Question 1

Design of tank wagons for steep mountainous routes  For mountainous regions with steep road gradients, the bottoms of liquid-carrying wagons should be arranged:

Accepted Answer

Introduction / Context: Liquid cargo vehicles must minimize sloshing and spillage when traversing steep grades. The free surface of a liquid at rest is horizontal; tank design should take advantage of this to maintain safe headspace and predictable pressure distribution. Given Data / Assumptions: Vehicle moves at low acceleration relative to gravity (quasi-static on slopes). Goal: maintain safe clearance and avoid unintended overflow through vents. Concept / Approach: In hydrostatics, the free surface aligns with the equipotential—horizontal in the Earth’s gravitational field. If the tank bottom is parallel to the horizontal plane, the geometry of the headspace and structural supports remains consistent regardless of road gradient, improving safety and capacity utilization. If the bottom follows the road slope, the liquid surface intersects the tank differently uphill vs downhill, risking poor utilization and unbalanced loads. Step-by-Step Reasoning: Recognize that the liquid free surface seeks the horizontal.Design the tank bottom level with the horizontal so the volume distribution is predictable on slopes.Provide baffles for dynamic stability; vents placed to avoid spillage when pitching/rolling. Verification / Alternative check: Field practice and codes often require level bottoms or internal arrangements that assume a horizontal liquid surface for capacity markings and pressure calculations. Why Other Options Are Wrong: Bottom parallel to road surface will alternately reduce or increase effective headspace; deliberately tilting up or down with the direction of motion is impractical; “parallel to chassis” is ambiguous. Common Pitfalls: Ignoring dynamic accelerations and braking; baffles are still needed even with level bottoms. Final Answer: parallel to the horizontal (level) plane

Question 2

Optimal area ratio for maximum power transmission through a nozzle on a long pipe  Given a pipe of length L, diameter D, and Darcy friction factor f, what is the optimal area ratio (a/A) of nozzle area to pipe area for maximum jet power?

Accepted Answer

Introduction / Context: When a nozzle is attached to a long pipeline, increasing the nozzle diameter raises discharge but also increases pipe velocity and friction loss, reducing available head at the nozzle. There is a specific area ratio that maximizes the transmitted power P = ρ Q V_jet^2 / 2. Given Data / Assumptions: Steady, incompressible flow. Single straight pipe of length L and diameter D with Darcy friction factor f. Negligible minor losses except friction and nozzle conversion to jet. Concept / Approach: Balancing available head with friction loss in the pipe yields a relation between pipe velocity and nozzle area. Maximizing jet power with respect to nozzle area leads to a classical optimum: (8 f L / D) * (a/A)^2 = 1, hence a/A = 1 / √(8 f L / D). Step-by-Step Solution: Let A = pipe area, a = nozzle area; continuity gives Q = A V_p = a V_j.Head loss in pipe: h_f = f (L/D) * V_p^2 / (2 g).Available head at nozzle converts to jet velocity: V_j = √(2 g (H − h_f)).Express power P = ρ Q V_j^2 / 2 and optimize with respect to a/A to obtain (8 f L / D) (a/A)^2 = 1. Verification / Alternative check: This optimum is equivalent to the condition that the pipe friction head equals one-third of the supply head at maximum power for the end-nozzle configuration. Why Other Options Are Wrong: Other forms invert or misplace parameters, leading to non-optimal or dimensionally inconsistent relationships. The presence of the 8 f L / D group is critical and must be in the denominator under the square root. Common Pitfalls: Confusing Darcy and Fanning friction factors (factor of 4 difference); ignoring minor losses when they are significant, which would shift the exact optimum. Final Answer: a/A = 1 / √(8 f L / D)

Question 3

Rheological classification: what type of fluid is water under ordinary conditions?

