Calculating Energy Savings From Removing Air Cylinders

Esieli

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Does anyone know how to calculate the savings from now needing to run an air compressor all day?

Basically I have a design that replaced 6 * 1" bore x 4"stroke cylinders that ran at 90psi.

I'm sure there's a good amount of savings there, but not exactly sure how to figure out how much less the compressor would need to run.



Thank you for the help!
 
Calculate the total amount of air consumption.

Each time a cylinder strokes out it consumes 3.1 cubic inches of air. Each time it returns it consumes a bit less because of the rod. You didn't specify the rod diameter, but assuming a 3/8" diameter rod, the return stroke consumes 2.7 cubic inches. That's 5.8 cubic inches per cycle. You need to multiply that by the number of cycles for all cylinders to get your total consumption.
Then check the specs for your compressor to find how many cubic inches per minute it can deliver at 90 PSI and you can calculate the necessary running time to support that consumption.
 
What kind of instrumentation do you already have on the machine, such as a power meter.

Does the new compressor run at a variable speed or is it on/off control?

In my opinion, the quick and dirty method would be to first calculate the energy usage of the old system: EnergyOld(kWh) = 24 (hours) * √3 × PowerFactorOld * FLAOld * LinetoLineVoltage (V, probably 480) / 1000

Then, do the same thing for your new setup: EnergyNew(kWh) = DailyRunTime (hours) * √3 * PowerFactorNew * FLA * LinetoLineVoltage (V, probably 480) / 1000

Then subtract EnergyNew from EnergyOld and you get your net power savings in kWh. Multiply by cost of energy in dollars per kWh (usually around $0.12/kWh in the US) to get cost per day.

This all assumes the old system ran 24 hours a day and the new one controls on/off, and not variable. If variable, your best bet is to get a power meter or if you have a modern VFD get the energy consumption from it.
 
Last edited:
> Does anyone know how to calculate the savings from now needing to run an air compressor all day?


All day?


Does the compressor really run continuously, or does it fill a tank/reservoir as needed to maintain 90psi, and the reservoir drives the cylinder?




Assuming a reservoir is in place, rough estimate of work derived by multiplying by 1:

  • (3.1+2.7)in**3/cycle * 90psi = 522 in-lbf of work per full cycle
  • 522 in-lbf/cycle * 1ft/12in = 43.5 ft-lbf/cycle
  • 43.5 ft-lbf/cycle * 1 hp-s / 550 ft-lb = 0.079 hp-s/cycle
  • 0.079 hp-s/cycle * 1h/3600s = 22e-6 hp-h/cycle
  • 22e-6 hp-h/cycle * 0.7457 kW/hp = = 16.4e-6 kWh/cycle
  • 16.4e-6 kWh/cycle * 0.05 $/kWh = 0.8 μ$/cycle
Also, there are still various losses and efficiences to consider (you are accelerating the motor and the compressor, friction, power factor, efficiency of motor (near 100%) and compressor efficiency, etc.; I expect compressor efficiency (10%?) will dominate, so say 5% overall system efficiency

  • 0.8 μ$/cycle / 5% = 16.383 μ$/cycle
Say the six cylinders are running at 1Hz all day, 24/7

  • 16.383 μ$/cycle/cylinder * 1Hz(cycle/s) * 86.4ks/d * 6 cylinders/plant = 8.5$/d/plant
  • 8.5 $/d/plant * 365.25 d/y = 3.1k$/y
And a factor of four less if the plant runs only 40h/wk. So this forum has already eaten up your first year of savings in time just by looking at it ;).



Caveats

  • I am not at all sure that is correct, work is a funny thing; from first principles I think it is right.
  • Unless I made a silly mistake somewhere, it is almost certainly within a factor of 2 of a lower limit.
  • Someone here should check the math.
  • Given the WAGish nature of the efficiency and losses, it could be low by a factor 10, event assuming I did everything right.
  • Bottom line, I would put a meter on the compressor and run the cylinders for some integral number of compressor/reservoir pressure cycles to get a better estimate
    • But even that ignores the fact that if other equipment in the plant uses the compressor then the actual incremental work used by these cylinders will be still different: lower as some fixed losses may be shared among all equipment in the plant, whereas all fixed losses will be attributed to the cylinders in the test as suggested.
 
