Voltage Drop for street lights

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Bob NH

Senior Member
RUWired said:
Bob, In my earlier post i said that the wire sizes seemed a little high for being so close to the source.I used 3 different methods to figure vd and came up ith the same results.Method 1 was using the 2kli/vd=cm. Method 2 was using your theory,but plugging in different numbers, and method 3 was using the online calculator. The online calculator matched your theory but plugging in different numbers.See if this table looks right.

footage,amps,max vd,ohms,ohms/kft,awg,ohms/kft,multiplier,vd,calculator
1200,-- 1.66,- 7.2,-- 4.33,-- 1.8,--- 10,-- 1.2,----- 2.88,- 4.78,- 4.9.--
1100,-- 3.22,- 7.2,-- 2.16,-- .98,--- 8, -- .78,------1.71,--5.67,--5.7,--
1000,-- 4.98,- 7.2,-- 1.44,-- .72,--- 6,--- .49,------.98,---4.88,--4.9,--
900,--- 6.66,- 7.2,-- 1.08,-- .60,--- 6,--- .49,------.88,---5.86,--5.9,--
800,--- 8.33,- 7.2,--- .86,-- .54,--- 6,--- .49,------.78,---6.49,--6.5,--
700,--- 9.96,- 7.2,--- .75,-- .53,--- 6,--- .49,------.68,---6.83,--6.9,--
600,---11.62,- 7.2,---.61,-- .51,--- 6,--- .49,------.58,---6.83,--6.9,--
500,---13.33,- 7.2,---.54,-- .54,,---6,--- .49,------.49,---6.53,--6.6,--
400,---15.00,- 7.2,---.48,-- .60,---6,----.49,------.39,---5.85,--5.9,--
300,---16.66,- 7.2,---.43,-- .71,---6,----.49,------.29,---4.83,--4.9,--
200,---18.33,- 7.2,---.39,-- .97,---8,----.78,------.31,---5.68,--5.8,--
100,---20.00,- 7.2,---.36,-- 1.8,---10,---1.2,------.24,---4.8,---5.2,--
column1=footage of light bases
column2=amps at each base(coming back to the source)
column3=maximum voltage drop
column4=ohms(vd divided by amps)
column5=ohms/Kft(ohms divided by total footage(2-legs)x 1000)
column6=awg per table 9 pvc pipe
column7=ohms/Kft(per table 9)
column8=Multiplier(ohms/kft divided by 1000 x footage(2-legs)
column9=VD((multiplier x amps)
column10=Vd using online calculator using 75 degree c
Rick

My analysis calculated the total voltage drop at the end of 1200 ft at 7.2 Volts or less.

Your calculation appears to be calculating a voltage drop of 4.8 to 6.83 Volts per 100 ft section.

It is apparent that you are doing things differently than I was where your segment that has 20 Amps has 4.8 Volts drop. If you are trying to get 3% voltage drop for the circuit, which I was trying to do, you need to use larger wire.

Check your total voltage drop by using the calculator to add all of the voltage drops in each section.
 

RUWired

Senior Member
Location
Pa.
Bob, My voltage drops listed are the total voltage drop up to that point. For instance, if i ran a #6 up to the 1000' mark with the 4.98amp load i would have a total drop of 4.88. Checking the online calculator confirms it with a 4.9.Likewise if i check the 300' mark with a 16.66 amp load i get a total drop 4.83 using a # 6 awg.Doing random checks at seperate points double checks my math for total drop at that point.
Rick
 

Smart $

Esteemed Member
Location
Ohio
ramsy said:
Smart, I believe a length multiplier a=1, rather than a=2 has understated your 0.14 voltage drop, from 0.29 ohms.

STD V-drop models ask for 1-way distance, then double that for total circuit length. This total is also adjusted by 0.866 for 3? loads, and this 2x distance conversion * 0.866 = (1.732 times 1-way distance).