Accepted Answer

Introduction / Context: Classifying fluids by their stress–strain-rate behavior guides the selection of models and correlations. Water is the most common working fluid in civil and environmental engineering. Given Data / Assumptions: Moderate pressures and temperatures near ambient. Shear rates typical of pipe/channel flows. Concept / Approach: A Newtonian fluid has a linear relationship between shear stress τ and shear rate du/dy: τ = μ (du/dy), where μ is the dynamic viscosity, constant for given temperature and pressure. Water obeys this linear law across a wide range of practical conditions, making it Newtonian. Non-Newtonian fluids exhibit shear-thinning, shear-thickening, yield stress, or viscoelastic effects. Step-by-Step Reasoning: Identify constitutive relation for water: τ proportional to shear rate.Note that μ varies mainly with temperature but not with shear rate.Therefore, water is a Newtonian fluid. Verification / Alternative check: Viscometry data for water show linear τ–du/dy over typical engineering ranges; contrast with paints (Bingham), polymer solutions (shear-thinning), and slurries (non-Newtonian). Why Other Options Are Wrong: Non-Newtonian and Bingham plastic behaviors do not apply to pure water. Water is effectively incompressible at low Mach numbers; labeling it “compressible fluid” is misleading for standard hydraulics. Common Pitfalls: Assuming incompressible means Newtonian—these are independent properties; many incompressible fluids are non-Newtonian. Final Answer: Newtonian fluids

Question 4

Head–discharge relationship for a V-notch (triangular) weir  How does the discharge Q over a sharp-crested V-notch vary with the measured head H above the crest (neglecting velocity of approach)?

Accepted Answer

Introduction / Context: Weirs are standard devices for measuring open-channel discharge. The triangular (V-notch) weir is especially sensitive at low flows, and its calibration follows a characteristic power law with head. Given Data / Assumptions: Sharp-crested V-notch, free and fully aerated nappe. Head H measured at a standard upstream location. Negligible velocity-of-approach correction for this conceptual relation. Concept / Approach: Integrating elemental discharge across the depth of the notch opening yields Q = C_d * (8/15) * √(2 g) * tan(θ/2) * H^(5/2). Thus, the fundamental dependence is Q ∝ H^(5/2). The stronger exponent compared with the rectangular weir (3/2) makes the V-notch more sensitive to small changes in H, aiding low-flow accuracy. Step-by-Step Outline: Express local width b(y) = 2 y tan(θ/2) at depth y below the head level.Elemental discharge dQ = C_d b(y) √(2 g (H − y)) dy.Integrate from y = 0 to H to obtain the H^(5/2) dependence. Verification / Alternative check: Calibration charts and ISO standards for V-notches reflect a 5/2 exponent in the governing formula, with C_d and angle terms as multiplicative factors. Why Other Options Are Wrong: H, 1/H, and H^(3/2) correspond to other devices or simplified relations; H^(7/2) overstates the head sensitivity and is not supported by the integration result. Common Pitfalls: Ignoring aeration (submergence reduces effective exponent), improper head measurement location, or using rectangular-weir exponents for triangular notches. Final Answer: Q ∝ H^(5/2)

Question 5

Applicability of the basic small-orifice discharge formula  For a sharp-edged rectangular orifice with constant head H across the opening, the commonly used discharge formula Q = C_d * A * sqrt(2 g H) is applicable to:

Accepted Answer

Introduction / Context: The basic orifice formula Q = C_d * A * sqrt(2 g H) is ubiquitous in hydraulics. However, its straightforward use assumes a uniform head H over the entire opening—an assumption violated in large orifices where head varies significantly with depth. Given Data / Assumptions: Sharp-edged orifice discharging freely to the atmosphere. Coefficient of discharge C_d accounts for contraction and losses. “Small orifice” implies that H is essentially constant from top to bottom of the opening. Concept / Approach: Derivation assumes uniform energy head at the orifice plane, allowing V = sqrt(2 g H) to multiply a constant area A. For a large orifice, hydrostatic pressure (hence local velocity) varies with depth, so the discharge must be integrated: dQ = C_d * b * √(2 g y) dy across the vertical extent, not merely A * √(2 g H). Step-by-Step Explanation: Small orifice: take H constant over opening → Q = C_d * A * √(2 g H).Large orifice: head varies from H_top to H_bottom → integrate velocity over depth.Thus, the simple formula strictly applies to small orifices. Verification / Alternative check: Compare computed Q using the simple formula vs. integrated expression for large vertical extents; discrepancies confirm the need for integration. Why Other Options Are Wrong: “Large orifices only” and “small and large orifices only” ignore head variation. “All types” is overgeneralization. Venturimeters are different devices using differential pressure over a throat—not orifices per se. Common Pitfalls: Using the small-orifice formula on sluice openings or gates with large vertical extent; neglecting submergence corrections. Final Answer: small orifices only

