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Thread: Voltage drop for new feeder in an existing building

  1. #11
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    Jan 2016
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    Quote Originally Posted by Grouch1980 View Post
    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
    v load 480
    i = 100
    Z = 5/100 = 0.05 Ohm

    if 250 kcmil rated ampacity = 255
    v drop = 0.05 x 255 = 13 or 2.7%
    if loading is 80% (cond sized 1.25 x load)
    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

  2. #12
    Join Date
    Jul 2015
    Location
    New York, NY
    Posts
    180
    Quote Originally Posted by Ingenieur View Post
    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
    v load 480
    i = 100
    Z = 5/100 = 0.05 Ohm

    if 250 kcmil rated ampacity = 255
    v drop = 0.05 x 255 = 13 or 2.7%
    if loading is 80% (cond sized 1.25 x load)
    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.

  3. #13
    Join Date
    Jul 2005
    Location
    LA basin, CA
    Posts
    1,809
    Quote Originally Posted by Ingenieur View Post
    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
    v load 480
    i = 100
    Z = 5/100 = 0.05 Ohm

    if 250 kcmil rated ampacity = 255
    v drop = 0.05 x 255 = 13 or 2.7%
    if loading is 80% (cond sized 1.25 x load)
    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

    Quote Originally Posted by Ingenieur View Post
    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.
    Roger Ramjet NoFixNoPay

  4. #14
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    Jan 2016
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    Quote Originally Posted by Grouch1980 View Post
    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

  5. #15
    Join Date
    Jul 2015
    Location
    New York, NY
    Posts
    180
    Quote Originally Posted by Ingenieur View Post
    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.

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