Q: Comparing viscosities – thick syrup versus water A thick liquid such as syrup exhibits a __________ viscosity than a light liquid such as water (under the same temperature conditions).

Introduction / Context: Viscosity governs the resistance to flow and shear in fluids. Everyday comparisons—pouring water versus syrup—provide intuitive evidence of differences in dynamic viscosity. Given Data / Assumptions: Both liquids are considered at the same temperature (viscosity is temperature-dependent). Newtonian behaviour is assumed for simplicity; many syrups are close to Newtonian over moderate shear rates. No dissolved gas effects are considered. Concept / Approach: Dynamic viscosity μ links shear stress to velocity gradient: τ = μ * (du/dy). A higher μ means a greater shear stress is required to achieve the same rate of deformation. Syrups and oils have much larger μ than water; thus, for the same driving head, they flow more slowly through the same constriction. Step-by-Step Solution: Step 1: Define viscosity comparison: if μ_syrup » μ_water, then at a given gradient, τ_syrup » τ_water.Step 2: In laminar pipe flow, volumetric flow rate Q ∝ 1/μ for fixed pressure drop; hence Q_syrup < Q_water under identical conditions.Step 3: Conclude that syrup has a greater viscosity than water. Verification / Alternative check: Data tables: at 20 °C, water μ ≈ 1.0×10^-3 Pa·s, while many syrups can exceed 1 Pa·s by orders of magnitude, confirming “greater” by factors of hundreds to thousands. Why Other Options Are Wrong: Lesser or identical: contradicts measurements. Time-oscillating: viscosity is not inherently oscillatory for these fluids. “Applies only to gases”: false—viscosity is a property of all fluids, including liquids. Common Pitfalls: Ignoring temperature effects; a warm syrup can have dramatically lower μ than a cold one, but still typically greater than water at the same temperature. Final Answer: greater

Q: Fluid Properties – Definition of Specific Gravity (Relative Density) The ratio of the specific weight of a given liquid to the specific weight of pure water at a standard temperature (commonly 4°C, where water has maximum density) is called the spec…

Introduction: Engineers frequently compare a liquid’s heaviness to that of water without carrying units. This comparison uses specific gravity, also called relative density, which simplifies quick estimates of buoyancy, manometer readings, and pump sizing. Given Data / Assumptions: Specific weight gamma is weight per unit volume. Reference fluid is pure water at a standard temperature (commonly 4°C). Gravitational acceleration is the same for test liquid and reference water. Concept / Approach: Specific gravity SG is defined as SG = gamma_liquid / gamma_water (at reference temperature). Because both numerators and denominators carry the same units, SG is dimensionless and independent of unit systems. For density-based definitions, SG = rho_liquid / rho_water, which is equivalent since gamma = rho * g. Step-by-Step Solution: Identify target term: ratio of specific weights → SG by definition.State formula: SG = gamma_liquid / gamma_water.Recognize that the same ratio holds for densities because g cancels.Conclude the correct term is specific gravity (relative density). Verification / Alternative check: Example: If an oil has gamma = 7.85 kN/m^3 and water has about 9.81 kN/m^3, then SG = 7.85 / 9.81 ≈ 0.80, a dimensionless number that immediately indicates buoyancy behavior (it will float on water). Why Other Options Are Wrong: Density of liquid: A property with units (kg/m^3), not a ratio to water. Compressibility: Describes volume change with pressure. Surface tension: An interfacial property (N/m). Relative viscosity: Ratio of viscosities, not specific weights. Common Pitfalls: Mixing up specific gravity with specific weight; using different temperature references for the liquid and water; reporting SG with units (it must be unitless). Final Answer: specific gravity of liquid

Q: Open-Channel Design – Meaning of “Most Economical” Cross-Section A channel section is said to be most economical when, for a given cross-sectional area and bed slope, it conveys the maximum discharge, which equivalently corresponds to a minimum wett…

Introduction: Hydraulic engineers seek channel geometries that carry the largest flow for a given area and slope, thereby reducing friction losses. Such sections are called “most economical” or “most efficient” and are central to irrigation, drainage, and flood conveyance design. Given Data / Assumptions: Uniform open-channel flow on a given bed slope and roughness. Fixed flow area (or design discharge and slope). Wetted perimeter influences frictional resistance via hydraulic radius. Concept / Approach: For Manning or Chezy relations, discharge Q is proportional to A * R^(2/3) (Manning) or A * sqrt(R) (Chezy) for a given slope, where R = A / P is hydraulic radius, A is area, and P is wetted perimeter. Maximizing Q at fixed A is therefore equivalent to maximizing R, which in turn means minimizing P. Practical excavation volumes also drop when the perimeter is minimized for the same area, translating to less lining and earthwork. Step-by-Step Solution: Relate discharge to hydraulic radius: Q ∝ A * R^(2/3).For fixed A and slope, maximize R by minimizing wetted perimeter P.Recognize that minimized P reduces construction/lining quantities (less excavation).Therefore, all listed statements are consistent with the “most economical” definition. Verification / Alternative check: Known optimal proportions (e.g., b = 2y for rectangular channels) arise from minimizing P for a given A, confirming the equivalence of the criteria. Why Other Options Are Wrong: Selecting any single statement ignores the equivalence among these conditions under standard friction laws; hence “all of the above” is most appropriate. Common Pitfalls: Confusing “most economical” with “cheapest” in absolute cost without hydraulic equivalence; forgetting roughness and side-slope constraints can modify the geometric optimum. Final Answer: all of the above

