Hi all !

After consulting with engineers and manufacturers I still end up with the same problem. I cannot find a way to calculate this:

An elevated load goes down at a fix speed. The motor connected to it is barely able to hold it. After a few cycle we get a Buss overvoltage error
on the ac drive. We add an external braking resistor to dissipate the regenerated energy and all is o.k. BUT ...

How can we calculate this theoriticaly regenerated energy (not calculating the losses).

At 60 Hz the drive lasted 45 minutes
At 15 Hz the drive lasted 5 minutes
At 90 Hz the load crashes at the bottom for the drive cannot brake it...

The problem I have with the calculation is that Energy is divided by time to give Power and this way when I slow it more it regen. less energy altough the motor fights gravity for a longer time...

One EE sugested that I take the Tork of the system needed to equal the force (F) from gravity and with the motor speed and the kynetic energy of the load, add them up and compare to the force from gravity. Whatever is left must be the power regenerated... I end up with squared seconds... There must be a way...

Any suggestions will be greatly appreciated.

Pierre Desrochers
System Integrator
[email protected]

 

Try accounting for dissipation and work seperately.

cww

 

Hi,
try at http://www.reliance.com/prodserv/standriv/appnotes/ "Breaking and Regen Energy with AC Drives"

Marian

 

At 13:51 20/07/00 -0400, Pierre Desrochers wrote:
<clip>
>I cannot find a way to calculate this:
>An elevated load goes down at a fix speed. The motor connected to it is
>barely able to hold it. After a few cycle we get a Buss overvoltage error
>on the ac drive. We add an external braking resistor to dissipate the
>regenerated energy and all is o.k. BUT ...
>How can we calculate this theoriticaly regenerated energy (not calculating
>the losses).
<clip>

The load energy dissipated is simply the potential gravitational energy of the suspended mass. If the drive cannot regenerate this energy
back into the hydro lines, then it must be dissipated in the drive (or regen resistor). Some energy will also be dissipated in the gear system, slides, etc.

>The problem I have with the calculation is that Energy is divided by time to
>give Power and this way when I slow it more it regen. less energy altough
>the motor fights gravity for a longer time...

No, the load energy which must be dissipated is the same regardless of the time (you said it yourself in the previous sentence). It is the *power* which changes with speed. This matters because the drive is limited in the power it can absorb.

>One EE sugested that I take the Tork of the system needed to equal the force
>(F) from gravity and with the motor speed and the kynetic energy of the
>load, add them up and compare to the force from gravity. Whatever is left
>must be the power regenerated...
>I end up with squared seconds... There must be a way...

I dealt with a similar problem a year or so ago (except it concerned repeatedly braking a rotating load). You can twist the numbers up in knots unless you go back to the basic physics. You have a certain potential energy, and it must be dissipated somewhere. The faster the mass falls, the more peak *power* the drive must absorb. The duty cycle of the process matters as well, as if the thermal capacity of the regen resistor is the limiting factor, then it may have time to cool down between cycles.
The drive is limited by power, not energy considerations. Beyond this, you have to deal with any special characteristics of the drive and
motor. If this is an induction motor for example, then if it is driven too slowly it may overheat.

Any electrical energy which is used by the drive while the load is being lowered is not being turned into useful work - it all is being
dissipated in the drive and motor somehow. It obviously isn't going into the load, since the load is also providing energy to the motor and drive (which they must get rid of).

In our case I used these basic principles to eliminate a number of drive and motor options. We had to accelerate and deccellerate a load
frequently, and found that it was the regenerative characteristics of the drive which was the limitation, not its ability to accelerate the load. We ended up using a DC motor and a four quadrant regenerative DC drive since the brushless motor/drives we looked at all had a problem with regeneration at the power we needed (we also needed a very compact motor). The DC drive was approximately $300 or so, and has proved quite satisfactory for our particular application.

You mentioned that "The motor connected to it is barely able to hold it." I'm not sure I understand this - does the motor not at some time raise the load also?

I think that someone with elevator experience could help you more than I have, but I hope that the above is of some use.


**********************
Michael Griffin
London, Ont. Canada
[email protected]
**********************

 

Visit www.MotionOnline.com and download the Handbook of AC Servo Systems (see page 35) and download the PE Servo System Manual (see page 47,48,49 on external regen resistors).

Let me know if you would like to discuss by phone. I'd be happy to help.

Best regards,
George Kaufman
770.497.8086 x2071

 

Hi
Try do it like this (it's very simplified):
Pb = (sb/tb) * (m*g + Ek/sb - Fo)

where:
Pb - average breaking power (watts),
sb - breaking distance (meters),
tb - breaking time (seconds),
m - weight of the load (kilograms),
g = 9,81 m/s2,
Ek - kinetic energy of the load (kg*m2/s2)
Fo - resistance to movement force (kg*m/s2) derived from moment of friction of gear boxes, motor etc.

Regen power is about 90% of Pb (because loses in motor).

If you need more detailed information I can send you a PDF-file about this item.

Regards,
Jacek Dobrowolski

Software Eng.
Electrical and Software Design Department
Secondary Division
International Tobacco Machinery Ltd. Poland

 

Use a counterweight or a counterweight air cylinder with dual stage precision regulator (high exhaust) if the size of the load allows. Then the load will be symmetrical (up vx. down).

 

I'm not sure how much it adds, but your system does not take into account motor inertia. In servo systems, the rotor intertia, as well as other rotary transmission components can add significantly to the overall system kinetic energy.