A delta motor running in star config.

[not done it yetParticipant @notdoneityet]

A question for the electricians on the forum.

I have a 2HP motor which I would like to use on a 1HP duty. I also prefer 3 phase motors to single phase.

It is a dual voltage 230/415V 3 phase motor and I have a 1.5 kW VFD with 3 phase output at 230V (well, same output voltage as the input).

How well would it run connected star? Would I get one HP? Would it overheat? Any other downsides?

Running 2HP on the machine would likely destroy the gearbox, as it is only really rated at 3/4HP, and old, so not a good option unless the motor power could be turned down without causing cooling problems.

Would this work, please?

[not done it yetParticipant @notdoneityet]

[MuzzerParticipant @muzzer]

Don't mess about deliberately connecting it up the wrong way. Connect it up in delta (230V) and if you want to reduce the max output, simply reduce the phase current settings in the VFD by the required ratio. If you want 0.75kW, enter half the nameplate current given for 230V.

Murray

[Colin WhittakerParticipant @colinwhittaker20544]

As Muzzer says, it's probably easier to dial things down on the variable frequency drive.

However, if you want to work in star mode then to a first order approximation I would expect the power to be reduced to 1 over root 3 or 58% of the delta configuration. Motor currents will also fall (so less torque and less heating) while the rotational speed will barely change and consequently the fan cooling will remain the same.

Time for the nostalgia moment … Many years ago in the Western Desert of Egypt I had to pull the same trick to get a workshop 60Hz air compressor to run on the town's 50Hz power supply without tripping the over-current breaker.

[not done it yetParticipant @notdoneityet]

Muzzer,

Wouldn’t reducing the current on the VFD would simply encourage trips, as the current in delta would like to be twice as much? Or is there a setting on the VFD for limiting current without tripping? I thought there was not.

Colin,

60% of the delta (or 415V star) would be acceptable as the speed could be turned down a tad, to lower the power a bit more? Would the motor be at least slightly less efficient, so the power would be reduced a bit further? As both delta and star are supplied by a 3 phase supply, wouldn’t the power output be proportional to the voltages, within reasonable tolerance (makes not a lot of difference, anyway)?

As torque, power and speed are interrelated, wouldn’t the torque remain about the same (or a bit more) than a normal delta 1HP motor?

I was thinking of the overheating occurring as the speed was reduced by turning down the frequency, but I suppose the 2HP motor will be equipped with a much larger fan than a 1HP motor anyway. That, in turn, will mean a slightly reduced motor efficiency as that fan will still be shifting enough air to cool a motor of nearly twice the power, so will absorb a few more Watts of the gross motor power.

Thanks, looks like a good option in the circumstances, unless there are other reasons not to do it….

[John HaineParticipant @johnhaine32865]

Isn't there a misapprehension here? Though the motor is capable of 2 hp if you wire in in delta (its 230V option), it only delivers the power/torque that the load draws from it. The operating current is approximately a proxy for delivered power (P = V x I) so if you are worried set the current limit lower so that if the mechanical load on the gearbox stalls the motor will cut out.

[Jim Young 2Participant @jimyoung2]

I too was thinking that although the motor was CAPABLE of 2HP that is a factor of the load imposed upon it. Thus use correctly wired, and just be aware that if too high a load is placed on the drive train, your motor will go over the 1HP ideal!

[Colin WhittakerParticipant @colinwhittaker20544]

Not,

The scaling factor of 58% is pretty good approximation to the reduced power and torque. Speed and efficiency will essentially be left unchanged.

If 58% is too high for you then it becomes a lot more difficult to adjust the power down further. Reducing the VFD frequency reduces the speed but increases the current and torque. As the power is proportional to the product of frequency and torque the power remains pretty well constant. Increasing the frequency sees the opposite with the power again left unchanged.

If you can't live with 1.16 HP then the only other passive technique to reduce the motor power down to 0.75 HP would be adding resistors into the power leads. My back of the envelope calculations suggests that three 64 ohm resistors would do the job while dissipating 80 Watts each. But this is not a route I am drawn to.

