I am posting again to see who may reply, I posted yesterday It seems by the time several people chime in ( thank you) the original question gets lost or watered down.
Here the input for my voltage drop Calc.
1. 277/480V 3P 4W
2. K value 12.9 for CU
3. distance is 500' from sub panel to service( underground in PVC)
4 load at the end of the circuit is 177 amps per phase
5. wire selected 300MCM
Here is my math VD= 1.73 . 12.9 . 177 . 500/300,000
VD=6.58 or 2.3% line to N
VD=6.58 or 1.3% line to line
Question 1 is this valid?
Question 2 is the method the project engineer uses (below) more accurate?
How he does it. Thank you all for your input! Happy holidays!
1. Using the voltage and current and the formula R = V / I, calculate the resistance in your circuit that would result in 100% VD.
2. Determine what voltage drop is acceptable as a percentage and multiply the value from Step 1 by that number. For example 2% would be 0.02 x R = Total Allowable Circuit Resistance
3. In order to get to the ohms/1000ft given in Table 8 of Chapter 9 in the NEC, we then need to figure out how many 1000 ft sections we have in our circuit. If, as in this situation, we?re estimating the distance to be 500 ft from line to load, the math is easy. Our circuit is 500 ft out the line and 500 ft back along the neutral. That?s one 1000 ft section. Had it been 250 ft we?d be at only half of a 1000 ft section and so the resistance in the wire could be twice as much and have the circuit resistance stay the same.
4. Then all that?s left is do divide the number from Step 2 by the number from Step 3 and compare it to the NEC.
In our case then,
R Total = 277 / 175 = 1.583 ohms
R Allowable = 1.5% * 1.583 = 0.0237 ohms
D = 530 * 2 = 1060 ft.
L = D / 1000 = 1.06 thousand ft
Conductor Resistance = 0.0237 / 1.06 = 0.0224 ohms/1000ft
Looking at Table 8, we then see that we need a wire of 600 MCM.
Digging a bit deeper, Table 8 is really for DC. The AC table is Table 9, but it?s for uncoated conductors only. Looking at the impact coating has on DC conductors of this size, there appears to be about a 4.2% kicker. If we can translate that over to the AC table, it looks like our 600 MCM coated becomes 0.02397 ohms/1000 ft and is therefore too high for our needs. We?d either need to go to a 700 MCM or allow the higher voltage drop (1.61% by my calculations).
Please either find fault in my calculation or provide your own to this detail. I?m afraid I still can?t determine where your numbers are coming from.
Here the input for my voltage drop Calc.
1. 277/480V 3P 4W
2. K value 12.9 for CU
3. distance is 500' from sub panel to service( underground in PVC)
4 load at the end of the circuit is 177 amps per phase
5. wire selected 300MCM
Here is my math VD= 1.73 . 12.9 . 177 . 500/300,000
VD=6.58 or 2.3% line to N
VD=6.58 or 1.3% line to line
Question 1 is this valid?
Question 2 is the method the project engineer uses (below) more accurate?
How he does it. Thank you all for your input! Happy holidays!
1. Using the voltage and current and the formula R = V / I, calculate the resistance in your circuit that would result in 100% VD.
2. Determine what voltage drop is acceptable as a percentage and multiply the value from Step 1 by that number. For example 2% would be 0.02 x R = Total Allowable Circuit Resistance
3. In order to get to the ohms/1000ft given in Table 8 of Chapter 9 in the NEC, we then need to figure out how many 1000 ft sections we have in our circuit. If, as in this situation, we?re estimating the distance to be 500 ft from line to load, the math is easy. Our circuit is 500 ft out the line and 500 ft back along the neutral. That?s one 1000 ft section. Had it been 250 ft we?d be at only half of a 1000 ft section and so the resistance in the wire could be twice as much and have the circuit resistance stay the same.
4. Then all that?s left is do divide the number from Step 2 by the number from Step 3 and compare it to the NEC.
In our case then,
R Total = 277 / 175 = 1.583 ohms
R Allowable = 1.5% * 1.583 = 0.0237 ohms
D = 530 * 2 = 1060 ft.
L = D / 1000 = 1.06 thousand ft
Conductor Resistance = 0.0237 / 1.06 = 0.0224 ohms/1000ft
Looking at Table 8, we then see that we need a wire of 600 MCM.
Digging a bit deeper, Table 8 is really for DC. The AC table is Table 9, but it?s for uncoated conductors only. Looking at the impact coating has on DC conductors of this size, there appears to be about a 4.2% kicker. If we can translate that over to the AC table, it looks like our 600 MCM coated becomes 0.02397 ohms/1000 ft and is therefore too high for our needs. We?d either need to go to a 700 MCM or allow the higher voltage drop (1.61% by my calculations).
Please either find fault in my calculation or provide your own to this detail. I?m afraid I still can?t determine where your numbers are coming from.