I see that most of us (including Mike Holt) use the old formula for computing voltage drop. VD = I*K*2*L/CM (change 2 to 1.732 for 3 phase). "K" has been defined equal to from 10 to 12 for copper and from 17 to 21 for aluminum. Wouldn't it make sense for us to change to the charts used in the NEC for these computations? Chapter 9, Tables 8 & 9 give a hundred different values for "K" (you have to do the math). Could it be there is a more accurate method of doing our computations? VD = I*2*L*R/1,000 the value of "R" would be determined from the Table. Use Table 8 for DC, and Table 9 for AC, (using Effective Z at 0.85 PF for Uncoated Copper Wires as "R") BTW, coated wires are varnished wires used in manufactured transformers and coils and the like.
My understanding of 'coated' is also those conductors coated with varnish for use in coils. The 'tinning' is a process required for copper wire used with rubber-insulated cable. Just my understanding; some of this information is in the 'American Electrician Handbook', but there is no definite statement saying what exactly 'coating' refers to in the tables.
Re: Voltage drop calculations#86925 01/09/0411:07 AM01/09/0411:07 AM
With a little algebra, I have re-written the formula for voltage drop so we can determine what size wire would be required for a long run, rather than trial and error. For single phase:
R = 15*E/I*L, where
R = the resistance per 1,000 feet of wire found in Table 8 or 9. Choose the next larger size of wire from the calculated value. E = the voltage I = the expected current draw L = the distance from source to load in feet
For three phase:
R = 17.32*E/I*L
Re: Voltage drop calculations#86926 01/09/0401:39 PM01/09/0401:39 PM