I am doing this right (series voltage drop calculation)

[hidroela]

Ok I got this scenario 120/208V 3Φ 4 wire
Panel ----------------------------light---------------------light---------------------light
|------100 ft-----------------------|---------50 ft ----------|---------50 ft ----------|
------------------------------------- 8 A ---------------------- 8 A ---------------------- 8 A

And these are my calculations
Branch #1
VD = (√3/2 * K * load * length) / cmil
VD = (√3/2 * 12.9 * 24 *100) / 10380
VD = 2.58V
Voltage at load # 1 =117.42 V
VD% = 2.15%
Branch #2
VD = (√3/2 * K * load * length) / cmil
VD = (√3/2 * 12.9 * 16 *50) / 10380
VD = 0.86V
Voltage at load # 2 =116.56 V
VD% = 0.72%
Branch #3
VD = (√3/2 * K * load * lenght) / cmil
VD = (√3/2 * 12.9 * 8 *50) / 10380
VD = 0.43V
Voltage at load # 3 =116.13 V
VD% = 0.36%
Total Voltage Drop = 3.225%

i am in the right track
Best regards

 

[gar]

110521-1059 EDT

I have no idea. You have not clearly defined your question.

If you do not know the derivation of your equation and how it relates to the problem, then your result may be quite wrong.

By implication you have a 3 phase Y source. Further you are connecting loads from line to neutral. I have no idea what the loads are on the three phases.

Where does sq-root of 3 divided by 2 come from, where does K come from and what assumptions, what is length, where does voltage enter in the equation? Just blindly plugging numbers into an equation can produce quite wrong results..

In its most basic form and assuming only resistance, then Vd = I*R. You really need to find out what this R is, and define what Vd is.

Maybe your calculations are correct, but I classify them as wrong.

A quick analysis for a single phase circuit with 120 V as the source, #10 copper wire, and assuming 1 ohm/1000 ft for #10 copper, then the resistance produces an approximate voltage drop of 0.2*8 = 1.6 volts to the first load with no others connected. If each of the other loads are constant current loads, which likely they are not, then the voltage drop to the first load is 4.8 V. Map this to your three phase circuit and if somehow the neutral current was balanced out, then the drop to the first would be 2.4 V .

You should understand the value of K is a function of the absolute temperature of the wire as well as other factors. The temperature of the wire is a function of the ambient temperature and the current thru the wire.

.

 

[hidroela]

sorry but i thought the diagram was explicit enough

all i want is calculate the voltage drop on a circuit where 3 lamps 120/208v multi tap are feed as show
first lamp 8A 100ft from the panel
second 8a 50ft from the first one
third 8A 50ft from second
all on the same 10-AWG conductor

Best regards

 

[dkarst]

sorry but i thought the diagram was explicit enough

all i want is calculate the voltage drop on a circuit where 3 lamps 120/208v multi tap are feed as show
first lamp 8A 100ft from the panel
second 8a 50ft from the first one
third 8A 50ft from second
all on the same 10-AWG conductor

Best regards

What isn't clear to me is whether this is just 3 phase system with all the lights tied from a single phase to neutral? If it is and you sketch an equivalent circuit, you don't need any √3 and I think it will become clear.

 

[hidroela]

so it would be like this

Branch #1

VD = (2 * K * load * length) / cmil

VD = (2 * 12.9 * 24 * 100) / 10380

VD = 5.97V

Voltage at load # 1 =114.03 V

VD% = 4.98%

-----------------------
Branch #2

VD = (2 * K * load * length) / cmil

VD = (2 * 12.9 * 16 * 50) / 10380

VD = 1.99V

Voltage at load # 2 =112.04 V

VD% = 1.66%

----------------------------------

Branch #3

VD = (2 * K * load * length) / cmil

VD = (2 * 12.9 * 8 * 50) / 10380

VD = 0.99V

Voltage at load # 3 =111.05 V

VD% = 0.83%

------------------------------

Total Voltage Drop = 7.46%

Best regards

 

[gar]

110521-2125 EDT

hidroela:

Neither your circuit or description makes the question clear.

If you specified that you had a single phase circuit, with its own neutral (not shared with anything else), and with a source voltage of 120 V, then what dkarst described is approximately correct.

Where there is some small error is that in the real world K varies from one load segment to another because of the change of current, also ambient temperature changes the value of K, and each of the load currents are different because each of the loads has a different voltage.

