O.T. CFM costs

About 5 weeks I started a new job in Hopkinsville, Ky.
Maintenance tech 1

We have a machine that makes 3" long parts and is capable of making a 36" long part. The stroke of the length is a 2.5" x 37" air cylinder. It cycles 35pcs per minute. My idea is to
put a 32" UHMW ballast inside the cylinder with a .5" hole bored in the center. This machine never makes any other length part (actually there are two machines). It currently
peaks at just over 85 SCFM while this cylinder is being charged. I've labeled these machines A GROSS WASTE of energy.
I ran this idea by our Engineering Manager and he thinks this would save only pennies, I strongly disagree. These machines have been running for seven years like this. I can't believe
this has gone on like this for so long.

Now to my question, what would be the cost savings if this ballast reduced the peak SCFM to say 45 ?

Sorry about this long rant.

Jeff K.

The US Department of Energy agrees with you, Jeff. Here is some info:
http://www.energystar.gov/ia/business/industry/compressed_air1.pdf

http://www.energystar.gov/ia/business/industry/compressed_air3.pdf

I think there is info at the EPA site as well.

http://www.energystar.gov/index.cfm...xml_no_dtd&oe=UTF-8&ie=UTF-8&q=compressed+air

So that 36 inch cylinder extends and retracts 35 times per minute? (That's one fast cylinder, which is why I'm assuming this isn't the case)

Or does it just extend to 36 inches and stay there the whole time they are running their 36 inch cut length part?

I guess the savings is going to be based on how often they have to exhaust/fill this cylinder. By the sounds of it, it stays at the 36 inch cut position almost all the time.

Caveman said:

The stroke of the length is a 2.5" x 37" air cylinder. It cycles 35pcs per minute. My idea is to
put a 32" UHMW ballast inside the cylinder with a .5" hole bored in the center.

Click to expand...

My concern is in the ballast. My presumption is that a 36 stoke cylinder that cycles 35 times a minute will have "cushions" at the end of the stroke. Your ballast might interfere with the normal operation of the cushion. Also if it's not relieved for air flow it might restrict the air flowing in and out of the cylinder. I'm concerned that you might get a "black eye" from your cylinder modification if it fails due to impact or does not move fast enough.

A couple of years ago I worked in a sawmill that had a horizontal resaw. The challenge with a horizontal is to separate the board on top from the board on the bottom. This is done by accelerating the two boards sideways and then stopping the bottom board only with an adjustable height shear. the top board flies off and falls in the correct sorting area. The manufacturer of the equipment used air cylinders to raise and lower the shear. These were (2) 6" inch bore 18" stroke and (2) 6" bore 2 inch stroke. If the sorting process required the 18" stoke cylinders to actuate 5 or 6 times in a row, the sawmill ran low on air and the shear did not rise to the proper height. When this happened the wood that should have gone to the trimmer ended up in the waste conveyor.

After some testing I discovered that the conventional way of controlling the air cylinders was not needed. I connected the hoisting (up) side of the cylinders to the air supply full time (no exhausting) and just used the control valve to supply and exhaust air from other side (lowering) of the cylinder. To the sawmill owner this was a win win situation. Almost no extra cost of materials, just some hose and pipe plugs, and he didn't run out of air and throw away product.

I do not know if what I'm about to suggest will work for your situation.

Look into using a uhmw ballast like your original thought, and using a smaller bore cylinder to return the stroke of the original cylinder. The ballast would be needed to limit the stroke of the larger and heftier cylinder, the smaller bore cylinder may not be able to limit the stroke before destructing. This based on the concept that you need the displacement (power) in one direction only. Another thought would be to use an external stop and cushion along with the smaller return stroke cylinder.

Ok, using the info you supplied, the formula from Tom's second link, and a couple of assumptions, here is what I come up with:

Assuming you have a 4" diameter x 37" long cylinder (if you said the diameter, I missed it) and are completely exhausting each side of the cylinder when cycling (which only really happens if you move the full length - if you move only 3" you certainly do have to add some air to re-compress the remaining 34" of air, but I don't know that formula - so I will use the full 37" as a worst-case calculation) you use 0.269 cubic feet of air for each stroke or twice that for a full cycle (ignoring the volume of the cylinder rod, of course).

area x length x 2 = cubic feet, or

[PI x (4^2) / 4 / (12^2)] x 37 x 2 = .538 cubic feet

If you are cycling this cylinder 35 times per minute, your actual CFM usage is:

0.538 cf * 35 per min = 18.835 cfm

Consequently, assuming 7000 hours per year running and your cost for power is $0.05/kWH:

18.835 cfm x 0.61 kW/cfm x 0.05 $/kWH * 7000 hr/yr = $4,021 per year.

