Voltage Drops???

How do you calculate voltage drops?

Ex. 540 foot run of 480 volt 4/0 copper with a load of 200amps.

Thanks in advance for any help you can give

Reply to
Sports One
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single phase, 3 phase or 3 phase 4 wire??????

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Reply to
SQLit

Be careful. Most tables and so called internet calculators are for MILD environments only. Resistance is directly related to temperature.

Reply to
Johan Lexington

And some of that temperature is just your ambient environment, and some of it is your heat rise due to a combination of power loss dissipation and the thermodynamics of the heat removal environment (is it open wire, buried, in a conduit, or wrapped in big R-factor insulation).

Reply to
phil-news-nospam

Table 8 in the NEC does take a conservative route, giving the resistance at 75C and still not making it a hard rule, only a guideline. If you are designing a circuit operate beyond 75c you are either into some serious engineering or you are planning to fail.

Reply to
Greg

Look up the Ohms/foot for #4 wire, multiply its length by 2 to account for the return path voltage drop, then apply Ohm's Law (V=I*R). This will give you your voltage drop.

Harry C.

Reply to
Harry Conover
4/0 Cu has a resistance of .0490 ohms/1000 feet.

So....540 ft/1000ft = .54

.54 x .0490 = .02646 ohms resistance per 540 foot chunk.

Vd = I R

Vd = 200 x .02646

= 5.292 volts drop. or 1.1%

Reply to
Cougercat

But the current goes there and back. So if the source and destination are

540 feet apart, you need to double the voltage drop because its going through 1080 feet of wire.

-- Mark Kent, WA

Reply to
SueMarkP

For a single phase circuit, yes. For a three phase circuit with a balanced load, no. And if its a three phase circuit, the basis of the voltage for the purpose of calculating a percentage drop should be 277V.

But then the OP didn't specify which it is. And the OP didn't specify the installation, which will affect the circuit reactance, which is really what you need to use for a voltage drop calculation. Unless this is a DC circuit, which the OP didn't specify either.

NEC Chap 9, Table 9 gives a reactance of 0.080 ohms per 1000 ft installed in steel conduit with a 0.85 power factor.

Lots of things to think about.

Reply to
Paul Hovnanian P.E.

Is that right. Somehow you missed the class on 3 phase delta loads; like heaters.

Reply to
Rigged

The configuration of the load doesn't matter. If its a balanced 3 phase load, a line current of 200A gives the same voltage drop. Since the OP didn't specify single or three phase, we can only guess. The load current was defined as "a load of 200A", not three individual phase to phase loads. When specified for a single three phase load, load currents refer to the circuit load, not the currents in the delta.

Reply to
Paul Hovnanian P.E.

destination are

For Power Factor =1.0 What about VD = 2 K L I / CMA for two wire single phase AC or DC Then we multiply by .866 for three phase, 3-wire and by .5 for 3-Phase,

4 -wire. For copper K = 11 for the circuit loaded less than 50 per cent K = 12 for a circuit loaded 50 percent or more L = one way circuit length in feet I is in amperes CMA is circular mil area from Table 8 of the NEC. K is the resistance of one foot of copper that is .001 inches in diameter. As I recall, Joe McPartland put this out for electricians about 30 years ago in the McGraw Hill books.

For other Power Factors we could use Table 9 of the NEC and the formula in the notes to calculate impedance: where Ze = R × PF + XL sin[arccos(PF)].

Where R and XL are taken from Table 9. However, R as given in the table is for 75 degrees C only.

There also the Drop Factor Method and the Mershom Diagram Method as discussed in the American Electrician's Handbook.

Maybe because of all these methods the NEC does not have too many mandatory voltage drop rules.

Reply to
Gerald Newton

That may be. The calculation of voltage drop to the end of the last branch circuit requires a certain amount of engineering judgement. Quite a few rule of thumb formulas exist to simplify matters for electricians. But these formulas all make certain assumptions and have limitations that the user needs to understand. If the NEC were to mandate maximum voltage drops it would require an engineering review of practically everything (which would suit me just fine financially ;-)).

As long as formulas are published with guidelines as to their range of valid conditions, a further engineering review should only be required in the event that they produce an answer that falls too close to or beyond these limits.

Reply to
Paul Hovnanian P.E.

I agree. If there enough engineers to design all circuits I am sure there would be a set of engineered drawings for just about everything an electrician installs. The fact is there are not enough engineers so we electricians have to use our rules of thumb and take our best shot for many installations. Perhaps this is why there is so much overkill in the NEC, our primary rule book.

Reply to
Gerald Newton

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