Expected voltage drop with a VFD 3

[Freshsurfer]

Hi,

Is possible to determine how could be the voltage drop during a start of a 550kW motor (crusher) by a MV VFD (AFE)?

 

[GroovyGuy]

Yes.

"I have not failed. I've just found 10,000 ways that won't work." Thomas Alva Edison (1847-1931)

 

[jraef]

You can program the drive current to be whatever it is capable of, but generally that is going to be either 110% for 30 seconds or 150% for 60 seconds. So that will depend on how the drive was selected for the type of load. If it's a centrifugal pump or fan, it will likely be the 110% current for 30 seconds maximum. With that as the max current, you can plug that value into a Trandient Starting Analysis software program like SKM or ETAP if you want an exact value.

On the other hand, you can, if necessary, program the VFD to never exceed FLA of the motor and, assuming you never need more than Full Load Torque to accelerate, take all the time in the world to do so and not cause any more voltage drop than you would running the motor in the first place.


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

 

[itsmoked]

It's a crusher Jeff.

Keith Cress
kcress -

http://www.flaminsystems.com

 

[Freshsurfer]

Thank you Jraef!, I'm still with a doubt about what is the expecting voltage drop when using a VFD, on the worst case when we need 150% for 60 seconds. I know that on the VFD we have the DC link that is equipped with a capacitor bank that holds the rush during the start, but I don't know if it is enough to state that the voltage drop on the incoming AC bus is going to be close to 0%. In this particular case we have to ensure that this voltage drop (on the AC incoming bus) will be less than 10%

 

[LionelHutz]

What voltage drop do you get running the motor at FLA???

Remember that the VFD is a power conversion device. Requiring 150% current to the motor at zero/low speeds doesn't mean the VFD will draw 150% current from the line. Real power in = Real power out + VFD losses (approximately). Inductive current going to the motor doesn't come from the line at all. You can probably supply 150% current to the motor up to around 70% speed before the VFD reaches 100% current draw from the line.

 

[jraef]

It's a crusher Jeff.

D'oh! It was right there...

Fresh surfer,
Lionel and I are not saying it will be zero % voltage drop, we are telling you that under the right circumstance the voltage drop will be the same as if the motor is simply running at all. That may or may not be zero %, we have no way of knowing! but you can use that.

The ABSOLUTE worst case scenario is that once your motor approaches slip speed, it MIGHT need to go higher than FLA, but will not be more than 150% of that value because the drive will not deliver it. And if you can't live with 150% for even a short time, simply tell the drive "Limit current to 100%" in programming and it will. That will force the drive to override the commanded speed/ramp time settings and extend the acceleration time, but if your motor can generate enough to torque for the load (meaning the original designer was not planning on overloading the motor on purpose to accelerate it), the drive can get you there, eventually, without going over FLA.


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

 

[buzzp]

I think he is talking about the voltage drop on the line side. This should not be a concern since this drop should be negligible assuming the drive is programmed properly. If it is programmed right, the drive will trip out or it will limit the current so there should be no Vdrop on the line side. The capacitors are for supplying the voltage to the switching devices, which in turn feeds your motor. These are charged at peak voltage not RMS.

 

[controlsdude]

If you current limit a vfd drive, the drive will go to some percentage of full speed hertz till the current limit kicks in, then the drive is only partially ramped up depending on the load since its a crusher, not sure how that look like? On a conveyor the current limit would just be a reduction in speed of the conveyor.

 

[LionelHutz]

Yes of course he's asking about the line side voltage drop, and no you can't just say it will be negligible since you have no clue about the system.

 

[waross]

Determine the current limit that the drive will be set at. That current will determine the voltage drop on both the line side and from the VFD to the motor. Take cable reactance into account. The drop will be more than that calculated on cable resistance alone.

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