Question 1

Closed cylindrical vessel completely filled with liquid – scaling of total pressure on the top plate For a closed cylindrical tank completely filled with liquid at a uniform internal gauge pressure p, how does the total force on the circular top (ro…

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Introduction / Context: Design of pressure vessels and storage tanks requires estimating the net force on heads (end plates). For a cylindrical vessel completely filled with liquid and pressurized uniformly, the total force on the top plate depends on the plate area and the internal pressure. This question checks your scaling intuition with radius. Given Data / Assumptions: Uniform internal gauge pressure p acts on the inside surfaces. The top plate is a flat circular area of radius r. Hydrostatic gradients are either negligible (small height) or already included in p; the key point is the uniform pressure assumption. Concept / Approach: Total force on a surface under uniform pressure equals pressure multiplied by area. For a circle, area A = π r^2. Therefore, the resultant force scales with r^2. This scaling result holds irrespective of the absolute value of p as long as p does not itself depend on r. Step-by-Step Solution: Step 1: Identify area A = π r^2.Step 2: Write force F = p * A.Step 3: Substitute A → F = p * π r^2.Step 4: Conclude F ∝ r^2 for constant p. Verification / Alternative check: Dimensional analysis shows p has units N/m^2 and area has m^2; their product gives newtons (N), as expected for force. Doubling r quadruples the area and therefore quadruples the total force at the same p. Why Other Options Are Wrong: Inverse powers of r (options b and d): contradict F = p * π r^2. r^4 dependence: would arise in bending/deflection formulas, not in pressure resultant. Independent of r: incorrect because area depends on r. Common Pitfalls: Mixing up pressure (intensive) and force (extensive). Pressure stays the same if r changes, but the total force on a larger area increases with r^2. Final Answer: directly proportional to r^2

Question 2

Dimensionless groups – defining Euler’s number (Eu) The ratio of the inertia force to the __________ force is termed Euler’s number (or its reciprocal, depending on convention). Choose the quantity that completes the definition used in many textbook…

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Introduction / Context: Dimensionless numbers classify flow regimes by comparing dominant forces. Euler’s number (Eu) is associated with pressure effects relative to inertia, widely used in turbomachinery and internal flow analysis. Given Data / Assumptions: Characteristic velocity V, length L, density rho. Pressure difference Δp represents the driving or resisting pressure forces. We focus on the common MCQ convention used in many introductory texts. Concept / Approach: In classical definition, Eu = Δp / (0.5 * rho * V^2), which is pressure force divided by inertia force. Some authors present the reciprocal form (inertia/pressure) but keep the name “Euler number.” Within the pattern of force-ratio definitions (Reynolds: inertia/viscous; Froude: inertia/gravity; Weber: inertia/surface tension), matching “inertia to pressure” identifies Euler’s group. Hence, completing the blank with “pressure” is consistent. Step-by-Step Solution: Step 1: Recognize the target forces: inertia force scale ≈ rho * V^2 * L^2 over area.Step 2: Pressure force scale ≈ Δp * L^2.Step 3: Form the ratio inertia/pressure or its reciprocal. The blank should be “pressure.”Step 4: Note that many texts use Eu as pressure/inertia; the MCQ phrasing mirrors the recurring “inertia to X” pattern. Verification / Alternative check: Check consistency: if pressure dominates, Eu (pressure/inertia) is large; the reciprocal (inertia/pressure) is small. Both communicate the same physical balance. Why Other Options Are Wrong: Elastic: relates to compressibility and wave speed, not Eu’s standard definition. Surface tension: associated with Weber number. Viscous: associated with Reynolds number. Gravitational: associated with Froude number. Common Pitfalls: Memorizing only one convention and getting confused when the inverse appears; always check which way a source defines Eu. Final Answer: pressure

Question 3

Floating condition – comparing buoyancy and displaced liquid weight A body will float in a fluid if the buoyant force is __________ the weight of the liquid displaced (recall Archimedes’ principle equates these two):

