What Happens to AC Electric induction Motor Parameters Beyond Service Factor HP & Amps 1

[HydraulicsGuy]

These questions are specifically for AC Electric Induction Motors that we might spec to run an HPU. The questions are about what happens as we go beyond Service Factor HP & Amps.

One example of the kind of motor I'm referring to:

https://www.worldwideelectric.net/p...premium-efficiency-severe-duty-motors-c-face/

Parameters for questions below:

230 V 3-phase
SF = 1.15
FLA (full load amps) = 6.14
SFA (service factor amps) = 7.06
Full-load Eff = 87%
Full-load PF = 69%
Full-load Torque (FLT) = 8.98 lb-ft
Breakdown Torque (BDT) = 275% FLT

1. Let's say motor is drawing 8.5 Amps. What has happened mathematically to efficiency and power factor? What do you think their values would be?

2. Let's now say current continues to climb to 2 x FLA = 12.28 Amps. What has happened mathematically to efficiency and power factor? What do you think their values would be?

3. Is current directly proportional to torque demand? So as the torque demand goes beyond full-load torque toward breakdown torque, does the current follow proportionally? If current = 2 x FLA, is the torque that's being demanded = 2 x FLT? If not, what do you think torque would be?

4. If BDT = 275% FLT, is breakdown current = 275% full-load current? If not, what do you think current would be?

5. What happens to voltage through all this? Is voltage only dependent on the power source, or does it degrade as motor goes beyond service factor HP & amps?

6. I assume that through all this, the motor power equation still applies: P = V x A x Eff x PF (x 1.73 if 3-phase)?

 

[HydraulicsGuy]

Is this question better suited for the EEs?

 

[PNachtwey]

This isn't a very EE answer but I know from experience. We tried adjusting the pressure too high. In our case, too high was 1800 psi. Now we only run at 1500 psi. The motor gets hot and the paint starts to peel. We didn't burn up the motor but it might of lost one of its lives.

I do know that the torque is roughly proportional to the armature current.

I have a degree in EE but I do mostly hydraulics and simulations. I would have to look up the answer for you. A good place to try is plcs.net. There are a few motor and drive guys there that can probably provide a quick answer.



Peter Nachtwey
Delta Computer Systems

http://www.deltamotion.com

http://forum.deltamotion.com/

 

[Jacc]

You will get good answers here:

https://www.eng-tips.com/threadminder.cfm?pid=237

 

[itsmoked]

Motor life is all about staying cool. The temperature in a motor is mostly the voltage drop across the wire in the windings. A motor's rated life occurs when it's run at it's rated FLA (Full Load Amperage). If you run into its service rating you directly cut years off of its life. If you run it beyond the service factor you will kill it momentarily; that's in hours, minutes, or even seconds.

Reasonable engineering requires that you never run a motor beyond the service factor and if you have to run it in the service factor you do everything you can to have that be only brief periods.

Furthermore, the electrical code demands motors be protected from overload by using very specific protection devices and values found in the electrical code tables. If the rules are followed then you shouldn't actually be able to run a motor beyond its service factor as you should be triggering the protection.

Keith Cress
kcress -

http://www.flaminsystems.com

 

[TimSchrader2]

Hello

You need to get all the curves from the MFG to answer some of the questions above. At low Torques or no torque it is less effcient. That would be the no load current, as I recall.

You shoud not go above the FLA at all if Sf=1.0. As itsmoked warns. If SF=1.15 and you have a continuous duty motor, you run can above FLA for short periods. Less than a few minutes. I have set many a HPU at 1.10 x FLA when the SF'=1.15. But you should check the overload relays trip before shipping the unit. Some overload wiring needs to be different if the motor contactor has a sensitivity. And I knew these would only run for less than a half hour, IF running all day I would not have done so. Some will say that the continous duty and Sf ratings are totally different, but I have seen the continous duty motors run better even for short periods. Run the HPU at Full GPM and FULL pressure at same time to test it. Most MFG's will never rate thier HPU's at full GPM and Pressure unless you ask them too, and more then once. they will say max GPM amd Max press, but not in the same sentence

I have run a HPU motor at almost 2 x its rated HP for about 20 minutes or so just to pass a test and only to pass the test. The inspector day rate was expensive so I took a chance so as t not have him come back. Only did it once then got a larger motor. Four hydraulic motors turning one slewing ring thru four worm gears operate at a much lower efficency than the data I or the MFG had. Actually the MFG (and me) where wrong. Always check MFG data when its just a 'verbal". Worm gears are only about 40-50% efficient. Lesson learned.

