determining voltage drop for 1-phase lines 1

[jfriddell]

Gang:

I'm a "lurker" on this board and have found many helpful tips and knowledge. Thanks to all who make these boards so interesting and helpful.

Now for my question: I'm an agricultural engineer who is quite rusty on my elementary electricity. Could somebody refresh me on the proper calculations to properly determine voltage drop along a supply wire for a 1-phase motor.

As I understand, the equation for a 3-phase drop is:
dV = 1.732 * I * K (conductor materials) * L/CM (cir mil)

For 1-phase, is the 1.732 factor simply removed? Please advise on the correct formula so that I can sharpen my electrical pencil.

Thank you.

 

[sreid]

For all electrical wiring the voltage drop is simply Ohm's Law, Volts = Current x Resistance.

A simole rule for wiring resistance is

10 ga. wire has a resistance of one Ohm per 1000 ft. Every three gages the resistance doubles (or halves). So

13 ga. is 2 Ohms/1000 ft.
16 ga. is 4 Ohms/1000 ft.
7 ga. is 0.5 Ohms/1000 ft.

Remember that there are two wires in the circuit. For example; 500 ft. from source to load with ten ga. wire carrying ten amps.

Resistance (R) = 1 Ohm/1000 ft. x 500 ft x 2/1000 = 1 Ohms

E = IR = 10 x 1 = 10 Volts drop.

 

[peebee]

I agree with sreid, except I would have used Z, impedance, equal to sqrt(R^2+X^2). Note that R is a conductor property, but X is dependant on conductor installation (particularly, whether or not you have magnetic steel conduit). You can get R and X out of the NEC.

Note that a key difference between 3-phase and 1-phase circuits, which sreid alluded to, is that you often have very low neutral current on 3-phase (or maybe no neutral at all), and therefore you don't get voltage drop on the neutral. That means that a 3-phase circuit, given the same conductor size and current, will have less voltage drop than a single-phase circuit.

 

[jbartos]

Suggestion: The posted equation in the original posting is for voltage drop across the conductor resistance R in Ohms. This is especially good for dc power distribution. When it comes to AC power distribution and voltage drop Peebee posting above addresses the issue

 

[jbartos]

Suggestion: Specifically, AC single phase 2-wire:
1. Approximately:
Vo=Esending,end-Ereceiving,end=Eo'-Eo=IxZx2xL/1000, in Volts
where
I is conductor loop current
Z=R+jX, in Ohms/1000ft, e.g. from NEC
L is Length of Line
2. Accurately:
Vo=Esending,end-Ereceiving,end=Eo'-Eo={sqrt[(Eo x cos(theta)+IxR)^2 + (Eo x sin(theta) + IxX)^2] - Eo}
Where:
theta is power factor angle of the load
X is 60cycle Reactance of line in Ohms/1000ft per conductor
R is AC resistance of line in Ohms/1000ft per conductor
I line current in Amps
Eo', Eo sending and receiving end voltages to neutral in Volts

 

[busbar]

Using line-to-line voltage, §3.11.1 in IEEE Std 141-1993 …Electric Power Distribution for Industrial Plants mentions a mutliplier of 1.732 for 3ø circuits and 2 for 1ø circuits.

 

[jbartos]

Thanks busbar, I have to add a factor 2 x L / 1000ft
namely:
2. Accurately:
Vo=Esending,end-Ereceiving,end=Eo'-Eo={sqrt[(Eo x cos(theta)+IxR)^2 + (Eo x sin(theta) + IxX)^2] - Eo}x 2xL/1000ft

 

[rock1]

Here are a few formulas for you to calculate voltage drop:

Voltage drop = 2 x K x I x L divided by circular mil

cir mil = 2 x K x I x L divided by desired volts dropped

max lengeth = cmils x desired volt dropped divided 2 x k x L

max current= cmils x desired volt dropped divided2 x k x L


Three Phase voltage drop:

Volt drop = 1.73 x K x I x L diveded by circular mils

circular mils = 1.73 x K x I x L divide by desired volt drop

max length= cir mils x desired volt drop divde 1.73 X K x I

max current= cirmils x desired volt drop divide 1.73 x K x L

I = current in conductor
K = is approximately 12 for copper and 18 for aluminum
L = length of circuit from supply point to load
cir mils = circular mil area of condutor from table 8 of NEC
1.73 = square root of 3

you multiply single phase loads by 2 because you have the wire out to the load and power coming back

you only use the 1.73 for three phase and you DO NOT multiply by 2

NEC wants no more than 3% volt drop for branch circuits and 3% for feeder but not more than 5% combind branch and feeder total

sample problem: single phase
150 amp load, 240 volt, 120 feet from source, desired voltage drop is 3% using copper conductors

2 x K X I x L divide by desired volt drop

2 x 12(for copper) x 150 amps x 120 feet
divide by 7.2 volts (3% of 240volts)

equals 60,000 cirmils

table 8 copper conductor THHN = #2 copper

Good luck with your problem

 

[jfriddell]

Gang:

Thanks for all the great answers. I feel more confident in my ability to ensure that I'm recommending the proper wire size.