The NEC value for K is based upon an assumptions of wire absolute temperature that should be an upper limit for most applications. So using that value of K will over estimate the voltage drop by a small amount.

What is not clear is whether you understand how the voltage drop equation is derived, and whether you could derive it. Without an understanding you may not be able to properly apply the equation.

.

 

[mivey]

hidroela,

You are starting to get on the right track but let's go back to the basics so you can understand what is going on with the formula you are using.

Some background info:

To start with, you are using a formula with a K-value (ohms*cmil/ft) for wire at 75?C(167?F) so don't forget to adjust for whatever temperature you are using for your wire. Also, the K-values normally used are 12.9 for copper and 21.2 for aluminum. These are roughly the highest DC resistances based values you will get for the wires we use. The other variables that come into play besides temperature are bigger wire sizes and conduit type. Because we have AC and not DC, we also consider the conduit type (PVC, Aluminum, Steel) and skin effect (significant in the larger wire sizes). So for AC, we not only consider resistance but reactance as well so we look at the wire's impedance, Z.

For example, here is what I calculate for K at 75?C:
AWG,kcmil-strands | Z_CU in PVC | Z_CU in Al | Z_CU in St | Z_AL in PVC | Z_AL in Al | Z_AL in St | R_CU (dc) | R_AL (dc)
18-1 | 12.59 | 12.59 | 12.59 | 20.74 | 20.74 | 20.74 | 12.59 | 20.74
18-7 | 12.88 | 12.88 | 12.88 | 21.22 | 21.22 | 21.22 | 12.88 | 21.22
16-1 | 12.62 | 12.62 | 12.62 | 20.77 | 20.77 | 20.77 | 12.62 | 20.77
16-7 | 12.88 | 12.88 | 12.88 | 21.18 | 21.18 | 21.18 | 12.87 | 21.18
14-1 | 12.74 | 12.74 | 12.74 | 20.80 | 20.80 | 20.80 | 12.62 | 20.80
14-7 | 12.74 | 12.74 | 12.74 | 21.25 | 21.25 | 21.25 | 12.91 | 21.25
12-1 | 13.06 | 13.06 | 13.07 | 20.90 | 20.90 | 20.90 | 12.60 | 20.77
12-7 | 13.06 | 13.06 | 13.07 | 20.90 | 20.90 | 20.90 | 12.93 | 21.22
10-1 | 12.47 | 12.47 | 12.47 | 20.77 | 20.77 | 20.77 | 12.56 | 20.76
10-7 | 12.47 | 12.47 | 12.47 | 20.77 | 20.77 | 20.77 | 12.87 | 21.18
8-1 | 12.91 | 12.91 | 12.92 | 21.48 | 21.48 | 21.49 | 12.61 | 20.80
8-7 | 12.91 | 12.91 | 12.92 | 21.48 | 21.48 | 21.49 | 12.84 | 21.13
6-7 | 12.93 | 12.93 | 12.97 | 21.30 | 21.30 | 21.32 | 12.88 | 21.20
4-7 | 13.09 | 13.09 | 13.18 | 21.38 | 21.38 | 21.43 | 12.86 | 21.20
3-7 | 13.39 | 13.39 | 13.52 | 21.19 | 21.72 | 21.28 | 12.89 | 21.21
2-7 | 12.96 | 13.60 | 13.80 | 21.44 | 21.44 | 21.57 | 12.87 | 21.17
1-19 | 13.13 | 13.93 | 14.21 | 21.27 | 22.10 | 21.46 | 12.89 | 21.17
1/O-19 | 13.50 | 14.49 | 13.94 | 21.63 | 22.66 | 21.90 | 12.88 | 21.23
2/O-19 | 14.49 | 14.49 | 15.13 | 22.05 | 22.05 | 22.48 | 12.87 | 21.16
3/O-19 | 14.72 | 15.46 | 15.87 | 22.92 | 22.92 | 23.49 | 12.85 | 21.14
4/O-19 | 15.73 | 16.62 | 17.15 | 22.87 | 24.84 | 23.75 | 12.87 | 21.16
250-37 | 16.55 | 17.55 | 18.74 | 23.59 | 24.72 | 25.12 | 12.88 | 21.18
300-37 | 18.04 | 19.17 | 20.40 | 24.60 | 25.91 | 26.47 | 12.87 | 21.21
350-37 | 19.31 | 20.55 | 22.19 | 25.53 | 27.01 | 28.15 | 12.85 | 21.18
400-37 | 20.74 | 22.07 | 24.09 | 26.88 | 28.51 | 29.46 | 12.84 | 21.16
500-37 | 23.72 | 25.22 | 28.04 | 29.03 | 30.92 | 32.90 | 12.90 | 21.20
600-61 | 27.17 | 28.81 | 32.47 | 31.85 | 33.95 | 36.73 | 12.84 | 21.18
700-61 | 30.44 | 32.22 | 37.03 | 34.63 | 36.94 | 40.81 | 12.88 | 21.21
750-61 | 31.86 | 33.71 | 39.29 | 35.85 | 38.24 | 42.86 | 12.83 | 21.15
800-61 | 33.49 | 35.30 | 41.36 | 37.41 | 39.77 | 44.79 | 12.88 | 21.20
900-61 | 36.69 | 38.42 | 45.38 | 40.45 | 42.74 | 48.55 | 12.87 | 21.15
1000-61 | 39.92 | 41.59 | 49.40 | 43.57 | 45.80 | 52.35 | 12.90 | 21.20
1250-91 | 48.57 | 50.07 | 60.39 | 51.89 | 53.96 | 62.84 | 12.88 | 21.13
1500-91 | 56.59 | 57.99 | 69.95 | 59.73 | 61.70 | 72.09 | 12.87 | 21.15
1750-127 | 64.58 | 65.89 | 79.31 | 67.56 | 69.45 | 81.21 | 12.86 | 21.18
2000-127 | 72.53 | 73.78 | 88.52 | 75.39 | 77.21 | 90.22 | 12.86 | 21.20


