MechE trying to understand VFD operating numbers!!

[cnuk]

We have an application with a 20 HP/1800 RPM motor hooked up driving a pump. The pump is operating at a flowrate and pressure that is equivalent to approx 10 HP. The drive is a Dura-Pulse from Automation direct which is a sensorless vector drive (don't know what sensorless vector means). It is wired at 460V input. We are operating the motor with an input frequency of 30Hz to get the speed we want. When we look at the parameters from the drive we read the following:
Motor Volts: 230
Motor Amps: 32
This part makes sense because P=VI and those numbers work out to 10HP just like the pump is demanding.

My question is why is the voltage 230V and not 460V? My basic understanding is that the frequency for AC motors is what controls the speed, not the voltage (as is DC motors). When we change the speed of the motor the voltage stays fixed at 230V. The current responds to the torque requirements as expected, just the voltage doesn't make sense to me.

Can anyone shed some light on how this works?

Thank You

 

[buzzp]

In a nutshell, when changing the frequency that an induction motor operates, you must maintain the V/Hz relationship, ie 460/60 or 7.67. So if you need the motor rpm to be whatever at 30Hz the voltage must be reduced by the same factor, 2 in your case. That is why the voltage changes with frequency. If it dont the motor could saturate and stall. This is for a standard drive. Just seen the last part of your post where the voltage was not changing when the speed was increased/decreased.

In your case, this is normal operation for the type of drive you have. Google for 'sensorless vector' drive (I believe this is for variable torque applications) and you will get some good sights for an explanation. Try this one for starters:

http://www.rockwellautomation.co.uk/downloads/technical/sensorless_vector_tech.pdf

 

[weh3]

The ratio of voltage to frequency must be constant. If rated voltage and frequency are 460 V and 60 Hz, respectively, then you have

460/60 = 230/30.

Since a Hertz is an inverse second, Volts/Hertz works out to volt-seconds, or Webers, which is a unit of magnetic flux. The iron core of the motor is designed for a certain maximum flux to keep it from saturating magnetically and to keep it from overheating. Keeping V/Hz constant means the flux is constant.

Regards,
William

 

[buzzp]

WEH3, I believe V/Hz relationship is maintained for standard drives, not the one the poster has.

 

[Skogsgurra]

It is true that the V/Hz ratio is kept constant in a scalar drive and not in a vector drive. A lightly loaded motor on a vector drive will have a lower than expected voltage. But, as load increases, the voltage goes up. And when you run the motor at its rated speed and load, the vector controller (if correctly designed) will see to it that the motor gets the correct voltage - which is the same as a scalar controller would output under the same speed and load.

 

[weh3]

The reasoning behind V/Hz is the same, although a vector drive introduces nuances into the relationship that make it nonlinear. That is to say, the vector drive senses the load requirements, whereas a scalar drive just outputs a constant V/Hz and whatever current is required. The current changes because the power factor requirement changes with the load. I would be interested to see how the relationship changes over the entire 0-100% speed range.

Regards,
William

 

[jraef]

There you have it cnuk. If you want some further reading on what Vector Control means, look at this FAQ written by myself and skogsgurra.
faq237-1062

The fact that yours is a Sensorless Vector Control drive is probably irrelevant because a Scalar drive (standard V/Hz) usually works just fine for centrifugal pumps, and in fact, there are some energy saving benefits available that cannot be done with most SVC drives unless you turn off the SVC algorithm and run them in Scalar mode. You didn't say what kind of pump application you have, so this may not be relevant, but I thought I'd throw it out there.

"Venditori de oleum-vipera non vigere excordis populi"

 

[hoy7t]

on most VFD the voltage is a user input. You tell the drive your incoming voltage. It then operates to control current to spin the motor. VFDs operate by switching current on or off. this produces an average current through the drive that results in motion, etc.

are you sure that this drive is designed for 460VAC input?

sensorless flux vector: to you this means that the drive can control speed and torque at a lower minimum than if it did not have flux vector capability. this is a matter of sophistication of the math within the drive. sensor based flux vector can operate at even lower minimums.

 

[DickDV]

The fundamental underlying issue causing the V/Hz ratio to be constant for constant motor torque production is that a motor is largely an inductor.

Since the reactance of an inductor is equal to 2 x pi x frequency x the inductance, and further, since the current in an AC circuit is equal to voltage divided by reactance

I = E/(2*pi*f*h)

you can see that, to keep the current constant, when the frequency is reduced, the voltage needs to be reduced in the same proportion.

In many drives, you can deliberately customize the V/Hz curve to modify the ability of the motor to produce torque. This is most commonly done on variable torgue (actually INCREASING torque) loads such as centrifugal fans and pumps matching the V/Hz curve to the torque curve of the load. This drastically limits available motor torque at low and medium speeds but, since the fan or pump doesn't require much torque at those speeds, it results in a small amount of energy savings.

Sensorless vector drives will automatically vary the V/Hz ratio depending on motor load requirements but will not usually increase the ratio above the motor nameplate value except at very low speeds where copper conductor resistance begins to modify the total impedance of the motor and, as a result, additional voltage may be required for torque production above the normal ratio.

