Question 1

Reynolds number — identify the force ratio it represents Reynolds number (Re) in fluid mechanics is the ratio of inertia force to which resisting force?

Accepted Answer

Introduction / Context:Reynolds number is a fundamental dimensionless parameter that predicts flow regimes (laminar, transitional, turbulent) in internal and external flows. It arises from nondimensionalizing the Navier–Stokes equations and compares the strength of inertia effects to viscous effects in a flow.Given Data / Assumptions:Characteristic velocity, length, and fluid properties are well defined.Re = (rho * V * L) / mu where rho is density, V is velocity, L is length scale, mu is dynamic viscosity.Concept / Approach:Reynolds number measures the ratio of inertia force to viscous force. High Re implies inertia dominates and turbulence is likely; low Re implies viscous forces dominate and flow tends to be laminar. Other forces (pressure, gravity, surface tension) form different dimensionless groups (e.g., Euler, Froude, Weber numbers).Step-by-Step Solution:Start from Re = rho * V * L / mu.Recognize that numerator scales with inertia (rho * V^2 * L^2 divided by characteristic area), while denominator contains viscosity μ which characterizes viscous resistance.Hence Re represents inertia force / viscous force.Select “viscous force.”Verification / Alternative check:Other force ratios: Euler number relates pressure to inertia; Froude compares inertia to gravity; Weber compares inertia to surface tension. This confirms Reynolds uniquely targets viscous effects.Why Other Options Are Wrong:Pressure force: linked to Euler number, not Reynolds.Elastic force: associated with Mach number and compressibility effects.Gravity force: captured by the Froude number, not Reynolds.Common Pitfalls:Interchanging roles of different dimensionless groups and mislabeling the force balance each represents.Final Answer:viscous force

Question 2

Definition check — specific volume In fluid properties, the “specific volume” is defined as which of the following?

Accepted Answer

Introduction / Context:Thermodynamics and fluid mechanics frequently use intensive properties that are independent of system size. Specific volume is such a property and is the reciprocal of density. Understanding its definition aids in state calculations and property table lookups.Given Data / Assumptions:Standard SI units: volume in m^3, mass in kg.Density ρ = mass/volume; specific volume v = volume/mass.Concept / Approach:Specific volume v is defined as volume per unit mass: v = V / m. Since density ρ = m / V, it follows that v = 1 / ρ. It applies to liquids and gases alike and is especially convenient for compressible substances where density changes appreciably with pressure and temperature.Step-by-Step Solution:Start with density ρ = m / V.Take the reciprocal: V / m = 1 / ρ.Recognize that V / m is the specific volume v.Therefore, the correct definition is “volume per unit mass.”Verification / Alternative check:Typical SI unit check: If ρ is kg/m^3, then v has units m^3/kg, confirming the definition. Property tables (e.g., for steam) list specific volume in m^3/kg.Why Other Options Are Wrong:Mass per unit volume → density, not specific volume.Weight per unit volume → specific weight (γ), different property.Volume per unit weight → sometimes called specific volume per weight (rare), not the standard definition used in SI property tables.Common Pitfalls:Confusing density, specific volume, and specific weight because of similar names; always check dimensional units.Final Answer:Volume per unit mass of a liquid or gas

Question 3

Stability of floating bodies — classify the equilibrium A floating body is given a small angular displacement. If it does not return to its original position and, instead, heels further away, what is the correct classification of its equilibrium sta…

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Introduction / Context:The response of floating bodies (ships, buoys) to small angular disturbances is characterized by metacentric height and the resulting restoring or overturning moments. Correctly identifying stable, neutral, or unstable equilibrium is crucial for naval architecture and safety.Given Data / Assumptions:Body floats in a fluid and is slightly heeled.Observation: it moves farther away from the original position rather than returning.Small-angle behavior is considered (initial stability).Concept / Approach:Equilibrium types: (1) Stable — a small angular displacement generates a restoring moment that brings the body back; (2) Neutral — no restoring or overturning moment, so it remains indifferent at the new angle; (3) Unstable — an overturning moment increases the heel. The described behavior (heels further away) clearly matches unstable equilibrium.Step-by-Step Solution:Apply a small heel to the floating body.Observe the direction of the resultant moment due to displaced buoyancy vs. weight line of action.If the moment increases the heel, the equilibrium is unstable.Therefore, select “Unstable equilibrium.”Verification / Alternative check:For initial stability, metacentric height GM > 0 → stable; GM = 0 → neutral; GM Why Other Options Are Wrong:Neutral equilibrium: would stay at the new position, not move further away.Stable equilibrium: would return toward upright.“Metacentric equilibrium only at zero heel” is not a standard classification and does not reflect the observed behavior.Common Pitfalls:Mislabeling neutral as unstable; neutral implies neither restoring nor overturning moment.Final Answer:Unstable equilibrium

Question 4

Venturimeter design — relative lengths of cones In a standard venturimeter, the length of the divergent (diffuser) cone is typically how much compared to the convergent cone?

