|Bob Worsley||09/08/2020 17:04:39|
|53 forum posts|
Let us have some basics.
All motors are rated on their output power, nothing else.
A VFD is rated on its output power, plus about 2000 other not important factors.
An induction motor is an amazingly complex electromagnetic machine. Pick up any book on motors from any charity shop to convince yourself. An induction motor is essentially a transformer where power is transferred from primary, stator, to secondary, rotor. With the huge air gap between the power factor of the motor just rotating, nothing whatever attached to the spindle, will be about 0.2. At full load the power factor of a 2.2kW motor will be 0.85. If you measure the current, note current, taken at these two points it will be near identical, about 5%, possibly 10%. Read up about circle diagrams.
To determine the current taken by an induction motor you take the rating plate Watts divided by voltage equals current, then divide by 0.8 for efficiency and again by 0.8 for power factor. This is pretty worst case. Note that efficiency and power factor improve as the motor gets larger. If you have a large machine with multiple kW motors then divide by 0.85.
This leads onto a little known fact about induction motors, they are very fussy about their voltage. The power output, that is torque, is proportional to the voltage squared. So if your mains is 216 volts, and your motor is old and rated at 240V then this is 0.9 down, so squared is 81% power!
Conduction angle. Oh, this has hurt many people! The VFD has a bridge rectifier stuck across the mains with the electrolytic caps after, the idea being to charge the caps. But the caps are also being discharged by the motor as it is driven. Now, the maximum voltage the caps can be charged to is the RMS line voltage times 1.41, root 2, the difference between the RMS and peak voltages. The conduction angle is the actual part of the sinewave that the mains is more positive than the cap, so current flows through the rectifier to charge the cap. If the cap is at 300V then it can be seen that this charging period is not 180 degrees, more like 40 degrees. The problem here is that all the power needed to run the motor, 370W, has to be passed through the rectifier in these 40 degrees. So whilst the motor is happily delivering 370W with a current of 1.1/2A for 180 degrees, the poor rectifier has to pass that power in a quarter of the time, so 6A nominally flows. We then have the problem that the current is actually peaky, more like 10A peak for very few degrees. This is why they are so infernally noisy. Need very very fast switching diodes to do it with minimum loaa and noise. Now factor in the usual case where the workshop is run from some lighting flex and it is obvious that current drop due both to the wire resistance and impedance start to become important.
Just to keep you all happy there were some posts about single phase in to 400V three phase out VFDs. Now these really are pushing the limits because they use a voltage doubling circuit to generate the 400V or so for the caps. So what is in the single phase VFD a full wave bride, in these it is half wave. One half wave charges the cap to 240V, the next half wave pushes the 240V on the cap to 450V. Not at all nice. Now you see what three phase is used, and wired to create 6 phases.
For SoD, more load means more REAL amps, not inductive amps. Power factor REALLY does vary widely as the current taken shifts from inductive to real amps. I don't mind at all, I make as many mistakes as anyone else!
Foe AJ, same as SoD, at no load the motor is a large leaky inductor with non0existant [power factor.
I said buy a VFD same size as the motor.
Buying a larger VFD really does nothing to improve the conduction angle or charge current.
The bottom line is that the electronics, particularly the rectifiers and caps, are seriously stressed. One 4kW VFD I have actually has a programme option to measure the impedance of the caps.
|Andrew Johnston||09/08/2020 21:53:11|
5635 forum posts
I measured the current on one phase of my lathe motor this evening. First measurement was with the lathe on, but motor stationary. Current reading was 0.98A. That's partly the lathe control gear and possibly other things in the workshop. Next measurement was with the motor running but no load. Current reading was 1.90A. Lastly I took a reasonable cut in steel (0.1" DOC, 8 thou/rev feed and 540rpm) but nowhere near full power. As the cut came on the current increased to 2.84A. Funny sort of 5 to 10%.
|Robert Atkinson 2||09/08/2020 22:17:00|
753 forum posts
While your conclusion ".. buy a VFD same size as the motor. " is correct, your explanation is seriously flawed.
I'm not even going to bother addressing the rest of your explanations flaws.
|1650 forum posts|
I have always used the details on the motor plate as indicating stated output power and the flc indicated for each operating voltage if a dual voltage motor. When installing switchgear for any motor the motor plate details were used for selecting the relevant starter and overload unit, the same method was used for selecting VFD units when required.
Edited By Emgee on 09/08/2020 22:48:03
|David Davies 8||09/08/2020 23:13:41|
121 forum posts
To elaborate on Emgee,s post, always use the full load current to determine the inverter size not motor power. Six and eight pole motors may have higher magnetizing currents than the usual two or four pole motors and hence will have higher flc which the vfd must be able to provide.
I found this out the hard way in work.
|Mike Poole||09/08/2020 23:26:20|
2699 forum posts
In a discussion with a well known VFD supplier it was pointed out that the output current capability of one of the Jaguar drives could manage a motor larger than the quoted horsepower for the VFD, he must be confident the drive would be reliable to promote running this close to the limit. In the home workshop for most of the time our drives have an easy life. Some drives in the factory I worked in would do hundreds of starts an hour with only a 3 hour rest in every 24, the motors would only be comfortably warm after this workout. A decent quality motor with a quality VFD is virtually bulletproof. Failure is usually catastrophic, I was passing by when a line stopped due to the VFD tripping, I opened the panel and reset the breaker, the bang was spectacular as was my leap backwards, instant diagnosis was get this drive out and get a spare from the stores, we were soon back in production.
|old mart||10/08/2020 16:49:22|
|1911 forum posts|
The only valid reason for buying a larger VFD than its associated motor is if you intend to fit a larger motor in the future. Otherwise it is just a waste of money. Spend the money on a good make from a reputable supplier who can supply user friendly installation and setting up instructions. If you don't do this, you will be running to this forum pleading for advise on how to get it working, and that advise will be as confusing as the instructions supplied with the VFD.
|not done it yet||10/08/2020 19:11:55|
|4877 forum posts|
Your Schneider VFD is almost certainly of better quality than some of the cheaper chinese offerings on the market. Due allowance needs to be made where appropriate.
