Answer:pressure head + kinetic head + potential head

Question 10

Archimedes’ law of buoyancy – magnitude of the buoyant force For a body immersed wholly or partially in a fluid at rest, the buoyant force acting on the body is always __________ the weight of the fluid displaced by the body.

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Introduction:Buoyancy is a cornerstone concept in fluid mechanics and naval architecture. The question tests the precise statement of Archimedes’ law regarding the magnitude of the buoyant force on a body immersed in a static fluid. Understanding this equality is essential for predicting floatation, apparent weight loss, and stability analyses.Given Data / Assumptions:Fluid at rest (hydrostatics).Body may be fully or partially submerged.Gravitational field is uniform; fluid density is constant.No fluid motion–induced dynamic effects (no lift or drag).Concept / Approach:Archimedes’ principle states: the buoyant force on a submerged body equals the weight of the fluid displaced by that body. This arises from integrating the hydrostatic pressure over the body's surface. Pressure increases with depth, producing a net upward resultant equal to rho * g * V_displaced, whose weight is rho * g * V_displaced. Hence, the magnitude of buoyant force equals the weight of fluid displaced, independent of the body's own material (though floatation depends on comparing forces).Step-by-Step Solution:Step 1: Recognize hydrostatic pressure p = p0 + rho * g * h acts normally on surfaces.Step 2: Integrate pressure over the body surface; vertical components at different depths do not cancel.Step 3: Net upward force obtained equals rho * g * V_displaced, which is the weight of the displaced fluid.Step 4: Conclude equality: F_buoyancy = weight of displaced fluid.Verification / Alternative check:For a floating object in equilibrium, weight of body W equals buoyant force, so W = rho * g * V_displaced. For a fully submerged, tethered body, the scale reading equals W − F_buoyancy, again showing F_buoyancy equals the displaced fluid weight irrespective of the body's density in still fluid.Why Other Options Are Wrong:Less than / more than: Contradict the exact equality from Archimedes’ law.Dependent on body shape only: Shape affects displaced volume but not the equality relation.Zero unless floating: Even a fully submerged object experiences buoyancy.Common Pitfalls:Confusing buoyant force with net force; net force depends on comparing buoyancy with the object’s weight. Also, mixing hydrostatic buoyancy with dynamic lift that occurs in moving fluids.Final Answer:equal to

Question 11

Orifices – dependence of jet velocity on head In a tank discharging through a sharp-edged orifice, the exit velocity of the liquid varies with the available head over the orifice. Do you agree with this statement?

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Introduction:Discharge through orifices is a staple topic because it links hydrostatics (available head) with kinematics (jet speed). The question checks whether the learner recognizes the dependence of exit velocity on the head of liquid in the reservoir feeding the orifice.Given Data / Assumptions:Large reservoir feeding a sharp-edged orifice of small area.Head H measured from the orifice centre to the free surface (or the upstream–downstream head difference for a submerged orifice).Steady conditions during the instant of measurement; coefficient of velocity Cv approximately constant.Concept / Approach:Torricelli’s relation gives the ideal jet speed V_ideal = sqrt(2 * g * H). Real jets satisfy V = Cv * sqrt(2 * g * H), showing that velocity varies with the square root of the available head. As H increases, the velocity increases according to the square-root law. The same head variable appears in the discharge Q = Cd * a * sqrt(2 * g * H), where Cd is the discharge coefficient and a is orifice area.Step-by-Step Solution:Step 1: Identify the head H controlling energy at the orifice.Step 2: Write V = Cv * sqrt(2 * g * H).Step 3: Conclude that velocity varies with H (specifically with sqrt(H)).Step 4: Note that Cv Verification / Alternative check:Dimensional analysis shows velocity has units of sqrt(length * acceleration); only a relation like sqrt(2 * g * H) produces correct dimensions from head. Laboratory orifice experiments confirm the square-root trend.Why Other Options Are Wrong:Disagree: Ignores Torricelli’s law and extensive experimental evidence.Viscosity-zero caveat: Real fluids have viscosity; Cv incorporates non-ideal effects while retaining sqrt(H) scaling.Submerged-only or gases-only: Head dependence applies to both free and submerged orifices (with appropriate H) and to liquids in general.Common Pitfalls:Confusing velocity dependence with area dependence (which controls discharge, not velocity), or forgetting that for a drowned orifice H is the free-surface difference across the opening.Final Answer:Agree

