Chapter 7 Energy of Hydraulic System

This law can be stated as follows: Pressure applied to a confined fluid is transmitted undiminished in all directions throughout the fluid and acts perpendicular to the surfaces in contact with the fluid. 7.1)PASCALS LAW Pressure in a fluid acts equally in all directions Pascals Law Applied to Hand Operated Hydraulic Jack Hand operated hydraulic jack 1. When the handle is pushed up, the piston rises and creates a vacuum in the space below it.2. Atmospheric pressure forces oil to leave the oil tank and flow check valve 1 to fill the void created below the pump piston. This suction process. 3. When the handle is pushed down, oil is ejected from the small diameter pump cylinder and flow through check valve 2 and enterthe bottom end of the large diameter load cylinder 4. Pressure builds up below the load piston as oil ejected from the pump cylinder.5. The pressure acting on the load piston= pressure generated by the pump piston. 6. The bleed valve is a hand operated valve, when opened allows the load to be lowered by bleeding oil from the load cylinder back to the oil tank. Pascals Law Applied to Hand Operated Hydraulic Jack http://www.docin.com/p-37717993.html Pascals Law Applied to Hand Operated Hydraulic Jack Pascals Law Applied to Hand Operated Hydraulic Jack 7.2)CONSERVATION OF ENERGY This energy laws states that energy cannot be created or destroyed. The total energy of system remains constant. Total energy = potential energy (elevation & pressure) +kinetic energy ( velocity) 7.2.1Potential energy due to elevation (PE) Figure shows chunk of fluid PE = WZ 7.2.2Potential energy due to pressure (PPE) PPE= pW7.2.3Kinetic energy due to velocity/ (KE) g

212KE 7.2.4Total energy(TE) possessed by W lb chunk of fluid flows through a pipeline of a hydraulic system: constant2W W2 gv pWZ ETCONTINUITY EQUATION CONTINUITY EQUATION CONTINUITY EQUATION A hydraulic cylinder is to compress a car body to bale size in 10 sec. The operation requires a 10m stroke and a 8000N force. If a 10bar pump is selected, find: 1. The required piston area 2. The necessary pump flow rate 3. The hydraulic power delivered by the cylinder 7.3)BERNOULLIS EQUATION Bernoulli's equation is one of the most useful relationships for performinghydraulic circuit analysis.

Its application allows us to size components such as pumps, valves, and piping for proper system operation.

The original Bernoulli equation can be derived by applying the conservation of energy lawto a hydraulic pipeline, as shown in next Figure. At station 1 we have W lb of fluid possessing an elevation Z1, a pressure P1 and a velocity V1. When this W lb of fluid arrives at station 2, its elevation, pressure, and velocity have become Z2, P2 and V2, respectively. 7.3.1Derivation ofBernoullis Equation Type of energy Station 1Station 2 Elevation WZ1WZ2 Pressure Kinetic 1pWgvW2212pWgvW222The total energy possessed by the W Ib of fluid at station 1equals to the total energy possessed by the same W Ib of fluid at station 2,provided frictional losses are negligibly small: Simplified, thus,= gv pZ221 11 gv pZ222 22 gv pWZ2W W21 11 gv pWZ2W W22 22 = Elevation head Pressure headVelocity head 7.3.2The Energy Equation By considering ; 1. HL (Head Loss) due to frictional losses take place between station 1 to station 2. 2. HP (Pump Head) represent energy of fluid added by a pump 3. HM (Motor Head) represent energy of fluid removed by a hydraulic motor The energy equation is given as follows where each term represents a head and has units of length; gv pH H Hgv pZL m p2Z222 2221 11 QxSGxHPft Hp3950) ( Where :HP= pump horse power, Q= pump flowrate (gallon per minute (gpm) SG = specific gravity Example 7.1 For the hydraulic system below, the following data are given below;a. The pump is adding 5 hp to the fluid ( pump horsepower=5) b. Pump flowrate is 30 gpm c. The pipe has1 in pipe diameter d. Specific gravity of the oil is 0.9Find the pressure available at the inlet to the hydraulic motor (station 2). Known the pressure at station 1 is atmospheric( 0 psig) and frictional losses between station 1 and 2 is 30 ft of oil Solution: Since no hydraulic motor between station 1 and 2, Hm = 0,V1 is negligibledue to cross section of an oil tank is large, so V1=0 ft 201 2 Z Zftx xft Hp7329 . 0 305 3950) ( ft 30 LH) / ( ) ( ) / (2 3s ft v x ft A s ft Q From To find; gv222gv pH H Hgv pZL m p2Z222 2221 11 2 2 2 200546 . 0 )121(4ft) (4) A(ft ft ft D Known 0668 . 044930449) () / (3 gpm Qs ft Qs ftftftft A ft Qs ft v / 2 . 12) ( 00546 . 0) ( 0668 . 0) () () / (23232 fts ft s ftgv4 . 2/ 4 . 64) / 2 . 12 (2Thus,2222 gv pH H Hgv pZL m p2Z222 2221 11 0 0 0 gv pH H ZL p2Z22 22 1 gv pZ H HL p2Z22 21 2 22 . 679pft ; find To 3water/ 2 . 56 ) 4 . 62 )( 9 . 0 ( x(SG)ft lbknown 2 32/ 200 , 38 lb/ft 56.2 x2 . 679 ft lb ft p Change to units of psi; psi in lb p 265 / 26514438,20022 gv p220 30 73222 2 The End