Voltag Drop

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Re: Voltag Drop

Your branch circuit or feeder conductors must be increased by 25% for continuous loads. Once you determine that conductor size, you would base your voltage drop off of that.

So for example, if I have a 45A continuous load, I initially would need #6 conductors. If the voltage drop exceeds my desired value, I would base it on the #6 conductors.
 
Re: Voltag Drop

If I understand your question correctly, no you don't use 125% of the continuous loads. For example, if you have a 45 amp continuous load, you would use 45 amps for the calculation. You would not use 45 *1.25 amps.

Steve
 
Re: Voltag Drop

NEC requirements do not effect voltage drop calculations.

If you are using this formula:

Voltage drop = 2 x K x L x I/cmil

You do not multiply I by 1.25

Edit: or any formula

[ February 17, 2005, 02:52 PM: Message edited by: physis ]
 
Re: Voltag Drop

Your right, you don't multiply the "I" by 125% because the "cm" is the conductor size required by increasing the load by 125%.
 
Re: Voltag Drop

quote by Bryan
Your right, you don't multiply the "I" by 125% because the "cm" is the conductor size required by increasing the load by 125%.
Your post makes me wonder if you are treating continuous loads differently than non-continuous.
Does a 50 amp continuous load on a #4 conductor present a different voltage drop than a 50 amp non-continuous load on the same #4? I believe the answer is no, they are the same. Forgive me if I misunderstand your thoughts.
Jim T
 
Re: Voltag Drop

No, what I am trying to state is that sizing the conductor for the continuous load is required, the volatge drop calculation is not. So, you first need to determine what size conductor is required for the circuit. Once you determine that, you can check to see if other circuit conditions present a voltage drop issue.

So again using the example of a 45-ampere continuous load, the "I" would be 45-ampere, but the "cm" would be based on 56.25-ampere rated conductor.

Am I making more sense?
 
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