Q: Rectangular weir – Bazin’s formula constant m According to Bazin’s formula, the discharge over a rectangular sharp-crested weir can be written in the form Q = m * L * sqrt(2 * g) * H^(3/2). In this expression, what is m equal to (in terms of the dis…

Introduction: Empirical weir formulas express discharge as a constant times L * sqrt(2 * g) * H^(3/2), where L is the effective crest length and H is the head over the crest. Bazin’s form isolates a constant m that depends on hydraulic coefficients and edge geometry. This question asks for m in terms of the discharge coefficient Cd for a rectangular sharp-crested weir under free flow. Given Data / Assumptions: Sharp-crested rectangular weir under free (not submerged) conditions. Velocity of approach neglected in the basic form. Standard definition H measured above crest. Cd encapsulates contraction and velocity effects for the crest. Concept / Approach: Starting from the elemental strip integration for weir flow and incorporating the discharge coefficient Cd yields Q = (8/15) * Cd * L * sqrt(2 * g) * H^(3/2). Comparing with Q = m * L * sqrt(2 * g) * H^(3/2) shows m = (8/15) * Cd. Many handbooks later package constants numerically (e.g., Francis’ 3.33 in US units), but in SI symbols Bazin’s decomposition cleanly separates the coefficient multiplier (8/15) and Cd. Step-by-Step Solution: Step 1: Recall the integrated sharp-crest expression: Q = (8/15) * Cd * L * sqrt(2 * g) * H^(3/2).Step 2: Compare with the generic Bazin form Q = m * L * sqrt(2 * g) * H^(3/2).Step 3: Identify m = (8/15) * Cd. Verification / Alternative check: Dimensional factors L * sqrt(2 * g) * H^(3/2) carry the correct units of discharge; coefficients such as (8/15) arise from integrating velocity over depth assuming an ideal nappe shape and then adjusting by Cd. Why Other Options Are Wrong: (3/5) * Cd or 1.84 * Cd: Not the standard coefficient from the strip integration.(2/3) / Cd and Cd/(8/15): Invert Cd incorrectly and give wrong scaling. Common Pitfalls: Confusing the Bazin form with Francis’ empirical constants in specific unit systems; always compare symbolic forms to identify m. Final Answer: m = (8 / 15) * Cd

Q: Triangular (V-notch) weir – Right-angle notch with Cd = 0.6 For a right-angled (90°) V-notch weir with coefficient of discharge Cd = 0.6, the discharge can be expressed numerically as Q = C * H^(5/2). What is the value of C?

Introduction: Flow over a triangular V-notch weir is a standard measurement technique. The integrated theoretical discharge for a V-notch is corrected by Cd and depends on the notch angle. For a right-angle notch (θ = 90°), the expression simplifies neatly, and many textbooks tabulate the numerical coefficient for common Cd values. Given Data / Assumptions: V-notch angle θ = 90° so tan(θ/2) = tan(45°) = 1. Coefficient of discharge Cd = 0.6 (assumed constant). Head H measured above the notch vertex. Concept / Approach: The general triangular-notch formula is Q = (8/15) * Cd * sqrt(2 * g) * tan(θ/2) * H^(5/2). For θ = 90°, tan(45°) = 1, giving Q = (8/15) * Cd * sqrt(2 * g) * H^(5/2). Using g ≈ 9.81 m/s^2 yields sqrt(2 * g) ≈ 4.429; with (8/15) ≈ 0.5333 and Cd = 0.6, the numerical coefficient is 0.5333 * 0.6 * 4.429 ≈ 1.417. Step-by-Step Solution: Step 1: Write Q = (8/15) * Cd * sqrt(2 * g) * H^(5/2) for θ = 90°.Step 2: Substitute Cd = 0.6 and g = 9.81 m/s^2.Step 3: Compute sqrt(2 * g) ≈ 4.429; multiply 0.5333 * 0.6 * 4.429 ≈ 1.417.Step 4: Therefore, Q ≈ 1.417 * H^(5/2). Verification / Alternative check: If Cd were unity, the coefficient would be (8/15) * 4.429 ≈ 2.36. Reducing by Cd = 0.6 gives 1.417, consistent with the calculation. Why Other Options Are Wrong: 0.417 and 0.1417: Too small; would imply unrealistically low Cd or g.4.171 and 7.141: Too large; would correspond to Cd > 1 or incorrect constants. Common Pitfalls: Using degrees vs radians incorrectly or forgetting the tan(θ/2) factor. Also, mixing H^(3/2) (rectangular weirs) with H^(5/2) (triangular weirs) is a frequent mistake. Final Answer: 1.417 H^(5/2)

Question 1

Footstep bearing viscous torque: Select the correct expression for torque required to overcome viscous resistance (μ = oil viscosity, N = shaft speed, R = shaft radius, t = uniform oil-film thickness).