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Introduction / Context:A venturimeter measures discharge using pressure difference between an upstream section and a throat. The geometry of the convergent and divergent cones is chosen to minimize head loss: a short, steep convergent (to accelerate flow) and a longer, gentle divergent (to recover pressure without separation).Given Data / Assumptions:Steady, incompressible flow through a classical venturimeter.Goal is good pressure recovery with minimal separation in diffuser.Concept / Approach:Adverse pressure gradients in diffusers cause boundary layer separation if the divergence angle is too large. To avoid this, the diffuser is made longer with a small half-angle (often around 3°–7°). Empirically, the diffuser length is about three to four times the length of the convergent cone for good performance and low permanent head loss.Step-by-Step Solution:Recognize that the diffuser must be longer to allow gradual area expansion.Typical design guidance: L_diffuser ≈ 3 to 4 * L_convergent.Select option “three to four times.”Verification / Alternative check:Handbooks and standard fluid machinery texts consistently recommend longer diffusers than convergents to avoid flow separation; field designs often fall within this 3–4× range.Why Other Options Are Wrong:Equal to or double: too short for reliable pressure recovery.Five to six times: unnecessarily long, increases size and cost without proportional benefit.Common Pitfalls:Confusing diffuser length with area ratios; the key is gentle expansion angle, which implies longer length for a given area change.Final Answer:three to four times

Question 5

Equivalent single pipe for a compound (series) pipe system Three pipes in series have diameters d1, d2, d3 and lengths l1, l2, l3. They are to be replaced by one equivalent pipe of uniform diameter d and total length l = l1 + l2 + l3, carrying the s…

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Introduction / Context:Designers often replace a series of pipes of different diameters with a single “equivalent” pipe to simplify calculations. For turbulent flow modeled with the Darcy–Weisbach relation and equal friction factor, the total head loss must be preserved at a given discharge to determine the equivalent diameter.Given Data / Assumptions:Pipes in series: lengths l1, l2, l3; diameters d1, d2, d3.Equivalent single pipe: length l = l1 + l2 + l3; diameter d.Same discharge Q through all series segments.Same friction factor f and same fluid; use Darcy–Weisbach head loss.Concept / Approach:For a segment i: h_fi = f * (li/di) * (vi^2 / (2g)). With Q constant, vi = 4Q/(π di^2). Thus h_fi ∝ li * Q^2 / di^5. Total head loss is the sum over segments. The equivalent pipe must satisfy h_f(eq) = Σ h_fi at the same Q, yielding the classical 1/d^5 relationship.Step-by-Step Solution:Write h_f(total) = K * [ (l1 / d1^5) + (l2 / d2^5) + (l3 / d3^5) ] * Q^2, where K collects constants.For the equivalent pipe: h_f(eq) = K * (l / d^5) * Q^2.Equate: l / d^5 = (l1 / d1^5) + (l2 / d2^5) + (l3 / d3^5).Solve for d: d = [ l / ( Σ li / di^5 ) ]^(1/5).Verification / Alternative check:Dimensional reasoning: as diameters increase, head loss reduces rapidly (power 5); combining smaller pipes increases the equivalent resistance, hence decreases equivalent d, consistent with the derived formula.Why Other Options Are Wrong:Power 4 relationships (options with d^4) arise in laminar (Hagen–Poiseuille) conditions, not in this Darcy–Weisbach turbulent-equivalent assumption.Option using weighted sum l1 d1^5 etc. in the numerator inverts the correct dependence.Although option (a) is an intermediate relation, the asked “size of equivalent pipe” requires the explicit expression for d, given in option (c).Common Pitfalls:Mixing laminar and turbulent formulas; ensure consistency of friction model and exponent.Final Answer:d = [ l / ( (l1 / d1^5) + (l2 / d2^5) + (l3 / d3^5) ) ]^(1/5)

