Question 1

Water Hammer in Pipes – Cause of pressure surge Water hammer (hammer blow) in a pipeline occurs under which operating condition?

Accepted Answer

Introduction:Water hammer is a transient hydraulic phenomenon characterized by a rapid pressure rise (or fall) and associated elastic waves in the pipe-fluid system. It is a key topic in fluid transients, affecting design of valves, pipes, supports, and surge protection devices. The question asks the practical trigger condition for hammer blow.Given Data / Assumptions:Incompressible liquid approximation with finite compressibility and pipe wall elasticity for transient analysis.Long, rigid-supported pipeline with valves and possibly pumps.Sudden closure implies closure time shorter than or comparable to the pipeline wave travel time.Concept / Approach:When a flowing liquid column is abruptly decelerated, momentum change generates a pressure wave that travels at the acoustic wave speed of the fluid-pipe system. The classical Joukowsky relation gives the initial surge: delta_p = rho * a * delta_v, where a is the wave speed and delta_v is the sudden velocity change. Therefore, rapid closure of a downstream valve (or rapid pump trip causing sudden velocity change) produces the largest surge magnitudes.Step-by-Step Solution:Step 1: Recognize that “suddenly brought to rest” implies delta_v is large over a very short time.Step 2: Apply the transient concept: Large |delta_v| in short time produces a strong pressure wave.Step 3: Compare to gradual closure: slower events allow the disturbance to spread, reducing peak surge.Step 4: Conclude that sudden closure is the operative condition for water hammer.Verification / Alternative check:Field experience and transient modeling show highest surge when valve closure time is less than 2L/a (round trip time), confirming the risk from rapid operations.Why Other Options Are Wrong:Excessive leakage: Causes pressure loss, not surge.Pipe burst: It is a possible consequence, not the cause.Gradual closure / slow pump start: These mitigate transients by minimizing delta_v per unit time.Common Pitfalls:Confusing symptoms and causes, or assuming hammer occurs only on closure; rapid opening can also create negative pressure waves, but the strongest classic case is sudden stop of flow.Final Answer:the flow of fluid through the pipe is suddenly brought to rest by closing of the valve

Question 2

Flat Plate in a Stream – Orientation vs pressure/drag When a thin plate is immersed in a liquid with its plane parallel to the flow direction, the pressure force experienced is __________ that for the same plate oriented perpendicular to the flow.

Accepted Answer

Introduction:For bluff and streamlined bodies, the orientation relative to the oncoming flow strongly affects pressure distribution and drag. A flat plate parallel to the flow presents a very small projected area normal to the flow, so its form (pressure) drag is much smaller than when the same plate is broadside (perpendicular) to the stream. The question probes this qualitative dependence.Given Data / Assumptions:Steady, uniform approach flow of a single-phase liquid.Identical plate, two orientations: parallel vs perpendicular to flow.Comparing pressure (form) drag component, not skin friction alone.Concept / Approach:Drag on immersed bodies has two principal components: pressure drag from flow separation and wake, and friction drag from viscous shear along the surface. With the plate perpendicular to the flow, the projected frontal area is maximum, separation is large, and form drag dominates. With the plate parallel to flow, the frontal area is minimal, separation is minor, and the main resistance is skin friction along its surfaces, which is typically much less than the broadside form drag for the same speed and fluid properties.Step-by-Step Solution:Step 1: Consider the projected area A_n normal to the flow. It is maximal in perpendicular orientation, minimal in parallel orientation.Step 2: Pressure drag is roughly proportional to dynamic pressure * A_n * drag coefficient; with small A_n the pressure component is small.Step 3: Compare the two cases: Perpendicular orientation yields greater pressure force than the parallel orientation.Step 4: Therefore the pressure force for the parallel orientation is less than that for the perpendicular orientation.Verification / Alternative check:Wind-tunnel and water-tunnel measurements consistently show orders-of-magnitude difference in form drag between broadside and edge-wise orientations of thin plates at similar Reynolds numbers.Why Other Options Are Wrong:More than / approximately equal to / independent: Contradict well established dependence on frontal area and separation behavior.Reynolds-dependent reversal: While coefficients vary with Reynolds number, the qualitative ordering of broadside vs edge-wise form drag does not invert.Common Pitfalls:Confusing total drag with only skin friction; in perpendicular orientation, form drag dominates, leading to much higher overall force.Final Answer:less than

Question 3

Open Channel – Most economical trapezoidal section The discharge through a trapezoidal channel is maximum (most economical section) when which geometric condition holds?

