Cable voltage drop

[tlp23]

Dear All,

I am trying to find out the voltage drop in a Copper cable of size less than 16sq.mm. Basically it is a 3 phase circuit & i have got the unit voltage drop mV/A/m value from BS 7671.
As all of you know the formula for voltage drop calculation is

Vd=(mV/A/m)*Length*Current/1000.

now my question is, since mine is a 3 phase circuit, should i multiply the above formula by Sqrt(3) or just use the above formula for both 1-phase circuits & 3-phase circuits.

please clarify.

 

[7anoter4]

For phase-to-phase voltage drop calculation it is correct only Vd=(mV/A/m)*Length*Current/1000
But for phase-to-neutral you have to divide by Sqrt(3)
See:
"Electrical Installation Calculations by B.D. Jenkins BSc, CEng, FIEE and other"

http://www.andazyar.org/doc/0632064854.pdf

 

[waross]

I disagree, but it may be a matter of semantics.
>A single phase load on a three phase circuit, use the tables.
>Three equal loads on the three phases, use 1.73
>A single phase line to neutral load use the line to neutral voltage and the tables. (Use 1.73 to derive the line to neutral voltage from the line to line voltage.)

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

 

[7anoter4]

What you said waross it is very logical.
But depends on which table we shall use in this BS.
The Table 4D4B [Column 4] it seems to be for three phase distribution cable. If we shall follow the Example 2.4 from page 38 [of the above link] we shall see how it was calculated using the Table 4D4B.
The Table 4D2B [Column 3] is for single phase cable[2 conductors], indeed.
That remember me in U.K. you have to drive on the left side...

 

[waross]

Self correction:
Actually, the voltage drop depends on the cable and the current. We only need the voltage to calculate the percentage voltage drop.
I agree with 7anoter4 when using three phase tables for single phase. Apologies.

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

 

[dpc]

The power factor of the load must also be considered in ac circuit voltage drop.

 

[7anoter4]

I agree with you dpc. This is a very simplified method for an approximate calculation. It does not take into consideration the actual pf, conductor temperature, resistance proximity factor and reactance [load factor?]and other.
If we'll take the simplified equation DV=sqrt(3)*I*(R*cos(Fi)+X*sin(Fi)) one needs to know a lot of data.
But this is the role of a standard: to state an illogical [oversized] limit [very unpleasant some time!] in order to overcome the unknown.