Feeder calcualtion with combination loads and VFD

[Rickyghart]

I need to deterimine the minimum size of a power source (diesel generator) . The loads are a mixture of things and I know how to get to a normal service size calculated. What is throwing me is I have several good sized motors ranging from 100HP down to 10 HP that are served by VFD and motors not VFD supplied. 430.122 says I take the input current rating at 125% of the nameplate of the drive for conductor sizing. So the questions are:

1) is the drive input current rating used instead of the HP FLA of the VFD supplied motor in 430.250 since this essentially a branch circuit to the drive not to a motor.
2) Does the 125% carry over to the service calculation for each drive or just the input rating?
3) Does the 25% of the largest motor fall to the next largest motor not supplied by a drive or to the largest motor which is served by a drive?

Any help would be appreciated. I am not asking if the real world reduced load seen on the generator is acceptable. I need the hard NEC requirement.

 

[Smart $]

Article 430, Part X has no bearing on service and feeder calculations?220.50, by reference therein 430.24, 430.25, 430.26, 440.6, which have no additional references to Article 430 Part X.

Welcome to the forum :thumbsup:

 

[petersonra]

I need to deterimine the minimum size of a power source (diesel generator) . The loads are a mixture of things and I know how to get to a normal service size calculated. What is throwing me is I have several good sized motors ranging from 100HP down to 10 HP that are served by VFD and motors not VFD supplied. 430.122 says I take the input current rating at 125% of the nameplate of the drive for conductor sizing. So the questions are:

1) is the drive input current rating used instead of the HP FLA of the VFD supplied motor in 430.250 since this essentially a branch circuit to the drive not to a motor.
2) Does the 125% carry over to the service calculation for each drive or just the input rating?
3) Does the 25% of the largest motor fall to the next largest motor not supplied by a drive or to the largest motor which is served by a drive?

Any help would be appreciated. I am not asking if the real world reduced load seen on the generator is acceptable. I need the hard NEC requirement.

The 125% is for figuring the conductor size. It has no bearing on what you would size the generator at.

I think you will find that the problem is that you could well be fine during normal running but be unable to start multiple motors simultaneously, as the start-up current is much higher than the normal running current.

add up the fla of all the loads. add 5X the fla of all the motors that might start simultaneously. plus 5X the largest non-vfd motor. that will give you a good idea of the bare minimum size generator you need.

vfd supplied motors will require less startup current.

 

[Rickyghart]

Article 430, Part X has no bearing on service and feeder calculations?220.50, by reference therein 430.24, 430.25, 430.26, 440.6, which have no additional references to Article 430 Part X.

Welcome to the forum :thumbsup:

430-24 is what is throwing me.

Is the drive a motor load or a non-motor load? In my mind the motor feeder is on the output of the drive and the feeder to the drive is not a motor feeder since the requirements of 430 Part X are somewhat different from a motor- nameplate based. It feeds an electronic device which has a higher input current than the FLA in 430.250 for the same size motor. I am supposed to use the largest of the two, I think.

What has gotten dicey for me is a 200HP drive was installed for a 75 HP motor. Motor FLA is lower that the rated input current for the drive. But the drive's input takes what it needs for a 200HP motor. I am just trying to understand what I am looking at with a drive.

Thanks for the welcome! Happy to be visibale to the forum. Usually I can search my way to an answer.

 

[Rickyghart]

The 125% is for figuring the conductor size. It has no bearing on what you would size the generator at.

I think you will find that the problem is that you could well be fine during normal running but be unable to start multiple motors simultaneously, as the start-up current is much higher than the normal running current.

Thanks. This starting to come together.

 

[petersonra]

430-24 is what is throwing me.

Is the drive a motor load or a non-motor load? In my mind the motor feeder is on the output of the drive and the feeder to the drive is not a motor feeder since the requirements of 430 Part X are somewhat different from a motor- nameplate based. It feeds an electronic device which has a higher input current than the FLA in 430.250 for the same size motor. I am supposed to use the largest of the two, I think.

What has gotten dicey for me is a 200HP drive was installed for a 75 HP motor. Motor FLA is lower that the rated input current for the drive. But the drive's input takes what it needs for a 200HP motor. I am just trying to understand what I am looking at with a drive.

Thanks for the welcome! Happy to be visibale to the forum. Usually I can search my way to an answer.

again, this is for sizing the wire, not the generator.

