static var as a means to stop droop
static var as a means to stop droop
(OP)
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.





RE: static var as a means to stop droop
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.
RE: static var as a means to stop droop
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.
RE: static var as a means to stop droop
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
RE: static var as a means to stop droop
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.
RE: static var as a means to stop droop
Keith Cress
Flamin Systems, Inc.- http://www.flaminsystems.com
RE: static var as a means to stop droop
RE: static var as a means to stop droop
RE: static var as a means to stop droop
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
RE: static var as a means to stop droop
As I said, I'm ready to hire a EE, but I'm trying to sort out the flaky advice from vendors.
RE: static var as a means to stop droop
RE: static var as a means to stop droop
RE: static var as a means to stop droop
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
RE: static var as a means to stop droop
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
RE: static var as a means to stop droop