Voltage Drop Clarity

[ElecEngbw]

Good morning. I am trying to determine if an existing circuit will handle the voltage drop for a new increased load. We are replacing a chiller and the existing circuit is 2 parallel runs of 4/0. The new chiller will have a FLA of 381A at 480V, 3-phase and is located 400' away from a new 450A circuit breaker. Using several different voltage drop calculators online, I am getting different values, but somewhere in the neighborhood of 2%. I believe the circuit is fine, but my question is more for my understanding of the voltage drop. I am confused how to apply VD to parallel runs.

Is this the correct equation:

VD = [2(.866)(L=400')(R=0.063 ohm/1000')(I=381A)] / 1000' = 16.63V

% = VD / System Voltage = 16.63 / (480V * Square Root (3)) = 0.0200 = 2.0%

How does the parallel runs play into the equation?

I appreciate your help.

 

[charlie b]

How does the parallel runs play into the equation?

Two parallel runs will reduce the VD by half. You can simply divide your results by 2, and get a VD of 8.32 volts.

% = VD / System Voltage = 16.63 / (480V * Square Root (3)) = 0.0200 = 2.0%

The square root of 3 does not come into this equation. 16.63 volts would be about 3.5% of a 480 volt system. But with two parallel runs of 4/0 the real voltage drop is 8.32, which is about 1.7% of a 480 volt system.

 

[ElecEngbw]

The square root of 3 does not come into this equation. 16.63 volts would be about 3.5% of a 480 volt system. But with two parallel runs of 4/0 the real voltage drop is 8.32, which is about 1.7% of a 480 volt system.

Thank you. So for a 3-phase circuit, the system voltage will always be Line-to-Line?

 

[ramsy]

Thank you. So for a 3-phase circuit, the system voltage will always be Line-to-Line?

Charlie, is he bundling 6-8 ccc's?

Me calc sees 190.5A on 4/0 burns up at 96?C with 7 ccc's, 80?C with 4 ccc's, 62?C with 3 ccc's.

Raceway isn't mentioned, so shouldn't we assume worse case, or ask how its bundled?

 

[charlie b]

So for a 3-phase circuit, the system voltage will always be Line-to-Line?

Yes. We are not dealing with a question of power, so the square root of three does not come into play.

 

[charlie b]

Raceway isn't mentioned, so shouldn't we assume worse case, or ask how its bundled?

Since the question was dealing with VD, I hadn't thought of ampacity issues. My instinct would not be to assume worst case. We were told that there were parallel 4/0s, and I took that to mean in separate conduits.

 

[bob]

The new chiller will have a FLA of 381A at 480V. 381 x 1.25 = 476 amps

4/0 at 75C ampacity = 230 amps x 2 = 460 amps. Looks like the ckt is not adequate without any derating.

 

[ramsy]

381 x 1.25 = 476 amps

Looks like the ckt is not adequate

Good call. Inductive load PF~85% also invalidates Voltage-drop calc.

 

[ptonsparky]

The new chiller will have a FLA of 381A at 480V. 381 x 1.25 = 476 amps

4/0 at 75C ampacity = 230 amps x 2 = 460 amps. Looks like the ckt is not adequate without any derating.

To small all the way. Asking an awful lot of a 450 amp breaker to start a 381 amp motor.

 

[ramsy]

Proper circuit size seems to demand expert experience, or an expert system, so far unavailable to those relying on the NEC. By the time any one individual can recall & master every critical NFPA ? table scheme, they are likely retired from the industry.

The NFPA ? practice of avoiding internationally-recognized formulas (IEEE), and standards (ICC), relies instead on memorizing an encyclopedia of proprietary references for multiple adjustments and derating schemes. The marginal result seems to repeatedly leave conductor temperature or inductive loads unchecked.

Rather than recognize installers are better served by having access to skilled engineering & fee schedules for estimating, NFPA ? tables for idiots becomes the great equalizer, in regard to overheated conductors, as Contractors and their unskilled labor become indistinguishable from unlicensed hacks, owner builders, or other Do-It-Yourselfers (DIY).

 

[david luchini]

The new chiller will have a FLA of 381A at 480V. 381 x 1.25 = 476 amps

4/0 at 75C ampacity = 230 amps x 2 = 460 amps. Looks like the ckt is not adequate without any derating.

You would have to know the MCA of the chiller to determine this. The MCA is 125% of the largest motor plus 100% of the other loads, not 125% of the chiller's FLA.