Voltage drop help

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Unbridled

Unbridled

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Happy Sunday evening everyone,
I have a voltage drop calculation question.
I have 5 LED 208V 1 PH Area lights going to be installed in a parking lot 100 ft. apart from each other, the 208V source also being 100 ft. from the first light pole.
Each LED light draws 3 Amps.
I plan on installing #6 Al USE, but not sure which end of the circuit to start the calculation.
Can someone show me the procedure and which end of the circuit to start on?
Thanks in advance
 
iwire

iwire

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Location
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I would do it point to point.

You have 5 segments, if you want to limit voltage drop to say 3% total you can allow about 0.6% drop per segment. So for segment one you would use the current of 5 fixtures and only the length of segment one and come up with a wire size to stay under 0.6% drop. Next do the second segment only using its length and the current of four lamps.

However, there are many ways to do it and you can decide to live with more than 3% or you may even be fine with 8% drop.
 
C

Carultch

Senior Member
Location
Massachusetts
Unbridled said:
Happy Sunday evening everyone,
I have a voltage drop calculation question.
I have 5 LED 208V 1 PH Area lights going to be installed in a parking lot 100 ft. apart from each other, the 208V source also being 100 ft. from the first light pole.
Each LED light draws 3 Amps.
I plan on installing #6 Al USE, but not sure which end of the circuit to start the calculation.
Can someone show me the procedure and which end of the circuit to start on?
Thanks in advance
Click to expand...


Call the final fixture, fixture #1.
Call the fixture nearest the source, fixture#5.

Resistance from fixture 1 to 2 is R1. Corresponding current is 1*I. And the volts Voltage drop, I*R1.
Resistance from fixture 2 to 3 is R2. Corresponding current is 2*I. And the volts Voltage drop, 2*I*R2.
Resistance from fixture 3 to 4 is R3. Corresponding current is 3*I. And the volts Voltage drop, 3*I*R3.
Resistance from fixture 4 to 5 is R4. Corresponding current is 4*I. And the volts Voltage drop, 4*I*R4.
Resistance from fixture 5 to to the source is R5. Corresponding current is 5*I. And the volts Voltage drop, 5*I*R5.

Assuming that you've uniformly spaced each fixture according to circuit length, then all resistances are identical. If not, they depend on length.
Rn = 2*r*Ln/(1000 ft/kft)

where
Rn is one of the above resistances, and n is the number 1 thru 5
r is the ohms/kft from the NEC.
Ln is the length of the portion of the circuit in particular. You need to account for the total length of the circuit, and not just the line-of-sight distances.
Note that the factor of 2 applies for single phase and DC circuits. If this is 3-phase, replace this with sqrt(3).

Add it up:
Vdroptotal = I*(R1 + 2*R2 + 3*R3 + 4*R4 + 5*R5)

Or, in terms of percentage:
%Vdroptotal = I*(R1 + 2*R2 + 3*R3 + 4*R4 + 5*R5)/Vnom

L1 thru L5 all equal 100 ft. The resistance of #6 AL wire is r=0.808 ohms/kft. Thus for single phase, each resistance Rn is 0.1616 Ohms.


%Vdroptotal = 3A*0.1616 Ohms *(1 + 2+ 3 + 4 + 5)/Vnom
%Vdroptotal = 3A*0.1616 Ohms *15/208V
%Vdroptotal = 3.5%
 
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Unbridled

Unbridled

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iwire said:
I would do it point to point.

You have 5 segments, if you want to limit voltage drop to say 3% total you can allow about 0.6% drop per segment. So for segment one you would use the current of 5 fixtures and only the length of segment one and come up with a wire size to stay under 0.6% drop. Next do the second segment only using its length and the current of four lamps.

However, there are many ways to do it and you can decide to live with more than 3% or you may even be fine with 8% drop.
Click to expand...

Thank You
 
Unbridled

Unbridled

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Carultch said:
Call the final fixture, fixture #1.
Call the fixture nearest the source, fixture#5.

Resistance from fixture 1 to 2 is R1. Corresponding current is 1*I. And the volts Voltage drop, I*R1.
Resistance from fixture 2 to 3 is R2. Corresponding current is 2*I. And the volts Voltage drop, 2*I*R2.
Resistance from fixture 3 to 4 is R3. Corresponding current is 3*I. And the volts Voltage drop, 3*I*R3.
Resistance from fixture 4 to 5 is R4. Corresponding current is 4*I. And the volts Voltage drop, 4*I*R4.
Resistance from fixture 5 to to the source is R5. Corresponding current is 5*I. And the volts Voltage drop, 5*I*R5.

Assuming that you've uniformly spaced each fixture according to circuit length, then all resistances are identical. If not, they depend on length.
Rn = 2*r*Ln/(1000 ft/kft)

where
Rn is one of the above resistances, and n is the number 1 thru 5
r is the ohms/kft from the NEC.
Ln is the length of the portion of the circuit in particular. You need to account for the total length of the circuit, and not just the line-of-sight distances.
Note that the factor of 2 applies for single phase and DC circuits. If this is 3-phase, replace this with sqrt(3).

