Question

wers with necessary diagrams and equations b) 15 Two balanced loads are connected to a 240 V, 50-H line as shown in Fig. 7(b) below. Load I draws 30 kW at a power factor of 0.6 Ingrins, while load 2 draws 45 KVAR at a power factor of 0.8 lagging. Assuming the absequence, determine the line curents Hallalla, in and ic as indicated in Fig. 7(b). Balanced load! Balanced load 2 Fig. 7(b) 8 a Forbalanced delta-wye system determine the p e

(Appropriate and detailed solution will certailny help a lot. Thaanks for your support.)

Answer 1

Ans)

Given

For load 1

$V_{line}=240\: \: kV$

$P_{1}=30\: \: kW,\: \: pf_{1}=cos\varphi _{1}=0.6\: \: lagging$

now

$I_{a1}=\frac{P_{1}}{\sqrt{3}V_{line}*pf_{1}}\angle -cos^{-1}pf_{1}=\frac{30k}{\sqrt{3}*240k*0.6}\angle -cos^{-1}0.6$ negative sign in angle because power factor is lagging current lags behind voltage

$\mathbf{I_{a1}=0.12\angle -53.13^{o}\: \: A}$

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For load 2

$Q_{2}=45\: \: kVAR,\: \: pf_{2}=cos\varphi _{2}=0.8\: \: lagging$

$P_{2}=Q_{2}cot\varphi _{2}=45k*cot(cos^{-1}0.8)=60\: kW$

$P_{2}=60\: kW$

now

$I_{a2}=\frac{P_{2}}{\sqrt{3}V_{line}*pf_{2}}\angle -cos^{-1}pf_{2}=\frac{60k}{\sqrt{3}*240k*0.8}\angle -cos^{-1}0.8$

$\mathbf{I_{a2}=0.18\angle -36.87^{o}\: \: A}$

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now

$I_{a}=I_{a1}+I_{a2}=0.12\angle -53.13^{o}+0.18\angle -36.87^{o}$

$\mathbf{I_{a}=0.298\angle -43.36^{o}\: \: A}$

as the given circuit is balanced all the line currents are same in magnitude with phase difference of 120 degrees as below

$I_{b}=I_{a}\angle -120^{o}=0.298\angle -43.36^{o}-120^{o}$

$\mathbf{I_{b}=0.298\angle -163.36^{o}\: \: A}$

$I_{c}=I_{a}\angle +120^{o}=0.298\angle -43.36^{o}+120^{o}$

$\mathbf{I_{c}=0.298\angle 76.64^{o}\: \: A}$