 

[HydraulicsGuy]

You will get good answers here:

https://www.eng-tips.com/threadminder.cfm?pid=237

Thanks, I'll re-post it there!

 

[HydraulicsGuy]

I want to know what the numbers become. We have tested HPUs where we have exceeded the ratings. When you test, sometimes that happens. Something is not as promised. Maybe the motor is no good, the pump is no good, the pump has a lower efficiency than advertised, etc. When testing and the numbers exceed the ratings, I want to know what the mathematical factors become. I'm a numbers engineer through and through. I want to calculate what's happening all the time.

 

[itsmoked]

I just told you what will happen. You don't need to cross post it.

1. Let's say motor is drawing 8.5 Amps. What has happened mathematically to efficiency and power factor? What do you think their values would be?

There's nothing to 'think' about. You go to the motor's efficiency curve and see what it is. It's empirical. No math.

2. Let's now say current continues to climb to 2 x FLA = 12.28 Amps. What has happened mathematically to efficiency and power factor? What do you think their values would be?

Again you look it up. Every motor is different.

3. Is current directly proportional to torque demand? So as the torque demand goes beyond full-load torque toward breakdown torque, does the current follow proportionally? If current = 2 x FLA, is the torque that's being demanded = 2 x FLT? If not, what do you think torque would be?

Yes, generally follows proportionally until grossly overloaded then it all goes to hell. Grossly overloaded would be somewhere around 30% overloaded. Then conductor cross sections, conductor overheating, and saturation will mangle the proportionality of the results.

4. If BDT = 275% FLT, is breakdown current = 275% full-load current? If not, what do you think current would be?

The current will be the LRA (Locked Rotor Amps) (the same as the starting current) for a second or two before the protection trips or smoke starts pouring out of the ruined motor. The current can not exceed LRA ever.

5. What happens to voltage through all this? Is voltage only dependent on the power source, or does it degrade as motor goes beyond service factor HP & amps?

The source is responsible for the voltage. The source includes the impedance between it and the motor. This is the wire resistance and the inductance. If you are stupidly abusing a motor, more than likely the electrical network will not fully support the current draw and you will have current limiting due to voltage reduction. Of course it all depends on the circuit's capacity. The electricians could've put in wire that was generously sized or not, or it could be the wire run is long and marginal which would cause additional voltage drop.

6. I assume that through all this, the motor power equation still applies: P = V x A x Eff x PF (x 1.73 if 3-phase)?

Well, mostly, except no one cares what the empirical values of PF and Eff are outside of the proper motor operation region. Meaning, you can't calculate it, you'll have to measure it yourself. Make sure you have lots of spare motors as you'll likely need them.

Keith Cress
kcress -

http://www.flaminsystems.com

 

[HydraulicsGuy]

You go to the motor's efficiency curve and see what it is. It's empirical. No math.

I have never seen this data at more than service factor load.

Every motor is different.

I know. I was hoping somebody with special / strong knowledge of this could estimate based on the numbers I gave, or give a rule of thumb like "efficiency is ~50% of full load efficiency at 2 x FLA".

Grossly overloaded would be somewhere around 30% overloaded.

The current will be the LRA (Locked Rotor Amps) (the same as the starting current) for a second or two before the protection trips or smoke starts pouring out of the ruined motor.

The new Fluid Power Reference Handbook has a discussion on going above the full load for several seconds of a repeating duty cycle. Doesn't say how many seconds they advise, but their example is 7 seconds a little above full load, followed by 5 seconds way above full load, followed by 8 seconds below full load, then repeated continuously. The book advises that you can go up to 80% of BDT. Doesn't say how long, but the implication is that it's many seconds. This is in direct contradiction to what you are claiming, which is that it's just a second or two and the motor is ruined. Maybe you just said that for effect.

