Power Generation MCQs Part 14

Contents

- 1 Overall efficiency of a steam power station is given by
- 2 The penstock, turbine and generator efficiencies of a hydroelectric power plant are 95, 90, and 85 respectively. The overall efficiency of plant is:
- 3 The ideal load factor should be
- 4 One atomic mass unit is equivalent to:
- 5 A power system has a connected load of 120 kW, maximum load of 100 kW, Base load of 25 kW and avera

## Overall efficiency of a steam power station is given by

Overall efficiency of a steam power station is given by:

- Overall efficiency = Thermal Efficiency * Electrical efficiency
- Overall efficiency = Thermal Efficiency * Mechanical efficiency
- Overall efficiency = Thermal Efficiency * Prime mover efficiency
- None of these

Correct answer: 1. Overall efficiency = Thermal Efficiency * Electrical efficiency

## The penstock, turbine and generator efficiencies of a hydroelectric power plant are 95, 90, and 85 respectively. The overall efficiency of plant is:

The penstock, turbine and generator efficiencies of a hydroelectric power plant are 95, 90, and 85 respectively. The overall efficiency of plant is:

- 62.7%
- 72.7%
- 82.7%
- 92.7%

Correct answer: 2. 72.7%

Solution:

Overall η = η (penstock) * η (turbine) * η (generator)

Overall η = 0.95 * 0.9 * 0.85 = 0.72675 = 72.7%

## The ideal load factor should be

The ideal load factor should be:

- 0.3
- 0.5
- 0.8
- 1

Correct answer: 4. 1

## One atomic mass unit is equivalent to:

One atomic mass unit is equivalent to:

- 210.3 MeV
- 455.6 MeV
- 822.4 MeV
- 931.4 MeV

Correct answer: 4. 931.4 MeV

## A power system has a connected load of 120 kW, maximum load of 100 kW, Base load of 25 kW and avera

A power system has a connected load of 120 kW, maximum load of 100 kW, Base load of 25 kW and average load of 50 kW. The load factor is:

- 0.4
- 0.5
- 0.6
- 0.7

Correct answer: 2. 0.5

Explanation: Load factor = Average load/Maximum demand = 50 kW/100 kW = 0.5