# 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.

Given information:

• 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.

Frank Stilwagner, CFPE, 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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