Question 6

Siphon with air vessel at the summit — purpose of the vessel In a siphon line, an air vessel is provided at the summit primarily to:

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Introduction / Context:Siphons can experience very low pressure at the summit, risking air separation or cavitation that breaks the continuous liquid column. Fitting an air vessel (or vacuum breaker/air chamber) near the summit helps stabilize operation and prevents flow interruption.Given Data / Assumptions:Siphon conveys liquid over a summit higher than the free surfaces.Risk: pressure at the summit may drop below vapor pressure, causing separation.Device: air vessel installed at or near the summit.Concept / Approach:The air vessel acts as a cushion and helps maintain continuity by smoothing pressure fluctuations and preventing the absolute pressure from falling too low. It also allows entrained air to collect and be managed so that the siphon does not lose its prime.Step-by-Step Solution:Recognize the failure mode: separation at low summit pressure.Introduce air vessel to stabilize pressure and avoid loss of prime.Primary function: avoid interruption and maintain continuous flow.Hence choose “avoid interruption in the flow.”Verification / Alternative check:Practical siphon designs specify maximum allowable elevation of the summit; adding a vessel reduces pressure swings and sustains operation, which confirms the purpose.Why Other Options Are Wrong:Increase discharge/velocity: discharge is governed by head difference and losses; the vessel is not a pump.Maintain fixed pressure difference: while it moderates fluctuations, it does not fix a specific Δp like a regulator.Common Pitfalls:Confusing air vessels in reciprocating pumps (which reduce pulsations) with siphon summit vessels; both smooth flow but serve different systems.Final Answer:avoid interruption in the flow

Question 7

Origin of viscosity in liquids — conceptual check The viscosity of a liquid primarily arises due to cohesion between its molecules. Is this statement correct?

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Introduction / Context:Viscosity is a measure of internal resistance to flow. Its microscopic origin differs between liquids and gases. Recognizing this difference helps explain why liquid viscosity generally decreases with temperature, while gas viscosity increases.Given Data / Assumptions:Continuum fluid behavior is considered, with molecular interpretation for trends.Liquids possess strong intermolecular forces; gases have widely spaced molecules.Concept / Approach:In liquids, adjacent layers resist relative motion largely because of molecular cohesion forces; raising temperature weakens cohesion, typically lowering viscosity. In gases, viscosity mainly results from molecular momentum exchange during collisions; as temperature rises, collision frequency and momentum transfer increase, so gas viscosity increases.Step-by-Step Solution:State the claim: liquid viscosity due to cohesion.Recall temperature trends: μ_liquid decreases with T; μ_gas increases with T.These opposite trends are consistent with the different physical mechanisms.Therefore, the statement is correct for liquids.Verification / Alternative check:Empirical correlations (e.g., Andrade’s equation for liquids) show exponential decrease of μ with temperature; Sutherland’s law for gases shows increase, matching the mechanistic explanation.Why Other Options Are Wrong:Incorrect: contradicts accepted molecular interpretation for liquids.“Only at high temperatures” and “only for ideal liquids” add conditions that are unnecessary and misleading.Common Pitfalls:Assuming a single mechanism for both liquids and gases; their behaviors differ fundamentally.Final Answer:Correct

Question 8

Units — dynamic viscosity (SI) Choose the correct SI unit for dynamic viscosity μ.

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Introduction / Context:Correct units are critical for dimensional checks and converting between property tables. Dynamic viscosity μ and kinematic viscosity ν often get mixed up due to the variety of unit systems used in practice (SI and CGS).Given Data / Assumptions:Dynamic viscosity μ relates shear stress to velocity gradient: τ = μ * (du/dy).Shear stress τ has units N/m^2; velocity gradient du/dy has units s^-1.Concept / Approach:From τ = μ * (du/dy), we get μ = τ / (du/dy). Therefore μ has units (N/m^2) / (s^-1) = N·s/m^2. In SI, this is also equal to Pa·s. CGS units define 1 poise = 0.1 Pa·s and 1 stoke = 1 cm^2/s for kinematic viscosity, not dynamic viscosity.Step-by-Step Solution:Start with τ in N/m^2 and du/dy in s^-1.Compute μ units: (N/m^2) / (s^-1) = N·s/m^2.Match this with the options → N-s/m2.Verification / Alternative check:Use ν = μ / ρ. If μ is N·s/m^2 and ρ is kg/m^3, then ν has units (N·s/m^2)/(kg/m^3) = (kg·m/s^2 * s / m^2) * (m^3/kg) = m^2/s, confirming consistency for kinematic viscosity units.Why Other Options Are Wrong:N-m/s^2 simplifies to kg·m^2/s^4, which is not viscosity.Poise: CGS unit for dynamic viscosity (1 P = 0.1 Pa·s), not SI base listing here.Stoke: CGS unit of kinematic viscosity (1 St = 1 cm^2/s), not dynamic viscosity.Common Pitfalls:Confusing dynamic and kinematic viscosity units or mixing SI and CGS without conversion.Final Answer:N-s/m2

