voltage drop question

[Electron_Sam78]

I'm rusty at calculating for voltage drop so I need a lil refresher. I'm tasked with evaluating some existing circuits for upgrade or repair if necessary. How do I do the calcs for segmented loads on a long run?

The circuit is 707' from supply to the first outlet, 105' to the next, 206' to the third, and 147' to the last one for a total circuit length of 1165'. Each outlet is a240V/16.8a load for a total of 67.2 Amps. It's run with 4/0 aluminum from supply to the final outlet.

Supply _________707'________o___105' o 206' o 147' o

My formula for determining max length with given load and distance is

L= 1000*Vd/2*R*I

R = .100 (for 4/0 AL so you don't have to look it up)

Using this formula it's a max of 893' under full load (assuming a 5% Vd of 12 Volts) but since this is a segemented circuit how do I figure it?

 

[winnie]

The easiest method:
Just calculate the voltage drop for each segment, starting with the voltage at the end of the segment before. At the end of each segment, you just use the total current of the remaining loads.

First you need to flip your equation around, and calculate the voltage drop given the length and current, thus:
Vd = 2 * L/1000 * R * I

Thus:
9.5V = 2 * 707/1000 * 0.1 * 67.2
1.1V = 2 * 105/1000 * 0.1 * 50.4
...
Total up all the segments and you get the voltage drop at the last load.

Something to keep in mind when you do this calculation: cooler wire is a better conductor than warmer wire, and the resistance tabulated in tables 8 and 9 assume a conductor temperature of 75C. For conductors as oversized as these, the temperature will be much lower than 75C...my guess is that you should be using an R value of 0.085 to 0.09 ohms per 1000 feet for your calculation.

You should also consider how your load responds to reduced voltage; does the current go up or down. Many loads draw less current when the input voltage goes down; you get reduced performance but the reduced current helps with your voltage drop. Some loads compensate for reduced input voltage by increasing the current they consume; the additional current draw makes voltage drop worse, but at the same time these loads can tolerate greater voltage drop.

-Jon

 

[GMc]

Electron,

Here is what I came up with. I'm sure someone will correct me if I'm wrong

@ 707' 67.2 A 8.1 VD 3.4%

@812 50.4 A .9 VD .4%

@1018 33.6 A 1.2 VD .5%

@1165 16.8 A .4 VD .2%

Total = 10.6 voltage drop 4.5 %


GMc

 

[Electron_Sam78]

First you need to flip your equation around, and calculate the voltage drop given the length and current, thus:
Vd = 2 * L/1000 * R * I

Actually it's Vd= 2*L*R*I/1000

 

[Electron_Sam78]

Electron,

Here is what I came up with. I'm sure someone will correct me if I'm wrong

@ 707' 67.2 A 8.1 VD 3.4%

@812 50.4 A .9 VD .4%

@1018 33.6 A 1.2 VD .5%

@1165 16.8 A .4 VD .2%

Total = 10.6 voltage drop 4.5 %


GMc

What was your method?

 

[GMc]

I just used an online voltage drop calculator.

 

[Electron_Sam78]

xxxxxxxxxxxxxxxxxxxxxx

 

[Electron_Sam78]

ok

9.5V @ 707'
1.1V @ 105' further
1.4V @ 206' more and
0.5V @ 147' more
________________
12.5V = 5.2%


I can live with that

 

[Electron_Sam78]

most accurate method?

most accurate method?

What'sthe most accurate method for calculating Vd? I notice that the calculator used by GMc comes up with different values.

 

[mivey]

What'sthe most accurate method for calculating Vd? I notice that the calculator used by GMc comes up with different values.

There are assumptions that must be made about the conduit, temperature, power factor, load type, etc that can lead to different results. The exact formula is from ANSI/IEEE Std 141.

Here is what you get for some different assumptions:

Constant amps, unity p.f., 60F, PVC conduit
Sect 1: 16.800A added, 67.200A section, 707.0 ft, 240.00V source, 232.35V load, 7.65V drop (3.2%)
Sect 2: 16.800A added, 50.400A section, 105.0 ft, 232.35V source, 231.50V load, 8.50V drop (3.5%)
Sect 3: 16.800A added, 33.600A section, 206.0 ft, 231.50V source, 230.38V load, 9.62V drop (4.0%)
Sect 4: 16.800A added, 16.800A section, 147.0 ft, 230.38V source, 229.98V load, 10.02V drop (4.2%)

Constant amps, 85% p.f., 60F, PVC conduit
Sect 1: 16.800A added, 67.200A section, 707.0 ft, 240.00V source, 231.45V load, 8.55V drop (3.6%)
Sect 2: 16.800A added, 50.400A section, 105.0 ft, 231.45V source, 230.50V load, 9.50V drop (4.0%)
Sect 3: 16.800A added, 33.600A section, 206.0 ft, 230.50V source, 229.26V load, 10.74V drop (4.5%)
Sect 4: 16.800A added, 16.800A section, 147.0 ft, 229.26V source, 228.81V load, 11.19V drop (4.7%)

