MGVCL Exam Paper (30-07-2021 Shift 2) Which of the following is an application layer service? File transfer, access and management Network virtual terminal Mail service Mail service File transfer, access and management Network virtual terminal Mail service Mail service ANSWER DOWNLOAD EXAMIANS APP
MGVCL Exam Paper (30-07-2021 Shift 2) પાણી : પાણીગ્રહણ :: હસ્ત : ??? છુટાછેડા પ્રેમાલાપ હસ્તમેળાપ વિવાહ છુટાછેડા પ્રેમાલાપ હસ્તમેળાપ વિવાહ 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) હુ ગીત ગાઈશ' ને કર્મણીમાં ફેરવો. ગીતને હુ જ ગાઇશ હુ ગીતને ગાઈશ મારાથી ગીત ગવાશે ગીતને હુ ગાઈશ ગીતને હુ જ ગાઇશ હુ ગીતને ગાઈશ મારાથી ગીત ગવાશે ગીતને હુ ગાઈશ ANSWER DOWNLOAD EXAMIANS APP
MGVCL Exam Paper (30-07-2021 Shift 2) A circuit breaker is rated 2500 A, 2000 MVA, 33 kV, 3 sec, 3-phase oil circuit breaker. Determine the breaking current. 25 kA 2500 A 350 kA 35 kA 25 kA 2500 A 350 kA 35 kA ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Breaking capacity of circuit breaker = √3*VL*IbVL - line voltage in voltIL - Line current/breaking in ampereVL = 33 kVS = 2000 MVAIb = 30*1000 (in kVA)/(√3*33)Ib = 34.99 kA
MGVCL Exam Paper (30-07-2021 Shift 2) A 1000 kVA ONAN cooled transformer has a load of 500 kVA throughout the day except for a period of 4 hours. What is the permissible overload for a duration of four hours. Assume the permissible load kVA as a fraction of rated kVA is 1.24. 2480 kVA 930 kVA 620 kVA 1240 kVA 2480 kVA 930 kVA 620 kVA 1240 kVA ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Permissible overload = Full load kVA*factor for permissible load= 1000*1.24= 1240 kVA
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