# Thread: Voltage drop for new feeder in an existing building

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Tagged Originally Posted by Grouch1980 Got it. So there's no rule of thumb / formula, other than to actually take a field measurement of the voltage at the point where the new connection is being made, or find as-builts / design drawings and calculate it.
you need 2 v readings
supply and load at the same time (2 guys, do each ph and avg)
take i readings at same time, should be steady, avg 3 phases
feeder Z = (v supply - v load) / i Ohms

max v drop = Z x cond ampacity (x 80% depending on application)

eg
v supply 485
i = 100
Z = 5/100 = 0.05 Ohm

if 250 kcmil rated ampacity = 255
v drop = 0.05 x 255 = 13 or 2.7%
v drop = 0.05 x 0.8 x 255 = 10 v or 2.1%

the closer the i is to the rated ampacity the better
>30% of rated ampacity with good measurements yeilds good results

you can also remove power and actually measure a loop by shorting one end
or calc off as-buits  Reply With Quote

2. Originally Posted by Ingenieur you need 2 v readings
supply and load at the same time (2 guys, do each ph and avg)
take i readings at same time, should be steady, avg 3 phases
feeder Z = (v supply - v load) / i Ohms

max v drop = Z x cond ampacity (x 80% depending on application)

eg
v supply 485
i = 100
Z = 5/100 = 0.05 Ohm

if 250 kcmil rated ampacity = 255
v drop = 0.05 x 255 = 13 or 2.7%
v drop = 0.05 x 0.8 x 255 = 10 v or 2.1%

the closer the i is to the rated ampacity the better
>30% of rated ampacity with good measurements yeilds good results

you can also remove power and actually measure a loop by shorting one end
or calc off as-buits
Thanks. ..... "the closer the i is to the rated ampacity the better"... why is that? i is your calculated demand amps.  Reply With Quote

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Tagged Originally Posted by Ingenieur you need 2 v readings
supply and load at the same time (2 guys, do each ph and avg)
take i readings at same time, should be steady, avg 3 phases
feeder Z = (v supply - v load) / i Ohms

max v drop = Z x cond ampacity (x 80% depending on application)

eg
v supply 485
i = 100
Z = 5/100 = 0.05 Ohm

if 250 kcmil rated ampacity = 255
v drop = 0.05 x 255 = 13 or 2.7%
v drop = 0.05 x 0.8 x 255 = 10 v or 2.1%

the closer the i is to the rated ampacity the better
>30% of rated ampacity with good measurements yeilds good results

you can also remove power and actually measure a loop by shorting one end
or calc off as-buits Originally Posted by Ingenieur you need 2 v readings..
feeder Z = (v supply - v load) / I "CLamp meter"
Field measurements with Logged impedance (Z) can be assumed constant so Voltage Drop is solved with any Load.

While (Z) and temperature will vary with load, any conservative error is on the side of caution.

Z = VD/I

If Logged Feeder VD = 5v with 100A Load, then Z = 0.05

Now Solve Voltage Drop for any amperage,

VD = Z*I

If 250 kcmil rated ampacity = 255 then 80% = 204A

0.05*204 = 10.2v of 485v or VD = 2.1%

Where actual termination temperature is needed, Temp must be logged.  Reply With Quote

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Tagged Originally Posted by Grouch1980 Thanks. ..... "the closer the i is to the rated ampacity the better"... why is that? i is your calculated demand amps.
the higher the i the higher the v drop
so the lower the instrument messurement error

i is the measured amps when tested  Reply With Quote

5. Originally Posted by Ingenieur the higher the i the higher the v drop
so the lower the instrument messurement error

i is the measured amps when tested
got it, thanks.  Reply With Quote

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