What's the load? Will it ever exceed 0.75 HP?

Colin

[larry phelan 1Participant @larryphelan1]

Might be easier just to use a 3/4 HP 3 Phase motor to begin with. There should be plenty of them around,and they can,t be all that dear,new or secondhand. Keep the 2 HP for something better.

[not done it yetParticipant @notdoneityet]

Thanks Colin,

The 1.2HP route will be the one I will persue, as there seems to be no other downside other than the motor being a little larger than a 1HP and a tiny loss of efficiency due to the larger cooling fan. It is not a continuous use, so efficiency is not an important factor – and the VFD will clearly have an efficiency also.

I was under the impression that most speed controllers reduced the available torque as the motor speed reduces, or one has to pay a premium for a controller that maintains the torque at lower speed operation.

Thanks for the insight. Job will now proceed.

[Gary WoodingParticipant @garywooding25363]

As a general rule-of-thumb, most VFDs give constant torque as the speed reduces (hence reduced power), and constant power as the speed increases (hence reduced torque).

[Phil WhitleyParticipant @philwhitley94135]

Do remember that as you reduce the speed below the motors syncronous speed, you are also reducing the output of its cooling fan, and therefore watch out for overheating, or provide some extra cooling if you use it at lower speeds than it was designed for.

[MuzzerParticipant @muzzer]

You can fit a small motor to large VFD and it will work fine. But if you overload it so it overheats, the VFD won't know any different. That's why you have to enter the nameplate values, particularly for the rated current. The VFD will limit the phase currents accordingly. So if you tell the VFD that the phase currents are what you'd expect to see on a 1HP motor, you will have limited the output power to that level. The other key information is the base speed which it can get from either the number of poles or by entering the base speed directly (depending on the VFD).

Very unlikely many of us will ever risk overheating a motor (unless we stand next to it discussing motors), not least if the thing has been derated by 50%. The efficiency of a typical VFD will be in the region of 97-98% or so, ie pretty much insignificant compared to the motor.

As I said, just connect it in delta (230V) and set the VFD's motor nameplate data to half of the motor values. Don't overthink it.

Murray

[not done it yetParticipant @notdoneityet]

Guys, it is not going to overheat, if running at one horsepower, with a fan capable of cooling a two horse motor is it?

If about the same efficiency (as Colin predicts) it could run at less than the half speed (recommended reduced speed limit for continuous running) for slowing a normal delta connected motor because it already has a fan much larger than a normal one horse motor.

Edited to add that they don’t add cooling fans to VFDs for no good reason.  I think I read in one manual that they were only about 85% efficient, so need a rather larger input than output.

Edited By not done it yet on 27/05/2018 16:35:10

[Nick HulmeParticipant @nickhulme30114]

Posted by Colin Whittaker on 27/05/2018 10:12:15:

Not,

Given that Torque is constant (and within rated speeds it is) and that Power output in induction motors is proportional to (RPM x Torque), this renders any suggestion of constant power with varying speed erroneous.

The actual case is therefore that a conventional induction motor produces it's rated power at the rated RPM when driving a load at maximum torque, reduction in speed results in a reduction in power, the exact result of an increase in speed depends upon the electrical characteristics of the individual motor.

[Colin WhittakerParticipant @colinwhittaker20544]

Nick,

My understanding, and it is some 40+ years out of date, is that to a first order approximation torque is proportional to current at small values of slip. Lowering the AC power frequency allows more current to flow giving more available torque at the same time it reduces the rotational velocity for the same value of slip.

For a conventionally wired induction motor operating at full power with the steel core close to saturating the effect of reducing the frequency would be to saturate the steel with a marginal increase in torque and a large increase in motor current that will rapidly over heat the motor. Because "Not" is working with a 430V motor driven with 230V the steel is nowhere near saturating and so this unusual constant power behaviour with variable frequency would be observed.

