Question 1

Equation of continuity for incompressible, steady flow  Let a₁, v₁ be area and mean velocity at section 1, and a₂, v₂ at section 2 of the same streamtube. Select the correct continuity relation.

Accepted Answer

Concept/Approach For steady incompressible flow, volume flow rate is constant: Q = a v = constant. Hence a₁v₁ = a₂v₂. Final Answer a₁v₁ = a₂v₂

Question 2

Definition of a broad-crested weir  A weir is termed broad-crested when the width of its crest is ____ half of the head (water depth) over the crest. Choose the correct completion.

Accepted Answer

Concept/Approach In a broad-crested weir the crest width is sufficiently large so that the flow over the crest becomes nearly parallel; this requires crest width greater than about half the upstream head over the crest. Final Answer more than

Question 3

Capillary rise in a small tube dipped in water: identify the cause  A narrow glass tube is dipped into water and the water rises in the tube. State whether the rise is due to viscosity or another effect.

Accepted Answer

Concept/Approach Capillary rise is caused by surface tension and the adhesive attraction between liquid and tube material, balanced by weight of the column. Viscosity influences the rate of rise, not the final height. Final Answer Disagree — the rise occurs due to surface tension (capillarity), not viscosity.

Question 4

Piezometer reading: what quantity does it measure?  A simple piezometer tube is connected to a pipeline to measure pressure. Choose the correct form of the measured quantity.

Accepted Answer

Concept/Approach A piezometer indicates the static pressure head as a liquid column height above the tap point. It directly measures head; conversion to N/m² requires multiplying by ρg. Final Answer Head of liquid (height column)

Question 5

Experimental determination of coefficient of viscosity  Which of the following methods can be used to determine a liquid's dynamic viscosity? Select the correct option.

Accepted Answer

Concept/Approach Viscosity can be found by several standard methods: capillary tube (Poiseuille), orifice viscometers (efflux time), and rotating cylinder or cone-and-plate viscometers (shear measurement). Final Answer All of these

Question 6

Energy grade line (TEL) vs hydraulic grade line (HGL)  In pipe flow, the total energy line lies above the hydraulic gradient line by what amount? Choose the correct quantity.

Accepted Answer

Concept/Approach HGL = z + p/γ. TEL = z + p/γ + v²/(2g). The difference (TEL − HGL) is the velocity head v²/(2g). Final Answer Velocity head

Question 7

Euler's equation of motion for inviscid flow: differential form  Select the correct differential relation along a streamline for steady, frictionless flow of a fluid of density ρ.

Accepted Answer

Concept/Approach Euler's equation (along a streamline, inviscid): the sum of differential pressure head, velocity head and elevation head changes is zero. Statement dp/ρ + V dV + g dz = 0Integrating gives Bernoulli's relation: p/ρ + V²/2 + g z = constant. Final Answer dp/ρ + V dV + g dz = 0

Question 8

Fluid mechanics — buoyancy versus displaced liquid: sinking condition  A body is immersed in a liquid. Consider the buoyant force compared to the weight of the displaced liquid as per Archimedes’ principle. Choose the correct relation that (as state…

Accepted Answer

Given data / concept Archimedes’ principle states that the buoyant force always equals the weight of the displaced liquid: Fb = Wdisplaced liquid. Clarification The float/sink criterion compares the body’s weight (Wbody) to the buoyant force: a body sinks if Wbody > Fb, floats if equal, and rises if less. However, relative to the weight of displaced liquid, the buoyant force is always equal by definition (independent of sink or float). Final Answer equal to

Question 9

Kinematic viscosity: definition check for fluids  Select the correct definition of kinematic viscosity (ν) in terms of absolute (dynamic) viscosity (μ) and density (ρ). Choose the correct relationship.

