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# GATE 2015 EE - SET 1 - Question 43

Solution :

$\delta$= $\frac{P_{a}t^{2}}{2M}$ , where M = $\frac{H}{f*180}$

where H is the ratio of the stored KE to machine rating

H = 2 , f = 50 ,t = 0.02 ,P=1 pu

$\delta$= $\frac{P_{a}t^{2}}{2M}$ =  $\frac{Pt^{2}}{2*\frac{H}{f*180}}$

$\delta$= $\frac{1*(0.02)^{2}}{2*\frac{2}{50*180}}$ = 0.9

$\delta$ = 5 +0.9 = $5.9^{0}$

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