Safely Starting a 10,000 HP Fan

[mesutphen]

I have a problem with our 7200V system that I want to fix. Our 7200V system is split into 2 busses. While starting a 10,000 HP Induced Draft (ID) fans, we must shut the tie breaker. If we do not, we will drop our 480V power centers on undervoltage. The problem here is that with the tie breaker shut, the available short circuit current (Isc) is 113kA, and the largest-rated breaker I have been able to find is 77 kA. With the tie breaker open, available Isc is only 56kA.

I’m looking at the solution from 2 perspectives: 1. Find a way to start an ID fan without having to shut the tie breaker, or 2. Find a way to limit the Isc with the tie breaker shut.

For limiting Isc with the tie shut, I have calculated the size of CLR I would need to limit the Isc while starting the fan, and 26MVA is the minimum size. Mainly due to physical size constraints, and also due to the huge cost associated with buying 2 of these, I would rather not go this path.

For starting the fan with the tie open, I have thought that maybe a capacitor bank in parallel with the fan may help, and would be cheaper and smaller than the CLR. However, I am not sure whether shifting the power factor will actually help, and am not sure about being able to switch the bank in and out fast enough to not have a bad effect on the rest of the system (specifically the generator). I know getting a different supply transformer, with a lower impedance, will help, but the cost for this makes me not want to go that path either.

The 7200V busses are supplied from the generator through our Unit Auxiliary Transformer (UAT) which is a 40/20/20MVA, supplying 7200V bus A through X winding at 14.054% impedance, and bus B through Y winding at 14.218% impedance. The 480V power centers are all supplied from the 7200V busses through individual transformers.

Does anyone have any other suggestions to:
a. Minimize voltage drop on the 7200V busses during ID fan starting
OR
b. Limit Isc while the tie breaker is closed

 

[dpc]

Regarding the CLR - you only need one - on the tie. That way it is not carrying current during normal operation.

As for the motor - is it starting across the line now? A reduced voltage starter can limit the current significantly, provided you can accelerate the fan at reduced torque.

An adjustable frequency drive would definitely solve the problem, but that's going to be an expensive drive.

 

[TurbineGen]

A soft starter could potentially work, however there could be a start up time issue. I don't know what the heating curve for the motor is or how much inertia the motor must overcome to turn the fan. You could end up having excessive start up times. Also there is the problem when the reactor is switched out for motor run, the voltage could drop significantly anyway.

A possibility is a pony motor. Perhaps you could use a smaller motor to get the fan rolling before you attempt to start it. Of course you'll only be able to get it rolling to a slow speed with the smaller motor. You can combine this with a soft starter and that might solve your problem.

Just my thoughts.

------------------------------------------------------------------------
If it is broken, fix it. If it isn't broken, I'll soon fix that.

 

[edison123]

At that HP and that voltage (IIRM, the max recommended HP at 7.2 KV is around 5000 HP ?), a soft starter like VFD or a fluid coupling would have been a wise choice.

Muthu

http://www.edison.co.in

 

[jraef]

You can get a solid state soft starter that size and voltage now. That would eliminate the transition spike issue. Have you calculated the maximum kVA you can pull without the tie breaker closed? If so, you can get a pretty good idea on the ability to start it. But better yet, hire someone with a software program called SKM Power Tools or ETAP and have them do what is called a "Transient Motor Starting Analysis" on this motor and load with your supply system. They can very accurately predict whether a particular starting method will or will not work. Something that large and critical should not be done with guess work.


"If I had eight hours to chop down a tree, I'd spend six sharpening my axe." -- Abraham Lincoln
For the best use of Eng-Tips, please click here -> faq731-376

 

[dpc]

When I mentioned reduced-voltage I was assuming a solid-state soft starter. But the problem with the soft starter will be accelerating the load. ID Fans can be tough.

I would not assume automatically that the soft starter will work - a study would be needed.

I guess another question is why the 480 V power centers are fitted to trip on undervoltage and what is the setting?

What is the actual voltage drop seen at the 480 V substations? The 480 V system can tolerate voltage down to 85% or so for short durations. Below that, the motor starters start to drop out.

 

[alehman]

It seems like a capacitor switched on during starting could help with the voltage problem. It should be a reasonably simple study.

Alan
----
"It’s always fun to do the impossible." - Walt Disney

 

[apowerengr]

Your short circuit values don't make sense to me given the system description. I'm assuming you have 56kA at 480 volts and that the fan is connected to the 7200 volt bus? Closing the 7200 volt bus tie won't cause the 480 volt fault current level to double.

What am I missing? Can you post a one-line?

 

[mesutphen]

I appreciate the input so far. To answer questions up to this point:

The motor is started across the line. I had thought about a soft starter before, but worried about the physical size. Anyone have an idea how big one might be?

The power center source breakers have 2 undervoltage relays. The first set at 90V with a 0.5 second time delay which will trip the source breaker and close the tie breaker (480V). The second is set at 80V with a 2 second time delay, which control UV lockout. I'm not sure how much voltage is actually seen at the power centers. It's been a long time since we tried to start an ID fan with the tie open.

We actually have 56kA at the 7200V bus, with the tie open. Once the tie goes shut, we have 111kA at the same 7200V bus. I've attached a breaker-open and breaker-closed picture from our Easypower software. The bus short circuit values are at the upper right of the bus, with symmetrical current in kA and asymmetrical current underneath that in parentheses. i.e. 70.715 (111.675)

Picture with tie closed:

http://files.engineering.com/getfil...a1-99ef-081a8ad37534&file=7200_Tie_Closed.BMP

Picture with tie open:

http://files.engineering.com/getfil...4024-80f8-102a563062a9&file=7200_Tie_Open.BMP

 

[jraef]

... I had thought about a soft starter before, but worried about the physical size. Anyone have an idea how big one might be?

Generally, solid state soft starters are about the same size as an across the line starter. But you cannot get them in two-high stacks; they will be one per section. 7200V 10,000HP will most likely be two bays, so 72" wide, including the switchgear. If you use your existing gear as a feeder / isolator, you may be able to add just a soft start controller down stream that would be 36" wide.


"If I had eight hours to chop down a tree, I'd spend six sharpening my axe." -- Abraham Lincoln
For the best use of Eng-Tips, please click here -> faq731-376

 

[apowerengr]

Your model still looks funny. 20 MVA at 14% on an infinite source at 7.2kV is only 11.45kA. Four of them in parallel is only 45.8kA.

I'm guessing you don't have two fully-independent sources.

How are you getting > 100kA?

 

[mesutphen]

There is 33.334kA coming from the source transformer, 55.398kA coming across the tie, and 23.259kA from various motor/transformer contributions.

 

[racookpe1978]

What about attacking the starting load mechanically?

In addition to looking at the electrical supply part of the problem, could you "unload" the backpressure on the fan (with a damper or vent in the ID fan output so there is much less backpressure against the motor as it starts? Then, as soon as the electrical "trip point" has passed - when the fan has begun to spin up - close the damper/vent and let fan backpressure come up to normal.

 

[apowerengr]

fault currents look too high to me based on the impedances you stated.

 

[LionelHutz]

This transformer uses the 40MVA winding as the input and the 2x 20MVA windings as the outputs, one connected to each buss???

If so, your fault current seems almost 3X too high to me as well. My quick calculation says the transformer would supply 24kA from both busses combined or 12kA per buss, which compares well with apowereng' calculation of 11.5kA for one buss.

 

[racobb]

Or if the system is actually 7200/12470 the infinite source duty would only be about 6.6 kA at 20 MVA.

Alan