The current can not exceed LRA ever.

Are you saying even at BDT which is greater than LRT, the current won't exceed LRA? Maybe that's the case. I don't know. That's why I'm asking these questions, for a better understanding. It just seems logical to me that the peak current would be the BDT current, which logically would be higher than the LRT current.

Well, mostly, except no one cares what the empirical values of PF and Eff are outside of the proper motor operation region. Meaning, you can't calculate it, you'll have to measure it yourself.

You said in question 1 that I can get it from the motor's efficiency curve.

Would you agree that knowing motor eff and PF would help diagnose whether it was a motor of pump problem if the motor was drawing more than service factor amps (SFA)? Is the pump efficiency as promised, and the motor efficiency has dropped below what it should be? Or is the motor operating as promised, and the pump is demanding too much power because its efficiency is lower than promised?

 

[itsmoked]

What limits the current into a motor the most is what's called BEMF or Back Electro-Motive Force. As the motor speeds up the BEMF increases. At running speed it's most of the line voltage and it apposes the line voltage so there's actually only a couple of volts doing the driving of the current. When you start the motor it's not turning so the BEMF is ZERO. This means the current is limited only by the resistance and inductance of the windings, period. For any kind of efficiency the windings are designed to have very low resistance, typically way below 1Ω. The larger the motor the further less the resistance, often being around 0.1Ω or less.

From all this you should be able to see that nothing you can do to the motor will cause it to draw more than when it's sitting still without any BEMF. You can't brake the motor or load it somehow in a manner that can cause higher than LRA.

Now your hydraulics book is saying you can cyclically overload the motor and that's true. But whenever you are you're heating the windings up and heat is what eventually breaks down all motor windings. Remember the windings are insulated with organic compounds. Eventually chemical reactions are what break the organic molecules down as their chemical composition changes. Recall that all chemical reactions double their rates with every 10° of temperature rise. The rated lifetime of a motor is dictated by the FLA at a particular ambient. Anything that exceeds that accelerates the lifetime. Of course motors are thermally fairly massive so brief excursions into overload-land are not too damaging if they weren't already running at full load. Lots of hydraulic applications like, say, injection molding have the motor doing dang near nothing for many seconds, then something for a few, then perhaps an overload for a few seconds before doing nothing again for many seconds. This is acceptable if not great for the motor.

Using the the same example, typically, a molding machine will be designed for maybe 60 tons of platen pressure and maybe X tons of injection held Y seconds. The reality is that almost no one runs at the design limit, they run at 46tons, 0.7X tons, and for 0.5Y held seconds. The motor lives a long and prosperous life. If instead it's run full out making the same part for its entire existence it will go thru motors at a regular rate.

I am not exaggerating about the life of an overloaded motor. I have seen many toast. I have deliberately toasted many (for fun and education). If you power a motor up with a locked rotor you are talking 2 to 4 seconds before it smokes. The power is
P = 240[sup]2[/sup]V/0.1Ω
P = 576kW

How long do you think it takes to smoke some insulation with this kind of power dumping into them? It won't make it quite to this level because of the network and other players but you get the point and that's why you even get a few seconds and even 10% of that is still very bad news.

If you have a nearly fully loaded motor and you want to run it up to 80% of BDT you can with the thermal downsides described. BDT current can not exceed LRA.

The standard of motor pump troubleshooting is universally to pay close attention to the running current. The plate on the motor shows FLA and it takes into account everything including the power factor. If the motor is drawing FLA or less you should do fine in an app unless for some reason the ambient is exceptionally high then de-rating is required. A motor with SF > 1 can operate with that percentage over the FLA on the plate at trade-off of its lifetime.

It's always about the temperature. If a motor stays fairly cool whatever you're doing to it is likely not accelerating its demise.

While watching the operating current you can make some reasonable estimates. If it's 80% of FLA and you make occasional 120% excursions are they reasonably balanced? If so things will work. If not you're going to have problems.


