Question 1

Dimensionless groups in fluid–particle systems: which of the following is NOT primarily used to characterize fluid–particle interaction phenomena?

Accepted Answer

Introduction: Fluid–particle interactions (settling, fluidization, drop/bubble breakup) are often analyzed using appropriate dimensionless numbers. Knowing which groups are relevant helps select proper correlations for drag, terminal velocity, and regime transitions. Given Data / Assumptions: Single particle or dispersed phase in a continuous fluid. Interest in drag, buoyancy, viscous, and surface tension effects. Concept / Approach: Galileo (and Archimedes) numbers combine gravity, density difference, viscosity, and size to correlate particle motion and drag. Weber number compares inertial to surface tension forces and is crucial for droplets/bubbles. The drag coefficient C_d directly characterizes particle–fluid momentum exchange. The Froude number compares inertia to gravity for free-surface or open-channel flows and is not central to single-particle interaction correlations in quiescent fluids. Step-by-Step Solution: List particle-relevant groups: Ga, Ar, Re_p, C_d, We (for interfacial phenomena).Identify Fr: more relevant to wave speed and open-channel flow rather than single-particle drag/settling.Therefore, Fr is the least relevant among the options. Verification / Alternative check: Empirical drag and terminal velocity charts use Re_p with C_d and Ga/Ar. Studies on drop breakup use We and Ohnesorge numbers; Fr is uncommon for isolated particle correlations. Why Other Options Are Wrong: C_d, Ga, Ar: standard in particle drag correlations. We: standard for droplets/bubbles dealing with surface tension effects. Common Pitfalls: Assuming any dimensionless group involving gravity applies equally; context matters (free surface versus dispersed particle systems). Final Answer: Froude number (Fr)

Question 2

Packed beds under laminar conditions: for laminar liquid flow through a packed bed, the pressure drop varies proportionally to which combination (V_s = superficial velocity, D_p = particle diameter)?

Accepted Answer

Introduction: Estimating pressure drop in packed columns is vital for absorber/stripper design. In the laminar regime, viscous forces dominate and the dependence on particle size and superficial velocity follows a specific scaling captured by the Ergun (or Kozeny–Carman) relations. Given Data / Assumptions: Single-phase, incompressible liquid. Laminar flow through a uniformly packed bed. Constant porosity and sphericity for particles. Concept / Approach: The Ergun equation splits viscous and inertial contributions. In laminar flow the viscous term dominates: Δp/L ∝ (μ V_s)/(D_p^2) * function of porosity. Thus, pressure drop is directly proportional to superficial velocity and inversely proportional to the square of particle diameter. Step-by-Step Solution: Start with Ergun viscous term: Δp/L = 150 (1 − ε)^2 μ V_s / (ε^3 D_p^2).For fixed packing (ε constant), Δp ∝ μ V_s / D_p^2.Therefore, the proportionality is V_s / D_p^2. Verification / Alternative check: Kozeny–Carman form also yields the same D_p^−2 scaling in creeping flow through porous media. Why Other Options Are Wrong: Terms with V_s^2 imply inertial effects (turbulent/pre-loading), not laminar regime. D_p^−3 scaling does not arise from the laminar viscous term. D_p / V_s^2 is dimensionally inconsistent with Δp scaling. Common Pitfalls: Using the full Ergun equation without checking Reynolds number; overestimating Δp by applying the inertial term in the laminar range. Final Answer: V_s / D_p^2

Question 3

Internal laminar pipe flow: how does shear stress vary across the cross-section of a circular pipe carrying a Newtonian fluid?

