Transformer sizing

[jfox]

Hello everyone,

A homework question is asking me if you have a 50 hp 480v motor and a 30A 277v load to calculate the transformer size required.

The FLA of the motor is 65A from table 430.250.
My kVA for the motor would be 65*480*1.732= 54 kVA

My single phase kVA would be 30*277= 8.3 kVA

If i understand this right i now add 54 + (8.3*3) = 78.9 kVA as my highest leg would be 54/3=18kVA per leg and 8.3kVA is added to a leg for a total of 26.3kVA. As my highest leg is 26.3kVA this is multiplied by 3 to give me my total 3 phase load, 78.9kVA.

My question is am i done at this point, meaning my answer is 78.9kVA or do i need to add 125% to this value? Someone also told me transformers could be sized up to 300% however im not seeing anything in the code book that supports either of these percentages for sizing the transformer itself.

 

[augie47]

Welcome to the Forum.
I find your question interesting and challenging.
The correct answer would depend on some info not given. For one, how is the single phase load distributed ? Normally you would try to balance it so that you would only have 10 amps per phase and a 57 kva transformer.
I may be incorrect, but I find no NEC requirement to figure the motor load at 1.25%. That said, in the real world you would also need to obtain some transformer data and additional motor data so as to the size transformer needed to handle the starting current.

Hopefully we will get some input from others.

 

[jfox]

For one, how is the single phase load distributed ? Normally you would try to balance it so that you would only have 10 amps per phase and a 57 kva transformer.
I may be incorrect, but I find no NEC requirement to figure the motor load at 1.25%. That said, in the real world you would also need to obtain some transformer data and additional motor data so as to the size transformer needed to handle the starting current.

Hopefully we will get some input from others.

The question doesnt state anything about the type of single phase load so i was assuming it was a single device pulling the 30 amps.

 

[david luchini]

I may be incorrect, but I find no NEC requirement to figure the motor load at 1.25%. That said, in the real world you would also need to obtain some transformer data and additional motor data so as to the size transformer needed to handle the starting current.

Hopefully we will get some input from others.

Gus, I think you may be incorrect about the motor load. 220.14(C) says motor load shall be calculated in accordance with 430.22...

430.22(A) says the conductors shall have an ampacity of at least 125% of the flc (unless it's other than continuous duty.)
I would interpret this as the "motor load" is based on 125% of flc.

Interestingly, I don't find anything in the code that says that the transformer has to sized larger than the load it will carry. You would certainly be limited by the primary and/or secondary OCPDs, though.

 

[bob]

Quote "If i understand this right i now add 54 + (8.3*3) = 78.9 kVA". No you just add 54 + 8.3 = 62.3 kva. I looks like ypu need a 75 kva transformer.
If possible you divide 8.3 kva to get the load per phase. So 54/3 = 18 kva and 8.3/3 = 2.8 kva. However, depending on the load you may not be able to split the load as shown. You may end up with all of in on one phase which is ok.

 

[Jraef]

Just my opinion, but to me it's either a trick question or the intent of the assignment is only to challenge your code knowledge. There is nowhere near enough information here to select the proper transformer.

Remember that the National Electric Code comes from the NFPA; National FIRE Protection Agency. "The Code" is not a design specification document, it is a list of installation rules and MINIMUM requirements meant primarily to prevent fires. It tells you what the minimum size of a transformer can be so that you don't create a hazardous situation, but "the Code" doesn't care if it works or not. You will not likely be able to start a 50HP motor with a 75kVA transformer ahead of it, but "the Code" is not the tool for determining what will work. That takes a much more detailed analysis of the motor, power system, acceptable voltage drop, starting method, load acceleration characteristics etc. etc. etc.

Either that or use a "rule-of-thumb" such as I do and make SURE it's going to be sufficient by over sizing it. For example if the motor is starting across-the-line, I would not use anything smaller than a 112.5kVA transformer (rule-of-thumb: For X-line start, transformer kVA = 2X motor HP, round up to next). But depending on what your instructor is looking for, you might get marked incorrect for that. So you need to try to psych out the intent of the question here. Then maybe if you answered with the word "minimum" that would be less incorrect.

Good luck.

 

[jfox]

Quote "If i understand this right i now add 54 + (8.3*3) = 78.9 kVA". No you just add 54 + 8.3 = 62.3 kva. I looks like ypu need a 75 kva transformer.
If possible you divide 8.3 kva to get the load per phase. So 54/3 = 18 kva and 8.3/3 = 2.8 kva. However, depending on the load you may not be able to split the load as shown. You may end up with all of in on one phase which is ok.