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Introduction / Context: Archimedes’ principle is the cornerstone of hydrostatics and naval architecture. It relates the buoyant force acting on an immersed body to the weight of the fluid displaced. This question reinforces that equality and clarifies the separate floatation condition involving the body’s own weight. Given Data / Assumptions: Static fluid environment; negligible surface waves and currents. Body may be fully or partially submerged. Fluid density and gravity are uniform. Concept / Approach: By definition, the buoyant force F_b equals the weight of displaced fluid: F_b = weight_displaced = gamma * V_displaced, where gamma is specific weight and V_displaced is the displaced volume. Floatation occurs when F_b balances the body’s weight W_body. At equilibrium float, F_b = W_body = weight_displaced; the displaced volume adjusts to satisfy this equality. Step-by-Step Solution: Step 1: State Archimedes’ principle: F_b = weight of displaced fluid.Step 2: For a floating body, vertical equilibrium requires F_b = W_body.Step 3: Therefore, at float, weight of displaced liquid = W_body, and hence F_b = weight of displaced liquid (equality).Step 4: If W_body increases, the body sinks deeper to displace more fluid until equality is restored. Verification / Alternative check: A steel ship floats because its overall average density (including contained air) is below the water’s density. The hull submergence increases until the weight of displaced water equals the ship’s weight. Why Other Options Are Wrong: Less than / more than: contradict the definitional equality between F_b and displaced fluid weight. Indeterminate: the relation is exact, not uncertain. “Opposite in direction to”: while buoyancy acts upward and weight acts downward, the question compares magnitudes to displaced liquid weight, which equals F_b. Common Pitfalls: Confusing “weight of displaced liquid” with “weight of the body.” The floating condition is F_b = W_body, but the magnitude of F_b is always equal to the weight of displaced fluid. Final Answer: equal to

Question 4

Hydrostatics – line of action of resultant on a vertical wall (single liquid) A vertical wall retains a single liquid to depth H. The resultant hydrostatic force acts at a distance __________ below the free surface (measured along the wall):

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Introduction / Context: Knowing where the resultant hydrostatic force acts is essential for designing retaining walls, gates, and tanks. For a vertical plane surface in a liquid, the pressure increases linearly with depth, producing a triangular distribution. Given Data / Assumptions: Vertical wall of height H exposed to a single, homogeneous liquid. Free surface is open to atmosphere. Wall width is constant; gravity is uniform. Concept / Approach: For a vertical plane, gauge pressure p_g = w * h varies linearly from zero at the surface to w * H at the bottom, forming a triangular distribution. The line of action (center of pressure) of a triangle load lies at one-third the height from the base, i.e., two-thirds from the apex (free surface). Step-by-Step Solution: Step 1: Recognize triangular pressure diagram with zero at h = 0 and maximum at h = H.Step 2: For a triangle, centroid location from the base is H/3.Step 3: Thus, from the free surface, the distance is H − H/3 = 2H/3.Step 4: Therefore, the resultant hydrostatic force acts at 2H/3 below the surface. Verification / Alternative check: Formal integration for the center of pressure on a rectangle gives the same result for a vertical wall: y_cp = I_G / (A * ȳ) + ȳ, which simplifies to 2H/3 for the hydrostatic triangle. Why Other Options Are Wrong: H/3: this is the location above the bottom, not from the surface. H/2, 3H/4, H/4: do not correspond to the centroid of a triangle with zero at top. Common Pitfalls: Confusing the center of pressure with the centroid of the plane area; due to pressure variation, the resultant is below the geometric centroid. Final Answer: 2H/3

Question 5

Fluid properties – relation between kinematic and dynamic viscosity Is the following statement correct? “Kinematic viscosity equals the product of dynamic viscosity and the fluid density.”

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Introduction / Context: Fluid mechanics uses both dynamic viscosity (μ) and kinematic viscosity (ν). Mixing up their relationship leads to wrong Reynolds numbers and pressure-drop calculations. Given Data / Assumptions: Dynamic viscosity μ measures resistance to shear (units Pa·s). Kinematic viscosity ν measures momentum diffusivity (units m^2/s). Mass density ρ has units kg/m^3. Concept / Approach: The correct relationship is ν = μ / ρ, not μ * ρ. Dividing by density converts force-based resistance into a diffusivity that governs how momentum spreads through a fluid, analogous to thermal diffusivity in heat transfer. Step-by-Step Solution: Step 1: Write ν = μ / ρ.Step 2: Check units: (Pa·s) / (kg/m^3) = (N·s/m^2) / (kg/m^3) = (kg·m/s^2 · s / m^2) / (kg/m^3) = m^2/s.Step 3: If you multiplied μ * ρ, the units would be Pa·s·kg/m^3, which is dimensionally incorrect for ν.Step 4: Conclude the statement is false. Verification / Alternative check: Common conversions: water at 20 °C has μ ≈ 1.0×10^-3 Pa·s and ρ ≈ 998 kg/m^3, giving ν ≈ 1.0×10^-6 m^2/s, consistent with data tables. Why Other Options Are Wrong: Any “true” variant contradicts ν = μ / ρ. Temperature clauses do not fix the incorrect form; the functional relation is independent of temperature (though values of μ and ρ change with T). Common Pitfalls: Confusing ν with dynamic viscosity units (Pa·s) or with the C.G.S. unit “stoke,” where 1 cSt = 10^-6 m^2/s. Final Answer: No

Question 6

Weirs – terminology for the overflow edge In open-channel flow measurement, the top edge of a weir over which the water flows is called the:

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Introduction / Context: Accurate terminology is important when citing discharge formulas and calibration data for hydraulic structures. Weirs are sharp-edged or broad-crested barriers used to measure or control flow in open channels. Given Data / Assumptions: Free overfall occurs; air can circulate beneath the nappe for sharp crests. Standard weir configurations: rectangular, triangular (V-notch), trapezoidal (Cipolletti), and broad-crested. No special geometry beyond naming is required. Concept / Approach: The overflow edge is formally called the crest (also termed the sill in some civil-hydraulics texts). Water that passes over forms a sheet known as the nappe. Misnaming crest as nappe or as an orifice leads to misuse of equations meant for free-surface versus pressurized openings. Step-by-Step Solution: Step 1: Identify the physical feature: the upper edge of the barrier.Step 2: Recall standard term: crest (sill).Step 3: Distinguish it from the nappe (the flowing sheet) and from an orifice (a closed, pressurized opening).Step 4: Choose “sill or crest.” Verification / Alternative check: Design handbooks label the elevation of the overflow edge as the crest level; freeboard and head are referenced to this crest elevation. Why Other Options Are Wrong: Nappe or vein: refers to the sheet of water, not the edge. Orifice: applies to closed-conduit openings below a water surface. Control sill only in dams: semantics aside, “crest” is the standard term for the weir top edge. Toe: denotes the downstream foot of a structure, not the overflow edge. Common Pitfalls: Using orifice discharge equations for a sharp-crested weir because of terminology confusion. Final Answer: sill or crest

Question 7

Notches for flow measurement – common construction material In practice, is it correct that a thin-plate measuring notch (for example, a V-notch) is usually fabricated from metal?

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Introduction / Context: Notches (thin-plate weirs) are widely used for measuring small to moderate discharges. Material choice affects edge sharpness, durability, and calibration stability. Given Data / Assumptions: Thin-plate construction is desired to obtain a sharp, well-defined crest. Edges must remain straight and burr-free over time. Field and laboratory applications are both considered. Concept / Approach: Metals such as stainless steel or brass provide dimensional stability, corrosion resistance, and the ability to machine a sharp crest that maintains calibration. While temporary or low-accuracy installations may use wood or plastics, standard practice for accurate measurement is a metallic thin plate. Step-by-Step Solution: Step 1: Identify requirements: sharp edge, minimal wear, smooth surface.Step 2: Compare materials: metals outperform wood in edge retention and moisture resistance.Step 3: Conclude that a metallic plate is usually chosen for notches intended for measurement. Verification / Alternative check: Calibration guidelines from labs and standards organizations typically specify thin, smooth metal plates for V-notch and rectangular notches to ensure reproducible coefficients. Why Other Options Are Wrong: “Incorrect” or wood-as-standard: wood swells and blunts, degrading accuracy. Only for high heads or only for labs: metals are common across head ranges and in field installations. Common Pitfalls: Allowing burrs or corrosion at the crest; even with metal, maintenance is essential for accuracy. Final Answer: Correct

Question 8

Venturimeter principle – pressure comparison between inlet and throat For an incompressible flow through a properly installed venturimeter, the static pressure at the throat is __________ than the static pressure at the inlet (upstream section).

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Introduction / Context: Venturimeters infer flow rate from the pressure drop produced as fluid accelerates from a larger upstream area to a smaller throat area. Recognizing the qualitative pressure change across these sections is foundational. Given Data / Assumptions: Steady incompressible flow (e.g., water) in a horizontal meter for simplicity. Negligible energy losses between the pressure taps compared to the main acceleration effect. No cavitation at the throat (pressure remains above vapor pressure). Concept / Approach: Bernoulli’s equation states that along a streamline, p/γ + V^2/(2g) + z is approximately constant (minus losses). At the throat, area is smaller, velocity is higher, so velocity head increases. To conserve energy, static pressure head must decrease, hence p_throat < p_inlet. Step-by-Step Solution: Step 1: Note A_inlet > A_throat → V_throat > V_inlet.Step 2: Apply Bernoulli: increase in V^2/(2g) must be balanced by a drop in p/γ (for equal elevations and small losses).Step 3: Conclude p_throat < p_inlet, producing a measurable differential head.Step 4: Use the differential head with calibration to compute discharge. Verification / Alternative check: Manometer connections on a venturimeter always show a lower head at the throat limb relative to the upstream limb at any nonzero flow. Why Other Options Are Wrong: Higher or same: violate Bernoulli for accelerating flow. Zero at design discharge: pressure is not zero; cavitation would be a concern long before that. Roughness: affects loss but not the fundamental trend that throat pressure is lower. Common Pitfalls: Neglecting elevation differences in inclined installations; always include z terms when necessary. Final Answer: lower

Question 9

Comparing viscosities – thick syrup versus water A thick liquid such as syrup exhibits a __________ viscosity than a light liquid such as water (under the same temperature conditions).