Accepted Answer

Introduction: Understanding radial distributions of velocity and shear stress in laminar pipe flow is essential for predicting pressure drop and wall shear stress, which affect energy consumption and scaling/erosion risks. Given Data / Assumptions: Newtonian, incompressible, steady, fully developed laminar flow in a circular pipe. No-slip at the wall. Concept / Approach: For fully developed laminar flow, the velocity profile is parabolic and the shear stress varies linearly with radius. Shear stress is zero at the pipe centerline (symmetry) and reaches a maximum at the wall, equal to the wall shear τ_w. Step-by-Step Solution: Momentum balance gives dP/dx = constant; τ(r) = −(r/2)(dP/dx).Because dP/dx is constant, τ(r) ∝ r.At r = 0 (centre), τ = 0; at r = R (wall), τ = τ_w = −(R/2)(dP/dx).Hence linear increase from centre to wall. Verification / Alternative check: The parabolic velocity profile u(r) = U_max (1 − r^2/R^2) differentiates to du/dr ∝ −r; for a Newtonian fluid τ = μ du/dr ∝ r, confirming linear variation. Why Other Options Are Wrong: Parabolic variation: applies to velocity, not shear stress. Uniform: contradicts the linear relation. Zero at wall: violates no-slip driven gradient; shear is maximal there. Maximum at centre: opposite of reality. Common Pitfalls: Mixing up velocity and shear stress shapes; forgetting symmetry implies zero shear at the centreline. Final Answer: Zero at the centre and increases linearly to a maximum at the wall

Question 4

Transition criterion in circular pipes: above which Reynolds number does laminar flow of a Newtonian fluid typically cease to exist (onset of transition)?

Accepted Answer

Introduction: The Reynolds number quantifies the ratio of inertial to viscous forces and is the primary criterion for flow regime identification in internal flows. Correctly identifying the laminar–turbulent boundary is essential for choosing correlations for friction and heat/mass transfer. Given Data / Assumptions: Circular pipe, Newtonian fluid, steady flow. Re = ρ v D / μ, based on pipe diameter. Concept / Approach: Experiments indicate laminar flow is generally maintained for Re < 2100. Between about 2100 and 4000 lies the transition region, and for Re > 4000 the flow is usually fully turbulent (subject to entrance effects and disturbances). Therefore “laminar ceases” is commonly associated with Re ≈ 2100. Step-by-Step Solution: Identify standard thresholds: laminar Re < 2100; transition 2100–4000; turbulent Re > 4000.Hence, laminar regime ceases to exist beyond about Re ≈ 2100. Verification / Alternative check: Classic experiments by Osborne Reynolds and subsequent standard texts adopt these nominal limits for smooth pipes. Why Other Options Are Wrong: 4000/3000/1500/1000: do not match widely used criterion for laminar cessation in pipes. Common Pitfalls: Confusing entrance length or surface roughness effects with the fundamental regime thresholds; exact transition can vary, but 2100 is the conventional design cut-off. Final Answer: Re ≈ 2100

Question 5

Settling regimes and size dependence: the terminal velocity of a particle moving in a fluid varies as d_p^n. What is n for Newton’s law regime (constant drag coefficient, high Re)?

Accepted Answer

Introduction: Terminal settling velocity depends on particle size differently in different drag regimes. Recognizing the exponent n in v_t ∝ d_p^n helps choose the correct correlation for classifier and cyclone design when Reynolds numbers are high and drag coefficient is nearly constant (Newton regime). Given Data / Assumptions: Single rigid particle settling in a fluid. Newton’s law regime (high particle Reynolds number). Drag coefficient C_d ≈ constant for the relevant Re range. Concept / Approach: At terminal velocity, drag balances apparent weight: (1/2) C_d ρ_f A v_t^2 = (π/6) d_p^3 (ρ_p − ρ_f) g. Solving for v_t gives v_t ∝ sqrt(d_p). Hence n = 0.5 in the Newton regime, contrasting with Stokes regime (n = 2) at low Re. Step-by-Step Solution: Write force balance: (1/2) C_d ρ_f (π d_p^2/4) v_t^2 = (π d_p^3/6) (ρ_p − ρ_f) g.Cancel common factors and rearrange: v_t^2 ∝ d_p.Therefore v_t ∝ d_p^0.5, i.e., n = 0.5. Verification / Alternative check: Dimensional analysis at high Re with constant C_d leads to v_t ∝ [ (ρ_p − ρ_f) g d_p / ρ_f ]^0.5, again yielding the square-root dependence. Why Other Options Are Wrong: n = 1, 1.5, 2, 3: correspond to other, non-Newtonian or low-Re situations; Stokes gives n = 2, not applicable here. Common Pitfalls: Applying Stokes’ n = 2 beyond its creeping-flow limit; always check Reynolds number regime before choosing n. Final Answer: 0.5