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Introduction:The “most economical” (most efficient) channel section provides maximum discharge for a given cross-sectional area and bed slope, or equivalently minimizes wetted perimeter for a given area. For a trapezoidal section, there is a classical geometric condition for the optimum in which the side length relates to the top width in a specific way.Given Data / Assumptions:Prismatic trapezoidal channel with bed width b, depth y, and side slope z (horizontal:vertical).Uniform flow with a given roughness and slope; resistance modeled by Manning or Chezy.Comparisons at the same area and slope, seeking maximum hydraulic radius R = A / P.Concept / Approach:For a trapezoid: top width T = b + 2z y; wetted perimeter P = b + 2 y sqrt(1 + z^2); area A = y (b + z y). Optimization to maximize R or minimize P for a given A yields two hallmark relations: (i) one of the sloping sides equals half of the top width (s = T / 2), and (ii) the hydraulic radius equals y / 2 at optimum. These relations uniquely characterize the most economical trapezoidal section.Step-by-Step Solution:Step 1: Express A and P in terms of b, y, z.Step 2: Set up the optimization to maximize R = A / P under fixed A or equivalently minimize P.Step 3: Derive the optimum conditions; standard calculus results lead to s = T / 2.Step 4: Map to the options: “the sloping side is equal to half the width at the top” matches s = T / 2.Verification / Alternative check:Substituting the optimum relations into R confirms R = y / 2, consistent with known textbook results. Geometric sketches also show that equal partition of the top width by the side length occurs at the optimum.Why Other Options Are Wrong:Top width twice bottom width: Not a general optimum condition.Depth equal to bottom width: No basis in the optimization.Side equal to bottom width: Dimensional equality does not ensure minimum perimeter.Hydraulic radius equals the depth: At optimum R = y / 2, not y.Common Pitfalls:Memorizing one relation (like R = y / 2) without remembering the geometric condition s = T / 2 that is often tested directly in exams.Final Answer:the sloping side is equal to half the width at the top

Question 4

Basic Property – Volume change with pressure The variation of the volume of a liquid with variation of pressure is defined as its what property?

Accepted Answer

Introduction:Continuum mechanics distinguishes between properties governing shape change under shear (viscosity) and volume change under pressure (compressibility). This question asks for the term that specifically quantifies how a liquid volume changes with pressure.Given Data / Assumptions:Isotropic liquid at standard temperature and moderate pressures.Small, reversible volumetric changes under pressure increments.No phase change or cavitation.Concept / Approach:Compressibility is defined as beta = − (1 / V) * (dV / dP), the fractional change in volume per unit increase in pressure. Its reciprocal K = 1 / beta is the bulk modulus of elasticity. For most engineering calculations with water at modest pressures, compressibility is small and the flow may be modeled as incompressible, but in transients and acoustics, finite compressibility is essential.Step-by-Step Solution:Step 1: Identify the response type: volumetric change due to hydrostatic or dynamic pressure.Step 2: Recall the property name: compressibility beta, or bulk modulus K = 1 / beta.Step 3: Match to options: “compressibility” is the correct terminology.Verification / Alternative check:Standard fluid property tables list compressibility and bulk modulus for liquids; water’s bulk modulus is about 2 * 10^9 Pa at room temperature, implying small volume changes for moderate pressure variations.Why Other Options Are Wrong:Surface tension: Relates to interface energy and curvature, not bulk volume change.Capillarity: Phenomenon arising from surface tension in small tubes.Viscosity: Measures resistance to shear deformation, not volume change.Elastic modulus of rigidity: Pertains to solids under shear, not fluids under pressure.Common Pitfalls:Assuming “incompressible” means zero compressibility; it means volume change is negligible for the application, not mathematically zero.Final Answer:compressibility

Question 5

Unsteady outflow – Time to empty a hemispherical tank A hemispherical tank of radius R has a small orifice of cross-sectional area a at its bottom and is initially full of liquid. Assuming a constant coefficient of discharge Cd and neglecting viscou…