Here are the values at 48.89?C(120?F):
AWG,kcmil-strands | Z_CU in PVC | Z_CU in Al | Z_CU in St | Z_AL in PVC | Z_AL in Al | Z_AL in St | R_CU (dc) | R_AL (dc)
18-1 | 11.53 | 11.53 | 11.53 | 18.95 | 18.95 | 18.95 | 11.53 | 18.95
18-7 | 11.79 | 11.79 | 11.79 | 19.39 | 19.39 | 19.39 | 11.79 | 19.39
16-1 | 11.55 | 11.55 | 11.55 | 18.98 | 18.98 | 18.98 | 11.55 | 18.98
16-7 | 11.79 | 11.79 | 11.79 | 19.36 | 19.36 | 19.36 | 11.79 | 19.36
14-1 | 11.67 | 11.67 | 11.67 | 19.01 | 19.01 | 19.01 | 11.55 | 19.00
14-7 | 11.67 | 11.67 | 11.67 | 19.42 | 19.42 | 19.42 | 11.82 | 19.42
12-1 | 11.96 | 11.96 | 11.97 | 19.10 | 19.10 | 19.10 | 11.54 | 18.98
12-7 | 11.96 | 11.96 | 11.97 | 19.10 | 19.10 | 19.10 | 11.84 | 19.39
10-1 | 11.42 | 11.42 | 11.42 | 18.98 | 18.98 | 18.98 | 11.50 | 18.97
10-7 | 11.42 | 11.42 | 11.42 | 18.98 | 18.98 | 18.98 | 11.79 | 19.35
8-1 | 11.82 | 11.82 | 11.84 | 19.63 | 19.63 | 19.64 | 11.55 | 19.01
8-7 | 11.82 | 11.82 | 11.84 | 19.63 | 19.63 | 19.64 | 11.76 | 19.31
6-7 | 11.85 | 11.85 | 11.89 | 19.47 | 19.47 | 19.50 | 11.80 | 19.38
4-7 | 12.02 | 12.02 | 12.11 | 19.56 | 19.56 | 19.61 | 11.77 | 19.38
3-7 | 12.30 | 12.30 | 12.44 | 19.39 | 19.87 | 19.48 | 11.80 | 19.38
2-7 | 11.92 | 12.51 | 12.73 | 19.63 | 19.63 | 19.77 | 11.79 | 19.34
1-19 | 12.12 | 12.85 | 13.16 | 19.50 | 20.25 | 19.71 | 11.80 | 19.35
1/O-19 | 12.50 | 13.40 | 12.98 | 19.85 | 20.79 | 20.16 | 11.80 | 19.40
2/O-19 | 13.46 | 13.46 | 14.15 | 20.29 | 20.29 | 20.75 | 11.79 | 19.34
3/O-19 | 13.77 | 14.44 | 14.95 | 21.14 | 21.14 | 21.76 | 11.77 | 19.32
4/O-19 | 14.82 | 15.61 | 16.29 | 21.19 | 22.97 | 22.14 | 11.78 | 19.34
250-37 | 15.71 | 16.59 | 17.94 | 21.96 | 22.97 | 23.56 | 11.79 | 19.35
300-37 | 17.24 | 18.23 | 19.67 | 23.03 | 24.20 | 24.97 | 11.78 | 19.38
350-37 | 18.56 | 19.64 | 21.51 | 24.01 | 25.33 | 26.69 | 11.76 | 19.35
400-37 | 20.05 | 21.21 | 23.42 | 25.41 | 26.85 | 28.08 | 11.76 | 19.34
500-37 | 23.09 | 24.39 | 27.43 | 27.68 | 29.35 | 31.60 | 11.81 | 19.37
600-61 | 26.59 | 28.00 | 31.91 | 30.61 | 32.45 | 35.55 | 11.76 | 19.35
700-61 | 29.90 | 31.43 | 36.49 | 33.49 | 35.49 | 39.71 | 11.79 | 19.38
750-61 | 31.34 | 32.92 | 38.78 | 34.75 | 36.81 | 41.80 | 11.74 | 19.33
800-61 | 32.98 | 34.53 | 40.84 | 36.32 | 38.36 | 43.75 | 11.79 | 19.37
900-61 | 36.21 | 37.69 | 44.86 | 39.40 | 41.37 | 47.53 | 11.78 | 19.33
1000-61 | 39.47 | 40.89 | 48.86 | 42.55 | 44.47 | 51.36 | 11.81 | 19.37
1250-91 | 48.16 | 49.44 | 59.85 | 50.96 | 52.73 | 61.92 | 11.79 | 19.30
1500-91 | 56.22 | 57.41 | 69.40 | 58.86 | 60.53 | 71.20 | 11.78 | 19.33
1750-127 | 64.23 | 65.34 | 78.75 | 66.74 | 68.34 | 80.34 | 11.78 | 19.35
2000-127 | 72.20 | 73.26 | 87.96 | 74.61 | 76.14 | 89.38 | 11.78 | 19.37

