Fluid Dynamics

Overview

Fluid dynamics studies fluids in motion. Key principles include conservation of mass (continuity) and conservation of energy (Bernoulli's equation).

Ideal Fluid Assumptions

• Incompressible (constant density)
• Non-viscous (no internal friction)
• Steady flow (velocity constant at each point)
• Irrotational (no turbulence)

Flow Rate

Volume Flow Rate

$$Q = Av$$

Where:

• $Q$ = volume flow rate (m³/s)
• $A$ = cross-sectional area
• $v$ = fluid velocity

Mass Flow Rate

$$\dot{m} = \rho Av = \rho Q$$

Continuity Equation

Conservation of mass for incompressible flow:

$$A_1 v_1 = A_2 v_2$$ $$Q = \text{constant}$$

This means:

• Narrow sections have faster flow
• Wide sections have slower flow

Bernoulli's Equation

Conservation of energy along a streamline:

$$P + \frac{1}{2}\rho v^2 + \rho gh = \text{constant}$$

Or between two points:

$$P_1 + \frac{1}{2}\rho v_1^2 + \rho gh_1 = P_2 + \frac{1}{2}\rho v_2^2 + \rho gh_2$$

Terms

• $P$ = static pressure (pressure energy per volume)
• $\frac{1}{2}\rho v^2$ = dynamic pressure (kinetic energy per volume)
• $\rho gh$ = hydrostatic pressure (potential energy per volume)

Special Cases of Bernoulli's Equation

Horizontal Flow ($h_1 = h_2$)

$$P_1 + \frac{1}{2}\rho v_1^2 = P_2 + \frac{1}{2}\rho v_2^2$$

Faster flow → lower pressure

Static Fluid ($v = 0$)

$$P + \rho gh = \text{constant}$$

This gives the hydrostatic equation.

Torricelli's Theorem

For fluid escaping from a hole in a tank:

$$v = \sqrt{2gh}$$

Where $h$ is the depth of the hole below the surface.

Venturi Effect

In a constriction:

• Velocity increases (by continuity)
• Pressure decreases (by Bernoulli)

Applications: carburetors, aspirators, atomizers

Lift Force

Pressure difference creates lift:

$$F_{\text{lift}} = (P_{\text{bottom}} - P_{\text{top}}) \times A$$

By Bernoulli (if $v_{\text{top}} > v_{\text{bottom}}$):

$$F_{\text{lift}} = \frac{1}{2}\rho(v_{\text{top}}^2 - v_{\text{bottom}}^2) \times A$$

Viscosity

Real fluids have internal friction (viscosity):

Viscous Force

$$F = \eta A \frac{dv}{dy}$$

Where $\eta$ = coefficient of viscosity (Pa·s)

Poiseuille's Law (Flow in Pipes)

$$Q = \frac{\pi r^4 \Delta P}{8\eta L}$$

Flow rate strongly depends on radius ($r^4$)!

Reynolds Number

Predicts laminar vs turbulent flow:

$$Re = \frac{\rho v D}{\eta}$$

Where $D$ = characteristic length (pipe diameter)

• $Re < 2000$: Laminar flow
• $2000 < Re < 4000$: Transition
• $Re > 4000$: Turbulent flow

Examples

Example 1: Continuity

Water flows through a hose (radius 1 cm) at 2 m/s. It exits through a nozzle (radius 0.5 cm). Find exit velocity.

$$A_1 v_1 = A_2 v_2$$ $$\pi(0.01)^2 \times 2 = \pi(0.005)^2 \times v_2$$ $$v_2 = 2 \times \left(\frac{0.01}{0.005}\right)^2 = 8 \text{ m/s}$$

Example 2: Bernoulli - Water Main

Water flows at 1.5 m/s in a 10 cm pipe at ground level ($P = 300$ kPa). Find pressure at 5 m height in a 5 cm pipe.

By continuity:

$$v_2 = v_1 \frac{A_1}{A_2} = 1.5 \times \left(\frac{10}{5}\right)^2 = 6 \text{ m/s}$$

By Bernoulli:

$$P_2 = P_1 + \frac{1}{2}\rho(v_1^2 - v_2^2) - \rho gh$$ $$P_2 = 300000 + \frac{1}{2}(1000)(1.5^2 - 6^2) - 1000(9.8)(5)$$ $$P_2 = 300000 - 16875 - 49000 = 234{,}125 \text{ Pa} \approx 234 \text{ kPa}$$

Example 3: Torricelli's Theorem

A tank has a small hole 3 m below the water surface. Find exit velocity.

$$v = \sqrt{2gh} = \sqrt{2 \times 9.8 \times 3} = 7.67 \text{ m/s}$$

Example 4: Airplane Wing

Air flows at 250 m/s over a wing and 200 m/s under it. Wing area = 20 m². Find lift. ($\rho_{\text{air}} = 1.2$ kg/m³)

$$\Delta P = \frac{1}{2}\rho(v_{\text{top}}^2 - v_{\text{bottom}}^2) = \frac{1}{2}(1.2)(250^2 - 200^2)$$ $$\Delta P = 0.6 \times 22500 = 13{,}500 \text{ Pa}$$ $$F = \Delta P \times A = 13500 \times 20 = 270{,}000 \text{ N} = 270 \text{ kN}$$

Example 5: Venturi Meter

A venturi meter has diameters 10 cm and 5 cm. Height difference in manometer is 15 cm (mercury). Find flow rate.

$$\Delta P = \rho_{Hg} \times g \times h = 13600 \times 9.8 \times 0.15 = 19{,}992 \text{ Pa}$$

Using Bernoulli and continuity:

$$v_1 = \sqrt{\frac{2\Delta P}{\rho[(A_1/A_2)^2 - 1]}}$$ $$v_1 = \sqrt{\frac{2 \times 19992}{1000 \times 15}} = 1.63 \text{ m/s}$$ $$Q = A_1 v_1 = \pi(0.05)^2 \times 1.63 = 0.0128 \text{ m}^3/\text{s}$$

Applications

• Wings and airfoils: Generate lift
• Pitot tubes: Measure airspeed
• Carburetors: Mix fuel and air
• Blood flow: Diagnose arterial blockages
• Chimneys: Create draft