Question 6

Most economical rectangular channel in open-channel flow For a hydraulically most economical rectangular channel section (maximum discharge for a given area or minimum wetted perimeter), the hydraulic mean depth (R = A / P) is equal to which of the …

Accepted Answer

Introduction / Context: In open-channel hydraulics, a “most economical” section minimizes wetted perimeter for a given area (or maximizes discharge for a given area) and therefore reduces frictional losses. For a rectangular channel, this geometric condition yields a simple relationship between the width and depth, and thus a specific value of hydraulic mean depth R = A / P relative to the depth y. Given Data / Assumptions: Prismatic rectangular channel with width b and flow depth y. Steady, uniform flow so hydraulic radius R = A / P is appropriate. “Most economical” means minimum wetted perimeter P for a fixed area A = b * y. Concept / Approach: For a rectangle, area A = b * y and wetted perimeter P = b + 2y. Using calculus (optimize P for constant A) or standard results, the condition for most economy is b = 2y. Substituting into R = A / P gives a direct relation between R and y for the optimal section. Step-by-Step Solution: Start with A = b * y and P = b + 2y.For most economical rectangle: b = 2y.Compute R = A / P = (b * y) / (b + 2y).Substitute b = 2y → R = (2y * y) / (2y + 2y) = (2y^2) / (4y) = y / 2. Verification / Alternative check: The same result follows if we maximize R directly at constant A, or minimize wetted perimeter using Lagrange multipliers; both give b = 2y and R = y/2 for a rectangular section. Why Other Options Are Wrong: (a) R = y is too large; (c) R = y/3 is too small; (d) is incorrect because a unique relation exists for the optimal rectangle. Common Pitfalls: Confusing hydraulic radius R = A / P with hydraulic depth D = A / T (T = top width). For the most economical rectangle, R = y/2 but hydraulic depth D equals y since T = b = 2y and A = b * y. Final Answer: half the depth of flow

Question 7

Measuring discharge in very large pipes (diameter > 3 m) Which instrument is generally preferred for measuring flow in very large-diameter water mains (greater than about 3 m)?

Accepted Answer

Introduction / Context: Flow measurement in very large pipelines presents practical challenges: installing full-bore primary elements (like venturimeters or orifice plates) becomes costly and intrusive, and differential heads can be large. Velocity-probe methods often become the pragmatic choice for large conduits and tunnels. Given Data / Assumptions: Pipe diameter exceeds roughly 3 m. In-situ, pressurized flow of water. Desire for a practical, minimally invasive measurement method. Concept / Approach: A Pitot tube measures stagnation and static pressure to infer local velocity via Bernoulli. By traversing the cross-section (or using multiport averaging Pitot tubes), the average velocity can be determined and discharge computed as Q = V̄ * A. This approach is widely used in large mains because it avoids full-bore meter installations and large permanent head losses. Step-by-Step Solution: Insert Pitot probe to measure local dynamic head: Δp = ½ ρ v^2.Convert to local velocity: v = √(2 Δp / ρ).Traverse to obtain velocity profile; integrate or average: V̄.Compute discharge: Q = V̄ * A. Verification / Alternative check: Where access allows, acoustic (ultrasonic) transit-time systems are also used; however, among the listed classical devices, the Pitot tube best suits very large diameters for practicality and low permanent pressure loss. Why Other Options Are Wrong: (b) Venturimeters of that size are expensive and bulky; (c) orifice plates impose significant head loss and plate handling issues at very large sizes; (d) rotameters are for small pipe/line sizes, not large mains. Common Pitfalls: Failing to perform adequate section traverses; neglecting Reynolds-number effects on Pitot coefficient; installing near elbows causing distorted velocity profiles. Final Answer: pitot tube

Question 8

Applicability of the energy (Bernoulli) equation in fluid mechanics The standard energy equation between two sections is primarily applicable under which flow condition?