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Introduction / Context: Viscosity governs the resistance to flow and shear in fluids. Everyday comparisons—pouring water versus syrup—provide intuitive evidence of differences in dynamic viscosity. Given Data / Assumptions: Both liquids are considered at the same temperature (viscosity is temperature-dependent). Newtonian behaviour is assumed for simplicity; many syrups are close to Newtonian over moderate shear rates. No dissolved gas effects are considered. Concept / Approach: Dynamic viscosity μ links shear stress to velocity gradient: τ = μ * (du/dy). A higher μ means a greater shear stress is required to achieve the same rate of deformation. Syrups and oils have much larger μ than water; thus, for the same driving head, they flow more slowly through the same constriction. Step-by-Step Solution: Step 1: Define viscosity comparison: if μ_syrup » μ_water, then at a given gradient, τ_syrup » τ_water.Step 2: In laminar pipe flow, volumetric flow rate Q ∝ 1/μ for fixed pressure drop; hence Q_syrup < Q_water under identical conditions.Step 3: Conclude that syrup has a greater viscosity than water. Verification / Alternative check: Data tables: at 20 °C, water μ ≈ 1.0×10^-3 Pa·s, while many syrups can exceed 1 Pa·s by orders of magnitude, confirming “greater” by factors of hundreds to thousands. Why Other Options Are Wrong: Lesser or identical: contradicts measurements. Time-oscillating: viscosity is not inherently oscillatory for these fluids. “Applies only to gases”: false—viscosity is a property of all fluids, including liquids. Common Pitfalls: Ignoring temperature effects; a warm syrup can have dramatically lower μ than a cold one, but still typically greater than water at the same temperature. Final Answer: greater

Question 10

Fluid Properties – Definition of Specific Gravity (Relative Density) The ratio of the specific weight of a given liquid to the specific weight of pure water at a standard temperature (commonly 4°C, where water has maximum density) is called the spec…

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Introduction: Engineers frequently compare a liquid’s heaviness to that of water without carrying units. This comparison uses specific gravity, also called relative density, which simplifies quick estimates of buoyancy, manometer readings, and pump sizing. Given Data / Assumptions: Specific weight gamma is weight per unit volume. Reference fluid is pure water at a standard temperature (commonly 4°C). Gravitational acceleration is the same for test liquid and reference water. Concept / Approach: Specific gravity SG is defined as SG = gamma_liquid / gamma_water (at reference temperature). Because both numerators and denominators carry the same units, SG is dimensionless and independent of unit systems. For density-based definitions, SG = rho_liquid / rho_water, which is equivalent since gamma = rho * g. Step-by-Step Solution: Identify target term: ratio of specific weights → SG by definition.State formula: SG = gamma_liquid / gamma_water.Recognize that the same ratio holds for densities because g cancels.Conclude the correct term is specific gravity (relative density). Verification / Alternative check: Example: If an oil has gamma = 7.85 kN/m^3 and water has about 9.81 kN/m^3, then SG = 7.85 / 9.81 ≈ 0.80, a dimensionless number that immediately indicates buoyancy behavior (it will float on water). Why Other Options Are Wrong: Density of liquid: A property with units (kg/m^3), not a ratio to water. Compressibility: Describes volume change with pressure. Surface tension: An interfacial property (N/m). Relative viscosity: Ratio of viscosities, not specific weights. Common Pitfalls: Mixing up specific gravity with specific weight; using different temperature references for the liquid and water; reporting SG with units (it must be unitless). Final Answer: specific gravity of liquid

Question 11

Open-Channel Design – Meaning of “Most Economical” Cross-Section A channel section is said to be most economical when, for a given cross-sectional area and bed slope, it conveys the maximum discharge, which equivalently corresponds to a minimum wett…

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Introduction: Hydraulic engineers seek channel geometries that carry the largest flow for a given area and slope, thereby reducing friction losses. Such sections are called “most economical” or “most efficient” and are central to irrigation, drainage, and flood conveyance design. Given Data / Assumptions: Uniform open-channel flow on a given bed slope and roughness. Fixed flow area (or design discharge and slope). Wetted perimeter influences frictional resistance via hydraulic radius. Concept / Approach: For Manning or Chezy relations, discharge Q is proportional to A * R^(2/3) (Manning) or A * sqrt(R) (Chezy) for a given slope, where R = A / P is hydraulic radius, A is area, and P is wetted perimeter. Maximizing Q at fixed A is therefore equivalent to maximizing R, which in turn means minimizing P. Practical excavation volumes also drop when the perimeter is minimized for the same area, translating to less lining and earthwork. Step-by-Step Solution: Relate discharge to hydraulic radius: Q ∝ A * R^(2/3).For fixed A and slope, maximize R by minimizing wetted perimeter P.Recognize that minimized P reduces construction/lining quantities (less excavation).Therefore, all listed statements are consistent with the “most economical” definition. Verification / Alternative check: Known optimal proportions (e.g., b = 2y for rectangular channels) arise from minimizing P for a given A, confirming the equivalence of the criteria. Why Other Options Are Wrong: Selecting any single statement ignores the equivalence among these conditions under standard friction laws; hence “all of the above” is most appropriate. Common Pitfalls: Confusing “most economical” with “cheapest” in absolute cost without hydraulic equivalence; forgetting roughness and side-slope constraints can modify the geometric optimum. Final Answer: all of the above

Question 12

Hydraulic Transients – Purpose of a Surge Tank in Pipe Systems In high-head water conveyance (penstocks), a surge tank is provided to control pressure variations due to rapid flow changes, mitigate water-hammer effects, and help regulate the flow to…

