Question
A 20 MVA, 11.2 kV, 4-pole, 50 Hz alternator has an inertia constant of 15 MJ/MVA. If the input and output powers of the alternator are 15 MW and 10 MW, respectively, the angular acceleration in mechanical degree/ s2 is _______. (round off to nearest integer)
Answer :
75 (Range : 74 to 76)
Given :
(i) S = 20MVA
(ii) Voltage (V) = 11.2kV
(iii) Number of poles, P=4
(iv) Inertia constant, H=15 MJ/MVA
(v) Input power, Pin= 15MW= Pm
(vi) Output power, Pe= 10MW
Angular acceleration______ mech.degree/s2
Swing equation, Md2δ/dt2=Pm-Pe
α=d2δ/dt2=Pm-Pe/M elect.degree/s2 ..........(i)
So equation (i),
d2δ/dt2=Pm-Pe/M elect.degree/s2
d2δ/dt2=Pm-Pe/M × 2/P mech.deg./s2
d2δ/dt2= Pm-Pe/(GH/πƒ) × 2/4 mech.deg./s2
d2δ/dt2= 75 mech.deg./s2
Hence, the correct answer is 75.
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