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Associate

Join Date: May 2010
Posts: 31

# V/F ratio, Power and Torque

05/25/2010 10:25 PM

A 4 KW motor is connected to a conveyor Belt via Gear unit taking 70% of FLA (8A).We want to is drive it through a VFD (ABB) and operates it at 10 Hz (120*10/4 = 300 RPM).We have 400 V 50 Hz system. The V/f ratio will be 8. It means that 80 Volt will be applied to the Stator winding at a frequency of 10 Hz.

Motor speed and torque reduces if voltage applied reduces (In case of Conventional Starter). How this is being compensated by V/f ratio in a VFD,

Will at that voltage (80 V) and speed (300 RPM) the drive will provide enough torque to drive the load?

(Name Plate: 4 KW, 400 V (Star), 8A, 1470 RPM, P.f =0.86)

Also at 10 Hz what will be the power consumption? (P = 1.732*400*8*0.86*0.8392 is used at 400 V 50 Hz case but here voltage and Frequency is changed)

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Guru

Join Date: Feb 2009
Location: Houston, USA
Posts: 973
#1

### Re: V/F ratio, Power and Torque

05/25/2010 11:41 PM

For VFD driven motor, ideally the

(1) Torque is constant up to the rated frequency and then decreases

(2) Power increases up to the rated frequency and then constant

So, even you apply reduced voltage to the motor terminal with reduced proportionate frequency, the motor will produce enough torque. The consumed power will be proportional to the applied voltage.

See these links for further help:

http://www.kilowattclassroom.com/Archive/VFDarticle.pdf

- MS

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#2

### Re: V/F ratio, Power and Torque

05/26/2010 12:46 PM

Torque reduces at the square of the voltage reduction when the frequency remains the same. In a VFD, you are changing not only the voltage but the frequency at the same time; maintaining that same V/Hz ratio, so torque remains the same throughout the speed range. That is the fundamental criteria for using a VFD.

As the speed reduces and torque remains the same, the total shaft power is of course reduced, being that the power is a function of torque and speed. So at 20% speed (assuming you have 50Hz primary power), the motor power will be 20% of what it would be at full speed. The losses increase slightly at lower power outputs, so the relationship between shaft power and consumed power changes, but if you assume 10% additional losses you will have a good number to estimate. So in your case, a 4kW motor, assuming it is using all of the 4kW at full speed (50Hz), will consume approximately .88kW at 10Hz.

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Location: Flanders (Belgium)
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#3

### Re: V/F ratio, Power and Torque

05/27/2010 5:22 AM

GA from me JRaef.

I want to add, that if frequency is higher than 50 Hz, you can get more mechanical power out of your motor ....

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#4

### Re: V/F ratio, Power and Torque

05/27/2010 1:59 PM

'Motor speed and torque reduces if voltage applied reduces (In case of Conventional Starter). How this is being compensated by V/f ratio in a VFD'

Please note the effect of frequency as well as voltage in the following, reducing Voltage on its own reduces flux and potential torque capability, when reduced accordingly with Hz you can maintain constant torque output performance. So, from a previous post,

Magnetic Flux = V/Z (where Z = Stator winding impedance, Ω)

Z = √(R2+ XL2 ) (where XL = Stator winding reactance, Ω) and

XL = 2πFL (where F = supply frequency, Hz and L = inductance of winding, H)

The impedance at low frequencies (starting) is more a function of the resistance than reactance and so when a low voltage at low Hz is applied you are trying to keep a constant magnetic flux in the air gap between stator and rotor. This can be done with a manual starting voltage setting but generally automatic control is done (Vector) on harder applications as you need to vary voltage to counter drop when the load increases to maintain the magnetising current part. At higher frequencies XL becomes more important. Smaller motors have more significant resistance. More simply, the motor demands current according to its load and the applied voltage and frequency, the VFD software calculates this load (open or closed loop) and adjusts the voltage to keep the current optimised for the calculated load all the time giving good control characteristics and efficient operation. Many people are familiar with motor starting curves. If you imagine starting at low Hz and increasing generating slip in the motor, the torque generated rises with slip similar to being at full speed with no load then applying load incrementally, the current rises accordingly from no load current to full load current etc.

'Will at that voltage (80 V) and speed (300 RPM) the drive will provide enough torque to drive the load?'

Very likely since at 5.6A (70% of 8A) the motor is only lightly loaded, maybe 25-50% and the pf is low, and at 10Hz the voltage drop against stator resistance affecting torque is not so dramatic. If you run in automatic or vector mode, no problem. If the conveyor load is significantly higher, you may need to ensure sufficient cooling since the fan is now hardly working to blow cooling air over the motor, but, from your description, I would expect no problems.

'Also at 10 Hz what will be the power consumption? (P = 1.732*400*8*0.86*0.8392 is used at 400 V 50 Hz case but here voltage and Frequency is changed)'

Yes the voltage and Hz are changed as you say, the pf will depend on the load. If it was full torque at 1/5 speed, you would expect 1/5 power approximately. I didn't understand the 0.8392 factor you added? so, √3 x 80v x Amps x pf, if A=5.6A and pf is about 0.6 (??) then electrical power would be 465Watts Motor efficiency looks about 84% at full load from your nameplate data.

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Associate

Join Date: May 2010
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#5

### Re: V/F ratio, Power and Torque

05/27/2010 9:40 PM

0.8392 is the efficiency of the Motor

Thanks for Ur Help. Most of my misconception regarding V/F are clear.Again Thanks for all participated

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