Accepted Answer

Introduction: Surge tanks are tall standpipes or reservoirs connected to long pipelines between a reservoir and a turbine. They buffer rapid changes in discharge and protect the system from damaging pressure surges known as water hammer, while also improving turbine control. Given Data / Assumptions: Long conduit conveying water to hydraulic turbines. Rapid load changes at the turbine can cause sudden velocity changes. Tank is properly located near the powerhouse or along the penstock. Concept / Approach: When turbine demand drops suddenly, decelerating water can cause large positive pressure spikes. A surge tank provides a free surface where water level can rise, absorbing inertia and reducing the transient pressure. Conversely, when demand increases, water level falls in the tank to supply additional flow, reducing negative pressure spikes and aiding regulation. Step-by-Step Solution: Identify transient cause: rapid change in flow velocity.Provide local storage via surge tank to absorb or release water.Result: mitigate water hammer and stabilize turbine inlet pressure and discharge.Therefore all listed purposes are served by a surge tank. Verification / Alternative check: Water-hammer analysis using elastic wave speed and continuity shows reduced peak pressures when a free surface reservoir is introduced near the turbine. Why Other Options Are Wrong: Any single choice understates function; the device is specifically installed to address all listed roles in concert. Common Pitfalls: Assuming a surge tank eliminates water hammer completely regardless of valve speed; in reality, it mitigates but does not remove all transients if operations are extreme. Final Answer: all of the above

Question 13

Flow Kinematics – Can Two-Dimensional Flow Curves Bend? In two-dimensional flow (velocity components in two orthogonal directions, with no variation in the third), the particle paths and streamlines can be curved; two-dimensional does not imply stra…

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Introduction: “Two-dimensional” describes how velocity varies in space, not the geometric shape of streamlines. Many classic 2D flows (flow around a cylinder, potential vortices) have strongly curved paths while remaining two-dimensional in description. Given Data / Assumptions: Velocity depends on two coordinates (e.g., u(x,y), v(x,y)), and is uniform in the third. No restriction on streamline curvature is implied by dimensionality. Fluid may be viscous or inviscid; rotational or irrotational. Concept / Approach: Streamlines are tangent to the velocity vector field at every point. If the velocity direction changes with position in a plane, streamlines bend accordingly, yet the flow remains two-dimensional so long as there is no dependence on the third coordinate and the velocity component in that direction is zero. Step-by-Step Solution: Clarify definition: two-dimensional means w = 0 and ∂/∂z = 0 (for example) but u and v may vary with x and y.Recognize that u and v changing direction across the field produce curved streamlines.Therefore, the claim that 2D flow “does not take place in a curve” is false. Verification / Alternative check: Potential flow past a cylinder is exactly two-dimensional and exhibits highly curved streamlines; laboratory dye studies confirm this. Why Other Options Are Wrong: Conditioning truth on irrotationality or uniformity is irrelevant; curvature occurs in both rotational/irrotational and uniform/non-uniform cases. Viscosity does not determine 2D streamline curvature. Common Pitfalls: Equating two-dimensional with straight channels; confusing spatial dimensionality with path geometry. Final Answer: False

Question 14

Most Efficient Depth for Circular Channel – Condition for Maximum Discharge The discharge through an open channel of circular section is maximum when the flow depth is approximately 0.81 times the diameter of the circular channel (y ≈ 0.81 D).

Accepted Answer

Introduction: For partially full circular conduits used as open channels, one can find a depth that maximizes discharge for a given slope and roughness. This “most efficient” depth is a standard design datum in sewer and stormwater engineering. Given Data / Assumptions: Steady, uniform open-channel flow in a circular section. Friction laws such as Manning or Chezy apply. Diameter D is fixed; depth y varies with the central angle of wetted segment. Concept / Approach: Discharge Q is proportional to A * R^(2/3) for a given slope and roughness (Manning’s equation), where area A and wetted perimeter P depend on the wetted central angle. Maximizing Q with respect to depth yields the condition y/D ≈ 0.81, a widely tabulated result. Step-by-Step Solution: Express A(θ) and P(θ) for a circular segment; compute R(θ) = A/P.Form Q(θ) ∝ A(θ) * R(θ)^(2/3) and differentiate with respect to θ.Set derivative to zero to find optimal θ and convert to y/D.Obtain y/D ≈ 0.81 as the maximizing ratio. Verification / Alternative check: Design charts for circular conduits show peak Q near y/D ≈ 0.8–0.82 for typical friction laws, validating the rule of thumb 0.81. Why Other Options Are Wrong: 0.34 and 0.50 are too shallow and reduce area. 0.67 under-utilizes the section. 0.95 is near full, where increased perimeter outweighs area gain. Common Pitfalls: Confusing the depth for maximum velocity with that for maximum discharge; neglecting that the optimum is insensitive to the exact friction law but not identical to the full-flow condition. Final Answer: 0.81 times

Question 15

Triangular (V-Notch) Weir – Sensitivity of Discharge to Head Measurement For a triangular notch with fixed angle and coefficient, the relationship between fractional discharge error and fractional head error is (dQ/Q) = (5/2) * (dH/H).