Question 6

Fluid classification by constitutive relation: a fluid in which shear stress is directly proportional to the velocity gradient (rate of shear) is called a

Accepted Answer

Introduction: Engineering correlations for pressure drop and heat/mass transfer depend on the constitutive behavior of the working fluid. This question checks recognition of the Newtonian model, the simplest and most widely used relation between shear stress and rate of shear. Given Data / Assumptions: Shear stress τ is proportional to velocity gradient du/dy. Constant proportionality (dynamic viscosity μ) independent of shear rate. Concept / Approach: A Newtonian fluid obeys τ = μ (du/dy). Examples include water, air, and many light oils at moderate conditions. Non-Newtonian fluids deviate from linearity or exhibit yield stress/time dependence (e.g., Bingham plastics, pseudoplastics, dilatants, thixotropic fluids). A perfect (inviscid) fluid has τ = 0 regardless of shear rate, an idealization not realized in practice. Step-by-Step Solution: Match definition: linear τ–(du/dy) with constant μ → Newtonian.Exclude models with yield stress (Bingham: τ = τ_y + μ_p du/dy), or shear-dependent μ (dilatant/pseudoplastic).Exclude inviscid: τ ≡ 0, not proportional to du/dy. Verification / Alternative check: Plotting τ versus du/dy gives a straight line through origin for Newtonian fluids; other models show offsets (yield) or curvature (shear-thinning/thickening). Why Other Options Are Wrong: Bingham plastic: needs a yield stress before flow begins. Perfect fluid: no viscous stresses at all. Dilatant/thixotropic: non-linear or time-dependent behavior. Common Pitfalls: Assuming all common liquids are Newtonian at all conditions; many slurries, polymer solutions, and pastes are non-Newtonian. Final Answer: Newtonian fluid

Question 7

Fluid mechanics — dimensional analysis refresher What is the correct fundamental dimensional formula for absolute (dynamic) viscosity μ in terms of mass (M), length (L), and time (T)?

Accepted Answer

Introduction / Context: Absolute (dynamic) viscosity μ links shear stress to velocity gradient in Newtonian fluids via Newton's law of viscosity. Knowing its dimensions is essential for checking equations, converting units (Pa·s, N·s/m^2, kg·m^-1·s^-1), and building dimensionless groups such as Reynolds number. Given Data / Assumptions: We are using fundamental dimensions: mass (M), length (L), and time (T). Newton's law of viscosity: τ = μ * (du/dy). Shear stress τ has units of force/area. Concept / Approach: Start from τ with dimensions of pressure: M L^-1 T^-2. The velocity gradient du/dy has dimensions T^-1. Solving μ = τ / (du/dy) gives the dimensional formula for viscosity. This approach prevents memorization errors and is reliable for cross-checking. Step-by-Step Solution: Write τ dimensions: τ = Force/Area = (M L T^-2)/L^2 = M L^-1 T^-2.Velocity gradient du/dy = (L T^-1)/L = T^-1.Compute μ: μ = τ / (du/dy) = (M L^-1 T^-2)/(T^-1) = M L^-1 T^-1. Verification / Alternative check: Use SI units: 1 Pa·s = 1 N·s/m^2 = (kg·m·s^-2)·s·m^-2 = kg·m^-1·s^-1, which maps to M L^-1 T^-1. Why Other Options Are Wrong: MLT^-1, ML^-1T, MLT: do not match the derived expression; they misplace the exponents or omit the negative exponent on time or length. Common Pitfalls: Confusing dynamic viscosity with kinematic viscosity ν, which has dimensions L^2 T^-1. Dropping the negative sign on the length exponent when expanding pressure units. Final Answer: ML-1T-1

Question 8

Open channel design — most economical trapezoidal section For the most economical trapezoidal channel, what is the ratio of flow depth y to hydraulic radius R (i.e., y : R)?

