Question

The following table gives the total real and reactive power losses for the secondary network given in Example 6.5. Explain the circumstances which cause minimum and maximum losses. 6.8 Bear in mind that the total P+jQ power delivered to the loads is identical in all cases. ZQu Mvar EP, MW Case No. 0.38807 0.33142 0.46648 0.82477 0.16379 0.14160 0 19263 0.36271

Answer 1

Case	P	Q	power factor= Cos($\o$)	Losses
1	0.16739	0.38807	0.3960	MEDIUM
2	0.14160	0.33142	0.3928	MEDIUM
3	0.19263	0.46648	0.3816	HIGHEST
4	0.36271	0.82477	0.4025	LOWEST

NOTE ---------------------->>>>>>>>>> pf angle $\o$ = Tan inverse ( Q/ P ) in above calculations

So

case 3 will have highest losses as power factor is lowest among all given cases

case4 will have lowest losses as power factor is highest among all given cases

*********************This is because , lower the power factor , more will be the reactive power , so more current will be drawn from the lines causing more losses and less use of active power .

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