transformer sizing

[baldie0217]

I just took the oregon master test, I had trouble with one question on sizing transformer.

Size a transformer for 208/120 wye with the loads of;

12-25 hp motors 3phase
5-15hp motors single phase
250-rec outlets 20amp 120v
1-10kva industrial heat single phase

what is the proper way to do this,and what do I do with the single phase loads, do I add for 1-10kva across all three phases ?

 

[JoeStillman]

Welcome to the forum!

You have unbalanced, 1Ø loads, so you need to calculate the worst phase leg kVA and multiply by 3.

Calculate the current on the heaviest loaded phase leg. That, multiplied by the line-neutral voltage is the heaviest loaded phase leg's kVA. Multiply that kVA by 3 to get the minimum transformer size.

12 @ 25 HP, fla = 74.8, 12 x 74.8 = 898A
5 @ 15 HP, fla = 82.5 A (assumed at 208V, 1Ø) maximum phase leg has 2 motors so 2 x 82.5 = 165A
250 receptacles, (assumed 180W, 1.5A each, balanced as much as possible) maximum phase leg has 84, 84 x 1.5 = 126A
10 kVA, 1Ø (assuming 208V, 1Ø again) = 10,000/208 = 48A

898+165+126+48 = 1,237A; 1,237 x 120 x 3 = 445.3 kVA = 500 kVA transformer.

If you wanted to do the math with the line-line voltage instead, it would be 1,237 x 208 x 1.732 = 445.6 = 500 kVA = same answer.

 

[kwired]

There is no 15 HP single phase motor in table 430.248.

That said you can find some in the field, but would have to go to motor nameplate to get a FLA value.

Just one problem I have with the question as OP worded it.

You could go with 746 watts per horsepower - but also would need to know power factor and efficiency to come up with a realistic VA figure.

 

[Carultch]

There is no 15 HP single phase motor in table 430.248.

That said you can find some in the field, but would have to go to motor nameplate to get a FLA value.

Just one problem I have with the question as OP worded it.

You could go with 746 watts per horsepower - but also would need to know power factor and efficiency to come up with a realistic VA figure.

Given a motor of a non-standard HP, is there a way to know what the equivalent FLC value would be as opposed to FLA?

As I understand it, the reason for the difference between FLC and FLA, is that FLC considers the amperes of a generic motor of a given mechanical power rating, so that the circuit can remain in place if you ever do replace the motor. Even if the replacement motor is less efficient than the original. FLC consideres a baseline efficiency, while FLA is the manufacturer's info of how much better they can do.

 

[kwired]

Given a motor of a non-standard HP, is there a way to know what the equivalent FLC value would be as opposed to FLA?

NEC requires us to use values for most common size motors per the tables near the end of art 430. Those values in those tables generally use worst case power factor and efficiency for the motors listed and are higher current levels then you will find for most motors you will ever encounter.

Problem is when a motor is not in those tables you don't have much other choice then to go with the motor nameplate. The test question failed to provide such information to use to find the answer - not a fair question IMO unless the examiners want to have varying answers and check to see if they were done correctly with varying inputs to the equations, or have multiple choice answers with only one answer that is within a range of reasonably possible outcomes.

 

[baldie0217]

thanks joe

thanks joe

thanks joe that was really good it will help very much