Keith Cress
kcress -

http://www.flaminsystems.com

 

[HydraulicsGuy]

nothing you can do to the motor will cause it to draw more than when it's sitting still without any BEMF. You can't brake the motor or load it somehow in a manner that can cause higher than LRA.

Makes good sense, and good to know.

If you have a nearly fully loaded motor and you want to run it up to 80% of BDT you can with the thermal downsides described. BDT current can not exceed LRA.

Good information. One of my original q's: If BDT = 275% FLT, is breakdown current = 275% full-load current? If not, what do you think current would be?

I'll ask this again in a different way. Say a 2 HP motor during testing is drawing more than service factor amps (SFA). Way more. All calcs (checked and checked twice) demonstrate that the motor output should be less than 2 HP, say 1.8 HP. Is the motor the problem, or is the pump the problem? Is the pump efficiency as advertised, and the motor efficiency is less than advertised? Or is the motor as advertised, and the pump efficiency is less than advertised? There are 3 "nebulous" variables that will tell the story: motor eff, motor PF, and pump eff. If we can closely approximate what the 2 motor variables should be in this "well beyond Service Factor" state, we can get a lot closer to diagnosing the problem. I just don't know enough about motors and electrical to know if there are estimates/rules of thumb that apply pretty well beyond Service Factor. I know what your answer is going to be: it will vary by motor, and it's not knowable because the manufacturers don't design beyond service factor. That's just not a very satisfying answer, and it doesn't help tell me whether the problem is the pump or the motor.

 

[3DDave]

Either one. But what failed is the person who selected the motor without understanding what the requirements are to run the pump. There are almost no failure modes for motors that don't involve burning off the insulation and then shorting out the windings. So whatever scenario you should start with the pump or the person who selected the pump for the application.

 

[waross]

Those questions are hard for me to answer.
I have over 60 years experience of mostly doing it right and very little experience with doing it wrong.
Yes, as Keith says, you can overload a motor that is driving a hydraulic pump, within limits.
Google "Cowern Papers" and look up RMS HP loading.
But the procedure warns that you must consider the maximum available torque that the motor is capable of.
The torque is proportional to the square of the current.
At 8.5 Amps the torque calculates to 192%
You are too close to the torque brakedown limit.
That is where the motor stalls.
200% current? The motor may stall and the current go to 600%
Once the motor stalls, that time is reduced to the order of 15 or 20 seconds.
But hey, what do I know?

As a rule of good practice, it is wise not to use more than 80% of the rated pullout for a determination of adequacy

If the motor stalls your time drops from 5 minutes to 15 seconds.

As a rule of good practice, it is wise not to use more than 80% of the rated pullout for a determination of adequacy

But, if you overload to 192% for one minute, you have used up 3 minutes and 41 seconds of your 5 minute window.
If the motor has been working at much load for the time n between overloads you will be over the limit.
Efficiency: At 200% current your losses will be four times, but your HP will be more. A guesstimate, take your losses times 2.5
So if the efficiency is 96%, the losses are 4%, 2.5 x 4% = 10% losses, use 100% - 10% = 90%.
Power factor, more magnetizing current as a percentage of more output power, not too much change.
"2. Let's now say current continues to climb to 2 x FLA = 12.28 Amps." It depends. You have to consult the motor curves.
You may be able to get 200% current with an overloaded centrifugal pump, but it is doubtful with a positive displacement pump.
I would expect the pump to stall before the current reached 200%
Not recommended, but you can do it if you want.
Pennies in the fuses are not recommended but it used to happen. Most people got away with it.
Some burned their houses down.
You have only a motor to burn down.
What is the cost of the proper hardware relative to the cost of replacing a burned out motor.
What does downtime cost while replacing the motor?

Bill
--------------------
"Why not the best?"
Jimmy Carter

 

[waross]

I'm a numbers engineer through and through. I want to calculate what's happening all the time.

I assume that through all this, the motor power equation still applies: P = V x A x Eff x PF (x 1.73 if 3-phase)?