Look at the graphic below and tell me if these 1?, L-to-L lights should get a distance adjustment of a=2 or a=1.732?
866 Demand Factor.gif
A length multiplier a=1 is correct for this calculation, as it is set up. Standard voltage drop models use one-way distance then a multiplier of 2 or 1.732 assuming only one [balanced, in the case of a 3? circuit] load on the circuit. In reality, there is quite often more than one load on the circuit and for a 3? circuit, load currents are not the same as line currents for multiple single-phase utilization of the same lines.

Voltage drop is a result of line current. Therefore my VD calculation as depicted earlier uses point-to-point line currents to determine voltage drop. For example, there is a 0.14 voltage drop for the last light on both Line A and Line C. This yields a total 0.28 voltage drop from fixture 11 to fixture 12. Additionally, compensation for being a 3? circuit is made when line currents are calculated. For instance, there are three (3) single-phase lighting loads?fixtures 10, 11, and 12? of 0.98A each on the wire segments from fixture 9 to fixture 10. The line currents are 0.98A x √3 (or 1.732) = 1.697A per line. My "Line _ Current" columns show 1.698A because it is calculating to maximum precision in the background. I generally do not set my SS's for "Precision as displayed".
 

RUWired

Senior Member
Location
Pa.
Bob,I see what you mean now.You only are dropping .6 volts at each base along the way so at the end 7.2 has only been dropped .Sorry for the confusion.
Rick
 

Smart $

Esteemed Member
Location
Ohio
ramsy said:
This total is also adjusted by 0.866 for 3? loads...
After further evaluation, perhaps the issue you put forth has merit. However, if there is an error in my calculation, 1) it would reduce the calculated voltage drop, at least across some segments of the circuit, and 2)I'm not sure how to correctly calculate such voltage drop. Looking for ideas...

First, we know in a balanced 3? load,
VDCIRCUIT = VDLINE ? √3

3VD.gif


So how do we calculate voltage drop on a 3? circuit when 1? loads are distributed at different lengths from the source?

3VDsegmented.gif
 
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ramsy

Roger Ruhle dba NoFixNoPay
Location
LA basin, CA
Occupation
Service Electrician 2020 NEC
If your length adjustment a=1 requires using 1.732 or 2 somewhere else, this brave new convention may be brilliant, but perhaps more difficult to prove without peer review or IEEE consideration, which currently favors this adjustment at VD formula length.

It may be easier to design the circuit for STD models first, then compare new formats against conventional results. Below, I divided the circuit among 3 phases, using STD 1?,2w methods, and compared the Cost Effect against 2 phases. I chose Ta=21c,.assuming engineering supervision is not available.

Code:
480v 1?,2w  a=2  Ta=21c  4800W/3?	Voltage Drop
Lamp#	Watts/PF=VA	Feet	AWG#	Ab	Bc	Ca
12	0400/.85=0471	300	12			0.84
11	0400/.85=0471	300	12		0.84
10	0400/.85=0471	300	12	0.84
09	0800/.85=0941	300	12			1.69
08	0800/.85=0941	300	12		1.69
07	0800/.85=0941	300	12	1.69
06	1200/.85=1412	300	12			2.53
05	1200/.85=1412	300	12		2.53
04	1200/.85=1412	300	12	2.53
03	1600/.85=1882	300	12			3.38
02	1600/.85=1882	200	12		2.26
01	1600/.85=1882	100	12	1.13
	====     ====	====	====	====	====	====
Total	1600     1882	1200*3		6.19	6.56	7.68 Volts
Cost Effect= 23,508 kcm/ft + 3 CB	1.3%	1.4%	1.6%

