Q: Specific energy in a rectangular open channel: the minimum specific energy occurs at the critical depth hc. Which expression is correct for this minimum value?

Introduction / Context:Specific energy E in open-channel flow is a key concept for analyzing transitions, critical flow, and controls such as weirs and flumes. For a rectangular channel, the minimum specific energy occurs at the critical depth hc, establishing an important design and diagnostic criterion. Given Data / Assumptions: Rectangular channel of width b carrying discharge Q. Specific energy: E = y + V^2/(2 g) with V = Q/(b y). Critical depth hc gives minimum E for fixed discharge. Concept / Approach:Write E(y) = y + (Q^2)/(2 g b^2 y^2). Differentiate E with respect to y and set dE/dy = 0 to find hc. The condition leads to V_c^2 = g hc and hence hc = (Q^2/(g b^2))^(1/3). Substituting back into E gives Emin = (3/2) hc. Step-by-Step Solution:E(y) = y + (Q^2)/(2 g b^2 y^2).dE/dy = 1 − (Q^2)/(g b^2 y^3) = 0 at y = hc.Thus Q^2/(g b^2) = hc^3; and V_c^2 = g hc.Emin = hc + (V_c^2)/(2 g) = hc + (g hc)/(2 g) = (3/2) hc. Verification / Alternative check:Plotting E versus y at constant Q shows a U-shaped curve with its minimum at y = hc and E_min/Evaluated ratio of 1.5, confirming the formula. Why Other Options Are Wrong:hc, 2 hc, (1/2) hc, hc/3 do not satisfy the extremum condition derived from E(y); they misrepresent the balance between depth and velocity head at critical flow. Common Pitfalls: Confusing specific energy with specific force (momentum function) used in hydraulic jump analysis. Using total head instead of specific energy (which is measured relative to the channel bottom). Final Answer:Emin = (3/2) * hc

Question 1

The ratio of inertia forces to viscous forces, used to characterize flow regime (laminar vs turbulent), is known as which dimensionless number?

Accepted Answer

Introduction / Context:Dimensionless numbers collapse complexity and guide similarity between model and prototype. The Reynolds number is paramount in internal and external flows for predicting laminar–turbulent transition and friction behavior. Given Data / Assumptions: Characteristic length L, velocity V, density ρ, and dynamic viscosity μ. Newtonian fluid. Concept / Approach: Reynolds number Re = (inertia forces)/(viscous forces) = ρ V L / μ. Large Re implies inertia dominates and turbulence is likely; small Re implies viscous dominance and laminar flow. Other dimensionless numbers represent different force balances (e.g., Froude: inertia/gravity; Weber: inertia/surface tension; Euler: pressure/inertia). Step-by-Step Solution: Write Re = ρ V L / μ = V L / ν, where ν is kinematic viscosity.Identify regimes in a pipe: laminar (Re <~ 2000), transitional (≈ 2000–4000), turbulent (>~ 4000).Use Re to select correlations (e.g., Hagen–Poiseuille vs. Blasius/Colebrook). Verification / Alternative check: Check units: ρ V L / μ → (M L^-3)(L T^-1)(L)/(M L^-1 T^-1) = dimensionless, as required. Why Other Options Are Wrong: Euler number relates pressure to inertia; Froude compares inertia to gravity; Weber compares inertia to surface tension; Mach compares flow speed to speed of sound. Common Pitfalls: Misusing hydraulic diameter for non-circular ducts; forgetting temperature dependence of μ and ν. Final Answer: Reynolds number

Question 2

In the CGS unit system, “poise” is the unit of which fluid property?

Accepted Answer

Introduction / Context:Engineering data often report viscosities in different unit systems. Knowing whether a given unit refers to dynamic or kinematic viscosity prevents order-of-magnitude errors in Reynolds number and pressure-drop calculations. Given Data / Assumptions: CGS (centimetre–gram–second) system. Dynamic viscosity μ and kinematic viscosity ν are distinct. Concept / Approach: Dynamic viscosity μ in CGS is measured in poise (P). Kinematic viscosity ν = μ/ρ has unit stoke (St) = cm^2/s. In practice, centipoise (cP) and centistoke (cSt) are frequently used (1 P = 100 cP; 1 St = 100 cSt). Step-by-Step Solution: Identify unit: 1 poise = 1 g/(cm·s).Relate to SI: 1 P = 0.1 Pa·s; 1 cP = 0.001 Pa·s.Water at ~20°C: μ ≈ 1 cP (0.001 Pa·s), ν ≈ 1 cSt. Verification / Alternative check: Dimensional breakdown: μ has units of shear stress divided by velocity gradient, which yields M L^-1 T^-1, consistent with g/(cm·s) in CGS. Why Other Options Are Wrong: (b) Velocity gradient has units s^-1, not poise. (c) Density has g/cm^3. (d) Kinematic viscosity uses stokes, not poise. (e) Surface tension uses dyn/cm. Common Pitfalls: Confusing cP and P (factor 100); mixing dynamic and kinematic viscosity when computing Reynolds number Re = V L / ν. Final Answer: Viscosity (dynamic viscosity)

