transformer size and derating wire

[jimmy_mcquade]

Hi everyone,

I am studying the industrial side of wiring and i have a disagreement with
a class mate.

part a.
(3) 100 kva single phase xfmrs are wired with the output in the "Y" configuration with the output voltage being 208 volts. What is the rated output current? Using the 2008 ugly's book, i see that a 100 kva single phase transformer is rated at 480.8 amps, but, when they are wired as a 3 phase system, you must divide by the sqrt(3) getting 277.6 amps. My classmate says that the current is the same since they are single phase transformers.
Which is correct?

I am stuck on this one.
part b.
Assuming that the loads are not balanced, what is the ampacity of the wire in the following drawing. You have (2) rigid conduits, each having (4) 350 MCM wires. Part of the wire is above ground, but mostly ungerground. The summer temperature is 104 degrees f. The rating of the terminal lugs to the connecting panel board are 75 degrees c.

From this is get that there are (2) 350 MCM wires per phase. Looking up the ampacity in the table i get 310 amps per conductor or 620 amps total. I then multiply that number by the temperature factor of 0.88 because the terminal lugs are 75 degrees C to get 545.6 amps. I then multiply this number by 0.8 because of the 4 wires in one conduit to get 436.48 amps for both conductors or 218.24 amps per cable.

Looking at Table 310.15(B)(2)(a) it says that the multiplyer is adjusted for ambient temperature. Did I take a double correction?

thanks in advance,
jimmy

 

[charlie b]

For part (a), neither of you is correct. And try not to rely on Ugly?s alone. You need to be able to figure it out yourself, and you should use the book as a backup.

If all you had was one single-phase 100 KVA transformer with a 208 volt secondary, then the secondary current would be found by dividing 100,000 by 208. That is the basis for Ugly?s answer of 480 amps. But you don?t have one such transformer, you have three of them. That gives you a total of 300 KVA. Also, if you wire three of these guys in a WYE configuration, the line-to-line voltage would be 208 times the square root of three, or about 360 volts. That is not a common voltage, but this is an academic exercise, so we will go with that.

To find the rated current, divide 300,000 VA by 360 volts, and then divide again by the square root of three. That gives you a rated current of 481 amps. So your friend had the correct answer. But his method was all wrong, so I would not give him a passing grade on this problem.

 

[jimmy_mcquade]

Charlie B,

Thanks,

But what about part b?

Jimmy

 

[charlie b]

For part (b), let me begin by saying that you don?t use a temperature correction factor based on the rating of the terminal lugs. You only correct for temperature if the ambient temperature (i.e., the dirt or air that surrounds the conduits) is above 30C.

This particular problem is badly worded. The person who wrote it is trying to direct the student toward failure, by providing misleading clues, by creating a situation that can be interpreted several different ways, and by leaving out important information. Here is what I mean:
? For starters, the ampacity of a wire has nothing to do with whether the loads are balanced. So that is a misleading clue.
? Also, saying that part of the run is above ground is leading the reader to conclude that you need to apply the ?summer temperature? as if it were the ?ambient temperature.? That is open to interpretation.
? Next I would ask what is meant by ?part of the wire is above ground?? Does that mean that part of the wire is outside the conduit, running in free air? Also, how much is ?part of the wire?? Is 99% of the wire above ground, or is 99% of the wire underground, or is it somewhere in the middle of these two values?
? Also, you didn?t give us a wire type. So we don?t know whether its insulation system is rated for 60C, for 75C, or for 90C. That would make a major difference in the starting point, when we do any required derating.
? Finally, the fact that there are four wires in a conduit does not imply that all four are current-carrying conductors. Therefore, we don?t know whether to apply the adjustment factors of table 310.15(B)(2)(a).

Sorry, but I can?t come up with an answer to this problem, without a great deal more information being supplied by the person who wrote it.

 

[wasasparky]

To find the rated current, divide 300,000 VA by 360 volts, and then divide again by the square root of three.

I would challenge the wording of the question:

(3) 100 kva single phase xfmrs are wired with the output in the "Y" configuration with the output voltage being 208 volts.

This tells me it's a 300kVA transformer with 208/120 secondary. (833A)

 

[jimmy_mcquade]

Sorry,

the wire is thhn.
the drawing shows that the wire comes from the transformers (about 6 ft)
into the rigid conduit which has about 4 ft above the ground. it then is burried in the ground.

the loads are (a) 215 amps, (b) 165 amps, (c) 195 amps and 42 amps if
i don't make a typo or took bad notes. the 3 phases are unbalanced and the neutral does have current.

Question for you, How do you determine the temperature rating of wire?

In regards to the amp rating of the wire, Wouldn't 90 degree rated wire be
rated for the 75 degree rating if the terminal lugs were rated for 75 degrees.
for example, 90 degree 350 mcm thhn wire is rated at 350 amps, but because the terminals are only rated for 75 degrees, wouldn't you rate the cable at 310 amps which is the 75 degree rating?

thanks again,
jimmy

 

[charlie b]

Question for you, How do you determine the temperature rating of wire?