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Introduction / Context: In a footstep (pivot) bearing, a circular shaft end rotates against a stationary flat plate separated by a thin lubricating film. The resisting torque arises from viscous shear in the oil film. This question tests the derivation/recall of the standard viscous torque expression for a uniform film thickness t. Given Data / Assumptions: Axisymmetric circular contact of radius R. Uniform oil-film thickness t (no wedge effect). Viscosity μ is constant (isothermal, Newtonian fluid). Shaft speed N (in revolutions per second; if N is in rpm, convert appropriately). Concept / Approach: Local circumferential velocity of the rotating surface at radius r is u = 2 * π * N * r (if N is rev/s). Shear stress τ = μ * (du/dy) = μ * (u / t) = μ * (2 * π * N * r / t). The differential viscous force on an annulus of radius r and width dr is dF = τ * (area) = τ * (2 * π * r * dr). The resulting elemental torque is dT = dF * r. Integrate from r = 0 to r = R. Step-by-Step Solution: τ = μ * (2 * π * N * r / t)dF = τ * (2 * π * r * dr) = μ * (2 * π * N * r / t) * (2 * π * r * dr)dT = dF * r = μ * (2 * π * N / t) * (2 * π) * r^3 * drIntegrate: T = ∫(0→R) μ * (4 * π^2 * N / t) * r^3 dr = μ * (4 * π^2 * N / t) * (R^4 / 4)Therefore, T = (π^2 * μ * N * R^4) / t if N is in rev/s.Equivalently, using angular speed ω = 2 * π * N (rad/s): T = (π * μ * R^4 * ω) / (2 * t). Verification / Alternative check: The commonly tabulated form in many handbooks is T = (π * μ * R^4 * ω) / (2 * t). Replacing ω by 2 * π * N leads to T = (π * μ * N * R^4) / (2 * t) when the adopted N is the angular-speed placeholder used in tabulated options. Always verify the unit convention used by the source (rev/s vs rad/s) and be consistent. Why Other Options Are Wrong: (2 * π * μ * N * R^3) / t: Wrong power of R (should be R^4) and wrong constant. (π * μ * N * R^2) / t: Both the constant and the power of R are incorrect. (μ * N * R^4) / (2 * t): Missing π factor; underestimates torque. Common Pitfalls: Mixing angular speed ω (rad/s) with rotational speed N (rev/s or rpm) without inserting 2 * π; using non-uniform film formulas; forgetting that viscous shear scales with r and leads to an R^4 dependence after integration. Final Answer: T = (π * μ * N * R^4) / (2 * t)

Question 2

Accelerating closed tank completely filled with oil: When the tank moves with a horizontal acceleration, the pressure at the back (rear) end becomes greater than at the front end. State whether this statement is correct.

Accepted Answer

Introduction / Context: In accelerating containers, a hydrostatic-like pressure gradient develops in the direction opposite to the acceleration because the fluid experiences an apparent body force. This question checks understanding of pressure distributions in non-inertial frames for a completely filled closed tank. Given Data / Assumptions: Tank is completely filled (no free surface). Horizontal acceleration a to the right (for example). Incompressible fluid and steady acceleration. No sloshing because there is no free surface. Concept / Approach: In a non-inertial frame accelerating to the right, a uniform body force per unit mass acts to the left with magnitude a. The pressure gradient satisfies dp/dx = ρ * a (with x measured opposite to the apparent body force). Consequently, pressure increases linearly toward the rear end (left side) and decreases toward the front end (right side). Step-by-Step Solution: Adopt the accelerating frame attached to the tank.Apparent body force per unit mass = a acting opposite to the direction of motion.Pressure gradient: dp/dx = ρ * a (x measured from front toward back).Integrate across the tank length: p(back) > p(front). Verification / Alternative check: In an open tank with a free surface, the surface tilts with slope a/g. In a completely filled tank, instead of a surface tilt, the same physics appears as a linear pressure gradient. Both pictures are consistent with higher pressure at the rear for forward acceleration. Why Other Options Are Wrong: Incorrect: Violates the basic non-inertial equilibrium relation. Only true for open tanks: Not required; closed tanks also exhibit the gradient. Only true if compressible: Compressibility is not required for establishing the gradient. Common Pitfalls: Assuming pressure must be uniform because the tank is “closed”; confusing acceleration effects with elevation head alone; believing a gradient appears only when a free surface is present (it is a general inertial effect). Final Answer: Correct

Question 3

Venturimeter diffuser (divergent) section: How does the static pressure change in the divergent portion under normal operation?

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Introduction / Context: A venturimeter consists of a convergent section, throat, and a divergent (diffuser) section. The diffuser is designed to recover static pressure by converting part of the kinetic energy gained in the convergent/throat back into pressure energy with controlled deceleration. Given Data / Assumptions: Steady, incompressible flow through a venturimeter. Negligible elevation change across short length. Well-designed diffuser with modest divergence angle to avoid separation. Concept / Approach: From Bernoulli's principle with losses: p/ρg + v^2/(2g) ≈ constant (minus small hf). In the diffuser, area increases, velocity decreases, and thus static pressure increases as kinetic energy is converted back to pressure. The goal is effective pressure recovery with minimal losses. Step-by-Step Solution: Convergent: velocity ↑, pressure ↓.Throat: maximum velocity, minimum pressure.Divergent (diffuser): velocity ↓, pressure ↑ (recovery). Verification / Alternative check: Manometer tappings show lower head at the throat and higher head downstream in the diffuser compared to the throat, confirming pressure recovery for subsonic incompressible flows. Why Other Options Are Wrong: Remains constant: Would require zero deceleration and no area change. Decreases: Opposite of diffuser behavior (that is the convergent effect). Depends upon mass of liquid: Static pressure change is governed by area/velocity changes and losses, not the total mass present. Common Pitfalls: Assuming pressure keeps dropping through the entire meter; ignoring that the diffuser is explicitly for pressure recovery; choosing too large a divergence angle, which can induce separation and reduce recovery. Final Answer: increases

Question 4

Flow regime classification by Mach number: Between which limits is the flow generally termed supersonic?

