Problem: Size An Accumulator
Size an accumulator that will supply the total flow to a cylinder for one cycle in and out.Give the answer in gallons.
- 4″ x 20″ x 2″ cylinder
- Minimum required force on the cylinder is 18,849 pounds when fully retracted and at end of cycle.
- Maximum system pressure is 3,000 psi.
- Slow charge and discharging system with no change in temperature
- Pre-charge is 1,500 psig.
Find Out The Solution
Area to extend cylinder is 4 x 4 x 0.7854 = 12.566 sq. in.
Volume to extend cylinder is 12.566 x 20″ = 251.32 cubic inches.
Area to retract cylinder is area of piston of
12.566 – area of rod 3.14 = 9.42 sq. in.
Volume to retract cylinder is 9.42 x 20 = 188.4 cubic inches.
Total volume needed is 251.32 + 188.4 = 439.72 cubic inches.
Minimum required pressure is F = PA.
18,849# / 9.42 sq. in. (EREA) = 2000 psi
Sizing of accumulator is P1V1 = P2V2 = P3V3. Then V3 – V2 = useable volume of oil.
In order to solve this problem with “shop math” and not have two unknown in the problem, I will solve for the amount of useable oil for a one-gallon accumulator and then compare to the total needed.
P1V1 = P2V2
1514.7 x 231 = 3014.7 x V2 V2 = 116.06 cu. in.
P2V2 = P3V3
3014.7 x 116.06 = 2014.7 x V3 V3 = 173.67 cu. in.
V3 – V2 =
173.67 – 116.06 = 57.6 cu. in. per one-gallon accumulator
Total needed is 439.72 / 57.6 (cu. in. per gallon) = 7.63-gallon accumulator.
Deadline past. Not available for submissions.
Winning Answer Submitted By:
弗兰克•StilwagnerCFPE, CFPMM, CFPS, CFPAI, Catching Fluid Power, Inc., Bolingbrook, IL
By Ernie Parker, AI, AJPP, AJPPCC, S, MT, MM, MIH, MIP, MMH,
Fluid Power Instructor, Hennepin Technical College,EParker@Hennepintech.edu
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