# Power Generation MCQs Part 17

Power Generation MCQs Part 17

## A hydroelectric power station has catchment area of 4.5 * 10^8 m². The average rainfall per annum in this area is 130 cm. Assuming that 35% of rainfall is lost due to evaporation, the availability of water is

A hydroelectric power station has catchment area of 4.5 * 10^8 m². The average rainfall per annum in this area is 130 cm. Assuming that 35% of rainfall is lost due to evaporation, the availability of water is:

1. 4.8m³/sec
2. 9.8m³/sec
3. 10.7m³/sec
4. 11.6m³/sec

Solution:

Total quantity of water available in whole year = Area in m² * Rainfall in meters * Yield factor = 4.5 * 10^8 * 130 * 10^-2 * 0.65 = 365625000 m³

—————–

Discharge Q = Total quantity of water in year/Number of seconds in year = 365625000/(365*24*3600) = 11.6 m³/sec

## 1 kWh = X kcal. Here X implies:

1 kWh = X kcal. Here X implies:

1. 400
2. 860
3. 1022
4. 1365

## Highest point on daily load curve represents

Highest point on daily load curve represents:

2. Average demand
3. Maximum demand
4. None of these

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

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

1. 25000 kW
2. 25250 kW
3. 25500 kW
4. 25750 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

## A hydroelectric power plant has head of 325 m and has an average flow of 1400 m³/s. The reservoir has an area of 6400 km². The available hydroelectric power is

A hydroelectric power plant has head of 325 m and has an average flow of 1400 m³/s. The reservoir has an area of 6400 km². The available hydroelectric power is:

1. 25451MW
2. 25901MW
3. 26351MW
4. 26801MW