MGVCL Exam Paper (30-07-2021 Shift 2) Which of the following is an application layer service? Network virtual terminal File transfer, access and management Mail service Mail service Network virtual terminal File transfer, access and management Mail service Mail service ANSWER DOWNLOAD EXAMIANS APP
MGVCL Exam Paper (30-07-2021 Shift 2) A three-phase, 4-pole, 50 Hz induction motor runs at 1425 rpm. Find out the percentage slip of the induction motor. 4% 3% 6% 5% 4% 3% 6% 5% ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Slip, s = (Ns - Nr)/NsNs = 120f/PNs = 1500 rpms = (1500 - 1425)/1500s = 0.05%s = 5%
MGVCL Exam Paper (30-07-2021 Shift 2) Find the word which is correctly spelt from the given options. Wrinkel Sentimant Maximum Absurrd Wrinkel Sentimant Maximum Absurrd ANSWER DOWNLOAD EXAMIANS APP
MGVCL Exam Paper (30-07-2021 Shift 2) A 50 Hz synchronous generator is connected to an infinite bus through a line. The p.u. reactances of generator and the line are j0.3 p.u. and j0.2 p.u. respectively. The generator no load voltage is 1.1 p.u. and that of infinite bus is 1.0 p.u. The inertia constant of the generator is 4 MW-sec/MVA. Determine the frequency of natural oscillations if the generator is loaded to 60% of its maximum power transfer capacity and small perturbation in power is given. 1.42 Hz 1.52 Hz 1.62 Hz 1.32 Hz 1.42 Hz 1.52 Hz 1.62 Hz 1.32 Hz ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Frequency of natural oscillation is given by,fn = {((dPe/dδ)at(δo))/M)}dPe/dδ = ((V1*V2)/X*(cosδ))= (11/05)*cosδ= (11/0.5)*0.5M = (H*s)/(πf)= 4/(50π)fn = (1.76*(3/50π))= 9.4 rad/sec = 1.32 Hz
MGVCL Exam Paper (30-07-2021 Shift 2) A three-phase, 12 kV, 50 Hz and 300 kW load is operating at 0.8 pf lag. If the pf is to be improved to unity, using star connected capacitor, the per phase value of the reactive VAR required is: 25 kVAR 50 kVAR 75 kVAR 100 kVAR 25 kVAR 50 kVAR 75 kVAR 100 kVAR ANSWER EXPLANATION DOWNLOAD EXAMIANS APP The reactive VAR required = kW*(tanφ1 -tanφ2)Cosφ1 = 0.8φ1 = 36.87 deg.Cosφ2 = 0φ2 = 0 deg.kVAR required = kW*(tan(36.87) - tan(0))= 300*1000(0.75)= 225.00 kVARPer phase value = 225.00/3= 75 kVAR
MGVCL Exam Paper (30-07-2021 Shift 2) In a short circuit test on circuit breaker, time to reach the peak re-striking voltage is 50 µs with the peak restriking voltage of 100 kV. Determine average RRRV (Rate of rise re-striking voltage)? 4 x 10⁶ kV/sec 3 x 10⁶ kV/sec 2 x 10⁶ kV/sec 1 x 10⁶ kV/sec 4 x 10⁶ kV/sec 3 x 10⁶ kV/sec 2 x 10⁶ kV/sec 1 x 10⁶ kV/sec ANSWER EXPLANATION DOWNLOAD EXAMIANS APP RRRV = Peak value of restriking voltage/time to reach the peak re-striking voltageRRRV = 100*1000/(50 μs)RRRV = 2*10⁶ kV/sec
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