Dear all,
How can i calculate a conductor size based on derated ampacity if i have distance from source ( 17M) , source voltage (110VAC), Current (34.7A), single phase. As per NEC Voltage drop should not exceed more than 3%.
Kindly give one example for reference.
Different people use different methods. I use the method I was told to use by the City of Los Angeles.
Start by going to NEC Chapter 9, Tables 8 and 9. For conductors #2 AWG or smaller, use the DC resistance from Table 8. For conductors larger than #2 AWG, use the Effective Z Table 9. I'm guessing you use metric distances, so you'll use Ohms/km. Choose the wire with the minimum ampacity for your application, and find the corresponding Ohms/km from the tables.
Now it's just a matter of applying Ohm's Law, V=IR. In this case, V is the voltage drop over the length of the conductor, I is the current flowing through the conductor, and R is the resistance of the conductor. You'll need to calculate R using the Ohms/km from the tables in Chapter 9, and the length of your conductors. Remember that for single phase circuits the length of the conductor is twice the distance from source to load.
Once you have calculated the voltage drop across your conductor, you can calculate the percent voltage drop by dividing by your nominal voltage and then multiplying by 100%.
So as not to just do this for you, I'll provide an example using different numbers. Suppose the distance from source to load is 12 meters, the load is 56.1 amps connected to a 60 amp circuit breaker, and the voltage is 230 volts, single phase. Also assume we are using equipment with 75 degree terminations.
Using the 75 degree column in NEC Table 310.16, we can see that the smallest wire that can be protected by our 60 amp breaker is #6 copper. Going to NEC Chapter 9 Table 8, we see that uncoated #6 copper has a resistivity of 1.608 Ohms/km. We now determine the resistance of our wire by multiplying the resistivity by the length of the wire.
R = (1.608 Ohms/km) x (.012 km) x (2) = 0.038592 Ohms
Using this, we can calculate the voltage drop using V=IR.
I x R = (56.1 amps) x (0.038592 Ohms) = 2.165 volts. This is the amount of voltage dropped over the length of the wire. Now we convert it to a percentage of the nominal voltage:
( 2.165V / 230V ) x 100% = 0.94%
In this case, our voltage drop is an acceptably small percentage, so we can choose to use #6 AWG copper and we're done. If we had come up with a percent voltage drop that was unacceptably large, we would choose a larger wire size and start over.
Hopefully this has been helpful.