Question

(a) Comment on the relative comparison in performance among gear, vane and piston pumps. (1 mark) (b) The following data are given for a fluid power system: Cylinder piston diameter = 0.2 m Cylinder rod diameter = 0.1 m Extending speed of cylinder = 0.07 m's External load on cylinder = 178 KN Pump volumetric efficiency = 92% Pump mechanical efficiency = 90 % Pump speed = 1800 rpm Pump inlet pressure = -27.6 kPa The total pressure drop in the line from the pump discharge port to the blank end of the cylinder is 517 kPa. The total pressure drop in the return line from the rod end of the cylinder is 345 kPa. Determine the Volumetric displacement of the pump, 1) Input power required to drive the pump, (m) Input torque required to drive the pump, and (iv) Percentage of pump input power delivered to the load. (8 marks) (c) If the diameter of cylinder piston increase, what happens to the input power required to drive the pump? (1 mark)

Answer 1

by to determine the inpert power required to drive the pump proceed as follows. write the formula of overall efficincy as no = (actual power delivered by pump acterol power delivered to pump Input power = 4PXQ no the force acting on the piston is equal to 178 KN pressure trop from rod side upto the deserrious comes out as (645+27.6 JKPG because the keserviour pressure is -27.6kPa write the formula of Pressure from the piston side during extension is P fext A (dpiston) here p is pressure is acting forwards piston area from rod end dpistons is the deameter of the piston area pressure from the rod end due to fluid in Cylindre Prod side = (345+27.6) kPa - force exerted by fluid from the rod endon pistons is as follows

F (dpistons)? f(drod)? seekstitue 345 for p 0.2 for episton & 0.1 for dood in the above expression to obtain force exerted on the back side of piston by the fluid 32 372.6 = F 10.212-10-12 F. -10.3)[0.1) X 372.6 - 8-7 ] the total force exterted at the rod side of piston is comes out as F: 178+8.77 |186.77kw obtain total pressure at the piston side p. 186.77X103 4 x 0.25- P = 5.948.23 kPa] therefore the output pressure of the pump comes out as Ppump - (517+5948.27.) - 6465.23 kPa overall efficeny no 20.90X0.92 - 0-828

Input power 216465.23X0.0022 0.828 17.17 kw c) to determine the input torque required to drive the pump proceed as follow write the formula of mechanical engi efficency? nm - 5948.23X0.0022 TX1800/60 T: 5948.2 3X0.0022X 60 0.9 X 1800 T: 484 Nm as to determine the percentage of pump input power delovered to wad proceed as follow- Y pump enpet delivered = - pumpperwerdeliverd to the wad Input power of the pump power delivered to the wod is Calculated as delivered pump perver (PxQr) Here p is the pressure at the piston by the flow 8 Qr is theoretical flow substitue 5# 5988.23 for p $ 0.0022 for Orto obtain pump power delwerd. pump power delivered - (5948.23 X0.0022) +13.08614 pump

y peemp input delivered = 13.086 17.17 ²100 -176.217.