voltage drop

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jrock0305 said:
left code book in office could someone help me out with the voltage drop formulas from nec thanks
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1ph Distance = CM x VD/2 x K x I
3ph Distance = CM x VD/1.732 x K x I
CM= circular mils
K = 12.5 at 75C and 10 at 20C
VD = VD you want
I - amps
IEEE Formula I(RCos theata + XSin theata) R and X form Table 9 NEC
Theata angle from the power factor.
 
What would the end voltage of a 50 amp 120/240 volt STW cord ( RV trailer cord) run 250 ft. be, assuming the load is 50 amps? Can anyone help me find it this way? I know the caluculator tells me that I need a # 3 for this length, but I sorta need this info so I can convince the customer that his 6/4 cable going 250 ft. may not be sufficient to efficiently carry the load of his medical RV. Thanks, Steve
 
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#6 is 0.491 Ohm per 1000 ft so 50 Amps round trip on a 250 ft cable is 0.491x2x250x50/1000=12.3 Volts. That is about 5% for a 240 volt circuit.

If it is powering balanced 120 Volt loads the voltage drop will also be about 5%.

It all depends on what his real load is. It probably isn't 50 Amps.

#4 would have about 60% of the voltage drop for the same loads, or about 3%. There is not much to gain between #4 and #3 because #3 is 2.5% voltage drop for the same condition.

It would be a big help if he doesn't really need 250 ft. If he can cut the cable in half it would cut the VD in half.
 
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