Also, a better approximate voltage drop formula is
V =IRcos@ + IX sin@
where
V = voltage drop in one conductor (line to neutral)
I = current in the one conductor
R = line resistance for one conductor in ohms
X = line reactance for one conductor in ohms
@ = the angle whose cosine is the load power factor
cos@ = load power factor in decimals
sin@ = load reactive factor in decimals

This is the approximate line-neutral voltage drop and is for one conductor. You generally multiply by 2 to get the total drop including the neutral drop (or calculate the neutral based on its current). You would generally multiply by sqrt(3) to get the voltage drop in a three-phase circuit where the current given is the same for each conductor (i.e. a balanced circuit with no neutral current).

The exact line-neutral voltage drop formula is:
eS + IRcos@ + IXsin@ - sqrt[eS^2 - (IXcos@ - IRsin@)^2]
where eS = sending or source voltage

 

[mivey]

My previous post was background information for you. There is nothing wrong with using an approximation as long as you understand its limits. But even so, there is some information that is not clear.

Unclear information:

First, you said you have a multi-tap fixture but then make a calculation based on 120 volts so I am not sure if you are connecting line-neutral (120 volt) or line-line (208 volt). You also say you have a 3-phase circuit but then you say the lights are all on the same wire so it is not clear so maybe we can assume they are all on one circuit, either 120 volt or 208 volt? Your posts and diagrams are not as clear as you think. I am going to assume one circuit with the lights connected line-neutral. If I were using one circuit, and had a light that could use 208, I would at least use a 208 volt circuit.

Another issue is the load amps. Is this at 120 volts or 208 volts? Are you sure they will all draw 8 amps even at a reduced voltage? Maybe you should consider them to be resistive loads instead.

Also, the target temperature is not specified.

Finally, you did not specify whether you were using solid or stranded #10.

