%3Cp%3E%3Cem%3EA%20power%20source%20feeds%20a%2010kVA%201%2F2.5kV%20transformer%

Question

%3Cp%3E%3Cem%3EA%20power%20source%20feeds%20a%2010kVA%201%2F2.5kV%20transformer%20through%20a%0Afeeder%20impedance%20of%2042%2Bj150%20ohms.%20The%20transformer's%20equivalent%0Aseries%20impedance%20referred%20to%20its%20low-voltage%20side%20is%200.12%2Bj0.5%0Aohms.%20The%20excitation%20branch%20can%20be%20neglected.%20The%20load%20on%20the%0Atransformer%20is%2095kW%20at%200.85%20PF%20lagging%20and%202400%20V.%3C%2Fem%3E%3C%2Fp%3E%0A%3Cp%3E%3Cbr%20%2F%3E%3C%2Fp%3E%0A%3Cp%3E%3Cem%3EDetermine%20the%20source%20voltage%2C%20regulation%20of%20the%20transformer%0Aand%20the%20overall%20efficiency%20of%20the%20power%20system.%3C%2Fem%3E%3Cbr%20%2F%3E%3C%2Fp%3E%0A

Explanation / Answer

Total impedanc from souce to load=42.12+150.5j ohms (Adding transformer series impedance and feeder impedance

Impedance magntiude =156.2829 = abs(z)

The total active power trasnferred    (= Vs*Vr*sin(lpha) / abs(Z))   

Assuming a infinte power bus at source,

Phase at source =0

Phase at load= cosinvers(0.85) =31.7883

alpha = (31.7883-0)=relative phase difference

P=9.5*1000=Vs*2400*sin(31.7883)/156.2829

Solving for Vs =9.5*1000*156.2829/(2400*sin(31.7883)) = 1174.3 V

Here i think the power trasmitted cannot be 95 kw as the KVA rating of transormer is just 10 KVA.Assuming the load is 9.5 kw the solution is obtained

B. V.R = I*Re*Cos(alpha)/ Vs+I*Xe*sin(alpha)/Vs

I = 2400-1174.3/ abs(z)

=4.3124 A

V.R=4.3142(42*cos(31..7883)+150*sin(31.7883))/(1174.3)

=0.4214= 42.14%

C. Losses = I*I * Req= (4.3124*4.3124)*(42.12)=783.296 w

Efficiecny = o/p power/ (I/P power+ losses)=95/(95+0.7832)=0.9918

=99.182 %

Only copper losses are considered as there is no data on other losses

  

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