液壓專業(yè)英語教程2

1、unit 2 pressure, work, and powerbasic term that are commonly used in the field of hydraulics and pneumatics must be discussed and understood.pressurethe word pressure is defined as force per unit area although other units may be used, pressure is commonly expressed in such units a pounds per square

2、inch. the abbreviation psi is usually employed to indicate pounds per square inch.fig.l shows an arrangement of two cylinders that are connected by a pipe or tube. a close-fitting piston is placed in each cylinder. in each cylinder (under the piston), the liquid and the connecting tube are shown, if

3、 it is assumed that there is no movement of each piston and that there is no leakage past each piston, the liquid and all the parts are at rest- a static condition. it is also assumed that a force f)of 100 pounds acts on piston acts on piston no. q and that there is no friction between each piston a

4、nd its cylinder wall. if piston no. 1 had a flat or face area of 2 square inches that is in direct contact with the liquid, the pressure in the liquid under piston no. 1 is equal to the force divided by the area (100 divided by 2), or 50 pounds per squareinch. thus, the liquid pressure at the face o

5、f piston no. 1 is 50 psi.piston area100 pounds100 poundspiston area 2 square inches 2 square inchespiston no jpiston no jpressure-50 psicylindern0.1pressure-50 psicylinder no.l“300 poundspiston no. 2tubing or pipecylinder no. 2piston area 6 square inchesfig. 1. illustrating a hydraulic jysfem having

6、 two cylinders and two pistons of different sizes.assuming that pistonno.2 is essentially at the same level as piston no.l, the liquid between the pistonsserves as a medium to transmit the pressure from one piston face to the other piston face. thus, the liquid pressure at the face of piston no.2 is

7、 50 psi, if the area of piston no.2 is 6 square inches, the force f2 on the face of piston no.2 is (6x50), or 300 pounds, thus, a force of 100 pounds at piston no.l develops a force of 300 pounds at piston no.2; this is accomplished by making the area of piston no.2 equal to three times the area of

8、piston no.l. in a sense, the arrangement (see fig. 1) is a fluid lever, similar to a mechanical lever using a metal bar and pivot.equal pressure at every point and in every direction in the body of a static liquid (a liquid at rest) is characteristic of all static fluids, liquids, or gases. this is

9、called pascal's law, after an early experimenter in this field of study. this law of pressure is very useful, and can be sued to advantage in countless applications.atmospheric pressurea blanket of air surrounds the earth; this is called the atmosphere. at the surface of the earth, atmospheric p

10、ressure, which is due to the weight of the air above the surface of the earth, can be measured. atmospheric pressure is commonly measured with a mercury baromete匸 thus , atmospheric pressure is often called barometric pressurefig. 2 illustrates the basic principle of a mercury barometer. the glass t

11、ube is open at the lower end and closed tat the upper end. initially, the tube is completely filled withpure mercury; then it is inverted, with the open end submerged, in a small vessel ormercury-mercury-glass tubeglass tubepressurepressureheight offig. 2. illustrating the bos/c parti0/ o mercury ba

12、romttr.height ofmercurymircury-=cistern-atmosphericpressure-=cistern-atmospheric pressurecistern containing mercury. the height of the columnofmercury gives the direct reading of the barometer; the weight of the air abovethebarometer balances the weight of the mercury column. barometric pressure is

13、usuallyexpressed in inches of mercury. a barometric height of 29.32 inches of mercury corresponds to an atmospheric pressure of about 14.7 pounds per square inchpressure measurementmany instruments or gauges that are used for measuring pressure employ a bourdon tube is a hollow metal tube that is ma

14、de of brass or a similar material; it is oval or elliptical in cross section, and is bent in the form of a circle. one end of the bourdon tube is fixed to the frame at point a ( where the fluid enters )the other end b (closed) is free to move the free end b actuates a pointer through a linkage syste

15、m. as fluid pressure inside the tube changes, the elliptical cross section changes, and the free end b of the bourdon tube moves inward or outward, depending on the character of the change. a convenient pressure scale or dial can be arranged from a calibration of the gaugefig. 3. cutaway vhw o/ a bo

16、urtlon tub# fyp« of pressure gauge (mt), and the indicator dial (right).the position of the free end b of the bourdon tube depends on the difference in fluid pressure between the inside and the outside of the tube. if the outside of the bourdon tube is exposed to atmospheric air pressure, the i

17、nstrument reading is a measure of so-called gauge pressure. for example, if the pressure reading at the outlet of an air compressor is 100 psi gauge, this indicates that the outlet air pressure is 100 psi above atmosopheric pressure. in this book, all pressures referred to are gauge pressures.in som

18、e instances, the pressure in a piece of equipment may be below atmospheric pressure; this condition is designated as vacuum for example, if the air in a tank is ata pressure that is below atmospheric pressure, the pressure gauge indicates a certain vacuum, or negative gauge pressure,definition of wo

