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 Post subject: Calculating clearance times on upstream protection?
PostPosted: Fri Dec 31, 2010 9:19 am 
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Posts: 1
Hello,

I am new to arc flash studies and so any help you can give me would be appreciated.

I have a query regarding the clearance times for upstream protection devices.

I understand that to calculate clearance times you must first calculate the bolted short circuit and then use the IEEE1584 spreadsheet to determine the arcing current.

I have noticed from the spreadsheet that the arcing current in LV systems is much more reduced compared to the arcing current for HV systems.

The system I am investigating has an 11kV and 400V switchboard with a transformer providing the voltage reduction. The 400V incomer does not have any electrical protection or fuses. The 400V board relies on the upstream protection relay on the 11kV side of the transformer.

For a 3ph fault on the 400V switchboard, the bolted short circuit values are: 11kV=1.57kA, 400V=40kA.

However, when I put the above current values into the IEEE1584 spreadsheet the 11kV current is only slightly reduced while the 400V current sees a heavy reduction

My worry is that the impedance of the arc at 400V will result in a much lower current value than 1.57kA flowing at 11kV through the protective device and hence a longer clearance time.

Therefore, I think I have two options:

1) Use the 1.57kA to work out the relay clearance time at 11kV.

2) Put the 40kA value into the IEEE1584 spreadsheet and get the heavily reduced current at 400V. Then use the ratio of the transformer to work back up and find the corresponding 11kV current to calculate the relay clearing time.


I hope that this is clear - any advice or opinions would be welcome.

Regards,

ukengin


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PostPosted: Fri Dec 31, 2010 11:30 am 
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Location: Scottsdale, Arizona
ukengin wrote:
Hello,

I am new to arc flash studies and so any help you can give me would be appreciated.

I have a query regarding the clearance times for upstream protection devices.

I understand that to calculate clearance times you must first calculate the bolted short circuit and then use the IEEE1584 spreadsheet to determine the arcing current.

I have noticed from the spreadsheet that the arcing current in LV systems is much more reduced compared to the arcing current for HV systems.

The system I am investigating has an 11kV and 400V switchboard with a transformer providing the voltage reduction. The 400V incomer does not have any electrical protection or fuses. The 400V board relies on the upstream protection relay on the 11kV side of the transformer.

For a 3ph fault on the 400V switchboard, the bolted short circuit values are: 11kV=1.57kA, 400V=40kA.

However, when I put the above current values into the IEEE1584 spreadsheet the 11kV current is only slightly reduced while the 400V current sees a heavy reduction

My worry is that the impedance of the arc at 400V will result in a much lower current value than 1.57kA flowing at 11kV through the protective device and hence a longer clearance time.

Therefore, I think I have two options:

1) Use the 1.57kA to work out the relay clearance time at 11kV.

2) Put the 40kA value into the IEEE1584 spreadsheet and get the heavily reduced current at 400V. Then use the ratio of the transformer to work back up and find the corresponding 11kV current to calculate the relay clearing time.


I hope that this is clear - any advice or opinions would be welcome.

Regards,

ukengin


You are perfectly clear.

Method 2 is correct. There is a greater reduction from bolted to arcing short circuit current both at lower voltages and also at higher short circuit currents at lower voltages. It has more to do with the magnitude of the arc impedance relative to the equivalent impedance of the bolted fault condition.

i.e. higher bolted current has a lower equivalent impedance so the arcing impedance is more dominant and provides a greater reduction.

For lower bolted currents, the equivalent impedance is greater and the additional arc impedance has less of an effect - less of a drop.

Calculating the arcing current on the secondary (400 V) side and then reflecting it through the turns ratio to the primary side is the correct method.

Let me know if I can be of any further help and best wishes!

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Jim Phillips, P.E.
Brainfiller.com


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