Page 1 of 2 12 LastLast
Results 1 to 10 of 11

Thread: Short Circuit MVA

  1. #1
    Join Date
    Mar 2012
    Location
    Minneapolis, MN
    Posts
    245

    Short Circuit MVA

    A question about short circuit MVA (MVAsc). If we have a generator of 2 MVA, then what will be its MVAsc ?

  2. #2
    Join Date
    Feb 2003
    Location
    Connecticut
    Posts
    3,038
    Quote Originally Posted by timm333 View Post
    A question about short circuit MVA (MVAsc). If we have a generator of 2 MVA, then what will be its MVAsc ?
    Not enough information to answer...You need to know the generator's subtransient reactance.

    Generator MVAsc = MVA/X''d

  3. #3
    Join Date
    Mar 2012
    Location
    Minneapolis, MN
    Posts
    245
    Thanks. So if a 2 MVA generator has Xd'' = 0.2, then MVAsc = 2/0.2 = 10 MVAsc, this makes sense. I am just being curious that why MVA and MVAsc are not the same. In other words, if the generator is capable to produce 10 MVAsc, then why its useful output power is only 2 MVA (and not 10 MVA)?

  4. #4
    Join Date
    Jan 2016
    Location
    Earth
    Posts
    6,049
    Quote Originally Posted by timm333 View Post
    Thanks. So if a 2 MVA generator has Xd'' = 0.2, then MVAsc = 2/0.2 = 10 MVAsc, this makes sense. I am just being curious that why MVA and MVAsc are not the same. In other words, if the generator is capable to produce 10 MVAsc, then why its useful output power is only 2 MVA (and not 10 MVA)?

    the 2 is into a load
    the 10 is limited only by its reactance, the windings will be destroyed if sustained

    similar to a xfmr
    rated mva 1
    impedance pu is 5%
    sc mva = 1/0.05 = 20

  5. #5
    Join Date
    Feb 2003
    Location
    New York, 40.7514,-73.9925
    Posts
    4,568
    The subtransient (x") time period lasts from about one to three cycles, which is from 16.7 milliseconds to 50 milliseconds (assuming a power frequency of 60 Hz). So if you consider useful power output for about 20-50 milliseconds, then you are good to go.
    Ron

  6. #6
    Join Date
    Jan 2016
    Location
    Earth
    Posts
    6,049

  7. #7
    Join Date
    Dec 2012
    Location
    Placerville, CA, USA
    Posts
    19,537
    Quote Originally Posted by timm333 View Post
    Thanks. So if a 2 MVA generator has Xd'' = 0.2, then MVAsc = 2/0.2 = 10 MVAsc, this makes sense. I am just being curious that why MVA and MVAsc are not the same. In other words, if the generator is capable to produce 10 MVAsc, then why its useful output power is only 2 MVA (and not 10 MVA)?
    Yet another reason is that the running MVA may well be limited by the size of the prime mover (engine or turbine for example), while the rotating mass of the system will allow the MVAsc to be limited only by the generator portion.
    A vendor of cheap wind turbines once famously rated his wind generator at roughly twice the sustained production for the reference wind speed on the grounds that once it got up to speed and you closed the output circuit it would deliver the higher power "for awhile" as the system slowed down to a new equilibrium speed.

  8. #8
    Join Date
    Jan 2016
    Location
    Earth
    Posts
    6,049
    generators produce heat and need cooled (air/fan, liquid, gas)

    losses (cooling load) = i^2 R
    assume rated i = 1 pu
    therefore sc i = 10 pu
    R is constant
    so cooling load is 100 times greater
    smoke lol

  9. #9
    Join Date
    Dec 2012
    Location
    Placerville, CA, USA
    Posts
    19,537
    Quote Originally Posted by Ingenieur View Post
    generators produce heat and need cooled (air/fan, liquid, gas)

    losses (cooling load) = i^2 R
    assume rated i = 1 pu
    therefore sc i = 10 pu
    R is constant
    so cooling load is 100 times greater
    smoke lol
    Also why large generators were once considered impractical because the misinterpretation of energy transfer conditions caused the designers to try to make source impedance and load impedance equal!

  10. #10
    Join Date
    Mar 2012
    Location
    Minneapolis, MN
    Posts
    245
    The first page of the link above from the website of Cummins states that R for generators is negligible. So does it mean that the generators have no X/R ratio like transformers? When we say, for example, the X/R on the primary side of the transformer is 15, does it mean that all of this X/R is due to the cable between generate and transformer (without any contribution from generator)?

Posting Permissions

  • You may not post new threads
  • You may not post replies
  • You may not post attachments
  • You may not edit your posts
  •