pressure slides
TRANSCRIPT
Unit 7:PressureAt the end of this unit, a student should be able to
define the term pressure in terms of force and area recall and apply the relationship pressure = force /area to new situations or to solve related problems describe and explain the transmission of pressure in hydraulic systems with particular reference to the hydraulic press recall and apply the relationship pressure due to a liquid column = height of column x density of the liquid x gravitational field strength to new situations or to solve related problems describe how the height of a liquid column may be used to measure the atmospheric pressure describe the use of a manometer in the measurement of pressure difference.
1. Pressure• Pressure is the force acting normally
per unit area• Pressure = Force
F
P AArea
• P = F/A• SI Unit: Pascal (Pa) or
Newton per square metre (N/m2)
Example 1
• The weight of a man is 600 N. Calculate the pressure he exerted on the floor if he is wearing a pair of track shoes and the area of contact of each shoe with the ground is 0.02 m2.
• Pressure = F/A = 600/(2x0.02) =15 000 Pa
Example 2
• A rectangular block of dimension 50cm by 30cm by 20 cm has a mass of 4 kg. Calculate the maximum and minimum pressure it can exert on the floor.
20 cm
30 cm50 cm
30 cm20 cm
50 cm Example 2
• Weight of block • = mg = 6 x 10 = 60 N• Maximum pressure exerted• = F/A(min)• = 60/(0.2 x 0.3) = 1000 Pa
Example 230 cm
20 cm
50 cm
• Minimum pressure exerted• = F/A(max)• = 60/ (0.5 x 0.3)• = 400 Pa
2. Liquid Pressure
• The pressure inside a volume of liquid depends on the depth belowthe surface of the liquid.
• The deeper it is, the greater the weight of the overlying liquid, and thus the greater the pressure.
Example 1: Liquid pressure increases with depth
Example 2: Liquid pressure increases with depth
• The thickness of the wall of dam increases downwards because the deeper it is, the greater the water pressure.
Damwaterland
Liquid Pressure: Formula
• The pressure at any point in a liquid at rest is given by:
• Pressure = hpg• where h = height of liquid in metre• p = density of liquid kg/m3
• g = gravitational acceleration in N/kg or m/s2
Proof Area A
height h Density p
• Consider a cylindrical container of of area, A, filled with liquid of density, p, to a height, h
Proof
• Volume of liquid = Ah• Weight of liquid• = mg = Vpg = Ahpg• Pressure on base• = Force/ Area • = Ahpg/ A = pgh
Area A
height h Density p
Proof
• Try to prove using a cuboid….
Liquid Pressure
• Pressure, P = hpg• Pressure at any point in a liquid is
independent of the cross-sectional area
Liquid Pressure
• A liquid always find its own level• (Liquid will flow to equalise any
pressure difference)
Liquid Pressure
• All points on the same level in a liquid have the same pressure
Liquid Pressure
liquid
• Pressure at any one depth in a liquid acts equally in all directions
Water tank
12 m
Example 1
• The water level in a water tank is 12 m above the tap. What pressure forces water out from the tap? (Density of water = 1000 kg/m3).
Water tank
12 m
Example 1
• Pressure at the tap is due to the water in the pipe above it.
• Pressure = hpg• = 12 x 1000 x 10 = 120 000 Pa
0.2 m0.5 m object
Example 2
• A regular shaped object is immersed in water of density 1000 kg/m3.
• (a) Calculate the water pressure at the top and the bottom of the object.
• (b) What is the resultant pressure on the object?
Example 20.2 m
0.5 m object
(a) Pressure exerted by water at the top surface of the object = h1pg
=0.2 x 1000 x 10
= 2000 Pa
0.2 m0.5 m object Example 2
• Pressure exerted by water at the bottom surface of the object = h2pg= 0.5 x 1000 x 10= 5000 Pa
0.2 m0.5 m object Example 2
• Resultant pressure on the object= 5000 - 2000= 3000 Pa (The object experiences an
upward force) Sometimes we call this the upthrust.
3. Atmospheric Pressure
• The atmosphere is the layer of air surrounding the Earth. It extends up to 1000 km above the Earth surface.
