PRESSURE - Form 1 Physics Notes

• Introduction and Definition
• Maximum and minimum pressure
• Pressure in Liquids
  • Liquid Levels
  • Fluid Pressure Formula
  • Transmission of Pressure in Liquids
  • Hydraulic Machines
• Atmospheric Pressure
  • Maximum Column of Liquid that can be Supported by Atmospheric Pressure
  • Measurement of Atmospheric Pressure
• Application of Pressure in Liquids and Gases
  • The Bicycle Pump
  • The Lift Pump
  • The Force Pump
  • Advantages of a Force Pump over a Lift pump
  • The Siphon

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Introduction and Definition

Pressure is the force acting normally (perpendicularly) per unit area.

The SI unit of pressure is N/M² or Nm^-2, which is also called pascal (Pa).

Pressure in solids depends on two main factors i.e. force and area

Example

A force of 100N is applied to an area 100mm². What is the pressure exerted on the area in Nm^-2.

Solution

Area; 100mm² = 0.0000001m² and Force = 100N

Pressure =F/A =100/0.0000001 = 1.0 x 10⁹Nm^-2

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Maximum and minimum pressure

Maximum pressure = ^Force/_{minimum area}

P_max = ^F/_Amin

Minimum pressure = Force/maximum area

P_min = ^F/_Amax.

Example

A block of wood measures 2cm by 3cm by 4cm and has a mass of 6 kg. Calculate its pressure when;

1. Area is minimum (maximum pressure)
2. Area is maximum (minimum pressure).

 

Solution

Possible Areas
2 x 3 =6cm²
2 x 4 =8cm²
3 x 4 =12cm²

1. A_min =6cm² =0.006m² and F =60N
   P_max =⁶⁰/_0.006 =100,000Nm^-2
2. A_max =12cm²=0.0012m² and f = 60 N
   P_min = ⁶⁰/_0.0012 =50,000Nm^-2

Exercise

1. A block of wood measures 3m by 6m by 2m and mass 3kg. Calculate;
   1. Maximum pressure
   2. Minimum pressure
2. A brick 20cm by 10cm by 5cm has a mass of 500g. Find maximum and minimum pressure. (take g = 10N/kg)
3. How much force must be applied on a blade of length 4cm and thickness 0.1mm to exert pressure of 5,000,000 Pa.

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Pressure in Liquids

Pressure in liquids depends on the following;

• Depth of the liquid
• Density of the liquid

Pressure in liquids increases with depth and density.

EXPERIMENT: To show variation of pressure in liquids

APPARATUS: A tall tin, nail and water

PROCEDURE

• Using the nail, make 3 holes A,B,C of the same diameter on a vertical line of one side of the tin
• Fill the tin with water as shown below.
  
• Observe water jets from the holes A,B,C.

OBSERVATION

The lower hole, A, throws water farthest, followed by B and lastly by c

EXPLANATION

The pressure of water at A is greatest than pressure at B and pressure at B is greater than pressure at C. Hence , pressure increases with depth.

QUESTION

Explain why a diver at the bottom of the dam experiences greatest pressure.

At the bottom of the dam depth is greatest and therefore the diver experiences greatest pressure due to the weight above him.

Liquid Levels

When a liquid is poured into a set of connected tubes with different shapes, it flows until the level are the same in all tubes as shown. 

This shows that the liquid flows to find its own level.

Liquid Levels in a U-Tube

When water is poured into a u-tube, it will flow into other arm. Water will settle in the tube with the levels on both arms being the same.

When one arm is blown into with the mouth, the level moves downwards, while on the other arm it rises. This is caused by pressure difference between the two arms as shown,

Pressure in liquids increases with depth below its surface

Pressure in a liquid at a particular depth is same in all directions.

Pressure in a liquid increases with density of the liquid.

 

Fluid Pressure Formula

Consider a container containing a liquid as shown below;

If A is the cross-section area of the column, h the height of the column and the density of the liquid then;

Volume of the liquid=cross-section area x density

=Ah

Mass of the liquid = volume of the liquid x density

=Ahρ

Therefore, Weight of the liquid = mass x gravitational force

=Ahρg

From definition of pressure = ^force/_area

Pressure=^Ahρg / _A

=ρhg

From the formula (p=ρhg) pressure is directly proportional to;

• Height of the column
• The density of the liquid

NOTE: Pressure in liquids does not depend on the cross-section area of the container.