The OP has not indicated the quality of the VFD he might be considering.
|Bob Worsley||11/08/2020 10:55:22|
|53 forum posts|
You are correct, the first three lines need some changes.
Motors are rated on their output power, correct.
Missing is that the motor will also have stated a full load current. This was what set the overload back in the days of contactors.
A VFD is rated on its output power. Looking at some VFDs the output current rating is not particularly clear what it means. I have got an IMO Jaguar, 3ph 14.9A 380/480V in and 3ph 4.0kW 9.0A 150% 1min out. So what do the figures mean? They don't seem to make sense correcting phase current for full line voltage, or phase voltage for full line current. There is a root 3 factor lurking in there.
I would repeat my comments on conduction angles, they really are important.
Can't see anything wrong with the rest of it.
All the books on induction motors aren't wrong. If you measure and get current readings that don't seem to match with the understanding that the motor has a very poor power factor then think. Read paragraph 6 and wonder. Even with an average or an RMS reading voltmeter and ammeter, what does the waveform look like?
Not certain what AJ's setup is, straight 3ph tp the motor or is this from a VFD?
|Andrew Johnston||11/08/2020 11:11:02|
5635 forum posts
Standard 3-phase supply.
|Robert Atkinson 2||12/08/2020 09:02:24|
753 forum posts
The Jaguar figures make sense to me.
Output rating 9A continuous 13.5A (150%) for 1 minute
|old mart||12/08/2020 19:24:05|
|1911 forum posts|
NDIY, that Schnieder ATV12 costs £124.20 at the moment, it is 0.75Kw driving a 1hp motor. I would not call that excessively expensive, especislly compared with the price of a recently mentioned brand on this forum.
|not done it yet||12/08/2020 21:57:09|
|4877 forum posts|
Neither do I for a better quality VFD, but there are some examples, on the net, at less than half that price - in that size. The OP wants one to drive a 370W motor, btw. Different people have different ideas on what is “excessive”. I think some ‘head space’ is likely worthwhile, if making a purchase at the bottom end of the market.
6186 forum posts
Bob's posts have me left in the dust. May I ask Bob to take us gently through the subject, with diagrams and formulae? With the exception of Conduction Angles, I can't bridge the gap between Bob's assertions and my references. What are your sources Bob?
I'm not claiming any expertise in this area and don't have a good book on Induction Motors, so a good explanation should be helpful.
Before embarking on the deeper aspects, can we be careful of statements like:
Motors are rated on their output power. As it's not difficult to find counter-examples, I'm not sure if Bob's made a schoolboy howler here, or a useful simplification. Another example suggests Bob doesn't get the difference between Power and Torque, but it may just be that technical English is imprecise.
Can we start with 'A motor will take pretty much the same current irrespective of its load, the power factor alters to produce the output power.' Can Bob explain Circle Diagrams and the difference between real amps and inductive amps in that context? I don't doubt power factor alters when an AC induction motor is loaded, I'm uncomfortable with the notion of current staying the same irrespective of load and it's the phase shift that produces power.
|Bob Worsley||13/08/2020 10:53:41|
|53 forum posts|
Make a start.
If a motor isn't rated on its output power, then what is it rated on? If you are driving a grain elevator then you work out the speed, weight of grain etc etc to get possibly a torque figure. Power is torque times speed, you know how fast the elevator should rotate so now have a power, measured in Watts.
A diagram of the motor characteristics form no to full load.
What is an album? I assumed it would open a window to search the computer for, in this case, photos?
|not done it yet||13/08/2020 11:09:50|
|4877 forum posts|
If a motor isn't rated on its output power, then what is it rated on?
For the uninitiated, some drives are advertised with the gross input power quoted. Bandsaws are a typical example, but other items can be similarly ‘over-hyped’ (an example is Arceuro always provide output power for their machines but some competitors quote the higher figure). No real difference than some compressors being ‘rated’ as so many units of volume ‘displacement’, not free air delivery - which is the real figure required by the user.
The problem that may arise here is where the driving and driven values do not overlap, or barely so
One example that I will always remember is the following ‘spherical items’ up. An industrial process stage had a 1000tph specification with an absolute minimum of 600tph. The clever consultants advised a transport system of... ...600tph maximum! The system never ever worked properly.🙂
|John Haine||13/08/2020 11:47:18|
|3270 forum posts|
This graph may be of use.
It shows that the current increases from the magnetising current of around 5A for this particular motor at synchronous speed to about 30A at full power. At synchronous speed the load is pretty well pure inductive, slip is zero, no current induced in the rotor, power factor near zero. As the power increases so does the current, its magnetising component stays about the same but the in-phase component greatly increases. At max power the PF works out at about 0.95 I think. At max torque the current increases to about 35A.
|Dave Halford||13/08/2020 12:07:40|
|870 forum posts|
Very true, using the input power avoids exposing the excessive losses some budget motors have. Books are technical based, quoting dodgy power claims is a sales pitch or 'puff' designed to catch the unwary. It's similar to quoting Total System Power or PMPO power in the audio world.
One real reason to overrate a drive is when someone already has an old (as in 1980's) 3ph motor he wants to use. The effcy is lower and it will pull more current than the modern version.
Buying the drive and motor from the same source should prevent any miss matches happening.
|Tim Stevens||13/08/2020 18:29:24|
1259 forum posts
And I thought that 3 brush dynamos were a bit complicated ...
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