Question 3

Chezy’s resistance constant C used in the Chezy equation for open-channel flow was empirically correlated/estimated by which classical investigator?

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Introduction / Context:The Chezy equation V = C √(R S) uses an empirical resistance coefficient C. Historically, engineers proposed correlations to estimate C from channel roughness and hydraulic radius. Knowing the names helps track which formula you are applying in design problems. Given Data / Assumptions: Steady uniform flow in open channels. Chezy form V = C √(R S), where R is hydraulic radius and S is slope. Concept / Approach: Kutter proposed a widely used formula to compute Chezy’s C in terms of a roughness coefficient and hydraulic radius. Bazin also developed a relation for C, but “Kutter’s formula” is the classic association students are often tested on. Manning, meanwhile, introduced an alternative equation V = (1/n) R^(2/3) S^(1/2) with roughness n rather than C. Step-by-Step Solution: Recognize the equation form: Chezy uses C; Manning uses n.Associate names: Kutter’s formula → Chezy’s C; Manning’s equation → n; Bazin also proposed a C-relation but is less cited in modern curricula.Thus the best single-choice match here is Kutter. Verification / Alternative check: Many handbooks list “Kutter’s formula for C” near the Chezy equation, reinforcing this pairing. Why Other Options Are Wrong: (a) Bazin did propose a C-relation, but exam convention often associates Kutter more directly. (c) Manning corresponds to a different formula using n. (d) Powell is unrelated in this context. (e) Darcy pertains to closed conduit head loss, not Chezy’s C. Common Pitfalls: Confusing the Chezy and Manning formulations; assuming the same roughness parameter applies across them without conversion. Final Answer: Kutter

Question 4

In orifice hydraulics, the term “hydraulic coefficients” commonly refers to which coefficient(s)?

Accepted Answer

Introduction / Context:Real orifice jets deviate from ideal theory due to contraction and energy loss. Engineers therefore use empirical coefficients to relate actual to theoretical values, collectively termed the hydraulic coefficients of an orifice. Given Data / Assumptions: Sharp-edged orifice, steady flow. Coefficients treated approximately constant over small head ranges. Concept / Approach: Key coefficients are: coefficient of contraction Cc (area reduction to vena contracta), coefficient of velocity Cv (actual jet speed/theoretical), and coefficient of discharge Cd = Cc * Cv (actual discharge/theoretical). Sometimes a “coefficient of resistance” is discussed via head loss relations, but standard trio remains Cc, Cv, and Cd—together called hydraulic coefficients. Step-by-Step Solution: Define Cc = A_vena / A_orifice.Define Cv = V_actual / V_theoretical.Define Cd = Q_actual / Q_theoretical = Cc * Cv. Verification / Alternative check: Laboratory calibration of an orifice plate gives Cd close to 0.6 for sharp-edged small orifices in water, consistent with typical Cc ≈ 0.62 and Cv ≈ 0.97. Why Other Options Are Wrong: (a), (b), and (c) each name part of the set; the most complete answer is (d) “All of the above.” (e) is false. Common Pitfalls: Using Cd alone without understanding its decomposition; assuming coefficients are constants across all heads and Reynolds numbers. Final Answer: All of the above

Question 5

For a two-dimensional velocity field with components u = a x and v = b y, the streamline family satisfies which curve type?

