FORCES 6. Pressure in fluids and hydraulic systems

Doc Brown's Physics Revision Notes

Suitable for GCSE/IGCSE Physics/Science courses or their equivalent

What is a fluid? What is the formula for pressure?

What causes pressure in liquids?

How do you calculate pressure in a liquid?

How do hydraulic systems work?

What do we use hydraulic systems for?

Particle theory, Fluids and Pressure

Gases and liquids are fluids because the particles are free to move around.

 low density FLUID high density FLUID NOT a FLUID

In fluids the inter-particle attractive forces are sufficiently weak to prevent a solid forming, allowing free random movement of the molecules of a liquid or gas.

Because of the weaker inter-particle force, the particles of gas will spread out to fill any space available.

The particles in a liquid are held much closer together because of greater inter-particle forces giving liquids a much greater density than gases.

Apart from water, liquid densities are usually a few % less than that of their solid form.

The density of liquids is so high, with so little space between the particles, they are almost impossible to compress to a smaller volume.

Gases have so much space between the particles that they are readily compressed to a smaller volume.

The closer together the particles are, the more compact the substance is, the greater density of it.

See also Particle theory models of gases, liquids and solids and the particle diagram above.

Liquids have a uniform density (same throughout its bulk) which only increases very slightly under extremely high pressures because there is so little free space.

However, there is considerable space between gaseous particles and its is relatively easy to compress the particles closer together e.g a bicycle pump filled with air or water. The air is easily compressed and the water isn't. The water can ruin the pump but you can't compress the water in it. High pressure water pistols rely on compressed air NOT compressed water.

In moving around the particles of a fluid collide with each other and with any surface they are in contact with.

Although the mass of an individual particle is minute and each collision involves the transfer of an equally minute amount of kinetic energy, collectively the trillions of collisions cause a pressure to be exerted in both gases and liquids.

The combined effects of these particle collisions produces a net resultant force at right angles to the surface of contact.

e.g. the pressure of gases in a container or the atmospheric air pressure around you.

The maximum pressure exerted in a fluid is considered to be due to the collective force of the particle collisions acting at right angles (normal, 90o, perpendicular) to the surface on which the collisions take place i.e. any surface in contact with the fluid.

Pressure is defined as force per unit area and is calculated from the simple formula

pressure = force ÷ area,    P = F / A,  F = P x A, A = F / P

P, pascals (Pa);   F, force in newtons (N);   A, area on which force acts in square metres (m2)

A force of 1 N acting on 1 m2 creates a pressure of 1 Pa

Ignoring the different weights of people (the 'force'), you should realise from the formula why it is better to be trodden by a broad shoe sole than a stiletto heel !!!! (CLUE !!! P = F / A, no exam pressure here !!!)

Liquid fluids have a much greater density than gaseous fluids, so for similar depths (or heights) of gas and liquid, liquids will create a much greater pressure because of the greater weight (due to gravity) of substance acting on the same surface area.

The greater the density of a material, the greater the number of collisions can take place, creating a greater pressure. This concept is most applicable to gases, which are so easily compressed under pressure, so considerably increasing their density.

Also, the greater the column of fluid e.g. water, the greater the pressure created - the greater weight acting on a given area.

Example calculations

Q1.1 If a weight of 200 N acts on a surface of 5 m2, calculate the pressure created.

pressure = force ÷ area,    P = F / A = 200 / 5 = 40 Pa

Q1.2 What force must be applied to a surface area of 0.0025 m2, to create a pressure of 200,000 Pa?

P = F / A, rearranging gives: F = P x A = 200000 x 0.0025 = 500 N

Q1.3 In a hydraulic lift system, what must the surface area of a piston be if a pressure of 300 kPa is used to give a desired upward force of 2000 N?

P = F / A, rearranging gives: A = F / P = 2000 / 300000 = 0.00667 m2

What is the piston surface area in cm2?

1 m2 = 100 cm x 100 cm = 10000 cm2, so the area of the piston = 10000 x 0.00667 = 66.7 cm2

Q1.4

Pressure in a liquid - density and depth factors

Density is a measure of how close the particles are together. The more compact they are, the greater the density.

As already mentioned, in liquids the density is uniform throughout and because there is so little space between the particles the density only slightly decreases with increase in temperature with the increased kinetic energy of the particles. However, the volume shows almost no change with increased pressure (so here you can consider liquids to be virtually incompressible).

All liquids expand on heating - observe a mercury or alcohol thermometer.

The density in a fluid varies with depth - it doesn't matter whether you are dealing with gases like the atmosphere or liquids like the water of a lake or ocean.