Question 12

Dynamics to statics transformation – name the principle A moving fluid mass can be analyzed at a notional static equilibrium by applying an imaginary inertia force equal in magnitude and opposite in direction to the real accelerating force. This sta…

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Introduction:Converting a dynamics problem into an equivalent statics problem by introducing inertia forces is a classic technique in mechanics and fluid dynamics. The question asks for the name of the principle embodying this equivalence for accelerating systems.Given Data / Assumptions:Finite mass of fluid undergoes acceleration.We conceptually add an inertia force equal and opposite to m * a.Analysis is then performed as if forces sum to zero (static equilibrium).Concept / Approach:D’Alembert’s principle reformulates Newton’s second law as an equilibrium condition by adding the fictitious inertia force F_inertia = − m * a to the system’s external forces. In fluid mechanics, this is used in accelerating containers (e.g., tilting free surfaces), rotating systems, and unsteady Bernoulli derivations, enabling static-like pressure field computations in non-inertial frames.Step-by-Step Solution:Step 1: Start from Newton’s second law: ΣF = m * a.Step 2: Add − m * a to both sides to form ΣF + (− m * a) = 0.Step 3: Interpret the augmented force system as static equilibrium in a non-inertial frame.Step 4: Use the static balance to compute pressure distributions or resultant forces.Verification / Alternative check:Applications include deriving pressure fields in linearly accelerated tanks (free surface tilts with tan(theta) = a/g) and rotating fluids (parabolic free surface), both obtained via D’Alembert’s construct.Why Other Options Are Wrong:Pascal / Archimedes / Bernoulli: Respectively deal with pressure transmission, buoyancy, and mechanical energy conservation, not the statics–dynamics transformation.None of these: Incorrect because the statement exactly matches D’Alembert’s principle.Common Pitfalls:Calling the method “fictitious forces” and assuming they are arbitrary. They are systematic constructs that allow equilibrium methods in accelerating frames.Final Answer:D'Alembert's principle

Question 13

Basic properties – numerical value of the specific gravity of water For engineering calculations, the specific gravity (relative density) of pure water is conventionally taken as:

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Introduction:Specific gravity, also called relative density, is a dimensionless ratio comparing the density of a substance to that of a reference fluid (for liquids, typically water at standard conditions). The question asks for the conventional value used for water itself.Given Data / Assumptions:Reference temperature near 4°C for exact unity or room temperature for approximate unity.Reference fluid for liquids is water unless stated otherwise.Dimensionless ratio SG = rho_substance / rho_water.Concept / Approach:By definition, when the substance is water and the reference is water, the ratio is unity. Although water density varies slightly with temperature and pressure, for most engineering problems SG(water) is taken as 1. This simplifies conversions between mass density and specific gravity in tabulations and calculations.Step-by-Step Solution:Step 1: Write SG = rho / rho_water.Step 2: For water as the substance, rho = rho_water → SG = 1.Step 3: Conclude the conventional value used in practice is 1.Verification / Alternative check:At 4°C, water density is approximately 1000 kg/m^3 by definition in SI, making SG precisely 1. At 20°C, density is about 998 kg/m^3; SG is still effectively 1 for routine work.Why Other Options Are Wrong:0.001 / 0.01 / 0.1 / 10: These values correspond to unrealistic relative densities for water under normal conditions.Common Pitfalls:Confusing specific gravity with specific weight or mass fraction. Specific gravity is unitless and compares densities directly.Final Answer:1

Question 14

Rotating liquid in a closed, completely filled cylindrical vessel In a closed cylindrical vessel completely filled with liquid and rotated about its vertical axis at constant speed, the total resultant force on the bottom equals the sum of the total…

