The pressure
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Chapter (four) The fluids ο· They are material that have the ability to flow. ο· The fluids include liquids and gases. ο· Gases are compressible and take the space of any container (have no definite volume). ο· Liquids are incompressible and have definite volume (take shape of container). Denisty and Relative denisty Density (ρ): it’s the mass per unit volume of a substance. π¬πππππππ π = π ππ π π , πππππ 3 ππ ππ , π«πππππππ πππππππ π. πΏ−3 . π π ππ3 πππ‘ Relative Density or specific density or specific weight or specific gravity: it’s the ratio between the density of the material to the density of the water at same temperature. π¬πππππππ ππ ππ ππ π πππ π£πππ’ππ, π πππ π‘πππππππ‘π’ππ ππ€ ππ€ ππ π πππ πππ π , π πππ π‘πππππππ‘π’ππ ππ π πππ π‘ππππππ‘ππ’ππ ππ ππ€ Relative has no unit or dimessional formal because it’s a ratio between same physical quantity. (unitless – dimesionless). The desity of water = 1000 ππ π3 =1 π ππ3 . so the numerical value of relative density of substance = the numerical value of its density. Application of density: Measuring the density of electrolytic solution of the car’s battery: During discharge, the sulphric acid (concenterate H2So4) during its reaction with lead plates forming lead sulphate, the density of the electrolytic solution decreases (Dilute H2So4). 1 On recharging, the sulphate radical is realeased from the lead plates back to the solution increase its density (conc H2So4) So we can identify the charging level of the battery by measuring the density of its electrolytic solution. Measuring density of the blood and urine in clinical medicine: The blood density in normal person ranges from 1040 If the blood density below 1040 ππ π3 ππ π3 to 1060 ππ π3 . , it refers to Anemia disease. The noraml denisty of urine is 1020 ππ π3 . in some diseases this value may increase due t the increasing if the secretion of salt. Pressure The pressure at a point (p):- it’s the average normal force acts on a unit surface area around the point. πΉ π π¬πππππππ π = where F is force (N), A area (π2 ). Or π = . W work, V volume π΄ π −1 π«ππππππππππ πππππππ π. πΏ . π −2 , πππππ π π2 , ππ, π½ π3 … … . . ππ‘π. Pressure at point inside a liquid (fluid): Suppose that a horizontal plate (X) of Area (A) in π2 is placed at a depth (h) in m melow the surface of a liquid of density (ρ) in ππ π3 . The force (F) exerts by the liquid cloumn on the plate equal its weight. π =π∗π =π∗π∗π =π∗π΄∗β∗π π= π π΄ = π∗π∗π΄∗β π΄ = π ∗ π ∗ β (the liquid pressure) π = ππ + π ∗ π ∗ β (absoulte pressure) 2 Liquid pressure: it’s the pressure exerts by a liquid at a point inside it. It is equal to the weight of the liquid colomn whose cross section area is 1 π2 at this point and its height is the vertical distance between that point and the free liquid surface. Items Liquid Pressure Absoulte pressure (total pressure) π =π∗π∗β π∗π Zero π = ππ + π ∗ π ∗ β π∗π Pa Graph Equation Slope Int.Y-axis Conclusion: The pressure is the same at same horizontal level inside same fluid. The liquid pressure doesn’t depend on the area. Notes: The liquid pressure acting on the base of the tank doesn’t depend on its area. Because the pressure is the noraml force acting on the unit area, and Area is not a factor the pressure depends on according to π = π ∗ π ∗ β. The pressure at any point in a liquid acts in all direction with same magnitude Because the pressure is the noraml force acting on the unit area. 3 The pressure acting on one of the walls of a liquid container = π1+π2 2 . Where P1, P2 the pressure of the start and end point. P, h realtionships Items Dir.h From the surface From the bottom Graph Relation Pa Slope Pa=0 Direct Int. Y axis π∗π Inverse The end point (max h) π∗π The atmospheric pressure The atmospheric pressure: it’s the weight of the air column over unit area of the earth’s surface. Or the pressure arising from the weight of mercury column of height 76 cm. Or the weight of mercury column with height equal =76 cm, cross 4 section area of 1 π2 , the density of mercury at 0 β°C is 13595 π π 2 ππ π3 and gravity of 9.8 . Measuring the atmospheric pressure: Mercury Barometer It consists of a glass tube