- Location
- Chapel Hill, NC
- Occupation
- Retired Electrical Contractor
I am new at this so forgive me if I don't get his correct. This is my second attempt.
I am having a problem with the VD calculation using VD=2xKxIxD/CM (assuming copper and single phase). Solving this equation for CM , we get CM=2xKxIxD/VD. Assuming a 230 volt system, using copper wire I want to calculate the wire size I need with a maximum 6.9volt drop (3% of 240 volts). Sounds simply enough-- if we use 300 amps over a distance of 100 feet we get an answer of 11274 CM. Looking that up in Table 8 of Chapter 9 we get 2/0 copper. We all know that 2/0 is not suitable for 300 amps. Of course, if we use the reverse, 100 amps for 300 feet we get the same answer since they are multiplied together in the equation. What gives here? The second result seems feasible but I am reluctant to use it. Does the amperage have to be a lower number than the distance in order for the problem to work. I am missing something somewhere--please help.
I am having a problem with the VD calculation using VD=2xKxIxD/CM (assuming copper and single phase). Solving this equation for CM , we get CM=2xKxIxD/VD. Assuming a 230 volt system, using copper wire I want to calculate the wire size I need with a maximum 6.9volt drop (3% of 240 volts). Sounds simply enough-- if we use 300 amps over a distance of 100 feet we get an answer of 11274 CM. Looking that up in Table 8 of Chapter 9 we get 2/0 copper. We all know that 2/0 is not suitable for 300 amps. Of course, if we use the reverse, 100 amps for 300 feet we get the same answer since they are multiplied together in the equation. What gives here? The second result seems feasible but I am reluctant to use it. Does the amperage have to be a lower number than the distance in order for the problem to work. I am missing something somewhere--please help.