480v 1?,2w  a=2  Ta=21c  4800W/1?	Voltage Drop
Lamp#	Watts/PF=VA	Feet	AWG#	Ab
12	0400/.85=0471	100	14	0.43
11	0800/.85=0941	100	14	0.86
10	1200/.85=1412	100	14	1.30
09	1600/.85=1882	100	14	1.73
08	2000/.85=2353	100	14	2.18
07	2400/.85=2824	100	12	1.70
06	2800/.85=3294	100	12	1.99
05	3200/.85=3765	100	12	2.28
04	3600/.85=4235	100	12	2.58
03	4000/.85=4706	100	10	1.72
02	4400/.85=5176	100	10	1.90
01	4800/.85=5647	100	10	2.08
	====     ====	====	====	====
Total	4800     5647	1200*2		20.75 Volts
Cost Effect= 15,362 kcm/ft + 2 CB	4.3%
 
sceepe said:
Anybody got a good Vdrop calculator for a string of light poles. I have used a simple calculator I made. However, it assumes the load is at the end of the wire. e.g. 1200' long circuit, with 400w load every 100' I need a good way to figure out what size wire I need. Assuming 4800 watts at the end of 1200' wire is a bit of overkill.

http://www.electricwest.com/files.php has a Series Voltage Drop Calculator.xls. I use this expanded out to cover multiple branch circuits serving street lights

As typically street lighting I do is fed from service pedestals I calculate for a maximum of 5% instead of 3%.
 
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Smart $

Esteemed Member
Location
Ohio
ramsy said:
If your length adjustment a=1 requires using 1.732 or 2 somewhere else, this brave new convention may be brilliant, but perhaps more difficult to prove without peer review or IEEE consideration, which currently favors this adjustment at VD formula length.
The current std 3?-circuit VD formula does not account for staggered 1? loads. Call me unconventional, but I would like a formula having greater accuracy.

I believe I have determined the basis for an adjusted formula. The current std formula uses the following vectorial diagram:

3VDvector.gif


Note the conductor voltage drop (red lines with green nodes) follows the angle of the current—which is at an angle to the applied voltage. Therefore the voltage drop would be:
VDCIRCUIT = 2 ? VDLINE ? cos 30? = VDLINE ? √3

The following is a vectorial diagram adjusted for 3 staggered, equally-spaced 1? loads along 3? circuit conductors:

3VDsegmentedvector.gif


Again, note the conductor voltage drop follows the angle of the current, somewhat. However, in using this model, it appears the voltage angle has changed on the B-C and C-A loads. I'm not certain if this in fact occurs. In addition, for multiple sets of staggered loads along the circuit, the angle of the current is not influenced as much near the source end of the circuit as it is at the end of the circuit. My theory needs further evaluation!
 
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ramsy

Roger Ruhle dba NoFixNoPay
Location
LA basin, CA
Occupation
Service Electrician 2020 NEC
Smart $ said:
The current std 3?-circuit VD formula does not account for staggered 1? loads. Call me unconventional, but I would like a formula having greater accuracy.

Yes, I agree. While it's not the STD formula's fault, using the same line voltage completely ignores the cumulative voltage drops in each segment. Accumulating prior current also ignores the higher impedances between segments.

I believe I have determined the basis for an adjusted formula.

If we want to test adjusted models against a proper baseline for segments, we really should use a metered circuit, carefully calibrated & controlled for reactance and power factor.

From now on, attempting to join segments with the STD formula should at least adjust each segment voltage and current for cumulative losses.
 

Smart $

Esteemed Member
Location
Ohio
ramsy said:
Yes, I agree. While it's not the STD formula's fault, using the same line voltage completely ignores the cumulative voltage drops in each segment. Accumulating prior current also ignores the higher impedances between segments.
Exactly.

ramsy said:
If we want to test adjusted models against a proper baseline for segments, we really should use a metered circuit, carefully calibrated & controlled for reactance and power factor.
Currently I'm using circuit simulation software to perform the analysis. I have already verified the voltage drop does in fact follow the angle of the current, and the voltage angle does change across loads where the current is unbalanced in the preceding segment. Unfortunately the "freeware" I'm using only has a plotted waveform output so I'm having to extract the numerical data and angle relationships manually :mad:

As the basic std formula does not take reactance and power factor into account, my first steps are using resistance only. Once a basic adjusted formula is developed, I'll most likely simulate the circuit with reactances to see if there is a variation in voltage drop due to power factor outside that which is adjusted for by the enhanced VD formula I use now.