The information your telling me seems to go against what i've seen elsewhere on this forum. I want to make sure i have at least this part of it right. Remeber the 8.3kVA is a single phase load and the 54kVA is a 3 phase load. So if i'm trying to figure out what my highest leg is would i not add the 18kVA (54/3) to the 8.3 and then multiply it by 3 for a total of 78.9kVA? Why would i need to divide a single phase load by 3 as you suggest?

 

[Volta]

... you have a 50 hp 480v motor and a 30A 277v load ...

The FLA of the motor is 65A from table 430.250. Agreed.
My kVA for the motor would be 65*480*1.732= 54 kVA Yes but, per the Note to T430.22(E), we need to consider the application continuous duty if we have no further information.
So add 25% per 430.22(A) and 210.20. 67.5kVA.

My single phase kVA would be 30*277= 8.3 kVA Agreed. No reason to think that it's continuous...

If i understand this right i now add 54 + (8.3 *3) =78.9 kVA as my highest leg would be 54/3=18kVA per leg and 8.3kVA is added to a leg for a total of 26.3kVA. As my highest leg is 26.3kVA this is multiplied by 3 to give me my total 3 phase load, 78.9kVA

I think your method is right: 67.5 + 8.3 = 75.8 total load on transformer. 67.5/3 = 22.5, 22.5+8.3 = 30.8, 30.8*3 = 92.4 kVA.

My question is am i done at this point, meaning my answer is 78.9kVA or do i need to add 125% to this value? Nope, we did on each individual load that needed it. Someone also told me transformers could be sized up to 300% however im not seeing anything in the code book that supports either of these percentages for sizing the transformer itself. They can be sized as large as you want, no limit to that.

So I'd figure for 92.4 kVA as the minimum NEC size. I don't know what the primary voltage is, but guess 112.5 would still be the next standard.

Lookin' good. :thumbsup:

 

[augie47]

Gus, I think you may be incorrect about the motor load. 220.14(C) says motor load shall be calculated in accordance with 430.22...
........................................

I would agree with that.

The information your telling me seems to go against what i've seen elsewhere on this forum. I want to make sure i have at least this part of it right. Remeber the 8.3kVA is a single phase load and the 54kVA is a 3 phase load. So if i'm trying to figure out what my highest leg is would i not add the 18kVA (54/3) to the 8.3 and then multiply it by 3 for a total of 78.9kVA? Why would i need to divide a single phase load by 3 as you suggest?

Purely in terms of "load" you woudl only add 8.3kva, however, if you were actually sizing the transformer you would have to take into account that the load is single phase and make sure that one transformer phase was not overloaded. For example you should not look at the load as 75 kva since one section of the transformer will be loaded higher. That may be where the question is leading. If you took the motor load of 65 amps x 1.25 (thanks, david) for 67.5 kva and added 8.3 you might assume a 75kva transformer would be sufficient (ignoring Jraef's facts momentarily) but the 8.3 would be only on one phase so the 75 would not be sufficient.

 

[Ralph Michaels]

The information your telling me seems to go against what i've seen elsewhere on this forum. I want to make sure i have at least this part of it right. Remeber the 8.3kVA is a single phase load and the 54kVA is a 3 phase load. So if i'm trying to figure out what my highest leg is would i not add the 18kVA (54/3) to the 8.3 and then multiply it by 3 for a total of 78.9kVA? Why would i need to divide a single phase load by 3 as you suggest?

Exactly. That seems to be the intent of the question; understanding that the 480V motor is a three-phase load and the 277V load is single-phase. To ensure you are not overloading any of the windings of the transformer, solve it exactly as you have:

54 kVA / 3 = 18 kVA per phase.
18 kVA + 8.3 kA = 26.3 kVA (the highest load on any one phase)
26.3 kVA x 3 = 78.9 kVA (the transformer minimum rating)
In the real world, the next standard size transformer would be 112.5 kVA (for low voltage transformers). We have no information on the primary voltage.)

Note that while motor wiring and overcurrent protection may need to be sized at 125%; this question is not about sizing either of these. There is no NEC requirement to over-size a transformer in this application. Meeting the load is good enough.

 

[david luchini]

To ensure you are not overloading any of the windings of the transformer, solve it exactly as you have:

Good advice, but where is there a code "requirement" that you don't overload transformer windings?

Note that while motor wiring and overcurrent protection may need to be sized at 125%; this question is not about sizing either of these. There is no NEC requirement to over-size a transformer in this application. Meeting the load is good enough.