The greater the height/depth of fluid, the greater the weight of particles that gravity is pulling down, hence the increase in force per unit area at a particular level, hence the increase in pressure.

The pressure in a liquid acts in all directions because the particles are colliding in all directions.

Liquid pressure increases with depth as the weight of the column of liquid increases.

You can calculate the pressure at a given depth created by the weight of liquid in the earth's gravitation field using the following formula:

pressure in a liquid = depth of liquid x density of liquid x gravitational field strength

P = hρg

P, pascals (Pa);   h = depth in metres (m);   ρ = density (kg/m3),  g = 9.8 N/kg (on the Earth's surface)

Unit connections

Taking the formula P = h x ρ x g 'apart' in terms of units.

pressure = force per unit area = height of column of material x density of material x gravitational constant

N / m2  =  m  x  kg/m3  x  9.8 N/kg, on the right the kg cancel out, m/m3 = 1/m2, you are left with N/m2 !!

Note: Upthrust force in fluids and flotation etc. are covered in

Example calculations   (the gravitational field effect is taken as 9.8 kg/N in these questions).

Q2.1 Calculate the pressure created by a 30 m depth of water given the density of water is 1000 kg/m3 and gravity 9.8 N/kg.

P = hρg

P = 30 x 1000 x 9.8 = 294000 Pa (2.94 x 105 Pa, 294 kPa)

Note: Atmospheric pressure is about 101 kPa, so a diver at these depths will experience a much greater pressure than on the surface. Increase in pressure causes more gases to dissolve in the blood stream. This can have serious consequences if time isn't allowed for the body pressure to adjust to the new external pressure. The bends, also known as decompression sickness disease, occurs in divers when dissolved gases (mainly nitrogen) come out of solution in bubbles and can affect any body area including joints, lung, heart, skin and brain. The effects can be fatal.

Q2.2 The density of sea water is ~1025 kg/m3, the maximum depth of the Atlantic ocean is ~8500 m (8.5 km).

(a) Calculate the water pressure at this depth.

P = hρg

P = 8500 x 1025 x 9.8 = 85400000 Pa (to 3 sf, 85400 kPa, 8.54 x 107 Pa, 8.54 x 104 kPa)

(b) By what factor is the pressure greater at these depths compared to the ocean surface?

Atmospheric pressure is ~101 kPa

Pressure at bottom of ocean ÷ pressure at surface = 85400 ÷ 101 = 846 (3 sf).

Note: This extraordinary increase in pressure mean to explore this 'alien' world you need a very strong submersible craft. However, evolution has allowed all sorts of creatures to live down at these depths, all fully pressure adjusted over time! If you (theoretically) brought any such creatures rapidly to the surface and exposed them to normal pressure, it would kill them!

Q2.3 At what depth in water is the increased pressure five times greater than atmospheric pressure (101 kPa)?

5 x 101 = 505 kPa, 505000 Pa, density of water 1000 kg/m3

P = hρg, rearranging gives h = P/ρg = 505000/(1000 x 9.8) = 51.5 m

Note: The pressure increase in water increases by about the value of atmospheric pressure for every 10 m.

Q2.4 At a depth of 12.5 m of a chemical solvent the pressure at the bottom of the storage tank due to the solvent was 306 kPa

Calculate density of the solvent.

P = hρg, rearranging gives ρ = P/hg = 306000/(12.5 x 9.8) = 2498 kg/m3

Q2.5 ?

Hydraulic systems - mechanical devices for the transmission of forces in liquids

Hydraulics is how we can use fluids (e.g. air or oil) to transfer force to achieve some useful work e.g. operating machinery like car jacks, car brakes or JCB digging machines.

As we have said above, pressure in fluids is caused by particle collisions with themselves and the surface of a container.

These collisions cause a net force at right angles to all surfaces the fluid is in contact with. Hence the equation

pressure = force ÷ area,    P = F / A

P, pascals (Pa);   F, force in newtons (N);   A, area on which force acts in square metres (m2)

A force of 1 N acting on 1 m2 creates a pressure of 1 Pa

(I've repeated these details because there are more calculations coming up!)

However, there is an important new idea to understand the applications of hydraulics.

The pressure in a fluid is transmitted equally in all directions.

AND, for the purposes of calculations, you consider the force involved to act at right-angles to the surface.

Since liquids are incompressible, if a force is applied to any point in an enclosed fluid system the net force is transmitted to any other point in the fluid (gas or liquid). This is the basis of all hydraulic systems.

The principles and calculations involving hydraulic systems.