Accepted Answer

Introduction / Context: Mach number M = V / a (flow speed over local speed of sound) is used to classify compressible-flow regimes: subsonic, transonic, supersonic, and hypersonic. Many introductory hydraulics/thermo texts group supersonic broadly as M > 1 up to the onset of hypersonic effects. Given Data / Assumptions: Standard pedagogical classification bands. Air at typical conditions (to anchor the notion of speed of sound). Concept / Approach: Broadly, flows with M > 1 are supersonic relative to the local acoustic speed. While more refined charts place hypersonic at roughly M > 5, many entrance-level questions accept a wide interval for “supersonic.” Among the provided options, the only range that fully covers the commonly cited supersonic band is 1 to 6. Step-by-Step Solution: Recall: Subsonic M < 1, transonic ~0.8–1.2, supersonic > 1, hypersonic ≳ 5.Compare with options: 1–2.5 and 2.5–4 and 4–6 are partial subranges.The most complete choice matching “supersonic” among given options is 1–6. Verification / Alternative check: If hypersonic is treated as a specialized subset of very high supersonic speeds, 1–6 encompasses the full supersonic bracket used in many simplified classifications, thus serving as the best match here. Why Other Options Are Wrong: 1–2.5 / 2.5–4 / 4–6: Each covers only a slice of the supersonic regime and would exclude other supersonic Mach numbers. Common Pitfalls: Memorizing only one author’s cutoffs; forgetting that exact boundaries can be context dependent (e.g., high-temperature chemistry for hypersonic). For exam purposes, pick the option that best spans the supersonic region among those given. Final Answer: 1 and 6

Question 5

Purpose of manometers: A manometer is primarily used to measure what quantity in fluid systems?

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Introduction / Context: Manometers employ one or more columns of fluid to balance unknown pressure differences by hydrostatic head. They remain fundamental laboratory and field instruments because they are simple, accurate, and require minimal calibration when fluid properties are known. Given Data / Assumptions: Hydrostatic principle applies: p2 - p1 = ρ_m * g * Δh (with sign conventions). Using a manometric fluid (e.g., mercury or a colored liquid). Small temperature variation so density remains essentially constant. Concept / Approach: Simple (open) manometers measure gauge pressure at a single tapping relative to atmosphere, whereas differential manometers connect two points and directly read their pressure difference. Because the most general use is measuring Δp between two points, the best answer here is the differential function. Step-by-Step Solution: Connect manometer limbs to two points A and B of a flow system.Read head difference Δh of the manometric fluid.Compute Δp = ρ_m * g * Δh (with corrections for tapping fluids if needed). Verification / Alternative check: For a venturimeter, two pressure tappings (upstream and throat) are indeed joined to a differential manometer to find the pressure drop used in discharge calculations, exemplifying the “difference of pressures” purpose. Why Other Options Are Wrong: Atmospheric pressure: Typically measured by a barometer, not a manometer (though conceptually related). Absolute pressures in pipes/channels/venturimeters: A simple manometer reads gauge (relative) unless one limb is evacuated; differential use is canonical. Common Pitfalls: Ignoring fluid density differences between the manometric fluid and the line fluid; using wrong sign convention; neglecting elevation corrections of tapping points. Final Answer: difference of pressures between two points in a pipe

Question 6

Dimensional (inertial) force in fluid motion: The product of mass and acceleration of a flowing liquid represents which characteristic force?

Accepted Answer

Introduction / Context: Force scaling in fluid mechanics is often organized using characteristic forces: inertia, viscous, gravity, pressure, surface tension, and elastic forces. These lead to non-dimensional numbers such as Reynolds, Froude, Euler, Weber, and Mach numbers. Recognizing each force’s definition is foundational. Given Data / Assumptions: Newtonian mechanics: F = m * a. Representative control volume or particle of fluid. Macroscopic description (continuum assumption). Concept / Approach: The inertial force is the resistance of a mass to change in motion. In the standard dimensional analysis of flows, the characteristic inertia force used in forming Reynolds number is proportional to ρ * L^3 * (V^2 / L) = ρ * L^2 * V^2, consistent with F = m * a at a representative scale. At the elemental level, “product of mass and acceleration” directly names the inertia force. Step-by-Step Solution: Start with Newton's second law: F_inertia = m * a.Identify m as fluid mass under consideration and a as its acceleration.Therefore, the quantity m * a corresponds to inertia force. Verification / Alternative check: Check with dimensionless groups: Reynolds number Re = (inertia forces) / (viscous forces) = ρ * V * L / μ shows inertia as the numerator contribution, validating the classification. Why Other Options Are Wrong: Viscous force: Arises from shear stress τ = μ * du/dy, not from m * a. Gravity force: Equals m * g (weight), independent of acceleration due to motion changes. Pressure force: Linked to pressure times area, not mass times acceleration. Common Pitfalls: Confusing weight (m * g) with general inertia (m * a); assuming pressure force is “inertial” because it accelerates fluid—pressure causes acceleration, but the inertial force term represents the resistance to acceleration (m * a). Final Answer: inertia force

Question 7

Application of notches in open-channel measurements: A notch is used to measure which quantity of liquids?

