Voltage drop

A

Alex_rg15

Member
Location
Texas
Occupation
Electrician
A 120/208, three phase system feeds a detached building with a load of 24,960VA using 1 AWG copper conductors. What is the maximum distance the feeder can be to not have a voltage drop of more than 5%?
I calculate the amps first
24,960 /(208*1.73) =69.364
Then I calculate the 5% of the VD
208 * 5% =10.4
Then apply the VD formula for L
L = VD * cmil / 1.73*K* I
L = 10.4 * 83690 / 1.73(12.9)(69.364)
L= 870376/ 1548
L= 562
My answer is 562 ft
But the book says 325ft
 
Idk what book you’re reading from. 1.732x12.9x69.364x562 divided by 83690 is 10.4 which fits your 5%


Sent from my iPhone using Tapatalk
 
joshfromoshoto said:
Idk what book you’re reading from. 1.732x12.9x69.364x562 divided by 83690 is 10.4 which fits your 5%


Sent from my iPhone using Tapatalk
Click to expand...
I know, I’m right on my math and in the process to get the voltage drop
I’m in the process to get my journeyman, and things like this get me all confuse
 
I've only checked once, but I think @Alex_rg15 used the correct equation and got the correct result.

Your number and the book number are off by a factor of 1.73.

My guess is that the person writing the book hit '3' and forgot to hit the 'sqrt' button when they did the calculation; they divided by 3 instead of dividing by 1.73 (sqrt(3))
 
Alex_rg15 said:
I calculate the amps first
24,960 /(208*1.73) =69.364
Then I calculate the 5% of the VD
208 * 5% =10.4
Then apply the VD formula for L
L = VD * cmil / 1.73*K* I
L = 10.4 * 83690 / 1.73(12.9)(69.364)
L= 870376/ 1548
L= 562
My answer is 562 ft
But the book says 325ft
Click to expand...
ramsy said:
24,960 / 208 / 1.732 = 69.2841
Click to expand...


L👀k what happens we he adds the (2)
24960/(208*1.732) =69.2840646651
24960/(208/1.732) = 69.2840646651
Or
24960/208/1.73=69.3641618497



ptonsparky said:
Well, that's enough to make a guy fail at life in general.
Click to expand...
🤣😂

Why the 325🤔
winnie said:
I've only checked once, but I think @Alex_rg15 used the correct equation and got the correct result.

Your number and the book number are off by a factor of 1.73.

My guess is that the person writing the book hit '3' and forgot to hit the 'sqrt' button when they did the calculation; they divided by 3 instead of dividing by 1.73 (sqrt(3))
Click to expand...

562/1.732= 324.480369515👏👏👏
 
Winnie's explanation in post #4 seems most likely, but I also think it is worth noting that problems like this should really specify the nature of the loading.

If all of the loads is balanced 3 phase loads, then the VD calculation in the OP is applicable. And assuming balanced loading seems reasonable.

But if the loading is instead balanced L-N loading, and the number of different loads and manner of their use admits the possibility that only the L-N loads on one phase will be in use at some given time, then a different VD calculation controls. Same current, but the voltage is 120V, so a factor of 1/sqrt(3) less allowable absolute voltage drop. And the voltage drop formula has a factor of 2 instead of sqrt(3). The upshot is that the allowable circuit length is half that of the balanced 3 phase loading case.

I'm not seeing a loading distribution assumption and associated worst case loading that would change the answer by a factor of sqrt(3), though.

Cheers, Wayne
 
You must log in or register to reply here.
Top