Question 9

Critical velocities in pipe flow — meaning of Re ≈ 2800 The flow velocity that corresponds to a Reynolds number of about 2800 in a circular pipe is commonly termed the:

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Introduction / Context:In internal pipe flow, the Reynolds number indicates whether the flow is laminar, transitional, or turbulent. Two reference values are often used to bracket the transition region: a lower critical Reynolds number and a higher critical Reynolds number.Given Data / Assumptions:Reynolds number Re = ρ V D / μ in a smooth circular pipe.Typical thresholds: laminar for Re ≲ 2000, transitional roughly between Re ≈ 2000 and Re ≈ 2800–4000, turbulent above the higher critical value.Concept / Approach:The lower critical Reynolds number marks the limit below which laminar flow is sustained (often cited near 2000). The higher critical Reynolds number marks the point above which fully developed turbulence is reliably established (often cited near 2800–4000 depending on sources and disturbances). Thus, associating Re ≈ 2800 with the “higher critical velocity” is a standard textbook convention in many question banks.Step-by-Step Solution:Recognize Re ≈ 2800 lies above the lower threshold (~2000).This value signifies the upper end of the transition window (higher critical).Therefore, choose “higher critical velocity.”Verification / Alternative check:Practical laboratory observations show that the exact transition range depends on inlet disturbances and pipe roughness. Nonetheless, exam conventions commonly use ~2000 (lower critical) and ~2800 (higher critical) as nominal markers.Why Other Options Are Wrong:Sub-sonic and super-sonic relate to Mach number in compressible flow, not Reynolds number.Lower critical velocity corresponds to Re around 2000, not 2800.Common Pitfalls:Assuming a single universal value for the higher critical Reynolds number; it varies with conditions, but the naming in this question follows common instructional values.Final Answer:higher critical velocity

Question 10

Fluid mechanics – identifying flow regime where viscous forces dominate over inertia Choose the correct option that best describes such a flow.

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Introduction:In fluid mechanics, flows are classified by the relative importance of viscous forces compared to inertia forces. This question tests whether you can identify the flow regime in which viscous effects dominate the motion, smoothing out velocity differences and preventing mixing across adjacent layers.Given Data / Assumptions:Single-phase, Newtonian fluid is considered.Internal flow in a conduit is the typical mental model, but the concept holds broadly.Low Reynolds number conditions are implied (Re much less than the critical value).Concept / Approach:Reynolds number Re = (rho * V * L) / mu compares inertia forces (rho * V^2 * L^2 scale) to viscous forces (mu * V * L scale). When Re is small, viscous forces dominate inertia. In such conditions, fluid moves in smooth, orderly layers with minimal mixing—this is laminar flow. When Re is large, inertia overwhelms viscosity, and turbulence can occur.Step-by-Step Solution:1) Identify the regime where viscous forces are most significant: low Re.2) Low Re implies streamlined motion with layered velocity profile.3) The technical term for this regime is laminar flow.Verification / Alternative check:A hallmark of laminar flow is a predictable velocity distribution (e.g., parabolic in fully developed pipe flow) and the absence of eddies; pressure drop follows linear relations with flow rate in the Hagen–Poiseuille regime, reinforcing viscous dominance.Why Other Options Are Wrong:Steady flow: time-invariant, not about viscous vs inertia dominance.Unsteady flow: time-varying; again not defining force balance.Turbulent flow: inertia dominates; strong mixing and eddies.Uniform flow: velocity same across a section; not guaranteed by viscous dominance.Common Pitfalls:Confusing “steady” with “laminar” or assuming low velocity automatically means steadiness. The key discriminator is the Reynolds number and the viscous–inertia balance, not whether the flow changes with time.Final Answer:laminar flow

Question 11

Laminar flow and velocity scale Fill the blank: Laminar flow __________ place at very low velocities (low Reynolds number conditions).