The extreme example of this is at zero speed. If the ratio were enforced at this speed, the voltage would have to be zero since the frequency is zero. Clearly, since a sensorless vector drive is often quite good at developing torque at zero speed, there must be some voltage in order to produce the current necessary to develop the torque.

 

[afterhrs]

cnuk,
I agree with hoy7t, something is definitely NOT correct with your application. The first sign is that you
state you have a 20 HP motor running at 30 HZ, 230 VAC,
and 32 AMPS. The approximate full load amp rating for a 20HP motor at 460 VAC is 27 AMPS(see Ugly's,page 39), 32 AMPS is a lot closer to half(non-linear curve) of the 54 AMPS required for the same 20 HP motor wired for 230 VAC running at 60 HZ
Here are three different things you can do to verify your application: 1-Check the part number of the VFD to ensure you have the correct model(most manufacturers sell VFD'S at a specified voltage). 2-Verify that no one has adjusted your factory preset or default parameters.
3-Disconnect the motor from the load and manually increase to 60HZ, if it is a 460 VAC set up, then you should be pulling about 15 to 18 AMPS with the motor unloaded at 1800 RPM. I would also verify incoming and outgoing voltage with a voltmeter at this point.
Let us know what you find.

Best regards,
afterhrs

 

[afterhrs]

cnuk,
Check your wiring also, I have witnessed a 230 VAC VFD
run a motor wired for 460 VAC. The motor would stall
under heavy load and be cool to the touch.
I added a few grey hairs on that problem.

Regards,
afterhrs

 

[Skogsgurra]

cnuk,

There is one "detail" that worries me: You are saying "When we look at the parameters from the drive we read the following:
Motor Volts: 230
Motor Amps: 32
This part makes sense because P=VI and those numbers work out to 10HP just like the pump is demanding."

In my world, your numbers do not work out to 10 HP, but 12.7 kVA electrical input. That input corresponds to around 10 kW power (cos(phi) assumed to be 0.8). Given a typical efficiency of 90 percent, the motor shaft then outputs a little more than 12 HP.

12 HP is close enough to the 10 HP you say that the pump needs at half speed. What makes me wonder is the way you arrived at the number. Your calculation is valid for a DC motor and half voltage at half speed makes good sense also. But you cannot use that calculation when dealing with a three-phase inductive load.


Also, I must warn you not to measure the motor voltage with a voltmeter (be it TRMS or not). The voltmeter will always show too high a value since it measures all components in the motor voltage and it is only the fundamental component that you shall measure. Measuring motor voltage will probably throw you into another loop of doubt and confusion - a loop that an ordinary electrician will not be able to break.


Your original posting did not say that anything is wrong with your VFD and/or motor. Just that you needed an explanation why motor voltage was lower than expected. You have had "some light shed" and I do not think that your application needs more investigation. Just leave it as is. If the pump pumps and the motor doesn't get very hot, then don't worry.

 

[cnuk]

Thanks for all the good feedback. The volts/hertz ratio is something I never knew about so that makes sense to me now. I have some more information that may help clarify or further confuse the issue. Before I get into that though I wanted to say that I did check and the drive is a 460V drive and it is wired for 460V, etc. We had an electrician in an everything checks out fine. Motor full load amps at 460V is 27A.

Ok, under the initial operating conditions (wired for 460V, but operating at 230V, 32 Amps, 30 Hz) the drive display listed that it was operating at 101 to 103% rated load. Because of this is kept thermally tripping. In order to solve the problem I changed the sheave ratio (doubled it) the electric motor drives the pump through. In the "new" configuration the electric motor is now operating at 60Hz to get the same pump operating conditions (10 HP output). The voltage reading at the drive display is now 460V and the current draw is approx 20 Amps. I say approx. because I am at home and don't have the exact current numbers in front of me. The drive display also says that it is now operating at approx 65% full load.

The voltage going up now makes sense based on what I've learned in the previous email, but what I don't understand is the % load numbers. The pump load has not changed so I would have assumed that the drive load should be constant as well. It appears that the drive is significantly less efficient at lower frequencies (101% load at 30Hz and 65% load at 60 Hz for same power output). Is it really an efficiency difference or is the drive just looking at the output current and comparing it to its rated value to get a % of full load? Any more input would be greatly appreciated.

Thank You very much for all your help!! A lot of smart people providing a lot of good information.

 

[Skogsgurra]

OK cnuk, I understand your concern about load now. Your question cannot be answered without knowing what drive you have. Some drives show current (I think yours does) and some show actual power. If you make the make and type known, I am sure that someone can tell what your drive is showing.

 

[afterhrs]

cnuk,
If you still have issues with this drive, do not hesitate to call the local rep.
A VFD can become your worst enemy or best friend,
the key is to choose the correct parameters for your application.

Good luck,
Afterhrs

 

[Tmoose]

Some applications need constant torque. Others need constant power. Sometimes the same drive can be configured to provide one or the other. Configuring V/Hz ratio is one way. The drive does not know if the motor cooling is sufficient at low rpm (fan on shaft or independent cooling fan), so the responsibility of not overheating the motor at low speeds must be taken by you.