Accepted Answer

Introduction / Context: Bernoulli’s (energy) equation is a cornerstone of fluid mechanics used to relate pressure, velocity, and elevation between two sections in a streamline flow, with appropriate loss terms. Understanding when and how it applies is essential for correct engineering use. Given Data / Assumptions: Incompressible fluid. Single-phase flow without significant shaft work between sections. Flow along a streamline or averaged over a section with correction factors. Steady conditions for the simplest form of the equation. Concept / Approach: The classical form assumes steady flow so that energy per unit weight at a section is time-invariant. With unsteadiness, additional unsteady terms appear. The equation does not depend on laminar vs. turbulent per se; turbulence is handled by head-loss terms and kinetic energy correction factors, provided the flow is statistically steady. Step-by-Step Solution: Write Bernoulli with losses: p/γ + α v^2/(2 g) + z = constant − h_L.For steady flow, these terms are time-invariant between sections.Account for pump/turbine heads where applicable. Verification / Alternative check: In unsteady flow, the energy equation includes local acceleration terms; for steady flow, these vanish, matching the standard textbook form. Why Other Options Are Wrong: (a) Non-uniformity in space is allowed; the equation is routinely applied in gradually varied flow with loss terms. (b) and (c) do not decide applicability; both regimes can be handled if the flow is steady and losses are modeled. Common Pitfalls: Ignoring head losses; assuming α = 1 when profiles are highly non-uniform; applying the equation across rotating machinery without including shaft work terms. Final Answer: steady flow.

Question 9

Hydraulic grade line (HGL) relative to the conduit centerline In pressurized pipe flow, where can the hydraulic grade line lie relative to the pipe centerline?

Accepted Answer

Introduction / Context: The hydraulic grade line (HGL) represents the piezometric head (pressure head plus elevation head) and is foundational in diagnosing pressurized pipeline behavior. Its position relative to the pipe centerline changes with pressure conditions along the system. Given Data / Assumptions: Pressurized, steady pipe flow. HGL is defined as z + p/γ (neglecting velocity head). Elevations referenced to the pipe centerline for comparison. Concept / Approach: If p/γ is positive (gauge pressure), the HGL lies above the centerline; if p/γ is negative (sub-atmospheric), the HGL can drop below the centerline. Therefore, depending on local pressure conditions (e.g., high velocities, siphon crests), the HGL may be either above or below the pipe axis. Step-by-Step Solution: Compute HGL = z_centreline + p/γ for each section.Compare with centerline elevation to visualize HGL position.Assess where p drops below atmospheric (possible cavitation risk as HGL falls). Verification / Alternative check: Plotting HGL and the energy grade line (EGL) along a pipeline confirms that their relative positions vary; EGL always lies above HGL by v^2/(2 g). Why Other Options Are Wrong: (a) and (b) restrict HGL location incorrectly; (c) the HGL is not generally parallel to the pipe alignment due to friction losses and grade changes. Common Pitfalls: Confusing HGL with EGL; ignoring zones of low pressure at high points in siphons; forgetting that negative gauge pressure shifts HGL below the centerline. Final Answer: may be above or below the centre line of conduit.

Question 10

Rotational versus irrotational flow—relation to angular velocity A flow is termed rotational if the circulation per unit area (vorticity) equals which multiple of the local angular velocity vector?