Accepted Answer

Concept/Approach Kinematic viscosity is defined as ν = μ / ρ where μ is dynamic viscosity and ρ is density. Units: m²/s. Final Answer ratio of absolute viscosity to the density of the liquid

Question 10

Large rectangular orifice: discharge formula in terms of heads H₁ and H₂ For a large rectangular orifice of breadth b with liquid heads H₁ (above top) and H₂ (above bottom), pick the correct discharge equation (C_d is coefficient of disch…

Accepted Answer

Concept/Approach For a large orifice (appreciable head variation over height), integrate elemental discharge along depth to obtain:Q = Cd b √(2g) (H2^{3/2} − H1^{3/2}). Step-by-step (outline) dQ = Cd b v dy, with v = √(2g h), integrate h from H1 to H2.∫ √h dh = (2/3) h^{3/2} ⇒ Q = Cd b √(2g) [(2/3)(H2^{3/2} − H1^{3/2})] × (3/2) absorbed into standard form. Final Answer Q = Cd × b √(2g) × (H2^{3/2} − H1^{3/2})

Question 11

Fluid mechanics – hydraulic gradient line (HGL) relative to pipe centerline Consider steady internal flow in a pressurized pipeline. Can the hydraulic gradient line lie above or below the pipe’s centerline, depending on pressure distribution and dat…

Accepted Answer

Introduction / Context: The hydraulic gradient line (HGL) represents the piezometric head line: z + p/(γ). For closed, pressurized pipes, its position relative to the physical pipe can vary with local pressure and elevation. Understanding HGL and the energy grade line (EGL) is essential for analyzing head losses and pump/siphon behavior. Given Data / Assumptions: Incompressible liquid (e.g., water), steady flow. HGL is defined as z + p/γ (velocity head excluded). Centerline elevation is a convenient datum for reference. Concept / Approach: At any pipe section, the piezometric head can be higher or lower than the pipe centerline elevation. If pressure is sufficiently high, p/γ raises HGL above the centerline; if the local absolute pressure approaches or drops below atmospheric, the computed HGL can fall to the centerline or below (e.g., siphons or high points where pressure decreases). Step-by-Step Solution: Define HGL: HGL = z + p/γ.For a pipe centerline at elevation z_c, compare HGL − z_c = p/γ.If p > 0 gauge, p/γ > 0 ⇒ HGL sits above the centerline. If p is small or negative (gauge), HGL can be at or below the centerline.Velocity head v^2/(2g) lies above HGL to form the energy grade line (EGL). Verification / Alternative check: Typical system profiles show HGL descending along flow due to head loss; at siphon summits, HGL may drop near or below the pipe crown, illustrating positions both above and below the centerline. Why Other Options Are Wrong: False / “only above” / “only below” contradict practical cases such as pumped lines and siphons. Common Pitfalls: Confusing HGL with EGL. HGL excludes velocity head; EGL = HGL + v^2/(2g), and always lies above HGL by the velocity head amount. Final Answer: True

Question 12

Hydraulic structures – material of construction for weirs In water resources engineering, a permanent weir used for flow measurement or control is generally constructed from which materials?

Accepted Answer

Introduction / Context: Weirs are overflow structures that raise upstream water level and/or measure discharge. For accuracy and durability, they must maintain crest geometry under hydraulic and environmental loading. Materials influence stability, service life, and maintenance. Given Data / Assumptions: Permanent installation on streams/canals for long-term service. Need for dimensional stability of crest and abutments. Exposure to weathering and hydraulic forces. Concept / Approach: Masonry and concrete provide mass, rigidity, and resistance to erosion—critical for maintaining crest shape and energy dissipation arrangements. While thin-plate (sharp-crested) weirs can be fabricated from metal for lab or temporary field use, permanent river training and measurement weirs are commonly masonry or reinforced concrete. Step-by-Step Reasoning: Identify performance requirements: dimensional stability, durability, scour resistance.Match materials: masonry/concrete meet massiveness and durability needs; timber/steel are more common in temporary or laboratory structures.Therefore, stating that a weir is usually made of masonry or concrete is correct for permanent works. Verification / Alternative check: Standard hydraulic texts and design manuals illustrate ogee, broad-crested, and sharp-crested permanent weirs in masonry or RC for field use. Why Other Options Are Wrong: No: contradicts typical civil engineering practice. Timber only / steel plate only: applicable mainly to temporary or experimental setups, not the usual permanent case. Common Pitfalls: Equating laboratory weir plates with field weirs; the latter require robust civil works. Final Answer: Yes

Question 13

Fluid properties – comparing dynamic viscosity At ordinary temperatures, the dynamic viscosity of water is __________ that of mercury.

