How do you compute voltage drop when given a specified distance and cable size?
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How do you compute voltage drop when given a specified distance and cable size?
Hank,
I use the voltage drop calculator from this site:
electrician2.com
Good site from Don. Mike Holt also has helpful calculators.
Since the Cable is already run, and distance is known, the Calculator will tell maximum current usable with the allowable Voltage Drop.
Another way to use the calculator is Know the load and distance to determine wire size.
To do the calculation yourself use this formula
VD=(2KLI)/cmil
VD = voltage drop in volts
K = K factor, this is approximate resistivity per foot, use 12.9 for copper or 21.2 for aluminum
L = distance one way
I = actual number of amps being used
cmil = wire size in circular mils, you can use chapter 9 table 8 of the NEC to find the cmil of standard wire gages.
You have the distance and K factor, all that is left is to find the cmil of the wire in the table and know how many amps.
Edit- this formula is for single phase, for 3 phase replace the constant 2 with the square root of 3
Thanks Don, Strat and Tev.
Apparently the software requires Visual Basic, which I don't have. But with the info from Tev, and my fingers and toes, I can work it out. OK, I have to use a couple of my wife's toes but she doesn't seem to mind.
No, it will ONLY run in Internet Explorer.
It does require JAVA which is a free download from Sun.
Need more info, holler.
Psst Tev:
The 2 is there because the wire goes to the load and back, hence the distance the electrons must travel from breaker to load is twice that distance. The 2 stays.
It is true that for say a 30 KW, 3 phase load, each leg shares =I/sqrt(3); so if if P=VI; 30 KW = 208*I
I(phase) = (30 * 1000)/208/sqrt(3)
Hope I did it right.
Things were going good, but found something to clarify.
This in bold is correct:
Everything else Tev states is correct also, just focusing on the 3 phase explanation, as a segway to:Quote:
Originally Posted by Tev
This statement is incorrect:
Quote:
Originally Posted by KeepItSimpleStupid
See backup here from Mike Holt as a reference, approximately mid way down the page:
Voltage_Drop_Calculations_03-26-2003
Find:
"Three-phase VD = 1.732 x K x I x D/CM. The difference between this and the single phase formula is you replace the 2 with 1.732 "
For more in depth formulas, since the above formula is technically for DC circuits, and does not take AC impedance, capacitance, or inductive loads, into consideration, review this:
IEEE Voltage Drop Formulae - Volts Electrical Design Software
These are the formulas used by NEC in the Handbook, as recommended by IEEE.
tk:
Time for an aside:
First, I used a 30 KW load. Take a look at. I'm not saying that he's right either. Going against Mike is a problem though.
I started with a load in Watts, not amps. Normally we would use amps. So, because I took watts and I is sqrt(3) of the single phase current. I should be right on this account. I's not higher, it's lower, right, so divide by sqrt(3). Say 100 V, 10 A
P = 1000 W
P = sqrt(3)*100*10 + sqrt(3)*100*10 + sqrt(3)*100*10
P= 1 * 100 * 10
P = 1000
Each phase is drawing sqrt(3) * 10 A
The problem, is that one is 120 deg out of phase out of phase with the other.
So if one phase is drawing 10 A, the other phase is drawing 10 * cos(120 deg) or 10 A* 0.866
So while one wire is drawing I, the other is drawing I/sqrt(3)
Vd = I * R + R * sqrt(3)/2 * I
Vd = R( I * sqrt(3)/2)*I)
So you would take the per phase I and divide by the sqrt(3)/2 or 0.866
so you take the single phase answer and multiply by sqrt(3)/2 0r replace the 2 with sqrt(3).
OK, makes sense. Thanks.
Yipes!
But, we need to start with the per phase I which is P/(V*I)/sqrt(3) with an example in watts.
Thanks, TK!
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