Power Generation MCQs Part 28

Contents

- 1 A large hydropower station has head of 350 m and an average flow of 1400
- 2 The knowledge of diversity factor helps in computing
- 3 The load factor (LF), average load (AL) and maximum demand (MD) are related mathematically by the formula
- 4 Demand factor is usually
- 5 A power station has maximum demand of 20000 kW

## A large hydropower station has head of 350 m and an average flow of 1400

A large hydropower station has head of 350 m and an average flow of 1400 m³/s. The reservoir is composed of series of lakes covering an area of 6400 km². The number of days the the available hydraulic power could be sustained if level of impounded water were allowed to drop by 1 m is:

- 32 days
- 39 days
- 46 days
- 53 days

Correct answer: 4. 53 days

Solution:

t = 6400 * 10^6/1400 = 4571428.57142857 s = 1269.8 h = 53 days

## The knowledge of diversity factor helps in computing

The knowledge of diversity factor helps in computing:

- Average load
- Peak load

- Units generated in kWh

Correct answer: 3. Plant capacity

The load factor (LF), average load (AL) and maximum demand (MD) are related mathematically by the formula:

- LF = AL * MD
- LF = AL/MD
- LF = AL + MD
- LF = AL – MD

Correct answer: 2. LF = AL/MD

## Demand factor is usually

Demand factor is usually:

- More than 1
- Equal to 1
- Less than 1
- None of these

Correct answer: 3. Less than 1

## A power station has maximum demand of 20000 kW

A power station has maximum demand of 20000 kW. The annual load factor is 50% and plant capacity factor is 40. What is reserve capacity?

- 4750 kW
- 5000 kW
- 5250 kW
- 5500 kW

Correct answer: 2. 5000 kW

Solution:

Energy generated per annum = Max * LF * Hourse in year = 20000 * 0.5 * 8760 kWh = 87.6 * 10^6 kWh

——————–

Plant capacity factor = Units generated per annum/(Plant capacity * Hours in a year)

——————–

Plant capacity = 87.6 * 10^6/(0.4*8760) = 25000 kW