Accepted Answer

Introduction / Context: Viscosity measures a fluid’s resistance to shear. Comparing fluids helps anticipate flow regimes and head losses. Water and mercury are common reference liquids with different molecular interactions and viscosities. Given Data / Assumptions: Room temperature conditions (around 20 °C). Dynamic viscosity μ is considered (not kinematic). Both liquids are pure and standard. Concept / Approach: At ~20 °C, water has μ ≈ 1 mPa·s, while mercury has μ ≈ 1.5–1.6 mPa·s. Hence, water is less viscous than mercury. Kinematic viscosity ν = μ/ρ would differ even more due to mercury’s very high density, but the question explicitly refers to viscosity (μ). Step-by-Step Reasoning: Identify property: dynamic viscosity μ (units Pa·s or N·s/m^2).Typical values: μ_water ≈ 1×10^-3 Pa·s; μ_mercury ≈ 1.5×10^-3 Pa·s (approx.).Compare: μ_water < μ_mercury ⇒ water is less viscous. Verification / Alternative check: Handbooks provide tabulated μ for common fluids; values confirm the same ordering across a reasonable temperature range near ambient. Why Other Options Are Wrong: Higher / same: contradict empirical data. “Undefined because viscosity has no units”: incorrect; dynamic viscosity has SI units Pa·s. Common Pitfalls: Confusing dynamic viscosity with kinematic viscosity; mercury’s ν is smaller than water’s because of its very high density, but μ is larger. Final Answer: lower than

Question 14

Atmospheric pressure at sea level – choose the most precise SI value Which of the following best states the standard atmospheric pressure at mean sea level in SI units?

Accepted Answer

Introduction / Context: Standard atmosphere (1 atm) is a reference pressure used in many engineering calculations. It has several equivalent representations, but SI practice prefers expressing it in kilonewtons per square meter or kilopascals. Given Data / Assumptions: Standard atmosphere at sea level, 0 °C, approximate standard conditions. Equivalences: 1 atm ≈ 101.325 kPa ≈ 10.33 m of water ≈ 760 mmHg at 0 °C. Concept / Approach: The most precise SI representation among the options is 101.3 kN/m^2 (i.e., 101.3 kPa). Head of water and mmHg are convenient equivalents but not SI base pressure units, and “103 kN/m^2” is an overestimate. Step-by-Step Reasoning: Recall: 1 atm = 101.325 kPa ≈ 101.3 kN/m^2.Water head equivalence: ~10.33 m; often rounded to 10.3 m.Mercury head equivalence: 760 mm at 0 °C. Verification / Alternative check: Converting 760 mmHg to SI pressure gives approximately 101.325 kPa using ρ_Hg ≈ 13,600 kg/m^3 and g ≈ 9.81 m/s^2. Why Other Options Are Wrong: 10.3 m of water and 760 mm of mercury are correct equivalences but are not SI pressure units; the question asks for the best SI value. 103 kN/m^2 is a rounded-up value and not the standard. Common Pitfalls: Mixing head units (length) with pressure units; always choose kPa or kN/m^2 for SI pressure unless specifically asked for head. Final Answer: 101.3 kN/m^2

Question 15

Open-channel/Weir flow – discharge over a rectangular sharp-crested weir with velocity of approach Given H1 = H + Ha, where H is the head over the crest and Ha is the velocity-of-approach head, select the correct expression for discharge Q over a re…

Accepted Answer

Introduction / Context: Discharge over a sharp-crested rectangular weir depends on the effective head causing flow. When approach velocity is non-negligible, an additional head term appears. This question checks recognition of the correct functional dependence on head. Given Data / Assumptions: Rectangular sharp-crested weir with coefficient Cd. H is static head over crest; Ha is velocity-of-approach head. H1 = H + Ha represents total effective head at approach section. Standard, free, fully aerated nappe; neglect end contractions beyond effective L. Concept / Approach: For a sharp-crested weir, the theoretical discharge is proportional to the integral of velocity across depth, yielding the 3/2 power law in head. Accounting for velocity of approach, the effective head terms enter as H1^(3/2) − Ha^(3/2). Step-by-Step Solution: Base relation (without approach velocity): Q = (2/3) Cd L √(2g) H^(3/2).With approach velocity: replace H by H1 and subtract the contribution of Ha to avoid double counting.Thus, Q = (2/3) Cd L √(2g) (H1^(3/2) − Ha^(3/2)).The constant 2/3 is often absorbed into empirical Cd or presented explicitly; options use the compact √(2g) form with the 3/2 exponents. Verification / Alternative check: Dimensional analysis confirms Q has units of L^3/T when head is to the 3/2 power multiplied by L √(2g). Linear or quadratic head dependences (options a and c) do not match the classical derivation. Why Other Options Are Wrong: (a) Linear in head: incorrect functional form. (c) Quadratic in head: inconsistent with weir theory. (d) 5/2 power: belongs to broad-crested or other profiles under different assumptions, not the sharp-crested rectangular case with this correction. Common Pitfalls: Forgetting to subtract Ha^(3/2), which leads to overestimation of Q when approach velocity is significant. Final Answer: Q = Cd · L · √(2g) · (H1^(3/2) − Ha^(3/2))

Question 16

Fluid properties – does specific gravity have units? Evaluate the statement: “Specific gravity has no units.”

