Transformer Questions

Discussion in 'Math' started by powdermonkey, Aug 6, 2012.

  1. powdermonkey

    Thread Starter New Member

    Aug 6, 2012
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    0
    Gday

    I'm tackling AC for the first time in 20 something years so that I can re-educate myself, and its been really hard going!! I'm stuck on a problem and I thought I understood the theory but cant make the connection to join all the dots - tbh I'm not sure that this is the right forum but you all seem to know your stuff so I thought I'd give it a punt....

    I'm struggling to find an example similar on the web so I can work through it and understand what I have to do to solve the problem but heres the question:

    A 250kVA transformer is at full load with a lagging power factor of 0.75. To correct the power factor to 0.9 a capacitor is placed in parallel to the secondary of the transformer. What is the required KVA of the capacitor pack Also how many KVAs is the transformer providing at the new power factor?

    I'm aware of the trigonometry bits and pf=cos θ and thus can calculate an angle to then work out KW and KVAR but apart from that and calculating that I think its a 17% increase in pf. I cant join the dots in my knowledge to get past having 2 unkowns in my equations to do more calculations. I know I'm missing something but cant work out what it is... any worked example help with explanations would be awesome. Really want to be able to get to grips with this.

    Cheers

    Simo
     
  2. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    5,448
    782
    The KVAR for the uncompensated load is determined first.

    With lagging pf=0.75 the load power phase angle is acos(0.75)=41.41°

    The uncompensated load KVAR is then 250*sin(41.41)=165.36KVAR [lagging]

    The load power is 250*0.75=187.5kW.

    A resulting compensated 0.9 pf would require a compensated kVA of 187.5kW/0.9=208.33 kVA.

    The compensated kVAR would then be √{(208.33)^2-(187.5)^2} kVAR or 90.8 kVAR [lagging]

    The change in KVAR required is 165.36-90.8=74.56 kVAR which would equate to the required compensation capacitance kVAR [kVA].
     
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  3. powdermonkey

    Thread Starter New Member

    Aug 6, 2012
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    Thank you!! Could you just explain why the 0.9pf is multiplied by the 187.5kw and not the 250 as the 0.75 is? And where is the compensated kVAR equation √{(208.33)^2-(187.5)^2} derived from?
     
    Last edited: Aug 7, 2012
  4. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    5,448
    782
    The 0.9pf isn't multiplied by 187.5kW - re-check what I wrote.

    The new apparent power is given by {Real Load Power} / {new power factor} since Real power = apparent power * pf.

    In this instance I assumed the change in pf was not accompanied by any change in real load power.

    From the power triangle we have using Pythagoras ...

    The Apparent Power S [kVA] =√{P^2+Q^2}

    where P = real power, Q = reactive power

    Hence Q = √{S^2-P^2}

    and the actual values are:-

    S = 208.33 kVA
    P = 187.5 kW
    Q = 90.8 kVAR
     
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  5. powdermonkey

    Thread Starter New Member

    Aug 6, 2012
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    0
    I understand it - which is sweet, dots joined really appreciate your help!!
     
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