DC Voltage Drop

[jsinclair]

I have a big problem on my hands that I can't seem to find a definite solution too, atleast that others will accept. There is much debate as to what the proper equation for voltage drop is when dealing with DC power. I have consulted various references (including this site) and cannot find a difference between AC single phase and DC voltage drop calculation. Heres the equations that I have been using:

1-Phase

VD = [2 x K x (length) x (load)] / CM

3-Phase

VD = [1.73 x K (length) x (load)] / CM


The people I am dealing with insist the equation is wrong for DC and cannot be applied. Can anyone give me any insight. Here's some sample data that I have for calcs if anyone is interested.

K = 11.1

L = 60 ft

I = 1200 A

CM = 211600cm (4/0)


VD = [2 x 11.1 x 60 x 1200] / 211600 = 7.55 volts


This all leads to my next question. Does the Code assume that the input voltage of the system is the same as the output voltage of the system even under load? Is there a voltage drop percentage in the code other than for Branch Circuits (2%) and Feeders (3%)?

As always, thanks in advance.


John

[ February 17, 2004, 06:19 PM: Message edited by: jsinclair ]

 

[roger]

Re: DC Voltage Drop

John, the VD calculations we all use are basically DC calculations with a generalized K factor.

To find a true K factor you would need to use these methods

75 deg = R x CM / 1000

90 deg = R x CM x 1.05 / 1000

60 deg = R x CM / 1000 x 1.05

Now, the code is note really involved with assuming whether or not the input and output voltages are the same, and the NEC only recommends voltage drop percentages 2% and 3% for a total of 5%, it does not mandate these numbers.

In our every day VD calculations we are only getting close, to be acurate in an AC circuit everything would have to be considered, Inductance, Capacitance, and Reactance.

Of course you know all this.

Roger

[ February 17, 2004, 06:58 PM: Message edited by: roger ]

 

[dereckbc]

Re: DC Voltage Drop

John, I have been designing DC battery plants for 25 years. Here is the accepted formula used by all Telco's. You had it right.

CM = (22.2 * I * D) / VD

CM = Circular Mills needed
22.2 = 2 * 11.1, or 2K of copper
I = Max load current in amps
D = 1-way cable distance in feet
VD = Voltage Drop desired.

After years of data logging and measuring I use 21.5 for K, never missed target yet, always over-killed it.