voltge drop

[tacomafc]

When this comes up in a test question I know how to solve for the correct answer. But this is a calculation I need to do in the field. Im not sure what constant to use 10.4 or 12.9.
If anyone could point me in the right direction I would appreciate it. Thank You

 

[bphgravity]

12.9 is the standard for copper at 75?C. Actual K = R x cm / 1000 (R is taken from Table 8 of the NEC)

 

[tacomafc]

I have 16 landscape lighting fixtures spread out along a 600 pipe run. The fixtures are 50 watts so I would say they are .41 amps each. If I calculate the total amps it is 6.5 amps.
Is this correct: VD = 12.9 x 6.5 x 600 x 2 / cma 8 awg 16,510 = 6.09 volts. This will be a violation because it is greater than 3 % witch is 3.6 volts. So I think there should be several switch legs to control the fixtures so the load are reduced. Or should I use a larger circuit and install in line fuses at the fixtures.

 

[dereckbc]

Cut down to 11.1 from 12.9.

I do a lot of data center and telecom design and find 12.9 to be too liberal with the copper and over-kill.

 

[LarryFine]

Remember, the load drops as you go down the circuit; the last length only carries a single light. You won't have the entire load along the entire 600ft.

I'd use a single 15a circuit and stagger the wire sizes. For example, run #10 for the home run and first four lights, #12 for the next six, and #14 for the last six.

I'm just making up nembers here, but you see what I mean. You can actually calculate the current for each section of the circuit.

 

[tacomafc]

Larry that is kind of what I was thinking but I would like to calculate the exact wire size at each location. What are the figures that I would want to use? Would I want to use all 16 fixture for the first box then use 12 fixtures for the second box and so on?

 

[LarryFine]

Larry that is kind of what I was thinking but I would like to calculate the exact wire size at each location. What are the figures that I would want to use? Would I want to use all 16 fixture for the first box then use 12 fixtures for the second box and so on?

Exactamundo. Us the resulting voltage at the first box as the starting voltage for the second run, etc.

 

[tacomafc]

ok so let me give it a try:

TO pull box 1
16 x .41 amps = 6.5
CMA = 12.9 x 6.5 x 200 x 2 / 3.6 vd
CMA = 9,316 # 10awg 10,380

To pull box 2
10 x .41 amps = 4.1
CMA = 12.9 x 4.1 x 200 x 2 / 3.6 vd
CMA = 5,876 #12awg 6,530

To pull box 3 (final box)
4 x .41 amps = 1.64
CMA = 12.9 x 1.64 x 200 x 2 / 3.6 vd
CMA = 2,350 # 14awg 4,110

Im sorry but I realy think I did something wrong here. Should I not use 3.6 and I should look for the vd instead.

 

[suemarkp]

Not to derail your question, but how are these questions asked? There is no code limit for voltage drop. If that is what they're asking, then its a trivial question. If they're asking what size wire to use to limit to 3% Vd, then you're on the right track.

In your last calculation series, you ate up all your voltage drop on the first segment. You'll need to break it up. Say 1.6V to the first pull box, 1V to the second, and 1V to the third.

 

[tonyou812]

Not a violation but enough voltage drop can affect the color of the light. So you will have bulbs giving off diffent shades of light. Ive seen this first hand. But for the most part most people dont even notice.
And your right, Voltage drop is not a code violation.

 

[LarryFine]

Taco, when you said 'first box,' I took it to mean there's a box at every fixture.

The entire (16 x .41a) 6.56a will appear in the home run to the first fixture, the run to the second fixture will start at that voltage, with (15 x .41a) 6.15a load, the third run would have (14 x .41a) 5.74a, the fourth (13 x .41a) 5.33a, etc.

If you only have 3 J-boxes, you'll have to do the same thing, but with the load current connected at each box plus the rest, plus the single-fixture wire length from the box to each fixture (which would only be 0.41a for each one.)

 

[NYHigh]

Remember, the load drops as you go down the circuit; the last length only carries a single light. You won't have the entire load along the entire 600ft.

I'd use a single 15a circuit and stagger the wire sizes. For example, run #10 for the home run and first four lights, #12 for the next six, and #14 for the last six.

I'm just making up nembers here, but you see what I mean. You can actually calculate the current for each section of the circuit.

Just out of curiousity, if you had 120v at the first light, why would u want to up size the wire there? Shouldnt u upsize the wire between the 2 fixtures where the actual voltage drop starts to happen? For instance, if the first 3 lights read 120v, 14awg wire would be fine. If the 4th light reads 110v, then u should start calculating what size wire u will need to run between this, and maintain the correct voltage from there? Isnt this correct?

 

[BAHTAH]

VD for multiple loads

VD for multiple loads

I have 16 landscape lighting fixtures spread out along a 600 pipe run. The fixtures are 50 watts so I would say they are .41 amps each. If I calculate the total amps it is 6.5 amps.
Is this correct: VD = 12.9 x 6.5 x 600 x 2 / cma 8 awg 16,510 = 6.09 volts. This will be a violation because it is greater than 3 % witch is 3.6 volts. So I think there should be several switch legs to control the fixtures so the load are reduced. Or should I use a larger circuit and install in line fuses at the fixtures.

I would use the LCL calculation (Load-Center-Length) to determine the length to use for this calculation. If I assume 50ft to the first light and then 37.5ft to each of the other lights I get a LCL of 331ft. Using 6.56 amps and #8 I get a drop of 2.83% (3.39 volts). As mentioned in another post the NEC does not specify voltaage drop although it woul be nice to actually be able to see the "light" when the installation is complete.

 

[Smart $]

Just out of curiousity, if you had 120v at the first light, why would u want to up size the wire there? Shouldnt u upsize the wire between the 2 fixtures where the actual voltage drop starts to happen? For instance, if the first 3 lights read 120v, 14awg wire would be fine. If the 4th light reads 110v, then u should start calculating what size wire u will need to run between this, and maintain the correct voltage from there? Isnt this correct?

The voltage drop starts at the source (0 volts dropped) then gradually increases as the distance increases. If you use the same size wire throughout the circuit, the rate (of increased voltage drop) lessens after each successive load's current is no longer on the conductors. The only way to change the rate on a typical parallel circuit per section is to change the size of wire.

So as posed, the voltage at the first load will not be 120V (unless you use superconductors :grin. It will have dropped accordingly for the size wire used and the current on that section.

One way to make the voltage drop equal for each load is to run one conductor "untapped" all the way to the end then backtrack it to make the load taps. This method makes the circuit distance across each load equal to the total distance... and of course requires much more and larger wire.

 

[LarryFine]

One way to make the voltage drop equal for each load is to run one conductor "untapped" all the way to the end then backtrack it to make the load taps. This method makes the circuit distance across each load equal to the total distance... and of course requires much more and larger wire.

Another way is to wire them in series. Just gotta be careful about using identical bulbs.