Compressor Shaft Power Vs. Actual Compressor Hp Rquirements

Reading your numbers I noted one thing:

You are in first place calculating your compressor power consumption, and finding it to be 7.4hp nominal.

Then, you measured your power input in the motor to be 9.3hp, this is your power consumption, not the power applied to the compressor.

If you check the numbers, you'll find a motor efficiency of around 0,79% (motor power consumption/nominal power consumption in compressor). THat's a number that makes sense. Anyway, did you do the power consumption calculation for the voltage you're applying?(P=Ui...) and checked tha coefficient for the 3f installation?

It seems that its ok. I'd like to read some answer from a colleague from electrics.

Similar to previous answers, the motor nameplate gives a power rating based on the maximum work it can be expected to do without overheating. If you ran a motor on overload (in excess of the nameplate rating) and it worked a treat I am sure you would be quite happy.

In round figures, your 7 hp machine is giving you 9 hp - great - a nice little workhorse.

Now look at your compressor (at a guess) rated at 37 scfm and fitted with a 7.4 hp (nameplate) motor - a quick calculation shows you get 5 scfm per hp - and that is very very very efficient (for any 100 psig machine let alone yours).

Compare this to another 35 scfm compressor but this time fitted with a 9.3 hp (nameplate) motor. Another quick calculation shows now you only get 4 scfm - which is a lot less efficient) - so which one is the best? - do you pay extra for the more efficient one?

Being power conscious, you buy the more efficient machine. You now connect up your 7.4 hp motor and measure the power actually consumed. You find it is running at 9.3 hp. Well there's a surprise - you don't get 5 scfm per hp after all. You only get 4 scfm which is typical of a compressor that size.

Your compressor is not consuming more power per scfm so to speak, it is simply running in excess of the nameplate rating. You just have a very nice little 7.4 hp motor capable of pushing out 9.3 hp.

But the nameplate says 7.4 hp - and that is fact. One up to the salesman.

horace40

If you installed a 7.5 Hp motor to this unit, you would exceed the the motor nameplate FLA and Service Factor Amps, which in turn would make the motor run hotter and greatly reduce the life of the motor. Am I missing something?

To N6377B. No, you are spot on. I was being a bit cynical in my old age.

If you bought a motor rated at 7.4hp, and connected it up to a machine that needed 9.3hp, you would probably burn the motor out after a while. Or if you had a decent starter it would shut down to protect itself.

But what if it was a 7.4hp high performance motor capable of running continuously at 9.3hp - you would no doubt be pleased. And being an engineer it might occur to you that if it was capable of running continuously at full load at 9.3hp - then why not rate it at 9.3 hp. Good question.

In actual fact you have not bought a 7.4hp motor. You have bought a 28.5 cfm compressor with a 7.4hp motor. What is the difference you might ask.

Now for the smoke and mirrors. A compressor with a motor rated at 7.4hp (on the label) is much cheaper to run (on paper) than a 9.3 hp motor. So you buy the economic compressor. Why not, if you can get 28.5 scfm from a 7.4hp motor why spend more money on a 9.3hp machine.

It is only because you have actually done some tests that you find the 7.4hp motor is running at 9.3hp. Which is fine as long as it does not burn. But it wont because it is a 9.3 hp motor with a 7.4hp label. And as long as the compressor pumps out 28.5 cfm who cares.

In other words, the 7.4 hp rating of the motor, taken at face value, has been built into your selection criteria to an extent enough to favour the purchase of that compressor. It always pays to double check the rating. Multiply the rated full-load amps by the rated voltage to get the actual hp.

In a less cynical mode, and if other calcs confirm, you have a genuine 7.4 hp motor running at 7.4 hp and delivering 28.5 cfm - which at about 4 cfm per hp is a typical expectation. Your precise tests measuring electrical input power and output airflow will confirm - or otherwise.

As previous question, more data needed.

No air flow is given, so can't check the 7.4 hp shaft power.

Perhaps I'm missing something, but I can't see where the 9.3 hp comes from. The motor (full-load) rating is not given, but at 12.5 amp the actual output power = motor rating x 12.5/13.4. If this = 7.4 hp, problem solved.