19、rk, energy, and poweras shown in fig.4, body weighing 20 pounds at a given level is indicated in position no.l. if the body is moved vertically through a distance or displacement of 9 feet, the action involves work. the technical term work is defined as the product of force times displacement, with

20、the force in the direction of the displacement. as the body moves from position no. 1 to position no.2, a force of 20 pounds moves the 20-ib. body through a displacement of 9 feet. this equals (9x20), or 180 foot-pounds of work.fig. 4. an illustration of "work."energy is defined as the cap

21、acity to do work. energy refers to a possibility. a body resting at position no.2 has certain capacity, or a certain energy. if the body is moved to the level 9feet below, (9 x 20), or 180foot-pounds is available to do work.the term work in itself does not involve a time element. rate of movement, o

22、r speed, is often important. power is defined as the time rate of doing work. if the body weighing 20 pounds were moved at a constant speed and in a vertical direction through a vertical distance of 9 feet in a time of 2 seconds, the “power” can be calculated as follows:20pounds x 9feet _ 90foot 一 p

23、oundspower= 2secondssecondthus, the power required is 90 foot-pounds per second one horsepower has been arbitrarily defined as equal to 550 foot-pounds per secondforce and work in a fluid devicefig. 5 is an illustration of a pump or compressor delivering fluid (either oil compressed air) to the left

24、-hand side of a piston in a cylinder. let p represent the fluid pressure, in psi, and a represent the piston area in sq. in. (abbreviation for square inches).cyund6rpistonf-forcecontrol valvectforce pressure (psi) x area (sq.in.)f4>a： pa： in which p fluid pressure in psi a-piston area is sq.in. p

25、ounds.«xsq.in.pistonpump orcompressorexample fpa50psi x 2 sq. in. 100 pounds5. llluttrtjting the relation between presture, area, and force.then the force f acting on the left-hand face of the piston is pa. for a pressure p of 50 psi and an area a of 2 sq. in, the force acting on the left-hand

26、face of the piston is equal to (50x2), or 100 pounds. assuming no friction due to the cylinder wall, the force f at the piston rod is equal to 100 pounds-fig. 6 is another illustration of a pump or compressor delivering fluid to the left-hand side of a piston is a cylinder. a s in t he previous exam

27、ple, let p represent thefluid pressure (psi) and let a represent the left-hand piston area (sq. in.). then the fluid force f acting on the left-hand piston face is f=pa if this force remains constant while pushing the piston through a displacement of distance l (inches), the work done by the fluid o

28、n the left-hand face of the piston is equal to the force f times the displacement l or the work w=pal=pl. for a fluid pressure of 50 psi, a piston area of 2 sq. in., and displacement of 3 inches, the work w con be determined as follows.w=(50 x 2 x 3)=300inch-poundsdisplacement actionfig.7 is a diagr

29、am of a pump delivering hydraulic fluid through a pipe or tube into a cylinder, the piston is shown at position no.l at a given time or part of the stroke. if the pump continues to deliver fluid into the cylinder, the fluid pushes the piston to the right-hand side a distance of 3 inches (position no

30、.2 ). the linear displacement or movement of the piston to the right is equal to 3 inches. for a piston area of 2 square inches and a piston displacement of 3 inches, the volumetric displacement of the piston is equal 20 (2 sp in x 3 in), or 6 cubic inches. assuming no leakage of fluid across the pi

31、ston from the left-hand side of the piston to the right-hand side of the piston, the total amount of fluid added to the cylinder is equal to 6 cubic inches; in other words, 6 cubic inches of fluid was admitted to the cylinder and pushed the piston, for a volumetric displacement equal to 6 cubic inch

32、es. with no leakage, this action is frequently called a positive-displacement action. the amount of fluid entering the cylinder is equal to the volumetric displacement of the pistonpfluid pressure (psi) 50 psicontrol valvea piston area (sq.inj 2 sq in.fforce (pa) -(50x21 100l2 piston displacement (i

33、nj 3 inches w * work <f x i) (50 x 2) (3) 300 inch-poundswork force x displacement (i)w-flpump orcompressorinch-pounds-pounds x inchesexample p-50 psi a *2sq. in. f-50x2 100 pounds l-3 in. w work 100 pounds x 3 inches 300 inch- p0un0sfig 6. illustrating nworku in a fluid device.in reference to th

34、e system illustrated in fig.l (two cylinders and two pistons), the area of piston no. 1 is 2 square inches and the area of piston no.2 is 6 square inches (3 times that of piston no.l). if a force of 100 pounds is applied at piston no. 1, there is corresponding force of 300 pounds at piston no.2. for

35、 a given fluid pressure, there is a force multiplication because of the difference in piston areas, assuming no leakage if it is assumed that piston no. 1 moves downward a distance of 0.03 inch for a positive-displacement action of the incompressible hydraulic fluid, the volume of fluid displaced by