So what does all that mean for you? Is the approximation close enough? I'll let you decide because it is your job. Here are the results for each section:

Constant current loads (and assuming the 8 amps is at 120 volts):

If we use the exact method and assume line-neutral (120 volts) loads at 8 amps each, #10 solid copper, 75?C(167?F) temperature, 100% power factor we get:
Source V | Vdrop | % drop | Load V | Ttl drop | Total %
120.00 | 5.76 | 4.80% | 114.24 | 5.76 | 4.80%
114.24 | 1.92 | 1.68% | 112.32 | 7.68 | 6.40%
112.32 | 0.96 | 0.85% | 111.36 | 8.64 | 7.20%

If we change the temperature to 29.44?C(85?F) we get:
Source V | Vdrop | % drop | Load V | Ttl drop | Total %
120.00 | 4.91 | 4.09% | 115.09 | 4.91 | 4.09%
115.09 | 1.64 | 1.42% | 113.45 | 6.55 | 5.46%
113.45 | 0.82 | 0.72% | 112.63 | 7.37 | 6.14%


If we use the exact method and assume line-line (207.85 volts) loads at 4.6188 amps each, #10 solid copper, 75?C(167?F) temperature, 100% power factor we get:
Source V | Vdrop | % drop | Load V | Ttl amps | Ttl drop | Total %
207.85 | 3.33 | 1.60% | 204.52 | 13.86 | 3.33 | 1.60%
204.52 | 1.11 | 0.54% | 203.41 | 9.24 | 4.43 | 2.13%
203.41 | 0.55 | 0.27% | 202.86 | 4.62 | 4.99 | 2.40%

If we change the temperature to 29.44?C(85?F) we get:
Source V | Vdrop | % drop | Load V | Ttl amps | Ttl drop | Total %
207.85 | 2.84 | 1.36% | 205.01 | 13.86 | 2.84 | 1.36%
205.01 | 0.95 | 0.46% | 204.06 | 9.24 | 3.78 | 1.82%
204.06 | 0.47 | 0.23% | 203.59 | 4.62 | 4.25 | 2.05%

Constant resistance loads (and assuming the 8 amps is at 120 volts):

If we use the exact method and assume line-neutral (120 volts) loads at 15 ohms each, #10 solid copper, 75?C(167?F) temperature, 100% power factor we get:
Source V | Vdrop | % drop | Load V | Ttl amps | Ttl drop | Total %
120.00 | 5.44 | 4.53% | 114.56 | 22.67 | 5.44 | 4.53%
114.56 | 1.80 | 1.57% | 112.76 | 15.03 | 7.24 | 6.04%
112.76 | 0.04 | 0.03% | 112.72 | 7.51 | 7.28 | 6.07%

If we change the temperature to 29.44?C(85?F) we get:
Source V | Vdrop | % drop | Load V | Ttl amps | Ttl drop | Total %
120.00 | 4.68 | 3.90% | 115.32 | 22.85 | 4.68 | 3.90%
115.32 | 1.55 | 1.35% | 113.77 | 15.17 | 6.23 | 5.19%
113.77 | 0.04 | 0.03% | 113.73 | 7.58 | 6.27 | 5.22%


If we use the exact method and assume line-line (207.85 volts) loads at 45 ohms each, #10 solid copper, 75?C(167?F) temperature, 100% power factor we get:
Source V | Vdrop | % drop | Load V | Ttl amps | Ttl drop | Total %
207.85 | 3.26 | 1.57% | 204.58 | 13.59 | 3.26 | 1.57%
204.58 | 1.09 | 0.53% | 203.50 | 9.04 | 4.35 | 2.09%
203.50 | 0.02 | 0.01% | 203.48 | 4.52 | 4.37 | 2.10%

If we change the temperature to 29.44?C(85?F) we get:
Source V | Vdrop | % drop | Load V | Ttl amps | Ttl drop | Total %
207.85 | 2.79 | 1.34% | 205.06 | 13.63 | 2.79 | 1.34%
205.06 | 0.93 | 0.45% | 204.13 | 9.07 | 3.72 | 1.79%
204.13 | 0.02 | 0.01% | 204.11 | 4.54 | 3.74 | 1.80%

Hope this helps.