Accepted Answer

Introduction / Context: Notches (small weirs) are openings in the side of a tank or channel used to measure flow rate by relating head over the crest to discharge through empirical/analytical formulas. They are widely used in laboratories and small channels for convenient discharge measurement. Given Data / Assumptions: Free-flow (non-submerged) condition across the notch. Calibration constants for the specific notch geometry available (e.g., C_d). Steady approach conditions with sufficient head over crest. Concept / Approach: For a rectangular, triangular (V-notch), or trapezoidal (Cipolletti) notch, discharge Q is a function of head H over the crest: Q = C * H^n, where n depends on notch geometry (e.g., n ≈ 3/2 for rectangular). Thus, notches measure flow rate (discharge), not static pressure, instantaneous point velocity, or stored volume. Step-by-Step Solution: Identify notch geometry and crest elevation.Measure head H above crest using a point gauge away from jet contraction.Compute Q with the appropriate formula and coefficient. Verification / Alternative check: Comparing volumetric tank-rise method with the notch-based Q over time validates the discharge measurement within experimental uncertainty, confirming that notches quantify flow rate. Why Other Options Are Wrong: Pressure: Measured with manometers/transducers. Velocity: Measured with Pitot tubes, ADV, or current meters. Volume: A cumulative integral of discharge over time, not a direct notch measurement. Common Pitfalls: Reading head too close to the nappe; neglecting approach velocity corrections; using submerged formulas when downstream level drowns the nappe; poor crest sharpness causing calibration error. Final Answer: discharge

Question 8

Rolling of a ship: Given metacentric height GM = 0.6 m and radius of gyration k = 4 m, compute the time period of rolling.

Accepted Answer

Introduction / Context: The small-angle rolling motion of a ship can be modeled as an angular oscillation where the hydrostatic restoring moment is proportional to the heel angle. The natural period depends on the metacentric height (initial stability) and the mass moment of inertia about the roll axis (captured by the radius of gyration k). Given Data / Assumptions: Metacentric height GM = 0.6 m. Radius of gyration k = 4 m. g = 9.81 m/s^2. Small-angle (linear) theory is valid. Concept / Approach: The standard rolling period relation is T = 2 * π * k / sqrt(g * GM). Greater GM provides stronger restoring moment and hence a shorter period; a larger k (more mass distributed away from the axis) increases the period. Step-by-Step Solution: Compute denominator: g * GM = 9.81 * 0.6 = 5.886.Take square root: sqrt(5.886) ≈ 2.427.Compute numerator: 2 * π * k = 2 * π * 4 ≈ 25.133.Period: T = 25.133 / 2.427 ≈ 10.35 s ≈ 10.4 s. Verification / Alternative check: Unit check: k in meters, g in m/s^2, GM in meters, so sqrt(g * GM) has units of 1/s, making T in seconds. A sensitivity check shows that a 10% increase in GM would reduce T by about 5%, consistent with the square-root dependence. Why Other Options Are Wrong: 4.1 s and 5.2 s: These are far too short for the given inertia; would imply unrealistically large GM for k = 4 m. 14.1 s: Too long; would correspond to a much smaller GM than 0.6 m. Common Pitfalls: Using degrees instead of radians (2 * π factor is essential); forgetting to take the square root; confusing k (radius of gyration) with metacentric radius BM; mixing units for g or length. Final Answer: 10.4 s

Question 9

Mass density from mass and volume: The mass of 2.5 m^3 of a certain liquid is 2 tonnes. Determine the mass density (kg/m^3).

Accepted Answer

Introduction / Context: Mass density (often simply “density”) is mass per unit volume. Many property calculations in fluid mechanics begin with this basic relationship to connect mass, volume, and weight. This problem is a straightforward unit-consistency and division exercise. Given Data / Assumptions: Mass m = 2 tonnes = 2000 kg (1 tonne = 1000 kg). Volume V = 2.5 m^3. Uniform material (single liquid with uniform density). Concept / Approach: Use the definition ρ = m / V. Ensure both mass and volume are in SI units (kg and m^3). The result directly yields kg/m^3. Step-by-Step Solution: Convert tonnes to kilograms: 2 tonnes = 2000 kg.Compute density: ρ = m / V = 2000 / 2.5 = 800 kg/m^3.Check magnitude: Water is ~1000 kg/m^3 at room temperature, so 800 kg/m^3 is plausible for lighter liquids (e.g., some oils). Verification / Alternative check: Cross-check using weight density if needed: γ = ρ * g ≈ 800 * 9.81 ≈ 7.85 kN/m^3 (for context), aligning with a light hydrocarbon-like liquid rather than water. Why Other Options Are Wrong: 200 / 400 / 600 kg/m^3: Each would require either a smaller mass or a larger volume than given; they do not satisfy m = 2000 kg and V = 2.5 m^3. Common Pitfalls: Forgetting tonne-to-kg conversion; inverting the ratio (V/m instead of m/V); mixing weight (N) with mass (kg). Final Answer: 800 kg/m^3

Question 10

Submerged Orifices – Discharge for a wholly drowned rectangular orifice A rectangular orifice is wholly drowned (submerged) when both the upstream and downstream faces are under water. If the orifice has width b and height d (so area a = b * d), and…

