Power Generation MCQs Part 21

Contents

- 1 Which of following power plants will take the least time in starting from cold conditions to full load conditions:
- 2 A power supply has domestic load having maximum demand of 2000 kW. If diversity factor and demand factor are 1.15 and 0.8 respectively, then connected domestic load is
- 3 Which of the following power plant has no standy losses
- 4 The maximum demand of a consumer is 5 kW and daily energy consumption is 40 units. The load factor is
- 5 A hydroelectric station is supplied from a catchment area of 200 km² with annual rainfall of 210 cm and has an effective head of 300 m. The yield factor is 65%. The available power is

## Which of following power plants will take the least time in starting from cold conditions to full load conditions:

Which of following power plants will take the least time in starting from cold conditions to full load conditions:

- Hydroelectric power plant
- Gas power plant
- Steam power plant
- Nuclear power plant

Correct answer: 1. Hydroelectric power plant

## A power supply has domestic load having maximum demand of 2000 kW. If diversity factor and demand factor are 1.15 and 0.8 respectively, then connected domestic load is

A power supply has domestic load having maximum demand of 2000 kW. If diversity factor and demand factor are 1.15 and 0.8 respectively, then connected domestic load is:

- 2200
- 2250
- 2300
- 2350

Correct answer: 2. 2250

Solution:

Sum of max demands of different domestic consumers = Max domestic demand * Diversity factor = 2000 * 1.15 = 2300 kW

——————–

Connected domestic load = 2300/Demand factor = 2300/0.8 = 2875 kW

## Which of the following power plant has no standy losses

Which of the following power plant has no standy losses:

- Steam power plant
- Diesel power plant
- Nuclear power plant
- Hydroelectric power plant

Correct answer: 4. Hydroelectric power plant

## The maximum demand of a consumer is 5 kW and daily energy consumption is 40 units. The load factor is

The maximum demand of a consumer is 5 kW and his daily energy consumption is 40 units. The load factor is:

- 0.33%
- 9.33%
- 21.33%
- 33.33%

Correct answer: 4. 33.33%

Solution:

Load factor = {Daily consumption/(Max. demand * 24)} * 100 = (40/(5*24))*100 = 33.33%

## A hydroelectric station is supplied from a catchment area of 200 km² with annual rainfall of 210 cm and has an effective head of 300 m. The yield factor is 65%. The available power is

A hydroelectric station is supplied from a catchment area of 200 km² with annual rainfall of 210 cm and has an effective head of 300 m. The yield factor is 65%. The available power is:

- 25451kW
- 25901kW
- 26351kW
- 26801kW

Correct answer: 1. 25451kW

Solution:

Quantity of water available per annum = Catchment area * Rainfall * Yield factor = 200 * 10^6 * 2.1 * 0.65 = 273000000 m³

——————–

Rate of flow of water = 273000000/365*24*3600 = 8.657 m³/sec

——————–

Power available (P) = 9.8 QH kW = 9.8 * 8.657 * 300 = 25451 kW