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The power in a 3 - phase
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written 3.5 years ago by |
Given:
$W_1+W_2=50\ kW..................(1)\\ $
$pf=0.6\ (lagging)$
$\phi=cos^{-1}(0.6)=53.13^{\circ}\\ tan\phi =\sqrt 3\dfrac{W_1-W_2}{W_1+W_2}\\ tan(53.13^{\circ})=\sqrt3\dfrac{W_1-W_2}{50}\\ W_1-W_2=38.49\ kW............................(2)$
Solving equations (1) and (2),
$W_1=44.25\ kW; \ \ \ \ \ W_2=5.75\ kW$