Add it up:
Vdroptotal = I*(R1 + 2*R2 + 3*R3 + 4*R4 + 5*R5)

Or, in terms of percentage:
%Vdroptotal = I*(R1 + 2*R2 + 3*R3 + 4*R4 + 5*R5)/Vnom

L1 thru L5 all equal 100 ft. The resistance of #6 AL wire is r=0.808 ohms/kft. Thus for single phase, each resistance Rn is 0.1616 Ohms.


%Vdroptotal = 3A*0.1616 Ohms *(1 + 2+ 3 + 4 + 5)/Vnom
%Vdroptotal = 3A*0.1616 Ohms *15/208V
%Vdroptotal = 3.5%
Click to expand...

You guys are great!!
 
I

Ingenieur

Senior Member
Location
Earth
Calc the drop for 100' @ 3 A = pu Vd, then;
5 x pu Vd
4 x pu Vd
3 x pu Vd
2 x pu Vd
1 x pu Vd
add them for the total
or 15 x drop for 100' 3 A

Total Vd = (2 x 100/1000 x R ohm/1k ft x 3 A) x 15
= 9 x R ohm/1k ft = 9 x 0.808 = 7.3 V or 7.3/208 ~ 3.5%
 
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junkhound

junkhound

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Renton, WA
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EE, power electronics specialty
re: Each LED light draws 3 Amps.

Is the supply for the LEDs a constant power load or a constant current load or a constant impedance load. They are different, will give a slightly different result.

Note:
a. Commercial LED lights probably a PFC circuit with a constant current output to the LED, which will will mean that the draw on the ac line is close to constant power, but will vary slightly with temperature. If you LED supply is this type, you don't really care about ac voltage drop, the LED poser supply will adjust the LED current by changing eh pulse width of the power supply switching.

b. lower quality LED power supply may be a simple rectifier, in which case the current will change with the voltage slightly depending on the diode temperatures in operation.

Need P/N of LED supply to calculate an accurate voltage drop - and you may not care unless drop over 25 volts!
Then you may have satisfactory operation of the LEDs with 12 or 10 AWG copper! 15 A total, 624 W each, for an accurate set of voltages at each LED power supply you need to draw a mesh circuit and calculate the currents.
 
T

topgone

Senior Member
Carultch said:
Call the final fixture, fixture #1.
Call the fixture nearest the source, fixture#5.

Resistance from fixture 1 to 2 is R1. Corresponding current is 1*I. And the volts Voltage drop, I*R1.
Resistance from fixture 2 to 3 is R2. Corresponding current is 2*I. And the volts Voltage drop, 2*I*R2.
Resistance from fixture 3 to 4 is R3. Corresponding current is 3*I. And the volts Voltage drop, 3*I*R3.
Resistance from fixture 4 to 5 is R4. Corresponding current is 4*I. And the volts Voltage drop, 4*I*R4.
Resistance from fixture 5 to to the source is R5. Corresponding current is 5*I. And the volts Voltage drop, 5*I*R5.

Assuming that you've uniformly spaced each fixture according to circuit length, then all resistances are identical. If not, they depend on length.
Rn = 2*r*Ln/(1000 ft/kft)

where
Rn is one of the above resistances, and n is the number 1 thru 5
r is the ohms/kft from the NEC.
Ln is the length of the portion of the circuit in particular. You need to account for the total length of the circuit, and not just the line-of-sight distances.
Note that the factor of 2 applies for single phase and DC circuits. If this is 3-phase, replace this with sqrt(3).

Add it up:
Vdroptotal = I*(R1 + 2*R2 + 3*R3 + 4*R4 + 5*R5)

Or, in terms of percentage:
%Vdroptotal = I*(R1 + 2*R2 + 3*R3 + 4*R4 + 5*R5)/Vnom

L1 thru L5 all equal 100 ft. The resistance of #6 AL wire is r=0.808 ohms/kft. Thus for single phase, each resistance Rn is 0.1616 Ohms.


%Vdroptotal = 3A*0.1616 Ohms *(1 + 2+ 3 + 4 + 5)/Vnom
%Vdroptotal = 3A*0.1616 Ohms *15/208V
%Vdroptotal = 3.5%
Click to expand...

May we know which wire manufacturer did you get your 0.808 ohms/Kft from? It seems higher than what Okonite lists! (0.674 ohms/1000 ft.)
 
Smart $

Smart $

Esteemed Member
Location
Ohio
topgone said:
May we know which wire manufacturer did you get your 0.808 ohms/Kft from? It seems higher than what Okonite lists! (0.674 ohms/1000 ft.)
Click to expand...
Chapter 9, Table 8... at 75°C.

The value from Okonite is at 25°C (per http://okonite.com/okonite-wire-and-cable-product-brochures/Engineering-Handbook.pdf, page 3). Look at the correction factors on page 4.

This is also the one area of voltage drop calculations where it seems hardly anyone ever compensates for the conductor temperature...
 
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