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Introduction:Fluids in rigid-body rotation develop a characteristic pressure field. This question probes whether the integrated force on the vessel bottom accounts both for the hydrostatic weight and for the additional pressure due to rotation (often nicknamed centrifugal pressure) when the container is completely filled and sealed.Given Data / Assumptions:Cylindrical vessel of plan area A and liquid height H, completely filled.Rotation at angular speed omega about the vertical axis; steady rigid-body rotation (no relative shear).Density rho, gravity g, no free surface deformation due to the closed top.Concept / Approach:The pressure field in rigid-body rotation is p(r, z) = p0 + rho * g * (H − z) + rho * omega^2 * r^2 / 2, where r is radial distance and z is measured upward from the bottom. The force on the bottom is F_bottom = ∫ p dA evaluated at z = 0 over the circular area. The first vertical term integrates to rho * g * H * A, equal to the liquid's weight. The rotational term integrates to rho * omega^2 / 2 * ∫ r^2 dA, the resultant of the so-called centrifugal-pressure distribution. Their sum (plus any uniform reference pressure p0 * A that cancels with external support) is the total bottom force, validating the statement.Step-by-Step Solution:Step 1: Write p(r, 0) = p0 + rho * g * H + rho * omega^2 * r^2 / 2.Step 2: Integrate over area: F = ∫_A p dA = p0 * A + rho * g * H * A + (rho * omega^2 / 2) ∫_A r^2 dA.Step 3: Recognize rho * g * H * A = weight of liquid (W).Step 4: Identify the third term as the total centrifugal-pressure resultant on the bottom.Step 5: Conclude F_bottom = W + (centrifugal resultant) + p0 * A, confirming the statement.Verification / Alternative check:If the top were open, a parabolic free surface would form; still, integrating p on the bottom gives W plus a rotational contribution relative to the free-surface reference. Thus, the inclusion of both components is physically consistent.Why Other Options Are Wrong:Incorrect / side-walls only: Although radial pressure acts on side walls, the rotational term also appears in bottom pressure because p depends on r^2 at the bottom.Conditional options: The relation does not require very small rotation or an open top for its validity.Common Pitfalls:Assuming that centrifugal effects do not influence vertical forces; they do via the r^2 dependence of pressure at the bottom surface.Final Answer:Correct

Question 15

Compressible-flow regimes – classify by Mach number According to standard nomenclature, when the Mach number exceeds 6, the flow is called:

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Introduction:Mach number M = V / a (flow speed divided by local speed of sound) classifies compressible-flow regimes. Recognizing regime boundaries is crucial for aerothermodynamics, nozzle design, and high-speed testing.Given Data / Assumptions:Standard regime names: subsonic, transonic, supersonic, hypersonic.Approximate thresholds used in practice.Concept / Approach:Common conventions are: subsonic M 6, which clearly lies in the hypersonic regime. Therefore the correct label is hypersonic (often spelled “hypersonic”; some sources write “hyper-sonic” in options).Step-by-Step Solution:Step 1: Compare M with regime thresholds.Step 2: Since M > 6, select hypersonic category.Verification / Alternative check:Aircraft like the reentry vehicles and certain missiles operate at hypersonic speeds (M ≫ 5), confirming the classification.Why Other Options Are Wrong:Subsonic / sonic / supersonic / transonic: Do not cover M greater than 6.Common Pitfalls:Assuming an exact single cutoff everywhere; while literature varies between 5 and 6, M > 6 is unquestionably hypersonic.Final Answer:hyper-sonic flow

Question 16

Definition check – meaning of sonic flow When the Mach number equals unity (M = 1), the flow is called sonic flow. Is this statement true or false?

