Pressure Free Essays

Pressure Definition of Pressure Fluid pressure: Force per unit area exerted by a fluid in a solid wall. Force acts perpendicularly to the surface in contacts. Fluid is a co u d s common word for gas a d/o liquid. We will write a custom essay sample on Pressure or any similar topic only for you Order Now o od o and/or qu d Pressure is a scalar quantity. It has the units of: N/m2 or Pa (or kPa) in SI system of units psi in Imperial system of units Pressure can also be expressed in terms of height of a column of liquid List of units of pressure measurements conversion of units Pascal s Pascal’s law Scalar quantity Units of Pressure SM(2) Pressure Pressure measurements Absolute pressure Gauge Pressure †¦ divided into three different categories: 1. Absolute pressure – which is defined as the absolute value o pressure (force-per-unit-area) ac g o of p essu e ( o ce pe u a ea) acting on a surface by a fluid. su ace ud Abs. pressure = pressure at a local point of the surface due to fluid – absolute zero of pressure (see page 63 of lecture notes) 2. Gauge pressure – difference between abs. pressure and atmospheric pressure – is always positive 101. 325 kPa or 14. 7 psi Equations Pressure term relationships a –ve gauge pressure is vacuum ve vacuum. Pressure term relationships †¢ Abs pressure = gauge pressure + atm pressure Abs. †¢ Abs. pressure = – gauge pressure + atm pressure (vacuum) gt; atm lt; atm SM(3) Pressure Pressure measurements Relation between abs. , gauge and vacuum Absolute pressure Gauge Pressure gauge Equations gauge) Pressure t erm relationships SM(4) Pressure Pressure term relationships Hydrostatic pressure 3. Differential pressure – measurement of an unknown pressure minus the reference to a o e u e e e ce o another unknown p essu e o pressure. – it is used to measure differential pressure i. . pressure drop (? P) in a fluid system SM(5) Fluid systems and Fluid pressures Fluid systems Two types of fluid systems: 1. Static system – in which fluid is at rest Fluid pressures Pressure measured i thi system i called static pressure P d in this t is ll d t ti Static pressure system s stem ‘’The pressure at a given depth in a static liquid is The due to its own weight acting on unit area at that depth plus external pressure acting on the surface o the qu d of t e liquid’’ Gauge pressure = ? gh – which i d hi h is dependent j t only on fl id d d t just l fluid density ( ) it (? and distance between below the surface of the liquid h. External pressure – is generally the atmospheric pressure SM(6) Fluid systems and Fluid pressures Fluid systems Fluid pressures Example: A hydraulic pump used to lift a car: when a small force f is applied to a small area a of a movable piston it creates a pressure P = f/a. This pressure is transmitted to and acts on a larger movable piston of area A which is then used to lift a car. Static pressure p Lesson: Pressure along the horizontal line always remains the same for uniform singly fluid SM(7) Fluid systems and Fluid pressures Fluid systems Fluid pressures Example: If the height of the fluid’s surface above the bottom of the five fluid s vessels is the same, in which vessel is the pressure of the fluid on the bottom of the vessel the greatest ? The amount of liquid in each vessel is not necessarily the same. y Answer: The pressure P is the same on the bottom of each vessel. Gauge pressure =F Force/Area /A = ? (hA)g/A = ? gh ‘’For gases: the pressure increase in the fluid due to increase in height is negligible because the density (thus, weight) of the fluid is relatively much smaller compared to the pressure being applied to the system’’. In other words, p = ? gh shows pressure is independent of the fact that the wt. of liquid in each vessel is different. This situation is referred to SM(8) as HYDROSTATIC PARADOX. Static pressure p Fluid systems and Fluid pressures Pressure term relationships Two types of