How many conductors in 1 phase?

[goodoboy]

Hello,

I am confusing myself. I am new to electrical studies and trying to calculate voltage drop (and determine what size gauge wire to use) for 260ft, 120V 1phase, 12.5 amps. I want to use #12, but not sure it work.

If I have a 120V 1 phase load (12.5amps), how many wires/conductors are in the cable? Is it 3?

If its 3, do I enter 3 for date entry field 8 here http://www.mikeholt.com/documents/freestuff/Copy_of_voltage_drop_calculations_by_Bill_Bamford.xls ??

I am not sure what to select if it says "select the number of parallel wires"

http://www.calculator.net/voltage-drop-calculator.html If I have a 120V 1phase load, what do I select where it says Number of conductors?

THank you kindly. I am bit loss

 

[augie47]

For conductors to be "parallel", they connect to the same point (terminal) at both ends.
If you have a 120v 3 wire circuit, the conductors are not considered to be in parallel.
(Actually the Code does not allow paralleling conductors smaller than 1/0*)

* as almost always "with exceptions"

 

[goodoboy]

For conductors to be "parallel", they connect to the same point (terminal) at both ends.
If you have a 120v 3 wire circuit, the conductors are not considered to be in parallel.
(Actually the Code does not allow paralleling conductors smaller than 1/0)

Thanks for your help. But I am not sure I understand. I need help with what to enter on the voltage drop spreadsheet where it says "select number of parallel wires" or " Number of conductors" on the calculator.

I am thinking to enter 3, because I have 3 wires.

I am confused and lost

 

[gar]

130724-1839 EDT

goodoboy:

The problem is that you are just plugging numbers into an equation without understanding the fundamentals.

The voltage drop in a portion of a series circuit is the impedance of that portion of the circuit times the current thru the circuit. Your current is 12.5 A. For the purposes here you can assume that impedance is equal to DC resistance.

You have not specified the maximum voltage drop that you can tolerate. Suppose it is 6 V or 5% of the supply voltage. Using 6 V and 12.5 A this calculates to a maximum of 0.48 ohms.

For your circuit there are two wires that carry the load current. Since the source to load distance is 260 ft that means you have 520 ft of wire providing resistance to load current flow. Wire tables usually describe wire resistance as ohms per 1000 ft. #12 copper wire has a resistance of 1.588 ohms per 1000 ft at 20 C. Thus, the resistance would be 0.52*1.588 = 0.826 ohms. This is too high and thus you can not meet a 6 V drop criteria. Further the resistance and voltage drop would be greater because the current thru the wire will increase the wire temperature and also ambient temperature has to be considered greater. Consult NEC tables for the criteria to use for copper resistance.

Since copper #12 won't meet a 6 V drop criteria you can either change that criteria, or go to a larger wire.

.

 

[goodoboy]

130724-1839 EDT

goodoboy:

The problem is that you are just plugging numbers into an equation without understanding the fundamentals.

The voltage drop in a portion of a series circuit is the impedance of that portion of the circuit times the current thru the circuit. Your current is 12.5 A. For the purposes here you can assume that impedance is equal to DC resistance.

You have not specified the maximum voltage drop that you can tolerate. Suppose it is 6 V or 5% of the supply voltage. Using 6 V and 12.5 A this calculates to a maximum of 0.48 ohms.

For your circuit there are two wires that carry the load current. Since the source to load distance is 260 ft that means you have 520 ft of wire providing resistance to load current flow. Wire tables usually describe wire resistance as ohms per 1000 ft. #12 copper wire has a resistance of 1.588 ohms per 1000 ft at 20 C. Thus, the resistance would be 0.52*1.588 = 0.826 ohms. This is too high and thus you can not meet a 6 V drop criteria. Further the resistance and voltage drop would be greater because the current thru the wire will increase the wire temperature and also ambient temperature has to be considered greater. Consult NEC tables for the criteria to use for copper resistance.

Since copper #12 won't meet a 6 V drop criteria you can either change that criteria, or go to a larger wire.

.

Thank you for the effort. i understand what you mean now. I have a series circuit with a return and NO parallel wires for my setup. Thanks, so I select 1.

Do I have to follow the 210.19 FPN No 4 or is this just recommendation. If I use 3%, that means #6 wire. and 5% would be #8.

 

[jumper]

Do I have to follow the 210.19 FPN No 4 or is this just recommendation. If I use 3%, that means #6 wire. and 5% would be #8.

