Pressure and Pascal's Law

Overview

Fluid mechanics deals with the behavior of fluids (liquids and gases) at rest and in motion. Pressure is a fundamental concept that describes the force exerted by a fluid.

Pressure

Definition

$$P = \frac{F}{A}$$

Where:

• $P$ = pressure
• $F$ = force perpendicular to surface
• $A$ = area

Units

• SI unit: Pascal (Pa) = N/m²
• 1 atm = 101,325 Pa = 101.325 kPa
• 1 bar = 100,000 Pa
• 1 mmHg = 133.3 Pa
• 1 psi = 6895 Pa

Pressure in Fluids

Key Properties

• Pressure acts equally in all directions at a point
• Pressure is perpendicular to any surface
• Pressure increases with depth

Pressure at Depth

$$P = P_0 + \rho gh$$

Where:

• $P_0$ = pressure at surface
• $\rho$ = fluid density
• $g$ = gravitational acceleration
• $h$ = depth below surface

Pressure Difference

$$\Delta P = \rho g \Delta h$$

Atmospheric Pressure

Standard atmospheric pressure:

$$P_{\text{atm}} = 101{,}325 \text{ Pa} \approx 1.01 \times 10^5 \text{ Pa}$$

Equivalent to:

• 760 mmHg
• 10.3 m of water
• 14.7 psi

Gauge vs Absolute Pressure

Absolute Pressure

Total pressure including atmospheric:

$$P_{\text{abs}} = P_{\text{atm}} + P_{\text{gauge}}$$

Gauge Pressure

Pressure relative to atmospheric:

$$P_{\text{gauge}} = P_{\text{abs}} - P_{\text{atm}} = \rho gh$$

Pascal's Law

Pressure applied to an enclosed fluid is transmitted equally to all parts of the fluid.

$$\Delta P_1 = \Delta P_2$$

Hydraulic System

$$\frac{F_1}{A_1} = \frac{F_2}{A_2}$$ $$F_2 = F_1 \times \frac{A_2}{A_1}$$

Mechanical advantage:

$$MA = \frac{F_2}{F_1} = \frac{A_2}{A_1}$$

Work Conservation

$$W_1 = W_2$$ $$F_1 d_1 = F_2 d_2$$

The small piston moves farther than the large piston.

Pressure Measurement

Manometer

Open-tube manometer:

$$P = P_{\text{atm}} + \rho gh$$

U-tube manometer:

$$P_1 - P_2 = \rho gh$$

Barometer

Mercury barometer:

$$P_{\text{atm}} = \rho_{Hg} \times g \times h$$

At sea level: $h \approx 760$ mm = 0.76 m

Examples

Example 1: Pressure at Depth

Find pressure at 10 m depth in a lake. ($\rho_{\text{water}} = 1000$ kg/m³)

$$P = P_{\text{atm}} + \rho gh$$ $$P = 101325 + 1000 \times 9.8 \times 10$$ $$P = 101325 + 98000 = 199{,}325 \text{ Pa} \approx 2 \text{ atm}$$

Example 2: Hydraulic Lift

A hydraulic lift has pistons of radius 2 cm and 20 cm. Force on small piston is 100 N.

$$A_1 = \pi(0.02)^2 = 1.26 \times 10^{-3} \text{ m}^2$$ $$A_2 = \pi(0.20)^2 = 0.126 \text{ m}^2$$ $$F_2 = F_1 \times \frac{A_2}{A_1} = 100 \times 100 = 10{,}000 \text{ N}$$

This can lift a 1000 kg car!

Example 3: Mercury Barometer

What height of mercury corresponds to 1 atm? ($\rho_{Hg} = 13{,}600$ kg/m³)

$$h = \frac{P_{\text{atm}}}{\rho g} = \frac{101325}{13600 \times 9.8} = 0.760 \text{ m} = 760 \text{ mm}$$

Example 4: Water Barometer

What height of water corresponds to 1 atm?

$$h = \frac{P_{\text{atm}}}{\rho g} = \frac{101325}{1000 \times 9.8} = 10.3 \text{ m}$$

Example 5: Dam Pressure

Find average pressure on a 50 m tall dam.

Average depth = 25 m

$$P_{\text{avg}} = \rho gh = 1000 \times 9.8 \times 25 = 245{,}000 \text{ Pa}$$

Example 6: Hydraulic Brake

Brake pedal applies 50 N to a 1 cm² piston. Brake pads have 10 cm² pistons.

$$F_{\text{brake}} = 50 \times \frac{10}{1} = 500 \text{ N per wheel}$$

Applications

• Hydraulic jacks and lifts: Multiply force
• Hydraulic brakes: Distribute force to all wheels
• Blood pressure: Measured in mmHg
• Scuba diving: Pressure increases with depth
• Weather systems: High/low pressure areas