Accepted Answer

Introduction / Context: Specific gravity (SG) is frequently used in fluid mechanics and material science to compare densities relative to water or another standard. Whether SG has units is a common conceptual check. Given Data / Assumptions: SG is defined as the ratio of a material’s density to the density of a reference (commonly water at 4 °C for liquids/solids, air for gases in some conventions). Both densities are measured in the same units. Concept / Approach: Because SG is a ratio of two like quantities (density/density), units cancel. Thus, specific gravity is dimensionless. Its numeric value can depend on temperature and, for gases, reference conditions because density varies with these, but SG itself carries no units. Step-by-Step Reasoning: Write SG = ρ_material / ρ_reference.If both ρ are in kg/m^3 (or any consistent unit), units cancel.Hence SG is a pure number, often close to 1 for many liquids near water’s density. Verification / Alternative check: Specific weight or relative density can also be used; properly defined, they remain dimensionless ratios when like quantities are compared. Why Other Options Are Wrong: Disagree: incorrect; SG is not expressed in kg/m^3 or any other units. “Depends on reference temperature” or “local gravity”: these affect the numeric value of the reference density or weight, not the dimensional nature of the ratio. Common Pitfalls: Confusing specific gravity (dimensionless) with specific weight γ = ρ g, which does have units (N/m^3). Final Answer: Agree

Question 17

Rheology – shear stress versus shear strain rate for a Newtonian fluid For a Newtonian fluid, what is the shape of the graph when shear stress is plotted against shear strain rate (rate of shear deformation)?

Accepted Answer

Introduction / Context: The constitutive relation for Newtonian fluids connects shear stress to shear strain rate. Recognizing this relationship helps identify flow behavior and select appropriate models in pipelines, boundary layers, and lubrication. Given Data / Assumptions: Newtonian behavior: τ = μ (du/dy), where du/dy is the velocity gradient (shear strain rate). Constant dynamic viscosity μ under given temperature and pressure. Concept / Approach: Because τ is directly proportional to the strain rate with constant proportionality μ, a plot of τ versus du/dy is a straight line through the origin with slope μ. Step-by-Step Reasoning: Write the law: τ = μ * (du/dy).For constant μ, τ increases linearly with du/dy.Graphically, the line passes through the origin; slope equals μ. Verification / Alternative check: Non-Newtonian fluids deviate: shear-thinning (pseudoplastic) show concave-down curves; shear-thickening (dilatant) show concave-up; Bingham plastics have a yield intercept. None of these are straight through the origin with constant slope. Why Other Options Are Wrong: Parabolic, hyperbolic, elliptical: do not represent τ ∝ du/dy with constant μ. Common Pitfalls: Confusing shear stress versus time or strain; the Newtonian linearity is with strain rate, not total strain. Final Answer: Straight line

Question 18

Instrumentation – orientation sensitivity of a Pitot tube If a Pitot tube is pointed upstream, downstream, or sideways in a flowing fluid, will it give the same reading in every case?

Accepted Answer

Introduction / Context: A Pitot tube measures stagnation (total) pressure by bringing the flow to rest isentropically at the probe tip. Its reading depends critically on alignment with the flow direction. This question tests understanding of how orientation affects the measured pressure. Given Data / Assumptions: Standard Pitot tube with an impact opening at the nose. Steady flow with a well-defined mean direction. Gauge measures stagnation pressure when correctly aligned. Concept / Approach: When the tube faces directly upstream (into the flow), the fluid at the tip stagnates and the measured pressure equals stagnation pressure p0 = p + ρ v^2/(2). If the tube faces sideways, the opening senses something close to static pressure; if it faces downstream, separated flow and low total pressure at the tip lead to erroneous, often near-static or even fluctuating readings. Step-by-Step Reasoning: Upstream alignment → measures stagnation pressure correctly.Sideways alignment → flow does not stagnate at the opening; reading tends toward static pressure.Downstream alignment → wake and reversed streamlines yield unreliable readings, not stagnation pressure. Verification / Alternative check: Prandtl Pitot-static probes incorporate separate ports: the nose port for stagnation, circumferential ports for static, highlighting the orientation dependence by design. Why Other Options Are Wrong: True: contradicts basic operation; orientation matters. Other conditional options are irrelevant; the principle holds across Re and fluids, barring special multi-hole yaw-insensitive probes. Common Pitfalls: Assuming a Pitot tube alone gives velocity regardless of alignment; yaw errors must be corrected or minimized. Final Answer: False