• The weight of the air exerts a pressure on the surface of the Earth. This pressure is called the atmospheric pressure
Atmospheric Pressure
• The atmospheric pressure is about 1.03 x 105 Pa or 105 Pa
• or 10 m of water• or 0.76 m of mercury (760 mmHg or
76 cm Hg)Hg here is the chemical symbol for mercury just like H is the chemical symbol for hydrogen
Simple Mercury Barometer
• The atmospheric pressure can be measured using a simple mercury barometer
760 mmReservoir of mercury
vacuum
Simple Mercury Barometer
• Height of mercury column for the mercury barometer:
• Pressure = hpg• 103 000 Pa = h x 13 600 kg/m3 x 10 N/kg• h = 0.76 mHg
Water Barometer
• Water can be used in a barometer instead of mercury. However, the glass tube used need to be much longer. Why?
Find out what will be the height if water is used instead of mercury for 1 atmosphere.
[Hint: you need to know the density of pure water.]
Mercury BarometerA
B0.2 m
C
0.76 m
D0.3 m
• Determine the pressure at points A, B, C and D
Mercury BarometerA
B0.2 m
C
0.76 m
D0.3 m
• Pressure at A = 0 mmHg • Pressure at B = 0.2 mHg• = hpg = 0.2 x 13600 x 10• = 27200 Pa = 27.2 kPa
Mercury BarometerA
B0.2 m
C
0.76 m
D0.3 m
• Pressure at C = 0.76 mHg• Pressure at D = (0.76 + 0.3) mHg• = 1.06 mHg
4. Pressure Difference• Manometer• A manometer consists of a U-tube
containing liquid and it is used to measure differences in gas or liquid pressure
X Y
Mercury
To gas supply
ManometerX Y
Mercury
To gas supply
• The height difference XY tells how much the gas pressure is different from atmospheric pressure
X
Y
Mercury
To gas supply
Z
12 cm
Example 1
• There is no pressure difference between X and Z
• Pressure difference between XY = 12 cm Hg
X
Y
Mercury
To gas supply
Z
12 cm
Example 1
• If atmospheric pressure is 75 cm Hg• Then Gas Pressure = pressure of X or Z
= (75 + 12) cm Hg= 87 cm Hg
15 cm12 cm
water
methylatedspirit
mercuryAB
Example 2
• A U-tube with some mercury at the bottom is set up vertically and 12 cm of water is added into one arm of the tube. Methylatedspirit is then added carefully into the other arm of the U-tube until the mercury levels are the same in both arm.
15 cm12 cm
water
methylatedspirit
mercuryAB
Example 2
• It is observed that the level of methylatedspirit is higher. What is the density of methylated spirit if the methylated spirit column is 15 cm high? (Density of water = 1000 kg/cm3)
Example 2• Let• hw = height of water column from level A• hm = height of methylated spirit column
from level B• PA = pressure at level A• PB = pressure at level B• Pa = atmospheric pressure• pw =density of water• pm = density of methylated spirit
15 cm12 cm
water
methylatedspirit
mercury
AB
Example 2
• PA = Pa + hwpwg• PB = Pa + hmpmg• But PB = PA
• Pa + hmpmg = Pa + hwpwg• 0.15 x pm x g = 0.12 x 1000 x g• 0.15 pm = 0.12 x 1000• pm = 800 kg/m3
15 cm12 cm
water
methylatedspirit
mercury
AB
5. Hydraulic Systems
• Hydraulic system work by using liquid pressure. They make use of two properties of liquid
1. Liquids are incompressible.2. If pressure is applied to a trapped
liquid, the pressure is transmitted to all parts of the liquid.
Simple Hydraulic Systems
small area piston
Large area piston
Applied forceOutput force
Hydraulic Systems
• When a force of 20 N is applied to a small piston of 0.01 m2, the pressure exerted on the liquid is given by
• Pressure = Force/ Area= 20/ 0.01 = 2000 Pa
Hydraulic Systems
• This pressure is transmitted to the larger piston. If the larger piston has an area of 0.1 m2, the force on the large piston is
• Force = pressure x area= 2000 x 0.1 = 200 N