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Introduction / Context:Streamlines are curves tangent to the instantaneous velocity vector everywhere. Identifying their mathematical form from u(x, y) and v(x, y) is a foundational skill in fluid kinematics and helps recognize flow patterns such as sources, sinks, and saddles. Given Data / Assumptions: Two-dimensional steady field: u = a x, v = b y, with constants a and b. Streamline definition in 2D: dy/dx = v/u. Assume a and b are non-zero real constants. Concept / Approach: For streamlines in 2D, dy/dx = v/u = (b y)/(a x). This separable ODE integrates to ln y = (b/a) ln x + C, or y = C x^(b/a). This is a power-law family of curves. When b/a is negative, the curves are rectangular hyperbolas; when b/a is positive, they are power curves that are not parabolas or ellipses. Among standard conic labels provided, “hyperbolic” best characterizes the common saddle-type case b = −a often discussed in textbooks. Step-by-Step Solution: Write streamline ODE: dy/dx = (b y)/(a x).Separate variables: (1/y) dy = (b/a) (1/x) dx.Integrate: ln y = (b/a) ln x + C ⇒ y = C x^(b/a). Verification / Alternative check: Special case b = −a gives y = C / x, a rectangular hyperbola—commonly cited. If b = a, y ∝ x (straight lines through the origin). The option set does not include “power curve,” so hyperbolic is the closest canonical classification for the typical saddle case. Why Other Options Are Wrong: (a), (d) circular/elliptical require x^2 + y^2 type relations. (b) parabolic needs y ∝ x^2 + … . (e) is too restrictive; general solution is not purely radial straight lines. Common Pitfalls: Forgetting to use dy/dx = v/u; misclassifying the family without considering the sign of b/a. Final Answer: Hyperbolic

Question 6

In a closed cylindrical vessel completely filled with liquid and rotating as a forced vortex about its vertical axis, the total hydrostatic-dynamic thrust acting on the top (lid) varies with which power of the cylinder radius?

Accepted Answer

Introduction / Context:When a completely filled cylindrical vessel of liquid is rotated about its vertical axis, the liquid executes a forced vortex motion (rigid-body rotation). Engineers often need to estimate the total thrust on the top cover or bottom due to the combined hydrostatic and centrifugal effects. Understanding how this thrust scales with the vessel radius is essential for lid design and fastening strength. Given Data / Assumptions: Cylindrical vessel of radius R, fully filled with an incompressible liquid of density rho. Uniform angular velocity omega about the vertical axis; free surface does not exist (vessel is closed and full). Gravity acts downward; analysis on a horizontal top lid. Concept / Approach:In a forced vortex, the pressure varies radially according to dp/dr = rho * omega^2 * r. Integrating from the axis to radius r gives a pressure rise proportional to r^2. The total thrust on the lid is obtained by integrating this pressure over circular rings of area 2 * pi * r * dr, which introduces another factor of r. Consequently, the final thrust scales with R^4. Step-by-Step Solution:Radial pressure distribution: p(r) = p(0) + (rho * omega^2 / 2) * r^2.Elemental force on ring: dF = p(r) * dA = p(r) * 2 * pi * r * dr.Integrate from r = 0 to R: F = ∫[0→R] [p(0) + (rho * omega^2 / 2) * r^2] * 2 * pi * r dr.The variable part integrates as ∝ ∫ r^3 dr, yielding a term ∝ R^4. Verification / Alternative check:Dimensional reasoning confirms that pressure rise includes r^2; integrating over area adds another r^2 cumulatively, leading to R^4 dependence for the rotational contribution to thrust. Why Other Options Are Wrong:radius, (radius)^2, (radius)^3: each underestimates the radial build-up compounded by area integration.None of these: incorrect because (radius)^4 is correct. Common Pitfalls: Confusing static hydrostatic thrust scaling (which involves R^2) with rotational forced-vortex thrust scaling (R^4). Forgetting that the vessel is completely full, so no parabolic free surface forms; still, the forced vortex pressure field applies. Final Answer:(radius)^4

Question 7

In energy accounting for fluid flow, which forms of mechanical energy can a fluid particle possess?

Accepted Answer

Introduction / Context:Bernoulli-based energy accounting in fluid mechanics tracks mechanical energy per unit weight as the sum of pressure head, velocity head, and elevation head. Recognizing these terms is fundamental to applying the energy equation correctly to pipes, open channels, pumps, and turbines. Given Data / Assumptions: Incompressible, steady flow with negligible shaft work across a short reach. Mechanical (not thermal) energy forms are of interest. Friction/head losses may be present but are accounted as separate loss terms. Concept / Approach:The mechanical energy per unit weight is expressed as p/(gamma) + V^2/(2 g) + z (pressure head + velocity head + elevation head). These correspond to pressure energy, kinetic energy, and potential (elevation) energy respectively. There is no concept of an “initial energy” in the Bernoulli statement; internal energy belongs to thermal accounting and is not a mechanical head term for incompressible flows. Step-by-Step Solution:Identify mechanical energy heads: pressure head p/gamma, velocity head V^2/(2 g), and elevation head z.Relate heads to energy per unit weight to build Bernoulli’s equation.Exclude undefined or non-mechanical terms from the head balance. Verification / Alternative check:Any standard derivation of Bernoulli’s equation shows only the three heads above; additional “initial” energy terms do not appear in mechanical energy balance. Why Other Options Are Wrong:Elevation only / kinetic+elevation only: incomplete; they omit pressure energy.“Initial energy”: not a defined mechanical energy form in fluid mechanics.Internal energy only: relevant to thermodynamics, not Bernoulli’s mechanical energy heads. Common Pitfalls: Calling pressure head “potential energy” without distinguishing elevation head. Mixing internal energy and temperature effects into the Bernoulli equation for incompressible flow. Final Answer:Pressure, elevation, and kinetic energy