The diagram above illustrates the idea of using a hydraulic system to apply a small force to produce a large force - an example of a force multiplier system to effect a mechanical advantage (remember moments and lever systems!)

Therefore: P1 = F1input/A1 and output P2 = F2output/A2   (where = pressure, A = area, F = force)

BUT, P1 = P2 because the pressure at any given moment in time is the same throughout the hydraulic system.

Therefore: F1/A1 = F2/A2,  rearranging gives the output force F2 = F1 x A2/A1

So the ratio of the two piston areas (A2/A1) gives you the force multiplying effect.

Note:  In hydraulic systems, piston 1 is in the master cylinder and piston 2 is in a slave cylinder (can be several of them).

Q3.1 In a simple hydraulic system piston 1 has a cross-section area of 0.000050 m2, and piston 2 has a cross-section area of 0.00025.

If a force of 35 N is applied to piston 1,

(a) What is the force multiplying ratio?

force ratio = A2/A1 = 0.00025/0.000050 = 5.0

(b) What output force is generated by piston 2?

F2 = F1 x A2/A1 = 35 x 0.00025/0.000050 = 35 x 5 = 175 N

Q3.2 A medium sized car typically weighs 1000 kg.  (gravity force 9.8 N/kg)

A hydraulic system to jack the car up above an inspection pit has four slave cylinders whose pistons have a cross section area of 0.01 m2. The cross-section area of the master cylinder piston is 0.00125 m2.

(a) What is the weight of the car?

w = mg = 1000 x 9.8 = 9800 N

(b) What is the minimum force that must be applied to the master cylinder to raise the car upwards?

F1/A1 = F2/A2, rearranging gives F1input = F2output x A1/A2

The output force must at least match the weight of the car, so F2 = 9800 N

A1 = 0.00125 m2, A2 = 4 x 0.01 = 0.04 m2 (remember there are 4 connected cylinders).

Therefore F1 = 9800 x 0.00125/0.04 = 306 N (3 sf)

So relatively small force can be used to raise a much larger weight by means of a hydraulic system.

Q3.3 Solve the following problem with a little help from the diagram on the right of a simple hydraulic system of two pistons and cylinders connected together.

The cross-section area of piston A1 is 0.000400 m2.

The cross-section area of piston A2 is 0.00800 m2.

(a) If a force of 40 N (F1) is applied to piston 1, calculate the pressure created in the fluid.

P = F / a = 40 / 0.0004 = 100 000 Pa

(b) Calculate the force F2 created by the force of 40 N on piston A1.

Since the pressure is the same throughout the hydraulic system, we can say the pressure acting on piston 2 is also 100 000 Pa.

P = F / A, F = P x A = 100 000 x 0.008 = 800 N

(c) What is the multiplying or mechanical advantage factor in this hydraulic system?

force multiplying factor = final force / initial force = 800 / 40 = 20

(d) In order to create a force of 2000 N from cylinder 2, what should the cross-section area of piston 2 be if the force applied to piston 1 is still 40 N?

The pressure P2 remains the same at 100 000 N

F2 is now 2000 N, need to solve for A2

P = F / A,  A = F / P = 2000 / 100 000 = 0.02 m2

Q3.4 ?

Some applications of hydraulic systems

Hydraulic jack system.

Four hydraulic piston and cylinder systems are used to jack up the red van. The fluid can be oil or compressed air.

The Land Rover (left) and the red  van (right) both have hydraulic brake systems.

Key for the 5 photographic diagrams: S = suspension spring;  H = the pipe conveying the hydraulic brake fluid force.

D = the brake drum and disc on which the brake pads in casing P are forced into contact with the smooth disc by the hydraulic transmitted force when you press the brake pedal.

A conveys speed of rotation information for the correct functioning of the advanced ABS braking system.

How does the brake system of a motor vehicle work?

When you press on the brake pedal of motor vehicle, the force is transmitted through the brake fluid liquid by a pipe system (in modern cars it is aided by an electric pump system). The braking mechanism involves several pistons and cylinders, known as the master cylinder (activated by the pedal) and slave cylinders on the brake drums (4 of the latter in the case of 4 wheeled vehicles). The transmitted force pushes the brake pads onto the brake disc on the brake drum and the resulting friction reduces the speed of the vehicle. When you take you foot off the brake pedal, the force is no longer transmitted and springs move the brake pads away from the brake discs to avoid unnecessary friction.

'Hydraulic' photographs by courtesy of Mark Raw of M T R Autotech Ltd garage, Castleton, North Yorkshire, England

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