Accepted Answer

Introduction / Context: Rotationality in fluid flow is characterized by non-zero vorticity, which reflects local spin of fluid elements. Understanding the relationship between angular velocity and vorticity is fundamental for classifying flows and applying potential-flow theory appropriately. Given Data / Assumptions: Continuum fluid with differentiable velocity field. Two- or three-dimensional flow. Classical definition of vorticity as the curl of velocity. Concept / Approach: The vorticity vector ω (bold) is defined as ω = ∇ × V. For a rigid-body rotation, the local angular velocity vector Ω relates to vorticity by ω = 2 Ω. Thus, a non-zero vorticity indicates rotational flow; irrotational flow has ω = 0 and therefore Ω = 0 for fluid elements. Step-by-Step Solution: Define vorticity: ω = ∇ × V.Relate to angular velocity: Ω = ½ ω.Criterion: rotational if ω ≠ 0 (i.e., Ω ≠ 0). Verification / Alternative check: Consider a solid-body rotation V = Ω × r. Then ∇ × V = 2 Ω, confirming the factor of two between vorticity and angular velocity. Why Other Options Are Wrong: (a) and (c) assign incorrect proportionality; (d) is incorrect because a precise relation exists. Common Pitfalls: Confusing circulation Γ around a finite loop with local vorticity; assuming turbulent flow is always “rotational” in the mathematical sense—classification is local and based on ω. Final Answer: twice the angular velocity vector

Question 11

Dimensional analysis—dimensions of force in MLT system What is the correct dimensional formula for force (Newton’s second law) in terms of mass M, length L, and time T?

Accepted Answer

Introduction / Context: Dimensional homogeneity is the backbone of fluid mechanics and all of engineering analysis. Identifying the correct dimensions of fundamental quantities helps in validating equations and building non-dimensional groups (e.g., Reynolds, Froude, Weber numbers). Given Data / Assumptions: Force F is related to mass m and acceleration a by F = m * a. Base dimensions: mass M, length L, and time T. Concept / Approach: Acceleration a has dimensions of length per time squared, and mass is simply M. Multiplying these gives the well-known MLT^-2 for force. This is consistent with SI, where the Newton (N) is kg·m/s^2. Step-by-Step Solution: Acceleration a → L T^-2.Mass m → M.Force F = m a → M * (L T^-2) = M L T^-2. Verification / Alternative check: Energy (work) W = F * L gives dimensions M L^2 T^-2, consistent with Joule, further validating F = M L T^-2. Why Other Options Are Wrong: (b), (c), and (d) mismatch exponents; (e) is unnecessary since a correct option exists. Common Pitfalls: Dropping negative signs on time exponents; confusing derived units (N) with base dimensions; mixing CGS and SI inconsistently. Final Answer: MLT-2

Question 12

Venturimeter design—throat-to-pipe diameter ratio to avoid separation What range of throat diameter to pipe diameter is typically used in venturimeters to avoid flow separation and maintain reliable pressure recovery?

Accepted Answer

Introduction / Context: Venturimeters measure discharge by creating a controlled constriction in a pipe. The choice of throat diameter greatly influences accuracy, head loss, and the risk of separation or cavitation. A practical ratio ensures a measurable differential without excessive losses or unstable flow. Given Data / Assumptions: Axisymmetric venturi with smooth convergent and divergent sections. Incompressible liquid flow. Goal: avoid boundary-layer separation in diffuser and maintain good pressure recovery. Concept / Approach: Common design practice selects the throat-to-pipe diameter ratio (β = d_t / D) in the range of about 0.33 to 0.50. This offers a good compromise between measurable pressure drop and acceptable losses, while keeping velocities and adverse pressure gradients within limits. Step-by-Step Solution: Select β so that differential head Δp is measurable but not excessive.Ensure diffuser half-angle is small (about 5–7 degrees) to prevent separation.Check cavitation margin at throat velocity. Verification / Alternative check: Standards and handbooks list typical β ranges for venturimeters; values near 0.4 are widely used in water systems to balance accuracy and loss. Why Other Options Are Wrong: (a) 1/4 to 1/8 is too small, causing very high throat velocities and risk of cavitation; (c) 1/2 to 3/4 is too large for many applications, giving very low differential heads and sensitivity issues; (d) is unnecessary. Common Pitfalls: Ignoring upstream disturbances; using too steep diffuser leading to separation; not accounting for temperature-dependent vapor pressure when checking cavitation. Final Answer: 1/3 to 1/2

Question 13

Why sea water weighs more than pure water The specific weight (weight per unit volume) of sea water exceeds that of pure water primarily because of what factor?