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Introduction:This question probes the relationship between flow regime and velocity scale through the lens of the Reynolds number. It checks whether you recognize that laminar behavior is associated with conditions where viscous forces dominate, typically occurring at relatively low velocities for a given fluid and geometry.Given Data / Assumptions:Single-phase Newtonian fluid.Characteristic length and fluid properties are fixed.Comparisons are made by varying velocity.Concept / Approach:The Reynolds number Re = (rho * V * L) / mu increases with velocity V. For a fixed fluid (rho, mu) and geometry (L), decreasing V decreases Re. Laminar flow is found at low Re, where viscous forces dominate inertia. Therefore, “very low velocities” favor laminar flow, while “very high velocities” tend toward transitional or turbulent regimes.Step-by-Step Solution:1) Recognize laminar flow occurs at low Re conditions.2) For fixed fluid and pipe, low Re corresponds to low velocity.3) Hence, laminar flow takes place at very low velocities.Verification / Alternative check:Empirical critical Reynolds numbers for internal flows (e.g., Re_critical around 2000 for pipes) indicate that below this threshold, flows are generally laminar, which typically requires relatively small velocities for common fluids and diameters.Why Other Options Are Wrong:does not take: contradicts low-Re physics.takes only at very high velocities: high velocity raises Re, favoring turbulence.never takes: laminar is widely observed (microfluidics, small pipes, viscous oils).takes only when density is zero: unrealistic and unrelated to regime classification.Common Pitfalls:Assuming that “steady” equals “laminar,” or overlooking that geometry and viscosity strongly affect the velocity needed for laminar conditions.Final Answer:takes

Question 12

Effect of pressure on bulk modulus Choose the correct statement about how the bulk modulus of a fluid changes as pressure increases.

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Introduction:The bulk modulus (K) measures a fluid’s resistance to uniform compression. This question assesses understanding of compressibility trends with pressure, especially in liquids where compressibility is small but not zero.Given Data / Assumptions:Fluid is uniform and isotropic under hydrostatic compression.Temperature effects are neglected for simplicity.Small to moderate pressure ranges typical of engineering systems.Concept / Approach:Bulk modulus is defined as K = -V * (dP / dV) = dP / (dV/V). Physically, higher K means the fluid is “stiffer” (less compressible). For many liquids, as pressure increases, the incremental compressibility decreases, so the slope dP/d(ΔV/V) rises; equivalently, K increases with pressure. Although the rate of increase may be mild, the trend is well established for water and most engineering liquids over practical pressure ranges.Step-by-Step Solution:1) Consider K = dP / (dV/V): if additional pressure produces progressively smaller fractional volume changes, K must increase.2) Empirical correlations (e.g., Tait equation for water) capture this rise of K with pressure.3) Therefore, the correct qualitative statement is that bulk modulus increases with pressure.Verification / Alternative check:Engineering handbooks show water’s bulk modulus near 2.1 GPa at atmospheric conditions and larger at several hundred bars, confirming the positive dependence on pressure.Why Other Options Are Wrong:remains same: ignores observed pressure dependence.decreases: opposite of typical liquid behavior.first increases then decreases / first decreases then increases: no such general pattern for ordinary liquids in the common range.Common Pitfalls:Confusing compressibility (which decreases with pressure) with bulk modulus (its reciprocal trend). Also, extrapolating gas behavior directly to liquids without checking property models.Final Answer:increases

Question 13

Newton’s law of resistance – underlying idealizations The law assumes simplified conditions about the body and surrounding fluid. Select the statement that best captures these assumptions.

