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Topic 18 of 19

Fluid Mechanics

Fluid mechanics describes the behavior of liquids and gases — both at rest (hydrostatics) and in motion (hydrodynamics). The key results — hydrostatic pressure, Archimedes' principle, and Bernoulli's equation — have applications from ships and submarines to aircraft wings and blood flow.

Key Concepts

Pressure

Force per unit area:

P = F/A

(Pa = N/m²). Pressure in a fluid is isotropic (acts equally in all directions at a point). In a static fluid, pressure increases with depth:

P = P_0 + \rho g h

Pascal's Principle

A change in pressure applied to an enclosed fluid is transmitted undiminished to every part of the fluid. Basis of hydraulic lifts:

F_1/A_1 = F_2/A_2

Archimedes' Principle

A submerged or floating object experiences an upward buoyant force equal to the weight of the fluid it displaces:

F_b = \rho_\text{fluid}\,V_\text{submerged}\,g

Equation of Continuity

For incompressible steady flow, the volume flow rate is constant:

A_1 v_1 = A_2 v_2

. A narrower pipe means faster flow.

Bernoulli's Equation

For steady, incompressible, non-viscous flow along a streamline:

P + \frac{1}{2}\rho v^2 + \rho g h = \text{const}

. Higher speed → lower pressure (the venturi effect, airplane lift).

Viscosity

Internal friction in a fluid resisting flow. Real fluids (unlike the ideal Bernoulli fluid) exert viscous drag. Laminar vs. turbulent flow is characterized by the Reynolds number.

Key Equations

Hydrostatic pressure

P = P_0 + \rho g h

Pressure at depth h below the surface; P₀ is atmospheric (surface) pressure.

Buoyant force

F_b = \rho_\text{fluid}\,V_\text{displaced}\,g

Archimedes' principle. Object floats when F_b equals its weight.

Continuity equation

A_1 v_1 = A_2 v_2 \quad (\text{incompressible})

Conservation of volume flow rate Q = Av.

Bernoulli's equation

P + \tfrac{1}{2}\rho v^2 + \rho g h = \text{const}

Energy conservation per unit volume along a streamline for ideal (inviscid, incompressible) flow.

Torricelli's theorem

v_\text{efflux} = \sqrt{2gh}

Speed of fluid exiting a hole at depth h below the free surface (from Bernoulli with P equal on both sides).

Worked Example

Water Tank Draining

Problem

A large open water tank has a hole 2 m below the water surface. Find the speed of water exiting the hole.

Solution

Apply Bernoulli between the free surface (1) and the hole (2). Both at atmospheric pressure; surface speed ≈ 0 (large tank):

P_0 + 0 + \rho g h = P_0 + \tfrac{1}{2}\rho v^2 + 0

Simplify:

v = \sqrt{2gh} = \sqrt{2\times9.8\times2} \approx 6.26 \text{ m/s}

Answer Water exits at ≈ 6.26 m/s (Torricelli's theorem).

Practice

Exercises

7 problems

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2 free · 5 Pro

Exercise 1 / 7 Free

+10 XP

An object is fully submerged in water (ρ = 1000 kg/m³). The submerged volume is V = 0.001 m³. Find the buoyant force Fb = ρgV (Archimedes' principle).

Depth: 40%

Fb = N

Exercise 2 / 7 Free

+10 XP

Fluid flows through a narrowing pipe. A₁ = 4 m², v₁ = 2 m/s, A₂ = 1 m². By the continuity equation A₁v₁ = A₂v₂, find v₂.

v₂ = m/s

3 of 7

Water flows through a pipe with cross-sectional area $A_1 = 0.04$ m² at speed $v_1 = 2$ m/s. The pipe narrows to $A_2 = 0.01$ m². What is the speed (in m/s) at the narrow section?

Unlock Exercise 3