Question 19

Internal flow – head loss dependence in laminar regime For fully developed laminar flow in a circular pipe, the loss of pressure head is proportional to which power of the mean velocity?

Accepted Answer

Introduction / Context: Understanding how head loss scales with velocity helps select pumps and size pipes. The relationship differs markedly between laminar and turbulent regimes. This question focuses on the laminar case. Given Data / Assumptions: Steady, incompressible, fully developed, laminar flow in a straight circular pipe. Newtonian fluid with constant viscosity μ. No entrance or exit losses considered for the proportionality discussion. Concept / Approach: Hagen–Poiseuille law: Δp = 32 μ L V / D^2 for laminar pipe flow. Converting to head loss h_f = Δp/(ρ g) shows h_f ∝ V. Thus, in laminar flow, head loss varies linearly with mean velocity. In contrast, turbulent empirical correlations (e.g., Darcy–Weisbach with f depending on Re and roughness) often show approximately V^2 dependence over ranges. Step-by-Step Solution: Start: Δp = 32 μ L V / D^2.Divide by ρ g: h_f = Δp/(ρ g) = (32 μ L /(ρ g D^2)) * V.Therefore, h_f ∝ V (first power). Verification / Alternative check: Darcy–Weisbach with laminar friction factor f = 64/Re yields h_f = f (L/D) (V^2/(2g)) = (64/Re)(L/D)(V^2/(2g)) = (64 μ /(ρ V D))(L/D)(V^2/(2g)) ∝ V, confirming linear dependence. Why Other Options Are Wrong: V^2, V^3, V^4 correspond to turbulent or nonphysical scalings for this regime. Common Pitfalls: Memorizing V^2 from turbulent flow and applying it universally. Always check Reynolds number to identify regime. Final Answer: velocity (first power)

Question 20

Fluid mechanics — definition of lift on an inclined plate Whenever a thin plate is held immersed at some angle to the direction of a moving liquid stream, a pressure distribution acts on it. The component of this resultant pressure that is perpendic…

Accepted Answer

Introduction / Context: In fluid mechanics and aerodynamics, forces on immersed bodies are resolved into components. Two standard components are drag (along the free stream) and lift (perpendicular to the free stream). A flat or curved plate placed at an angle to the oncoming flow experiences a pressure and shear distribution that results in a net force with these two components. Given Data / Assumptions: An immersed plate is inclined at some angle to the direction of flow. The fluid exerts pressure and shear stresses over the plate surface. Resultant surface force can be decomposed into components parallel and perpendicular to the free stream. Concept / Approach: By definition, the component of the resultant fluid force perpendicular to the undisturbed flow direction is called lift. The component parallel to the undisturbed flow direction is called drag. This definition applies broadly, regardless of whether the plate is a wing, hydrofoil, or a simple immersed flat plate at an angle of attack. Step-by-Step Solution: Identify the resultant pressure and shear force on the plate due to the flow.Resolve this resultant into components along and normal to the free stream.Name the along-stream component as drag and the normal component as lift.Hence, the statement given is precisely the definition of lift → it is correct. Verification / Alternative check: In wing theory, lift L and drag D are often expressed using coefficients (C_L and C_D) relative to dynamic pressure and planform area. The orientation of these forces is always defined with respect to the free stream, not to the body surface, which confirms the perpendicular component is lift. Why Other Options Are Wrong: False: contradicts the standard definition. Only true for very small angles: lift exists for a broad range of angles until stall; the definition does not depend on angle size. Laminar vs. turbulent: flow regime changes magnitude, not the definition. Common Pitfalls: Confusing “normal to plate” with “normal to flow.” Lift is defined perpendicular to the free stream, not necessarily normal to the plate surface. Final Answer: True

Hydraulics and Fluid Mechanics Questions