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Thread: Voltage Dip Estimation

  1. #11
    Join Date
    Jun 2017
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    Quote Originally Posted by MyCleveland View Post
    Anyone.....?
    before this slips off the page.
    The answer to the light flicker question is outlined in IEEE 241 [Gray Book]. Visible flicker is seen with as little as 0.5% voltage change. Light output for say fluorescent lighting is directly proportional to the applied voltage.

    "Acceptable voltage flicker magnitude depends on the frequency of occurrence.... At 10 dips per hour, people begin to detect incandescent lamp flicker for voltage dips larger than 1% and begin to object when the magnitude exceeds 3%.... A typical motor may draw 5 kVA/hp and a transformer impedance may be 6%. The equation below estimates flicker while starting a 15 hp motor on a 150 kVA transformer: 15 hp * (5 kVA/hp) * (6%/150kVA) = 3% flicker"

    It goes on to say if you are on the boarder of 10 dips at this 3% flicker you can better review via locked rotor currents. Of course you have the conductor length to consider as we have outline also.

    Max VD for a motor is typically limited by the relays or contactors - subject to the CEBMA (or now ITI) curve. This value is as much as 35%, but target 30%.

    Max VD for a service depends. If there is a UPS or otherwise on the service...

  2. #12
    Join Date
    Oct 2016
    Location
    Cleveland, Ohio
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    132
    Quote Originally Posted by publicgood View Post
    The answer to the light flicker question is outlined in IEEE 241 [Gray Book]. Visible flicker is seen with as little as 0.5% voltage change. Light output for say fluorescent lighting is directly proportional to the applied voltage.

    "Acceptable voltage flicker magnitude depends on the frequency of occurrence.... At 10 dips per hour, people begin to detect incandescent lamp flicker for voltage dips larger than 1% and begin to object when the magnitude exceeds 3%.... A typical motor may draw 5 kVA/hp and a transformer impedance may be 6%. The equation below estimates flicker while starting a 15 hp motor on a 150 kVA transformer: 15 hp * (5 kVA/hp) * (6%/150kVA) = 3% flicker"

    It goes on to say if you are on the boarder of 10 dips at this 3% flicker you can better review via locked rotor currents. Of course you have the conductor length to consider as we have outline also.

    Max VD for a motor is typically limited by the relays or contactors - subject to the CEBMA (or now ITI) curve. This value is as much as 35%, but target 30%.

    Max VD for a service depends. If there is a UPS or otherwise on the service...
    No UPS considered, just wanted to put a flag in a spreadsheet....X (max value exceeded)

  3. #13
    Join Date
    Dec 2008
    Posts
    475
    I think it could be something like this one:
    Electrical Fault Level Calculations Using MVA Method by Lee Wai Meng
    http://www.jmpangseah.com/wp-content.../chapter-5.pdf

  4. #14
    Join Date
    Jan 2016
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    Quote Originally Posted by Julius Right View Post
    I think it could be something like this one:
    Electrical Fault Level Calculations Using MVA Method by Lee Wai Meng
    http://www.jmpangseah.com/wp-content.../chapter-5.pdf
    using that method assuming no initial load and values from op's link
    Sxfmr = 0.300 mva, pu Z 5.5%, Ssc = 5.4545 mva
    Smtr = 6 x 124 x 480 x sqrt3 = 0.6185 mva
    % load = (0.6185 x 5.4545) / (0.6185 + 5.4545) = 0.5555
    ( (Smtr x Ssc)/(Smtr+Ssc) )
    pu v at mtr = 5.4545/(0.5555 + 5.4545) = 0.9077 or 9.23%
    ( Ssc/(% load + Ssc) )

    the op uses Smtr/(Smtr + Ssc) = 0.6185/(0.6185+5.454)x100=10.18%

    it's very simple imo
    xfmr isc = 6561 A so Z = 480/6561 = 0.0732 Ohm
    vdrop thur xfmr = 6 x 124 x 0.0732 = 54.5 v or 11.34% at terminals



  5. #15
    Join Date
    Oct 2016
    Location
    Cleveland, Ohio
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    Quote Originally Posted by Ingenieur View Post
    using that method assuming no initial load and values from op's link
    Sxfmr = 0.300 mva, pu Z 5.5%, Ssc = 5.4545 mva
    Smtr = 6 x 124 x 480 x sqrt3 = 0.6185 mva
    % load = (0.6185 x 5.4545) / (0.6185 + 5.4545) = 0.5555
    ( (Smtr x Ssc)/(Smtr+Ssc) )
    pu v at mtr = 5.4545/(0.5555 + 5.4545) = 0.9077 or 9.23%
    ( Ssc/(% load + Ssc) )

    the op uses Smtr/(Smtr + Ssc) = 0.6185/(0.6185+5.454)x100=10.18%

    it's very simple imo
    xfmr isc = 6561 A so Z = 480/6561 = 0.0732 Ohm
    vdrop thur xfmr = 6 x 124 x 0.0732 = 54.5 v or 11.34% at terminals
    Ingenieur
    Is this not the same as my original attachment ?
    Downloaded, but no time for review as of yet.
    Method you showed....way back is what I used in spreadsheet I posted.

  6. #16
    Join Date
    Nov 2007
    Posts
    810
    I think it's late but here's a link to an old article that deals with estimating a voltage dip when you have a large motor connected to a utility.
    link

  7. #17
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    Jan 2016
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    PA
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    Quote Originally Posted by topgone View Post
    I think it's late but here's a link to an old article that deals with estimating a voltage dip when you have a large motor connected to a utility.
    link
    lol
    look at the first equation in the link
    I just cranked this out

    v drop = Smtrstart / Ssc x V (Smtrstart ~ LR)
    v drop % = Smtrstart / Ssc x 100

    from op paper
    v drop % = 618.5/5454.5 x 100 = 11.34%
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