Pole mounted utility transformer

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sparks1

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Example: A 25KVA transformer protected by a primary 15T fuse supplys 100 Amps to an over head service for a home. The secondary is 240 Volts.

What is the interrupt rating of the primary fuse?
Is it 22,000 Amps or Greater?

Can you explain how 22,000 Amps RMS SYM interrupt rating developes in about half a cycle to open the fuse?

The Primary fuse will carry up to 150% Or 156.25 Amps. Is this the available fault current?

When a short circuit occurs in the transformer secondary.The secondary winding can carry 25KVA/240 volts or 104.166 Amps. The secondary short circuit current will pass through the transformer. But, the secondary to primary current ratio will not be high enough to open the fuse. Is this correct?
When the secondary short circuit current exceeds 104.166Amp it will blow open the winding. Is this Correct?

Is 104.166 secondary Amps my available Short circuit current can you explain?


Charlie,
Thanks for you input!
Philip
 
Re: Pole mounted utility transformer

Philip, you need to sit down with an electric utility engineer, these answers would come easier while having a beer and dinner. :D

Example: A 25KVA transformer protected by a primary 15T fuse supplies 100 Amps to an over head service for a home. The secondary is 240 Volts.

What is the interrupt rating of the primary fuse?

The interrupting rating of the fuses we use is 10 kA. If the available fault current is greater than that, we have to do something different. Closer to the substation, we will use a 200 ampere frame or a sectionalizing device that will open the load after the substation relay drops it the 2nd time.

Is it 22,000 Amps or Greater?

No, see my answer above.

Can you explain how 22,000 Amps RMS SYM interrupt rating develops in about half a cycle to open the fuse?

The amount of fault current available is dependant upon the size of the transformer that is feeding the fault, the impedance of that transformer, and the impedance of the lines feeding the transformer. When we give the amount of fault current available at our transformers, we make the assumption of an infinite bus (that assumption is not true), and a bolted fault (almost every fault has dynamic impedance introduced into the circuit). The quick and dirty way of calculating fault current is to divide the full load ampere rating of a transformer by its impedance.

The Primary fuse will carry up to 150% Or 156.25 Amps. Is this the available fault current?

No, see the above discussion.

When a short circuit occurs in the transformer secondary.The secondary winding can carry 25KVA/240 volts or 104.166 Amps. The secondary short circuit current will pass through the transformer. But, the secondary to primary current ratio will not be high enough to open the fuse. Is this correct?

No. If the fault is solid enough, enough fault current is available, and close enough to the transformer, the primary fuse will open.

When the secondary short circuit current exceeds 104.166Amp it will blow open the winding. Is this Correct?

No. Transformers will stand up to a good sized fault for a short period of time. If enough current develops to take out the fuse fast enough, the transformer will not be damaged.

Transformers will carry up to 300% for a short period of time before they are damaged. It is common for us to overload transformers as much as 200% for periods of one hour or less.

Is 104.166 secondary Amps my available Short circuit current, can you explain?

No. See my answers above. :D
 
Re: Pole mounted utility transformer

You guys seem like you have the whole envelope under control. I enjoyed reading this Charlie. :)

When a short circuit occurs in the transformer secondary.The secondary winding can carry 25KVA/240 volts or 104.166 Amps. The secondary short circuit current will pass through the transformer. But, the secondary to primary current ratio will not be high enough to open the fuse.
Click to expand...
I'm not qualified to correct you but I have a question and a comment.

When you said:

short circuit occurs in the transformer secondary
Click to expand...
Did you mean "on" the secondary?

And I think you can expect a lot more than 300% for a short time.

Edit: That didn't sound right. What I meant was, I wonder how high the current can go for a short time.

[ January 25, 2005, 01:54 PM: Message edited by: physis ]
 
Re: Pole mounted utility transformer

Sam, the 300% is load current, not fault current. We use the same terminology by saying a fault occurs in the secondary or primary. I think of "in the secondary" as being in the coils of the transformer or in the secondary conductors. :D
 
Re: Pole mounted utility transformer

You're saying you can actually run at 300% for a while. I'm ok there.

And for a short "in" the secondary, the ratio changes. New transformer. Different figures? No?

:)
 
Re: Pole mounted utility transformer

I guess that's why Charlie is on a code making panel. He knows his stuff
 
Re: Pole mounted utility transformer

You are absolutely right Sam. I get up at 5:15 and I'll give you a call.
cheeky-smiley-013.gif
 
Re: Pole mounted utility transformer

Originally posted by physis:
I'm not qualified to correct you but I have a question and a comment.
Click to expand...
I think that you could make a correction.
But, your correction, might need to be corrected (just not by me). :)
 
Re: Pole mounted utility transformer

The trouble is, Charlie knows guys at PG&E. And those people send me bills for electricity. I don't want to look like this
speechless-smiley-040.gif
at the end of the month
 
Re: Pole mounted utility transformer

fault current calculations were a pain in the *** my second year in school...
fortunately that instructor was the only one who thinks they are very important..
 
Re: Pole mounted utility transformer

The real problem is that they are real important.

Suppose the serving electric utility tells you that the available fault current is 55 kA. The available service equipment has a bracing of 50 kA and 100 kA. The available circuit breakers are 50 kA and 100 kA. Without doing the calculations, you must use the 100 kA equipment.

After setting up your feeders, the same thing applies. Without doing the calculations, you must use the 100 kA equipment. Either you have to pay an engineer to do the calculations, lose the bid to another company who knows how to do the calculations, learn to do them yourself, or just guess and open yourself up to huge liability and become a danger to the public. :D
 
Re: Pole mounted utility transformer

They are also used extensively in the arc flash/blast calculations used for electrical safe work practices. In some ways these calculations must be more accurate then those used to select the correct equipment. With equipment, if the calculation is conservative, that is the calc shows a higher fault current then really exists, all that happens is there is some extra costs for the higher rated equipment. When selecting PPE for "on or near" energized work, the higher calculated fault current may result in the selection of a level of PPE that is too low to protect you. That is because the PPE is selected on the "incident energy" level and with a higher calculated fault current, the incident energy may actually be lower because the?higher current causes the OCPD to open quicker and quicker opening can mean less incident energy.
Don
 
Re: Pole mounted utility transformer

Re: The secondary side of the transformer. The general understanding is, "in' or "on" refer to the samething.

The primary side of the tranformer is connected to two phases. Each phase is 120 degrees apart. Each of the two secondary legs are also 120 degrees out of phase. Phase A and phase B going to your panel box with a center tapped neutral to give you 120 volts A-N or B-N or A-B or 240 volts to serve a balanced load.
This must be two phase with a neutral ???
 
Re: Pole mounted utility transformer

If you only have two line conductors (what you are calling phases) there can be no phase difference between them. It is the fact that you have taken only 1 pair of line conductors to your transformer primary that has created a single phase system, regardless the number of conductors on the secondary.

When we are discussing the number of phases in an electrical system we are always talking about the number of phase-phase voltages never ever the phase to neutral voltage. The result is: a single phase system has 2 hot/phase/line conductors and a single line-line voltage possibility (i.e L1-L2), a two phase system has 4 hot conductors and two line-line voltage possibilities (i.e. L1-L2 and L3-L4) and finally a three phase system has 3 hot/phase/line conductors and three line-line voltage possibilities (i.e. L1-L2, L2-L3, and L3-L1).
 
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