Voltage drop formula on 480V single phase site ltg.

[mike9999]

I saw this formula for voltage drop calc.

http://www.electrician2.com/calculators/vd_calculatoradv.htm

It does not have any option to select a 480v single phase that I need for site ltg. I know I would need #4 Alum. and PVC conduit for direct burial.

Does anyboy has a simple formula or an excel file that will do this?

Thanks.

 

[Wes G]

Try the basic Ohms law approach using the Table 9 impedances. VD= load amperage times resistance per foot times total distance there and back. If you know you have a power factor problem than divide your amperage by the PF. (Square root of three does not enter the picture unless you are using all 3 phase conductors)

There are some fancy AC formulas which are nice for engineering exactness, but for all practical applications don't come up with an answer that is significantly different.

 

[infinity]

You could try this calculator, it works on single phase 480:

http://www.elec-toolbox.com/calculators/voltdrop.htm

 

[don_resqcapt19]

When you have a calculator like this that does not have the actual voltage that you are using, you can use any voltage that has the correct type of circuit. In this case use the 2 wire 120 volt circuit. The actual drop in volts is not dependentant on the voltage, one on the type of circuit. The only thing that you have to do is to use the drop in volts divided by the supply voltage to get the percentage.
Don

 

[mike9999]

I tried this formula:

VD = (21.2 for Al)*(I)*(2L) / Circ. mills

1) Correct me if I am wrong; is VD the actual voltage drop and not a percentage?

2) I am confused since I do not have a neutral for a loop for 480V single phase; do I still need to multiply the lenght by 2?

3) Also don_resqcapt19 mentioned that "The actual drop in volts is not dependentant on the voltage". Isn't it the higher the voltage the less voltage drop because of the higher voltage punch?

 

[don_resqcapt19]

Mike,

Correct me if I am wrong; is VD the actual voltage drop and not a percentage?

That formula gives you the drop in volts. Divide the result by the circuit voltage to get the percent.

I am confused since I do not have a neutral for a loop for 480V single phase; do I still need to multiply the lenght by 2?

The current has to flow on both wires, so you need to multiply by 2.

Also don_resqcapt19 mentioned that "The actual drop in volts is not dependentant on the voltage". Isn't it the higher the voltage the less voltage drop because of the higher voltage punch?

The voltage drop in volts is only based on the current flow and the resistance of the circuit. 200' of #12 will have the same voltage drop for 12 volts as it does at 480 volts.
Don
Don

 

[mike9999]

Thank You So Much!