Voltage Drop Question - Building Wiring

[chevywaldo]

I'm not understanding something very basic here.

I know what voltage drop is. At least I thought I did.

Here is a question:

Assume you have a power supply in a control panel of a building. 500 feet away is the load device. When doing voltage drop calculations for the wire (500 feet each way) am I to add up the sum of the voltage drop along all 1000 feet of wire - or just 500 feet? All the voltage drop calculators and formulas I have seen only ask you for the distance from the power supply to the load (1/2 the total wire length). Then they go on to say how the voltage drop of the wire is calculated on that distance only (power supply to load device). What I am not understanding is what about the voltage drop on the return wire (the other 500 feet). Why is that voltage drop not included in the total voltage drop of the circuit? Or is it?

Lee

 

[berkeman]

I'm not understanding something very basic here.

I know what voltage drop is. At least I thought I did.

Here is a question:

Assume you have a power supply in a control panel of a building. 500 feet away is the load device. When doing voltage drop calculations for the wire (500 feet each way) am I to add up the sum of the voltage drop along all 1000 feet of wire - or just 500 feet? All the voltage drop calculators and formulas I have seen only ask you for the distance from the power supply to the load (1/2 the total wire length). Then they go on to say how the voltage drop of the wire is calculated on that distance only (power supply to load device). What I am not understanding is what about the voltage drop on the return wire (the other 500 feet). Why is that voltage drop not included in the total voltage drop of the circuit? Or is it?

Lee

You will indeed get voltage drop on both wires, determined by V=IR. If you are using canned calculators or tables, they will need to specify if they take the doubling of the wire length into account or not. Different sources treat it differently. You can also just do your own calculation as a double-check, to see if they are doing the doubling for you, or if you have to do it yourself.

I like to remember that 18AWG wire measures 1 Ohm for every 155 feet. And remember that the resistance doubles (or halves) for every 3AWG change. So with 12AWG wire, you can go 4x as far (620') before you get an Ohm.

 

[chevywaldo]

You will indeed get voltage drop on both wires, determined by V=IR. If you are using canned calculators or tables, they will need to specify if they take the doubling of the wire length into account or not. Different sources treat it differently. You can also just do your own calculation as a double-check, to see if they are doing the doubling for you, or if you have to do it yourself.

I like to remember that 18AWG wire measures 1 Ohm for every 155 feet. And remember that the resistance doubles (or halves) for every 3AWG change. So with 12AWG wire, you can go 4x as far (620') before you get an Ohm.

that helps a little bit - but I'm still not getting it.

checkout this website. They explain an example scenario where there is a 350 run of paired wired. Are they saying the total length is 350 feet, or half the distance is 350 feet? Then they calculate the voltage drop based on 350 feet, when I think there is 700 total feet. Am I wrong or right?

http://www.securityideas.com/howtocalvold.html

 

[berkeman]

that helps a little bit - but I'm still not getting it.

checkout this website. They explain an example scenario where there is a 350 run of paired wired. Are they saying the total length is 350 feet, or half the distance is 350 feet? Then they calculate the voltage drop based on 350 feet, when I think there is 700 total feet. Am I wrong or right?

http://www.securityideas.com/howtocalvold.html

Their table and my 18AWG number makes it clear that they are counting both lengths of wire in their calculation. So when they say a wire cable length of 350', they mean a total wire length of 700'.

Look at their table -- they list the voltage drop for 100 feet of 18AWG wire as 1.27V for 1A.

Look at my number that I like to keep memorized -- 155' of 18AWG wire is 1 Ohm.

100/155 = 0.65

2 * 0.65 = 1.29

Make sense?

 

[chevywaldo]

Their table and my 18AWG number makes it clear that they are counting both lengths of wire in their calculation. So when they say a wire cable length of 350', they mean a total wire length of 700'.

Look at their table -- they list the voltage drop for 100 feet of 18AWG wire as 1.27V for 1A.

Look at my number that I like to keep memorized -- 155' of 18AWG wire is 1 Ohm.

100/155 = 0.65

2 * 0.65 = 1.29

Make sense?

ok - much better. Now I get it. I thought I was loosing my mind. Plus, I've got a veteran guy at a jobsite in Texas telling me that they don't figure voltage drop on the total length of wire, rather they use the distance from power supply to load (half the total length) I was arguing with him that's not true - that they total length of wire is used in the final circuit voltage drop calculation. He said "they don't do it that way - you use only a one way distance, not the total loop distance" for which I replied "you're incorrect" and the conversation got heated.

For example if we had a 500 foot run, he was just figuring voltage drop for 500 feet of wire. I was telling him it's 1000 feet of wire (total round trip distance). He kept telling me I'm wrong and that "they don't figure it that way". I think what he meant is that the total wire length is already factored into all the voltage drop calculators and that total round trip length is already taken into account by the voltage calculator "doubling" the one way wire length. The whole thing was just a misunderstanding I guess.

Lee