So your cost of using this cylinder (at its full stroke) is $4K. Obviously you can only save a portion of this since you are not eliminating the process but only modifying it.

Hope this is right and I hope this helps.

Steve

Replace the air with an electric actuator. Most all of the actuator companies have price saving charts on the conversions.

Many thanks to all that read or replied to this problem, I feel
like I didn’t explain this very well.

This 2.5” diameter by 37” stroke cylinder remains fully extended.
It is retracted to an adjustable cushioned stop currently set at 3”.
The part is clamped to the stroke cylinder assembly, then extended
fully. The part is held in place and cut by another clamp while the
stroke cylinder unclamps and retracts.

As you can see we are not using 34” of this cylinder

While this machine is running the stroke cylinder around the piston
area is very hot to the touch (above 120 F) I’m guessing due to the
wasted energy and lack of lubrication.

If this works well I’m going to order a reseal kit with a shorter barrel.

Again, thanks

Jeff K.

Cylinder air supply should have a filter to trap particulate and an oiler for lubrication.

I would expect a cylinder approx 3 foot long cycling at 35 cycle per minute to get warm. I think you would get warm too moving 3 feet at almost 2 second interval.

A cylinder with a shorter length better matched to your load requirement will save compressed air and money.

Dan Bentler

Ok, let's try this again – with this new information and the realization that I did my first calculation based on standard atmosphere air, my first numbers were incorrect. Here is what I have now:

I did a little research and found that “standard” air at sea level is at a pressure of 14.7 psi and if you compress air to 100 psi, the same volume will hold 6.8 times the amount of air it does at 1 atmosphere:

100 / 14.7 = 6.803

Now, with a 2.5" diameter cylinder moving 3 inches, you are moving .00852 cubic feet of air each stroke. Furthermore, you are also re-compressing the air in the unused 34" of the one side of the cylinder. So this breaks up the air usage into two parts; first we have the volume we are actually using and the volume we are re-compressing. Since you are only interested in the saving the costs associated with the second part, we will ignore the first part.

Your 34” of cylinder can hold 0.0966 cubic feet of "standard" air, but because we are only exhausting the air rather than pushing it all out like we would if we were moving the piston, we only need to replace 5.8 times of 1 atmosphere air for each stroke rather than 6.8 (1 atmosphere of air stays in the cylinder). And unlike my previous calculation, this only happens on one side of the cylinder because you are only moving back the 3” after the cylinder has been fully extended.

Now, using the same calculation, the total cost of air when using this cylinder is:

[PI x (2.5^2) / 4 / (12^2)] x 37 = .0966 cubic feet of volume.

Multiply that by the compression factor of 5.8 gives us:

0.0966 x 5.8 = 0.5602 cubic feet of "standard" air at 100 psi.

Consequently:

0.5602 cf x 35 per minute = 19.607 cfm

Once again, assuming 7000 hours per year and $0.05/kWH:

19.607 cfm x 0.61 kW/cfm x 0.05 $/kWH x 7000 hr/yr = $4,186.09 savings per year

Steve

Thank you Steve, that is exactly the information I was looking for.

You are Awsome

Caveman,

My pleasure. It was fun and educational doing the research and figuring.

I came up with one more thought for you...I believe I was mistaken (favorably for your cause, actually) when I used the factor of 5.8.

Because we measure air pressure using gauges that display "zero" at 1 atmosphere (actually 14.7 psi) and "100" at system pressure (actually 114.7 psi), I believe the true calculation for the effective volume of compressed air should be:

114.7 / 14.7 = 7.803

Once again, since 1 atmosphere is left in the cylinder, we subtract 1 and get a factor of 6.803.

The rest of the calculation should hold true with this only this one factor changed and the consequent cost savings should be:

0.0966 x 6.8 x 35 x 0.61 x $0.05 x 7000 = $4,908.54 per year.

Steve