The formula is also used to determine pressure due to a gas column.

EXAMPLE

1. 1. A diver is 10m below the surface of water in a dam. If the density of water is 1000kg/m³. Determine the pressure due to the water on the diver. (take g=10N/Kg)
   Pressure = hρg = (10 x 1000 x 10) = 100,000 N/m²
2. The density of mercury is 13600Kg/m³. Determine the liquid pressure at a point 76cm below mercury level.
   Pressure =hρg =0.76 x 13600 x 10 = 103,360 N/m²
3. Calculate the pressure due to water experienced by a diver working 15m below the surface. (take g = 10N/kg and density of sea water = 1.03 g/cm³)

Transmission of Pressure in Liquids

Pressure applied at one part in a liquid is transmitted equally to all other parts of the enclosed liquid. (Plunger).

This is the principle of transmission of pressure in liquids called Pascal’s principle which states that pressure applied at a given point of the liquid is transmitted uniformly or equally to all other parts of the enclosed liquid or gas.

Gases may transmit pressure in a similar way when they are confined and incompressible.

Hydraulic Machines

The principle of transmission of pressure in liquids is made use in hydraulic machines where a small force applied at one point of a liquid produces a much larger force at some other point of the liquid.

Hydraulic Lift

The hydraulic lift consists of a small piston S of cross-section A₁ and a large piston L of cross-section area A2. When a force is applied on piston S, the pressure exerted by the force is transmitted throughout the liquid to piston L.

At the smaller piston S the force applied F₁ cause a pressure P₁ at the crosssection area A₁.

Therefore P₁=^F₁/A₁

The pressure is equally transmitted throughout the liquid to the larger piston.

Thus at small piston pressure is equal to the pressure at the large piston

F₂=P₁ x A₂

But P₁=^F₁/A₁

F₂=^F₁/A₁x A₂

^F₂/F₁=^A₂/A₁

NOTE; Equation applies if pistons are at the same level

EXAMPLE 1

Find F₂ if A₁ = 0.52m², A₂ = 10m² and F₁= 100N

^F₂/₁₀₀=¹⁰/_0.25

F₂= ^{(100 x 10)}/_0.25

=4000N

EXAMPLE 2

Determine f₂ in the figure below. Density of the liquid =800kg/m³ and g=10N/kg

PA = PB

^{(60 x 10)}/_0.008 = (^F₂/_0.00025)+ (0.15 x 800 x 10)

0.00025(7500 -1200) = F2

F₂=18.45N

Hydraulic Brake System

The force applied on the foot pedal exerts pressure on the master cylinder. The pressure is transmitted by the brake fluid to the slave cylinder. This causes the pistons of the slave cylinder to open the brake shoe and hence the brake lining presses the drum. The rotation of the wheel is thus resisted. When the force on the foot pedal is withdrawn the return spring pulls back the brake shoe which then pushes the slave cylinder piston back.

Advantage of this system is that the pressure exerted in master cylinder is transmitted equally to all four wheel cylinders.

The liquid to be used as a brake fluid should have the following properties;

1. Be compressible, to ensure that pressure exerted at one point is transmitted equally to all other parts in the liquid
2. Have low freezing point and high boiling point.
3. Should not corrode the parts of the brake system.

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Atmospheric Pressure

Atmosphere means the air surrounding the earth. The air is bound round the earth by the earth’s gravity.

The atmosphere thins outwards indicating the density of air decreases with the distance from the surface of the earth

The pressure exerted on the surface of the earth by the weight of the air column is called air pressure.

Atmospheric pressure can be demonstrated by crushing can experiment.

EXPERIMENT: To demonstrate the existence of the atmospheric pressure

APPARATUS: Tin container with a tight-fitting cork, water, tripod stand, Bunsen burner.

PROCEDURE

• Remove the cork from the container and pour in some little water.
• Boil the water for several minutes.
• Replace the cork and allow the container to cool or pour cold water to cool it faster.

OBSERVATION

During cooling, the container crushes in.

EXPLANATION

Steam from boiling water drives out most of the air inside the container. When heating, the steam pressure inside the container balances with atmospheric pressure outside.

On cooling the steam condenses. A partial vacuum is therefore created inside the container. Since pressure inside the container is less than the atmospheric pressure outside, the container crushes in.

NOTE: Steam inside the container condenses lowering the pressure. The outside atmospheric pressure exceeds the pressure inside the container thereby crushing it.