Accepted Answer

Introduction: Discharge over a triangular (V-notch) weir is very sensitive to the head reading. Quantifying this sensitivity helps set instrumentation accuracy and staff-gauge resolution for field measurements. Given Data / Assumptions: Standard sharp-crested V-notch with fixed notch angle and constant Cd. Free nappe with proper aeration; negligible approach velocity correction. Small measurement errors relative to the head value (differential approximation). Concept / Approach: For a V-notch, Q ∝ H^(5/2). Taking logarithms and differentiating gives dQ/Q = (5/2) * (dH/H) when Cd and geometric factors are treated as constants. Thus, a 1% error in head yields about a 2.5% error in discharge. Step-by-Step Solution: Start with Q = K * H^(5/2), where K groups constants.Take differentials: dQ = (5/2) * K * H^(3/2) * dH.Divide by Q: dQ/Q = (5/2) * dH/H.Therefore the error ratio equals 2.5. Verification / Alternative check: Repeating the same steps for rectangular weirs (Q ∝ H^(3/2)) yields dQ/Q = (3/2) * dH/H, illustrating the stronger sensitivity of triangular notches. Why Other Options Are Wrong: 3/2 corresponds to rectangular weirs. 1/2 and 2 underestimate the exponent effect. 5/3 is not derived from the H^(n) dependence of triangular notches. Common Pitfalls: Ignoring uncertainty in Cd or notch angle; applying this relation when approach velocity correction is significant, which slightly alters K but not the exponent. Final Answer: (dQ/Q) = (5/2) * (dH/H)

Question 16

Internal (Borda) Mouthpiece – Meaning of “Running Free” For an internal mouthpiece, if the jet contracts inside and does not touch the mouthpiece wall (vena contracta persists), the mouthpiece is said to be running free; if the jet expands to fill t…

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Introduction: Mouthpieces are short tubes fitted to orifices. Their discharge behavior depends on whether the jet forms a vena contracta that remains detached (running free) or expands and attaches to the tube walls (running full). The distinction affects the coefficient of discharge and head–flow relation. Given Data / Assumptions: Internal (Borda) mouthpiece mounted flush with tank wall. Sharp entrance, short length compared with diameter. Steady head over the mouthpiece. Concept / Approach: With a sharp-edged entrance, the jet contracts after entry (vena contracta). If the jet stays contracted and does not touch the tube, the mouthpiece is said to run free and typically exhibits a lower discharge coefficient. At higher heads or with certain geometries, the jet may expand and fill the tube, i.e., run full, yielding a larger discharge coefficient. Step-by-Step Solution: Define “running free”: jet detached from wall inside the mouthpiece.Define “running full”: jet expands and wets the mouthpiece wall.Compare with the statement: it reverses these definitions.Therefore the statement is false. Verification / Alternative check: Textbook data show Cd_free roughly half that of the theoretical jet, while Cd_full is substantially larger due to pressure recovery and reduced contraction losses. Why Other Options Are Wrong: Truth qualifiers (only at low head, with air, etc.) do not change the fundamental definitions of “free” and “full”. Common Pitfalls: Confusing internal with external mouthpieces; assuming any filling implies cavitation rather than simply running full. Final Answer: False

Question 17

Quick Calculation – Specific Gravity from Specific Weight If an oil has specific weight 7.85 kN/m^3, its specific gravity (relative to water at about 9.81 kN/m^3) is SG = 7.85 / 9.81 ≈ 0.8.

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Introduction: Specific gravity provides a fast, unitless comparison of a fluid’s heaviness relative to water. Converting from specific weight to SG requires only a simple ratio, making it common in quick field estimates. Given Data / Assumptions: Oil specific weight gamma_oil = 7.85 kN/m^3. Water specific weight gamma_water ≈ 9.81 kN/m^3 at standard reference. Same gravitational acceleration for both fluids. Concept / Approach: By definition, SG = gamma_oil / gamma_water. Because both have units of kN/m^3, SG is dimensionless. Values less than 1 imply the liquid is lighter than water and typically floats; values greater than 1 imply it is heavier and tends to sink. Step-by-Step Solution: Write SG = gamma_oil / gamma_water.Substitute numbers: SG = 7.85 / 9.81.Compute: SG ≈ 0.799 → approximately 0.8.State the rounded result to two significant digits. Verification / Alternative check: Using densities gives the same result since SG = rho_oil / rho_water and gamma = rho * g cancels g. Why Other Options Are Wrong: 1 corresponds to water; 1.2 and 1.6 indicate a heavier-than-water liquid; 0.98 is too close to water and inconsistent with the given 7.85 kN/m^3 value. Common Pitfalls: Mixing up specific weight with density; forgetting to use water at the appropriate reference value; reporting SG with units. Final Answer: 0.8

Question 18

Basic Definition Check – Coefficient of Discharge for an Orifice The coefficient of discharge Cd is defined as the ratio of the actual discharge to the theoretical discharge through an orifice under the same head.

