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# mv/A/m - Voltage drop

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## mv/A/m - Voltage drop

(OP)
mV/A/m - Voltage drop. How the cable manufacturers list the Voltage drop table in their catalogue. How do they calculate this mV/A/m?

NEC:
For three phase: VD = 1.732 x L x R x I / 1000
For single phase: VD = 2 x L x R x I / 1000
R = ohm/kM

IEE, BS 7671:
For single & three phase VD = (mV/A/m x L x I )/ 1000

This question arose because, NEC Voltage drop calculations include multiplier such as 1.732 and 2 for three phase and single phase respectively but IEE wiring regulations, BS 7671 does not include any multiplier?

### RE: mv/A/m - Voltage drop

The "I" is different in those formulas

### RE: mv/A/m - Voltage drop

Perhaps BS 7671 is referring to the Voltage drop in a single conductor, I seem to recall that the NZ code used to do it that way.

BTW. I assumed it was millivolt / Amp / meter, was I correct?
I couldn't imagine millivolts per mile

### RE: mv/A/m - Voltage drop

Cable manufacturers use impedance at 50 Hz or 60 Hz in the tables, not resistance. The table values are valid at a specified load power factor. Two cables may have the same resistance but differing impedances.

Multipliers:
NEC: L, the length is measured from the source to the load. The current must return to the source hence the 2x factor for single phase and the root three factor for three phase.

IEC: It is up to you, the user, to determine the total length of the circuit, (L), applying the three phase correction factor when required.
The NEC formula assumes that both conductors have equal currents and equal impedances and may not be valid for determining terminal voltages in unbalance three wire single phase circuit.
With the IEC formula, the voltage drop/rise may be calculated for each conductor and the terminal voltages may easily be determined.
With an unbalanced three wire circuit, each conductor will have a different current and the neutral may be reduced, resulting in a higher unit impedance. In such a case, the multipliers built into the formula are not valid.
Don't be surprised if the result obtained from an application of the formula does not agree with the value in the manufacturers tables.
If you are concerned with complying with the code, use the IEC or NEC methods.
If you are trouble shooting or planning an installation where the actual voltage drop may be an issue,

Here is an instance of an actual installation where the true voltage drop of cables was important.
A large mine mill had an incoming supply at 140 kV.
This was stepped down to 13kV.
The 13 kV fed a lineup of 13 kV breakers with appropriate ratings.
Each breaker fed a unit sub-station that in turn fed a power distribution center at 480 Volts.
The incoming breaker adjacent to each unit sub transformer was not rated for the full available short circuit current at the 13 kV lineup.
The designer calculated that 100 feet of feeder cable would introduce enough series impedance to reduce the available short circuit current at the unit subs to within the rating of the incoming breakers.
Accordingly the construction specifications called for a minimum length of feeder of 100 feet to all unit subs.

Bill
--------------------
"Why not the best?"
Jimmy Carter

### RE: mv/A/m - Voltage drop

I think the factor 2 or sqrt(3) is taken into consideration in the table data [it is built-in].
For instance on Table 4D1B for 2* 2.5 mm^2 the mV/A/m is 18.
From IEC 60228 Table II the resistance per meter is 7.41 ohm/km. or 7.41 mΩ/m [1 mΩ/m=1 mV/A/m].
For 70oC we shall multiply by the factor k=(234.5+70)/(234.5+20)=1.197.
Then 2*7.41*1.197=17.73 mV/A/m.
For 3 cables sqrt(3)*7.41*1.197= 15.36 [In the above mentioned table it is written 15].

(OP)

### RE: mv/A/m - Voltage drop

In my opinion, the voltage drop is calculated using the formula:
VD=√3(IRcos(φ)+IXsin(φ)) for I=1 A and cos(φ)=0.8.

### RE: mv/A/m - Voltage drop

I agree with 7anoter4.
Impedance times root three times 80% PF.

Bill
--------------------
"Why not the best?"
Jimmy Carter

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