Maximum Column of Liquid that can be Supported by Atmospheric Pressure

When water is sucked up a straw, the air inside the straw reduces. The atmospheric pressure acting on the surface is now greater than the pressure inside the straw. Water is thus pushed up the straw by atmospheric pressure.

If the straw was long enough and sealed at the top, it would be possible to estimate the height of water that would be supported by atmospheric pressure.

In case of water the column is too large.

At sea level the atmospheric pressure supports approximately 76cm of mercury column or approximately 10m of water column.

EXAMPLE

1. A girl in a school situated in the coast (sea level) plans to make a barometer using sea-water of density 1030 kg/m³. If atmospheric pressure is 103,000 N/m², what is the minimum length of the tube that she will require?
   P=hρg but p is atmospheric pressure
   103,000=h x 1030 x 10
   H=10m
2. A sea diver is 35m below the surface of sea water. If the density of the sea water is 1.03g/cm³ and g=10N/kg. Determine the total pressure on him.
   P_T =P_a + hρg
   =103,000 + (35 x 1030 x 10)
   =463,500N/m²
   
3. The air pressure at the base of a mountain is 75cm of mercury while at the top is 60cm of mercury. Given that the average density is 1.25kg/m³ and density of mercury is 13,600kg/m3. Calculate the height of the mountain.
   Pressure difference due to column of air = pressure difference due to mercury column
   h_aρ_ag = h_mρ_mg
   h_a = ^h_mρ_mg/ρ_ag
   ha=^{(0.15 x 13600 x 10)}/_{(1.25 x 10)}
   =1632m

EXERCISE

1. The barometric height at sea level is 76cm of mercury while that at a point on a highland is 74cm of mercury. What is the altitude (height) of the point? Take g =10N/kg, density of mercury =13600kg/m³ and density of air =1.25kg/m³.
2. A student in a place where the mercury barometer reads 75cm wanted to make an alcohol barometer, if alcohol has a density of 800kg/m³, what is the minimum length of the tube that could be used?

 

Measurement of Atmospheric Pressure

The U-Tube Manometer

Is an instrument used to measure fluid pressure.

It consists of a u-tube filled with water or any other suitable liquid or gas as shown.

Pressure at Z = Atmospheric pressure due to column of water.

Pressure at X = pressure at Z
P_g = P_g

Pressure at Z = atmospheric pressure + pressure due to column of water
P_g = P_a + hρg.

Since the density of water and gravitational force is known we can determine pressure of a gas if the atmospheric pressure is known.

EXAMPLE

Suppose h=20cm, P_a = 103,000N/m² and density=1000kg/m³, determine the total pressure (P_g)

P_g =103,000 + (0.2 x 1000 x 10)
=105,000N/m2

Simple Mercury Barometer

At sea level atmospheric pressure supports approximately 76cm of mercury column or 10m of water column.

This difference in height column between mercury and water is that mercury is much denser than water.

Mercury column forms a simple barometer, its height changing inside on the glass tube as air pressure outside changes.

The space above mercury in the barometer tube must contain air or water vapour since the barometer reading will be as shown above.

The space above in mercury in the tube when upright is called toricellian vacuum

The height h of the column is a measure of the atmospheric pressure.

At sea level, h=76cm since density of mercury = 13600kg/m³.

P_a = hρg
=0.76 x 13600 x 10
= 103,360N/m² (it is also referred as one atmosphere 1 atm)

Fortin Barometer

Is an improved version of a simple mercury barometer. Was designed by FORTIN

The ivory pointer acts as the zero mark of the main scale. The leather bag acts as reservoir of mercury height.

Before taking the reading, the level of mercury surface in the reservoir is adjusted by turning the adjusting screw until the surface of mercury just touches the tip of the ivory index.

The height is then read from the main scale and vernier scale.

The reading obtained from the barometer are in terms of the height of mercury column and written as mmHg or cmHg.

For example at sea level h=760mmHg and density of mercury=13600kg/m³

P_a =hρg =0.76 x 13600 x 10
=103,360Nm^-2

Aeroid Barometer

Is a portable type of barometer consisting of a sealed, corrugated metal box as shown below.

The pointer would indicate a particular value of atmospheric pressure of the surrounding so that any changes in pressure would be noticeable by movement of the pointer to either side of this atmospheric value on the scale.