The easiest way is to look at the conductor types at the top of table 310.16. For example, you will see type TW in the 60C column, so you infer that that wire has a temperature rating of 60C. Similarly, you can see that type RHW has a rating of 75C, and THHN has a rating of 90C. The letter ?H? in the conductor name adds 15 degrees to the insulation?s rating. For example, type ?THW? (with only one ?H?) has a rating of 75C, and type ?THHW? (with a second ?H?) has a rating 15 degrees higher, or 90C.

for example, 90 degree 350 mcm thhn wire is rated at 350 amps, but because the terminals are only rated for 75 degrees, wouldn't you rate the cable at 310 amps which is the 75 degree rating?

You are correct. However, you can use the value of 350 amps as the starting point for derating. For example, if you have to derate for having 4 current-carrying conductors in a conduit, then you can take the 350 times the 80%, instead of taking 310 times 80%. Then, since the result of 350x0.8 is 280, and since that is lower than the tabulated value of 310, you assign the ampacity a value of 280. As a different example, suppose that you had to derate the same conductors for an ambient temperature of 32C, but not for having more than 3 CCCs in the conduit. You take the value of 350 (ampacity at 90C), multiply that by 0.91 (the derating factor for 32C ambient and 90C wire), and get a result of 336 amps. That is higher than the 75C value of 310. So you have to assign an ampacity of 310.

 

[charlie b]

This tells me it's a 300kVA transformer with 208/120 secondary. (833A)

It is a good bet that that was the intent of the person who wrote the question. As you said, the wording was not at all clear.

 

[jimmy_mcquade]

Charlie B,

in regards to the last part of my question in part b.

Lets say that the conductors are in a room that's 104 degrees f in the summer
and not vented.

Wouldn't i derate the cables for temperature compensation?
and then wouldn't i have to derate the conductors again because of 4 conductors in a single conduit?

regards,
jimmy

 

[Cold Fusion]

...it's a 300kVA transformer with 208/120 secondary. (833A)

Slow typer, already discussed

cf

 

[charlie b]

Let?s say that the conductors are in a room that's 104 degrees f in the summer and not vented. Wouldn't I derate the cables for temperature compensation?

I think that the person who wrote this question was trying to get you to use the exception under 310.15(A)(2). Part of the circuit is in a 104 degree environment (let me mention that this is also a point of debate, but let?s leave it alone for the moment). The rest of the run is underground (i.e., protected from the summer heat). The exception talks about 10 feet and 10 percent. I don?t know whether I can use it, because without seeing the drawing, I am still not sure how long the total run is.

. . . and then wouldn't I have to derate the conductors again because of 4 conductors in a single conduit?

Those are two separate issues, and you might have to apply both. What we still don?t know is whether to count all four of the conductors as being ?current-carrying conductors.? If the fourth wire is a neutral wire, as it appears to be, and if it is only carrying the unbalanced load from the other conductors, and I think the current values you posted seem to bear out this possibility, then you don?t count the neutral wire. Thus, you only have three current-carrying wires in each conduit, and you don?t have to derate for that issue.

 

[jimmy_mcquade]

Charlie B,

I just spoke to another student and this problem came from a plant that he worked at a plant that had the following:
2300 volt primary and 208 volt secondary transformers with the output wired in the "Y" configuration.

using a book formula, if kva = (volts * amps * sort(3))/1000,
then wouldn't my (3) 100 kva single phase xfmers wired in a 3 phase manner have 277.6 amps?
100 kva = (208*277.6*sqrt(3))/1000 = 277.6 amps

thank you all,
jimmy

 

[charlie b]

. . . he worked at a plant that had the following: 2300 volt primary and 208 volt secondary transformers . . .

Let?s stop right there for the moment. Please clarify. Are you saying that (1) There are three single-phase transformers, and that (2) looking at just one of the transformers, if you put 2300 volts on the primary, you would get 208 volts on the secondary? That is an unusual secondary voltage for a single phase transformer. If you put three of them together, and connect as a WYE, then you will get line-to-line voltage of 360, as I said before.

Or are you saying . . .

. . . with the output wired in the "Y" configuration.

. . . that AFTER you wire the three together, to form a three-phase transformer bank, then and only then do you get 208 volts from phase to phase on the secondary of the entire bank? That would tell me that each individual single-phase transformer had a secondary voltage of 120. So which is it?

100 kva = (208*277.6*sqrt(3))/1000 = 277.6 amps

First of all, presuming that you have three single-phase transformers, each rated at 100 KVA, then the total KVA is 300. Therefore, when calculating secondary current, you have to use the value of 300 KVA, not 100 KVA. Secondly, I can't confirm or refute your calculation until I know the secondary voltage of the three-phase bank. So you will have to clarify that point before I can help any further.

 

[jimmy_mcquade]

Charlie B,

I ran back to class to clarify things.
1. There are three single phase transformers rated at 100 kva each.
The primary voltage is 2300 volts.
2. The transformers are wired together as a 3 phase system.
3. The output line to line voltage is 208.

Thanks everyone for all your help.
regards,
jimmy

 

[charlie b]

In that case, the secondary rated current is given by 300,000 divided by 208, and divided again by 1.732, or 833 amps. So it turns out that wasasparky had the right answer, after all.