Load sharing between motors 3

[Martin300]

We've got a conveyor that we're upgrading from 1 x 350HP motor to 2 x 250 HP Motors. Both motors will run from a single starter with individual overload protection.

The problem is that the motors are not identical; close but not exact. They are both Tatungs and 2300V 250HP but one is 1775 RPM, 59.9Amps, 92.6% efficiency. The other is 1780 RPM, 57.0 Amps, 94.5% efficiency.

Does anybody have an estimation of how much imbalance there might be in load sharing? The motors couple through identical gearboxes into separate drive pulleys close together - the 48" conveyor belt does the coupling.

Thanks in advance.

 

[waross]

They should work well together as long as the load is below maximum. At maximum load, the 1780 RPM motor will be dragged down to 1775 and can be expected to draw more than rated current. At less than full load, the slip will be less and both motors will be running above rated speed. Don't expect exact current sharing or exact load sharing. I anyone really wants an exact match on current, suggest matched 250 HP motors on his budget.
Your slips are 20 RPM and 25 RPM respectively. Expect one motor to be at about 80% load when the other is at 100% load. Total HP within rated current = (100%+80%)/2 = 90% = 450 Hp.
Approximately.

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

 

[itsmoked]

You could possibly run one DOL and the other with a VFD to eek out the full 500HP.

Keith Cress
kcress -

http://www.flaminsystems.com

 

[waross]

If the system has to start loaded that may problematical. However, if the system may be started unloaded, the VFD could bring both motors up to speed and work very well.
(Problematical doesn't mean that it couldn't start loaded with a VFD on one motor The VFD could start one and when it could not accelerate any further, the second could be connected DOL.)
It may take some tweaking.)

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

 

[controlsdude]

if you need them to work together than have an encoder on the conveyor to the drive to close the speed loop, also you can use the pulse train signal between the drives also to have one the master than the other is the slave.

 

[LionelHutz]

If you want to talk VFD's then it should be possible to just run the faster motor in torque limit so the VFD basically slows down the speed of the motor to keep the motor at full-load. Then, the other motor will eventually load up if the conveyor load is high enough.

This is assuming the use of a decent quality VFD. Of course, using an encoder requires a decent VFD too.

I'm not too sure on using 1 VFD and 1 full-voltage though. That presents a problem getting both motors to work together if the conveyor can be started loaded with both motors.

 

[roydm]

2300 V VFD, that sounds expensive.
Have you considered the belt slip it will help balance out the load won't it?
If you find the load is too far out you have the option of modifying either pulley slightly to get them back into line.
I worked with a similar conveyor, 2 pulleys, 2 gearboxes, 3 x 900 HP motors, there was insignificant difference between the motor by itself and the pair.
Roy

 

[Martin300]

As Roy noted, a 2300V VFD is expensive to the point where buying identical motors is more cost effective. A 600V VFD is cheaper and could be run is some form of current limit or droop mode, but having one motor on 2300v and the other on 600v becomes a major hassle for lockout procedures.

I graphed the two motors assuming slip and amps were a linear function and, as Bill stated, came up with the same kind of load sharing values. I believe this will be adequate for our needs, even starting up fully loaded.

Thanks for everybody's help. It's nice to bounce some ideas around.

 

[electricpete]

I think most of the practical aspects have been covered. As long as neither is overloaded, the exact sharing is probably not critical (is it?). And the 1780 rpm will reach it's full load first.

Nevertheless, the fraction of load sharing can be calculated based on the nameplate slip, by noting that torque starts at 0 torque with 0 slip and goes to full torque at full slip.

T1780 = Trated*Nslip/20
T1775 = Trated*Nslip/25

Ttotal = Trated*Nslip*(1/20 + 1/25)

T1780/Ttotal = 1/20 / (1/20 + 1/25)
T1780/Ttotal = 1 / (1 + 20/25)
T1780/Ttotal = 25/40 (slightly more than half)

Likewise you can find
T1775/Ttotal = 20/40 (slightly less than half)

Note that the nameplate speeds are ballpark numbers anyway (NEMA MG-1 allows 5 rpm deviation in these numbers). Also, losses cause increase in slip but do not contribute to output torque - was neglected in the above analysis.

=====================================
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[electricpete]

Whoops. Those fractions have to add up to 1, so must have had error:

The 1780 rpm motor takes fraction of total load:
(1/20)/(1/20 + 1/25) = 25/45

The 1775 rpm motor takes fraction of total load:
(1/25)/(1/20 + 1/25) = 20/45

20/45 + 25/45 = 1.
Order is restored to the universe.

=====================================
Eng-tips forums: The best place on the web for engineering discussions.

 

[ServiceEng]

What about a mechanical constant torque coupler added to both motor. Would that be too inefficient?

 

[UKpete]

As an aside, the rail traction people know all about load sharing of motors - all the motors being coupled via the wheels. In the old days of dc traction motors (not that long ago) it was a particular problem and great care had to be taken to adjust individual motor airgaps so that the load was shared. With ac motors, where they share inverters, the worry was that differing wheel diameters would cause poor load sharing but I think experience has shown that the problem wasn't that serious, even that it may be self-correcting.

The easy way to assess load-sharing, as has already been stated above, is to compare the motor currents.

 

[Martin300]

An interesting follow-up on this. Electrically everything worked OK but the load sharing was very poor. While most of the right questions were asked - identical motors and gearboxes it turns out the existing drive pulleys were a (slightly) different size.

Since the motors are high efficiency the rated speed is 1780 RPM (we can assume 1800 RPM no load) so only 20 RPM of slip. 20 RPM is 1.1%.

The existing conveyor drive pulleys are 30" with 3/4" lagging so a 31.5" diameter. If one the pulley diameter is worn by 1/8" that is a 0.4% difference so motor #1 will have to be at 36% load before motor #2 starts to do any work. Our case turns out even worse than that, so at full load with motor #1, motor #2 is only taking about 20% load.

The solution will be to put both motors & gearboxes on a single shaft and drive pulley.

The requirement for both pulleys to be identical is really basic in this setup, but nobody (including me) questioned it.

Interestingly, if we had a lower efficiency motor with higher slip (say a 1740 RPM motor) then the effects of the mismatched drive pulleys would be reduced somewhat.

So the conclusion is make sure you ask all the mechanical questions, even if your area of concern is only electrical. Don't trust the mechanical experts to consider all the factors.

 

[itsmoked]

We thank you for the update.

Keith Cress
kcress -

http://www.flaminsystems.com

 

[waross]

Yes I know. I worked in a plant years ago where 25% to 30% of the electrical problems we were faced with turned out to be mechanical problems.

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