Efficiency at 70% full load

Discussion in 'Homework Help' started by Tera-Scale, Feb 3, 2011.

  1. Tera-Scale

    Thread Starter Active Member

    Jan 1, 2011
    164
    5
    A transformer:

    -50KVA
    -11000/220V 50Hz
    -NO load 0.3Amps
    -No Load losses = 650W

    determine;
    a)
    i)iron loss current. = 59.1mA
    ii)magnetizing current = 294mA
    iii)no load pf. = 0.201

    b) a s.c. test gives Zeq =125Ω ref. to primary
    and full load Power losses = 850W

    determine:

    i) full load sec current = 227.275Amps
    ii)efficiency with 70% full load and 0.8 pf lagging = 97.41% (not sure)
    iii) the regulation for the condition of b ii)

    My problem is that I'm not sure if i should consider Inot(no load component) in part a) to workout b ii) and where to apply the 0.8 pf lag.
     
  2. Tera-Scale

    Thread Starter Active Member

    Jan 1, 2011
    164
    5
    I used the Power loss at full load and worked it out at 70% load to find the regulation at b ii) to be 0.9885

    Can anyone confirm it for me pls?
     
  3. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    5,448
    782
    The worst case efficiency would occur at rated load of 50kVA.

    The stated full load loss is 850 Watts.

    So the full load efficiency would be 100* (50,000-850)/50,000 expressed as a %age - i.e 98.3%. So your answer of 97.41% at 70% of rated load is not consistent.

    I did some rough calcs as follows.

    Primary side Full load current = 50,000/11,000=4.545A

    The primary series resistance loss is approximately the full load loss minus the no load loss - i.e. 850W-650W=200W

    The series resistance is then Rps=200W/(4.545^2)=9.68Ω

    The stated primary series impedance is Zeq=125Ω (complex)

    Hence Xps=√(125^2-9.68^2)=124.6Ω

    At 70% rated load and 0.8pf the current will be 0.7*4.545 or 3.182A @ -36.87° [2.546-j1.91].

    The primary series power loss would then be 3.182^2*9.68=98W.

    Assume the iron losses are about the same @ 650W.

    Total losses at 70% rated power would be 748W.

    Giving an efficiency of 100*(50,000-748)/50,000=98.5%

    The primary voltage drop at 70% rated load and 0.8pf would be

    (9.68+j124.6)*(2.546-j1.91)=262.5+j298.7=397.7 @48.68° Volts

    The regulation would then be

    100*(11,000-397.7@46.68°)/11,000=97.65%.

    So I also get different answer on that score.
     
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