of length one meter with a unit cross section and completely filled with mercury When the tube is inverted in a dish filled with mercury the surface of the mercury in the tube falls to a certain level. Torricellian vaccum: it’s the zone abthe mercury level in barometer tube it’s pressure is nearly zero. The atmospheric pressure acting to up the mercury and the weight of mercury acting to down it. When two forces are equals the mercury will stop moveing and the its height represents the atmopheric pressure (cm.Hg) ππ = ππβ + ππππ = 0.76 ∗ 9.81 ∗ 13595 π 1 ππ‘π = 1.013 ∗ 105 2 π Notes: The mercury surface in the barometric tube keeps having the same height above the free mercury surface in the dish whatever the inclination of that tube. As the inclination of the barometric tube increase the torricellian vacumm decrease till it disapeare when the vertical height of the closed end ≤Pa (76 cm.Hg) above the free surface. 5 Pa decrease with the increase height from the earth’s surface so the reading of the barometer will decrease too, the height of the building or mountain can be measured from the following: ππππ ∗ π ∗ β = πβπ ∗ π ∗ β2 − πβπ ∗ π ∗ β1 πβπ β= ∗ (β2 − β1) ππππ We use mercury insteade of water due too the high desnity as the high is inversly proportional with the density to make the process of measurement easier. Application on Pressure: Measuring the blood pressure: There are two values have to be measure, the first one is Systolic pressure, max pressure, the contraction of the heart , the blood is pushed from the left ventricle to the Aorta then to arteries, the second on is Diastolic pressure, min pressure, the relaxiation of the heart. Normal people pressure values are 120 Torr-80 Torr Adjusting the car tyres pressure: ……………… Pressure unit convert Unit Pa Pa 1 Bar ππ−π atm Bar πππ 1 π = π. πππ π. πππ Atm π. πππ ∗ πππ 1.013 1 mm.Hg (Torr) π = π. ππ ∗ ππ−π π π. πππ ∗ ππ 6 Take away Define Density, relative Density, Liquid pressure at a point, Atomspheric pressure. What are the factors affecting the following: 1- Pressure at a point inside the liquid. 2- Specific gravity of the liquid. What’s meant by: 1- The pressure at a point is 10 π π2 . 2- The atmoshperic pressure at the top of a mountain is 0.9 bar. 3- The specific density of Aluminium is 2.7. 4- The noramal force acting on the unit area = 9 ∗ 105 π. 5- The density of the oil is 970 ππ π3 . Geave reasons: 1234567- The relative density is unitless. The tyres of the military cars are broad. The needle has pointed end. The base of the dam is made thicker than the top. Mercury is prefered in barometer. The atmoshperic pressure is neglected in submarine pressure caluclation. The pressure exerted by a liquid on the base of the container doesn’t depend on the area of the base. Prove that: The pressure exertd by a liquid at a point inside it is π ∗ π ∗ β. Complete 1- 3 atm= …… π/π2 = ……… cm.Hg = ………. Torr= ……… Bar. 2- 700 mm.Hg = ………. millibar. 7 3- Force (F) acts on an area (A) makes an angle (θ) with the normal to the surface, the pressue equation is ………… Problems: 1- A swmming pool has a length of 30 m, width 10 m, height of water is 3 m given that Pa = 1.013 ∗ 105 π π2 , ππ€ = 1000 ππ π3 ,π = 10π π 2 . πΆππππ’πππ‘π. a- Water pressure acting on the base. b- The weight of the water in the pool. c- The pressure at a point 125 cm above its bottom. d- The water pressure acting on its walls. 2- Some fashions accessoriers are electroplated with gold coating 0.85 ∗ 10−4 ππ thick. Calculate the area that can be coated by 1 Kg of gold knowing that ρg= 19.3 π ππ3 . 3- A bottle of mass 0.03 kg when empty, 55g full of water, 370 g full of mercury. Calculate the relative density of mercury. 4- A submarine is designed to bear max pressure 12 atm. Find the max depth and the force acting on its windows, if its dimensions are 40*70 cm (given that g=9.8 π/π 2 , ρw=1000 ππ π3 , ππ = 1.013 ∗ 105 π/π3 ) 5- A layer of water of thickness 50 cm rests on a layer of mercury of thickness 20 cm. what is the difference in pressure between two points one at the interface between the water and mercury and the other at the bottom of π ππ π π3 the mercury layer (given that g=10 2, ρw=1000 , πβπ = 13600ππ π3 .) 8