It would be nice to have a field model to record actual values for comparison to the "paper" model. I don't have one readily accessible :roll:

ramsy said:
From now on, attempting to join segments with the STD formula should at least adjust each segment voltage and current for cumulative losses.
That's the idea. Perhaps it can even be expanded upon for mid-circuit branches :smile:
 

ramsy

Roger Ruhle dba NoFixNoPay
Location
LA basin, CA
Occupation
Service Electrician 2020 NEC
Smart $ said:
It would be nice to have a field model to record actual values for comparison to the "paper" model. I don't have one readily accessible :roll:

It appears 3 notable field studies were already published:
1) IEEE Std 399-1997, IEEE Recommended Practice for Industrial and Commercial Power Systems Analysis.

2) “Power System Control and Stability” by P.M. Anderson and A.A. Fouad.

3) “Computer Aided Power System Operation and Analysis” by R.N Dhar.

While IEEE pubs. are subscription only, it seemed we might have to check at local libraries, until I found some of this was made available to the public domain in the form of comparison to ETAP load-flow results. (PDF files)

If we can find the relevant study results among all these comparisons, ETAP_ComparisonResults.pdf, nothing is stopping us from using it to model our own application.

Smart $ said:
That's the idea. Perhaps it can even be expanded upon for mid-circuit branches :smile:

That would be nice, and practical.
 
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voltage drop calculations for street light segments

voltage drop calculations for street light segments

I have looked over the calculations for voltage drop for street lights with source voltage of 240v with equal segments and all lights listed at 400W as mentioned in the previous posts. My question now is for 120v source with unequal segment spacing and lights that are 100w and 200w. Can someone show me an illustration on how to calculate the Voltage drop? In this situation 12ov is th only option in my case.
 

Smart $

Esteemed Member
Location
Ohio
kasey_young said:
I have looked over the calculations for voltage drop for street lights with source voltage of 240v with equal segments and all lights listed at 400W as mentioned in the previous posts. My question now is for 120v source with unequal segment spacing and lights that are 100w and 200w. Can someone show me an illustration on how to calculate the Voltage drop? In this situation 12ov is th only option in my case.
Welcome to the forum, kasey.

Perhaps the following illustration will do...?

1VDsegmented.gif


Essentially, you will have to calculate the VD for each segment, then add them all together. In calculating the VD for each segment, you will have to use the cumulative current (I) at that stage of the circuit, and the actual length (L) of the segment. The illustration depicts the basic VD formula. It does not include any adjustments for conductor or ambient temperature, or power factor. The resistance value (R) can be obtained in Chapter 9, Table 9 as it varies with the size of the wire. Table 9 notes also contain a little information on using an impedance value (Z) instead of resistance. Let me know if you wish to use a more enhanced formula, which includes the adjustments noted above.

If you want to approximate the size wire needed:
cmil = 2 ? K ? I ? L ? VD​
where cmil is the conductor size in circular mils, K = 11 or 12, 50% or less loaded and 51-100% loaded, respectively, I is the total circuit current, L is the [one-way] total circuit length, and VD is the target voltage drop.

In being limited to 120V, are you limited to a one-pole circuit also? I'm asking because if you can use a two-pole common neutral circuit, you could effectively cut your voltage drop to nearly half that of a one-pole circuit, other things being equal. All the lights must reasonably be expected to always be on and off concurrently.

EDITED to add "[one-way]"
 
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Smart $

Esteemed Member
Location
Ohio
ramsy said:
...That would be nice, and practical.
In the meantime ;) a shortcut for approximating voltage drop for evenly distributed equal loads (using same gauge wire throughout the run of the circuit) is to use a length (L) in the std vd formula that is at the midpoint of the loads. Nothing astounding, but it seems to work quite well...