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Introduction:Terminology in compressible flow hinges on Mach number M = V / a. The label “sonic” corresponds to the special condition where the flow speed matches the local speed of sound, and it underpins critical-flow phenomena like choking in nozzles.Given Data / Assumptions:M is computed with the local speed of sound a, not a reference value.Medium can be gas or, in principle, any compressible fluid with a defined a.No additional thermodynamic restrictions are required for the definition.Concept / Approach:By definition, sonic conditions occur when V = a, i.e., M = 1. This is independent of whether the process is isothermal or adiabatic; those assumptions matter for evaluating a, not for the naming convention. Sonic points often correspond to critical sections in nozzles where mass flow becomes maximum for given upstream conditions.Step-by-Step Solution:Step 1: Write M = V / a.Step 2: For M = 1, V equals the local speed of sound.Step 3: This is, by definition, called sonic flow.Verification / Alternative check:Nozzle theory and shock-tube experiments consistently identify the M = 1 location as the sonic or critical condition, confirming the terminology.Why Other Options Are Wrong:False: Contradicts the standard definition.Conditional statements about isothermal, ideal gas, or incompressible flows are irrelevant to the naming convention.Common Pitfalls:Using a nonlocal value of a to compute M; always use the local thermodynamic state.Final Answer:True

Question 17

Units – identify the CGS unit of dynamic viscosity Poise is used as the unit of viscosity in the centimetre–gram–second (CGS) system. Do you agree?

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Introduction:Engineering uses both SI and legacy CGS units for viscosity. The question asks whether “poise” belongs to the CGS system and what type of viscosity it measures.Given Data / Assumptions:Two viscosity types: dynamic (mu) and kinematic (nu).CGS base units: centimetre, gram, second.Common submultiples: centipoise (cP) and centistokes (cSt).Concept / Approach:In CGS, the unit of dynamic viscosity is poise (P), where 1 P = 1 g/(cm·s). In SI, dynamic viscosity is in Pa·s (N·s/m^2). Kinematic viscosity's CGS unit is stokes (St), 1 St = 1 cm^2/s. Therefore, it is correct to say “poise is the unit of viscosity in CGS,” with the understanding it refers to dynamic viscosity. Water at 20°C has mu ≈ 1 cP (0.01 P) and nu ≈ 1 cSt.Step-by-Step Solution:Step 1: Distinguish mu (dynamic) vs nu (kinematic).Step 2: Recall CGS units: P for mu, St for nu.Step 3: Confirm the statement is correct for dynamic viscosity.Verification / Alternative check:Unit conversions: 1 P = 0.1 Pa·s; 1 cP = 1 mPa·s. These standard conversions confirm P is a CGS unit, not SI.Why Other Options Are Wrong:Kinematic-only or SI-only claims misidentify which viscosity and which system uses poise.Gas-only qualifier is unnecessary; units apply to any fluid.Common Pitfalls:Mixing up poise (dynamic) with stokes (kinematic) or reporting SI values with CGS unit names.Final Answer:Agree

Question 18

Dimensionless numbers – is the reciprocal of Euler number called the Newton number? The statement “The reciprocal of the Euler number is called the Newton number” is:

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Introduction:Dimensionless numbers help compare the relative importance of physical effects in fluid systems. Euler number (Eu) and Newton number (often used to mean power number, Np) describe different ratios. The question challenges a common confusion about their relationship.Given Data / Assumptions:Euler number Eu typically defined as Eu = Δp / (rho * V^2) (pressure forces to inertial forces).So-called Newton number in mixing literature is the power number Np = P / (rho * N^3 * D^5).Other contexts may use “Newton number” loosely, but it is not standardly the reciprocal of Eu.Concept / Approach:The reciprocal of Eu would be (rho * V^2) / Δp, which is not generally named the Newton number in standard fluid mechanics. Newton (power) number relates agitator power to inertial scaling and is independent of pressure-coefficient definitions like Eu. Therefore, the statement is incorrect.Step-by-Step Solution:Step 1: Define Eu = Δp / (rho * V^2).Step 2: Compute reciprocal 1/Eu = (rho * V^2) / Δp.Step 3: Note that standard terminology does not call 1/Eu the Newton number; Newton (power) number is Np = P / (rho * N^3 * D^5).Verification / Alternative check:Textbook tables of dimensionless groups list Eu separately from Np and do not present a reciprocal relation naming convention. Some fields use a “pressure coefficient” Cp similar to Eu; again, no standard Newton-number reciprocity exists.Why Other Options Are Wrong:Any “Yes/True” variants propagate a nonstandard and misleading equivalence.Common Pitfalls:Interchanging names of dimensionless groups across subdomains (e.g., mixing vs external aerodynamics) and assuming simple reciprocal relationships where none are defined.Final Answer:No