The aneroid barometer movement makes it adaptable to measure heights.

Aneroid barometers (Altimeters) are used in aircrafts to measure heights.

Its normally calibrated in millibars. 1 bar=100,000Nm^-2

1 millibar (mbar)=100Nm^-2

Pressure Gauges

They are portable and are used mostly for measuring gas pressure, tyre pressure, pressure of compressed air compressors and steam pressure.

It is made of coiled flexible metal tubes which uncoil when the pressure inside increases. The movement of the tube is made to drive a pointer across a scale, through a combined system of levers and gears.

EXAMPLE

The pressure of a car tyre, measured with a pressure gauge is 40Ncm^-2. What is the total pressure of the tyre in Nm^-2.

P_Total =p_a +gauge pressure

=103,360 + (40 x 10,000)
=503,360Nm^-2

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Application of Pressure in Liquids and Gases

The Bicycle Pump

A bicycle pump is a simple form of compression pump.

The pump is connected to a tyre which has a rubber valve in it. When the pump handle is drawn out air below the washer expands and its pressure is reduced below the atmospheric pressure.

Air from outside the pump the flows past the leather washer into the barrel. The higher air pressure in the tyre closes the tyre valve.

When the pump handle is pushed in, the air in the pump barrel is compressed.

The high pressure in the barrel presses the leather washer against the sides of the barrel. When the pressure of the compressed air becomes greater than that of air in the tyre, air is forced into the tyre through the tyre valve which now opens.

NOTE: There is an increase in temperature of the pump barrel during pumping because work is done during compressing the air.

The Lift Pump

It is used to raise water from wells. It consists of a cylindrical metal barrel with a side tube. It has two valves P & Q.

 

 

Upstroke

When the plunger moves during upstroke, valve A closes due to weight and pressure of water above it. At the same time, air above valve B expands and the pressure reduces below atmospheric pressure.

The atmospheric pressure on the water surface in the well below this pushes water up past valve B into the barrel.

The plunger is moved up and down until the space between A and B is filled with water.

Downstroke

During down stroke valve B closes due to its weight and pressure of water above its piston.

Limitations of Lift Pump

The atmospheric pressure support only 10m column of water, which is actually a theoretical value but practically this pump raises the water less than 10m because of;

• Low atmospheric pressure in places high above sea level.
• Leakages at the valves and pistons

The Force Pump

This pump can be used to raise water to heights more than 10m.

Upstroke

During upstroke, air above the valve B expands and its pressure reduces below atmospheric pressure. The atmospheric pressure on the water in the well below pushes water up past valve B into the barrel.

NOTE: Pressure above valve A is atmospheric hence the valve does not open.

Downstroke

During down stroke, the valve B closes. Increase in pressure in the water in the barrel opens valve A and forces water into chamber C so that as water fill the chamber air is trapped and compressed at the upper part.

During the next stroke, valve A closes and the compressed air expands ensuring continuous flow.

 

Advantages of a Force Pump over a Lift pump

1. Force pump enables continuous flow of water.
2. Height to which water can be raised does not depend on the atmospheric pressure. It depends on;
   • Amount of forces applied during down stroke.
   • Ability of the pump and its working parts to withstand pressure.

The Siphon

A tube can be used to empty tanks or draw petrol from petrol tanks in cars. When used in this way it is referred as a siphon.

Pressure on the surface of the liquid is atmospheric pressure. Since end C of the tube is below the surface A by height h, pressure at C is greater than that at the surface.

The tube is first filled with the liquid after which it will continue to run so long as end C is below the liquid surface.

Pressure at C = p_a + hρg. The excess pressure (hρg) cause the liquid to flow out of end C

The siphon will work only if;

• End of the tube C is below the surface of A of the liquid to emptied.
• The tube is first filled with the liquid, without any bubbles in it.
• The tube does not rise above the barometric height of the liquid from the surface A of the liquid to be emptied.
• One end of the tube is inside the liquid to emptied.

NOTE: A siphon can operate in a vacuum.

REVISION QUESTIONS

1. The atmospheric pressure on a particular day was measured as 750mmHg. Express this in Nm^-2. (density of mercury = 13600kg/m3 and g=10N/kg) P=hρg=0.75 x 13600 x 10
2. A roof has a surface area of 20,000cm². If atmospheric pressure exerted on the roof is 100,000Nm^-2, determine the force on it. (take g = 10N/kg) P =F/A

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