An interconnected cable link generating stations1 and 2 is shown in figure. The desired voltage profile is flat, i.e. $| V_{1} | = | V_{2} | = 1$ pu. The total demands at the two buses are
$S_{D_{1}} = (15 + j 5) p . u .$
$S_{D_{2}} = (25 + j 15) p . u .$

The station loads are equalized by the flow of power in the cable. Generator $G_{1}$ can generate a maximum of 20.0 p.u. real power. Sending end station power factor is_______ lagging.

0.9

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Solution

The correct option is A 0.9
Since there is no-real power loss in the cable
$P_{G 1} + P_{G 2} = P_{D 1} + P_{D 2} = 40 p . u .$
For equalization of station loads
$P_{G 1} = P_{G 2} = 20 p . u .$
Equalization means that,
$P_{s} = P_{R} = 20 p . u .$
$P_{s} = P_{R} = \frac{| V_{1} | | V_{2} |}{X} sin δ_{1}$
$5 = \frac{1 \times 1}{0.05} sin δ_{1}$
$δ_{1} = {14.5}^{\circ}$
$∴ V_{1} = 1 ∠ {14.5}^{\circ} p . u .$
Total load on station $1 = (15 + j 5) + (5 + j 0.64)$
=( 20+j5.64)
Power factor at station $1 = cos ({tan}^{- 1} \frac{564}{20})$
$= 0.9625$ lagging.

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Similar questions

| V_{1} | = | V_{2} | = 1 p u .

S_{D_{1}} = (30 + j 10) p u a n d S_{D_{2}} = (25 + j 10) p u .

G_{1} w h e n δ = 20^{\circ}

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