Question 19

Identify the principle – upward thrust on immersed bodies When a body is immersed wholly or partially in a liquid, it experiences an upward force equal to the weight of the liquid displaced by the body. This statement is known as:

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Introduction:The foundational rule governing buoyancy is a named principle that relates the upward hydrostatic thrust to the weight of displaced fluid. The question asks the learner to correctly identify that named statement used in almost every floatation and apparent-weight problem.Given Data / Assumptions:Body immersed fully or partially in a liquid at rest.Uniform gravity, constant fluid density.No flow-induced dynamic lift; purely hydrostatic forces are considered.Concept / Approach:Archimedes’ principle states: the buoyant force on a submerged body equals the weight of the fluid displaced by that body. This principle follows from integrating the hydrostatic pressure over the surface and is independent of the body’s own density. The principle of floatation is a related consequence: a floating body displaces an amount of liquid whose weight equals the body’s weight.Step-by-Step Solution:Step 1: Translate the statement “upward force equals weight of displaced fluid.”Step 2: Recognize it as exactly Archimedes’ principle.Step 3: Distinguish from other laws: Pascal's law deals with pressure transmission; Bernoulli with energy conservation; D’Alembert with inertia forces.Verification / Alternative check:Weighing an object in and out of water shows an apparent weight loss equal to rho * g * V_displaced, confirming Archimedes’ statement experimentally.Why Other Options Are Wrong:Pascal's law: Uniform transmission of pressure in a confined fluid.Principle of floatation: Consequence for floating bodies (buoyant force equals weight), not the general law for any immersion.Bernoulli's theorem and D'Alembert's principle: Address different phenomena (energy and dynamics-to-statics transformation).Common Pitfalls:Using “principle of floatation” as a synonym. It is related but not the general statement for any immersion depth.Final Answer:Archimedes' principle

Question 20

Units in fluid mechanics – identifying the correct unit of surface tension In engineering practice, what is the correct S.I. unit for surface tension (also called surface energy per unit area or line force per unit length)?

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Introduction / Context:Surface tension quantifies the energetic or mechanical effect associated with liquid–gas or liquid–liquid interfaces. It can be viewed either as energy per unit area (how much energy is needed to create new surface) or as an equivalent line force acting tangentially along a boundary. This question asks you to recall its correct S.I. unit and to distinguish it from commonly confused units such as pressure, viscosity, or torque.Given Data / Assumptions:S.I. base dimensions and units are used.Surface tension symbol: typically sigma or γ_s (not to be confused with specific weight).No numerical calculation is required; unit recognition is the focus.Concept / Approach:Two equivalent definitions are standard: (1) mechanical definition as line force per unit length acting at an interface, and (2) thermodynamic definition as surface energy per unit area. From the mechanical viewpoint, the unit is newton per metre. From the thermodynamic viewpoint, the unit newton metre per square metre reduces to newton per metre as well.Step-by-Step Solution:Step 1: Mechanical view → surface tension = force / length, so unit = N/m.Step 2: Thermodynamic view → surface energy / area = (N·m) / m^2 = N/m.Step 3: Compare with distractors: pressure is N/m^2 (Pa), viscosity is Pa·s, torque is N·m.Step 4: Conclude that the only correct S.I. unit consistent with both views is N/m.Verification / Alternative check:Typical values confirm the unit: water at about 20 °C has surface tension near 0.072 N/m; mercury is roughly 0.48–0.49 N/m. Expressing these in N/m is standard in handbooks and lab reports.Why Other Options Are Wrong:N/m^2 (Pa): unit of pressure or stress, not surface tension.N/m^3: no common fluid property uses this unit.N·m: unit of energy or torque; per area would reduce to N/m.Pa·s: unit of dynamic viscosity, unrelated to interfacial phenomena.Common Pitfalls:Confusing surface tension with pressure at a curved surface; while pressure jump across a meniscus depends on surface tension (Laplace equation), their units are different.Final Answer:N/m

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