fluid systems: 2. Dynamic pressure system Dynamic pressure system – more complex and diffi lt t measure l d difficult to – pressure measured in this system is called dynamic pressure – three terms are defined here 1. static pressure, 2. dynamic p p y pressure 3. total pressure SM(9) Fluid systems and Fluid pressures Dynamic pressure system Pitot tube Total pressure/Stagn p g ation press. Steady-state dynamic systems – Static pressure can be measured accurately by tapping into the fluid s ea (po A) e u d stream (point ) – total pressure (or stagnation pressure) can be measured by inserting Pitot tube into the fluid stream (point B) –;gt; total pressure (or stagnation pressure) = static pressure+ dynamic pressure SM(10) Fluid systems and Fluid pressures Dynamic pressure system Pitot tube Total pressure/Stagn p g ation press. SM(11) Problems 1. The diameters of ram and plunger of an hydraulic press are 200 mm and 30 mm, respectively. Find the weight by the hydraulic press when the force applied at the plunger is 400 N. Solution: Diameter of the ram, D = 200 mm = 0. 2 m Dia. of plunger, d = 30 mm = 0. 03 m p g , Force on the plunger, F = 400 N Load lifted, W: Area of ram, A = (pi/4)*D2 = 0. 0314 m2 Since the intensity of pressure will be Area of plunger, equally transmitted (due to Pascal’s Pascal s 4 a= ( i/4)*d2 = 7 068 * 10-4 m2 (pi/4)*d 7. 068 law), therefore the intensity of Intensity of pressure due to plunger, pressure at the ram is also = p = 5. 66 * 10-5 N/m2 p = F/a = 400 / 7. 068 * 10-4 But the intensity of pressure at the = 5. 6 * 105 N/m2 ram = Weight /Area of ram = W/A = Therefore, W/0. 0314 = 5. 66 * 10-5 W/0. 0314 or W = 17. 77 * 103 N = 17. 77 kN SM(12) Problems 2. For the hydraulic jack shown here find the load lifted by the large piston when a force of 400 N is applied on the small piston. Assume the specific weight of th li id i th j k i 9810 N/ 3. i ht f the liquid in the jack is N/m So lution: Diameter of small piston, d = 30 mm = 0. 03 m Area of small piston, piston a= (pi/4)*d2 = 7. 068 * 10-4 m2 Pressure intensity transmitted to the Diameter of large piston, D = 0. 1 m large piston, 5. 89 * 105 N/m2 Force on the large piston = Pressure intensity * area of large piston 5. 689 * 105 * 7. 854 * 10-3 = 4468 N Area of large piston, A = (pi/4)*D2 = 7. 854 * 10-3 m2 Force on small piston, F = 400 N F ll i t Hence, load lifted by the large piston = 4468 N Load lifted, W: Pressure intensity on small piston, p = F/a = 400 / 7. 068 * 10-4 = 5. 66 * 105 N/m2 Pressure at section LL LL, pLL = F/a + pressure intensity due to height of 300 mm of liquid = F/a + ? gh = 5. 66 * 105 + 9810 * 300/1000 = 5. 689 * 105 N/m2 SM(13) Problems 3. A cylinder of 0. 25 mm dia. and 1. m height is fixed centrally on the top of a large cylinder of 0. 9 m dia. and 0 8 m h i ht B th th cylinders d 0. 8 height. Both the li d are filled with water. Calculate (i) Total pressure at the bottom of the bigger cylinder and cylinder, (ii) Wt. of total vol. of water What is the HYDROSTATIC From the calculations it may b e observed that PARADOX between the two results? the total pressure force at the bottom of the cylinder is greater than the wt. of total volume Solution: Area at the bottom, of water contained in the cylinders. A = (pi/4)*0. 92 = 0. 6362 m2 (p ) This is hydrostatic paradox paradox. Intensity of pressure at the bottom p = rgh = 19620 N/m2 Wt. of total vol. of water contained Total pressure force at the bottom in the cylinders, y P = p*A = 19620 * 0. 6362 = W = rgh * volume of water 12482 N = 9810 ((pi/4)*0. 92 *0. 8 *(pi/4) *0. 252*1. 4) SM(14) = 5571 N References †¢Transport Phenomena by Bird, Stewart, Lightfoot †¢Fluid Mechanics and Hydraulic machines by R K Rajput R. K. †¢http://www. freescale. com/files/sensors/doc/app_note/AN1573. pdf (18 F 10) †¢http://www. ac. wwu. edu/~vawter/PhysicsNet/Topics/Pressure/Hydro Static. html (18 F 10) SM(15) How to cite Pressure, Papers