Question 8

For a rectangular sharp-crested weir, discharge varies as H^(3/2). If the head H is measured with a +1% error, what is the approximate percentage error in the computed discharge?

Accepted Answer

Introduction / Context:Hydrometric structures like sharp-crested weirs convert a head measurement H into discharge Q using Q ∝ H^(3/2). Understanding error propagation through power relations is critical for estimating measurement uncertainty in field hydrology. Given Data / Assumptions: Rectangular sharp-crested weir with standard flow conditions (aerated nappe, no submergence). Q ∝ H^(3/2) ignoring coefficient variations. Small relative error, linearized propagation applicable. Concept / Approach:If Q ∝ H^n, then a small relative error dH/H produces dQ/Q = n * dH/H. For n = 3/2, a 1% head error gives 1.5% discharge error. This simple rule of thumb is widely used in discharge rating analyses. Step-by-Step Solution:Write Q = k * H^(3/2), where k is constant for geometry and Cd.Differentiate: dQ/Q = (3/2) * dH/H.Substitute dH/H = 0.01 → dQ/Q = 0.015 → 1.5%. Verification / Alternative check:Finite difference check: H → 1.01 H; Q_new/Q_old = (1.01)^(1.5) ≈ 1.015, confirming 1.5%. Why Other Options Are Wrong:1.25%, 1.75%, 2.25%: do not match the 3/2 scaling; they over/understate sensitivity. Common Pitfalls: Applying the 3/2 exponent to suppressed or drowned weirs without checking conditions. Ignoring additional uncertainty from Cd calibration and head measurement devices. Final Answer:1.5%

Question 9

For a hydraulically long pipe run, which head-loss components are commonly neglected in comparison to the dominant losses?

Accepted Answer

Introduction / Context:In extended pipeline systems, designers must decide which loss terms significantly impact total head requirements. Over long distances, some localized losses are negligible when compared to frictional head loss distributed along the pipe length. Given Data / Assumptions: Steady, incompressible flow in a long, prismatic pipe. Minor losses (entrance, exit, fittings) are small relative to major (friction) losses over long lengths. Fully rough or transitional friction factors may apply, but the reasoning remains. Concept / Approach:Total head loss = friction loss (major) + sum of minor losses. For long pipelines, the Darcy–Weisbach friction term 4 f L/D * (V^2 / (2 g)) dominates. Entrance and outlet losses are finite and do not scale with L, so their relative contribution diminishes as L increases. Step-by-Step Solution:Estimate friction head: h_f ≈ 4 f (L/D) (V^2 / (2 g)).Entrance and exit losses: h_e = K_e V^2/(2 g), h_o = K_o V^2/(2 g), with K’s of order unity.For L/D large, h_f ≫ h_e + h_o → neglect entrance and exit in preliminary sizing. Verification / Alternative check:Numerical example shows that for L/D in the thousands, minor-loss contributions are below a few percent of total head, justifying their omission in first-pass design. Why Other Options Are Wrong:Entrance only / exit only: both are minor, and both are typically neglected together.Friction loss: cannot be ignored; it is dominant in long lines.“Minor losses in fittings”: may or may not be negligible depending on numerous fittings; the safe generalization is about entrance and exit specifically. Common Pitfalls: Neglecting significant valve and fitting losses in complex networks. Using this simplification for short pipes where minor losses can be comparable to friction. Final Answer:Both entrance and exit losses

Question 10

Flow classification by Mach number: if the Mach number M for a fluid flow is less than 1, the flow is termed as:

Accepted Answer

Introduction / Context:Compressible-flow regimes are categorized by the Mach number M = V / a, where V is flow speed and a is the local speed of sound. Correct regime identification guides which physical effects and equations are important (e.g., shock waves, compressibility corrections). Given Data / Assumptions: Single stream of a gas in a duct or external flow. Local Mach number M < 1. Concept / Approach:Standard ranges: subsonic (M < 1), sonic (M = 1), supersonic (M > 1 up to ~5), hypersonic (M ≳ 5), transonic (M near 1, typically 0.8–1.2). For subsonic flows, traditional incompressible methods can be adequate when M ≲ 0.3–0.5, with corrections as M increases toward unity. Step-by-Step Solution:Compare given M with regime thresholds.Since M < 1, the correct term is “subsonic.” Verification / Alternative check:Textbook charts of compressible flow consistently adopt these definitions and boundaries. Why Other Options Are Wrong:Sonic: applies only at M = 1.Supersonic / hypersonic: require M > 1, and much greater for hypersonic.Transonic: refers to around-unit Mach with mixed sub/supersonic regions. Common Pitfalls: Equating “incompressible” with all M < 1; in practice, incompressible assumptions are usually valid near M ≲ 0.3. Overlooking temperature variation effects on the local speed of sound. Final Answer:Subsonic

Question 11

Specific energy in a rectangular open channel: the minimum specific energy occurs at the critical depth hc. Which expression is correct for this minimum value?

Accepted Answer

Introduction / Context:Specific energy E in open-channel flow is a key concept for analyzing transitions, critical flow, and controls such as weirs and flumes. For a rectangular channel, the minimum specific energy occurs at the critical depth hc, establishing an important design and diagnostic criterion. Given Data / Assumptions: Rectangular channel of width b carrying discharge Q. Specific energy: E = y + V^2/(2 g) with V = Q/(b y). Critical depth hc gives minimum E for fixed discharge. Concept / Approach:Write E(y) = y + (Q^2)/(2 g b^2 y^2). Differentiate E with respect to y and set dE/dy = 0 to find hc. The condition leads to V_c^2 = g hc and hence hc = (Q^2/(g b^2))^(1/3). Substituting back into E gives Emin = (3/2) hc. Step-by-Step Solution:E(y) = y + (Q^2)/(2 g b^2 y^2).dE/dy = 1 − (Q^2)/(g b^2 y^3) = 0 at y = hc.Thus Q^2/(g b^2) = hc^3; and V_c^2 = g hc.Emin = hc + (V_c^2)/(2 g) = hc + (g hc)/(2 g) = (3/2) hc. Verification / Alternative check:Plotting E versus y at constant Q shows a U-shaped curve with its minimum at y = hc and E_min/Evaluated ratio of 1.5, confirming the formula. Why Other Options Are Wrong:hc, 2 hc, (1/2) hc, hc/3 do not satisfy the extremum condition derived from E(y); they misrepresent the balance between depth and velocity head at critical flow. Common Pitfalls: Confusing specific energy with specific force (momentum function) used in hydraulic jump analysis. Using total head instead of specific energy (which is measured relative to the channel bottom). Final Answer:Emin = (3/2) * hc

Question 12

An orifice meter installed in a pipeline is primarily used to measure which quantity reliably (after proper calibration)?

Accepted Answer

Introduction / Context:Differential head meters such as orifice meters, venturimeters, and flow nozzles are ubiquitous in industry. They create a known relationship between pressure drop and flow rate, enabling robust discharge measurements with appropriate coefficients and calibration. Given Data / Assumptions: Concentric orifice plate with known beta ratio (d/D). Steady, incompressible flow; pressure taps upstream and downstream of the plate. Discharge coefficient Cd determined by standards or calibration. Concept / Approach:Bernoulli plus continuity across the orifice section relates pressure drop to velocity through the throat area. The volumetric discharge Q is then Q = Cd * A2 * sqrt[2 * Δp / (rho * (1 − beta^4))], where A2 is the orifice area and beta is diameter ratio. With Cd known, Q follows from measured Δp. Step-by-Step Solution:Measure pressure differential Δp using taps and a manometer/transducer.Compute throat velocity and multiply by throat area.Apply discharge coefficient Cd to correct for losses and non-idealities. Verification / Alternative check:Comparison against volumetric tank tests confirms the orifice meter’s Q accuracy within prescribed ranges of Reynolds number and beta ratio when standards are followed. Why Other Options Are Wrong:Static pressure: a pressure gauge alone does this; orifice meter depends on Δp.Average speed independent of area: discharge requires area * velocity.Local vena-contracta velocity: not directly measured by the pressure taps.Density: not directly measured; it must be known or inferred. Common Pitfalls: Using incorrect Cd outside the calibration Reynolds number range. Ignoring temperature and fluid property variations affecting rho and Cd. Final Answer:Volumetric discharge (flow rate)

Question 13

For a forced vortex (rigid-body rotation) of a liquid in a horizontal plane with angular velocity ω, what is the radial pressure gradient relation?