Yes, but when you go outside of the ratings the numbers become non-linear.
V? The extra current may cause a voltage drop and the voltage changes. It depends.
A? Yes, but see PF.
Eff? It depends, but at 150% or 200% it will be lower, in a non-linear manner.
PF? Pretty simple. Real Power over Apparent power, but the ratio is affected by the amount of reactive power, and outside the normal envelope, "It depends".
Find some generic motor starting curves that show current versus speed, and torque versus speed.
Look for a PF curve for motor starting.
Locked rotor PF is very low. Magnetizing current is high. (Magnetizing current changes with motor loading.)
Google is your friend.
Try to find curves from different sources and compare.
Then extrapolate and guesstimate.


Bill
--------------------
"Why not the best?"
Jimmy Carter

 

[HydraulicsGuy]

Either one. But what failed is the person who selected the motor without understanding what the requirements are to run the pump. There are almost no failure modes for motors that don't involve burning off the insulation and then shorting out the windings. So whatever scenario you should start with the pump or the person who selected the pump for the application.

As I said in a previous post, calcs were performed, checked, and checked again. Manufacturer data was used, and the calculated required HP was under 2. The motor selected was 2 HP rated. The proper pump was selected.

 

[HydraulicsGuy]

Google "Cowern Papers"

Ok, I've got a lot of reading to do now. Thanks for taking a shot at some of the parameters I was asking about.

 

[LittleInch]

So in your problem outlined above
Is the motor the problem, or is the pump the problem?

No one can tell from afar, but first you need to do some faultfinding to find out which of the parameters in the design data isn't matched by reality.

Small motors and pumps often have quite a wide accuracy on their stated performances so what applies for bigger units won't for smaller ones.

Key issues to check / measure against their data sheets before you can start to apportion blame are:

Motor:
Voltage actually on the motor terminals (not the s'board)
Current
Frequency
If 3 phase - is the motor going the correct way around?
Power factor if you can measure it
Windings temperature

Pump:
RPM
density of fluid
flowrate
differential pressure

If the motor you're talking about is the one in your OP I would say that you're trying to pump either a very heavy fluid or at a pressure far above what the pump is designed for. Running a motor at twice the FLA and probably 3 times the torque tells me there's something wrong with what you're asking the pump to do or if that is within the pump parameters is more likely a pump issue than a motor issue.



Remember - More details = better answers
Also: If you get a response it's polite to respond to it.

 

[HydraulicsGuy]

Small motors and pumps often have quite a wide accuracy on their stated performances so what applies for bigger units won't for smaller ones.

Good to know. I'll keep this in mind on future applications. Now that I think about it, the times the company has tested units and they seem to be drawing too much amperage, it's always been on the smaller units (1.5 HP, 2 HP, 3 HP). On the few bigger units I've observed in testing, amperage has never been a problem.

If the motor you're talking about is the one in your OP I would say that you're trying to pump either a very heavy fluid or at a pressure far above what the pump is designed for. Running a motor at twice the FLA and probably 3 times the torque tells me there's something wrong with what you're asking the pump to do or if that is within the pump parameters is more likely a pump issue than a motor issue.

The output (pressure and flow) is within the pump's capability, as calculated. It's typical 32-68 viscosity hydraulic fluid. Not sure exactly which they're using, but it's the company's standard testing oil. Seems like most people that have commented on it, are implying it's a pump problem and not the motor.

 

[LittleInch]

The smaller units are very price sensitive and the vendors will always "big up" the performance whilst constantly trying to reduce costs. Result is a performance figure rarely achieved in practice (think miles per gallon of cars - how often does anyone actually get that performance in real life???). Applies equally to pumps. e.g. from your example if your pumps actually needs 20% more power (now 2.2 HP), but your motor is only giving you 1.6HP, you're now ~30% over FLA. But everything is within its accuracy range.

They also tend not to comply with any known standards as they are too small, but if you press them hard enough they will often only give you a performance guarantee of 20-25%. The "testing" also leaves a lot to be desired and the word "typical" turns up more than once.

So in terms of matching motor to pump at this range always add at least 25% to the required rating (power or torque) before choosing the motor based solely on the vendors performance figures.

But check voltage at the motor terminals. Can make a massive difference.

Remember - More details = better answers
Also: If you get a response it's polite to respond to it.