 

[mivey]

Also, here are the K values I get at 29.44?C(85?F):

AWG,kcmil-strands | Z_CU in PVC | Z_CU in Al | Z_CU in St | Z_AL in PVC | Z_AL in Al | Z_AL in St | R_CU (dc) | R_AL (dc)
18-1 | 10.74 | 10.74 | 10.74 | 17.62 | 17.62 | 17.62 | 10.74 | 17.62
18-7 | 10.98 | 10.98 | 10.98 | 18.03 | 18.03 | 18.03 | 10.98 | 18.03
16-1 | 10.76 | 10.76 | 10.76 | 17.65 | 17.65 | 17.65 | 10.76 | 17.65
16-7 | 10.98 | 10.98 | 10.98 | 18.00 | 18.00 | 18.00 | 10.98 | 18.00
14-1 | 10.87 | 10.87 | 10.87 | 17.67 | 17.67 | 17.67 | 10.76 | 17.67
14-7 | 10.87 | 10.87 | 10.87 | 18.06 | 18.06 | 18.06 | 11.01 | 18.05
12-1 | 11.14 | 11.14 | 11.15 | 17.76 | 17.76 | 17.76 | 10.75 | 17.64
12-7 | 11.14 | 11.14 | 11.15 | 17.76 | 17.76 | 17.76 | 11.03 | 18.03
10-1 | 10.64 | 10.64 | 10.64 | 17.65 | 17.65 | 17.65 | 10.71 | 17.64
10-7 | 10.64 | 10.64 | 10.64 | 17.65 | 17.65 | 17.65 | 10.98 | 17.99
8-1 | 11.02 | 11.02 | 11.04 | 18.26 | 18.26 | 18.27 | 10.76 | 17.68
8-7 | 11.02 | 11.02 | 11.04 | 18.26 | 18.26 | 18.27 | 10.95 | 17.96
6-7 | 11.05 | 11.05 | 11.09 | 18.11 | 18.11 | 18.14 | 10.99 | 18.01
4-7 | 11.22 | 11.22 | 11.32 | 18.20 | 18.20 | 18.26 | 10.96 | 18.02
3-7 | 11.49 | 11.49 | 11.64 | 18.05 | 18.50 | 18.15 | 10.99 | 18.02
2-7 | 11.16 | 11.71 | 11.93 | 18.29 | 18.29 | 18.44 | 10.98 | 17.99
1-19 | 11.38 | 12.05 | 12.38 | 18.19 | 18.88 | 18.41 | 10.99 | 17.99
1/O-19 | 11.76 | 12.60 | 12.27 | 18.54 | 19.41 | 18.86 | 10.99 | 18.03
2/O-19 | 12.71 | 12.71 | 13.44 | 18.98 | 18.98 | 19.47 | 10.98 | 17.98
3/O-19 | 13.08 | 13.69 | 14.28 | 19.83 | 19.83 | 20.49 | 10.96 | 17.96
4/O-19 | 14.16 | 14.88 | 15.68 | 19.96 | 21.60 | 20.97 | 10.97 | 17.98
250-37 | 15.10 | 15.90 | 17.37 | 20.76 | 21.69 | 22.42 | 10.98 | 17.99
300-37 | 16.67 | 17.56 | 19.15 | 21.88 | 22.95 | 23.89 | 10.98 | 18.02
350-37 | 18.02 | 18.99 | 21.02 | 22.91 | 24.11 | 25.64 | 10.95 | 17.99
400-37 | 19.56 | 20.59 | 22.95 | 24.35 | 25.65 | 27.08 | 10.95 | 17.98
500-37 | 22.65 | 23.80 | 27.00 | 26.72 | 28.21 | 30.68 | 11.00 | 18.01
600-61 | 26.19 | 27.44 | 31.51 | 29.74 | 31.38 | 34.71 | 10.95 | 18.00
700-61 | 29.53 | 30.87 | 36.12 | 32.68 | 34.45 | 38.94 | 10.98 | 18.02
750-61 | 30.98 | 32.37 | 38.42 | 33.97 | 35.80 | 41.06 | 10.94 | 17.97
800-61 | 32.63 | 33.99 | 40.48 | 35.55 | 37.36 | 43.02 | 10.98 | 18.01
900-61 | 35.88 | 37.17 | 44.49 | 38.67 | 40.41 | 46.82 | 10.98 | 17.97
1000-61 | 39.15 | 40.39 | 48.49 | 41.84 | 43.53 | 50.67 | 11.00 | 18.01
1250-91 | 47.88 | 49.00 | 59.49 | 50.32 | 51.87 | 61.27 | 10.98 | 17.95
1500-91 | 55.97 | 57.00 | 69.02 | 58.26 | 59.72 | 70.58 | 10.98 | 17.97
1750-127 | 63.99 | 64.96 | 78.37 | 66.17 | 67.56 | 79.74 | 10.97 | 17.99
2000-127 | 71.98 | 72.90 | 87.57 | 74.06 | 75.40 | 88.80 | 10.97 | 18.01

 

[Smart $]

Wow!