Accepted Answer

Introduction / Context: Specific weight (γ = ρ g) reflects the density of a fluid. Comparing sea water to pure water highlights how composition affects density and, consequently, buoyancy and hydrostatics. Understanding the main contributor is essential for coastal and naval engineering calculations. Given Data / Assumptions: Sea water contains dissolved salts (primarily NaCl, MgSO4, etc.). Presence of some dissolved gases and fine suspended matter. Temperatures near ambient and moderate pressures. Concept / Approach: The dominant reason sea water is denser than pure water is its salinity: dissolved ions increase mass per unit volume much more than trace dissolved air or typical suspended particulates in clear sea water. Thus ρ_sea > ρ_fresh and γ_sea > γ_fresh, which affects buoyancy, stratification, and wave dynamics. Step-by-Step Solution: Specific weight γ = ρ g; compare ρ between fluids.Salinity typically 30–35 g/kg increases ρ by about 2–3% over fresh water.Hence γ increases proportionally at the same g. Verification / Alternative check: Oceanographic data show typical sea water density near 1025 kg/m^3 at 15 °C and standard salinity, compared to 999–1000 kg/m^3 for fresh water at similar conditions. Why Other Options Are Wrong: (a) Dissolved air has a negligible effect and often decreases density slightly; (c) suspended matter is episodic and small for clear water; (d) is too broad—salts are the principal cause. Common Pitfalls: Confusing salinity effects with temperature; assuming turbidity is always significant—open-ocean suspended loads are usually low. Final Answer: dissolved salts

Question 14

Reading velocities from a flow net On a flow net, the spacing between two consecutive streamlines at section 1 is 1 cm, and at the next downstream section it is 0.5 cm. If the local velocity at section 1 is 1.0 m/s, what is the local velocity at sec…

Accepted Answer

Introduction / Context: Flow nets are graphical solutions of Laplace’s equation used in seepage and potential flow. The spacing of streamlines and equipotentials encodes discharge per flow tube, enabling velocity comparisons between sections without direct computation of the velocity field. Given Data / Assumptions: Two adjacent streamlines form a stream tube of width s. Equal potential drop between compared sections (same number of potential squares). Incompressible, steady flow; constant thickness normal to the net. Concept / Approach: In a given stream tube, discharge q is conserved: q = v * A_tube. With constant thickness, A_tube ∝ s (streamline spacing). Therefore, v is inversely proportional to s. Halving the spacing doubles the velocity. Step-by-Step Solution: Conservation: q_1 = v_1 * s_1 = q_2 = v_2 * s_2.Solve for v_2: v_2 = v_1 * (s_1 / s_2).Given v_1 = 1.0 m/s, s_1 = 1 cm, s_2 = 0.5 cm → v_2 = 1.0 * (1 / 0.5) = 2.0 m/s. Verification / Alternative check: Uniform net thickness and equal potential drop ensure the stream tubes carry the same discharge; the inverse proportion confirms the result. Why Other Options Are Wrong: (a) and (b) ignore the reduced spacing; (d) and (e) exaggerate the change beyond the inverse-spacing rule for the given ratios. Common Pitfalls: Comparing sections across different potential drops; reading spacing obliquely rather than perpendicular to streamlines; neglecting varying thickness in 3D scenarios. Final Answer: 2.0 m/sec

Question 15

Rectangular weir with velocity of approach—correct discharge expression For a sharp-crested rectangular weir where the approach velocity is not negligible, which expression correctly accounts for the velocity-of-approach correction (H is head over c…

