Compressor

Nope.

The job to be done has not changed, you still want to compress a certain volume of air. Think about it, you use a compressor to deliver pressure at a certain CFM. CFM = Cubic Feet per Minute. Where is power mentioned in there?

The motor is sized to do that job. If it could have been done with a smaller motor, it probably would have, because motor HP costs money for the OEM, not just the user. If by chance the motor was over sized, it would still only draw as much power as it needs to. There might have been a little extra overhead power consumed if it was over sized (extra iron hysteresis losses) but not much, maybe a few percent.

If you put a smaller motor on it, it will just overload. When it does, and it trips off, you do save energy but of course you don't compress air!

In addition (you didn't ask but I'll tell you anyway because this question always follows), if you change the stroke by changing a pulley and use a smaller motor, or you change the motor speed with a VFD (which also reduces HP), you still don't save energy on a compressor. You still need the same volume of air and compressing it takes a fixed amount of HP and time. Changing the compression volume just changes the time it takes to deliver the needed amount, so although you use less HP, you use it for longer. Remember, you are billed for kWh, not kW.

Time proven method for reducing energy consumption on compressors: fix the leaks. There is no better and cheaper way to cut the losses.

JRaef, I think you missing my point. I am not questioning the makeup of the 5 HP motor driving a one piston compressor with proper Pully ratios and what have you. What I am asking is: If one stroke = one revolution, deliveres lets say 36 cu in of air to a maximum of 140 psi tank pressure and the rotor delivers 6 cu in every 60 deg = 1 revolutin . My thinking is that 6 cu in of air are easier to compress than 36 cu in .

You are comparing 1/6 of the work to the whole. Sure its easier but its not done. A full rotation is involve to accomplish the whole. At that point they are equal. So you would need a motor of equal size.

The energy it takes to compress whatever it is your compressing fixed for your system. To save power your goal should to maximize the efficiency of the system you are inputting power into. So if you are asking if using a more efficient compressor which requires a smaller motor to drive it will save electricity the answer is yes. Improving the efficiency of power conversion from electrical to mechcanical is your only area to reduce electrical power requirements.

jgunders, let me rephrase what my question is. Do I save energy by compressing 6 cu in of air every 60 degees that equals 36 cu in per revolution, versus one piston stroke per revolution compressing 36 cu in. Tank pressure 140 PSI. Can you see the volume per rev is the same .Will the power requirement be the same .

In case 1, you are compressing 6 x 6cu in = 36 cu in per revolution to 140 psi.

In case 2, you are compressing 36 cu in per revolution to 140 psi.

It appears to be the same amount of work, therefore I believe it will require the same amount of power.

It sounds like to correctly answer your question we would need to be able to compare the efficiencies of the compressors. I am not a compressor expert but would guess it would be an easy question for a Mechanical Engineer to answer. The energy it takes to do the job will not change but a difference in mechcanical efficiency could save you some electrical power.

"had a 5 HP one piston compressor on top the 60 gal tank. If I had a rotary type compressor ,compressing only 1/6 of the volume every 60 deg , that would be the same volume per one revolution. Could I use a smaller motor and save electicity."

NO. Compressing the same volume of intake air to the same final pressure takes the same amount of power any way you slice it.

What you need to do is to replace the present compressor and install the "Neal Tank" compressor, which once started will produce all the compressend air you will ever need.

The Neal Tank