Calculating Energy Savings From Removing Air Cylinders

Whoops!


Ignore my earlier calculation, thermodynamically

• dE = dQ - pdV

but air is compressible and dQ is 0 (adiabatic; no change in entropy either).


So work is not

• (p1-p0) * v1

which assumes an incompressible fluid, which is more or less what my calculations were doing ...

I can see you guys are ignoring the power meter idea. It is easy to take readings and compute the difference. Oh well.


But since you guys want to go off the deep end I wrote this many years ago.


https://deltamotion.com/peter/Mathcad/Mathcad - EnergyInAir.pdf




The calculations are irrelevant because all that matters is the energy required to make the compressor and actuator motors work. I don't see where fussing with physics helps.

Peter Nachtwey said:

I don't see where fussing with physics helps.

Click to expand...

Peter please: it's thermodynamics, not physics.



Because it puts a lower limit, perhaps ballpark, estimate on the savings, to decide if doing it the right way, i.e. with a meter, is worth significant effort.


For example, measuring the incremental cost of some unknown number of 6" air cylinder cycles on a compressor that runs intermittently under bang-bang control into a reservoir tank serving other equipment, if that is the case, would be problematic at best. It's easier to calibrate torque screwdrivers.



And if we cannot trot them out now and again, then what is the point of the laws of thermo:

• you can't get ahead;
• you can't break even;
• you can't get out of the game?

Originally posted by Peter Nachtwey:

The calculations are irrelevant because all that matters is the energy required to make the compressor and actuator motors work. I don't see where fussing with physics helps.

Click to expand...

I think that exercise is generally referred to a "engineering". At least that is what we would consider it in the backwater that is Brillion, WI.

Keep in mind that the PROPOSED system hasn't been implemented yet. Yes, he can try measure the compressor energy being used by the system today. Whether he would be able to actually resolve the usage of the devices he is talking about relative to all the other varying loads in the plant is another matter. And yes, you can make the point that if he can't resolve the difference then it really doesn't matter, at least at his site with his electric rates. But I would say that, in a typical production environment where you typically aren't allowed to shut down everything other than what you are working on at the moment, if you really want to know this information calculation would be the better path in this case.

Keith

Physics includes many disciplines.
https://en.wikipedia.org/wiki/Thermodynamics
You stand corrected.

I would call this a fluid power problem to be more specific.

Simply buying a power meter would be cheaper than all the time wasted discussing this.

Peter Nachtwey said:

Physics includes many disciplines.
https://en.wikipedia.org/wiki/Thermodynamics
You stand corrected.

I would call this a fluid power problem to be more specific.

Simply buying a power meter would be cheaper than all the time wasted discussing this.

Click to expand...

If this machine was currently running, I could do so with a power meter, but this is a machine I basically I designed to function similarly, but with all electric motion in place of pneumatics.

I do not have the ability to hook the old design up and see what kind of power it takes for our compressor to keep up. That's why I'm trying to figure out an average cost per cfm a company may give toward a ROI..

First law of thermodynamics: Energy is neither created nor destroyed. The load will require the same amount of power regardless of whether the source is pneumatic or electric actuators. Therefore you need to determine the load, and then compare the efficiency of delivering that power by direct electric operators or by a compressor.

Your electric operators are probably on the order of 90% to 95% efficient.

Note that regardless of the pressure needed in the cylinder the compressor in a normal shop air system works at around 100 psig. Use that in your power calculations. Then you need to know the volume of ambient air compressed corresponding to the volume displaced by cylinder travel, often referred to as FAD - free air delivery.

This link may help you. https://www.engineeringtoolbox.com/horsepower-compressed-air-d_1363.html

Note that this is the "air power" and must be divided by the compressor efficiency to get consumed power. A rule of thumb is 4 CFM / hp https://www.engineeringtoolbox.com/air-compressor-types-d_441.html

Then you can calculate the power cost: https://www.engineeringtoolbox.com/compressed-air-power-costs-d_1013.html

Tom Jenkins said:

First law of thermodynamics: Energy is neither created nor destroyed. The load will require the same amount of power regardless of whether the source is pneumatic or electric actuators. Therefore you need to determine the load, and then compare the efficiency of delivering that power by direct electric operators or by a compressor.

Click to expand...

Bingo!

Tom Jenkins said:

A rule of thumb is 4 CFM / hp

Click to expand...

so 0.25 hp-min/ft**3:


So multiplying by unity


0.25 hp-min/ft**3 * 60s/min * 1/1278 ft**3/in**3 * 5.8 in**3/cycle


= 0.05 hp-s/cycle @ 100psi and presumably including compressor efficiency, vs. my self-discredited non-compressible calculation which put it at 0.079 hp-s/cycle at 90psi excluding compressor efficiency, so within a factor of 10-20. I'm happy with that, but it would still needs a meter if the process existed, which it does not apparently, and it is still doubtful if it could be measured even with a meter.


But it's still bogos because I have not considered the heat added.

Peter Nachtwey said:

Physics includes many disciplines.
https://en.wikipedia.org/wiki/Thermodynamics
You stand corrected.

Click to expand...

Touche`!

Peter Nachtwey said:

I would call this a fluid power problem to be more specific.

Click to expand...

tomayto tomahto ...

Peter Nachtwey said:

Simply buying a power meter would be cheaper than all the time wasted discussing this.

Click to expand...

You're not exactly the first to suggest that.


But not cheaper than building the plant with the air cylinders to measure, which is apparently the only way to use a meter at this point.




- [jes' pokin' the bar;-]

For those who are still interested and playing along at home, this is a page from the site that Tom Jenkins linked to earlier:

https://www.airbestpractices.com/te...electric-tool-calculations-and-considerations

It paints a fairly grim picture of pneumatics at the actuator end but, depending on your compressor technology, decreasing air usage might not have system benefits equivalent to the point-of-use benefits.

Keith

Very likely you are loosing more thru air leaks.

A company not to be named hired us to determine the specs for a new 100 hp compressor. It was to add to 3 similar existing.
We went in with headphones and pointed mikes.

We found and they repaired so many leaks that ultimately the existing third compressor was unnecessary.
I was at a company when they found a blind cap had come loose on an 8 in supply line.
Having said all that compressed air is shockingly expensive to produce and deliver.

When you find your savings please post them.

If this compressor is controlled by a VFD that feature is probably built into the VFD.

Just take the KwH used and multiply by what your electric cost is per KwH and you will know to the penny the Energy Cost.

I used this feature on most of my VFD's and they are dead on, according to my electric bill.