Voltage drop calcs and Q

[ibuildpower]

Hello:

This is my first post. I'd like to thank Mike Holt and our moderators for their time and effort in providing this forum. Thank you for taking the time to read my post.

I'll show my age and say that when I studied voltage drop the formula was (X) x K x I x L / MCM. Q was not a factor. I'm preparing to take the Master / Contractor test in a new state, and I want to make sure that I understand the new formula.

As I understand it, the new formula is: (X) x K x Q x I x L / KCMIL. (X) is 2 for single phase or 1.73 for three phase; K is 12.9 for Cu; 21.2 for Al; Q is from table 9 in chapter 9 NEC (c), I is current in amps, L is length of run; MCM or KCMIL is the cross sectional area of the conductor in circular mils.

If I have all that down correctly, then I'm ready to ask my question: For wire smaller than 2/0, for the purposes of a license test, do I disregard (enter zero for) Q in the equation, or does Q need to be applied for all wire sizes?

I think the answer is that I would disregard Q for wire smaller than 2/0. If there is more wrong with my thinking on this than simply what to do with Q, then, help me out here please. :-?

Thanks,
ibuildpower
"Opportunity is missed by many, because it comes wearing overalls and looks like work," Thomas Edison (quoted from memory).

 

[mivey]

...MCM or KCMIL is the cross sectional area of the conductor in circular mils.

well...ok, but use cmil without the "K" or "M"

If I have all that down correctly, then I'm ready to ask my question: For wire smaller than 2/0, for the purposes of a license test, do I disregard (enter zero for) Q in the equation, or does Q need to be applied for all wire sizes?

You can apply it for all but it is not real significant for the smaller wires and you should be close enough for the exam. Also, Q is found by taking Z in table 9 and dividing by ohms in table 8.


If you want to be more precise in the non-exam world, use the exact formula from IEEE std 141:

Vd = V + IRcos@ + IXsin@ - sqrt[V^2-(IXcos@-IRsin@)^2]

where:
Vd = L-N volt drop
V = source voltage
I = current
R = AC Resistance from NEC Table 9 (Ohms to Neutral)
X = AC Reactance from NEC Table 9 (Ohms to Neutral)
@ = phase angle

Good luck on the exam!

PS: Welcome to the forum!

Add: You would not substitute a 0 for Q but a 1 if you are not going to use it.

 

[drbond24]

If I have all that down correctly, then I'm ready to ask my question: For wire smaller than 2/0, for the purposes of a license test, do I disregard (enter zero for) Q in the equation, or does Q need to be applied for all wire sizes?

If you enter 0 for Q in that equation then you have made the numerator, and therefore the entire fraction, equal to 0.