Transformer Ratings, i.e. AA/AA/FA and Fault Current

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NS_EE06

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Location
Pennsylvania
What are the impacts of specifying multiple temp. rated transformers with respect to impedance and fault current? Here's an example:

80deg/150deg
AA/AA/FA
3000/3990/5000kVA
%Z = 5.75

What are the effects on fault current when operating above the 3000kVA base rating continuously? Will the impedance change greatly or only slightly?

Detailed calculations would be done later in design, but I would like to use approximation methods at the beginning of schematic design to anticipate equipment ratings.

Any thoughts or white papers that address this issue?

Thanks for the help.
 

richxtlc

Senior Member
Location
Tampa Florida
What are the impacts of specifying multiple temp. rated transformers with respect to impedance and fault current? Here's an example:

80deg/150deg
AA/AA/FA
3000/3990/5000kVA
%Z = 5.75

What are the effects on fault current when operating above the 3000kVA base rating continuously? Will the impedance change greatly or only slightly?
The ratings AA/AA/FA have to do with the cooling of the transformer and nothing to do with the impedance value. The more you can cool a transformer the greater the KVA loading it can have without overheating the unit. The AA is natural circulation and the FA is forced air or fans.
 
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charlie b

Moderator
Staff member
Location
Seattle, WA
Occupation
Electrical Engineer
I agree with Rich's comment, and will add that the point (i.e., load value) at which a transformer is operating, at the moment a fault takes place, will not have an effect on the fault current. The transformer is capable of putting out 5000 kVA, and that is the value to use in calculating a fault current.

Welcome to the forum.
 

jim dungar

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Staff member
Location
Wisconsin
Occupation
Retired Electrical Engineer - Power Systems
%Z is more properly called %IZ. This value is the % of rated voltage required to supply 100% of rated current through a short circuit.

The nameplate %Z is only valid at the nominal VA, unless noted otherwise.
 

richxtlc

Senior Member
Location
Tampa Florida
If you look at the transformer nameplate it should say %IZ at base value. As far as the Short Circuit Study goes you would use the %IZ and FLA value on the nameplate of the transformer.
 

NS_EE06

Member
Location
Pennsylvania
%Z is more properly called %IZ. This value is the % of rated voltage required to supply 100% of rated current through a short circuit.

The nameplate %Z is only valid at the nominal VA, unless noted otherwise.
Jim,

Using %IZ as the % of "rated" voltage to supply 100% of "rated" current through the short on the secondary side...here's my next question. If the transformer is rated at 3000kVA (100%) is the subsequent higher kVA values considered overload (< 3hrs in duration)? Most info I come across says "no".

Or is it considered to be e.g. 133% or 166% of rated value that can be used for continuous current? Wouldn't this require the nameplate to list all available kVA capacities? If so, all three kVA's would be considered base quantities depending on operating conditions.

I'm sure I'll find the solution to my questions through research, but I wanted to thank all of you for the help so far.
 

rbalex

Moderator
Staff member
Location
Mission Viejo, CA
Occupation
Professional Electrical Engineer
Yes. That was my point, really.
The 5.75% won't be applicaple to all three rating stated in the OP.
And it doesn't apply to any tap settings other than nominal voltage either. According to NEMA TR-1 the impedance (among other things) is "...based on the normal rating unless otherwise specified."
 

jim dungar

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Staff member
Location
Wisconsin
Occupation
Retired Electrical Engineer - Power Systems
Jim,

Using %IZ as the % of "rated" voltage to supply 100% of "rated" current through the short on the secondary side...here's my next question. If the transformer is rated at 3000kVA (100%) is the subsequent higher kVA values considered overload (< 3hrs in duration)? Most info I come across says "no".

Or is it considered to be e.g. 133% or 166% of rated value that can be used for continuous current? Wouldn't this require the nameplate to list all available kVA capacities? If so, all three kVA's would be considered base quantities depending on operating conditions.

I'm sure I'll find the solution to my questions through research, but I wanted to thank all of you for the help so far.
If the transformer nameplate lists multiple KVA's then it has multiple continuous current ratings. There would be a different %Z for each rating. If the nameplate only lists a single %Z it is, per NEMA standard, for the nominal (which is usually the lowest) KVA rating.
 

steve66

Senior Member
Location
Illinois
Occupation
Engineer
Jim,

Using %IZ as the % of "rated" voltage to supply 100% of "rated" current through the short on the secondary side...here's my next question. If the transformer is rated at 3000kVA (100%) is the subsequent higher kVA values considered overload (< 3hrs in duration)? Most info I come across says "no".

Or is it considered to be e.g. 133% or 166% of rated value that can be used for continuous current? Wouldn't this require the nameplate to list all available kVA capacities? If so, all three kVA's would be considered base quantities depending on operating conditions.

