Volyage Drop

[ukumar]

Dear All,

I need to know How to calculate the Voltage drop and what needs to be done to avoid voltage drop as per NEC.

e.g- I have a 1600Amps breaker ,480 Volt, 60hz and distance is approximate 1000 feet.

So how to calculate the voltage drop for this application.

I have seen formula on mikholt.com
VD= 1.732KID/CM
K- Constant for copper or Aluminium wire
D- Distance one way
CM- Circular miles

In my case K- 12.9 ohms
I -1600 Amps
D- 1000Feet
CM - how to take a value of CM here.

Kindly reply.

 

[laidman]

Re: Volyage Drop

Need to know wire size and numbers of conductors per leg. circular mills of conductors can be found in NEC

 

[electricman2]

Re: Volyage Drop

CM is circular mils and is a measure of the cross sectional area of the conductor. For conductors 4/0 and smaller, use table 8 to determine conductor AWG size from the CM. Also it is only necessary to use the actual load to calculate the voltage drop. Is the actual load 1600A?

 

[physis]

Re: Volyage Drop

Go here and you can avoid the inaccuracy you'd get if you use K=12.9.

If you're uncertain of your results I can do the calculation for you.

 

[don_resqcapt19]

Re: Volyage Drop

Use the formula that you have, but solve for CM and not for voltage drop. Pick the maximum amount of voltage drop that you want and the formula will tell you the CM of the wire that is required.
Don

 

[physis]

Re: Volyage Drop

Using two sets of parallel conductors for each calculation:

2000 MCM
Voltage drop at 1300 amps is 1.6%
Voltage drop at 1600 amps is 1.9%

1750 MCM
Voltage drop at 1300 amps is 1.8%
Voltage drop at 1600 amps is 2.2%

1500 MCM
Voltage drop at 1300 amps is 2.1%
Voltage drop at 1600 amps is 2.6%

Those values come from the actual resistance of each conductor from chapter 9 Table 8 2002 NEC.

Using Vd=(1.732KID)/cmil will give you a slightly lower result that is less accurate because the actual value of K varies with every conductor. But the difference isn't much and there's no harm in using it.

Here's how to apply it:

Vd = (1.732 x K x I x D) / cmil
Vd = (1,732 x 12.9 x 1600 x 1000) / cmil
Vd = (35,748,480) / cmil

Go to Chapter 9 Table 8.
The first column is AWG and past 4/0 is kcmil (thousand circular mils)
Above 4/0 cmil (circular mils) is in the third column.

To complete the equation: Vd = (35,748,480) / cmil for 1500 kcmil (1,500,000 cmil)

Vd = (35,748,480) / cmil
23.83 = (35,748,480) / 1500 kcmil (1,500,000)

The voltage drop is 23.83 volts across one 1500 kcmil conductor for three phase.

voltage drop divided by supply voltage is the percentage of voltage drop.

23.83/480=5%

For equal size parallel conductors you can divide the voltage drop of one conductor by the number of parallel conductors.

23.83/2=11.9

11.9/480=2.5%

I have 2.6% above for the same conductor because I used the actual resistance, not K.