I'll use the example of a 3? circuit with 12 lights operating at 480V, 1?, 400W, 0.85 PF spaced every 100 ft and balanced on the three lines and staggered A-B, B-C, and C-A, repeating four times. We have:

Code:
Fixture No.	Distance	Source	I_A-B	I_B-C	I_C-A
12		1200		C-A			0.980
11		1100		B-C		0.980	
10		1000		A-B	0.980		
9		900		C-A			0.980
8		800		B-C		0.980	
7		700		A-B	0.980		
6		600		C-A			0.980
5		500		B-C		0.980	
4		400		A-B	0.980		
3		300		C-A			0.980
2		200		B-C		0.980	
1		100		A-B	0.980
Calculate line current. In this case...
I_LINE A, B, and C = 4 ? 0.980 ? √3 = 6.792A​
Determine the load midpoint distance for each line of the last load...
L_LINE A = (100' + 1200') ? 2 = 650'
L_LINE C = (200' + 1200') ? 2 = 700' \\The first Line C load is the B-C load at 200'.
L_AVERAGE = (650' + 700') ? 2 = 675'​
Use the std voltage drop formula with an R of 2 for 12 AWG:
VD = √3 ? L ? I ? R ? 1000
VD = √3 ? 675' ? 6.792A ? 2 ? 1000 = 15.882V... ? 480 = 3.309%​
These calculations do not include adjustments for reactance or temperature. Using an Excel spreadsheet to calculate the VD for each segment, using the same VD formula, I get a summed result of 3.333%.

The reason for considering this approach is that in attempting to develop a more accurate formula for voltage drop in these types of circuits has indicated it will involve, by comparison, a quite complex calculation...
 

ramsy

Roger Ruhle dba NoFixNoPay
Location
LA basin, CA
Occupation
Service Electrician 2020 NEC
Smart, rather than ask to check your references on that 3.333% baseline, I've created a new thread using the only voltage drop example I have, Validated & Verified (V&V) against IEEE and several other standards.

It appears both of our earlier models only matched this V&V example using 420 Amps rather than the 400 provided. I suspect this ETAP example has done something important to the amperage, which we both are missing.
 
voltage drop for 120v. single pole circuit street lights

voltage drop for 120v. single pole circuit street lights

hi , thanks for the reply to my previous question. I wanted to clarify my question from previous post. Basically what I am trying to find is the formula used to find the unknown values of current, voltage, resistance and voltage drop at each segment of street lights. These values change as the distances increase and are not constant values. k factor is constant. I have tried to follow the other illustrations on the street light circuit with 240v source but can't seem to use the same method to make my calculations in my situation.
Excuse me for my ignorance so I will give my scenario as follows:

7 street lights distance between light poles:
pole#1- 200W source to pole#1-100'
pole#2-100W pole#1 to pole#2-275'
pole#3-100W pole#2 to pole#3-300'
pole#4-100W pole#3 to pole#4-320'
pole#5-200W pole#4 to pole#5-290'
pole#6-100W pole#5 to pole#6-315'
pole#7-200W pole#6 to pole#7-300'

I know at this point that my last segment run to EOL will be #10thw cu.,
already stipulated is the single pole circuit of 120v. Starting from scratch,how would you figure all of the unknown values at each segment in order to size your conductors from source to pole#6 ? I know it involves the use of ohms/kft but error at that point in my calculation. basically, I would like to see the formula in this scenario illustrated so that I can correctly calculate the Vd for street lights , I know I can use software for the answers but really interested in how to calculate it myself. Thanks again, really appreciate any input.
 

