Power Generation MCQs Part 17
Contents
- 1 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
- 2 1 kWh = X kcal. Here X implies:
- 3 Highest point on daily load curve represents
- 4 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?
- 5 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 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:
- 4.8m³/sec
- 9.8m³/sec
- 10.7m³/sec
- 11.6m³/sec
Correct answer: 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:
- 400
- 860
- 1022
- 1365
Correct answer: 2. 860
Highest point on daily load curve represents
Highest point on daily load curve represents:
- Load factor
- Average demand
- Maximum demand
- None of these
Correct answer: 3. Maximum demand
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?
- 25000 kW
- 25250 kW
- 25500 kW
- 25750 kW
Correct answer: 1. 25000 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:
- 25451MW
- 25901MW
- 26351MW
- 26801MW
Correct answer: 1. 25451MW
Solution:
Formula for available hydroelectric power P is: P = 9.8 QH kW
——————–
P = 9.8 * 1400 * 325 kW = 4459 * 10³ kW = 4459 MW