Basic Physics and Laws for Pneumatics Explained

Basic Physics for Pneumatics

Force

According to SI standards, the unit of force is the Newton (N). 1 Newton is the force that can give a 1 kg mass an acceleration of 1 m/s in 1 second.

N = 1 kg. m/s²

According to Newton’s law: Force = Mass . Acceleration

F = m . a

Pressure

Pressure, one of the most important units for pneumatics, is the force applied per unit area.
The unit of pressure (P) is the bar.

Pressure = Force / Area
P = F / A

P = pressure
F = force
A = area

According to the International System of Units (SI), the unit of force is the Newton (N), and the unit of length is the meter (m). Accordingly, the unit of area is (m²). Since pressure is defined as the ratio of force to area, the unit of pressure is N/m². 1 N/m² = 1 Pascal, represented by the symbol Pa.

The “bar” is used as the unit of pressure in pneumatics. For this reason, unit conversions must be done correctly wherever needed.

1 bar = 100000 Pa
1 bar = 10N/cm²
1 bar = 1 kg/cm²

Example Problem:

Calculate the pressure exerted by a solid object with a base area of 0.5 m² and a weight of 10 kg.

Given:

m = 10 kg
A = 0.5 m²

Find:

P = ?

Solution:

F = m . a = 10 . 10 = 100 N
P = F/A = 100 / 0.5
= 200 N/m²

Boyle’s Law

A gas’s pressure and volume are inversely proportional at constant temperature; this is one of the fundamental principles of pneumatic system design.

Pascal’s Law

Pressure applied at one point in a closed system is transmitted equally to every point in the fluid.

Bernoulli’s Principle

The fact that pressure decreases as flow speed increases is taken into account when sizing pneumatic line diameters and designing valves.

Laws Used in Pneumatics

Pascal’s Law

Pascal’s law states: A fluid compressed in a closed container applies the same pressure to every surface of the container.

Pnömatik için Temel Fizik ve Kanunlar — CHANGING FORCE

In the system shown in the picture, Pascal’s law states that the pressure at the input and output is the same.
Since we find pressure using the formula P = F/A;

P = F₁ /A₁ = F₂ / A₂

it’s correct to write it this way. Therefore, since the pressure is always the same, if we apply a different cross-sectional area in the first section versus the second section, the transmitted force will be different.

Boyle’s Law (Boyle-Mariotte Law)

The pressure and temperature of gases whose volume decreases as they’re compressed increase. If air is compressed so that its temperature remains constant, the product of its volume and pressure before compression equals the product of its volume and pressure after compression.

There’s an inverse relationship between pressure and volume. According to the equation below, when temperature is held constant, air pressure increases based on the compression ratio.

Boyle-Mariotte law pressure volume — Compressing air while holding temperature constant

For example: if air is compressed to twice its density, its absolute pressure will increase by a factor of two.

Example Problem:

At atmospheric pressure, free air has a volume of 1 liter. If its temperature is held constant and its volume is reduced to 0.5 liters, what will its pressure be?

Given:
P₁ = 1 atm
V₁ = 1 lt
V₂ = 0.5 lt

Find
P₂ = ?

Solution:
P₁ . V₁ = P₂ . V₂ →
1 x 1 = P₂ x 0.5
P₂ = 2 atm

Gay-Lussac’s Law

1 – Expansion Under Constant Pressure

Gases, provided their pressure remains constant, expand by equal amounts when heated by equal amounts. There’s a direct proportion between volume and temperature.

Example:
A gas at a certain pressure has a temperature of 10°C and a volume of 1 liter. If its pressure remains constant and its temperature is raised to 40°C, what will its volume be?

Solution:
T₁ = 10°C = 273 + 10 = 283 K T₂ = 40°C = 273 + 40 = 313 K
V₁ / V₂ = T₁ / T₂
1 x 313 = 283 x V₂
V₂ = 1.106 lt

2 – Expansion Under Constant Volume

Regardless of a gas’s initial temperature, pressure, or type, if it’s heated by a certain amount under constant volume, its pressure increases by an equal amount. There’s a direct proportion between pressure and temperature.

Example Problem:
The temperature of the air inside an air tank left outdoors was measured at night and found to be T₁ = 10°C, with pressure P₁ = 4 Bar. The temperature of the air inside the tank rises to T₂ = 50°C during the day. Calculate the pressure (P₂) at this temperature.

Solution:
T₁ = 10°C = 273 + 10 = 283 K T₂ = 50°C = 273 + 50 = 323 K

p₁ / p₂ = T₁ / T₂
p₁ . T₂ = T₁ . p₂
4 x 323 = 283 x p₂
p₂ = 4.56 Bar

General Gas Equation

In an environment with no heat transfer, if we compress any gas (reduce its volume), its temperature value will rise along with the pressure.

Example Problem:
Free air at atmospheric pressure has a volume of 10 liters and a temperature of 10°C. The air is compressed, reducing its volume to two liters. Its temperature becomes 50°C. Calculate the pressure of the compressed air.

Given:
T₁ = 10°C = 273 + 10 = 283 K
T₂ = 50°C = 273 + 50 = 323 K
V₁ = 10 lt = 0.010 m³
V₂ = 2 lt = 0.002 m³
P1 = 1 atm = 1.013 bar

general pressure rules general gas equation

Solution:
P₁ . V₁ / T₁ = P₂ . V₂ / T₂
1.013 . 0.010 /283
P₂ . 0.002 / 323
P₂ = 5.78 bar is the result for absolute pressure. The gauge pressure is 5.78 – 1.013 = 4.767 bar.

Compressed Air Flow Rate

The flow rate of a fluid flowing through a given cross-section varies based on the size of the cross-section and the flow speed (Q = A.V). As long as the cross-section doesn’t change, the speed of the air flowing through a pipe stays the same. As the cross-sectional area increases, flow speed decreases; as the cross-sectional area decreases, flow speed increases. In summary, we can say: air flows faster through smaller cross-sections.

Example:
The speed of air passing through a pipe with a small cross-section of 10 cm2 is 4.2 m/s. What will the air speed be if the pipe’s cross-section is enlarged to 25 cm2?

Given:
A1 = 10 cm2
V1 = 4.2 m/s = 420 cm/s
A2 = 25 cm2