Constant load resistance, unity p.f., 60F, PVC conduit
Sect 1: 16.284A added, 64.786A section, 707.0 ft, 240.00V source, 232.62V load, 7.38V drop (3.1%)
Sect 2: 16.226A added, 48.502A section, 105.0 ft, 232.62V source, 231.80V load, 8.20V drop (3.4%)
Sect 3: 16.151A added, 32.276A section, 206.0 ft, 231.80V source, 230.73V load, 9.27V drop (3.9%)
Sect 4: 16.125A added, 16.125A section, 147.0 ft, 230.73V source, 230.35V load, 9.65V drop (4.0%)

Constant load resistance, 85% p.f., 60F, PVC conduit
Sect 1: 16.226A added, 64.515A section, 707.0 ft, 240.00V source, 231.80V load, 8.20V drop (3.4%)
Sect 2: 16.162A added, 48.289A section, 105.0 ft, 231.80V source, 230.88V load, 9.12V drop (3.8%)
Sect 3: 16.079A added, 32.127A section, 206.0 ft, 230.88V source, 229.69V load, 10.31V drop (4.3%)
Sect 4: 16.049A added, 16.049A section, 147.0 ft, 229.69V source, 229.27V load, 10.73V drop (4.5%)

Constant load resistance, 85% p.f., 60F, Steel conduit
Sect 1: 16.194A added, 64.365A section, 707.0 ft, 240.00V source, 231.34V load, 8.66V drop (3.6%)
Sect 2: 16.126A added, 48.171A section, 105.0 ft, 231.34V source, 230.37V load, 9.63V drop (4.0%)
Sect 3: 16.038A added, 32.045A section, 206.0 ft, 230.37V source, 229.12V load, 10.88V drop (4.5%)
Sect 4: 16.007A added, 16.007A section, 147.0 ft, 229.12V source, 228.67V load, 11.33V drop (4.7%)

Constant load resistance, 85% p.f., 75F, Steel conduit
Sect 1: 16.179A added, 64.298A section, 707.0 ft, 240.00V source, 231.13V load, 8.87V drop (3.7%)
Sect 2: 16.110A added, 48.119A section, 105.0 ft, 231.13V source, 230.15V load, 9.85V drop (4.1%)
Sect 3: 16.020A added, 32.008A section, 206.0 ft, 230.15V source, 228.86V load, 11.14V drop (4.6%)
Sect 4: 15.988A added, 15.988A section, 147.0 ft, 228.86V source, 228.40V load, 11.60V drop (4.8%)

 

[kbsparky]

Actually it's Vd= 2*L*R*I/1000

Ummm ... one of the first principles taught in basic math is when dealing with multiplication or division, the order of the equation is irrelevant.

You will end up with the same result. :roll:

A*B*C/2 = A/2*B*C

 

[Hameedulla-Ekhlas]

Ummm ... one of the first principles taught in basic math is when dealing with multiplication or division, the order of the equation is irrelevant.

You will end up with the same result. :roll:

A*B*C/2 = A/2*B*C

I am not sure they are the same. They could be the same if u write like this

A*B*C/2 = (A/2)*B*C same result

what u have written it looks like this

A*B*C/2 = A/(2*B*C) just take care of parentheses

 

[mivey]

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I am not sure they are the same. They could be the same if u write like this

A*B*C/2 = (A/2)*B*C same result

what u have written it looks like this

A*B*C/2 = A/(2*B*C) just take care of parentheses

You are incorrect. With no parenthesis given, there are none implied.

 

[Hameedulla-Ekhlas]

ok, if parentheses are not required just write in Ms.excel both and see the result.

 

[mivey]

ok, if parentheses are not required just write in Ms.excel both and see the result.

Whether or not Excel follows the rules of math is immaterial. Excel has a history of erroneous calculations.

 

[mivey]

Whether or not Excel follows the rules of math is immaterial. Excel has a history of erroneous calculations.

FWIW, both of the formulas in Excel yielded 10.5:
=1/2*3*7
=1*3*7/2

Maybe you need an upgrade or service patch.

 

[Hameedulla-Ekhlas]

Ok friend you are right but I just wanted to make it clear the A/2*B*C whether the B and C is located in numerator or denominator.

What math rules are I accept that but writing clearly is very important.

 

[mivey]

Ok friend you are right but I just wanted to make it clear the A/2*B*C whether the B and C is located in numerator or denominator.

What math rules are I accept that but writing clearly is very important.

In that respect you are correct. If there is any doubt you can use extra parenthesis as they have a higher order precedence. Extra parenthesis are often used to group certain calculations so that the steps can be followed more clearly.

 

[GMc]

ok

9.5V @ 707'
1.1V @ 105' further
1.4V @ 206' more and
0.5V @ 147' more
________________
12.5V = 5.2%


I can live with that

I recalculated using paper and pencil and came up with the same numbers as you. Using the online calculator was probably quicker but it didn't take into account the new voltage (drop) at each segment so I was a little off by the time I got to the end.

GMc