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Introduction:Unsteady tank-draining problems combine hydrostatics (head–depth relation) with continuity through an orifice governed by Torricelli’s law corrected by the discharge coefficient Cd. For curved tanks like hemispheres, the surface area and cross-section change with depth, leading to integrals that must be evaluated to obtain the emptying time.Given Data / Assumptions:Hemispherical tank, radius R, orifice area a at the bottom centre.Coefficient of discharge Cd is constant over the draining process.Outflow obeys Q = Cd * a * sqrt(2 * g * h), with h measured from the orifice.Neglect additional head losses and velocity of approach in the tank (large plan area relative to orifice).Concept / Approach:Let h be instantaneous liquid height from the orifice. For a hemisphere, the cross-sectional area at height h is A(h) = pi * (2 * R * h − h^2). The unsteady mass balance gives A(h) * dh/dt = − Q = − Cd * a * sqrt(2 * g * h). Rearranging: dt = − A(h) / (Cd * a * sqrt(2 * g * h)) * dh. Integrate h from R down to 0 (or 0 up to R with a sign change).Step-by-Step Solution:Step 1: Write A(h) = pi * (2 * R * h − h^2).Step 2: Set dt = A(h) * dh / (Cd * a * sqrt(2 * g * h)).Step 3: Evaluate I = ∫[0→R] pi * (2 * R * h − h^2) / sqrt(h) dh = ∫[0→R] (2 * pi * R * h^(1/2) − pi * h^(3/2)) dh.Step 4: Compute antiderivatives: ∫ h^(1/2) dh = (2/3) h^(3/2); ∫ h^(3/2) dh = (2/5) h^(5/2).Step 5: Evaluate at R: I = 2 * pi * R * (2/3) R^(3/2) − pi * (2/5) R^(5/2) = (14/15) * pi * R^(5/2).Step 6: Thus t = I / (Cd * a * sqrt(2 * g)) = (14 * pi * R^(5/2)) / (15 * Cd * a * sqrt(2 * g)).Verification / Alternative check:Dimensional analysis: numerator has length^(5/2); denominator contributes length^(3/2)/time, yielding time. The result reduces to known flat-tank results when the geometry is changed appropriately.Why Other Options Are Wrong:Scaled variants with 8/15 or altered powers of R do not match the integral of A(h) / sqrt(h).Forms mixing R inside the square root in the denominator incorrectly combine variables.Expressions without R^(5/2) cannot capture the geometry of the hemisphere.Common Pitfalls:Using A = constant (as if the tank were prismatic) or forgetting the square root on 2 * g * h in Torricelli’s outflow law, both of which lead to incorrect time estimates.Final Answer:t = (14 * pi * R^(5/2)) / (15 * Cd * a * sqrt(2 * g))

Question 6

Pressure scales – Relation between absolute, gauge, and atmospheric Absolute pressure is related to gauge pressure and atmospheric pressure by which expression?

Accepted Answer

Introduction:Engineering practice uses multiple pressure references. Gauge pressure measures pressure relative to ambient atmosphere, while absolute pressure is measured relative to an absolute vacuum. Understanding the relation among these is essential for thermodynamics, fluid statics, and instrumentation.Given Data / Assumptions:Ambient atmospheric pressure is Patm (approximately 101.325 kPa at sea level).Gauge pressure Pg = Pabs − Patm by definition.Vacuum pressure Pvaccum refers to a pressure below atmosphere often reported as a magnitude beneath Patm.Concept / Approach:From Pg = Pabs − Patm, rearrange to Pabs = Pg + Patm. This identity holds regardless of whether Pg is positive (above atmosphere) or negative (below atmosphere). Conversions in instruments follow this identity to translate between scales.Step-by-Step Solution:Step 1: Start with the gauge definition: Pg = P − Patm.Step 2: Solve for absolute pressure P: P = Pg + Patm.Step 3: Match to option statements to select the correct relation.Verification / Alternative check:Consider a tire at 220 kPa absolute. With Patm ≈ 100 kPa, Pg ≈ 120 kPa. The identity Pabs = Pg + Patm is satisfied numerically, confirming correctness.Why Other Options Are Wrong:Gauge − atmosphere / atmosphere − gauge: Algebraically incorrect rearrangements.Gauge − vacuum: Terminology confusion; vacuum reading is not a linear subtractor to convert gauge to absolute.Vacuum + atmosphere: This can equal absolute only when “vacuum” is carefully defined as the deficit from absolute zero, which is not the usual industrial convention.Common Pitfalls:Mixing sign conventions for vacuum and gauge readings, or assuming Patm is constant across altitude and weather conditions without correction.Final Answer:gauge pressure + atmospheric pressure

Question 7

Flow visualization – Definition of streamline The imaginary curve drawn in a flowing fluid such that the tangent at any point gives the instantaneous direction of motion at that point is called a what?

