MGVCL Exam Paper (30-07-2021 Shift 3) The process of peeling of rocks into layers is called Delta Exfoliation Sublimation Brachans Delta Exfoliation Sublimation Brachans ANSWER DOWNLOAD EXAMIANS APP
MGVCL Exam Paper (30-07-2021 Shift 3) Match the following shown in table A = (ii), B = (i), C = (iii) A = (ii), B = (iii), C = (i) A = (iii), B = (ii), C = (i) A = (iii), B = (i), C = (ii) A = (ii), B = (i), C = (iii) A = (ii), B = (iii), C = (i) A = (iii), B = (ii), C = (i) A = (iii), B = (i), C = (ii) ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Bus Type - Known Parameter - Unknown ParameterLoad Bus -P, Q - V, phase angleGenerator Bus - P, V (magnitude) - Q, Voltage phase angleSlack Bus Voltage - magnitude and phase angle - P, Q
MGVCL Exam Paper (30-07-2021 Shift 3) Consider a solar PV plant with the following specific conditions:Analysis period: 1 yearMeasured average solar irradiation intensity in 1 year: 150 kWh/m²Generator area of the PV plant: 10 m² Efficiency factor of the PV modules: 15%Electrical energy actually exported by plant to grid: 135 kWhCalculate the performance ratio. 75% 60% 80% 50% 75% 60% 80% 50% ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Where,A - total solar panel area in meter square.r - efficiency of solar panelh - solar installation per meter square area.Solar panel performance ratio = Actual energy (kWh)/(A*r*h)= (135/(0.15*10*150))= 0.60= 60 % (in percentage)
MGVCL Exam Paper (30-07-2021 Shift 3) A 1000 kVA ONAN cooled transformer has a load of 500 kVA throughout the day except for a period of 2 hours. What is the permissible overload for a duration of two hours. Assume the permissible load kVA as a fraction of rated kVA is 1.43. 700 kVA 715 kVA 1050 kVA 1430 kVA 700 kVA 715 kVA 1050 kVA 1430 kVA ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Permissible overload for transformer = Full load kVA*factor for permissible overload= 1000*1.43= 1430 kVA
MGVCL Exam Paper (30-07-2021 Shift 3) The current flowing in to a balanced delta connected load through line 'a' is 10 A when the conductor of line ’b’ is open. With the current in line 'a' as reference, compute the symmetrical components of the line currents. Assume the phase sequence of ’abc’. Ia₀ = 10 A.Ia₁ = (-5 + j2.44) A.Ia₂ = (5 - j2.44) A. Ia₀ = 10 A.Ia₁ = (-1.66 + j2.88) A.Ia₂ = (1.66 + j2.88) A. Ia₀ = 0 A.Ia₁ = (-1.66 + j2.88) A.Ia₂ = (1.66 + j2.88) A. Ia₀ = 0 A.Ia₁ = (-5 + j2.88) A.Ia₂ = (5 - j2.88) A. Ia₀ = 10 A.Ia₁ = (-5 + j2.44) A.Ia₂ = (5 - j2.44) A. Ia₀ = 10 A.Ia₁ = (-1.66 + j2.88) A.Ia₂ = (1.66 + j2.88) A. Ia₀ = 0 A.Ia₁ = (-1.66 + j2.88) A.Ia₂ = (1.66 + j2.88) A. Ia₀ = 0 A.Ia₁ = (-5 + j2.88) A.Ia₂ = (5 - j2.88) A. ANSWER EXPLANATION DOWNLOAD EXAMIANS APP For the given question,Ib = 0Ia = -IcCalculation:Ia1 = 1/3*(Ia + a*Ib + a²*Ic)= 1/3*(10 + 10∟-120°)= 5 + j*2.88 AIa2 = 1/3*(Ia + a²*Ib + a*Ic)= (1/3)*(10 + 10∟120°)= 5 - j*2.88 AIao = (1/3)*(Ia + Ib + Ic)= (Ia - Ia)= 0
MGVCL Exam Paper (30-07-2021 Shift 3) Which state is the largest producer of wind energy in India? Andhra Pradesh Tamil Nadu Maharashtra Madhya Pradesh Andhra Pradesh Tamil Nadu Maharashtra Madhya Pradesh ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Rank - State - Capaciy1 - Tamilnadu - 7.5 GW2 - Maharastra - 5 GW3 - Karnataka - 4.8 GW4 - Rajsthan - 4.3 GW
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