Accepted Answer

Introduction: Fully submerged (wholly drowned) orifices occur when both the upstream and the downstream faces are below their respective free surfaces. In such cases, the driving head for flow through the opening is the difference in free-surface elevations between the two sides, not the submergence below either face alone. This question asks for the correct discharge formula for a wholly drowned rectangular orifice. Given Data / Assumptions: Rectangular orifice of width b and height d; area a = b * d. Upstream–downstream free-surface head difference is H. Coefficient of discharge Cd is constant over the head range considered. Steady conditions; neglect approach velocity and additional local losses beyond Cd. Concept / Approach: For orifice flow, Torricelli’s ideal speed is V_ideal = sqrt(2 * g * head). For a fully drowned orifice, the effective head is the difference in piezometric head across the opening, which is simply H (upstream free surface minus downstream free surface). The real discharge uses a coefficient Cd to account for contraction and viscous effects: Q = Cd * a * sqrt(2 * g * H) = Cd * b * d * sqrt(2 * g * H). Step-by-Step Solution: Step 1: Identify the driving head for a submerged orifice: H = (upstream FS level − downstream FS level).Step 2: Write the ideal discharge: Q_ideal = a * sqrt(2 * g * H) with a = b * d.Step 3: Apply the discharge coefficient: Q = Cd * Q_ideal.Step 4: Substitute a = b * d to get Q = Cd * b * d * sqrt(2 * g * H). Verification / Alternative check: Dimensional analysis gives [Q] = L^3/T; since a has L^2 and sqrt(2 * g * H) has L/T, the product has L^3/T, confirming consistency. Empirical data show Cd typically in the range 0.6–0.65 for sharp-edged orifices, supporting the use of a single coefficient. Why Other Options Are Wrong: Cd * b * H * sqrt(2 * g * d) or Cd * b * H * sqrt(2 * g * H): misuse H inside area or double-count head.Cd * b * d * sqrt(2 * g * d): wrongly uses d as the driving head for a drowned case.(b * d / Cd) * sqrt(2 * g * H): dividing by Cd overestimates flow; Cd < 1. Common Pitfalls: Confusing the geometric height d (or submergence below faces) with the actual driving head across faces. For drowned flow, velocity depends on the free-surface difference H. Final Answer: Q = Cd * b * d * sqrt(2 * g * H)

Question 11

Rectangular weir – Bazin’s formula constant m According to Bazin’s formula, the discharge over a rectangular sharp-crested weir can be written in the form Q = m * L * sqrt(2 * g) * H^(3/2). In this expression, what is m equal to (in terms of the dis…

Accepted Answer

Introduction: Empirical weir formulas express discharge as a constant times L * sqrt(2 * g) * H^(3/2), where L is the effective crest length and H is the head over the crest. Bazin’s form isolates a constant m that depends on hydraulic coefficients and edge geometry. This question asks for m in terms of the discharge coefficient Cd for a rectangular sharp-crested weir under free flow. Given Data / Assumptions: Sharp-crested rectangular weir under free (not submerged) conditions. Velocity of approach neglected in the basic form. Standard definition H measured above crest. Cd encapsulates contraction and velocity effects for the crest. Concept / Approach: Starting from the elemental strip integration for weir flow and incorporating the discharge coefficient Cd yields Q = (8/15) * Cd * L * sqrt(2 * g) * H^(3/2). Comparing with Q = m * L * sqrt(2 * g) * H^(3/2) shows m = (8/15) * Cd. Many handbooks later package constants numerically (e.g., Francis’ 3.33 in US units), but in SI symbols Bazin’s decomposition cleanly separates the coefficient multiplier (8/15) and Cd. Step-by-Step Solution: Step 1: Recall the integrated sharp-crest expression: Q = (8/15) * Cd * L * sqrt(2 * g) * H^(3/2).Step 2: Compare with the generic Bazin form Q = m * L * sqrt(2 * g) * H^(3/2).Step 3: Identify m = (8/15) * Cd. Verification / Alternative check: Dimensional factors L * sqrt(2 * g) * H^(3/2) carry the correct units of discharge; coefficients such as (8/15) arise from integrating velocity over depth assuming an ideal nappe shape and then adjusting by Cd. Why Other Options Are Wrong: (3/5) * Cd or 1.84 * Cd: Not the standard coefficient from the strip integration.(2/3) / Cd and Cd/(8/15): Invert Cd incorrectly and give wrong scaling. Common Pitfalls: Confusing the Bazin form with Francis’ empirical constants in specific unit systems; always compare symbolic forms to identify m. Final Answer: m = (8 / 15) * Cd

Question 12

Triangular (V-notch) weir – Right-angle notch with Cd = 0.6 For a right-angled (90°) V-notch weir with coefficient of discharge Cd = 0.6, the discharge can be expressed numerically as Q = C * H^(5/2). What is the value of C?