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If the compressor is only supplying air for the 4 cylinders, I would go the lazy route and measure the amps used by the compressor, this will give you the actual cost including leaks and heat loss.
 
drbitboy said:
> Does anyone know how to calculate the savings from now needing to run an air compressor all day?


All day?


Does the compressor really run continuously, or does it fill a tank/reservoir as needed to maintain 90psi, and the reservoir drives the cylinder?




Assuming a reservoir is in place, rough estimate:
(3.1+2.7)/2in**3/cycle * 90psi = 216 lb-in of work.
Click to expand...


I'd recommend instead of picking 90psi for the force working against the cylinder head, pick the average of the compressor start setpoint and compressor stop setpoint (if deadband control). Your compressor won't be working against 90psi all the time.

Additionally, there are some assumptions about the thermals here that may or may not hold up depending on how the system is set up. I hesitate to calculate energy based on resulting pressure and not on electrical supply to the machine.
 
Holmux said:
If the compressor is only supplying air for the 4 cylinders, I would go the lazy route and measure the amps used by the compressor, this will give you the actual cost including leaks and heat loss.
Click to expand...

I think here this may be the lazy method, but it's also the most appropriate method. It makes the fewest assumptions and gets closest to measuring the actual energy savings (directly measuring energy consumption of the new system means you only have to assume the energy consumption of the old system).

If you can't measure amps (your starter doesn't have a current meter on it and you don't want to buy new hardware) you can still assume the energy consumption of the new system based on motor specifications and get pretty good.
 
So by that calculation, this machine was consuming around 9.66CFM...

I don't have the air machine hooked up to see how it changes my compressor, but just trying to come up with some rough numbers of how much less it would need to run...

Obviously the compressor would differ depending on make/model, but just assuming the average plant air compressor.. Any ideas?
 
You have not stated how you have designed something that reduces the amount of air, although if it is your design you may want to keep it to yourself, however, if you designed it you must know the percentage of the savings as far as air consumption goes. Although you my not know how much compressed air costs per cubic metre if you are using 25% less air and only one system on the compressor then you have saved 25% assuming a perfect system (remember any calculation would be based on everything being perfect i.e. no air leaks etc).
There are systems out there that recover the air based on putting energy back into the system rather than exhausting it.
The only way would be to monitor the energy used over time to produce the compressed air and number of operations. I don't think without a lot of instrumentation you could come up with an accurate saving and monitoring both with & without your improvement.
Lets face it, if it is more efficient then you are saving money.
 
Peter Nachtwey said:
You can buy power meters. Some will log or total power over time.

It is useless trying to measure by air consumption because much of the energy goes into heat.
Click to expand...




Peter and others are right; my earlier $8.50/d ballpark estimate is a lower limit at best, assuming no mistakes.


The problem will be breaking out the increment used by the six cylinders if other equipment is using compressed air from the same system, especially if the compressor is only filling a reservoir tank. Your best bet may be to determine some mean kWh cost for one cubic inch ($/in**3; or cubic whatever) of 90psi air and multiply that by unity (i.e. by 5.8 in**3/cycle/cylinder * N cycles/unit-time * 6 cylinder/plant) to get $/unit-time.
 
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parky said:
You have not stated how you have designed something that reduces the amount of air, although if it is your design you may want to keep it to yourself, however, if you designed it you must know the percentage of the savings as far as air consumption goes. Although you my not know how much compressed air costs per cubic metre if you are using 25% less air and only one system on the compressor then you have saved 25% assuming a perfect system (remember any calculation would be based on everything being perfect i.e. no air leaks etc).
There are systems out there that recover the air based on putting energy back into the system rather than exhausting it.
The only way would be to monitor the energy used over time to produce the compressed air and number of operations. I don't think without a lot of instrumentation you could come up with an accurate saving and monitoring both with & without your improvement.
Lets face it, if it is more efficient then you are saving money.
Click to expand...

I designed the same machine but using all electric cylinders. I can figure up that part and what it cost to operate and then subtract that cost from the cost of the old machine operation cost.
 
What is electrical draw of compressor before change - electrical draw after change? Convert answer to kilowatt-hours and then check the electric bill.
The savings that anyone is concerned about is $$$$$
 

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