help me..

Discussion in 'The Projects Forum' started by amalina, Apr 12, 2012.

  1. amalina

    Thread Starter New Member

    Apr 12, 2012
    3
    0
    A 10 kVA 2400/240V 60Hz transformer was tested with the following results: Power input during short circuit test is 340W, power input during open circuit test is 168 W. Determine the efficiency of the transformer at full load. Given the load power factor is 0.8.
     
  2. jimkeith

    Active Member

    Oct 26, 2011
    539
    99
    Something is wrong with your power in during short circuit--way low
     
  3. #12

    Expert

    Nov 30, 2010
    16,284
    6,797
    Maybe that isn't a consideration for a HOMEWORK problem?
     
  4. amalina

    Thread Starter New Member

    Apr 12, 2012
    3
    0
    this is the question my lecturer give..how about this one can u help?
    -->A transformer 50Hz, 440/220V, the primary current, power and voltage are 1.25A, 50W and 440V under no load condition. Ignore the primary winding resistance and reactance. Calculate:

    a) Core current and resistance (Ic and Rc)
    b) Magnetizing current and reactance (Im and Xm)
    c) Power factor under no load condition
     
  5. Kermit2

    AAC Fanatic!

    Feb 5, 2010
    3,789
    945
    No one here will do your homework problem for you.

    Post YOUR answer and someone will tell you if it is right or wrong, and if you show your equations and math they may be able to tell you where it is wrong.

    Good luck on the test :D
     
  6. panic mode

    Senior Member

    Oct 10, 2011
    1,320
    304
    How can you say that? Why do you think so?



    Can you post entire question exactly as posted in your book? What model of transformer is assumed?
     
  7. panic mode

    Senior Member

    Oct 10, 2011
    1,320
    304
    hint - in general:

    Zs=Req+jXeq // series with high voltage side
    Zp=Rcu || jXm // parallel with high voltage side

    when testing Short circuit, transformer is tested with low voltage side shorted and reduced voltage applied at high voltage side at rated current (of the high voltage side).
    due reduced voltage, Zp is neglected so all losses are considered to be due Zs. Any power loss (power is real) must be due Req:
    Psc=Vsc*Isc*cos(theta_sc)
    Isc=rated current of high voltage side


    when testing Open Circuit, transformer is tested with high voltage side open, low voltage side at rated voltage and rated current.
    Any power loss (power is real) is due Rcu.

    When in operation, all power losses are lumped into Ploss. Power rating of transformer is given (S and PF so we can calculate P).

    what is left is to determine efficiency....

    hope this helps
     
  8. panic mode

    Senior Member

    Oct 10, 2011
    1,320
    304
    i will eventually. i didn't calculate it yet and for all i know, they could be right. i was asking what makes them think that answer should be so obvious.

    i suspect that they may have jumped to conclusion thinking that short circuit of a 10kVA transformer powered from 2400V must lead to higher power dissipation (which would be right if that was the case).

    however, transformer short circuit test is not carried out with full line voltage because obviously this is not safe. that is why i posted common test procedure and described model of transformer it applies to.

    happy?
     
  9. John P

    AAC Fanatic!

    Oct 14, 2008
    1,634
    224
    I don't see why this request for someone to do a student's homework appeared in Projects Forum rather than Homework Help.
     
  10. panic mode

    Senior Member

    Oct 10, 2011
    1,320
    304
    agreed, i hope moderators move this to homework section
     
  11. amalina

    Thread Starter New Member

    Apr 12, 2012
    3
    0
    sorry everyone i just try asking..sorry if i mistake posting this here..tq 4 help me..
     
  12. mlog

    Member

    Feb 11, 2012
    276
    36
    Let us know what you get for an answer. I came up with η=0.994.
     
Loading...