Fluid Mechanics

Fluid Mechanics

🟢 Lite — Quick Review (1h–1d)

Rapid summary for last-minute revision before your exam.

Fluid mechanics splits into fluid statics (fluids at rest) and fluid dynamics (fluids in motion). At rest, pressure at depth h in a liquid of density ρ is P = hρg, and it acts equally in all directions (Pascal’s law). An immersed body loses weight equal to the fluid it displaces (Archimedes’ principle). In motion, two equations dominate JEE Advanced: the continuity equation A₁v₁ = A₂v₂ (mass conservation for an incompressible fluid) and Bernoulli’s equation P + ½ρv² + ρgh = constant along a streamline (energy conservation per unit volume). For viscous flow, Stokes’ law F = 6πηrv governs drag on a sphere, while Poiseuille’s equation Q = πPr⁴/8ηl gives flow rate through a cylindrical pipe. Remember that the pressure jump across a soap bubble is 4T/R (two surfaces) but only 2T/R across a liquid drop (one surface) — a classic trap.

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🟡 Standard — Regular Study (2d–2mo)

Standard content for students with a few days to months.

Hydrostatics and Pascal’s Law

At a point inside a static fluid, the pressure is the same in every direction — this isotropy of pressure is Pascal’s law. In a gravitational field the pressure increases with depth according to P = P₀ + hρg, where P₀ is the pressure at the free surface. A consequence: a hydraulic lift multiplies force by the ratio of piston areas, F₂/F₁ = A₂/A₁, since pressure transmitted is unchanged.

Archimedes’ Principle

A body submerged (partially or fully) in a fluid experiences an upward buoyant force equal to the weight of displaced fluid: F_b = V_displaced · ρ_fluid · g. The body’s apparent weight is W − F_b. If ρ_body < ρ_fluid the body floats with fraction ρ_body/ρ_fluid of its volume submerged.

Equation of Continuity

For a steady flow of an incompressible fluid through a pipe of varying cross-section, mass conservation demands that the volume flow rate Q = Av is constant. Hence A₁v₁ = A₂v₂: speed is higher where the cross-section is smaller.

Bernoulli’s Equation

Applying the work–energy theorem to a fluid element moving along a streamline of a steady, incompressible, non-viscous flow yields P + ½ρv² + ρgh = constant. This is energy per unit volume conserved along one streamline. Practical applications include the Venturi meter (flow speed from pressure difference between a wide and a narrow section), Torricelli’s theorem v = √(2gh) for efflux from a tank, and the Pitot tube measuring airspeed from the difference between stagnation and static pressure.

Viscous Flow

Real fluids have internal friction characterized by viscosity η (Pa·s). For a sphere of radius r moving slowly through a fluid, viscous drag is F = 6πηrv (Stokes’ law), valid when flow is laminar. Through a cylindrical pipe, the Hagen–Poiseuille law gives flow rate Q = πPr⁴/8ηl, showing that Q depends on the fourth power of radius — a small change in r dramatically alters flow.

Surface Tension and Capillarity

Molecules at a surface experience a net inward pull, creating surface tension T (N/m). The pressure difference across a curved interface is given by the Young–Laplace equation: ΔP = 2T/R for a liquid drop or a jet (one surface), and ΔP = 4T/R for a soap bubble (two surfaces). In a capillary tube of radius r with contact angle θ, the liquid rises (or depresses) by h = 2T cosθ/(rρg) — capillarity.

Worked Example

A horizontal pipe of cross-section 20 cm² narrows to 5 cm². Water (ρ = 1000 kg/m³) flows at 2 m/s in the wide section. By continuity, v₂ = 2 × 20/5 = 8 m/s. Applying Bernoulli’s equation (h constant): P₁ − P₂ = ½ρ(v₂² − v₁²) = ½ × 1000 × (64 − 4) = 30 000 Pa.

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🔴 Extended — Deep Study (3mo+)

Comprehensive coverage for students on a longer study timeline.

When Bernoulli Fails

Bernoulli’s equation is inviscid and incompressible, valid strictly along a single streamline of steady, irrotational flow. It cannot be applied between two different streamlines because the constant can differ. In gas flow at speeds approaching the speed of sound, compressibility matters and the compressible Bernoulli equation with γ = cₚ/cᵥ must be used. For viscous internal flow, mechanical energy is dissipated and the Bernoulli equation with head loss is needed: P₁ + ½ρv₁² = P₂ + ½ρv₂² + ΔP_loss, where ΔP_loss is computed from Poiseuille’s law (laminar) or the Darcy–Weisbach equation (turbulent).

Reynolds Number — The Regime Decider

The dimensionless Reynolds number Re = ρvL/η (L is a characteristic length) determines flow regime. For flow in a pipe, Re < 2000 is laminar, Re > 4000 is turbulent; in between is the unstable transition region. Stokes’ law holds only at very low Re (≲ 1); for higher Re, drag transitions to a pressure-dominated form F ∝ ρv²A with a near-constant drag coefficient. JEE Advanced problems occasionally ask students to verify the regime before choosing the right law.

Edge Cases and Traps

1. Gauge vs absolute pressure. Manometer readings give gauge pressure; add P_atm for absolute. Bernoulli’s equation in its standard form uses absolute pressure.
2. Soap bubble vs liquid drop vs air cavity inside liquid. Drop: ΔP = 2T/R. Bubble: ΔP = 4T/R. Cavity (single gas–liquid interface): ΔP = 2T/R.
3. Atmospheric pressure in Torricelli problems. A small hole near the surface of an open tank: applied pressure difference is ρgH, not ρgH + P_atm, because P_atm cancels.
4. Stokes’ terminal velocity: A sphere falling under gravity in a viscous medium reaches v_t = 2r²(ρ − σ)g/(9η), where σ is fluid density.
5. Capillarity sign. If θ > 90°, cosθ is negative and the liquid depresses (e.g., mercury in glass).

Connection to Other Topics

Fluid mechanics links to work–energy theorem (Bernoulli is essentially W–E per unit volume), Newton’s second law (viscous force and pressure gradient in Navier–Stokes), and thermodynamics (compressible flow, Mach number). Capillarity is governed by molecular cohesive and adhesive forces — the same physics as wetting and contact angle.

Common Mistakes

• Using continuity for compressible flow without multiplying by ρ.
• Applying Stokes’ law when Re ≫ 1 (turbulent regime).
• Treating atmospheric pressure as negligible in barometer-style sub-problems where it is the only contribution.
• Confusing the velocity of efflux (Torricelli, v = √(2gh)) with the velocity of the receding free surface (v/2, by continuity for a tank of larger cross-section).

Practice Prompts

1. A Pitot tube on an aircraft measures a pressure difference ΔP between stagnation and static ports. Derive the airspeed in terms of ΔP and air density ρ. (Answer: v = √(2ΔP/ρ).)
2. A U-tube manometer with mercury connects two points in a horizontal water pipe. The mercury level difference is 5 cm. Find the pressure difference in pascals. (Use ρ_Hg g h = 13 600 × 9.8 × 0.05 ≈ 6664 Pa.)

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