Fault Current

Status
Not open for further replies.
jim dungar said:
This is not what protective devices do.

The Amps Interrupting Current or Rating (AIC or AIR) is the amount of fault current the protective device can interrupt/clear without damaging itself. The AIC rating has to be greater than the amount of fault current available on its load side terminal. So the selection of a transformer primary side protective device involves understanding the system parameters and pretty much has nothing to do with the secondary side at all.
Click to expand...

I think you're misreading his question, but that's probably my fault. His question only involved the 480V circuit breaker, but I asked about the secondary because a 200A c/b for a 225kVA transformer seemed odd.
 
In general the OCPD protecting the transformer and the primary conductors does not protect the secondary conductors, however with a delta-delta transformer the current on the secondary side is proportional to the current on the primary side (even in unbalanced conditions) so your primary OCPD can protect your secondary conductors. If the secondary conductor would be properly protected by a 400A OCPD (at 240V) in this case they would be considered properly protected by a 200A OCPD (at 480V on the primary side).

There is the open question of the 200A OCPD being large enough to handle the 'inrush' current of the transformer. It might be too small and subject to excessive tripping on startup.

But back to the original question; I do not believe you can come up with a good answer without more information. You know you need a device with a 200A trip rating, and you are trying to figure out what interruption rating it needs.

You could be really conservative, and assume infinite current available at the start of the 140 foot long feeder, and calculate what the available current is...but I get about 160kA. Since the feeder alone is not sufficient to reduce the available fault current acceptably, you need to know the real available fault current upstream.

You could _assume_ that the existing installation was properly engineered, and that the available fault current at the 800A breaker is less than its 30kA rating. In which case a 200A breaker with a 30kA rating or one that is series rated with the main to get a 30kA rating is sufficient. The OP has expressed doubt about this.

So you are left with getting the one line or getting the transformer impedance information.

-Jon
 
buffalonymann said:
We are a plastics extrusion company; extruders have long barrels with resistance heaters attached. We use 480/240 delta-delta XFMR to supply single phase heaters. They installed parallel secondary conductors to distribution blocks to power the 3 and 4 kw heaters. They installed a 300 amp breaker on the primary. I am eliminating one of the secondary sets of conductors and so I need to limit the primary current to limit the current on the now single secondary conductors.
Click to expand...

200 Amps is still too high, and here is why:

The circuit breaker protects the transformer and must be sized per 450.3(B), maximum 125% of rated transformer current for primary protection only. Note it says maximum, so while the OCPD might seem small, as others have pointed out, there is nothing to my knowledge that says this isn't allowed. The maximum OCPD size is 338.7 Amps (225,000/(480*1.73)*125%). However, we cannot round up to the next higher rating in this case due to what is being done on the secondary. this provides a 300 Amp, primary side, maximum OCPD.

The problem I see here is the secondary conductors must be protected according to 240.21(C). The primary protection extends to the secondary provided 240.21(C)(1) is satisfied: the rating of the OCPD does not exceed the secondary conductor ampacity multiplied by the secondary to primary transformer voltage ratio (240:480 or 1:2). your transformer must be single voltage, 3-wire secondary for this to be allowed. This means the secondary conductors must have an ampacity of at least 400 Amps (200 * 480/240) if you intend protect the secondary conductors with the primary OCPD. I suspect you have parallel sets of 350's currently installed, which would still be compliant with 240.21.

Unless someone can make the argument that protection of the secondary conductors is not required, removing one of these sets and replacing the 300 with a 200 Amp breaker would create a violation. Assuming there are parallel 350's, each good for 310 Amps, the primary side OCPD must be maximum of 150 Amps to ensure the single set of remaining 350's are protected.