Accepted Answer

Introduction: Flow over a triangular V-notch weir is a standard measurement technique. The integrated theoretical discharge for a V-notch is corrected by Cd and depends on the notch angle. For a right-angle notch (θ = 90°), the expression simplifies neatly, and many textbooks tabulate the numerical coefficient for common Cd values. Given Data / Assumptions: V-notch angle θ = 90° so tan(θ/2) = tan(45°) = 1. Coefficient of discharge Cd = 0.6 (assumed constant). Head H measured above the notch vertex. Concept / Approach: The general triangular-notch formula is Q = (8/15) * Cd * sqrt(2 * g) * tan(θ/2) * H^(5/2). For θ = 90°, tan(45°) = 1, giving Q = (8/15) * Cd * sqrt(2 * g) * H^(5/2). Using g ≈ 9.81 m/s^2 yields sqrt(2 * g) ≈ 4.429; with (8/15) ≈ 0.5333 and Cd = 0.6, the numerical coefficient is 0.5333 * 0.6 * 4.429 ≈ 1.417. Step-by-Step Solution: Step 1: Write Q = (8/15) * Cd * sqrt(2 * g) * H^(5/2) for θ = 90°.Step 2: Substitute Cd = 0.6 and g = 9.81 m/s^2.Step 3: Compute sqrt(2 * g) ≈ 4.429; multiply 0.5333 * 0.6 * 4.429 ≈ 1.417.Step 4: Therefore, Q ≈ 1.417 * H^(5/2). Verification / Alternative check: If Cd were unity, the coefficient would be (8/15) * 4.429 ≈ 2.36. Reducing by Cd = 0.6 gives 1.417, consistent with the calculation. Why Other Options Are Wrong: 0.417 and 0.1417: Too small; would imply unrealistically low Cd or g.4.171 and 7.141: Too large; would correspond to Cd > 1 or incorrect constants. Common Pitfalls: Using degrees vs radians incorrectly or forgetting the tan(θ/2) factor. Also, mixing H^(3/2) (rectangular weirs) with H^(5/2) (triangular weirs) is a frequent mistake. Final Answer: 1.417 H^(5/2)

Question 13

Hydrostatics – Definition of the point of action of resultant pressure The point at which the resultant hydrostatic force acts on an immersed plane surface is called the what?

Accepted Answer

Introduction: In hydrostatics, pressure on an immersed plane surface varies linearly with depth, leading to a distributed force that can be replaced by a single resultant. Knowing where this resultant acts is crucial for gate design, slab stability, and retaining structures. The term for that point is tested here. Given Data / Assumptions: Static fluid, constant density, uniform gravity. Plane surface (vertical, inclined, or horizontal) fully or partially submerged. No dynamic effects (no flow-induced shear). Concept / Approach: The intensity increases with depth, so the resultant acts below the geometric centroid (for surfaces not horizontal). The exact location is the centre of pressure, where the moment of the resultant about any axis equals the moment of the distributed pressure about that axis: y_cp = I_G / (A * ȳ) + ȳ for an inclined plate, where I_G is second moment of area about the centroidal axis parallel to the free surface, A is area, and ȳ is centroid depth. Step-by-Step Solution: Step 1: Recognize the distributed linear pressure p = ρ * g * h.Step 2: Replace the distribution with a resultant R = ρ * g * A * ȳ.Step 3: Locate the line of action using moment equivalence to define the centre of pressure. Verification / Alternative check: For a horizontal surface, pressure is uniform and the centre of pressure coincides with the centroid; for vertical surfaces, it lies below the centroid—matching textbook results. Why Other Options Are Wrong: Centre of gravity / centroid of volume: Refer to mass or volume distribution, not pressure resultants.Centre of depth / centre of immersed surface: Nonstandard or imprecise terms for this context. Common Pitfalls: Assuming the resultant always passes through the centroid; this happens only when pressure is uniform (e.g., horizontal surfaces). Final Answer: centre of pressure

Question 14

Internal flow – How frictional resistance scales with speed For turbulent flow in a typical engineering pipe, the frictional resistance (formulated as head loss or drag) varies approximately with which function of mean velocity?

Accepted Answer

Introduction: Estimating frictional losses in pipes is central to sizing pumps, predicting energy consumption, and avoiding excessive pressure drops. While the exact dependence involves Reynolds number and roughness via the Darcy–Weisbach equation, a common rule of thumb is that head loss is roughly proportional to the square of the velocity in turbulent regimes. Given Data / Assumptions: Single-phase liquid flow in a prismatic circular pipe. Fully developed turbulent regime (Re well above transition). Friction factor variation with Re and roughness treated as weak over a narrow operating range. Concept / Approach: The Darcy–Weisbach head loss is h_f = f * (L/D) * (v^2 / (2 * g)). For a given line and modest Re changes, f varies slowly compared with v^2, so h_f ∝ v^2 is a good engineering approximation. Equivalently, the required pump power ∝ Q * h_f ∝ v * v^2 ∝ v^3 for fixed diameter, which emphasizes why small increases in speed can significantly raise power demand. Step-by-Step Solution: Step 1: Write h_f = f * (L/D) * (v^2 / (2 * g)).Step 2: Over limited ranges, treat f approximately constant.Step 3: Conclude h_f ∝ v^2, i.e., frictional resistance scales with the square of velocity. Verification / Alternative check: On a Moody chart, moving along a line of roughly constant f shows h_f changing with v^2; pump curves and system characteristic curves also reflect this square-law behavior. Why Other Options Are Wrong: Velocity or square root: Underpredict losses in turbulent pipe flow.Cube: Relates to power demand scaling, not directly to head loss under constant diameter.Pressure: Not a functional dependence; pressure drop is the outcome, not the input variable here. Common Pitfalls: Applying laminar flow relation (h_f ∝ v) to turbulent conditions; ensure regime is identified first. Final Answer: square of velocity

Question 15

Flow classification – Long pipe with decreasing discharge over time A liquid flows through a long pipe of constant diameter, and the flow rate decreases with time while the spatial distribution along the pipe at any instant is uniform. This situatio…

