VFD and motor - Low RPM

W@ttson

W@ttson

Senior Member
Location
USA
Hello,

I have a general question regarding VFD operation with a motor.

Suppose a motor has an FLA of 100A. It is rated at 460V 60Hz.

Now suppose you want to run the motor at 6hz. To maintain you V/Hz ratio, since you load is the same, your voltage would decrease to 46V. Now if the same power is being delivered to the motor, I would assume that the current to the motor would need to increase.

P = I*V*sqrt(3)*PF
For same P, with decreased V, the I would be inversely proportional.

What if this increased current is greater than the FLA of the motor? Say the current measured to the motor is 130A. Is there risk to the motor? I assume not since the duration of operating the motor would only be 30 min or 60 min or if it were continuous it would have a blower on it.

Any insight on this where I might be missing something would be appreciated.
Thank you
 
The power delivered to the motor depends on the shaft load. What kind of load do you have that is constant HP as you slow it down?
You can (roughly speaking) get constant torque at lower than nameplate speeds. So available HP will drop proportional to the speed.
 
retirede said:
The power delivered to the motor depends on the shaft load. What kind of load do you have that is constant HP as you slow it down?
You can (roughly speaking) get constant torque at lower than nameplate speeds. So available HP will drop proportional to the speed.
Click to expand...
its a hoisting application. its a constant torque load. I guess I shouldn't be seeing constant KW going to the motor with difference in speed then.
 
This is true, but you normally don't deliver constant power at variable speed. Mechanically, the power delivered is the product of angular velocity (which you are turning down) and torque. To require constant power mechanically, that means the torque required would have to go UP. That normally doesn't happen.
 
Yeah, now I am confused.

My data shows the following:

full speed = 900rpm
Voltage = 460V
Current = 75A
KW = 22KW


10% Speed = 90RPM
Voltage = 52V
Current = 142A
KW = 11KW


if the power reduced (not sure why by 2), why is the current so high?
 
W@ttson said:
Yeah, now I am confused.

My data shows the following:

full speed = 900rpm
Voltage = 460V
Current = 75A
KW = 22KW


10% Speed = 90RPM
Voltage = 52V
Current = 142A
KW = 11KW


if the power reduced (not sure why by 2), why is the current so high?
Click to expand...
Are the hoist/motor/drive all part of a factory assembly?
Also, how are you measuring these data points?
 
W@ttson said:
Yeah, now I am confused.

My data shows the following:

full speed = 900rpm
Voltage = 460V
Current = 75A
KW = 22KW


10% Speed = 90RPM
Voltage = 52V
Current = 142A
KW = 11KW


if the power reduced (not sure why by 2), why is the current so high?
Click to expand...

In condition 1, assuming the motor is idealized as 100% efficient, I calculate an output torque of 233 Newton-meters.
In condition 2, with the same assumption, I calculate an output torque of 1167 Newton-meters. About 5 times the torque load.

This is what I've inferred, using P = omega*tau for mechanical power, where omega is the RPM translated to radians/second, and tau is the torque.

Condition 2's current/voltage/power figures are consistent with what I'd expect, which indicates a power factor of 86%. I'm skeptical of the 75A current figure in condition 1. Using 460V and 22 kW electrical, and the same power factor as condition 2, I calculate 25.6A as the current I'd expect for condition 1. Perhaps 75A is the sum of the currents on all 3 phases, rather than the current per phase. Standard arithmetic sum of RMS values, rather than vector sum.

This is about 1/5th of the current of condition 2. Current directly determines the output torque, so I'd expect 5 times the torque to imply 5 times the current.

This is the problem with too high of a Volts-to-Hertz ratio. The Hertz and motor geometry, directly determine the maximum possible RPM of an AC motor. The only other place for the power to go, is the motor torque, which draws more current as a direct consequence.
 
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