Fluid Power Journal

Problem: Calculate SCFM

Calculate SCFM:

  • 50 Pound Spring
  • 2″ Diameter Cylinder
  • 6″ Stroke
  • 1 Second to Extend
  • 1 Second to Retract
  • No Dwell Time
  • Hose is 3/8″ diameter and 6″ long between valve and cylinder
  • Neglect air in fitting or valve
See the Solution

Force is 200 lbs. + 50 lbs. of spring = 250 lbs.

Area of 2” diameter cylinder is 3.14 sq. in.

Pressure: F=PA  250 / 3.14 = 79.6 PSI

Compression Ratio =
(79.6 + 14.7) =6.42 C. R.
       14.7

Cycles / Minute = 60 / 2 sec. = 30 cpm

Volume of cylinder = 3.14 Sq. in. x 6” = 18.84 in³

CIM = 18.84 x 30 cpm = 565.2 CIM

Volume of hose = .11 Sq. in. x 6” = .66 in³

Total volume = 565.2 + .66 = 565.86 CIM

CFM = 565.86 / 1728 = .327 CFM

SCFM = .327 x 6.42 C. R. = 2.09 SCFM

Note – A gravity retracted cylinder (without the spring) would be:

200 lbs. / 3.14 = 63.69 PSI

C. R. (63.69+ 14.7) = 5.33 C. R.
                 14.7

Volume of cylinder and hose @ 30 CPM = .327 CFM (from above)

SCFM = .327 x 5.33 C. R. = 1.74 SCFM instead of 2.09 SCFM

Spring return cylinders do not use energy to retract, but more to extend. No free energy.

 

Deadline past. Not available for submissions.

By Ernie Parker, AI, AJPP, AJPPCC, S, MT, MM, MIH, MIP, MMH

This teaser is printed in the Fluid Power Journal. Those who submit the correct answer before the deadline will have their names printed in the Society Page newsletter and in Fluid Power Journal. The winners will also be entered into a drawing for a special gift.

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