Nope. FPNs are just recommendations and not enforceable.

Not saying that the recommendation is bad idea, just not code.

Personally, I try to follow that FPN if the budget allows, but that is only an opinion, not code.

 

[goodoboy]

Nope. FPNs are just recommendations and not enforceable.

Not saying that the recommendation is bad idea, just not code.

Personally, I try to follow that FPN if the budget allows, but that is only an opinion, not code.

Thanks, but how much of a voltage drop will stop the end-device from not working, due to the dropping voltage?

 

[RichB]

The depends on the equipment and manufacturers reccomendations--IMHO From what I have read in this thread I would go with a 5% drop and use the #8--5% of 120 volts is 6 volts taking you down to 114 volts--most equipment rated for 120 volts will work fine at 114 volts--again depending on manufacturers reccomendations--I would also reccomend reading up on voltage drop theory and how it works in relation to OCPD and wire sizing and equipment utilization to get a better handle on how and why the calculators and related equipment work--Hope this helps

 

[kwired]

130724-1839 EDT

goodoboy:

The problem is that you are just plugging numbers into an equation without understanding the fundamentals.



.

I have to agree, if you don't understand the fundamentals, how do you expect to get valid results?

Thanks, but how much of a voltage drop will stop the end-device from not working, due to the dropping voltage?

That depends on the characteristics of the load. A heating element just heats a little less when there is low voltage, but may still be enough to get the job done. Other loads may have major problems if voltage is too low.

 

[goodoboy]

The depends on the equipment and manufacturers reccomendations--IMHO From what I have read in this thread I would go with a 5% drop and use the #8--5% of 120 volts is 6 volts taking you down to 114 volts--most equipment rated for 120 volts will work fine at 114 volts--again depending on manufacturers reccomendations--I would also reccomend reading up on voltage drop theory and how it works in relation to OCPD and wire sizing and equipment utilization to get a better handle on how and why the calculators and related equipment work--Hope this helps

Thank you RichB for the help. I plan to read on voltage drop this weekend to gain more understanding. I'm sticking with #8 and will call the mfg to check their recommendations as well.

 

[ggunn]

Thanks for your help. But I am not sure I understand. I need help with what to enter on the voltage drop spreadsheet where it says "select number of parallel wires" or " Number of conductors" on the calculator.

I am thinking to enter 3, because I have 3 wires.

I am confused and lost

When the calculator is asking for the number of parallel conductors, it is asking what number to divide the current by to determine how much is flowing through each of them. If you have 100A through two parallel conductors, the solution is the same as if it were 50A through a single conductor.

 

[goodoboy]

When the calculator is asking for the number of parallel conductors, it is asking what number to divide the current by to determine how much is flowing through each of them. If you have 100A through two parallel conductors, the solution is the same as if it were 50A through a single conductor.

Thanks ggunn,

This makes sense, this is sort of asking if you I have a parallel circuit and if so how many conductors in parallel. For this case, its a straight run.

 

[ggunn]

Thanks ggunn,

This makes sense, this is sort of asking if you I have a parallel circuit and if so how many conductors in parallel. For this case, its a straight run.

It's not sort of the same thing, it's exactly the same thing.

 

[kwired]

For voltage drop, yes when asked number of conductors it is likely asking for number of conductors in parallel.

Otherwise you do multiply length by two for single phase circuit because there is one line out and one line back.

For three phase you multiply by square root of three (for a circuit with balanced loading anyway) because each conductor is carrying that portion of the total load.

 

[goodoboy]

Thank you all very much.

It will take me some time to learn the electrical piratical and theory. Good to have help.

 

[Strathead]

Thanks ggunn,

This makes sense, this is sort of asking if you I have a parallel circuit and if so how many conductors in parallel. For this case, its a straight run.

For terminology sake and to avoid confusion, a parallel circuit and the number of parallel conductors in a phase are not the same thing. A parallel circuit is a circuit where devices are wired in parallel to each other. As in the case with virtually all normal power utilizing pieces of equipment, lighting, receptacles etc. The rare occasion for loads is some lighting such as old christmas lights, airport runway lighting, etc. A switch is wired in series to the lights it controls to contrast. So, a parallel circuit is a circuit where the devices are wired in parallel. When you wire conductors in parallel, you are referring to the number of conductors that originate from the exact same terminal point (electrically) and end at the exact same terminal point. See the difference?