CFM FOR 6" BORE CYL WITH 50" STROKE

Well, hydraulics and pneumatics are different in trhat hydraulic fluid is, for all practical purposes, incompressible.

Air, on the other hand isn't incompressible.

You'll need to know the pressure in the cylinder to determine how many cubic feet of air at STP (standard temperature and pressure -- 72 deg. F, 14.7 psi) will be required to be pumped into the volume of the air cylinder.

You already know the volume of the cylinder, 9 X Pi X 50 = 1414 cu. in. = 0.82 cu. ft.

The rate is simple enough . . . When you know how many cubic feet of air you'll need, then all you have to do is either dictate how fast you need the cylinder to move, or put up with the speed you can move it with a given compressor.

VE = Volumetric efficency. Since the piston cannot completely touch the end of the cylinder without damage, there is a small space called the clearance. Then, there are valves that let gas in/out they have clearances that air is "trapped". finally, the air that is trapped in the clearance volume areas compresses, gets hot, expannds, cools somewhat.

there are formulas to estimate the VE from the clearance volume (typically stated in percent). http://www.arielcorp.com/application_manual/arieldbCompressor_Theory.htm

Typical cylinders have 15% cleance and an 75% volumetric efficency. Depends on manufacturer and gas compressed, and rations. The higher the rations, the lower the VE too.

Definitely use the volume equation for baseline volume times the rpm's. Also remember your clearances for the cylinder, (15% may be a bit high) and calculate your efficiencies. Vicini is right about the air being different temperatures and pressures. You will have to find a nominal or average state for your air to define what you are looking for.

here is a VE equation

VE = 100 -ratios - %clearance*(Zs/Zd* (Ratios ^ (1/k)) - 1)

Ratios = Pd/Ps (discharge Press in absolute/suction press in absolute)

Zs = supercompressibility suction

k = Cp/Cv Cp = molar heat capacity at constnt pressure, Cv = at constant volume

Probably so but immaterial to the case at hand. However a continous flow rate of 10.8 SCFM will extend two 6 x 50" cylinders with 80 PSIG and raise the gangway in 60 seconds if it requires less force than 4,000 lbf. Sort of keep it simple for us mental migets who just want to get the job done.

CFM (cubic feet per minute) = Pye x radius squared x stroke divided by 1728 (cubic inchs to cubic feet) x RPM.

Phil

THE TASK AT HAND IS THAT THE GANGWAY WITH TWO CYLINDERS HAS AN OPERATING PRESSURE OF 80 PSI AND ROD SIZE OF 1 3/8" DIAMETER. WHAT RPM AND VE SHOULD I USE FOR CALCULATING HOW CFM SHOULD IT TAKES TO OPERATE THIS GANGWAY EFFICIENTLY. THE GANGWAY IS PROBABLY LOWERING AT 1 RPM HYPERTHECTICALLY.

Hi Again

This forum is famous for not giving enough information. So to make the second mistake I will ASSUME:

1. The rod end of both cylinders is upward.
2. The load of the gang way is half the force that the cylinders provide.
3. Gravity works.

To lower the cylinder remove the pressure that is holding it (and the gangway) up.

Use a speed control for the exhausting air or the gangway may fall abruptly

To raise the gangway the flow rate is Q(scfm) = [.0019 x 36 x 50 x 94.7] / 60sec.

Flow rate per cylinder is 5.4 SCFM and 10.8 SCFM total for both cylinders in the upward direction only.

Further: The piston face in many if not most pneumatic cylinders do contact one or both ends and may or may not have cushions ("valves") in either end.

Break away might typically be about 10 or 15 PSI differential and the force provided by the cylinders for vertical lift should be at least 25% more than the load.

Not least: A gangway may have people or material that may be damaged in the event of a pressure loss free fall. I would suggest pilot operated check valves close connected to the cylinders to trap internal air if pilot pressure is lost. Also the gangway should not be lowered when there is no air to stop the decent in an emergency. Use a pressure switch.

One more consideration: A bind or mechanical blockage (Like forgetting to remove the pin) when the pressure is applied to raise the gangway could cause a violent launch if the blockage is suddenly cleared under high pressure. Various sensors can ascertain that the gangway is in motion within a short time after being initiated.

Improperly done one might create a catapult that would make the cow jumping over the moon a mere fairy tale or an elevator in free fall less spectacular.

Forget VE and concentrate on those who will walk the plank.

TOM,

FIRST OF ALL THANKS. I DON'T KNOW THE CONSTANTS THAT YOU ARE USING IN THE FLOW FORMULA. IS THIS FORMULA FOUND IN ROAKE'S HANDBOOK OF ENGINEERING FORMULAS? ARE THE NUMBERS .0019, 36 AND 94.7 CONSTANTS THAT I CAN GET FROM A TABLE?

