Chapter 7: Pressure

Overview

This chapter explores the fundamental concepts of pressure in fluids and gases, which are essential for understanding fluid mechanics, hydraulic systems, buoyancy, and many real-world applications. From the pressure experienced deep underwater to the principles behind airplane flight and hydraulic machinery, pressure concepts are crucial in both everyday life and advanced engineering.

Learning Objectives

After completing this chapter, you will be able to:

Calculate pressure in liquids using the hydrostatic pressure formula
Understand atmospheric pressure and its measurement
Apply Pascal's principle in hydraulic systems
Use Archimedes' principle to determine buoyant forces
Explain Bernoulli's principle and its applications
Solve problems involving pressure in various fluid systems

Pressure in Liquids

Main Concept

Pressure in liquids is caused by the weight of the liquid column above a point. It acts in all directions.

Key Principles

Liquid pressure increases with depth and liquid density
At the same depth, pressure is the same at all points
Pressure is independent of the shape of the container

Key Formulas

Liquid Pressure, P:

P = hρg

Where:

$P$ = Pressure in Pascal (Pa)
$h$ = Depth of liquid (m)
$ρ$ = Density of liquid (kg m⁻³)
$g$ = Gravitational acceleration (9.81 m s⁻²)

Important Terms

Density, $ρ$ : Mass per unit volume. $ρ = m/V$

Pressure in Liquids Diagrams

Pressure Depth Relationship

Pressure vs Depth Relationship

The graph of pressure vs depth is a straight line passing through the origin, with slope = ρg.

Example: At what depth in water will the pressure be 200,000 Pa?

P = 200,000 Pa
ρ_water = 1000 kg m⁻³
g = 9.81 m s⁻²

h = \frac{P}{ρg} = \frac{200,000}{1000 \times 9.81} = 20.39 \text{ m}

Atmospheric Pressure

Main Concept

Atmospheric pressure is the pressure caused by the weight of the air column acting on the Earth's surface.

Key Principles

Atmospheric pressure decreases with altitude (height)
Standard atmospheric pressure at sea level is approximately 101,325 Pa or 1 atm, which can support a mercury column of 76 cm height

Key Formulas

$P_atm ≈ 1 × 10⁵ Pa$
Measured using mercury barometer or aneroid barometer

Important Terms

Barometer: Instrument to measure atmospheric pressure
Manometer: Instrument to measure gas pressure

Atmospheric Pressure Diagrams

Atmospheric Pressure Visualization

Atmospheric Pressure Variations

Altitude (m)	Pressure (Pa)	Pressure (atm)
Sea level	101,325	1.00
1000	89,874	0.89
2000	79,495	0.78
5000	53,920	0.53
10000	26,436	0.26

Gas Pressure

Main Concept

Gas pressure in a closed container is caused by the random collision of gas molecules with the container walls.

Key Principles

Gas laws (Boyle, Charles, Pressure) describe relationships between P, V, and T
Pressure increases with temperature and decreases with volume

Key Formulas

Using Manometer:

$P_gas = P_atm + hρg$ (if liquid level in arm connected to gas is lower)

Important Terms

Bourdon Gauge: Instrument to measure high gas pressure

Pascal's Principle

Main Concept

Pressure applied to a confined liquid is transmitted uniformly to all parts of the liquid and also to the container walls.

Key Principles

This principle is the basis for hydraulic systems, which use liquids to transmit force
Hydraulic systems act as force multipliers. A small force on a small piston produces a large force on a large piston

Key Formulas

$P_1 = P_2$
$F_1/A_1 = F_2/A_2$
Force Multiplication Factor: $F_2/F_1 = A_2/A_1$

Where:

$F_1$ , $F_2$ = Force on small and large pistons
$A_1$ , $A_2$ = Cross-sectional areas of small and large pistons

Hydraulic Applications

Hydraulic Lift:

Small force creates large lifting force
Used in car jacks, hydraulic presses, excavators

Advantages:

Force multiplication
Smooth operation
High efficiency

Worked Example

Problem: A hydraulic system has pistons with areas 0.01 $m^2$ and 0.5 $m^2$ . If a force of 500 N is applied to the small piston, calculate the force on the large piston.

Solution:

$F_1$ = 500 N
$A_1$ = 0.01 $m^2$
$A_2$ = 0.5 $m^2$

Using Pascal's principle:

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

F_2 = F_1 \times \frac{A_2}{A_1} = 500 \times \frac{0.5}{0.01} = 25,000 \text{ N}

Answer: Force on large piston = 25,000 N

Archimedes' Principle

Main Concept

When an object is fully or partially immersed in a fluid, it experiences an upthrust force equal to the weight of the displaced fluid.