Accepted Answer

Introduction / Context:Vortex motions are common in rotating machinery and mixing vessels. In a forced vortex, the fluid rotates as a solid body with tangential velocity v = ω r. The radial pressure variation must balance the required centripetal acceleration of the fluid elements. Given Data / Assumptions: Steady rotation about a vertical axis; analysis in a horizontal plane (z constant). Incompressible fluid with density ρ. Negligible radial flow (pure rotation). Concept / Approach:Centripetal acceleration of a rotating element is v^2 / r = ω^2 r. The radial pressure gradient supplies this inward acceleration: (1/ρ) * dp/dr = v^2 / r = ω^2 r. Hence dp/dr = ρ * ω^2 * r, showing that pressure increases quadratically with radius. Step-by-Step Solution:Use equilibrium in the radial direction: dp/dr = ρ * v^2 / r.With v = ω r, substitute to obtain dp/dr = ρ * ω^2 * r.Integrate if needed: p(r) = p(0) + (ρ * ω^2 / 2) * r^2. Verification / Alternative check:Laboratory observations show parabolic free surfaces in open vessels, consistent with p ∝ r^2 in the liquid below. Why Other Options Are Wrong:ρ ω r: missing one power of ω.ρ ω^2 / r: incorrect dependence on r.ρ v^2: dimensionally inconsistent (missing 1/r).Zero gradient: contradicts required centripetal force. Common Pitfalls: Confusing forced vortex (v = ω r) with free vortex (v ∝ 1/r), where dp/dr = ρ k^2 / r^3. Ignoring gravity; here analysis is strictly in a horizontal plane where z is constant. Final Answer:dp/dr = ρ * ω^2 * r

Question 14

A partially filled water tank is carried on a truck accelerating forward at a constant rate on level ground. What happens to the water free surface as seen in the tank?

Accepted Answer

Introduction / Context:Free-surface behavior in accelerating containers is a classic application of dynamic equilibrium in non-inertial frames. Understanding the tilt direction is important in vehicle-tank design, slosh analysis, and safe transport of liquids. Given Data / Assumptions: Truck accelerates forward with constant horizontal acceleration a on level ground. Tank is partially filled; free surface is exposed to air. Quasi-steady condition, negligible wave dynamics during the instant considered. Concept / Approach:In the accelerating frame, a pseudo body force acts on the liquid opposite to the direction of acceleration. The effective gravity vector g′ is the vector sum of real gravity g (downward) and pseudo acceleration a (backward). The free surface aligns perpendicular to g′, so it tilts upward toward the direction of the pseudo acceleration (toward the rear of the truck). Step-by-Step Solution:Define g′ = g (down) + (−a) (backward).The free surface normal is along g′; therefore the surface slopes with tanθ = a/g.Since g′ leans backward, the liquid level is higher at the rear and lower at the front. Verification / Alternative check:Small-angle prediction tanθ = a/g matches measurements in accelerating lab carts and explains spillage risk at the front baffle during rapid starts. Why Other Options Are Wrong:Rise at front / fall at back: opposite to pseudo-force direction.Remains horizontal: only true when a = 0.None of these: incorrect because option stating rise at back, fall at front is correct. Common Pitfalls: Confusing truck braking (deceleration) with acceleration; braking would reverse the tilt. Forgetting that the free surface is always perpendicular to the resultant of gravity and inertial forces. Final Answer:It rises at the back side and falls at the front side

Question 15

In a venturimeter, why is the divergent (diffuser) cone made longer than the convergent cone?

