static var as a means to stop droop

[dcasto]

I'm being told by a vendor that if we use a static var device that we can start very large (1000HP) motors across the line and the power company would not see droop on their system. I can't imagine that the rack of capacitors could be big eneough or if big enough, it would be cheaper to install a soft start.

 

[dpc]

It's possible. Static Var compensation (SVC) is more involved than just a capacitor bank. But you need a lot of vars.

If the only concern is limiting voltage dip during motor starts, a soft starter or even a VFD will be less costly, and probably easier on the motor than starting across the line.

I've seen SVCs used on arc furnaces and they are amazingly effective once they are tuned up properly.

 

[davidbeach]

I think there may be some confusion. The SVC may be able to mitigate the transient voltage dip associated with the motor starting but it will not have a significant affect on the steady state "droop" associated with the motor running.

dcasto's use of "droop" does not suggest to me the transient sag/dip during motor starting but suggests a longer term condition, and a soft starter won't help much either.

If the only concern is the initial dip, the dpc has the answer.

 

[waross]

The Electric Machinery Mfg. Copmpany
Thees people used to publish a series of technical booklets titled E-M Synchronizer. Special issue #200-syn-59, copywrite 1961, discusses shunt capacitors for starting.
A quick look at an example motor.
400 hp
full load KVA = 360 KVA
Starting KVA = 2200 KVA This depends on the system characteristics as well as the motor characteristics.
voltage dip at start up = 30.5%
Starting KVAR =2100 KVAR

Shunt capacitors for starting.
1050 KVAR of capacitors are added to offset 50% of the starting KVARs.
The voltage dip at startup is reduced to 20.6%

1050 KVARs is about 900% of the amount of KVARs that it is safe to leave connected to the motor terminals continuously.
The extra KVARs must be cut out of the circuit in steps as the motor accelerates. A conventional Power Factor Controller may not act fast enough.
You may need a purpose built VAR controller to respond fast enough to the changing conditions as the motor accelerates.
respectfully

 

[dcasto]

Thanks all, as I suspected, vendor hype. There is no magic cure-all for starting large motors and correcting the running power factor with a single device without a custom computerized piece of equipment switching capacitors (which is what the original vendors system may be doing, but they were a little sketcy on how everything works, more question from me will help) . A good soft start and set of capacitors, each doing thier jobs is better and cheaper.

The exception is a vfd will start with nearly zero torque and little incurrent flow and have a PF of almost 1, but the cost is more than any combination of other devices, based on some limited quotes.

I guess it time to hire a EE to get the final sort on it all.

 

[itsmoked]

So you are concerned only by the starting transients, not the steady-state voltage drop correct?

Keith Cress
Flamin Systems, Inc.-

http://www.flaminsystems.com

 

[dcasto]

Yes, it comes down to can I start a 1000 HP motor across the line with no droop on the incoming line with capacitors alone.

 

[davidbeach]

dcasto, At least in North American usage, what you are describing would not be a droop, but rather a voltage dip.

 

[Marke]

The problem is that when you initially connect the motor to the supply, the current is very high, typically 600% - 900% of the motor rating, and the power factor is very low, typically 0.15 - 0.2. As the motor accelerates, the power factor rises and the current falls by a small amount. To achieve what you want, the correction capacitors must be dynamically controlled such that they are sequentially switched out as the motor accelerates.
This requires a lot of capacitor stages with rapid switching and a fast controller that takes them out. There will also be a major transient when you switch on both the motor and capacitors.

Incidently, although a VSD has a Cos(phi) of better than 0.95, the true power factor is much less and can be as ow as 0.7

Best regards,

Mark Empson

http://www.lmphotonics.com

 

[dcasto]

OK, I'm a ChemE, I surrender. will a static var stop a voltage drop so I do not need a reduced voltage starter or VFD . The incoming is 12.4 KV and the motor is 4160. The transfomer and incoming line are rated to 3000 KW and there will be 2 1000 HP motors plus about 500 HP of other motors.

As I said, I'm ready to hire a EE, but I'm trying to sort out the flaky advice from vendors.

 

[davidbeach]

A static var compensator will help some, see above. Enough static var compensator to equal the performance of a VFD will likely be far more expensive than the VFD.

 

[dcasto]

thanks davidbeach.

 

[itsmoked]

So you have about 2MW of 'running' motor and a 3MW source. Ouch. That is a tight squeeze. I see your concern and it's valid.

The cheapest solution is probably a pony motor. A 50HP motor that just gets the whole mess spinning. This requires a load type that can be spun up. Some sort of process that once the machine is running can be loaded by opening a valve, closing a valve, turning on a feed conveyor, etc.

What kind of load are you running with these motors?

Can you state an order that they will always start in? Can the two 1000's be started before the other 500?

Will the 1000s always be running together? There are some games you could play. Like getting one 1kHP started then use it to provide stored energy for starting the second unit since now you are closer to the 3MW limit. Don't know it depends on your processes.

Next cheapest is a Soft Starter. That can limit you to something like 200% rather than the aforementioned 8-900% hit.

Lets take a gander at the current. If we have the 500 and one 1k running, we have 1500hp x 746W/HP = 1.2MW
This translates to 1.2MW/4160 = 288A

The transformer is good for 3MW. 3MW/4160V = 720A

Now a 1kHP motor alone = 1000HP x 746W/HP = 0.746MW
Current 0.746MW /4160V = 180A

Now if that motor is direct-on-line(DOL) 180A x 8 = 1440A

So the already running 288A plus the DOL 1440A = 1728A. More than twice the transformer's ability. (I can't see! I can't see! Turn on the lights.)

But if you can limit that second 1kHP to 200%(depends on load type) then we have; 288A + (180A x 2) = 648A

720A - 648A = still a positive number.

If you can tell us about your loads others here can estimate your likelihood of a 200% start working for you.

Then you have your third option: VFD. Which realistically will probably provide a 125% start but will cost you a bundle. Now if your process would like a variable 1kHP then it could provide you with some nice flexibility.

Keith Cress
Flamin Systems, Inc.-

http://www.flaminsystems.com

 

[Marke]

Hello dcasto

Are you concerned about the voltage drop on the 4160 line, or the 12KV line?
If it is the 12KV line, then the transformer size is not the issue, it is the impedance of the 12KV supply.
If the impedance of the 12KV supply is low enough, you will be able to start those motors without causing a depression on the 12KV line, but it depends on what the impedance is. Typically, the impedance of the supply is indicated by the fault current capacity of the supply.
If your fault current capacity on the 12KV line was greater than 5KA at 12KV, then the voltage drop during start should be less than 5%.
The start current reduction that can be achieved using a soft starter is dependent on the start torque required by the machine and the start characteristics of the motor.
In most cases, I would expect to see the start current drop to less than 450%, but sometimes 500% or more is required to start the load.
If the concern is about the voltage drop on the output of the transformer, then you need to know the impedance of the transformer. This may be expected to be in the range of 4% - 6%. If the 12KV line was solid, then the voltage drop on the output of the transformer would be in the order of 10% from starting one machine. (Assuming 600% start current and 5% impedance for the transformer).

I think that a good EE would be able to give you the answers quite quickly provided that he has access to good information.

Best regards,

Mark Empson

http://www.lmphotonics.com

 

[dcasto]

Got the base data I need and areas for the EE to research. Thanks again.