36、 piston no. 1 is equal to the volume of fluid displaced by piston no. 2 . for a positive-displacement action, piston no.2 then moves upward 0.01 inch (piston area times displacement is equal to the displaced volume) for piston no/1, the total work is (100x0.03), or inch-pounds, for piston no.2, the

37、total work is (300x0.01), or 3 inch-pounds.rate of flow and piston travelthe volume rate of fluid flow through a device can be expressed in various. fox example, it is common to express volume rate of flow for liquids in gallons per minute (gpm). one gallon is equal to 231 cubic inches for a practic

38、al example, an oil pump may be said to deliver a flow of 10 gpm; this corresponds to a rate of (10x231), or 2310 cubic inches per minute. for air flow, it is common to use a volume rate stated as cubic feet per minute (cfm). fox example, it may be said that air enters an air compressor at a rate of

39、30 cubic feet per minute. the term capacity is used to designate the volume rate of flow through a pumpfig. 8 a illustrates the flow of hydraulic liquid from a pump. a volume rate q of 3 gpm indicates a flow rate of 11.55 cubic inches per second. as illustrated in fig. 8b, the piston position has mo

40、ved from position no.l to position no.2, and there is a given constant volume rate of flow q from the pump. let a represent the face area of the piston (square inches) and vp the constant velocity or speed of the piston(inches per second). assume no leakage of fluid past the piston; in other words,

41、there is a positive-displacement action, during a given time there is a given volume rate of flow from the pump (cubic inches). the volume leaving the pump must equal thevolumetric displacement of the piston .the constant volume rate q is the time rate of pump delivery. the produce avp is equal to t

42、he time rate of cylinder volume displacement, if there is no leakage of fluid, as shown by the formula: q=avp.if the volume rate q is equal to 3 gpm, the time rate of cylinder volume displacement corresponds to 11.55 cubic inches per second if the piston face area is 5 square inches, the speed of th

43、e piston is equal to:vp=q/a= 11.55/5=2.31 inches per secondthus, the speed of the piston depends on the volume rate of delivery from the pump.q from pumppistonpositionno jpistonpositionno. 2! *-vparea-aoutlft丄 (volume rate of flowpumpas cubic inches persecond or gallons (per minute. gpm1 gallon231 c

44、ubic inchesfluid in 廠 inletexample： q-3cpm q-3x231 <1135 cubic inches"60second aqa vpexampix： q3gpm q* 11.55 絞/a-sq.inches“ q h.55 . inchesvp.t._.uitkbamrea of piston sq. in.vp-velocity of piston, inchesq voujk ratl cubic inchessecond fig. b. ihuttratlng flow of hydraulic liquid from a pump：

45、 (a) pump outlet flow; (b) rate of piston travel.summarywhen a fluid acts on one side of a piston, as illustrated in fig. 7, let p represent the differential fluid pressure across the piston (in psi) and a the area (sq. in) the fluid force f (in pounds )on the piston can be determined by the formula

46、; f=pa. other units can be used; the foregoing is used to illustrate a common set of units.let l represent linear displacement or movement of the piston (in feel) to the right with a constant force f (see fig.7). for the movement from position no. 1 to position no.2, the work w? in foot-pounds, is g

47、iven by the relation: w= fl.as the piston moves from position no. 1 to position no. 2 during the time interval t (in seconds), assume that the piston velocity vp (in feet per second) is constant. the piston velocity vp can be expressed in the form: vp=l/t. during the displacement from position no. 1

48、 to position no.2, with no leakage, the volume rate of flow q is given by the relation: q=avp. if a is given in square feet and vp is given in feet per second, then q is in terms of cubic feet per second. various conversions are possible. as indicated in the appendix-conversion factors, for example,

49、 1 gallon per minute (gpm) is equal to 0.002228 cubic feet per second.power is defined as the time rate of doing work. assuming a constant linear velocity vpas the piston moves from position no. 1 to position no. 2, let r represent power (foot-pounds per second). then, power r can be expressed in th

50、e following forms as:r=fl/t=fvp; and r=pa vp=pqthe first power equation shows that power is proportional to the product of force times piston velocity. as shown in the second equation, power can also be expressed as the product of the pressure difference and the volume rate of flow.as an example, if

51、 the volume rate of flow q is 3 gpm (corresponding to 11.55 cubic inches per second) and the pressure differential is 500 psi, then the power becomes;r= 500 x 11.55=5775 inch-pounds per secondr= (500 x 11.55)/12=481 foot-pounds per secondsince 1 horsepower equals 550 foot-pounds per second, the powe

52、r becomes:r=(500x 11.55)/(12x550)=0.875 horsepowerthe term torque is used in connection with turning and twisting actions. an automotive internal combustion engine develops a torque at the crankshaft to turn the crankshaft and ultimately to turn the rear wheels of the car. an electric motor develops a torque at a rotating shaft. as illustrated in fig. 9, if a motion