Homework Help! Transformer Calculations and motor circuit

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SirCringe

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You are a newly hired electrician at chemical processing facility. You have been asked for a preliminary assessment of a proposed installation. Given a 3-phase 500KVA 480/240 Delta-Delta Transformer, determine both the primary and secondary side protection. The transformer will feed a service panel 15 ft away, what size THWN conductors are required?


The service panel for the transformer problem feeds 3-3phase motor loads.
208V, 10 hp, S.F. 1.0, 250ft away from panel
208V, 15 hp, S.F. 1.10 300 ft away from panel
230V, 7 1/2 hp S.F. 1.10 150 ft away from panel
What are required starter sizes?, What are required overload heater sizes?, What are required feeder sizes (conductors)-limit VD to 3% and note VD is per phase. Size required feeder protection. Provide conduits sizes for all feeders (EMT).

Thanks in advance!
 
We can help but you're going to have to show some of the work you've done on the solutions.

Welcome to the Forum. :)
 
infinity said:
We can help but you're going to have to show some of the work you've done on the solutions.

Welcome to the Forum. :)
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Thankyou,

So what I've got so far on the transformer is:
First the find the FLA by dividing
Primary: 500kVA / (1.732 x 480) = 602A
NEC 450.3(B) 250% x 602A = 1505A
Primary Max OCPD is 1600A

Secondary: 500kVA / (1.732 x 240) = 1204A
NEC 450.3(B) 125% x 1204 = 1505A
Secondary Max OCPD is 1600A

Am I on the right track?
 
Right track but I believe when you use the 250% primary protection rule the "next size up" doesn't apply.
(In 450.3 you will see "Note 1" only applies to the 125% primary rule.)
 
Why would you use the maximum permitted primary protection? That increases the costs as you must size the primary conductors to that OCPD. 125 to 150% of primary full load current will permit most transformers to be energized without tripping.
 
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augie47 said:
Right track but I believe when you use the 250% primary protection rule the "next size up" doesn't apply.
(In 450.3 you will see "Note 1" only applies to the 125% primary rule.)
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So your saying not to size up the OCPD to 1600 A? So the next rating would be 1200A, according to 240.6(A)
 
So continuing on the next step would be finding the conductor size for the secondary size on a 15 foot run. What is the allowed voltage drop and where can I find that in the NEC?
 
A answer to your direct question is that the voltage drop formula is not in the NEC. You can easily find it on line or in an UGLY's book among other places. Other than a few specific places, the NEC addresses voltage drop as an Informational Note in 215.2 so you are not required to address it by the NEC.
In your question they appear to want to limit the VD to 3% on the motor circuits (even though you state "feeder:) so you would need to look at each motor circuit and apply the formula based on the motor FLA from Table 430.250 I assume.
Your next step should involve at lot of Art 430 looking at each motor
 
SirCringe said:
Please show me how lol
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For example. Start with each motor and choose its branch conductors and protective devices. Ignore voltage drop.
Then select the feeders to the motor branch circuits and the feeder protective devices. Again ignore voltage drop.
Next select the transformer secondary conductors and protective devices, base these on the feeders you previously selected. Again ignore voltage drop.
Finally select the transformer primary side protective device and conductors considering the secondary protective device you previously chose.
Now you can perform the voltage drop calculations and adjust conductors as needed. This time start at the source and work towards the motors.
 
SirCringe said:
So continuing on the next step would be finding the conductor size for the secondary size on a 15 foot run. What is the allowed voltage drop and where can I find that in the NEC?
Click to expand...
There are no general code requirements in the NEC for voltage drop. There are two or three code rules that do specify a maximum voltage drop, but they do not apply through out the code. The are informational notes that suggest a maximum voltage drop, but those are not enforceable rules.
 
SirCringe said:
So continuing on the next step would be finding the conductor size for the secondary size on a 15 foot run. What is the allowed voltage drop and where can I find that in the NEC?
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It matters not what the NEC says about voltage drop, you need to follow the 3% in the specs.

You need to base VD on the load currents not the transformer maximum output or input.
 
SirCringe said:
How do I find the conductor size for the secondary? My FLA is 1204 A and the required conductor type is THWN Copper.
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You have to know the rating of the secondary overcurrent protective device. That device does not have to match the full secondary current, and if secondary protection is required for the transformer that device cannot exceed 1600 amps for this transformer.

There will be many combinations that will work, depending on how many parallel sets of conductors you want to use.

No idea why you were given a 500kva transformer for a total load that could be served by a 75kva transformer.
 
don_resqcapt19 said:
You have to know the rating of the secondary overcurrent protective device. That device does not have to match the full secondary current, and if secondary protection is required for the transformer that device cannot exceed 1600 amps for this transformer.

There will be many combinations that will work, depending on how many parallel sets of conductors you want to use.

No idea why you were given a 500kva transformer for a total load that could be served by a 75kva transformer.
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So here's what I got,
My over current protection device is 1600 A due to multiplying my FLA of 1204 by 125%.

Then I chose 2 parallel conductors of 1500 kcmil THWN Copper since one of them has a rating 625 A so by doubling it I get a amp rating of up to 1250A.
 
SirCringe said:
So here's what I got,
My over current protection device is 1600 A due to multiplying my FLA of 1204 by 125%.

Then I chose 2 parallel conductors of 1500 kcmil THWN Copper since one of them has a rating 625 A so by doubling it I get a amp rating of up to 1250A.
Click to expand...
It is a design choice for you to go with the maximum value of 1600A. Barring and project, or test question, specs you don't have to be any larger than connected load.

Your secondary conductors need to match your secondary protection. You chose 1600A so you will also need 1600A of conductors.

Have you ever tried to install 1500 kcmil conductors? My experience has been to not design above 500kcmil CU or 750kcmil AL.
 
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