Current in VFD 1

[buddy91082]

Trying to determine how current changes when frequency in changed using a VFD.

Voltage will change with frequency based on v/hz ratio.

Hp change will be the cubed of frequency change.

for simplicity, lets say HP = Volts x amps.

If frequency changes 10%, voltage will change 10% and HP will be reduced by .1%

If HP = Volts x amps, then

.1% = 10% x % change in current

% change in current = 1%. Make sense?

 

[Skogsgurra]

No, not really. You cant multiply 10% and 10% and expect a valid result.

It is 905 and 90%, so you get 81%

Then, if that is valid or not is a matter of a deeper discussion. But we leave that for now.

Gunnar Englund

http://www.gke.org

--------------------------------------
Half full - Half empty? I don't mind. It's what in it that counts.

 

[waross]

Look at the speed versus current curve for the motor on normal 60 Hz.
60 Hz minus the actual speed is the slip speed or the slip frequency.
The slip frequency determines the greater part of the real current.
So, at any nominal VFD speed or frequency, subtract the actual speed from the VFD speed to get the slip speed at that combination of frequency and loading.
Refer back to the original speed versus current curve. From the the motor synchronous speed subtract the slip speed and estimate the current.
The current depends mostly on the torque demanded by the load. As the frequency changes and the speed changes, the torque demanded by some loads changes.
It may not be a simple linear relationship between voltage and current.
For a true constant torque load, the real current may remain constant over a wide speed range.
As the frequency changes, the inductive reactance of the motor windings changes.
The reactive component of the current AKA the magnetizing current may be more dependant on the voltage.
You can not apply a simple formula for a resistive circuit to a motor.
A motor is much more complex than a resistor and has more moving parts than a resistor.

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

 

[itsmoked]

You know that you can direct the VFD to show you the power and current? Change the speed and watch the 'magic' directly.

Keith Cress
kcress -

http://www.flaminsystems.com

 

[LionelHutz]

"Hp change will be the cubed of frequency change."

No, HP is whatever the load requires. In theory, if the motor is fully loaded at all times then the HP changes linearly with frequency. In other words, the maximum HP available from the motor is linearly proportional to the frequency.

If you want to use a rough rule of thumb, you could ratio the current range between the motor no-load current and full-load current by the percentage of torque the load requires. For example, say the motor no-load current is 30% of FLA. Then, if the load requires 50% torque the motor would draw approximately 65% of FLA current.

If you want to be exact then you pretty much need to do real world motor measurements on the motor.

 

[waross]

The maximum safe HP that a motor may develop is limited by the current.
Also the maximum safe torque that the motor may develop is limited by the current.
The voltage is limited by the Volts per Hertz ratio. As the frequency is lowered by a VFD, the effective voltage is lowered proportionally.
So, 50% frequency equals 50% voltage times the same torque equals 50% HP.
This is the safe limit, subject to cooling concerns at lower speeds.
This works both ways.
In some applications 480 volts may be supplied to a VFD driving a motor connected for 240 Volts. (230 Volts rated)
Now the Volts per hertz ration may be maintained above base frequency.
The 240 Volt motor may be supplied with an effective voltage of 480 Volts or 200% at 120 Hertz or 200% for an available safe HP of 200%.
(Subject to mechanical issues such as bearing life.)


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

 

[jraef]

Also, when discussing current a VFD, you must differentiate between input current and output current, they are no longer the same. That's because the DC bus capacitors are effectively correcting the displacement power factor to .95 regardless of motor loading, whereas on the motor side that is not the case. All motor current, whether active or reactive, must come from the drive output, resulting in the output current being higher than the input current.

Output current is also going to almost directly follow torque in the motor. So if we think of the drive as a "constant torque" source, the current would be the same regardless of speed. In reality loads don't work that way, even "constant torque" loads are more dynamic than tht due to multiple factors such as friction, transmission losses etc.. In a "variable torque" (centrifugal/quadratic) load, as speed drops the power required by the load drops at the cube of the speed change, so that means torque required will drop as well. But at the same time, with load coming off of the motor as speed drops, the motor's power factor is dropping too, so more of the current going to the motor is reactive. That just means the current will change with speed in a variable torque load, but it is not linear.

Personally, other than for purposes of monitoring for motor overload, I find it pointless to get caught up in what motor current is or is not doing when a VFD is involved. It's too difficult to measure it accurately, and too complex to worry about it. The conventional wisdom on motor curves etc. are all based on fixed frequency, that all goes out the window when a VFD is in the loop.

"You measure the size of the accomplishment by the obstacles you had to overcome to reach your goals" -- Booker T. Washington