- Thread starter toughguy
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- Location
- Salt Lake City, Utah

Yes, the square root of three is part of the formula. I agree with Roger, is this for homework? If so, let us know what you are coming up with and we'll help you out. I think most of us (myself included) enjoy helping people, but we don't enjoy do homework for them.

Again, let us know what your up to, and we'll be happy to help

- Location
- New Jersey

Toughguy:

I don/t think you mean a150,000 kva transformer.this would be one hella-big transformer.

you must be thinking of a 150kva transformer.If so this the formula for a 3-phase transformer.

Current in @ phase =kva*1000

-------------

volts*1.732

Current=150,000

-------- =181 ampers

Hope this helps

phil c

830

- Location
- New Jersey

Toughguy:

IN my last post I don/t think my formula turned out right. IT should be

current=kva times 1000 divided by volts times the sq root of three

current=150,000

------ = 181 amp

480*1.732

- Location
- Seattle, WA

- Occupation
- Electrical Engineer

That is a valid formula, but it does not help answering this specific question. The ?power? to which your formula applies is measured in watts, and is called ?real power.? The type of power addressed in the question is measured in VA (or KVA), and is called ?apparent power.? The formula to convert KVA to amps does not contain the term ?power factor.? Phil C?s answer is correct.Originally posted by wyedelta:The correct formula is:

power= squareroot of 3*voltage*current*power factor.

- Location
- Seattle, WA

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- Electrical Engineer

In Phil C?s posting of 11:06, he divided by the square root of three. You could multiply by the inverse of the square root of three, and get the same answer. He tried to show the division in the normal manner, by having the 150,000 number shown above a line (to show division), and the 480*1.732 on the line below that. But the forum?s word processing program would not show the spacing the way he wanted. I?ll put the same information into a single line:

Current = 150,000 / (480*1.732) = 181.

toughguy:

what are you up to? phase current or line current?

Guys: if you compute the current using apparent power(kva) you also include the transformer loses.what i am trying to say when i wrote the formula was that we must compute the line current

for maximum current the transformer can deliver.

I believe every transformer were designed where the power factor is not greater than 0.8 because this is the ideal power factor in genrating plant.Of course pf varies but I believe it will not less tha 0.65 otherwise the efficiency of your generating set is questionable and generating cost of electricity will be higher.

- Location
- Seattle, WA

- Occupation
- Electrical Engineer

I may be misinterpreting your statement. But it seems to me that you are trying to associate ?transformer losses? with ?power factor,? as though the former is the cause of the later. That is not how things work. Power factor is the natural result of wires being wound in circles, and is not a function of the amount of any copper, eddy current, hysteresis, or other losses internal to the transformer.Originally posted by wyedelta:Guys: If you compute the current using apparent power (KVA) you also include the transformer loses. What I am trying to say when I wrote the formula was that we must compute the line current for maximum current the transformer can deliver.

If power is lost within a transformer (as is the case in all real components), then the voltage at its terminals will be lower, and the current delivered to the world will be lower as well. But for our purposes, we must presume that (losses or no losses), the terminal voltage will be the rated voltage, and the power delivered to the world will be the rated power.

As I said before, your formula is perfectly valid. But to use it, you must first convert the given value of KVA to an equivalent value of kW. That requires you to multiply by an assumed value of power factor (PF). Then, when you use your formula to calculate current, you must first ?solve for current.? That is, you alter the formula to get current on one side of the equation and everything else on the other. In this process, you will divide ?power? (i.e., kW, a value you computed by taking KVA times PF), by that same assumed value of PF. Your final answer will be the same 181 amps.

charlie

I fully understand what you have wrote.All iam saying is that line current is different for phase current in delta connection.and we all know that KVA is not the full power that the transformer can deliver. We were ask for the maximum current that the x'mer can deliver.when you devide kva by voltage you are simply geting the max. current in the phase and not the line current. Let me tell you that the PF does not varry because of transformer loser itseft.whatever is the pf your generating plant is delivering that's what you get it is the same with the frequency.we know that motors and x'mer are rated in KVA but it does not mean it can deliver that max. pwer because as we know that is apparent power ( in apparent power everything is included including those loses you have mention.)

as for me if iam going to compute for that given transformer i will take to consideration the pf before so that the x'mer will not be overloaded.

thanks anyway for your reply.i do appreciate it.

- Location
- Seattle, WA

- Occupation
- Electrical Engineer

Once current has made its way out of the generator, an observer looking only at the current in the wire cannot tell if the generator was a WYE or a DELTA. To that observer, it is ?line current.? I think we can agree that we are interested only in line current. Several methods are available to compute the line current:

</font>

- <font size="2" face="Verdana, Helvetica, sans-serif">* Use KVA and ignore power factor, and the result will be ?181 amps.?

* Take power factor into account, and the result will be ?181 amps.?

* If the transformer is connected WYE (line current = phase current), the result will be ?181 amps.?

* If the transformer is connected DELTA (line-to-line voltage = phase voltage, and line current > phase current), the result will be ?181 amps.?

* If the transformer is a theoretical ?ideal? (i.e., no internal losses), and is rated 150 KVA, the result will be ?181 amps.?

* If the transformer has unusually high internal losses, but is still rated 150 KVA, the result will be ?181 amps.?</font>

I applaud your desire to take any additional steps necessary to ensure the transformer is not overloaded. Bear in mind that the NEC limitations are based on a percentage of the transformer?s rated current. No matter what method you choose to calculate the rated current, the result should be the same.

charlie b;

Ya you are correct we are only up to line current and not the power. I am sorry, i am pushing about the max. power, I didn't realised that TOUGHGUY were after line current only.

Thanks anyway, I was refresh again about the theory. You are right as long as the magic SQ. RT of 3 is there we will come out with the same current event we are computing the KVAR, KVA, KW.

I apologized again for pushing for the power wherein our subject matter is line current.

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