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Introduction:Newton’s law of resistance, historically used to relate drag to velocity (often leading to drag proportional to V^2 at high Reynolds numbers), relies on idealized assumptions. This question checks recognition of the classical simplifications about surface smoothness, fluid occupancy, and particle interactions.Given Data / Assumptions:Classical, idealized formulation of resistance on a body moving through a fluid.Macroscopic viewpoint; molecular-scale interactions are not explicitly modeled.High-level assumptions used to derive simple resistance expressions.Concept / Approach:To obtain tractable expressions for drag, early formulations assume: (1) perfectly smooth body surfaces so local roughness effects on boundary layers are neglected, (2) the medium fully occupies the space (no voids), and (3) simplified interaction between fluid particles (no complex inter-particle forces beyond the continuum description). Grouping these together yields the combined alternative “all of the above.”Step-by-Step Solution:1) Identify each statement as a simplifying assumption used in classical drag laws.2) Recognize that standard derivations presume smooth bodies and a fully filled fluid domain.3) Conclude that all listed assumptions collectively underpin the ideal law.Verification / Alternative check:Modern drag correlations improve on these assumptions by incorporating roughness, boundary-layer behavior, and empirical coefficients, but the foundational idealizations remain historically recognized in the simplest treatments.Why Other Options Are Wrong:Individual statements (a), (b), (c) are incomplete alone; the question asks for the best capture of assumptions, which is the combined option.none of the above: contradicts the recognized simplifications.Common Pitfalls:Confusing this historical/ideal law with detailed boundary-layer theory, or assuming surface roughness and turbulence modeling are inherently included in the earliest formulations—they are not.Final Answer:all of the above

Question 14

Rotating liquid in a cylindrical vessel – identify the flow type When a vertical-axis rotation depresses the free surface at the center and raises it near the wall, what is this flow called?

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Introduction:Rotating fluids in a cylindrical container develop a characteristic free-surface shape: depressed at the axis and elevated near the wall, forming a paraboloid. This question assesses whether you can identify the underlying flow category—namely, a vortex flow—and recognize the physics of forced vortices.Given Data / Assumptions:Rigid-body rotation about the vertical axis (forced vortex).Incompressible fluid with a visible free surface.Steady rotation after spin-up transients have decayed.Concept / Approach:In a forced vortex, every fluid particle rotates with the same angular velocity. The radial pressure gradient balances the centrifugal effect: dp/dr = rho * r * omega^2. Integrating this hydrostatically in the vertical direction yields a paraboloid free surface z = (omega^2 * r^2) / (2g) + constant, depressed at r = 0 and raised at larger r. This is the textbook signature of vortex flow under forced rotation.Step-by-Step Solution:1) Observe the free-surface deformation: center down, rim up.2) Match with the forced vortex model with rigid-body rotation.3) Conclude the flow type is a vortex flow (specifically a forced vortex).Verification / Alternative check:Plotting z versus r^2 gives a straight line in a forced vortex, confirming the paraboloid form and the vortex nature of the motion.Why Other Options Are Wrong:steady flow: can be steady or unsteady; does not name the rotational nature.turbulent flow: turbulence may or may not occur; this label misses the rotational structure.uniform flow: implies uniform velocity over a section, not valid here.compressible flow: not a defining feature of this phenomenon.Common Pitfalls:Confusing free vortex (omega varies with r) with forced vortex (rigid-body rotation). The described paraboloid under container rotation is the forced vortex case.Final Answer:vortex flow

Question 15

Equivalent single pipe for a compound pipe Two pipes are said to be hydraulically equivalent when replacing a compound line by one new pipe. Choose the correct equivalence condition.

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Introduction:Pipeline design often replaces a series (compound) system with a single “equivalent” pipe for easier analysis. This question asks you to recall the precise criteria that define equivalence for head loss and flow rate.Given Data / Assumptions:Incompressible, steady flow.Same fluid in both configurations.Minor losses may be either included in the head loss accounting or neglected consistently.Concept / Approach:An equivalent pipe must transport the same discharge Q under the same total head loss H between the same endpoints as the original compound system. Matching Q ensures throughput equivalence; matching H ensures the same driving energy requirement. Length and diameter individually need not match; they adjust to meet the H–Q condition via friction relations (e.g., Darcy–Weisbach or Hazen–Williams).Step-by-Step Solution:1) Identify target variables to preserve: discharge Q and head loss H.2) Use a friction relation (e.g., H = f * (L/D) * (V^2 / (2g))) to select the equivalent L and D that yield the same H at the same Q.3) Conclude that equivalence means same head loss and same discharge.Verification / Alternative check:Hydraulic-grade line across endpoints remains unchanged if H and Q match, confirming energy and capacity equivalence.Why Other Options Are Wrong:Same length or diameter: not necessary individually.Same head loss and velocity: equal velocity does not guarantee equal Q unless diameters also match.Same Reynolds number: not required and may differ because D and V can change.Common Pitfalls:Over-focusing on geometric sameness rather than hydraulic performance (H–Q) equivalence.Final Answer:loss of head and discharge of both the pipes is same

Question 16

Footstep bearing with viscous lubrication If the shaft speed is doubled, how does the torque needed to overcome viscous resistance change?