I would also encourage OP to consider table 450.3(B), Note 2 with respect to the heaters. That note might help you prove that secondary protection is provided.
 
drktmplr12 said:
200 Amps is still too high, and here is why:

The circuit breaker protects the transformer and must be sized per 450.3(B), maximum 125% of rated transformer current for primary protection only. Note it says maximum, so while the OCPD might seem small, as others have pointed out, there is nothing to my knowledge that says this isn't allowed. The maximum OCPD size is 338.7 Amps (225,000/(480*1.73)*125%). However, we cannot round up to the next higher rating in this case due to what is being done on the secondary. this provides a 300 Amp, primary side, maximum OCPD.

The problem I see here is the secondary conductors must be protected according to 240.21(C). The primary protection extends to the secondary provided 240.21(C)(1) is satisfied: the rating of the OCPD does not exceed the secondary conductor ampacity multiplied by the secondary to primary transformer voltage ratio (240:480 or 1:2). your transformer must be single voltage, 3-wire secondary for this to be allowed. This means the secondary conductors must have an ampacity of at least 400 Amps (200 * 480/240) if you intend protect the secondary conductors with the primary OCPD. I suspect you have parallel sets of 350's currently installed, which would still be compliant with 240.21.

Unless someone can make the argument that protection of the secondary conductors is not required, removing one of these sets and replacing the 300 with a 200 Amp breaker would create a violation. Assuming there are parallel 350's, each good for 310 Amps, the primary side OCPD must be maximum of 150 Amps to ensure the single set of remaining 350's are protected.

I would also encourage OP to consider table 450.3(B), Note 2 with respect to the heaters. That note might help you prove that secondary protection is provided.
Click to expand...

What is the violation that you believe exists?
 
winnie said:
In general the OCPD protecting the transformer and the primary conductors does not protect the secondary conductors, however with a delta-delta transformer the current on the secondary side is proportional to the current on the primary side (even in unbalanced conditions) so your primary OCPD can protect your secondary conductors. If the secondary conductor would be properly protected by a 400A OCPD (at 240V) in this case they would be considered properly protected by a 200A OCPD (at 480V on the primary side).

There is the open question of the 200A OCPD being large enough to handle the 'inrush' current of the transformer. It might be too small and subject to excessive tripping on startup.

But back to the original question; I do not believe you can come up with a good answer without more information. You know you need a device with a 200A trip rating, and you are trying to figure out what interruption rating it needs.

You could be really conservative, and assume infinite current available at the start of the 140 foot long feeder, and calculate what the available current is...but I get about 160kA. Since the feeder alone is not sufficient to reduce the available fault current acceptably, you need to know the real available fault current upstream.

You could _assume_ that the existing installation was properly engineered, and that the available fault current at the 800A breaker is less than its 30kA rating. In which case a 200A breaker with a 30kA rating or one that is series rated with the main to get a 30kA rating is sufficient. The OP has expressed doubt about this.

So you are left with getting the one line or getting the transformer impedance information.

-Jon
Click to expand...

What do you believe the inrush current could be? Why would it exceed the connected resistive load?
 
buffalonymann said:
What is the violation that you believe exists?
Click to expand...

Unless someone can make the argument that protection of the secondary conductors is not required, removing one of these sets and replacing the 300 with a 200 Amp breaker would create a violation. Assuming there are parallel 350's, each good for 310 Amps, the primary side OCPD must be maximum of 150 Amps to ensure the single set of remaining 350's are protected.
Click to expand...

240.21(C)(1) would be violated.. the secondary conductors would not be protected.

edit:

buffalonymann said:
What do you believe the inrush current could be? Why would it exceed the connected resistive load?
Click to expand...

Transformer inrush is a function dominated by the magnetizing current necessary to establish an electromagnetic field. Generally speaking, it can be up to 15 times the transformer rated current for several cycles, or higher for special use transformers. It also depends at what point in the sine wave cycle you turn on the transformer. The worst case is when you switch on the transformer when the voltage crosses zero, with some exceptions. Worst case must be assumed since there is no control over what point in the cycle the transformer is switched on. What happens to that 200 amp inverse time breaker at 4,000 amps? How about the 150?

 
Last edited:
buffalonymann said:
What do you believe the inrush current could be? Why would it exceed the connected resistive load?
Click to expand...