Wow!

[You guys need to ease up a little on hidroela. Explain and assist rather than baffle and discourage. Mivey's approach isn't all that bad except all the data seems to be over the top for all that is involved. I have to wonder how often this much engineering goes into a simple 3-light outside circuit?]

hidroela,

First off, running a two-wire would require a circuit rating higher than 20A (your load is 24A if on one circuit). If residential, that would be a violation of 210.23, and you will have to run at least a three-wire circuit. If other than residential, you can run a two-wire circuit, but your lampholders must be heavy-duty rated.

Continuing with other than residential installation, if your luminaires are multi-tap 120/208, you could run the two-wire circuit at 208V and get less voltage drop, or perhaps even use #12... or #14 for a three-wire circuit... and not be required to have heavy-duty lampholders.

As for a residential installation, just let us know that is the case and we'll (I'll) go from there...

So for now we just need some clarification on the particulars of installation. For Mivey's sake, you may want to provide some data as to conductor and conduit or cable type(s).

 

[Dennis Alwon]

Looks like the OP is using 3 phase- if so, then

Single-phase VD = 2 x K x I x D/CM.
Three-phase VD = 1.732 x K x I x D/CM. The difference between this and the single phase formula is you replace the 2 with 1.732.

 

[mivey]

Here are the results for each section...

I just went back and looked at my calcs and for some reason I had put in a 0% power factor in the last line section with the constant resistance calcs. Here are the corrected values:

Constant resistance loads (and assuming the 8 amps is at 120 volts):

If we use the exact method and assume line-neutral (120 volts) loads at 15 ohms each, #10 solid copper, 75?C(167?F) temperature, 100% power factor we get:
Source V | Vdrop | % drop | Load V | Ttl amps | Ttl drop | Total %
120.00 | 5.43 | 4.52% | 114.57 | 22.62 | 5.43 | 4.52%
114.57 | 1.80 | 1.57% | 112.78 | 14.98 | 7.22 | 6.02%
112.78 | 0.90 | 0.79% | 111.88 | 7.46 | 8.12 | 6.77%

If we change the temperature to 29.44?C(85?F) we get:
Source V | Vdrop | % drop | Load V | Ttl amps | Ttl drop | Total %
120.00 | 4.67 | 3.89% | 115.33 | 22.81 | 4.67 | 3.89%
115.33 | 1.55 | 1.34% | 113.78 | 15.12 | 6.22 | 5.18%
113.78 | 0.77 | 0.68% | 113.01 | 7.53 | 6.99 | 5.82%


If we use the exact method and assume line-line (207.85 volts) loads at 45 ohms each, #10 solid copper, 75?C(167?F) temperature, 100% power factor we get:
Source V | Vdrop | % drop | Load V | Ttl amps | Ttl drop | Total %
207.85 | 3.26 | 1.57% | 204.59 | 13.58 | 3.26 | 1.57%
204.59 | 1.08 | 0.53% | 203.50 | 9.03 | 4.34 | 2.09%
203.50 | 0.54 | 0.27% | 202.96 | 4.51 | 4.88 | 2.35%

If we change the temperature to 29.44?C(85?F) we get:
Source V | Vdrop | % drop | Load V | Ttl amps | Ttl drop | Total %
207.85 | 2.79 | 1.34% | 205.06 | 13.62 | 2.79 | 1.34%
205.06 | 0.93 | 0.45% | 204.13 | 9.06 | 3.72 | 1.79%
204.13 | 0.46 | 0.23% | 203.67 | 4.53 | 4.18 | 2.01%

 

[mivey]

Mivey's approach isn't all that bad except all the data seems to be over the top for all that is involved.

Well, he did ask how to do it and to evaluate his formula method. As gar pointed out

110521-1059 EDT
Just blindly plugging numbers into an equation can produce quite wrong results..

I have to wonder how often this much engineering goes into a simple 3-light outside circuit?

It goes into every one of mine as I use the same program for every volt drop calc (at least the LV ones).

First off, running a two-wire would require a circuit rating higher than 20A (your load is 24A if on one circuit). If residential, that would be a violation of 210.23, and you will have to run at least a three-wire circuit. If other than residential, you can run a two-wire circuit, but your lampholders must be heavy-duty rated.

Also good points to make for hidroela's enlightenment.

So for now we just need some clarification on the particulars of installation. For Mivey's sake, you may want to provide some data as to conductor and conduit or cable type(s).

