Voltage Drop - So many choices....

[cppoly]

For a 3-phase circuit, Mike Holt uses the forumla V.D. = (1.732 * Q * I * D) / C.M.

Two questions:

1) Does this formula come from the NEC (I can't find it)? If someone asks how I do my calculation, I can tell them where I got.

2) The Q is determined from Table 9 using 12.9 for copper & 21.2 for aluminum and adjusted for AC resistance. But it looks like this formula doesn't include power factor, it only relies on the type of conduit the cables are in. Table 9 also includes "Effective Z at .85 PF" column. It looks like these AC resistances are higher in this column for cables above 1/0. Anyone have any comments on calculating voltage drop using PF vs. using the Mike's formula?

 

[bob]

For a 3-phase circuit, Mike Holt uses the forumla V.D. = (1.732 * Q * I * D) / C.M.

Q*D/CM is basicly the resistance of the conductor. So you have 1.73 * I * R. Does that look familar?

 

[cppoly]

Yes, but depending on the power factor, the R will change.

 

[iceworm]

... Does this formula come from the NEC (I can't find it)? If someone asks how I do my calculation, I can tell them where I got. ...

The NEC is not exactly in the mathematical model business. As for Mike's formula, you could tell them, "Physics 101, DC voltage drop"

... But it looks like this formula doesn't include power factor, it only relies on the type of conduit the cables are in. Table 9 also includes "Effective Z at .85 PF" column. It looks like these AC resistances are higher in this column for cables above 1/0. Anyone have any comments on calculating voltage drop using PF vs. using the Mike's formula?

Chapter 9, Table 9 is my preferred method. And you got to read the notes. The NEC likely got this from IEEE 141. The resistance and reactance are likely empherical data. I don't know where it came from.

There are about two types of VD calcs I do.
1. Check the terminal voltage at the end of a long line is sufficient to run the motors with out overheating. For this one, the Table 9, .85pf is fine.

2. Check the terminal voltage under motor starting conditions is sufficient. For this one I'll still use Table 9, but it will be a radically diferent pf.

ice

 

[cppoly]

If I use the "Effective Z at .85 PF" column, what formula applies:

The Q*I*D or V = I.R. (using the ohms per 1000 feet)?

Thanks for the IEEE PDF. I think it's too complex for a V.D. calculation though. It's impossible to know what phase angle to use in the formula.

 

[iceworm]

If I use the "Effective Z at .85 PF" column, what formula applies:
The Q*I*D or V = I.R. (using the ohms per 1000 feet) ....

Per table 9 the "Effective Z at .85 PF" column is ohms to neutral. So the voltage drop (line-to-neutral) is:

VD(l-n) = table value X (circuit length /100 ft) X circuit load (amps)

VD(l-l) = VD(l-n) X sqrt(3)

(Right out of the handbook)

... Thanks for the IEEE PDF. ...It's impossible to know what phase angle to use in the formula.

No, the phase angle is exactly specified by the power factor. One can make some pretty good guesses at the pf.

pf = cos(phase angle between V and I) for pf = .85, phase angle = 31.8deg

... I think it's (the IEEE PDF) too complex for a V.D. calculation though. ...

Possibly, but the concepts are good. In particular, The VD is not in-line with either the sending voltage nor the receiving voltage The IX component is 90deg out from the IR component.

For example, in small wires, say #12, the resistance is 40X the reactance. So at low power factors, (motor starting) the VD is low compared to high power factor loads. The formula in Table 9 accounts for that - with a bit of error, but still pretty well.

ice

 

[cppoly]

Thanks ice.

I meant the phase angle (which is derived from the PF) is impossible to know because of the unlikely case of knowing multiple loads' combined PF, but I think a .85 PF effective Z is a good estimate right?

I looked at the handbook notes at the bottom of table 9, and I like this method.

One thing I noticed though, it says:

VD(l-n) = table value X (circuit length /100 ft) X circuit load (amps) (Isn't this an error though? The V.D. should have a 2 in the formula for two wires line & neutral)

 

[iceworm]

... One thing I noticed though, it says:
VD(l-n) = table value X (circuit length /1000 ft) X circuit load (amps) Isn't this an error though? The V.D. should have a 2 in the formula for two wires line & neutral)

Good catch - but no.

The calculation is based on a balanced three phase circuit. So, there is no current on the neutral. The current you are measuring on the one phase lead is being returned on the two other phase leads - no voltage drop on the neutral.

Look at the neutral as though it is and isolated reference point connected back to the source.

ice

 

[iceworm]

... but I think a .85 PF effective Z is a good estimate right? ...

I think it has been the industrial standard for induction motor loading for maybe 150 years - every since Tesla and Westinghouse got together at Niagra Falls.

However, if the loads are not industrial standard induction motors, but rather large VFD, data center switching power supplies, then one would have to pick a different number.

I suspect that .85 pf is low for todays higher efficiency motors, but I have not seen any papers (such as IEEE) that gives better values.

For a motor at the end of a long line, .85 pf sounds fine.

ice

 

[cppoly]

For a single phase load, I would have to include the 2 in the formula since the neutral has return current:

VD(l-n) = table value X (circuit length * 2 /1000 ft) X circuit load (amps)

 

[Smart $]

For a single phase load, I would have to include the 2 in the formula since the neutral has return current:

VD(l-n) = table value X (circuit length * 2 /1000 ft) X circuit load (amps)

For a 2-wire circuit, yes.

Since you brought up figuring in pf, another consideration is temperature affect. Note the tables give resistance (or impedance) at 75?C. In most cases, your conductors will not be operating at 75?C. See Note 2 to C9T8. The problem with this is no method is provided to determine T2.