Fluid Dynamics

Fluid dynamics is a branch of physics that studies the behavior of liquids and gases in motion, exploring how they flow, interact with surfaces, and respond to forces. It underpins phenomena like lift in aerodynamics, pressure in hydraulics, and circulation in weather systems. This guide provides a comprehensive introduction to fluid dynamics, covering all basic formulas (Bernoulli’s principle, continuity equation, pressure-depth relation, buoyant force), detailed examples, and practical applications to illustrate its significance in engineering, meteorology, and everyday life.

Bernoulli’s Principle (with Continuity, Pressure-Depth, Buoyant Force)

Bernoulli’s principle states that for an inviscid, incompressible fluid, the total energy along a streamline is conserved:

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

Where:

\( P \): Pressure (Pa)
\( \rho \): Density (kg/m³)
\( v \): Velocity (m/s)
\( g = 9.81 \, \text{m/s}^2 \): Gravitational acceleration
\( h \): Height (m)

Related Formulas:

Continuity Equation: \( A_1 v_1 = A_2 v_2 \)
Pressure-Depth Relation: \( P = P_0 + \rho g h \)
Buoyant Force (Archimedes’ Principle): \( F_b = \rho_{\text{fluid}} V_{\text{submerged}} g \)

Examples

Example 1: Bernoulli’s Principle

Water (\( \rho = 1000 \, \text{kg/m}^3 \)) flows at \( v_1 = 2 \, \text{m/s} \), \( h_1 = 0 \), \( P_1 = 101325 \, \text{Pa} \). Find the total energy term:

\[ P_1 + \frac{1}{2} \rho v_1^2 + \rho g h_1 = \text{constant} \] \[ = 101325 + \frac{1}{2} (1000) (2)^2 + (1000) (9.81) (0) \] \[ = 101325 + \frac{1}{2} (1000) (4) + 0 \] \[ = 101325 + 2000 \] \[ = 103325 \, \text{Pa} \]

Example 2: Continuity Equation

Water flows through a pipe with \( A_1 = 0.05 \, \text{m}^2 \), \( v_1 = 3 \, \text{m/s} \), into a narrower section \( A_2 = 0.02 \, \text{m}^2 \). Find \( v_2 \):

\[ A_1 v_1 = A_2 v_2 \] \[ (0.05) (3) = (0.02) v_2 \] \[ 0.15 = 0.02 v_2 \] \[ v_2 = \frac{0.15}{0.02} \] \[ v_2 = 7.5 \, \text{m/s} \]

Example 3: Pressure-Depth Relation

Pressure at a depth \( h = 10 \, \text{m} \) in water (\( \rho = 1000 \, \text{kg/m}^3 \)), atmospheric pressure \( P_0 = 101325 \, \text{Pa} \):

\[ P = P_0 + \rho g h \] \[ = 101325 + (1000) (9.81) (10) \] \[ = 101325 + 98100 \] \[ = 199425 \, \text{Pa} \]

Example 4: Buoyant Force

Buoyant force on an object of volume \( V = 0.1 \, \text{m}^3 \) fully submerged in water (\( \rho = 1000 \, \text{kg/m}^3 \)):

\[ F_b = \rho_{\text{fluid}} V_{\text{submerged}} g \] \[ = (1000) (0.1) (9.81) \] \[ = 100 \cdot 9.81 \] \[ = 981 \, \text{N} \]

Applications

Fluid dynamics is essential in various fields, from engineering to environmental science. Below are examples with calculations:

Example 1: Airplane Wing Lift (Bernoulli’s Principle)

Air (\( \rho = 1.225 \, \text{kg/m}^3 \)) flows over a wing at \( v_1 = 100 \, \text{m/s} \), under at \( v_2 = 90 \, \text{m/s} \), \( h_1 = h_2 \). Pressure difference:

\[ P_1 + \frac{1}{2} \rho v_1^2 = P_2 + \frac{1}{2} \rho v_2^2 \] \[ P_1 - P_2 = \frac{1}{2} \rho (v_2^2 - v_1^2) \] \[ = \frac{1}{2} (1.225) (90^2 - 100^2) \] \[ = \frac{1}{2} (1.225) (8100 - 10000) \] \[ = \frac{1}{2} (1.225) (-1900) \] \[ \approx -1163.75 \, \text{Pa} \]

Example 2: Pipe Flow (Continuity Equation)

Water flows in a pipe with \( A_1 = 0.1 \, \text{m}^2 \), \( v_1 = 2 \, \text{m/s} \), into \( A_2 = 0.04 \, \text{m}^2 \). Find \( v_2 \):

\[ A_1 v_1 = A_2 v_2 \] \[ (0.1) (2) = (0.04) v_2 \] \[ 0.2 = 0.04 v_2 \] \[ v_2 = \frac{0.2}{0.04} \] \[ v_2 = 5 \, \text{m/s} \]

Example 3: Submarine Depth (Pressure-Depth)

Pressure on a submarine at \( h = 200 \, \text{m} \) in seawater (\( \rho = 1025 \, \text{kg/m}^3 \)), \( P_0 = 101325 \, \text{Pa} \):

\[ P = P_0 + \rho g h \] \[ = 101325 + (1025) (9.81) (200) \] \[ = 101325 + 2011050 \] \[ = 2112375 \, \text{Pa} \]

Example 4: Floating Object (Buoyant Force)

Buoyant force on a block (\( V = 0.05 \, \text{m}^3 \)) in oil (\( \rho = 850 \, \text{kg/m}^3 \)):

\[ F_b = \rho_{\text{fluid}} V_{\text{submerged}} g \] \[ = (850) (0.05) (9.81) \] \[ = 42.5 \cdot 9.81 \] \[ \approx 416.93 \, \text{N} \]

Example 5: Hydraulic Lift (Pressure)

Pressure at \( h = 1 \, \text{m} \) in hydraulic fluid (\( \rho = 900 \, \text{kg/m}^3 \)), \( P_0 = 101325 \, \text{Pa} \):

\[ P = P_0 + \rho g h \] \[ = 101325 + (900) (9.81) (1) \] \[ = 101325 + 8829 \] \[ = 110154 \, \text{Pa} \]

Example 6: Weather System (Bernoulli’s Principle)

Air (\( \rho = 1.225 \, \text{kg/m}^3 \)) at \( v_1 = 20 \, \text{m/s} \), \( h_1 = 0 \), \( P_1 = 100000 \, \text{Pa} \). Total energy term:

\[ P_1 + \frac{1}{2} \rho v_1^2 + \rho g h_1 = \text{constant} \] \[ = 100000 + \frac{1}{2} (1.225) (20)^2 + 0 \] \[ = 100000 + \frac{1}{2} (1.225) (400) \] \[ = 100000 + 245 \] \[ = 100245 \, \text{Pa} \]