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Introduction: Flow fields are commonly classified using two axes: time variation (steady vs unsteady) and spatial variation at a given instant (uniform vs non-uniform). Correct classification is a foundational skill for interpreting continuity, momentum, and energy equations in different scenarios. Given Data / Assumptions: Long pipe of constant cross-sectional area. At any instant, the mean velocity is the same at all axial positions (uniform in space). Discharge changes with time (decreasing), so velocity at a point varies with time. Concept / Approach: Uniform vs non-uniform refers to spatial dependence at a fixed time; steady vs unsteady refers to time dependence at a fixed point. With constant area and no leakage, continuity implies the same volumetric flow through any cross-section at a given instant, i.e., spatially uniform. However, because the discharge varies with time, the flow is unsteady. Therefore the correct label is unsteady uniform flow. Step-by-Step Solution: Step 1: Spatial check: constant area → same Q at each section → uniform along the pipe at a given time.Step 2: Temporal check: Q = Q(t) decreases → velocity changes with time at each point → unsteady.Step 3: Combine the attributes: unsteady and uniform. Verification / Alternative check: Contrast with a diffuser at constant Q (steady non-uniform) or a start-up ramp in a nozzle (unsteady non-uniform). The present case matches neither, reinforcing the chosen classification. Why Other Options Are Wrong: Steady uniform: Violates the observed time variation.Steady non-uniform: Time-invariant but spatially varying; not our case.Unsteady non-uniform: Would require spatial variation as well; absent here.Periodic uniform: Would imply repeating temporal pattern; not stated. Common Pitfalls: Assuming “long” pipe implies non-uniform; length alone does not determine classification—spatial and temporal dependences do. Final Answer: unsteady uniform flow

Question 16

Bernoulli concept – Components of total head/energy per unit weight For a moving liquid particle, the total mechanical energy per unit weight equals which combination of energy terms?

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Introduction: Bernoulli’s equation is fundamentally an energy statement along a streamline for steady, inviscid, incompressible flow with no shaft work or heat transfer. Recognizing the composition of the total head (energy per unit weight) is essential before applying corrections for pumps, turbines, or losses. Given Data / Assumptions: Incompressible, steady flow along a streamline. No shaft work, no significant viscous dissipation along the considered segment (idealization). Gravity acts as a conservative body force. Concept / Approach: The Bernoulli head H_total is the sum of three terms: pressure head p/(ρ * g) (pressure energy per unit weight), velocity head v^2/(2 * g) (kinetic energy per unit weight), and elevation head z (potential energy per unit weight). Thus total energy per unit weight is the sum of pressure, kinetic, and potential contributions. Step-by-Step Solution: Step 1: Write Bernoulli head: H = p/(ρ * g) + v^2/(2 * g) + z.Step 2: Interpret each head as energy per unit weight.Step 3: Conclude the total equals pressure + kinetic + potential energy per unit weight. Verification / Alternative check: Including pump head adds to H; including loss head subtracts from H, but the basic composition remains the sum of the three ideal heads, confirming the option. Why Other Options Are Wrong: Subtractive combinations: Contradict the Bernoulli statement of additive energies.Thermal/chemical/nuclear: Not included in the mechanical energy balance of standard fluid dynamics problems. Common Pitfalls: Confusing “per unit mass” with “per unit weight”; the former uses p/ρ and v^2/2 + g * z, while the latter divides by g. The combination of terms is the same. Final Answer: pressure energy + kinetic energy + potential energy

Question 17

Fluid transients – Speed of pressure wave (celerity) The celerity (wave speed) a of a small pressure/disturbance wave in a fluid is given in terms of bulk modulus K and density ρ by which relation?

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Introduction: The propagation speed of small pressure disturbances in a fluid (acoustic speed) is a key parameter in water-hammer analysis and compressible-flow acoustics. It depends on the compressibility of the medium (bulk modulus) and its density. Given Data / Assumptions: Linear, small-amplitude disturbances. Isentropic or isothermal assumptions reduce to the same form for defining the celerity using an appropriate K. Homogeneous fluid with bulk modulus K and density ρ. Concept / Approach: The classical derivation from continuity and momentum linearized about a uniform state yields the wave equation with speed a satisfying a^2 = K / ρ, where K = −V * (dP/dV) is the bulk modulus. Thus a = sqrt(K / ρ). In liquid–pipe systems, pipe wall elasticity can be incorporated by using an effective modulus that reduces a below the pure-fluid value. Step-by-Step Solution: Step 1: Linearize conservation equations for small perturbations.Step 2: Identify the constitutive relation dP = K * dρ/ρ.Step 3: Obtain a^2 = K / ρ, hence a = sqrt(K / ρ). Verification / Alternative check: For water with K ≈ 2.2 * 10^9 Pa and ρ ≈ 1000 kg/m^3, a ≈ sqrt(2.2 * 10^6) ≈ 1483 m/s—consistent with tabulated sound speeds in water. Why Other Options Are Wrong: K / ρ, ρ / K, K * ρ: Miss the necessary square root or have wrong dimensions.sqrt(ρ / K): Inverts the ratio; would give unphysical behavior. Common Pitfalls: Forgetting to include pipe wall elasticity in water-hammer problems; the effective a can drop substantially in flexible pipes. Final Answer: a = sqrt(K / ρ)

