Power Generation MCQs Part 28

Power Generation MCQs Part 28

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:

1. 32 days
2. 39 days
3. 46 days
4. 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:

1. Average load
2. 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

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

1. LF = AL * MD
2. LF = AL/MD
3. LF = AL + MD
4. LF = AL – MD

Correct answer: 2. LF = AL/MD

Demand factor is usually

Demand factor is usually:

1. More than 1
2. Equal to 1
3. Less than 1
4. 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?

1. 4750 kW
2. 5000 kW
3. 5250 kW
4. 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