Accepted Answer

Introduction:This question examines viscous shear in a lubricated thrust (footstep) bearing. The torque required to overcome viscous resistance depends on shear stress in the oil film, which scales with relative speed and viscosity for Newtonian lubricants.Given Data / Assumptions:Newtonian lubricant with constant viscosity mu.Uniform film thickness h under operating conditions.Laminar Couette-type shear between the rotating surface and stationary pad.Concept / Approach:Shear stress tau = mu * (du/dy). For a rotating disk over a stationary surface with film thickness h, the velocity gradient scales with omega * r / h. The elemental shear force scales with tau * area, and elemental torque scales further with radius r. Integration across the radius yields torque T proportional to mu * omega / h times a geometric factor. Thus, T ∝ omega (and ∝ speed N).Step-by-Step Solution:1) For Newtonian films, tau ∝ velocity gradient ∝ omega.2) Torque is the moment of shear force, so T ∝ tau and hence T ∝ omega.3) Doubling omega (or speed N) doubles T.Verification / Alternative check:Classical formulas for a flat, full-film thrust bearing give T = (pi * mu * omega * R^4) / (2 * h) for idealized geometry, explicitly linear in omega, confirming the doubling behavior.Why Other Options Are Wrong:four/eight/sixteen times: imply quadratic or higher power laws; not valid for linear viscous shear.no change: contradicts direct proportionality to speed.Common Pitfalls:Confusing viscous shear scaling with turbulent skin-friction laws or bearing hydrodynamic lift scaling; this question targets the linear Newtonian-shear regime with constant film thickness.Final Answer:double

Question 17

Triangular (V) notch discharge–head relationship Identify how discharge varies with head H for a sharp-crested triangular notch.

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Introduction:Flow measurement over notches and weirs relies on empirical–theoretical relations between discharge Q and head H. For triangular (V) notches, the exponent of H is a key discriminant. This question checks whether you know that Q varies with H to the power 5/2.Given Data / Assumptions:Sharp-crested, well-ventilated triangular notch.Free, fully contracted flow without submergence.Coefficient of discharge treated as approximately constant over small H ranges.Concept / Approach:Elementary derivation integrates elemental discharge strips across the varying width b(y) of the notch opening with local velocity proportional to sqrt(2gH_local). For a V-notch of angle theta, width b ∝ y, and integration yields Q = C_d * (8/15) * tan(theta/2) * sqrt(2g) * H^(5/2). Hence, the dependence is Q ∝ H^(5/2).Step-by-Step Solution:1) Express notch width versus depth: b = 2 * y * tan(theta/2).2) Local velocity v ∝ sqrt(2g * (H - y)).3) Integrate elemental discharge b * v dy from y = 0 to H to obtain Q ∝ H^(5/2).Verification / Alternative check:Standard hydraulics texts list Q = K * H^(5/2) for V-notches, distinguishing them from rectangular weirs where Q ∝ H^(3/2).Why Other Options Are Wrong:H^(3/2) (direct or inverse): characteristic of rectangular weirs, not V-notches.Inverse proportionalities: unphysical for increasing head producing more flow.Independent of H: contradicts basic weir theory.Common Pitfalls:Mixing up the H exponents between rectangular and triangular notches or forgetting the role of notch angle in the proportionality constant, not in the exponent.Final Answer:directly proportional to H5/2

Question 18

Continuity for incompressible steady flow in a pipe True or False: For an incompressible liquid flowing continuously through a pipe, the discharge (quantity per second) is different at different sections.