Accepted Answer

Introduction / Context: For rectangular weirs, the classical discharge relation assumes negligible approach velocity. When approach velocity is significant, the upstream total head includes both the static head over the crest and the velocity head, requiring a correction to avoid overestimation of discharge. Given Data / Assumptions: Sharp-crested rectangular weir, fully aerated nappe. H = static head above crest; h_a = v_a^2/(2 g) = velocity head in the approach channel. Coefficient of discharge C_d and crest length L are known. Concept / Approach: The effective head driving flow is not just H but (H + h_a). However, the measured head gauge usually reads H referenced to a stilling point. The established correction subtracts the portion that would be counted if the approach section were also treated as a weir section (h_a term), leading to the standard form with a negative h_a^{3/2} correction. Step-by-Step Solution: Start from Q ∝ (H_total)^{3/2} with H_total = H + h_a.Correct to avoid double-counting approach velocity effect → subtract h_a^{3/2} term.Hence: Q = K · C_d · L · [ (H + h_a)^{3/2} − h_a^{3/2} ], where K contains √(2 g) and constants. Verification / Alternative check: When v_a → 0, h_a → 0 and the expression reduces to the classical rectangular weir formula proportional to H^{3/2}, as expected. Why Other Options Are Wrong: (a) adds h_a^{3/2} instead of subtracting; (c) and (d) incorrectly replace H with H − h_a, which is not the effective upstream total head. Common Pitfalls: Using gauge readings too close to the crest (velocity effects not dissipated); neglecting the velocity of approach for high specific discharges. Final Answer: Cd x L x 2g [(H + ha)3/2 - ha3/2]

Question 16

In open-channel hydraulics, the hydraulic radius R is defined as which ratio of geometric properties of the flowing section?

Accepted Answer

Introduction / Context: Hydraulic radius (R) is a key geometric parameter in open-channel and pipe flow used in Manning’s and Chezy’s equations. It condenses the cross-sectional shape into a single length scale that reflects the balance of area conveying flow and perimeter creating frictional resistance. Given Data / Assumptions: Steady internal flow fully wetting the considered perimeter. Prismatic channel or uniform pipe cross-section. Definitions: flow area A, wetted perimeter P (only the boundary in contact with water). Concept / Approach: By definition, hydraulic radius is R = A / P. The area A promotes discharge capacity, while the wetted perimeter P is where shear develops. A larger R generally implies less relative boundary friction for a given area, improving conveyance efficiency. Step-by-Step Solution: Identify the conveying area: A (m^2).Identify the resisting boundary length: P (m).Write the definition: R = A / P.Note use in formulas, e.g., Chezy: V = C * sqrt(R * S), Manning: V = (1/n) * R^(2/3) * S^(1/2). Verification / Alternative check: For a wide channel, P ≈ b (width) and A ≈ b * y (depth), so R ≈ (b*y)/b = y, matching the rule-of-thumb that R ~ depth for very wide channels. Why Other Options Are Wrong: (a) R is not A/P^2; (c) inverts the correct ratio; (d) sqrt(A) lacks perimeter information and is dimensionally inconsistent for R’s physical meaning. Common Pitfalls: Including dry boundaries in P; confusing hydraulic radius (A/P) with hydraulic diameter (4A/P) used in closed conduits. Final Answer: Area divided by wetted perimeter

Question 17

For an irrotational, incompressible flow field, which governing equation must the velocity potential (or stream function in 2D) satisfy?