Inrush current might be 5-25x the normal full load current in worst case situations. See http://www.eaton.com/ecm/idcplg?IdcService=GET_FILE&dID=3558670 for examples and calculations.

You want an OCPD that is simultaneously small enough to protect the secondary conductors and large enough to not trip on this momentary (only a few cycles duration) inrush current.

-Jon
 
buffalonymann said:
What is the violation that you believe exists?
Click to expand...

This transformer divides the voltage by 2 and multiplies the current by 2. Thus the OCPD on the _primary_ side of the transformer must be selected on the basis of 1/2 the ampacity of the conductors on the secondary side, if you want the OCPD to protect those conductors.

-Jon
 
winnie said:
In general the OCPD protecting the transformer and the primary conductors does not protect the secondary conductors, however with a delta-delta transformer the current on the secondary side is proportional to the current on the primary side (even in unbalanced conditions) so your primary OCPD can protect your secondary conductors. If the secondary conductor would be properly protected by a 400A OCPD (at 240V) in this case they would be considered properly protected by a 200A OCPD (at 480V on the primary side).

There is the open question of the 200A OCPD being large enough to handle the 'inrush' current of the transformer. It might be too small and subject to excessive tripping on startup.

But back to the original question; I do not believe you can come up with a good answer without more information. You know you need a device with a 200A trip rating, and you are trying to figure out what interruption rating it needs.

You could be really conservative, and assume infinite current available at the start of the 140 foot long feeder, and calculate what the available current is...but I get about 160kA. Since the feeder alone is not sufficient to reduce the available fault current acceptably, you need to know the real available fault current upstream.

You could _assume_ that the existing installation was properly engineered, and that the available fault current at the 800A breaker is less than its 30kA rating. In which case a 200A breaker with a 30kA rating or one that is series rated with the main to get a 30kA rating is sufficient. The OP has expressed doubt about this.

So you are left with getting the one line or getting the transformer impedance information.

-Jon
Click to expand...

drktmplr12 said:
240.21(C)(1) would be violated.. the secondary conductors would not be protected.

edit:



Transformer inrush is a function dominated by the magnetizing current necessary to establish an electromagnetic field. Generally speaking, it can be up to 15 times the transformer rated current for several cycles, or higher for special use transformers. It also depends at what point in the sine wave cycle you turn on the transformer. The worst case is when you switch on the transformer when the voltage crosses zero, with some exceptions. Worst case must be assumed since there is no control over what point in the cycle the transformer is switched on. What happens to that 200 amp inverse time breaker at 4,000 amps? How about the 150?
Click to expand...

240.21(C)(1) provides for not OCPD on secondary
 
winnie said:
This transformer divides the voltage by 2 and multiplies the current by 2. Thus the OCPD on the _primary_ side of the transformer must be selected on the basis of 1/2 the ampacity of the conductors on the secondary side, if you want the OCPD to protect those conductors.

-Jon
Click to expand...

My secondary conductor is rated at 421 Amps, so I should be good with 200A on the primary
 
drktmplr12 said:
240.21(C)(1) would be violated.. the secondary conductors would not be protected.

edit:



Transformer inrush is a function dominated by the magnetizing current necessary to establish an electromagnetic field. Generally speaking, it can be up to 15 times the transformer rated current for several cycles, or higher for special use transformers. It also depends at what point in the sine wave cycle you turn on the transformer. The worst case is when you switch on the transformer when the voltage crosses zero, with some exceptions. Worst case must be assumed since there is no control over what point in the cycle the transformer is switched on. What happens to that 200 amp inverse time breaker at 4,000 amps? How about the 150?
Click to expand...

I don't think we'll see any issues with a few cycles of high current
 
kingpb said:
Why not post a one-line; a simple sketch would be worth at least two pages of posts going back and forth.
Click to expand...

The original issue is I didn't have the primary xfmr to calculate fault current, I have one coming so I'll post the info when I get it. Just want to be sure I'm using the right calculation
 
Status
Not open for further replies.
Top