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Introduction:Flow visualization uses several kinematic constructs: streamlines, pathlines, and streaklines. Although related, each has a distinct definition and physical interpretation. This question asks for the curve whose tangent everywhere gives the instantaneous flow direction field.Given Data / Assumptions:Steady or unsteady flow field v(x, t).We are considering an instantaneous snapshot of the velocity field.No specific body forces or boundaries are required to define the kinematics.Concept / Approach:A streamline is defined at an instant by the differential relation dy/dx = v_y / v_x in 2D, ensuring its tangent is aligned with the local velocity vector. A pathline is the actual trajectory traced by a single fluid particle over time. A streakline is the locus of all particles that have previously passed through a particular point (e.g., dye injected continuously). In steady flows, all three coincide; in unsteady flows, they differ.Step-by-Step Solution:Step 1: Identify the key phrase “tangent gives direction at that point”.Step 2: Map this to the definition of a streamline constructed from the instantaneous velocity field.Step 3: Conclude that the required curve is the streamline.Verification / Alternative check:Textbook diagrams of steady flow show streamlines coincident with particle paths; in unsteady flows, releasing dye from a point generates a streakline that may not align with streamlines at later times, reinforcing the distinction.Why Other Options Are Wrong:Path line: Tracks one particle over time, not the instantaneous field direction at every point.Streak line: Locus of particles that passed through a point, not necessarily tangent to the instantaneous velocity everywhere.Potential line: A contour of velocity potential; tangent is not generally the velocity direction unless specific conditions hold.Vortex line: Curve everywhere tangent to vorticity, not velocity.Common Pitfalls:Assuming all three lines are always identical. They coincide only under steady conditions.Final Answer:stream line

Question 8

Units – Kinematic viscosity in SI What is the SI unit of kinematic viscosity (nu)?

Accepted Answer

Introduction:Kinematic viscosity nu characterizes the ratio of viscous diffusion to momentum inertia and is defined as the dynamic viscosity mu divided by density rho. Dimensional clarity is important for unit conversion and checking equations in fluid mechanics and heat transfer.Given Data / Assumptions:Definition: nu = mu / rho.Base SI units: mass kg, length m, time s.Dynamic viscosity mu has SI unit Pa·s = N·s/m^2 = kg/(m·s).Concept / Approach:Compute units: nu = (kg/(m·s)) / (kg/m^3) = (kg/(m·s)) * (m^3/kg) = m^2/s. This unit emphasizes that kinematic viscosity acts like a diffusion coefficient for momentum, analogous to thermal diffusivity alpha with units m^2/s.Step-by-Step Solution:Step 1: Write nu = mu / rho.Step 2: Substitute mu units kg/(m·s) and rho units kg/m^3.Step 3: Cancel kg and simplify powers of m to obtain m^2/s.Verification / Alternative check:In CGS, the unit is cm^2/s (1 stokes = 10^−4 m^2/s). Converting confirms the SI result m^2/s.Why Other Options Are Wrong:N-m/s and N-m: Units of power or energy, not viscosity.N-s/m2: This is the unit of dynamic viscosity, not kinematic.kg/(m-s): Also dynamic viscosity, not kinematic.Common Pitfalls:Confusing mu and nu because both appear in the Navier–Stokes equations; remember nu multiplies the Laplacian in the momentum diffusion term and carries m^2/s.Final Answer:m2/s

Question 9

Hydrostatics – Pressure variation with depth At any point within a static liquid, the intensity of pressure varies how with depth measured below the free surface?

Accepted Answer

Introduction:Hydrostatic pressure distribution in an incompressible fluid at rest follows a simple linear law with depth under a uniform gravitational field. Recognizing this relationship is foundational for manometry, buoyancy, and design of storage tanks and dams.Given Data / Assumptions:Liquid at rest (no shear stresses) and constant density rho.Uniform gravitational acceleration g.Reference free surface at pressure P0 (often atmospheric).Concept / Approach:From hydrostatic equilibrium, dp/dz = − rho * g with z positive upward. Integrating between the free surface and a depth h gives p = P0 + rho * g * h. Therefore, the gauge pressure increases linearly and is directly proportional to depth h beneath the surface, independent of the container’s shape or plan area (Pascal’s hydrostatic paradox).Step-by-Step Solution:Step 1: Write the hydrostatic equation: dp/dz = − rho * g.Step 2: Integrate for constant rho and g: p(z) = P0 + rho * g * (z0 − z).Step 3: With depth h = (z0 − z), obtain p = P0 + rho * g * h.Step 4: Conclude the direct proportionality p − P0 ∝ h.Verification / Alternative check:Experimental manometer readings show equal increments of depth produce equal increments of pressure, regardless of vessel shape, confirming the linear law.Why Other Options Are Wrong:Proportional to area or length of the vessel: Pressure is an intensive property and does not depend on container dimensions.Inversely proportional to depth: Opposite to the hydrostatic law.Independent of depth: Only true at the free surface or in microgravity; not in normal hydrostatics.Common Pitfalls:Assuming a wide container has higher pressure because it holds more fluid; pressure depends only on vertical depth and density, not on total volume or shape.Final Answer:directly proportional to the depth of liquid from the surface

