EE MCQ

MGVCL Exam Paper (30-07-2021 Shift 2)
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Consider a solar PV plant with the following specific conditions: Analysis period: 1 year

Measured average solar irradiation intensity in 1 year: 120 kWh/m²

Generator area of the PV plant: 10 m²

Efficiency factor of the PV modules: 15%

Electrical energy actually exported by plant to grid: 110 kWh

Calculate the performance ratio.

Measured average solar irradiation intensity in 1 year: 120 kWh/m²

Generator area of the PV plant: 10 m²

Efficiency factor of the PV modules: 15%

Electrical energy actually exported by plant to grid: 110 kWh

Calculate the performance ratio.

The irradiation values measured on location yields an average solar irradiation for the entire analysis period of 120 kWh/m².

This irradiation value is extrapolated to the modular area of the PV plant as follows:

Irradiation value in kWh/m² x plant area in m² = 120 kWh/m² x 10 m² = 1,200 kWh

In order to subsequently calculate the nominal plant output, the irradiation value for the PV plant is multiplied by the modular efficiency:

1,200 kWh x 15 % = 1,200 kWh x 0.15 = 180 kWh

PR = 110 kWh/180 kWh = 0.61112 = 61.11 %

Related Question

MGVCL Exam Paper (30-07-2021 Shift 2)
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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.

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)*0.5

M = (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)
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According to Global Wind Energy Council in 2016, India ranked____ in the Global Wind Power Installed Capacity Index with cumulative installed wind power generation.

India ranked 4^th in the Global Wind Power Installed Capacity index with cumulative installed wind power generation capacity of 25088.

Wind power generation capacity in India has significantly increased in recent years. As of 28 February 2021, the total installed wind power capacity was 38.789 GW, the fourth largest installed wind power capacity in the world.

MGVCL Exam Paper (30-07-2021 Shift 2)
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The line currents in amperes in phases a, b and c respectively are 500 +j150, 100 – j600 and – 300 + j600 referred to the same reference vector with phase sequence of abc. Find the symmetrical component of currents.

Calculation:

Ia1 = 1/3*(Ia + a*Ib + a²*Ic)

= 500 + j150 + (- 0.5 + j*0.866)*(100 – j6000) + (-0.5 - j*0.866)*(– 300 + j600)

= (546.41 + j*156.47) A

Ia2 = 1/3*(Ia + a²*Ib + a*Ic)

= 500 + j150 + (- 0.5 - j*0.866)*(100 – j6000) + (-0.5 + j*0.866)*(– 300 + j600)

= (-146.41 - j*56.47) A

Iao = 1/3*(Ia + Ib + Ic)

= 1/3*(500 + j150 + 100 – j6000 – 300 + j600)

= (100 + j*50) A