Accepted Answer

Introduction:The continuity equation is a cornerstone of fluid mechanics. For incompressible steady flow, it asserts constancy of volumetric flow rate along a streamline (and through any cross-section of a pipe), regardless of local changes in area and velocity.Given Data / Assumptions:Incompressible liquid (density constant).Steady flow (no accumulation with time).No leakage or branching between the two sections being compared.Concept / Approach:Continuity for incompressible steady flow: A1 * V1 = A2 * V2 = Q (constant). If the pipe diameter changes, velocity adjusts inversely with area to keep Q the same. Therefore, the statement that discharge is “different at different sections” is false.Step-by-Step Solution:1) Write continuity: Q = A * V for incompressible steady flow.2) Between any two sections, Q1 = A1 * V1 and Q2 = A2 * V2.3) With no sources/sinks, Q1 = Q2, hence discharge is the same.Verification / Alternative check:Measurement devices placed at different diameters show different velocities but the same volumetric flow rate when the system is closed and steady, validating continuity.Why Other Options Are Wrong:True: contradicts continuity.Depends only on viscosity: viscosity affects losses and profiles, not mass conservation.True only in laminar/turbulent: regime does not change mass conservation.Common Pitfalls:Confusing velocity changes with discharge changes. Remember that velocity can vary with area while Q remains constant.Final Answer:False

Question 19

Head losses in pipes arranged in series Select the correct statement about the total head loss when pipes are connected in series.

Accepted Answer

Introduction:This question targets network hydraulics. When several pipe segments are connected in series carrying the same flow, their frictional and local losses accumulate. Understanding this additive nature is essential for sizing pumps and predicting pressures at nodes.Given Data / Assumptions:Steady, incompressible flow.Same discharge through each series segment.Head losses include friction and, if accounted, minor losses at fittings and entries/exits.Concept / Approach:Energy equation between inlet and outlet across a series of segments: H_total loss = Σ(H_f,i + H_m,i). Because the discharge is the same through each segment, each segment’s loss can be computed (e.g., via Darcy–Weisbach) and then summed. Therefore, the total head loss equals the sum of head losses in individual pipes.Step-by-Step Solution:1) Write Bernoulli with head-loss terms for each segment.2) Add the equations across the series to eliminate intermediate elevations/pressures.3) Conclude that total loss is the algebraic sum of segment losses.Verification / Alternative check:Equivalent single-pipe calculations reproduce the same overall head drop if the summed loss is used as the design constraint, confirming additivity.Why Other Options Are Wrong:No: contradicts the energy balance.Only if diameters are equal: additivity holds regardless of diameter differences.Only for laminar flow: valid for any regime; the formula for each segment changes but losses still add.Only if minor losses are neglected: minor losses also add when included.Common Pitfalls:Forgetting to include local (minor) losses or misapplying the friction factor regime; the summation principle remains valid either way.Final Answer:Yes

Question 20

Fluid Statics — Free-Surface Tilt in an Accelerating Open Tank An open tank containing a liquid starts from rest and moves with a uniform horizontal acceleration a. What is the relation between the inclination angle θ made by the free liquid surface…

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Introduction / Context: When a container with a free liquid surface accelerates horizontally, the surface reorients to remain perpendicular to the resultant of the real gravitational field and the inertial (body) acceleration field. This is a classic fluid statics-in-accelerating-frames problem used in marine, aerospace, and process engineering. Given Data / Assumptions: Open tank with a liquid and a free surface exposed to atmosphere. Uniform horizontal acceleration a; gravity acts vertically downward with magnitude g. Neglect surface tension and sloshing transients; consider steady inclined surface after acceleration is established. Concept / Approach: In a non-inertial frame attached to the tank, an inertial body force per unit mass acts opposite to the acceleration, of magnitude a. The free surface aligns normal to the resultant acceleration vector formed by g downward and a backward horizontally. The slope of the surface is given by tan θ = a / g, where θ is the angle with the horizontal. Step-by-Step Solution: Resultant body acceleration vector has horizontal component a and vertical component g. Free surface must be perpendicular to the resultant; equivalently, pressure must be constant along the free surface. Geometric relation gives tan θ = (horizontal component) / (vertical component) = a / g. Verification / Alternative check: If a = 0, tan θ = 0, so the surface is horizontal as expected. If a increases, θ increases smoothly; in the limit a ≫ g, the surface approaches a near-vertical orientation, consistent with intuition. Why Other Options Are Wrong: 2 a / g and a / (2 g) insert unjustified factors; a^2 / (2 g) has wrong dimensional form inside tan and no basis in this static equilibrium model. Common Pitfalls: Mixing up θ with the complement angle; forgetting that the surface is orthogonal to the resultant acceleration, not parallel. Final Answer: tan θ = a / g.

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