Voltage drop calculation and power factor

[buddy91082]

Checking voltage drop calculation using ohm's law and nec conductor properties table and it occurred to me the nec tables list impedancea at .85 power factor and only list the resistance of conductors as uncoated.

I did the calculations using the x and r values from the nec tables and am getting almost half of what a simple online voltage drop calculator would produce for a result of voltage drop. I checked length, current and the units of length I am using. The only conclusion I can make is that the online versions using the z value at .85 power factor from the nec tables.

Anybody run into this before?

Thanks.
B

 

[mikekilroy]

why not list your calcs here? it is easy to forget to double the length cuz what goes out must come back another wire

 

[7anoter4]

If it is about:
Table 9 Alternating-Current Resistance and Reactance for 600-Volt Cables, 3-Phase, 60 Hz, 75°C (167°F)— Three Single Conductors in Conduit.
Since it was calculated as "Ohms to Neutral per 1000 Feet", you have to take only one length-as the neutral it is only one point of all three phases' intersection.
Let's check z calculation for [for instance] 2/0 AWG:
Rac= (1+ycs+ysp+yp)*Rdc [as per Neher and McGrath "The Calculation of the Temperature Rise and Load Capability of Cable Systems" where:
ycs=skin effect factor; ycp =proximity effect factor and yp=pipe factor.
You may take it from Table 9 directly Rac=0.1 ohm/1000 ft.
X=k*2*pi*f*(0.1404*log10(2*S/dcond)+0.0153)/10^3[ ohm/1000 ft].
k=1.23 for pvc conduit for random lay and 1.5 for steel conduit. As for cradle configuration S=1.1*overall dia. Overall dia=0.668 inches [Table5 for RHH cable] and dcond= 0.418 [Table 8].We shall get 0.0426 for pvc and 0.0525 for steel conduit.
If we take R=0.1 and X=0.043 we'll get z=0.1*0.85+0.043*0.527=1.076 [close to 0.11 as per Table 9].

 

[buddy91082]

Gents, thanks for the response. I don't have my cacls handy with me. Will upload them tomorrow when at work.

But the gist of it is for 277v, single phase ac circuits. If i can remeber correctly, the following was the input data:

length of run (one way) - 400'
load - 23 amps.

The load is fluorescent lamps. When i ran the calcs using a simple online calculator, i was getting a result of #4 awg. I did the hand calcs, using ohms law to check using NEC table 8 (I think it was Table 8). I used the uncoated r (why don't they have coated values?)and the x as noted in table 8 and was coming up with (i think) # 8 awg as an acceptable voltage drop to 3%.

Also, the PF for the load is close to 1. Am I correct in assuming that using the z tables listed at a PF of .85 would have given a conservative result?

Thanks.
b