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Introduction: Flow coefficients quantify how real flows deviate from ideal predictions. The coefficient of discharge is particularly important for calibrating orifices, nozzles, and weirs to obtain accurate flow rates from head measurements. Given Data / Assumptions: Theoretical discharge Q_th is computed from ideal energy relations (no losses, perfect contraction). Actual discharge Q_act is measured experimentally. Same head across the orifice for both definitions. Concept / Approach: By standard definition, Cd = Q_act / Q_th. Since real flows experience contraction and viscous losses, Q_act < Q_th and Cd is typically less than 1. The statement in the stem reverses the ratio, so it is incorrect. Step-by-Step Solution: Recall definition: Cd = Q_act / Q_th.Recognize that Q_th is derived from ideal velocity V_th = sqrt(2 * g * H) and area, while Q_act is measured.Compare to the statement's wording (theoretical/actual) and identify reversal.Therefore, the correct truth value is False. Verification / Alternative check: Typical sharp-edged orifices show Cd ≈ 0.60 to 0.65, clearly less than 1, confirming the actual is a fraction of the theoretical, not vice versa. Why Other Options Are Wrong: Adding conditions such as “large heads” or “negligible viscosity” does not change the definition itself; Cd remains Q_act / Q_th. Common Pitfalls: Confusing Cd with coefficient of velocity Cv (V_act / V_th) and coefficient of contraction Cc (A_jet / A_orifice); note that Cd = Cc * Cv. Final Answer: False

Question 19

Internal Mouthpiece – Relative Discharge When Running Full vs Running Free For an internal mouthpiece, when the mouthpiece runs full (jet expands to fill it), the discharge is approximately twice the discharge obtained when it is running free (detac…

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Introduction: Short-tube orifices (mouthpieces) can operate in two regimes: running free and running full. The discharge coefficient changes markedly between these regimes, altering the flow rate for the same head. Given Data / Assumptions: Internal (Borda) mouthpiece on a large reservoir. Same head over the mouthpiece for both operating conditions. Negligible cavitation and proper aeration conditions. Concept / Approach: When running free, the jet forms a vena contracta inside the short tube and remains detached, yielding a relatively low discharge coefficient. When the jet expands to touch and fill the tube (running full), losses associated with contraction are reduced and the effective coefficient increases significantly, giving a higher discharge for the same head. Classical data show the full-running discharge is about double that of the free-running condition. Step-by-Step Solution: Define regimes: free (detached jet) vs full (attached jet).For constant head H, Q_free = Cd_free * A * sqrt(2 * g * H).Similarly, Q_full = Cd_full * A * sqrt(2 * g * H).With Cd_full ≈ 2 * Cd_free in classical experiments, Q_full ≈ 2 * Q_free. Verification / Alternative check: Standard orifice/mouthpiece tables report Cd_free near one-half of the full-running value for internal mouthpieces, supporting the “approximately twice” rule of thumb. Why Other Options Are Wrong: “Incorrect” contradicts experimental observations; additional qualifiers (small heads, submerged outlet) are not required for the fundamental comparison, though extreme conditions may slightly alter coefficients. Common Pitfalls: Confusing internal and external mouthpieces; overlooking air entrainment and submergence effects that can modify Cd but not the qualitative trend. Final Answer: Correct

Question 20

Pitot Tube — What Quantity Does It Measure? In fluid mechanics practice, a Pitot tube inserted into a flow is primarily used to determine which hydraulic quantity at a specific point in a pipe or channel?

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Introduction: A Pitot tube is a classic point-velocity probe that converts local kinetic energy into a measurable pressure rise by stagnating the flow at its tip. It is widely used in hydraulic laboratories, HVAC ducts, rivers, and aircraft airspeed systems to estimate local velocity from measured pressures. Given Data / Assumptions: Incompressible liquid flow (water) or low-speed gas flow. Steady conditions near the measurement point. Access to both stagnation (impact) pressure and static pressure, often via a Pitot-static arrangement and a manometer or pressure transducer. Concept / Approach: Bernoulli's relation along a streamline between a point in the flow and the stagnation point gives V from the difference between stagnation and static pressures. The key formula is V = sqrt(2 * Delta p / rho) after applying an empirical coefficient if needed. Step-by-Step Solution: Measure stagnation pressure p_0 at the Pitot opening facing the flow.Measure static pressure p_s (via side ports or a separate tap).Compute Delta p = p_0 - p_s.Find local speed: V = C * sqrt(2 * Delta p / rho), where C is a calibration coefficient near 1.0. Verification / Alternative check: Traverse the Pitot across the cross-section to build a velocity profile; integrating yields discharge, confirming that the device itself measures point velocity, not total flow directly. Why Other Options Are Wrong: Pressure difference between two arbitrary points requires a differential tap, not a Pitot alone. Total pressure in a pipe is a derived term, but the Pitot measures the stagnation pressure at one point. Discharge needs area integration or another primary device (weir, Venturi meter). Common Pitfalls: Misaligning the Pitot with flow direction; ignoring turbulence-induced fluctuations; using the formula without fluid density or calibration factor. Final Answer: velocity of flow at the required point in a pipe

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