RUWired

Senior Member
Location
Pa.
Casey, what you are missing is figuring the Vd per foot, and Vd for each segment.If you take the total Vd (3.6@ 120v),divided by your total footage(1900'),you will get the Vd per foot. In your example, the distance between the last 2 poles is 300' and the amperage is 1.66 (200w).Vd per foot is .00189474. 300' x .00189474 =.5684 Vd per segment.You can then use the ohm's per kft formula or the 2kli/vd to find circular mils.( i come up with a #6 for the last run).
Rick
 

Bob NH

Senior Member
kasey_young said:
I wanted to clarify my question from previous post. Basically what I am trying to find is the formula used to find the unknown values of current, voltage, resistance and voltage drop at each segment of street lights. These values change as the distances increase and are not constant values. k factor is constant. I have tried to follow the other illustrations on the street light circuit with 240v source but can't seem to use the same method to make my calculations in my situation.
I will give my scenario as follows:

7 street lights distance between light poles:
pole#1- 200W source to pole#1-100'
pole#2-100W pole#1 to pole#2-275'
pole#3-100W pole#2 to pole#3-300'
pole#4-100W pole#3 to pole#4-320'
pole#5-200W pole#4 to pole#5-290'
pole#6-100W pole#5 to pole#6-315'
pole#7-200W pole#6 to pole#7-300'

I know at this point that my last segment run to EOL will be #10thw cu.,
already stipulated is the single pole circuit of 120v. Starting from scratch,how would you figure all of the unknown values at each segment in order to size your conductors from source to pole#6 ? I know it involves the use of ohms/kft but error at that point in my calculation. basically, I would like to see the formula in this scenario illustrated so that I can correctly calculate the Vd for street lights , I know I can use software for the answers but really interested in how to calculate it myself. Thanks again, really appreciate any input.

There isn't a single formula.

1. Draw a diagram of your wiring from source to pole to pole . . . . to the end.

2. Calculate the current in each segment of wire from source to the end. That is independent of wire size. If it is a single pole 120 Volt circuit then you must calculate the current in both the feed and the neutral.

3. Decide your criteria for voltage drop (how many volts).

4. If the neutral and supply are the same size, then half of the drop will be in the neutral and half in the hot wire.

5. The voltage on the neutral at the end of the line will be the total voltage drop in the neutral.

6. The voltage on the hot wire at the end of the line will be the voltage at the source minus the voltage drop in the hot wire.

Use any of the formulas or table for calculating voltage drop in each segment.

I suggest you set it up in a spreadsheet.
 

acvolts

Member
Location
New York
Voltage Drop Calculation

Voltage Drop Calculation

Assuming 400W fixtures (HID lamp source) with a Power Factor of 0.95 and Efficiency of 0.95, an ambient temperature of 86 degrees, XHHW conductors in Schedule 40 PVC Raceway, 240V, 1P, 1200 feet long with fixture spaces 100 feet apart, I calculate the following results: Pole #1, 0.93V, 0.39 VD%, cdtr size #3, Pole #2, 1.79V, 0.75% VD, cdtr size #3, Pole #3, 2.58V, 1.08% VD, cdtr size #3, Pole #4, 3.94V, 1.65% VD, cdtr size #3, Pole #5, 4.51V, 1.89% VD, cdtr size #3, Pole #6, 5.01V, 2.10% VD, cdtr size #3, Pole #7, 5.53V, 2.32% VD, cdtr size #4, Pole #8,5.97V, 2.50% VD, cdtr size #4, Pole #9, 6.32V, 2.65% VD, cdtr size #4, Pole #10, 6.58V, 2.76% VD, cdtr size #4, Pole #11, 6.75V, 2.83% VD, cdtr size #4, Pole #12, 7.07V, 2.96% VD, cdtr size #10. The actual voltage at the end of the circuit is calculated at 232.93V. The calculation was performed using VOLTS form Dolphins-Software, which calculated the results using the IEEE Standard 141 Exact Formulae including ambient temperature considerations. The calculation was performed in less than one minute. If fact, it took longer to type this response than it did to perform the calculation.
 

sparkease

Member
tion, 208 volt street lightning circuit 600 foot length of run. Do I figure voltage drop at 1200 feet this is single phase . There are 18 lamps at 1.10 amps each.
 
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