Question 10

Buoyancy – line of action and center of buoyancy The line of action of the buoyant force (upthrust) on a submerged or floating body passes through the centre of gravity of the displaced liquid volume, i.e., the centroid of the displaced volume. Stat…

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Introduction:Buoyancy is the upward force exerted by a fluid on an immersed body. This question checks whether you know the exact point of application of the buoyant force and how it relates to the displaced fluid volume.Given Data / Assumptions:Rigid body partially or fully immersed in a static fluid.Fluid is continuous and at rest; free surface may exist.Body may be in equilibrium or constrained; stability effects (metacentre) are discussed conceptually.Concept / Approach:Archimedes’ principle gives magnitude: buoyant force equals weight of displaced fluid. The line of action of this force passes through the centroid of the displaced volume of fluid, known as the center of buoyancy. For floating stability, the center of buoyancy shifts with heel, and the metacentre concept becomes relevant, but the buoyant force still acts through the instantaneous centroid of displaced volume.Step-by-Step Solution:1) Consider a body displacing volume V_d of fluid of specific weight w.2) Magnitude of buoyant force: F_b = w * V_d.3) Pressure forces act normal to the surface and integrate to a resultant acting vertically upward.4) Resultant line of action passes through the centroid of the displaced fluid volume (center of buoyancy).5) Therefore the given statement is correct.Verification / Alternative check:Hydrostatic pressure distribution increases with depth. Integrating pressure over the wetted surface yields a resultant that must pass through the centroid of the displaced volume because that is the balance point of the distributed pressures.Why Other Options Are Wrong:False: contradicts the hydrostatic integration result.True only for incompressible liquids: gases also obey hydrostatics; the definition is geometric, not restricted to liquids.True only for gases at low pressure: unnecessary restriction; the result is general for fluids at rest.Common Pitfalls:Confusing center of buoyancy with center of mass of the body; mixing up metacentric height (stability) with buoyant line of action at a given heel.Final Answer:True

Question 11

Flow dimensionality – interpretation of streamlines A flow whose streamlines form a curve in a plane (i.e., velocity components vary in two spatial directions) is best described as:

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Introduction:Flow dimensionality classifies how many spatial coordinates are needed to describe variations in the velocity field. This question probes the distinction between one-dimensional, two-dimensional, and three-dimensional flow descriptions using streamlines.Given Data / Assumptions:Steady visualization of streamlines in a plane.Streamlines are curves rather than straight, indicating lateral variation.Out-of-plane variations are negligible (typical textbook 2D idealization).Concept / Approach:In one-dimensional flow, properties vary along only one coordinate and are uniform across cross-sections; streamlines are effectively parallel lines. In two-dimensional flow, velocity components depend on two coordinates (e.g., x and y), producing curved streamlines within a plane. Three-dimensional flow would require variations in all three spatial directions.Step-by-Step Solution:1) Observe that streamlines bend within a plane → lateral gradients exist.2) Lateral gradients imply dependence on two coordinates (say x and y).3) Therefore, the appropriate idealization is two-dimensional flow.Verification / Alternative check:Potential-flow solutions around cylinders or airfoils in a 2D section produce curved streamlines but ignore spanwise (z) variations, matching the 2D definition.Why Other Options Are Wrong:One-dimensional flow: would have straight, non-bending streamlines in the plane.Three-dimensional flow: requires out-of-plane variation not stated here.Four-dimensional flow: not a standard fluid-mechanics classification.Common Pitfalls:Assuming any curve means 3D; curvature in a single plane is a hallmark of 2D modeling.Final Answer:two-dimensional flow

Question 12

Units – the meaning of the stoke (St) “Stoke” refers to which of the following physical quantities and unit systems?

Accepted Answer

Introduction:Engineering problems often mix unit systems. This question checks whether you can associate the legacy C.G.S. unit “stoke” with the correct property in fluid mechanics.Given Data / Assumptions:C.G.S. (centimetre–gram–second) and S.I./M.K.S. systems are in view.Kinematic viscosity (nu) and dynamic viscosity (mu) are distinct.Standard relationships between units apply.Concept / Approach:Kinematic viscosity nu = mu / rho. In the C.G.S. system, nu is measured in stokes (St) where 1 St = 1 cm^2/s. In S.I., kinematic viscosity uses m^2/s (no special named unit). Dynamic viscosity uses poise (P) in C.G.S. and Pa·s in S.I.Step-by-Step Solution:1) Identify property: kinematic vs dynamic viscosity.2) Recall the C.G.S. unit name: 1 St = 1 cm^2/s for nu.3) Therefore “stoke” corresponds to kinematic viscosity in C.G.S. units.Verification / Alternative check:Conversion: 1 St = 10^-4 m^2/s. Typical light oils: nu ~ a few cSt (centistokes), consistent with practice.Why Other Options Are Wrong:M.K.S. kinematic viscosity: uses m^2/s without the name “stoke”.M.K.S. dynamic viscosity: unit is N·s/m^2 or Pa·s (S.I.).S.I. dynamic viscosity: Pa·s, not stoke.Common Pitfalls:Confusing poise (dynamic viscosity) with stoke (kinematic viscosity); mixing cSt with cP.Final Answer:kinematic viscosity in C.G.S. units