Have a great weekend.

Jeremiah

 

[peebee]

Note that ASHRAE-90.1 (which is well on its way to being adopted here and there and everywhere across the US as we speak) has tighter requirements for voltage drop than NEC:

Feeders: 2% at full connected load
Branches: 3% at full connected load.

Also, if I remember correctly, NEC states their voltage drop limits as guidelines only (except in some special circumstances, such as fire pump starting). ASHRAE-90.1 states their limits as absolute mandatory maximums.

 

[sreid]

At the distances likely to be wired in this post, I would think that voltage drops due to line reactance would be minimal. Any comments?

 

[jbartos]

Suggestion to peebee (Electrical) May 27, 2004 marked ///\\Note that ASHRAE-90.1 (which is well on its way to being adopted here and there and everywhere across the US as we speak) has tighter requirements for voltage drop than NEC:

Feeders: 2% at full connected load
Branches: 3% at full connected load.
///The overall voltage drop requirement is the same from both: ASHRAE and NEC, namely, 5% voltage drop maximum.\\\

 

[jbartos]

Suggestion: Reference: NFPA 70-2002 National Electrical Code Article 210.19(A)(1)(FPN No. 4):

FPN No. 4: Conductors for branch circuits as defined in
Article 100, sized to prevent a voltage drop exceeding
3 percent at the farthest outlet of power, heating, and lighting loads, or combinations of such loads, and where the
maximum total voltage drop on both feeders and branch
circuits to the farthest outlet does not exceed 5 percent,
provide reasonable efficiency of operation. See 215.2 for
voltage drop on feeder conductors.

 

[peebee]

Yeah, but 2% on feeders is still less than 3% on feeders, though, jbart -- and this would be an important difference on MOTOR FEEDERS, which was the original topic mentioned in this thread.

Also, please note that NEC does not include any *requirement* for max voltage drop. It simply includes a helpful little footnote mentioning that 5% max provides for reasonable efficiency of operation, just in case you happen to care about such things.

In contrast, ASHRAE states that 2+3% is an absolute code-mandated maximum permissible drop, and that's the max allowable even if you think it would be pretty slick to heat your place with your wiring. Too bad for you, 5% is all the heat you'll ever get out of those puppies.

 

[jbartos]

Suggestion to the previous posting: Every footnote in NEC counts since it is the Electrical Safety Code.

 

[peebee]

I never said the footnotes don't count. What I said, is that there's no requirement in the footnote!!! It only says, hey, by the way, if you want your installation to run efficiently, 5% is a nice voltage drop number to use, if you want to!!!

The ONLY place that I'm aware of NEC actually MANDATING a specific maximum permissible voltage drop is on FIRE PUMP STARTING. I admit, there may well be possible that there are other places where other wiring also is also specified by NEC. BUT -- GENERAL PURPOSE wiring is most definitely NOT limited to 5% drop by NEC.

Read those FPN's again and tell me I'm wrong.

 

[peebee]

By the way, not to get too far off on a tangent, but there's a strong argument that FPN's really DON'T count, at least not nearly so much as the BPTs (Big Print Text):

NEC "90.5(C) Explanatory Material. Explanatory material . . . is included in this Code in the form of fine print notes (FPNs). Fine print notes are informational only and are not enforceable as requiremetns of this Code."

So maybe they count, but they don't really count for all that much if they're not enforceable, do they?

 

[buzzp]

For voltage drop the NEC could care less, nothing is enforceable for standard installations (non fire).

Really, one should be concerned with NEMA MG standard for this. Motors designed around their guidelines are derated starting at 1% VUB (as defined by NEMA).

 

[jbartos]

Suggestion: The NFPA 70-2002 National Electrical Code does care about voltage drop more than the above posting suggest. Namely, the Article 647 includes:
"""(D) Voltage Drop. The voltage drop on any branch circuit shall not exceed 1.5 percent. The combined voltage drop of feeder and branch-circuit conductors shall not exceed 2.5 percent.
(1) Fixed Equipment. The voltage drop on branch circuits supplying equipment connected using wiring methods in Chapter 3 shall not exceed 1.5 percent. The combined voltage drop of feeder and branch-circuit conductors shall not exceed 2.5 percent."""

 

[peebee]

Article 647 covers 120-volt systems that run at 60-volts to ground (see 647.1). That's a pretty unusual installation, the only place I'm aware of that uses systems like that are some (not all) professional recording studios.

As I mentioned previously, "I admit, there may well be possible that there are other places" than FIRE PUMP STARTING where voltage drop is specified -- so we can add professional recording studios to this list now too.