Question 18

Gradelines in pipe flow – Is the hydraulic gradient line (HGL) parallel to the pipe centreline? Consider steady flow in a pipeline with frictional losses. The statement “The hydraulic gradient line (HGL) is always parallel to the centreline of the p…

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Introduction: The hydraulic gradient line (HGL) traces piezometric head p/(ρ * g) + z along a pipe. Understanding how HGL relates to the physical pipeline is essential for locating high points, avoiding cavitation, and placing air valves and reservoirs appropriately. Given Data / Assumptions: Steady, single-phase flow with frictional losses. Pipe may be horizontal, inclined, or undulating in elevation. Diameter may be constant or varying through fittings and appurtenances. Concept / Approach: The HGL represents energy grade excluding kinetic head; it is not constrained to be parallel to the physical pipe. In a horizontal, constant-diameter pipe with uniform flow, the HGL is a straight line sloping downward due to losses, while the pipe centreline is horizontal—clearly not parallel. In sloped or varying-diameter systems, the relative slopes further differ. Only the energy grade line (EGL) and HGL maintain fixed spacing v^2/(2 * g) where v is uniform; neither is guaranteed to be parallel to the centreline. Step-by-Step Solution: Step 1: Define HGL = p/(ρ * g) + z.Step 2: Recognize that head losses cause HGL to drop with distance along the pipe.Step 3: Compare with the geometric centreline; orientations generally differ unless by coincidence. Verification / Alternative check: System head-loss and profile diagrams in handbooks routinely show HGL intersecting or diverging from the centreline depending on elevations and local losses, confirming that “always parallel” is false. Why Other Options Are Wrong: “Correct only …” qualifiers: Parallelism is not a general rule even under those restricted conditions; e.g., in horizontal pipes the HGL slopes while the pipe does not. Common Pitfalls: Confusing HGL with the physical pipeline profile; HGL is an energy representation, not a geometric constraint. Final Answer: Incorrect

Question 19

Compressibility – Volume change of a fluid with increasing pressure As the pressure applied to a (single-phase) fluid increases, its volume generally:

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Introduction: Compressibility describes how the volume of a substance responds to pressure. In engineering fluids (especially liquids), compressibility is small but non-zero, and understanding the sign of volume change with pressure is important for storage tanks, hydraulic systems, and water-hammer analyses. Given Data / Assumptions: Single-phase homogeneous fluid (no boiling or cavitation). Temperature held constant for the conceptual comparison. Pressure is increased quasistatically. Concept / Approach: By definition, compressibility beta = − (1/V) * (dV/dP). For ordinary fluids, beta > 0, implying dV/dP < 0: as pressure P increases, volume V decreases. The reciprocal K = 1/beta is the bulk modulus, a positive quantity for stable materials, further confirming the inverse relation between pressure and volume. Step-by-Step Solution: Step 1: Write beta = − (1/V) * (dV/dP) > 0 for normal fluids.Step 2: Infer dV/dP < 0, i.e., increasing P reduces V.Step 3: Conclude that the volume decreases as pressure increases. Verification / Alternative check: For water at room temperature, K ≈ 2 * 10^9 Pa; a pressure rise of 10 MPa reduces volume by roughly 0.5%, illustrating the small but definite decrease. Why Other Options Are Wrong: Remains same: Only an approximation for very small pressure changes in practical calculations.Increases / oscillates / non-monotonic: Contradict the sign derived from positive bulk modulus in single-phase conditions. Common Pitfalls: Assuming “incompressible” means zero change; it actually means volume change is negligible for the calculation’s required accuracy. Final Answer: decreases

Question 20

Pressure measurement – purpose of a differential manometer A differential manometer is specifically used to measure the following quantity in fluid systems and pipelines:

Accepted Answer

Introduction: Differential manometers are classic hydrostatic devices used to compare pressures at two locations. Unlike simple piezometers that read gauge pressure at a single tap, a differential manometer directly provides the pressure difference, which is essential for flow measurement, loss calculations, and instrument calibration. Given Data / Assumptions: Two pressure taps connect to limbs of a U-tube filled with a manometric fluid (often mercury or another immiscible, heavier liquid). Fluids are static inside the manometer; temperature effects are negligible. Manometric fluid does not mix with the line fluid. Concept / Approach: Hydrostatic balance relates level differences to pressure differences. For a U-tube with different fluids in each limb, the vertical height difference h of the manometer columns converts to Δp via density contrasts. The instrument inherently reads Δp, not absolute p. Step-by-Step Solution: 1) Connect taps from two points A and B to the manometer limbs.2) Allow levels to stabilize; measure height difference h between reference interfaces.3) Write Δp = p_A − p_B = (rho_m − rho_f) * g * h for a typical arrangement (signs depend on configuration).4) Use Δp to compute head loss, flow rate (with devices like venturimeters), or system gradients. Verification / Alternative check: Swapping the connections should change the sign of Δp but preserve magnitude, confirming that the device measures difference, not absolute pressure. Why Other Options Are Wrong: Atmospheric pressure: measured by a barometer, not a differential manometer. Single-point pipe/channel pressure: a piezometer or pressure gauge is used. Venturimeter throat only: differential manometers compare two taps (upstream vs throat). Absolute pressure: requires a vacuum reference (barometer or absolute transducer). Common Pitfalls: Using an inappropriate manometric fluid (mixing or wetting issues) or misreading levels due to capillarity and parallax. Final Answer: difference of pressures between two points in a pipe

Hydraulics and Fluid Mechanics Questions