Question 13

Weirs – definition of a narrow-crested weir A sharp-edged weir is called a narrow-crested weir if the width of its crest is __________ half the head of water above the crest (measured upstream):

Accepted Answer

Introduction:Weir classification influences discharge equations and head–discharge calibration. This question tests the criterion that distinguishes narrow-crested weirs from broad-crested ones based on crest width relative to upstream head.Given Data / Assumptions:Steady open-channel flow over a sharp-edged obstruction (weir).Head measured as water level above crest.Negligible approach velocity correction for conceptual classification.Concept / Approach:If the crest width is small compared with the head, flow accelerates over a short distance and behaves like flow over a sharp edge (narrow-crested). If the crest is wide compared with the head, the flow attains near-critical conditions on the crest (broad-crested behavior).Step-by-Step Solution:1) Define criterion: crest width b compared to head H.2) Narrow-crested if b 3) Therefore the blank should read “less than”.Verification / Alternative check:Empirical discharge coefficients for narrow vs broad crests reflect different flow regimes; lab observations corroborate the width–head criterion.Why Other Options Are Wrong:Equal to / more than: move classification toward broad-crested behavior.Indeterminate: classification is standard based on b/H ratio.Common Pitfalls:Confusing “narrow-crested” with “sharp-crested” terminology; ignoring approach velocity effects (important for discharge calculation but not for this definition).Final Answer:less than

Question 14

Pressure measurement – meaning of vacuum pressure Is it correct to say that “vacuum pressure” is always the negative of the gauge pressure (i.e., vacuum pressure = − gauge pressure when pressure is below atmospheric)?

Accepted Answer

Introduction:Different reference frames for pressure—absolute, gauge, and vacuum—can be confusing. This question checks the algebraic relationship between gauge pressure and vacuum pressure when a system is below atmospheric pressure.Given Data / Assumptions:Atmospheric pressure p_atm is the local reference.Gauge pressure p_g = p_abs − p_atm.Vacuum pressure p_vac is the shortfall below atmospheric: p_vac = p_atm − p_abs when p_abs Concept / Approach:Combining the definitions gives p_vac = − p_g for sub-atmospheric conditions. Thus, a gauge reading of −30 kPa corresponds to a vacuum pressure of +30 kPa (of vacuum), while absolute pressure is p_abs = p_atm − 30 kPa.Step-by-Step Solution:1) Write gauge: p_g = p_abs − p_atm.2) Write vacuum: p_vac = p_atm − p_abs.3) Therefore p_vac = −(p_abs − p_atm) = −p_g.4) Hence the statement is correct as a definition for sub-atmospheric systems.Verification / Alternative check:Manometer readings referenced to atmosphere invert when switching from gauge to vacuum scale, consistent with p_vac = −p_g.Why Other Options Are Wrong:No: contradicts the algebraic relation.Only for ideal gases / only when temperature is constant: the relation is definitional and independent of equation of state or temperature.Common Pitfalls:Mixing absolute and gauge numbers; reporting “−30 kPa vacuum” (nonsensical)—vacuum is reported as a positive shortfall, while gauge may be negative.Final Answer:Yes

Question 15

Rotating liquid in a vertical-axis cylinder – depth of paraboloid A cylindrical vessel of radius r is rotated steadily at angular speed ω about its vertical axis. The free surface becomes a paraboloid. The depth difference between the rim and the ve…