Accepted Answer

Introduction / Context:A venturimeter measures flow by accelerating fluid through a convergent to a throat and then decelerating it in a diffuser (divergent). The geometry of the diffuser is crucial for recovering pressure with minimal energy loss and avoiding separation. Given Data / Assumptions: Incompressible flow through a standard venturimeter. Entrance (convergent) half-angle is typically steeper than the exit (diffuser) half-angle. Goal: minimize head loss while maintaining accurate pressure differential. Concept / Approach:Rapid diffusion (sudden flow deceleration) can cause boundary-layer separation in the divergent section, producing large energy losses. A longer diffuser with a small half-angle (often about 5–7 degrees) decelerates the flow gradually, maintaining attached flow and enhancing pressure recovery. Step-by-Step Solution:Flow accelerates in the convergent: velocity ↑, static pressure ↓.At the throat: maximum velocity, minimum static pressure.In the diffuser: velocity ↓, static pressure ↑; gradual expansion reduces adverse pressure gradient and separation. Verification / Alternative check:Empirical data show lower permanent head loss for venturimeters with carefully designed long diffusers compared to short, steep diffusers or abrupt expansions. Why Other Options Are Wrong:Shorter diffuser: increases separation risk and loss.Equal lengths purely for symmetry: not a performance criterion.“No reason” or “reduce throat area oscillations only”: do not address the fluid dynamic purpose of pressure recovery. Common Pitfalls: Using too large a diffuser angle, leading to separation and high losses. Neglecting surface finish and Reynolds-number effects on diffuser performance. Final Answer:To make the diffuser longer to achieve gradual pressure recovery and avoid flow separation

Question 16

Pitot tube usage in fluid mechanics  In flowing liquids, a Pitot tube is primarily used to measure which quantity at a point in the stream?

Accepted Answer

Introduction / Context:A Pitot tube is a classic point-velocity measuring device used in fluid mechanics and hydraulic engineering. It converts kinetic energy at a point into a measurable pressure rise (stagnation or total pressure), which when compared with static pressure gives the local flow speed. Understanding exactly what the Pitot tube measures helps avoid common selection and interpretation errors in field and laboratory work. Given Data / Assumptions: Incompressible, steady flow at the measuring location. Axis of the Pitot tube aligned with the local flow direction (to avoid cosine errors). Viscous and probe-tip losses incorporated in an empirical coefficient where needed. Concept / Approach:The Pitot tube senses total (stagnation) pressure p_t at its tip and static pressure p_s via side ports or a separate static probe. Bernoulli's principle along a streamline relates velocity head to the difference p_t − p_s. The point velocity V is determined from the pressure difference using V = C * sqrt(2 * (p_t − p_s) / rho), where C is a calibration factor close to unity. Step-by-Step Solution: Measure stagnation pressure p_t at the Pitot opening.Measure static pressure p_s from a static tap (or Pitot-static tube).Compute dynamic pressure: q = p_t − p_s.Relate to velocity: V = C * sqrt(2 * q / rho). Verification / Alternative check:In water channels, compare Pitot-derived velocity with average velocity Q/A from volumetric measurements; expect agreement after accounting for profile non-uniformity and calibration. Why Other Options Are Wrong:Discharge requires integrating velocity over an area (not measured by a single Pitot reading). Depth is a geometric water-level measure. Static pressure alone is obtained by a piezometer/static tap, not a bare Pitot tip. Therefore, the Pitot tube’s primary purpose is measuring point velocity. Common Pitfalls:Misalignment with flow, using the formula without calibration, and applying point velocity directly as section-average velocity without profile correction. Final Answer:velocity

Question 17

Terminology of sharp-crested weirs  In open-channel hydraulics, the upper edge or surface of a weir over which water flows is known as:

Accepted Answer

Introduction / Context:Correct terminology for weir components is essential for interpreting design manuals and test data. The part of the weir that the water just passes over controls head–discharge calibration and is central to weir classification. Given Data / Assumptions: A sharp-crested weir (rectangular or V-notch) under free, fully aerated flow. Standard definitions used in hydraulic structures. Concept / Approach:The sharp upstream edge of a weir that water passes over is called the crest or sill. After passing over the crest, the sheet of water that springs clear is called the nappe. Confusing crest (a fixed structural edge) with nappe (the flowing sheet of water) is a common mistake. Step-by-Step Distinction: Identify the structural element: the weir’s top edge = crest/sill.Identify the flowing sheet of water = nappe downstream of the crest.Ancillary components: approach channel, abutments, and downstream apron. Verification / Alternative check:Textbook definitions consistently use crest (or sill) for the top edge and nappe for the water sheet; field drawings label the crest elevation as the control datum for head measurement. Why Other Options Are Wrong:“Nappe” is the water sheet, not the structural edge. “Vein” is an outdated synonym for jet/stream; not the structural part. “None of these” and “apron” (downstream floor) are incorrect here. Common Pitfalls:Measuring head too near the crest; proper head measurement point is usually placed upstream to avoid velocity-of-approach effects. Final Answer:sill (crest)

Question 18

Instrument used to measure velocity of flow at a point  Which device is classically used to determine the local flow speed in a pipe or open channel?

