Cable Voltage Drop equa

[7JLAman4]

I'm looking for some assistance re: cable voltage drop of a 3-Ph Cu cable 416300CM 405' 450VAC with 260Amps. Normally for simplicity, the following formula has been used:

VD% = sqrt(3) (12 x I x L x 100) / (CM V)

What is the more precise method for determining cable voltage drop given the above information? I'm not quite certain of the estimated/expected operating temperature. In the above, I believe the temperature is assumed 25°C.

 

[waross]

For a precise solution you must consider the impedance of the cable to an alternating current. The reactance and skin effect of the cable may be more significant than the resistance.
Find a set of cable impedance tables and look up your conductor configuration.
The power factor of the load is often assumed to be unity.
A rigorous solution of voltage drop would require both temperature and power factor to be taken into consideration.
"It depends" on the reason that you want to calculate the power factor.


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

 

[jghrist]

A close approximation is given by:
VD = I·(R·cosø + X·sinø)
Where VD is voltage drop (phase-to-neutral) in volts, I is current in amperes, R and X are the total resistance and reactance of the cable in ohms, and cosø is the power factor. For single phase, include both phase and neutral wire lengths in the R and X. For three phase, include only the circuit length.

 

[raptor6458]

For cable resistance and reactance your temp to be considered is 90C., its permissible operating temp.
Your assume temp should be 20C.
You also have to look for the short circuit carrying capacity of your cable.

Hope this helps.

 

[7anoter4]

Hi jlamann
If you really want to know how to calculate the voltage drop on a cable
then pay attention to it as follows:
x=2*pi*Fq*Lm[ohm/1000 ft]
Lm=0.001404*log(2*a/dia)[H/1000ft]
a=approx.insulated core dia[inch]
dia=conductor dia[inch]
Fq=frequency [Hz] usually 60 or 50
Pi=3.14159 [approximate]
Rac=Rdc*(1+kskin+kproximity)[ohm/1000ft] ac resistance of one conductor.
kskin=skin effect factor in the conductor itself
kproximity= proximity effect between adjacent conductors
Rdc =dc resistence [ohm/1000ft]
Rdc =roT*Length[ft]/conductor cross section[inch^2]*kstranding
roT=resistivity measured at "T" temperature[oC]
usually this T temperature is taken according to maximum temperature permited
for cable core insulation for instance for PVC=70oC and for XLPE =90 oC
roT=[approx.]ro*(1+kT(T-20)
ro=resistivity[specific resistence] ohm/ft*inch^2 measured at 20 oC[68 oF]
ro for copper coductor soft drawn ro=1/57[ohm.m/sqr.mm]
for copper coductor hard drawn ro=1/56[ohm.m/sqr.mm]
kT =Temperature coefficient [1/degree C]
For copper is approximative 0.004
kstranding=a slight difference may be also from stranding process of the conductor and another from cable lay-out [insulated cores are also twisted togather with a certain pitch].
If you succeeded to calculated this then
DU= I*(cosFi-jsinFi)*(Rac+jx) in complex. That means you take the current as perfect sinusoidal without harmonics .
Further, you have a DUa active drop voltage in the same phase as supply voltage
and DUr reactive drop voltage 90 electric degrees before U.
DUa=I.(R.cosFi+x.sinFi)*kphno
DUr=I.(x.cosFi-R.sinFi)*kphno
Formally DU=sqrt(Dua^2+Dur^2)[volts]
Or DU%=DU*100/U
If U is L-N then kphno=1 if U is L-L kphno=sqrt(3)
But all you need is:
1)take the resistance ac and x from the cable catalog as waross indicated
2)DUa is what you need and forget Dur as jghrist indicated.
3)Forget all my complicated calculation from above.
Regards

 

[7anoter4]

Never the less I think that for more information you may see:

http://www.okonite.com/engineering/voltage-regulation.html#Reactance of conductors at 60Hz

Also:

http://www.okonite.com/engineering/index.html

Regards