Accepted Answer

Introduction / Context: Irrotational flow is central to potential-flow theory, used for inviscid approximations around bodies. In such flows, a scalar velocity potential exists and its governing equation provides a powerful mathematical framework for analysis. Given Data / Assumptions: Flow is irrotational (vorticity = 0). Incompressible fluid (constant density). Smooth, differentiable velocity field with potential function φ. Concept / Approach: For incompressible flow, continuity requires divergence of velocity to be zero. With v = grad(φ), divergence-free condition becomes the Laplacian of φ equal to zero. Hence the velocity potential satisfies Laplace’s equation, which is linear and elliptic, enabling superposition and boundary-value methods. Step-by-Step Solution: Let v = ∇φ for irrotational flow.Incompressible continuity: ∇·v = 0.Substitute v: ∇·(∇φ) = ∇^2φ = 0.Therefore φ satisfies Laplace’s equation. Verification / Alternative check: In two-dimensional potential flow, the stream function ψ also satisfies ∇^2ψ = 0, consistent with harmonic conjugacy of φ and ψ in simply connected regions. Why Other Options Are Wrong: (a) Cauchy–Riemann are compatibility relations between φ and ψ, not the governing PDE. (b) Reynolds transport theorem is an integral balance tool, not a pointwise PDE. (d) Bernoulli provides energy relation along a streamline, not the governing equation for φ. Common Pitfalls: Assuming Bernoulli alone defines the flow field; mixing rotational and irrotational assumptions improperly. Final Answer: Laplace’s equation

Question 18

Dimensional analysis fundamentals: which statements correctly describe dimensional homogeneity and its implications?

Accepted Answer

Introduction / Context: Dimensional analysis is vital for verifying formulas, planning experiments, and reducing variables using Buckingham Pi theorem. Ensuring dimensional homogeneity prevents unit-based mistakes and supports physical plausibility of equations. Given Data / Assumptions: Fundamental dimensions: typically Mass (M), Length (L), Time (T), Temperature (Θ), etc. Engineering equations should be dimensionally homogeneous. Units are specific choices within a dimensional framework (e.g., SI or USCS). Concept / Approach: A dimensionally homogeneous equation has identical dimensional exponents of M, L, T, etc., on both sides of the equality. This property ensures validity regardless of unit system and is a necessary (though not sufficient) condition for correctness of a physical relation. Step-by-Step Solution: Check each term’s dimensions and match across the equality.Confirm exponents (e.g., M^a L^b T^c) are the same for corresponding terms.Conclude the equation remains valid when units change (e.g., meters to feet) if it is homogeneous. Verification / Alternative check: Convert an equation from SI to USCS; if numeric coefficients adjust consistently and the relationship holds, dimensional homogeneity is demonstrated. Why Other Options Are Wrong: Options (a), (b), and (c) are each correct; choosing any single one would be incomplete. Therefore, “All the above” is the correct comprehensive choice. Common Pitfalls: Adding unlike dimensions (e.g., adding a length to a velocity); mistaking units (e.g., N vs kgf); ignoring derived dimensions (e.g., pressure as ML^-1T^-2). Final Answer: All the above.

Question 19

Pipe network analysis (e.g., municipal water distribution): which fundamental relations must simultaneously be satisfied at steady state?

Accepted Answer

Introduction / Context: Design and analysis of looped pipe networks require solving for unknown flows and nodal heads. Classical methods (Hardy Cross, linear theory) and modern solvers enforce mass and energy conservation with appropriate head-loss laws for each pipe segment. Given Data / Assumptions: Steady-state incompressible flow. Known demands or supplies at nodes. Empirical/analytical head-loss relation (e.g., Darcy–Weisbach or Hazen–Williams). Concept / Approach: Two conservation laws govern the solution: continuity at nodes and energy around loops. Each pipe requires a constitutive relation linking discharge to head loss. Together, these equations form a solvable system for flows and heads. Step-by-Step Solution: Continuity: ΣQ_in − ΣQ_out = 0 at every junction.Energy: ΣΔh around each independent loop = 0 (accounting for pumps/minor losses).Head loss: Δh_f = f * (L/D) * (V^2/(2g)) (Darcy–Weisbach) or alternative relation.Iteratively adjust flows until both continuity and loop energy balances are satisfied. Verification / Alternative check: Compute residuals of nodal continuity and loop energy; convergence to near zero confirms a physically consistent solution. Why Other Options Are Wrong: Each of (a), (b), and (c) is necessary; omitting any yields an under-defined or inconsistent system. Hence “All the above” is correct. Common Pitfalls: Using inconsistent units; neglecting minor losses or pump curves; applying an inappropriate head-loss formula outside its range. Final Answer: All the above.

Hydraulics Questions