Accepted Answer

Introduction / Context: “Most economical” sections minimize wetted perimeter for a given area (or maximize discharge for a given cross section) and are widely used in irrigation and stormwater channels. Recognizing the characteristic ratios lets you size channels quickly without lengthy calculus. Given Data / Assumptions: Trapezoidal open-channel flow, steady uniform conditions. Hydraulic radius R = A/P, where A is area, P is wetted perimeter. We ask for the optimal ratio y:R at the most economical condition. Concept / Approach: For the most economical trapezoidal (as for the rectangular optimal case), geometry leads to the result R = y/2. This stems from minimizing P for a fixed A using first-order optimality conditions on the geometric relations for trapezoids with side slope m (horizontal:vertical). Step-by-Step Solution: Recall optimal relations: (i) half the top width equals one sloping side; (ii) hydraulic radius at optimum is R = y/2.Therefore, y : R = y : (y/2) = 2 : 1.Hence the required ratio equals 2. Verification / Alternative check: Derivations via Manning/Chézy with Lagrange multipliers (minimizing wetted perimeter at fixed area) reproduce R = y/2 for the trapezoidal best section. Why Other Options Are Wrong: 0.5, 1, 1.5: inconsistent with the known optimum R = y/2. Common Pitfalls: Confusing the condition for maximum velocity with that for maximum discharge. Applying rectangular-channel results blindly; fortunately, both optimal rectangular and trapezoidal sections share R = y/2. Final Answer: 2

Question 9

Vortices in fluid mechanics — identifying correct statements Which statement about forced and free vortices, and flow regimes, is correct?

Accepted Answer

Introduction / Context: Vortical motion is common in mixers, drains, and atmospheric flows. Distinguishing forced vs. free vortices and recalling basic laminar/turbulent features are foundational for fluid engineers. Given Data / Assumptions: Definitions of forced vortex (solid-body rotation) and free vortex (irrotational swirl). Newton's law for Newtonian fluids. Characteristics of turbulent flow. Concept / Approach: Forced vortex: every fluid element spins with constant angular velocity, like a solid disc (v ∝ r). Free vortex: angular momentum is conserved and tangential speed varies as vθ ∝ 1/r. Newton's law of viscosity holds for Newtonian fluids, including in laminar regimes. Turbulence is replete with eddies and cross-currents. Step-by-Step Solution: Evaluate (a): matches definition of a forced vortex → true.Evaluate (b): false; Newton's law defines laminar shear behavior in Newtonian fluids.Evaluate (c): describes forced, not free, vortex → false.Evaluate (d): turbulence is characterized by eddies and fluctuations → false. Verification / Alternative check: Plotting vθ vs. r: solid-body linear profile for forced vortex; hyperbolic for free vortex confirms the definitions. Why Other Options Are Wrong: (b) Misstates Newtonian behavior. (c) Confuses forced and free vortex characteristics. (d) Contradicts the very definition of turbulence. Common Pitfalls: Assuming any rotating flow is a forced vortex; the driving mechanism matters. Final Answer: A forced vortex occurs when fluid rotates as a solid about an axis.

Question 10

Pumping systems — function of a foot valve on the suction line Foot valves fitted to pump suctions serve what basic valve function in the flow circuit?

Accepted Answer

Introduction / Context: Foot valves are installed at the end of suction lines (often with a strainer) to maintain pump priming. Recognizing the functional class of this device helps in troubleshooting loss of prime and backflow problems. Given Data / Assumptions: Foot valve sits at the suction inlet, typically submerged. It allows upward flow into the line but blocks reverse flow out of the line. No active actuation; it is a check-type device. Concept / Approach: A foot valve is essentially a check (non-return) valve. In valve-function taxonomy, any device that enforces one-way flow provides directional control. It prevents drain-back of liquid from the suction line and preserves prime when the pump is stopped. Step-by-Step Solution: Identify one-way function → non-return action.Map non-return to directional control category.Therefore, the best match among the options is “directional control.” Verification / Alternative check: Common pump manuals specify “foot valve (check valve with strainer)” for suction lift applications. Why Other Options Are Wrong: Relief: limits pressure; not its role. Three/four way: used for flow routing in circuits; not applicable. Pressure reducing: maintains downstream pressure; not applicable. Common Pitfalls: Confusing specific valve names with function classes. The functional category here is one-way (directional) control. Final Answer: directional control

Fluid Mechanics Questions