Accepted Answer

Introduction / Context:Hydraulic engineers employ various instruments to quantify flow. While Venturimeters and orifice meters infer discharge from pressure differences across constrictions, a Pitot tube directly senses local velocity through stagnation pressure. Given Data / Assumptions: Steady, incompressible flow at the measurement location. Probe aligned with flow; calibration factor near unity. Concept / Approach:The Pitot tube measures stagnation and static pressures, whose difference relates to the dynamic pressure. Using Bernoulli’s relation, the local velocity is obtained via V = C * sqrt(2 * (p_t − p_s) / rho). In contrast, Venturimeter and orifice meter provide section discharge Q; velocity is then Q/A and not strictly a point value. Step-by-Step Solution: Install Pitot or Pitot-static tube at the point of interest.Record stagnation and static pressures to get dynamic pressure.Compute local velocity from the dynamic pressure. Verification / Alternative check:Compare with a calibrated current meter (open channels) or with PIV/LDV in labs; results align within instrument uncertainty when properly used. Why Other Options Are Wrong:Venturimeter and orifice meter measure discharge over a section, not point velocity. “Current meter only” ignores that a Pitot tube is the canonical point-velocity device. “None of these” is incorrect. Common Pitfalls:Not correcting for velocity profile when converting point velocity to average velocity, and misalignment causing under-reading. Final Answer:Pitot tube

Question 19

Accelerating a water-filled rectangular tank horizontally  What constant horizontal acceleration is required so that the free surface of water in a rectangular tank makes a 45° slope to the horizontal?

Accepted Answer

Introduction / Context:When a tank of liquid is accelerated horizontally, the free surface tilts due to the superposition of gravitational and inertial accelerations. The angle of the surface is a basic result from hydrostatics in a non-inertial frame and is widely applied in tanker design and instrument calibration. Given Data / Assumptions: Rectangular tank, partially filled with water. Uniform horizontal acceleration a. No sloshing transients (quasi-static condition); steady tilt achieved. Concept / Approach:In a uniformly accelerating frame, the effective body force on the fluid is the vector sum of gravity (g vertically downward) and inertial acceleration (a directed opposite to motion). The free surface becomes perpendicular to the resultant. The tangent of the free-surface angle θ to the horizontal satisfies tan θ = a / g. Step-by-Step Solution: Given θ = 45°, so tan θ = 1.Use relation: tan θ = a / g.Therefore, a / g = 1 ⇒ a = g. Verification / Alternative check:For θ smaller than 45°, a < g; for θ larger than 45°, a > g. The 45° case is the midline where horizontal and vertical accelerations are equal in magnitude. Why Other Options Are Wrong:a = g/2 would give θ ≈ 26.6°, and a = 2 g, 2.5 g, 3 g give θ much steeper than 45°. Only a = g satisfies tan θ = 1. Common Pitfalls:Confusing degrees and radians; forgetting that the free surface is always perpendicular to the effective gravity vector in steady acceleration. Final Answer:g

Question 20

Dimensionless numbers in fluid mechanics  The ratio of inertia force to viscous force for a flowing fluid defines which standard dimensionless parameter?

Accepted Answer

Introduction / Context:Dimensionless numbers characterize competing physical effects in fluid flow. Recognizing which ratio corresponds to which number is foundational for similarity analysis, scaling, and regime identification (laminar vs turbulent). Given Data / Assumptions: Single-phase Newtonian fluid. Characteristic length L and velocity V defined for the flow. Concept / Approach:Reynolds number Re compares inertia to viscous effects: Re = (ρ V L) / μ. High Re indicates dominance of inertial effects and potential turbulence; low Re indicates viscous-dominated, laminar behavior. Other numbers represent different force ratios: Froude (inertia/gravity), Euler (pressure/inertia), Weber (inertia/surface tension), and Mach (inertia/compressibility via acoustic speed). Step-by-Step Association: Inertia force scale ~ ρ V^2 L^2 per area, viscous force scale ~ μ V L per area.Form their ratio to obtain Re ~ ρ V L / μ.Identify the corresponding name: Reynolds number. Verification / Alternative check:Common thresholds: in internal pipe flow, Re <~ 2100 (laminar), Re >~ 4000 (turbulent), with transition in between—confirming its role in regime prediction. Why Other Options Are Wrong:Froude involves gravity head; Euler relates pressure to inertia; Weber involves surface tension; Mach relates flow speed to speed of sound. Only Reynolds captures inertia/viscous competition. Common Pitfalls:Mixing Fanning and Darcy friction factors with Re definitions; forgetting to use consistent units for ρ, μ, V, L. Final Answer:Reynolds number

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