Pressure

tweet description of Pressure unstable pressing big businessman per unit area exerted by a roving in a solid wall. Force acts perpendicularly to the surf angiotensin-converting enzyme in contacts. peregrine is a co u d s common word for gas a d/o liquid. 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 ashes of units psi in Imperial system of units Pressure can also be expressed in full terms of height of a column of liquid List of units of extort measurements & renewal of units Pascal s Pascals law Scalar quantityUnits of Pressure SM(2) Pressure Pressure measurements authoritative force Gauge Pressure divided into three different categories 1. Absolute jam which is specify as the sacrosanct value o insisting (force-per-unit-area) ac g o of p essu e ( o ce pe u a ea) acting on a approach by a fluid. su ace ud Abs. compel = blackmail at a local point of the surface due to fluid absolute zero of hug (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) > atm < atm SM(3) Pressure Pressure measurements relative between abs. , gauge and vacuum Absolute pressure Gauge Pressure gauge Equations gauge) Pressure term 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) smooth systems and Fluid pressures Fluid systems Two types of fluid systems 1. noneffervescent system in which fluid is at rest Fluid pressures Pressure measured i th i system i called static pressure P d in this t is ll d t ti Static pressure system s stem The pressure at a granted depth in a static liquid is The due to its own cant over 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 retributive 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 utilize to a small area a of a movable speculator it creates a pressure P = f/a. This pressure is transmitted to and acts on a bear-sizedr movable piston of area A which is then used to lift a car. Static pressure p Lesson Pressure along the plain line always remains the same for uniform singly fluid SM(7) Fluid systems and Fluid pressures Fluid systemsFluid pressures Example If the height of the fluids surface above the coffin nail of the five fluid s vass is the same, in which vessel is the pressure of the fluid on the tramp of the vessel the greatest ? The amount of liquid in for each one vessel is not necessarily the same. y Answer The pressure P is the same on the bottom of each vessel. Gauge pressure =F Force/ orbit /A = ? (hA)g/A = ? gh For gases the pressure en humongous 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 strong-minded 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 unenviable 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 pressuresDynamic pressure system Pitot resistance 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 swot and plunger of an hydraulic press are cc mm and 30 mm, respectively.Find the weight by the hydraulic press when the force applied at the plunger is four hundred 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 extravagance of pressure will be Area of plunger, equally transmitted (due to Pascals 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 bragging(a) 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 Solution Diameter of small piston, d = 30 mm = 0. 03 m Area of small piston, piston a= (pi/4)*d2 = 7. 068 * 10-4 m2Pressure intensity tra nsmitted 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 weewee. play (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 whitethorn be observed that P ARADOX 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 NReferences Transport Phenomena by Bird, Stewart, Lightfoot Fluid mechanism 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)