Not just for my sake but for the purpose of verifying the accuracy of the simple volt drop calc. One must know the limitations of what they are dealing with. The difference in 7.46% (the simple calc) and 6.14% (one of my exact formula calcs) can be significant if it causes you to change a wire size. Remember that the simple K-formula method tends to be very conservative, up to a point.

 

[gar]

110522-1217 EDT

As I have tried to point out above I think the use of equations without an understanding of the basis of the equation being used is a mistake.

In the original post there was not an adequate description of the problem and associated assumptions. The diagram was basically OK but failed to describe what current was on the return path, whether the input was 120 or 208 (down in the attempt to calculate voltage drop it indirectly appeared the input was 120 V), assuming a 3 phase Y there was no indication of how to treat the neutral current.

Clearly an equation was being applied without an understanding of the relationship of the equation to the circuit. In my opinion the way to help educate someone on the solution to the problem is to get them to think about the problem in the most basic form. This does not mean just providing the answer.

I think the original question was basically what is the voltage drop at various points in a single phase circuit with several discrete loads with #10 copper wire in all segments of the circuit. The same load current was defined at each load. And the assumption of no other currents than those defined in any of the circuit. To get % voltage drop it is also necessary to define the source voltage. For simplicity it would be assumed that inductance, capacitance, and skin-effect effects were negligible. I may be wrong on this assumption, but without a clear definition I have no other choice than do a best guess.

There is a big error in the result depending upon whether or not the neutral current is canceled in some way. Meaning a different equation.

If we can get hidroela to clearly define the circuit and assumptions to be applied, then it is feasible to walk him thru a procedure to solve the problem.

.

 

[mivey]

Looks like the OP is using 3 phase.

But it was unclear what he meant by

all on the same 10-AWG conductor

If we assume 3-phase, then use the simple K-factor calculation with the 17.3 multiplier and compare that with:

Constant current loads (and assuming the 8 amps is for a balanced load):

If we use the exact method and assume 3-phase loads at 8 amps each, #10 solid copper, 75?C(167?F) temperature, 100% power factor we get:
Source V | Vdrop | % drop | Load V | Ttl amps | Ttl drop | Total %
207.85 | 4.99 | 2.40% | 202.86 | 24.00 | 4.99 | 2.40%
202.86 | 1.66 | 0.82% | 201.19 | 16.00 | 6.65 | 3.20%
201.19 | 0.83 | 0.41% | 200.36 | 8.00 | 7.48 | 3.60%

If we change the temperature to 29.44?C(85?F) we get:
Source V | Vdrop | % drop | Load V | Ttl amps | Ttl drop | Total %
207.85 | 4.25 | 2.05% | 203.59 | 24.00 | 4.25 | 2.05%
203.59 | 1.42 | 0.70% | 202.17 | 16.00 | 5.67 | 2.73%
202.17 | 0.71 | 0.35% | 201.46 | 8.00 | 6.38 | 3.07%

Constant resistance loads (and assuming the 8 amps is at 3-phase 120/208 volts):

If we use the exact method and assume 3-phase loads at 14.9889 wye-ohms each, #10 solid copper, 75?C(167?F) temperature, 100% power factor we get:
Source V | Vdrop | % drop | Load V | Ttl amps | Ttl drop | Total %
207.85 | 4.84 | 2.33% | 203.00 | 23.30 | 4.84 | 2.33%
203.00 | 1.61 | 0.79% | 201.39 | 15.48 | 6.45 | 3.10%
201.39 | 0.80 | 0.40% | 200.59 | 7.73 | 7.26 | 3.49%

If we change the temperature to 29.44?C(85?F) we get:
Source V | Vdrop | % drop | Load V | Ttl amps | Ttl drop | Total %
207.85 | 4.15 | 2.00% | 203.70 | 23.41 | 4.15 | 2.00%
203.70 | 1.38 | 0.68% | 202.32 | 15.56 | 5.53 | 2.66%
202.32 | 0.69 | 0.34% | 201.63 | 7.77 | 6.22 | 2.99%

 

[Smart $]

Here's a past post that may help:

http://forums.mikeholt.com/showthre...or-Parking-Lot-Lighting?p=1186589#post1186589

 

[Smart $]

110522-1217 EDT

...

There is a big error in the result depending upon whether or not the neutral current is canceled in some way. Meaning a different equation.

...

If he ends up running a three- or four-wire circuit, there most defintely will be.