Voltage drop calc clarifications

[NorthSparky]

Hope you're all having a great morning,

I was looking at Mike Holts voltage drop equations page and have a couple questions (This page).

For VD formula he uses:

[FONT=&quot]"VD = 2 x K x Q x I x D/CM - Single Phase[/FONT]
[FONT=&quot]VD = 1.732 x K x Q x I x D/CM - Three Phase[/FONT]
[FONT=&quot]“VD” = Volts Dropped: The voltage drop of the circuit conductors as expressed in volts.[/FONT]
[FONT=&quot]“K” = Direct Current Constant: This is a constant that represents the direct current resistance for a one thousand circular mils conductor that is one thousand feet long, at an operating temperature of 75º C. The direct current constant value to be used for copper is 12.9 ohms and 21.2 ohms is used for aluminum conductors. The “K” constant is suitable for alternating current circuits, where the conductors do not exceed No. 1/0.[/FONT]
[FONT=&quot]“Q” = Alternating Current Adjustment Factor: Alternating current circuits No. 2/0 and larger must be adjusted for the effects of self-induction (skin effect). The "Q" adjustment factor is determined by dividing alternating current resistance as listed in NEC Chapter 9, Table 9, by the direct current resistance as listed in Chapter 9, Table 8.[/FONT]
[FONT=&quot]“I” = Amperes: The load in amperes at 100 percent, not 125 percent for motors or continuous loads.[/FONT]
[FONT=&quot]“D” = Distance: The distance the load is located from the power supply, not the total length of the circuit conductors.[/FONT]
[FONT=&quot]“CM” = Circular-Mils: The circular mils of the circuit conductor as listed in Chapter 9, Table 8."

My questions are:

When my conductor is 2/0 or larger, do I use BOTH K and Q? Then just use K for <2/0 conductor?

For I (load current), could I just use the rating of the OCPD to get a worst case scenario?

For distance, just curious why conductor length isn't used? Because it is harder to measure with loops in boxes, and distance between panels is close enough?

Thanks for your guys time, I'm new here and will probably have a lot more stupid questions for you![/FONT]
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[Carultch]

“D” = Distance: The distance the load is located from the power supply, not the total length of the circuit conductors.

For distance, just curious why conductor length isn't used? Because it is harder to measure with loops in boxes, and distance between panels is close enough?

I'm not sure where you are getting your information that it is explicitly not the length of conductor. It is the length of the conductor that matters. If the conductor takes a significantly serpentine path, you need to account for that. The effective round trip length, which is why we double it for single phase and DC, and multiply it by sqrt(3) for three phase.

As you've mentioned, it is difficult to predict the exact length of conductor in advance of installing it, which is why we tend to settle on simply the total raceway distance, or a general length estimate. But for the physical nature of this equation, it is the conductor length that matters.

 

[NorthSparky]

That's from the website linked in original post, which is from the parent website of this forum. I was thinking along the same lines you are, I don't understand why they would want me to use distance between panels.

 

[Smart $]

...
When my conductor is 2/0 or larger, do I use BOTH K and Q? Then just use K for <2/0 conductor?

That's what the 'notes' for this formula say you should do.*

For I (load current), could I just use the rating of the OCPD to get a worst case scenario?

You could... but the actual long-time high load current seldom reaches the OCPD rating... and it shouldn't when the circuit is properly sized. Doing so would often lead to the result requiring one, two, perhaps more sizes larger wire.... and larger wire costs more $$$.

For distance, just curious why conductor length isn't used? Because it is harder to measure with loops in boxes, and distance between panels is close enough?

As Carultch noted, you use the circuit conductor length. The reason the 'notes' state it like that is that you only use theone-way length, i.e. source to load, not the entire circuit 'loop' length, i.e. not source-load-source.

One of the first things you should understand about any voltage drop calculation is that it is a preemptive aproximation. A truly accurate result is a fleeting concept.

*An alternative method is to substitute the impedance value from Chapter 9 Table 9 for the (K x Q/CM) portion of the equation, adjusting for the value representing 1,000ft.

 

[NorthSparky]

So if you are trying to do this calculation prior to an installation, to make sure your cable size is adequate before you buy the stuff, do you normally just make a conservative estimate of distance and base the current on the calculated VA of load/known nameplate current of any specific loads? Any rules of thumb for when this is worth the effort?

Do these equations come from an IEEE, or some other, standard?

 

[Smart $]

So if you are trying to do this calculation prior to an installation, to make sure your cable size is adequate before you buy the stuff, do you normally just make a conservative estimate of distance and base the current on the calculated VA of load/known nameplate current of any specific loads? Any rules of thumb for when this is worth the effort?

Do these equations come from an IEEE, or some other, standard?

If you use the NEC calculated load, you are already being conservative. For example, a typical residential service has a 100A calculated load. On most average days, the actual load will not exceed 70A (excluding high inrush events). The calculated load takes into consideration the not-so-average days and specific points in time, i.e. instances approaching the extreme. Weigh other factors in accordingly.

The one you ask about in the OP is a modified version (Q added) of the most basic formula that's existed since the electrical industry had its start. I don't believe it started as a standard or if it ever was, and IEEE didn't exist back then. IEEE now has what they call the exact voltage drop formula (google it)...

 

[NorthSparky]

If you use the NEC calculated load, you are already being conservative. For example, a typical residential service has a 100A calculated load. On most average days, the actual load will not exceed 70A (excluding high inrush events). The calculated load takes into consideration the not-so-average days and specific points in time, i.e. instances approaching the extreme. Weigh other factors in accordingly.

The one you ask about in the OP is a modified version (Q added) of the most basic formula that's existed since the electrical industry had its start. I don't believe it started as a standard or if it ever was, and IEEE didn't exist back then. IEEE now has what they call the exact voltage drop formula (google it)...

Thanks again for the info

 

[Smart $]

Thanks again for the info

Follow up...


Here's a pretty good treatise on voltage drop calculation:

https://pdhonline.com/courses/e426/e426content.pdf

 

[jcrawford]

Reactance?

Reactance?

Hi all,

I was just comparing the results from these calculations (http://www.mikeholt.com/technnical-voltage-drop-calculations-part-one.php, same as what you posted) to values used at Cooper and Okonite (http://www.cooperindustries.com/con...US_Ele_Tech_Lib_Voltage_Drop_Calculations.pdf, http://www.okonite.com/engineering/voltage-regulation.html) and at large conductor sizes, I'm seeing big differences (underestimation of voltage drop). I think it's because the MH equations aren't accounting for reactance. Anyone know how to properly incorporate reactance? I'm looking at 600MCM conductors and seeing differences of a factor of 2. Thanks!

 

[Smart $]

Hi all,

I was just comparing the results from these calculations (http://www.mikeholt.com/technnical-voltage-drop-calculations-part-one.php, same as what you posted) to values used at Cooper and Okonite (http://www.cooperindustries.com/con...US_Ele_Tech_Lib_Voltage_Drop_Calculations.pdf, http://www.okonite.com/engineering/voltage-regulation.html) and at large conductor sizes, I'm seeing big differences (underestimation of voltage drop). I think it's because the MH equations aren't accounting for reactance. Anyone know how to properly incorporate reactance? I'm looking at 600MCM conductors and seeing differences of a factor of 2. Thanks!

What part of my last post escapes you?