'Pressure\r'

' drive interpretation of shove liquified blackmail: strong point per unit of measurement area exerted by a unruffled in a self-colored wall. Force acts perpendicularly to the scratch in contacts. precarious is a co u d s common boundary for gas a d/o fluid. o od o and/or qu d embrace 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 get loafer also be expressed in terms of height of a towboat of liquid List of units of coerce measurements & axerophthol; conversion of units dada s Pascal’s law Scalar quantityUnits of Pressure SM(2) Pressure Pressure measurements Absolute press Gauge Pressure … divided into 3 different categories: 1. Absolute pinch †which is outlined as the absolute value o squash (force-per-unit-area) ac g o of p essu e ( o ce pe u a ea) acting on a open air by a silver. su ace ud Abs. draw = embrace at a local point of the surface overdue to liquid â € absolute home in of closet (see page 63 of nettle notes) 2. Gauge jam †difference between abs. squash and automatic teller machineospheric stateospheric drag †is invariably positive 101. 325 kPa or 14. 7 psi Equations Pressure term relationships a â€ve adjudicate coerce is make clean ve vacuum. Pressure term relationships • Abs atmospheric storm = gauge pressure + atm pressure Abs. • Abs. pressure = †gauge pressure + atm pressure (vacuum) > atm < atm SM(3) Pressure Pressure measurements Relation between abs. , gauge and vacuum Absolute pressure Gauge Pressure gauge Equations gauge) Pressure term relationships SM(4) Pressure Pressure term relationships Hydro unmoving pressure 3. Differential pressure †measurement of an inglorious pressure minus the reference to a o e u e e e ce o an separate unkn bear p essu e o pressure. †it is used to measure derivative pressure i. . pressure drop (? P) in a bland system SM( 5) fluent systems and Fluid pressures Fluid systems Two types of fluid systems: 1. nonoperational system †in which fluid is at rest Fluid pressures Pressure deliberate i thi system i called static pressure P d in this t is ll d t ti unchanging 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 cocksure 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(prenominal) on fl id d d t just l fluid absorption ( ) it (? and distance between below the surface of the liquid h. External pressure †is for the most part the atmospheric pressure SM(6) Fluid systems and Fluid pressures Fluid systems Fluid pressures representative: A hydraulic pump used to fig up a car: when a secondary force f is utilize to a small area a of a moveable piston it creates a pressure P = f/a. This pressure is transmittable to and acts on a declamatoryr movable piston of area A which is and so used to lift a car. atmospherics pressure p Lesson: Pressure on the horizontal line always corpse 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 butt of the five fluid s vass is the same, in which vessel is the pressure of the fluid on the stinkpot of the vessel the greatest ? The list of liquid in each vessel is not necessarily the same. y firmness: The pressure P is the same on the bottom of each vessel. Gauge pressure =F Force/ field of operation /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 use to the system’’.In other words, p = ? gh shows pressure is independent of the occurre nce that the wt. of liquid in each vessel is different. This situation is referred to SM(8) as hydrostatic puzzle. Static pressure p Fluid systems and Fluid pressures Pressure term relationships Two types of fluid systems: 2. propellent pressure system Dynamic pressure system †more knotty and diffi lt t measure l d difficult to †pressure measured in this system is called dynamic pressure †3 terms are defined hither 1. static pressure, 2. dynamic p p y pressure 3. supply pressure SM(9) Fluid systems and Fluid pressuresDynamic pressure system Pitot tube get along 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 stagnancy pressure) can be measured by inserting Pitot tube into the fluid stream (point B) â€;gt; total pressure (or stagnation pressure) = static pressure+ dynamic pressure SM(10) Fluid systems and Fluid pres sures Dynamic pressure system Pitot tube hit pressure/Stagn p g ation press. SM(11) Problems 1. The diameters of ram and under weewee diver 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. ascendent: 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 hindrance elevate, W: vault of heaven of ram, A = (pi/4)*D2 = 0. 0314 m2 Since the strong point of pressure forget be Area of plunger, equally hereditary (due to Pascal’s Pascal s 4 a= ( i/4)*d2 = 7 068 * 10-4 m2 (pi/4)*d 7. 068 law), therefore the metier of transport 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 * cv 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) Proble ms 2. For the hydraulic mother fucker shown here find the load lifted by the large piston when a force of 400 N is applied on the small piston. Assume the detail weight of th li id i th j k i 9810 N/ 3. i ht f the liquid in the jack is N/m resolvent: 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 transmitted to the Diameter of large piston, D = 0. 1 m large piston, 5. 89 * cv 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 piston chamber 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 big cylinder and cylinder, (ii) Wt. of total vol. of water What is the HYDROSTATIC From the calculations it may be 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 problem 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 • air Phenomena by Bird, Stewart, Lightfoot •Fluid M echanics and hydraulic machines by R K Rajput R. K. •http://www. freescale. com/files/sensors/ mercantilism/app_note/AN1573. pdf (18 F 10) •http://www. ac. wwu. edu/~vawter/PhysicsNet/Topics/Pressure/Hydro Static. html (18 F 10) SM(15)\r\n'