Best regards, Colin

P.S. You may find the following link on Induction Motor torque of interest, **LINK**

[not done it yetParticipant @notdoneityet]

Colin,

You have a PM.

[Carl Wilson 4Participant @carlwilson4]

Just get and fit a smaller motor. Right tool for the job and all that.

[Clive SteerParticipant @clivesteer55943]

I don't know if I'm stating the obvious but a star connected (415v) will work on 240V but won't be capable of producing its rated power. This is why on VFDs motors cannot produce max power at lower frequencies as they reduce the voltage applied to the motor. An induction motor has common characteristics to other types in that it produces a back EMF related to speed so at higher speeds more volts is required to get the current that gives the power (W=V.I). A 2 HP motor is rated to produce 2HP safely and continuously and a overload trip system is there to stop the motor catching fire and not necessarily there to protect the machine it is driving. For instance when a motor starts up the current drawn (locked rotor) can be 10 times the allowable full load current but only for a short time as the motor/machine accelerates to design speed and the motor protection device (overload trip) allows for this.

If a 2hp motor is fitted to a machine designed for a 3/4 HP motor then the user must ensure he doesn't overload the machine. Most lathes with gearing rarely use the full power the motor can provide due to cutting tool and workpiece limitations.

Regarding fan cooling I tend to run all my motors on VFDs set with the max speed at 100Hz so a 1400rpm motor will be running at 2800 rpm. So when I slow it down the fan is still producing sufficient cooling even at 50% full speed.

Clive

[Michael GilliganParticipant @michaelgilligan61133]

The note linked by Stuart Bridger, in this previous thread, may be of interest: **LINK**

http://www.model-engineer.co.uk/forums/postings.asp?th=97755

MichaelG.

[not done it yetParticipant @notdoneityet]

Thanks all. Colin has given the information I needed. There was no intention to purchase a smaller motor if this one will derate sufficiently, so any replies on that theme were completely off-topic. Clearly, if the motor would not be sufficiently derated, I would have retained the current single phase motor fitted, or resorted to obtaining a lower rated motor when one became available at the right price.

A larger motor will have no cooling problems, like slowing a lower powered motor (with a consequently smaller capacity fan), so those scenarios were already covered before I listed the thread.

Clive, I am sure that VFDs do not change the supply voltage. I may be wrong, but don’t think so – they alter the frequency – it is in the name and that seems to say it all. Also, reducing the motor power was the requirement from the start of the thread.

The gearbox was, I think, designed with plenty of surplus capacity and was later used with a 1HP motor, although the gearbox may have been fitted out with modifications for the later application – but the question was simply re running on 230V star, not of running the motor at 2HP.

I’m now going ahead to do it, so thread effectively closed.

[Nick HulmeParticipant @nickhulme30114]

Posted by not done it yet on 29/05/2018 11:12:46:

Clive, I am sure that VFDs do not change the supply voltage. I may be wrong, but don’t think so – they alter the frequency – it is in the name and that seems to say it all. Also, reducing the motor power was the requirement from the start of the thread.

Some Vector Drive VFDs increase current (and thus voltage) at any speed when RPM drop is detected in order to maintain the set speed

[Anonymous]

Posted by not done it yet on 29/05/2018 11:12:46:

Clive, I am sure that VFDs do not change the supply voltage. I may be wrong, but don’t think so – they alter the frequency – it is in the name and that seems to say it all. Also, reducing the motor power was the requirement from the start of the thread.

A VFD doesn't change the supply, ie, input, voltage but they most certainly can, and do, change the output voltage. The output of a VFD is actually a PWM waveform, not a sine wave per se. The peak to peak PWM voltage is fixed, usually the same as the rectified mains. The width of each sequential pulse varies, in order to create a waveform with a given frequency. However, a uniform percentage change of width across all pulses will leave the frequency the same, but will change the output voltage.

Clive is correct to say that as the frequency reduces the VFD can also reduce the effective applied voltage. This is needed as less voltage is required to drive a fixed phase current, as the back EMF is lower as the motor is spinning more slowly.

Andrew

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