While there is no NEC requirement to over-size the transformer in this application, there isn't any NEC requirement to "meet the load" either. The code requires that the feeder conductors and circuit breaker be sized to at least 125% of the continuous load plus 100% of the non-continuous load. The NEC calculated motor load is 81.25A, and the lighting load is 30A, so the feeder and OCPD must be at least 81.25+(1.25*(30))=118.75. So the minimum required OCPD size on the secondary feeder is 125A.

The 480/277V secondary of a 45kVA transformer cannot be protected by a 125A c/b (using primary & secondary protection per 450.3B), but a 75kVA transformer can be protected by a 125A c/b. So a 75kVA transformer would be the NEC "required" transformer size to support the load as stated. Of course, nothing in the code prevents using a larger (properly protected) transformer.

 

[Ralph Michaels]

Good advice, but where is there a code "requirement" that you don't overload transformer windings?

Its not a code question.

While there is no NEC requirement to over-size the transformer in this application, there isn't any NEC requirement to "meet the load" either. The code requires... so a 75kVA transformer would be the NEC "required" transformer size to support the load as stated. Of course, nothing in the code prevents using a larger (properly protected) transformer.

Again, its not a code question. I have done many of these problems, and IMHO, it looks like the question is simply asking for calculations based on the loads. It will be interesting for the OP to respond with the actual answer (and method of coming up with it), when & if he gets it from the instructor.

 

[david luchini]

Its not a code question.

I don't know if its a code question or not, but...

Someone also told me transformers could be sized up to 300% however im not seeing anything in the code book that supports either of these percentages for sizing the transformer itself.

The OP seems to be treating it as a code question.

Again, its not a code question. I have done many of these problems, and IMHO, it looks like the question is simply asking for calculations based on the loads. It will be interesting for the OP to respond with the actual answer (and method of coming up with it), when & if he gets it from the instructor.

I'd be interested in hearing the correct answer too. I don't see how a load calculation alone will give you a "required" transformer size. "Required" based on what?

 

[bob]

Exactly. That seems to be the intent of the question; understanding that the 480V motor is a three-phase load and the 277V load is single-phase. To ensure you are not overloading any of the windings of the transformer, solve it exactly as you have:

54 kVA / 3 = 18 kVA per phase
18 kVA + 8.3 kA = 26.3 kVA (the highest load on any one phase)

Thats ok to determine the higest load on one phase

26.3 kVA x 3 = 78.9 kVA (the transformer minimum rating)

That is not correct. You only have 8.3 kva on one leg. So the load is 26.3 + 18 + 18 = 62.3 for the transformer load

 

[kingpb]

Minimum 3 phase KVA required is 78.9KVA, the minimum standard transformer size would be 112.5KVA.

However, the motor may not start, depending on motor characteristics. The likely size transformer needed may be a 150KVA.

 

[bob]

Minimum 3 phase KVA required is 78.9KVA, the minimum standard transformer size would be 112.5KVA.

However, the motor may not start, depending on motor characteristics. The likely size transformer needed may be a 150KVA.

Show me the math. I show you mine.

 

[kingpb]

480V, 3P Mtr load = 54KVA
277V, 1P load = 8.3KVA

54KVA/3 + 8.3KVA = 26.3KVA (Worst case per pole)

26.3KVA x 3 = 78.9KVA (non-std size 3ph Xfmr)

Next Std size = 112.5KVA

May work, maybe not, depends on motor characteristics, so its possible that a 150KVA could be needed.

 

[bob]

480V, 3P Mtr load = 54KVA
277V, 1P load = 8.3KVA

54KVA/3 + 8.3KVA = 26.3KVA (Worst case per pole)
26.3KVA x 3 = 78.9KVA (non-std size 3ph Xfmr)
Next Std size = 112.5KVA

May work, maybe not, depends on motor characteristics, so its possible that a 150KVA could be needed.

I've already told you that is not correct. You only have 26.3 kva on one leg. Why do you keep putting the same incorrect
statements up. Did you not read my correction of your previous comment. The load is not 78.9 kva. Read the correction again
and see if you can understand why your information is wrong. Don't keep putting up incorrect information for others to read.

 

[IamJonscranium]

Just something graphic for the OP. Notice how the two single phase loads are connected to the three phase feeder. The feeds are distributed as evenly as possible, but the load on phase 2 will be double phase 1 & 3.

 

[david luchini]

Just something graphic for the OP. Notice how the two single phase loads are connected to the three phase feeder. The feeds are distributed as evenly as possible, but the load on phase 2 will be double phase 1 & 3.

That is not correct. Assuming the loads have the same pf, the load on phase 2 will be 1.732 times the load on phases 1& 3.