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Introduction:When a liquid rotates as a forced vortex, its free surface forms a paraboloid of revolution. This question asks for the depth difference between the center and rim, which is fundamental in vortex dynamics and rotating-tank problems.Given Data / Assumptions:Steady rigid-body rotation with angular speed ω.Cylindrical container of radius r; gravity g acts downward.Neglect evaporation, capillarity, and transient sloshing.Concept / Approach:In a forced vortex, all fluid particles have the same angular velocity ω. The free-surface profile satisfies z = (ω^2 * r^2) / (2 * g) + constant. Hence the height difference between the rim (radius r) and the center (r = 0) is Δh = ω^2 * r^2 / (2 * g).Step-by-Step Solution:1) Write energy balance in a rotating frame: pressure + potential + centrifugal potential is constant on the free surface.2) Along the free surface, p ≈ p_atm; thus elevation varies with r to balance centrifugal effects.3) Integrate centrifugal acceleration a_c = ω^2 * r to get head: (ω^2 * r^2) / (2 * g).4) Evaluate at r and at 0: Δh = [(ω^2 * r^2)/(2 * g)] − 0.Verification / Alternative check:Dimensional check: [ω^2 * r^2 / g] has units of length; the factor 1/2 arises from integrating r dr.Why Other Options Are Wrong:ω^2 * r^2 / g: missing the 1/2 integration factor.ω^2 * r / g: incorrect dependence; should be proportional to r^2.r^2 / (2 * g * ω^2): inverted dependence on ω.Common Pitfalls:Confusing free-vortex (irrotational) with forced vortex; forgetting that the profile is parabolic (∝ r^2), not linear in r.Final Answer:ω^2 * r^2 / (2 * g)

Question 16

Discharge coefficients – relation among Cv, Cc, and Cd For a sharp-edged orifice/mouthpiece, the coefficient of discharge (Cd) is __________ the coefficient of velocity (Cv):

Accepted Answer

Introduction:Practical orifice flow accounts for contraction and losses via three coefficients: coefficient of velocity (Cv), coefficient of contraction (Cc), and coefficient of discharge (Cd). This question asks about their ordering.Given Data / Assumptions:Sharp-edged orifice with a vena contracta.Steady incompressible flow.Losses and contraction are not negligible.Concept / Approach:Definitions: Cv = V_actual / V_ideal, Cc = A_jet / A_orifice, and Cd = Q_actual / Q_ideal. Since Q = A_jet * V_actual and Q_ideal = A_orifice * V_ideal, it follows that Cd = Cc * Cv. Because 0 Step-by-Step Solution:1) Write Cd = Q_actual / Q_ideal.2) Substitute Q expressions to get Cd = (A_jet/A_orifice) * (V_actual/V_ideal) = Cc * Cv.3) Since Cc Verification / Alternative check:Typical values for sharp edges: Cc ~ 0.62, Cv ~ 0.97, hence Cd ~ 0.62 * 0.97 ≈ 0.60, confirming Cd Why Other Options Are Wrong:Same as / more than: contradicts Cd = Cc * Cv with Cc Indeterminate: the inequality follows directly from definitions.Common Pitfalls:Confusing Cd with Cv or ignoring the contraction effect that reduces effective area.Final Answer:less than

Question 17

Flow regimes – velocity and turbulence “Turbulent flow takes place at high velocities.” Do you agree with this general statement for internal flows in pipes?

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Introduction:Turbulence is characterized by chaotic, eddying motion and enhanced mixing. In internal flows, the transition from laminar to turbulent regimes is commonly correlated with increasing flow speed through the Reynolds number.Given Data / Assumptions:Internal flow in a circular pipe.Reynolds number Re = (rho * V * D) / mu.Surface roughness and disturbances exist at realistic levels.Concept / Approach:For many engineering applications, laminar flow occurs at low Re, transitional at intermediate Re, and turbulent at sufficiently high Re. Since Re grows with velocity V, higher velocities favor turbulence. The simplified statement is broadly correct, though the precise threshold depends on geometry and disturbances.Step-by-Step Solution:1) Recognize that Re ∝ V; increasing V raises Re.2) Typical thresholds: laminar Re 3) Therefore, “high velocities” (yielding high Re) generally promote turbulence in pipes.Verification / Alternative check:Experimental observations show onset of puffs/slugs in the transitional band and fully developed turbulent spectra beyond the threshold, with friction factor following turbulent correlations (e.g., Colebrook).Why Other Options Are Wrong:Disagree: ignores the Re–velocity relationship.Agree only for very viscous fluids: viscosity affects Re, but the statement is generally valid across fluids.Agree only when Reynolds number exceeds a threshold: while true in detail, the question seeks a general judgment; “Agree” captures the broad engineering rule.Common Pitfalls:Treating the velocity threshold as universal; thresholds vary with roughness, disturbances, and entrance effects.Final Answer:Agree

Question 18

Dimensionless groups – force ratios in fluid flow The ratio of inertia force to viscous force in a flow is known as:

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Introduction:Dimensionless numbers help predict similarity and regimes in fluid mechanics. This question asks you to identify which group compares inertia effects with viscous effects.Given Data / Assumptions:Characteristic velocity V, length L (or diameter D), density rho, viscosity mu.Internal or external flows where both inertia and viscosity matter.Steady characteristic scaling.Concept / Approach:Reynolds number Re is defined as Re = (rho * V * L) / mu. It compares inertia force scale (rho * V^2 * L^2 over area) to viscous force scale (mu * V * L over area), leading to the ratio Re. Other numbers compare different effects (gravity, surface tension, pressure).Step-by-Step Solution:1) Identify target ratio: inertia / viscous.2) Recall Re = rho * V * L / mu.3) Conclude that Reynolds number is the required dimensionless group.Verification / Alternative check:Laminar–turbulent transition and drag correlations widely use Re, confirming its role as inertia–viscous comparator.Why Other Options Are Wrong:Froude number: inertia / gravity (V / sqrt(gL)).Weber number: inertia / surface tension (rho * V^2 * L / sigma).Euler number: pressure / inertia (Δp / (0.5 * rho * V^2)).Common Pitfalls:Mixing up which physical effect is in the denominator; memorize each group’s comparison for clarity.Final Answer:Reynolds number

Question 19

Pressure measurement with a piezometer tube The pressure indicated by a simple piezometer tube connected to a pipe (open to atmosphere at the tube top) is:

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Introduction:Piezometer tubes are the most basic devices for measuring fluid pressure by observing the static head of a liquid column. This question asks which pressure reference is obtained by a plain piezometer.Given Data / Assumptions:Transparent vertical tube connected to a pressurized pipe containing an incompressible liquid.Tube top is open to the local atmosphere.Capillarity and vapor pressure effects are negligible.Concept / Approach:The piezometer shows the height h such that p_pipe − p_atm = gamma * h. Because the tube is open to atmosphere, it inherently measures pressure relative to atmosphere—i.e., gauge pressure. Absolute pressure would require adding atmospheric pressure to this reading or using a sealed reference.Step-by-Step Solution:1) Read column height h above the connection centerline (with datum corrections).2) Compute p_gauge = gamma * h.3) Recognize that absolute pressure is p_abs = p_atm + p_gauge.Verification / Alternative check:Switching to a different local atmospheric pressure changes the absolute pressure but not the gauge reading principle, confirming the device measures relative to atmosphere.Why Other Options Are Wrong:Vacuum pressure: a plain piezometer cannot read gas/vacuum directly and cannot indicate below-atmospheric pressures for gases.Absolute pressure: requires reference to vacuum, not atmosphere.Atmospheric pressure: that is the reference, not the measured value.Common Pitfalls:Using a piezometer for gases (impractical); forgetting elevation corrections when comparing points.Final Answer:gauge pressure

Question 20

Open-Channel Flow Classification – Depth vs Critical Depth If the actual flow depth in an open channel is greater than the critical depth, the regime is called subcritical (also termed tranquil) flow. This state is gravity-dominated with Froude numb…

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Introduction:In open-channel hydraulics, flow is categorized by comparing the actual depth with the critical depth and, equivalently, by the Froude number Fr = V / sqrt(g * y_h). Recognizing subcritical (tranquil) versus supercritical (torrential) flow determines how disturbances propagate and which equations or control sections to use.Given Data / Assumptions:Prismatic open channel carrying water under gravity.Depth is greater than the critical depth for the given discharge and section.Hydraulic depth y_h is characteristic depth (equals flow area divided by top width for general sections).Concept / Approach:Critical depth is defined where specific energy is minimum for a given discharge. When actual depth is larger than the critical depth, gravitational potential effects dominate inertial effects, yielding a low Froude number (Fr Step-by-Step Solution:1) Identify the regime criterion: Fr = V / sqrt(g * y_h).2) For depth > critical depth, velocity is lower for the same discharge, so Fr 3) Fr 4) Therefore the correct classification is tranquil flow.Verification / Alternative check:On a mild channel slope with given discharge, backwater curves (M-profiles) occur when depth exceeds normal or approaches subcritical conditions, consistent with tranquil flow characteristics (slow, deep, gravity-dominated).Why Other Options Are Wrong:Torrential flow: Another name for supercritical, requires Fr > 1 and depth less than critical. Critical flow: Occurs only at exactly the critical depth (Fr = 1). Supercritical flow and rapid flow: Both indicate depth less than critical with high velocity.Common Pitfalls:Confusing the terms tranquil and torrential; mixing pipe flow criteria (Reynolds number) with open-channel criteria (Froude number); assuming high depth always implies high discharge without considering section shape and specific energy.Final Answer:tranquil flow

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