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Thread: IEEE std Voltage Drop

  1. #1
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    IEEE std Voltage Drop

    I have been trying to calculate the voltage drop using the IEEE std.141
    Vd = es + IRcos(theta) + IX sin(theta) + sqrt(es^2 - (IXcos(theta) - IRsin(theta))^2)

    Inputs:
    Number of conductors: Single
    Conductor size: 1/0
    Power factor: 80%
    Load Current(Amps) 91
    System voltage: 480
    Length of Run (feet) 460
    Conduit: EMT

    I am coming up with 1.62% Voltage Drop

    Am I correct?

    Using the simplified formula V = I Rcos(theta) + I Xsin(theta) I get 1.95% Voltage Drop
    Don't know which one is correct.


    Thanks a lot

  2. #2
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    Quote Originally Posted by schicco View Post
    I have been trying to calculate the voltage drop using the IEEE std.141
    Vd = es + IRcos(theta) + IX sin(theta) + sqrt(es^2 - (IXcos(theta) - IRsin(theta))^2)

    Inputs:
    Number of conductors: Single
    Conductor size: 1/0
    Power factor: 80%
    Load Current(Amps) 91
    System voltage: 480
    Length of Run (feet) 460
    Conduit: EMT

    I am coming up with 1.62% Voltage Drop

    Am I correct?

    Using the simplified formula V = I Rcos(theta) + I Xsin(theta) I get 1.95% Voltage Drop
    Don't know which one is correct.


    Thanks a lot
    At first the actual voltage drop formula is:
    Vd = es + IRcos(theta) + IX sin(theta) - sqrt(es^2 - (IXcos(theta) - IRsin(theta))^2)
    and the result is 1.9247% and from the approximate formula is 1.9233%.
    The R=0.12 ohm/1000 ft and X=0.055 ohm/1000ft from NEC Table 9 for steel conduit.

  3. #3
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    Julius
    Thanks a lot, I see where I was going wrong. Now I am getting your numbers.
    Again thanks

    Steve

    Quote Originally Posted by Julius Right View Post
    At first the actual voltage drop formula is:
    Vd = es + IRcos(theta) + IX sin(theta) - sqrt(es^2 - (IXcos(theta) - IRsin(theta))^2)
    and the result is 1.9247% and from the approximate formula is 1.9233%.
    The R=0.12 ohm/1000 ft and X=0.055 ohm/1000ft from NEC Table 9 for steel conduit.

  4. #4
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    Julius
    Just want to check again
    What numbers did you use for the cos(theta) and sin(theta)
    Thanks a lot

    Steve


    Quote Originally Posted by Julius Right View Post
    At first the actual voltage drop formula is:
    Vd = es + IRcos(theta) + IX sin(theta) - sqrt(es^2 - (IXcos(theta) - IRsin(theta))^2)
    and the result is 1.9247% and from the approximate formula is 1.9233%.
    The R=0.12 ohm/1000 ft and X=0.055 ohm/1000ft from NEC Table 9 for steel conduit.

  5. #5
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    Quote Originally Posted by schicco View Post
    Julius
    Just want to check again
    What numbers did you use for the cos(theta) and sin(theta)
    Thanks a lot

    Steve
    Theta is .8. Cos theta is 0.99990252400930421155123810793869. Sin theta is 0.01396218033914527162191138769678 (roughly )
    The more they overthink the plumbing, the easier it is to stop up the drain

  6. #6
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    Thanks a lot
    Quote Originally Posted by dkidd View Post
    Theta is .8. Cos theta is 0.99990252400930421155123810793869. Sin theta is 0.01396218033914527162191138769678 (roughly )

  7. #7
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    Since I am feeding a fire pump motor; I should then multiply by 1.73 to get the Line to Line?
    Then it would mean I have gone way over 3% voltage drop, right?
    The 2%(feeder)+3%(branch)=5%vd applies?
    Not sure if Fire Pump falls under branch or feeder (definition of branch is wiring that comes after the final protection device)

    Thanks

    Quote Originally Posted by Julius Right View Post
    At first the actual voltage drop formula is:
    Vd = es + IRcos(theta) + IX sin(theta) - sqrt(es^2 - (IXcos(theta) - IRsin(theta))^2)
    and the result is 1.9247% and from the approximate formula is 1.9233%.
    The R=0.12 ohm/1000 ft and X=0.055 ohm/1000ft from NEC Table 9 for steel conduit.

  8. #8
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    Dec 2010
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    1,268
    Quote Originally Posted by schicco View Post
    Since I am feeding a fire pump motor; I should then multiply by 1.73 to get the Line to Line?
    Yes
    Quote Originally Posted by schicco View Post
    Then it would mean I have gone way over 3% voltage drop, right?
    The 2%(feeder)+3%(branch)=5%vd applies?
    Not sure if Fire Pump falls under branch or feeder (definition of branch is wiring that comes after the final protection device)

    Thanks
    Codes limit the running voltage drop to 5%, and starting to 15%.
    The more they overthink the plumbing, the easier it is to stop up the drain

  9. #9
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    What confuses me about the code definition of branch ckt ("the conductors between the branch circuit final overcurrent device protecting the circuit and the outlets") is;
    Why does it only mention outlets?
    If I am going to hard wire a motor or heating, it cannot be branch?

    From the same panel you can have an OCPD feeding outlets and another OCPD feeding a motor. Are they all not branch ckts?

    Thanks

    Andy

    Quote Originally Posted by dkidd View Post
    Yes


    Codes limit the running voltage drop to 5%, and starting to 15%.
    Attached Images Attached Images  
    Last edited by aprice44; 05-16-17 at 11:56 AM.

  10. #10
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    Quote Originally Posted by aprice44 View Post
    What confuses me about the code definition of branch ckt ("the conductors between the branch circuit final overcurrent device protecting the circuit and the outlets") is;
    Why does it only mention outlets?
    If I am going to hard wire a motor or heating, it cannot be branch?

    From the same panel you can have an OCPD feeding outlets and another OCPD feeding a motor. Are they all not branch ckts?

    Thanks

    Andy
    Definition from Article 100

    Outlet. A point on the wiring system at which current is
    taken to supply utilization equipment.

    Don't confuse outlet with receptacle.
    The more they overthink the plumbing, the easier it is to stop up the drain

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