Key Principles

Buoyant Force: Upward force acting against the object's weight in fluid
Object will float if buoyant force equals its weight
Object will sink if its weight is greater than the buoyant force

Key Formulas

Buoyant Force, F_B:

F_B = Weight of displaced fluid = Vρg

Where:

$V$ = Volume of displaced fluid (equal to volume of submerged part of object)
$ρ$ = Density of fluid
$g$ = Gravitational acceleration

Important Terms

Hydrometer: Instrument to measure relative density of liquids based on Archimedes' principle
Plimsoll Line: Mark on ship hull to show safe loading level in water of different densities

Floatation Conditions

Condition	Result
Weight = Buoyant Force	Floating
Weight > Buoyant Force	Sinking
Weight < Buoyant Force	Rising

Real Applications

Ships:

Designed with large hull volume to displace enough water
Plimsoll line ensures safe loading

Submarines:

Use ballast tanks to control buoyancy
Can float, sink, or remain suspended

Balloons:

Hot air balloons displace more air than their weight
Helium balloons are less dense than air

Bernoulli's Principle

Main Concept

In steady fluid flow, fluid pressure is lower where the fluid flows with higher velocity, and vice versa.

Key Principles

This principle is a manifestation of energy conservation in moving fluid
As fluid velocity ( $v$ ) increases, pressure ( $P$ ) decreases

Key Formulas

No specific calculation formulas at SPM level, but the concept is qualitative.

Important Terms

Aerofoil: Cross-sectional shape of airplane wings designed to produce lift. Air flows faster over the curved surface, creating low pressure, while air underneath flows slower with higher pressure. This pressure difference produces lift.
Venturi Tube: Tube that narrows in the middle, showing fluid accelerates in the narrow section, and pressure decreases.

Applications of Bernoulli's Principle

Airplane Flight: Wings generate lift through pressure differences
Atomizers: Spray bottles use fast air flow to create low pressure
Carburetors: Mix fuel and air using pressure differences
Meteorology: Explains wind patterns and weather systems
Sports: Baseball curveballs, golf ball dimples

Venturi Effect

Venturi Tube:

Narrow tube with varying cross-section
Fluid accelerates through narrow section
Pressure decreases where velocity is high

Formula:

P_1 + \frac{1}{2}ρv_1^2 = P_2 + \frac{1}{2}ρv_2^2

SPM Exam Tips

Common Mistakes to Avoid

Pressure Direction: Remember pressure in fluids acts in all directions
Units: Use Pascals (Pa) for pressure, not atmospheres in calculations
Buoyancy: Remember buoyant force equals weight of displaced fluid, not just fluid weight
Hydraulic Systems: Remember pressure is transmitted, not force

Problem-Solving Strategies

Identify the System: Determine if liquid pressure, gas pressure, hydraulic, or buoyancy problem
List Given Data: Write down all known quantities with units
Choose Appropriate Formula: Select the right equation for the situation
Apply Principles: Use Pascal's, Archimedes', or Bernoulli's principles as needed
Verify Results: Check if answer makes physical sense

Important Formula Summary

Concept	Formula
Liquid Pressure	$P = hρg$
Pascal's Principle	$F_1$ / $A_1$ = $F_2$ / $A_2$
Buoyant Force	$F_B = Vρg$
Atmospheric Pressure	$P_atm ≈ 1 × 10⁵ Pa$

Practical Applications

Real-World Examples

Diving: Pressure increases with depth, limits diving depth
Weather Systems: High and low pressure areas affect weather
Hydraulic Machinery: Car jacks, construction equipment
Ships and Submarines: Buoyancy principles for navigation
Aviation: Bernoulli's principle for flight

Safety Considerations

Pressure Vessels: Must be designed to withstand internal pressure
Deep Sea Operations: High pressure requires special equipment
Hydraulic Systems: Can be dangerous if pressure fails
Gas Storage: Proper safety valves to prevent overpressure

Summary

This chapter covered essential pressure concepts:

Liquid Pressure: Increases with depth and density
Atmospheric Pressure: Caused by air weight, decreases with altitude
Pascal's Principle: Pressure transmission in fluids, basis for hydraulic systems
Archimedes' Principle: Buoyancy and floatation
Bernoulli's Principle: Pressure-velocity relationship in fluid flow

Master these concepts to understand fluid mechanics, hydraulic systems, buoyancy, and aerodynamics - fundamental to many engineering and everyday applications.

Practice Questions

Calculate the pressure at a depth of 50 m in seawater (density = 1025 kg m⁻³).
A hydraulic lift has pistons with diameters 2 cm and 20 cm. If a force of 100 N is applied to the small piston, calculate the lifting force on the large piston.
A block of wood with volume 0.2 $m^3$ and density 600 kg m⁻³ is placed in water. Calculate: a) The buoyant force acting on it b) Whether it will float or sink
Explain why atmospheric pressure is different at different altitudes.
Describe how Bernoulli's principle explains the lift generated by an airplane wing.