OUR GANGWAYS TYPICALLY HAS A LIFTING CAPACITY OF 500 POUNDS USING 2 CYLINDERS ON 3, 4,5 OR PERHAPS 6 STEP GANGWAYS. FOR FURTHER CLARIFICATION DO A GOOLGE SEARCH ON GANGWAYS AND THEIR OPERATION.

This formula is found in Tom Kreher's crib notes.

Their is only one constant, .0019 which we will come back to.

36 is the bore diameter a variable you stipulated, 6² [ 6 x 6 = 36]

94.7 is the pressure, a variable you stipulated plus 1 atmosphere. 80psig+14.7 = 94.7

All the constants were crunched to prevent doing them over and over and over and ~

The bore diameter ² x .7854 [a constant] = Area

The volume, bore area x stroke, is in cubic inches divided by 1728 = cubic feet.

The compression factor [# of scf per unit of volume at any given pressure] is found by (P _psig + 14.7 _psia) / 14.7 [constant]

The time is normally specified in seconds. T _sec / 60 _sec/min = T _min

The flow in Standard Cubic Feet per Minute _SCFM = Volume x compression factor/ time

We have crunched the constants (.7854 x 60) / (1728 x 14.7) = .001855 or .0019

Then Q _scfm = .0019 (or .001855) x D^2" x L" x (P+14.7) / T _sec

Using your variables Q = .0019 x 36 x 50 x 94.7 / 60 = 5.3979 _scfm for one cylinder 50" stroke in one minute or twice that, 10.8 _scfm for two cylinders.

In practical applications you will not notice changes in heat ratios, humidity or barometric pressure.

You have almost 57² inches of area and at 80 psig a theoretical force about 4,500 lbs. As your gangway nears verticle the cylinder force vector will be greatly reduced but so will the load.

I would cut back the pressure so that you don't toss people and things off the gang-way like flap jacks and don't neglect safety measures in the avent of a pressure loss.

Nobody answered what RPM meant. Rotations per minute. How many cycles the engine performs per minute.

Yeah RPM's: revolutions per minute.

Sometimes we assume that people know everything.

Hope you really wanted to know.

CFM (Cubic Feet per Minute) requires a TIME factor.

From the dimensions only the cylinder volume is determined.

CFM is an ambiguous term that requires that the pressure be specified. Most pneumatic calculations use SCFM (Std Cu. Ft. per Min) or if only the volume SCF (Std Cu Feet).

You might think of SCF or SCFM as the volume of atmospheric air the compressor must take in to deliver the amount required at the working pressure.

Here is a nifty formula with all the constants crunched that you may paste on your lamp. Books only give the long, time consuming, error prone fromulas.

Q (scfm) = .0019 x D"^2 x L" (strk) x [P (psig) + 14.7] / T (sec)

So if you wished to stroke the subject cylinder 50" in 30 seconds at 80 psig:

Q = [.0019 x 36 x 50 x 84.7] / 10 = 28.967 (30 SCFM) flow rate.

If you want to know how much compressed air the cylinder holds at a given pressure in SCF try V(scf) = .00003 x D^2 x L" x [P(psig) + 14.7]

Then at 50 psig (pounds per square inch gage pressure) the volume of compressed air(disregarding the rod) would be V = .00003 x 36 x 50 x 64.7 = 3.49 SCF

From another angle. The physical volume obviously v = .7854 x 36 x 50 /1728 = .818 Cubic Foot. If you multiply the physical volume by the compression factor [Cf =(P+14.7)/14.7] and for 70 psig the compression factor 84.7/14.7= 5.76 Then 5.76 x .818 = 4.7 SCF. For your amusement and amazement multiply by the fraction of a minute ie. 60sec/min / 10sec x 4.7 SCF = 28.2 SCFM. Pretty close to the flow, Q, above.

In summary the physical volume x the compression factor x the time to complete a stoke deals only with flow rate to drive the cylinder one stroke in the allotted time.

For cycles, a double acting pneumatic cylinder will also consume compressed air on the return stroke. Again disregarding the rod volume, if one stroke consumes 4.7 SCF then one cycle would consume 9.4 SCF and the actual consumption would be determined by the number of cycles per minute. However unless the time is of each stroke is much faster than 10 seconds each cycle would take 20 seconds and 3 cycles per minute x 9.4 SCF per cycle would consume 28.2 SCFM. If that sounds familiar it is. With continuous cycles the instantaneous flow rate per cycle and the continuous flow rate are the same.

You asked. Tom