I would use 125% of the rated input of the VFD for calculating the feeder size, although one could argue you should use the motor FLA from the tables.

 

[kingpb]

Caterpillar has a sizing program that is accessible on-line. It used to require the software to be installed on your computer but it has been switched to internet based.

Sizing DEG when motor starting gets tricky and the size can vary greatly based on the size of the motor and the load starting sequence, i.e. all at once or staged. Even the order of staging can make a huge difference.

The size will be based on not only current/KW but the DEG has to be able to provide the VARS that will be needed during motor start; a 200Hp (150kW) motor starting across the line is going to draw maybe 196KVAR give or take. The DEG has to supply that along with whatever other loads are attached, and also starting.

 

[petersonra]

Caterpillar has a sizing program that is accessible on-line. It used to require the software to be installed on your computer but it has been switched to internet based.

Sizing DEG when motor starting gets tricky and the size can vary greatly based on the size of the motor and the load starting sequence, i.e. all at once or staged. Even the order of staging can make a huge difference.

The size will be based on not only current/KW but the DEG has to be able to provide the VARS that will be needed during motor start; a 200Hp (150kW) motor starting across the line is going to draw maybe 196KVAR give or take. The DEG has to supply that along with whatever other loads are attached, and also starting.

not many 200HP are started across the line these days.

in any case where did you find that it will draw only 30% extra at startup?

 

[Rickyghart]

I am aware of the starting KVA of motor loads. The drives are the fuzzy part for me. I have seen large motors on large drives that are 50% loaded trip electronic breakers because of the massive instantaneous current draw makes it look like a instantaneous fault. All the while the maintenance guys clamp-on shows very little current.
All said, it is the fear that the excitation field may collapse on the alternator when these motors start AND the VFD does its voodoo that has me thinking too hard for a Friday.
Thanks all!

 

[petersonra]

I am aware of the starting KVA of motor loads. The drives are the fuzzy part for me. I have seen large motors on large drives that are 50% loaded trip electronic breakers because of the massive instantaneous current draw makes it look like a instantaneous fault. All the while the maintenance guys clamp-on shows very little current.
All said, it is the fear that the excitation field may collapse on the alternator when these motors start AND the VFD does its voodoo that has me thinking too hard for a Friday.
Thanks all!

Most meters average over some period of time unless you set it to display peak values.

I can't say I have ever seen a VFd trip a Cb like that but I almost never use CBs with electronic trips.

 

[kingpb]

not many 200HP are started across the line these days.

in any case where did you find that it will draw only 30% extra at startup?

Ooops, thanks Bob for catching that, my bad.

I forgot to increase by 6x starting current, which would mean the Gen would need to supply around 1176KVAR (@0.2 starting PF) 0.2 may be a little low, probably more like 0.24-0.26 but unless you have nameplate data its best to be safe.

So, the gen has to be able to carry the running kW, but handle the starting load. Gen manufacturers talk about starting KVA and KW (sKVA and sKW), but in reality the var capability of the machine has to be considered even though they don't talk abut that.

Looking on CAT specsizer, for a single 200Hp motor started ATL, they say you would need a 750KW machine.

 

[kingpb]

I am aware of the starting KVA of motor loads. The drives are the fuzzy part for me. I have seen large motors on large drives that are 50% loaded trip electronic breakers because of the massive instantaneous current draw makes it look like a instantaneous fault. All the while the maintenance guys clamp-on shows very little current.
All said, it is the fear that the excitation field may collapse on the alternator when these motors start AND the VFD does its voodoo that has me thinking too hard for a Friday.
Thanks all!

VFDs require larger generators. The current drawn by these drives is nonlinear (having harmonics), which causes a distorted voltage drop across the reactance of the generator. Since VFDs are nonlinear, you must include an additional generator capacity sizing factor to keep voltage distortion to a reasonable level of approximately 15%
total harmonic distortion (THD) or less. The larger the generator, the greater the reduction in impedance of the power source (generator), which in turn, reduces the effects caused by harmonic current distortion.

For six-pulse VFDs, a typical generator sizing factor would be twice the running kW of the drive. This offsets any reduction in starting kW and kVA. If it is the pulse width modulated (PWM) type (or includes an input filter to limit current distortion to less than 10%), then you can reduce the sizing factor down to 1.4 times the running kW of the drive.

 

[Rickyghart]

Fantastic! Thanks kingpb!

This is what my Spidey sense has been telling me.