I'm sure I'll find the solution to my questions through research, but I wanted to thank all of you for the help so far.
Sorry for asking, but from your last post, it's not really clear if you understand that the 3 different KVA ratings correspond to 3 different types of cooling. If that's the case, just let us know, & I'm sure someone will be glad to explain them.
 

NS_EE06

Member
Location
Pennsylvania
Sorry for asking, but from your last post, it's not really clear if you understand that the 3 different KVA ratings correspond to 3 different types of cooling. If that's the case, just let us know, & I'm sure someone will be glad to explain them.
I do understand, for the most part. What I'm having trouble with is the fact that these different temp/cooling ratings allow for more kVA to be supplied from the transformer. Then, with additional kVA, why doesn't fault current increase? From what I found for quick estimating purposes, I(sc) = I(fla)*(1/%Z). The more I dig into this topic, the more diverse of answers I find.

I don't want to settle for "it won't change the available fault current" when a transformer has multiple ratings as it relates to fault current. It's obvious that I have to dig further, and I will. This forum is a starting point.

And thanks for offering the assistance.
 

winnie

Senior Member
Location
Springfield, MA, USA
Occupation
Electric motor research
The impedance of a transformer depends only on the physics of the transformer, essentially the internal resistance and reactance of the device. The values of internal resistance and reactance will change very slightly at different temperatures, but remain essentially constant.

The different KVA ratings do not change the internal resistance or reactance of the transformer. The different KVA ratings will not change the amount of current available should you short the terminals of the transformer.

Because transformer impedance _ratings_ are given in terms of the nominal current rating of the transformer, the 'impedance number' will be different for each 'KVA number'; as Jim mentions above if a single impedance number is given, that applies to one of the KVA numbers.

-Jon
 

Besoeker

Senior Member
Location
UK
I do understand, for the most part. What I'm having trouble with is the fact that these different temp/cooling ratings allow for more kVA to be supplied from the transformer. Then, with additional kVA, why doesn't fault current increase? From what I found for quick estimating purposes, I(sc) = I(fla)*(1/%Z).
It's because the physical values in ohms of the R and the X that make up the Z don't change much.
 
I do understand, for the most part. What I'm having trouble with is the fact that these different temp/cooling ratings allow for more kVA to be supplied from the transformer. Then, with additional kVA, why doesn't fault current increase? From what I found for quick estimating purposes, I(sc) = I(fla)*(1/%Z). The more I dig into this topic, the more diverse of answers I find.

I don't want to settle for "it won't change the available fault current" when a transformer has multiple ratings as it relates to fault current. It's obvious that I have to dig further, and I will. This forum is a starting point.

And thanks for offering the assistance.
OK, let me try it.

The kVA 'increases' because the windings are allowed to operate at a higher current rating due tot he fact that either:
? allowed to operate at a higher temperature as a temporary measure
? allowed to operate at an even higher current rating because there is more heat removed.
The transformer does not 'supply' fault current to a significant degree although the latent magnetic flux does offer a slight contribution. The short circuit available on the secondary side of the transformer has to do with the short circuit available from the system that comprises:
? Generated or utility contribution
? Local contribution on the primary side, such as motors, etc.
The transformers?' characteristic of impedance and resistance will determine how MUCH of the available sources will be LET THROUGH as short circuit contributors. Since the resistance and impedance will not change significantly - as pointed out by others - due to the transformer is being operated at a higher load, the short circuit let-through will not change in a meaningful way.

Additional source of short circuit on the secondary side of the transformer would come from the generating sources - if any - and motor contributors on that side. Obviously they contribution level will remain the same. (In real terms the secondary contribution will be at the highest level since the transformer now operates at it's maximum load, but it should be taken into consideration at the calculations as the short circuit contributors are not demand based.)
 

steve66

Senior Member
Location
Illinois
Occupation
Engineer
I do understand, for the most part. What I'm having trouble with is the fact that these different temp/cooling ratings allow for more kVA to be supplied from the transformer. Then, with additional kVA, why doesn't fault current increase? From what I found for quick estimating purposes, I(sc) = I(fla)*(1/%Z). The more I dig into this topic, the more diverse of answers I find.

I don't want to settle for "it won't change the available fault current" when a transformer has multiple ratings as it relates to fault current. It's obvious that I have to dig further, and I will. This forum is a starting point.

And thanks for offering the assistance.
Use the %Z given along with the nominal KVA for the short circuit calcs. This will tell you how much current the transformer can provide under short circuit conditions.

Don't get too hung up on the KVA rating. The KVA rating is just a number not to exceed to keep the transfomer from overheating. Under short circuit conditions, the KVA number goes out the window.
 
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