Voltage drop calcs for power factor other than NEC .85 standard

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cppoly

Senior Member
Location
New York
Table 9 contains an example on how to calculate the voltage drop for a circuit with another power factor other than .85 as listed in the column 'Effective Z at .85 PF.'

The example uses 600 kcmil aluminum conductors with PF of .7, circuit length 250, and PVC conduit. After all said and done, the corrected impedance 'Zc' turns out to be .0531 ohms to neutral (per 1000 ft). It calculates this by adjusting the resistance and reactance using the load power factor of .7 and 'Xl' and 'AC Resistance' values from Table 9.

In this example, if I just simply take the 'Effective Z at .85 PF' value and divide by the new power factor of .7 I get a factor of .1214. If I take this correction factor and multiply by the impedance given for 'Effective Z at .85 PF', I get an ohms to neutral value of .0534!

This seems a lot easier. Basically take .85 and divide by the new power factor and multiply the impedance value found in the 'Effective Z at .85 PF' column and you have your adjusted power factor.

Does this make sense and can this be used in other examples? Oddly, it came it be really close in this example. .0534 vs .0531 is a .56 percent difference.
 
The formula generally used is VD = IR Cos ? + IX Sin ? . With a distance of only 250 ft you
will find that just using the R value will be accurate. Therefore the different PF does not make
a significant difference.































































 
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