No voltage is given, so can't check by that route, but just by coincidence (or maybe not) the first figures I tried, 400 volt, 12.5 amp and effy and P.F. both 0.8 came to spot on 7.4 hp.

Thanks for the reply. The information you are requesting is in includeed in my initial qustion at the top of the page. The voltage is 460V and the air delivery is 28.75 ACFM.

Reply to #6 - N6377B, I'm a bit puzzled. I can't see voltage and air delivery in original posting. Full text I've seen posted is below. But putting the figures in I agree with 7.4 hp compressor shaft power (assuming 65% efficiency) and 9.3 hp motor output power.

Also a point about statement "compressor shaft power = Motor Power x Motor Efficiency x Transmission Losses (.97%)." Motor power here of course is electrical input power, and if you can measure it direct, OK. But it sounds like you're measuring current (and either measure volts or take supplier's word!). In that case you need to bring power factor in (from motor data sheet if you only have basic instruments) and motor shaft power (hp) = (volts x amps x eff x PF x √3)/746

"CAGI and ISO 3857 and ISO 1217 refers to compressor shaft power = electrical input x motor efficiency/.746. Shaft power defined as the power required at the compressor drive shaft. It is the sum of mechenical losses and the internal power.

Now I have someone that determines compressor shaft power exactly as I have quoted. "Motor Power x Motor Efficiency x Transmission Losses (.97%).

Now I understand the above equation, however the results as per this equation (Given there is no errors in their Math) results in a much lower Hp requirement when actual motor readings are taken. I will illistrate this by example.

I have a report for a compressor which reflects a shaft power of 7.40 Hp at 574 RPM at 100 psig

Now when I operate and test the compressor at 574 RPM at 100 psig. I take amp readings for each of the three phases and average the three readings. In this case the data is 12.3, 11.8, and 13.5 which averages to 12.5 amps. The motor nameplate FLA is 13.4 at the voltage I am operating the compressor. I take 12.5/13.4 = .93 which equals 9.3 Bhp. I respond by reporting that the compressor is 9.3 Bhp not 7.40 Hp as they report.

What is your opion on this subject? Am I missing something or how do I explain my findings in a way they will understand.

Thanks for your input."

I worked mostly with larger screw compressors, but you can also have a contribution by the air temperature. The compressor may actually operate at 7.4 Bhp (or within the allowable tolerance to be able to print that on the dataplate) if the air entering the compressor is at the upper end of the allowable inlet temperature. It should be tested at a standard temperature, but if the unit was tested in a test cell at a temperature of 100 deg F and then you have it sitting outside sucking in 40 deg F air, the higher air density will cause variation in the overall package horsepower. I've seen it happen. On the flip side, you are probably putting out slightly more cfm at the 9.3 Bhp if it is due to the temperature.

Nate,

Your comment answers the question asked by the thread poster "Am I missing something?" - Yes its the temperature that he missed in his data to correct his findings at standard temperature to cmpare vs. the name plate rating normally calculated at standard temperature. (from a TCN Engineer)

Agree, and for Quality reporting, make sure that the measuring instruments were properly calibrated to within allowable tolerance. (M. E. MORA)

N6377B,

First of all, you cannot accurately determine your motor HP by just measuring the Amps per phase. There are special meters made for this express purpose, that take into account that the amps are out of phase with the volts, (this what the power factor refers to). However, the power factor is determined at rated load, and at lesser and greater loads, it varies from the stated value, as does efficiency. To take an average of the amp readings you obtained and reference that to the nameplate FLA to determine actual HP delivered is not a good way at all. You have to measure the actual voltage also. In addition I find the amperage spread between the phases to be larger than normal. I have no idea why you are measuring that big a difference unless the voltages between phases are really off too.

A rule of thumb HP calculation for a 3 phase motor is:

This assumes the motor is operating at or not too far from its rated HP, since the relationship is not linear the further away from rated HP you go. (At no load, it still draws power, when lightly loaded it is less efficient than its nameplate value, and at LRA, it produces none.)

(Total of (Amps x Volts per phase)/3) x Motor Efficiency x Power Factor x 1.73 / 746 Watts per HP (The 1.73 is the square root of 3 required for 3 phase calculations)

Greg

Thanks, I appreciate your help and thanks to everyone else.