Fluid Dynamics

Fluid dynamics studies the motion of liquids and gases, underpinning phenomena like airplane lift, hydraulic systems, and weather patterns. This MathMultiverse guide covers Bernoulli’s principle, continuity equation, pressure-depth relation, and buoyant force, with detailed examples, calculations, and applications in engineering, meteorology, and daily life, enhanced with interactive visualizations.

Key Principles

Bernoulli’s Principle

For an inviscid, incompressible fluid:

\[ P + \frac{1}{2} \rho v^2 + \rho g h = \text{constant} \]

Where \( P \): pressure (Pa), \( \rho \): density (kg/m³), \( v \): velocity (m/s), \( g = 9.81 \, \text{m/s}^2 \), \( h \): height (m).

Continuity Equation

Conservation of mass:

\[ A_1 v_1 = A_2 v_2 \]

Pressure-Depth Relation

Pressure increases with depth:

\[ P = P_0 + \rho g h \]

Buoyant Force

Archimedes’ principle:

\[ F_b = \rho_{\text{fluid}} V_{\text{submerged}} g \]

Examples

Bernoulli’s Principle

Water (\( \rho = 1000 \, \text{kg/m}^3 \)) at \( v_1 = 2 \, \text{m/s} \), \( h_1 = 0 \), \( P_1 = 101325 \, \text{Pa} \):

\[ P_1 + \frac{1}{2} \rho v_1^2 + \rho g h_1 = 101325 + \frac{1}{2} (1000) (2)^2 + 0 = 103325 \, \text{Pa} \]

Velocity vs. Pressure (Bernoulli)

Pressure decreases as velocity increases.

Continuity Equation

Water in a pipe: \( A_1 = 0.05 \, \text{m}^2 \), \( v_1 = 3 \, \text{m/s} \), \( A_2 = 0.02 \, \text{m}^2 \):

\[ A_1 v_1 = A_2 v_2 \implies v_2 = \frac{(0.05)(3)}{0.02} = 7.5 \, \text{m/s} \]

Pressure-Depth

Water (\( \rho = 1000 \, \text{kg/m}^3 \)) at \( h = 10 \, \text{m} \), \( P_0 = 101325 \, \text{Pa} \):

\[ P = P_0 + \rho g h = 101325 + (1000)(9.81)(10) = 199425 \, \text{Pa} \]

Buoyant Force

Object (\( V = 0.1 \, \text{m}^3 \)) in water (\( \rho = 1000 \, \text{kg/m}^3 \)):

\[ F_b = \rho_{\text{fluid}} V_{\text{submerged}} g = (1000)(0.1)(9.81) = 981 \, \text{N} \]

Applications

Airplane Wing Lift

Air (\( \rho = 1.225 \, \text{kg/m}^3 \)), \( v_1 = 100 \, \text{m/s} \), \( v_2 = 90 \, \text{m/s} \), \( h_1 = h_2 \):

\[ P_1 - P_2 = \frac{1}{2} \rho (v_2^2 - v_1^2) = \frac{1}{2} (1.225) (8100 - 10000) \approx -1163.75 \, \text{Pa} \]

Pipe Flow

Water: \( A_1 = 0.1 \, \text{m}^2 \), \( v_1 = 2 \, \text{m/s} \), \( A_2 = 0.04 \, \text{m}^2 \):

\[ v_2 = \frac{(0.1)(2)}{0.04} = 5 \, \text{m/s} \]

Submarine Depth

Seawater (\( \rho = 1025 \, \text{kg/m}^3 \)), \( h = 200 \, \text{m} \), \( P_0 = 101325 \, \text{Pa} \):

\[ P = 101325 + (1025)(9.81)(200) = 2112375 \, \text{Pa} \]

Floating Object

Block (\( V = 0.05 \, \text{m}^3 \)) in oil (\( \rho = 850 \, \text{kg/m}^3 \)):

\[ F_b = (850)(0.05)(9.81) \approx 416.93 \, \text{N} \]