This set of Irrigation Engineering Multiple Choice Questions & Answers (MCQs) focuses on “Hydropower”.
1. The minimum power which a hydropower plant can generate throughout the year is called as ________________
a) power plant capacity
b) power plant load
c) firm power
d) water power
Explanation: The firm power is the net amount of power which is continuously available from a plant without any break on a guaranteed basis. The consumers can always be sure of getting this power and this power is available under the most adverse hydraulic conditions.
2. If the peak load for a power plant equals the plant capacity then the ratio of the capacity factor to load factor will be _________________
c) < 1
d) > 1
Explanation: Load factor = Average load over a certain period / Peak load during that period
Capacity factor = Average load over a given time period / Plant capacity
Since, Plant capacity = Peak load
Ratio of CF / LF = 1.
3. If the peak load on a power plant having a capacity of 100 MW is 70 MW during a given week and the energy produced is 58, 80,000 kWh, the capacity factor for the plant for the week will be ___________________
Explanation: Capacity factor = Average load/ Plant capacity
Average load = Energy produced / Time (in hours) = 58,80,000 / 7 x 24 = 35000 kW
C.F = 35 MW / 100 MW = 0.35 (or 35%).
4. During a certain week a power plant turns out 84,00,000 kWh and the peak load during the week is 100,000 kW. What is the load factor during the week?
Explanation: Load factor can be defined as the ratio of average load to the peak load over a given period.
Average load = 84,000,000 / (24 x 7) = 50,000 kW
L.F = Average load / Peak load = 50,000 / 100,000 = 0.5 or 50%.
5. A canal drop is 6 meters and discharges available through the turbine is 50 cumec. Find the electrical energy available.
a) 2352 KW
b) 2352 MW
c) 2.352 kW
d) 235.2 kW
Explanation: The electrical energy (in kW) = 7.84 Q H (using 80% efficiency) where H is the design head in meters and Q is the design discharge in cumecs.
Electrical energy = 7.84 x 50 x 6 = 2352 kW.
6. The ratio of maximum power utilised to the maximum power available is _____________
a) power factor
b) plant use factor
c) reserve capacity
d) capacity factor
Explanation: Plant use factor is nothing but utilization factor which can be defined as the ratio of maximum power utilized to the maximum power available if the water head is assumed to be constant. The value usually varies from 0.4 to 0.9 for a hydel plant.
7. The net amount of power which is continuously available from a plant without any break is known as _____________
a) firm power
b) secondary power
c) power factor
d) utilization factor
Explanation: Secondary power is the excess power available over firm power during peak-off hours. It is supplied to consumers as and when available basis. Power factor is the ratio of actual power (in kW) to the apparent power (in kilo volt-ampere).
8. The value of power factor is generally _____________________
a) equal to unity
b) less than unity
c) greater than unity
d) equal to zero
Explanation: The value of power factor depends upon the relationship between the inductance and resistance in the load and it can never be greater than unity. The usual system load has a power factor varying from 0.8 to 0.9. But if various induction motors are installed in the load, the value will be 0.5.
9. To ensure maximum overall plant efficiency, the rated head should be equal to ______________
a) the design head
b) the gross head
c) the operating head
d) effective head
Explanation: The Rated head is the head at which the turbine functioning at full gate opening will produce a power output. It is nothing but the specified head that is in the nameplate of the turbine. This is equal to the design head of the turbine so as to ensure maximum overall plant efficiency.
10. The difference of head at the point of entry and exit of turbine is _______________
a) design head
b) gross head
c) effective head
d) rated head
Explanation: Design head is the net head under which the turbine reaches peak efficiency. The difference in the water level elevations at the point of diversion of water for the hydel scheme and the point of return of water back to the river is called a gross head. The rated head is the head that is specified in the name-plate of the turbine.
11. The load on a hydel plant varies from a minimum of 10,000 kW to a maximum of 35,000 kW. Two turbo-generators of capacities 20,000 kW each have been installed. Calculate Plant factor.
Explanation: Plant factor can be defined as the ratio of the energy actually produced (Average load) to the maximum energy that can be produced at a particular time.
Since two turbo generators are installed, the total installed capacity = 40000 kW
P.F = [(10000 + 35000) / 2] / 40000 = 56%.
12. Calculate utilisation factor if the maximum power utilised is 40,000 kW and two turbo generators installed each of capacity 23,000 kW.
Explanation: Utilisation factor = Maximum power utilised / Maximum power available
U.F = 40,000 / (2 x 23000) = 86.9% (since two generators are installed).
13. If the peak load on a power plant having a capacity of 100 MW is 70 MW during a given week. Calculate the utilization factor.
Explanation: The utilization factor is the ratio of the maximum power utilized to the maximum power available.
U.F = Max power utilised/ Max power available = 70 / 100 = 0.7.
14. 400 cumecs of water are being released from dam storage to meet the downstream demand through the turbines of the connected hydro plant. The effective head of water acting on the turbines is 50 m. The efficiency of the hydro plant is 0.8. The electrical power generated from this plant is __________________
a) 1,56,800 MW
b) 156.8 M kW
c) 156.8 MW
d) 156.8 kW
Explanation: Using 80% efficiency, Electrical energy = 7.84 QH where H is the design head in meters and Q is the design discharge in cumecs.
Electrical energy = 7.84 x 400 x 50 = 1, 56,800 kW = 156.80 MW.
15. What is the highest elevation of water level that can be maintained in the reservoir without any spillway discharge either with gated or non-gated spillway?
a) Normal Water level
b) Minimum Water level
c) Weighted average level
d) Operating head
Explanation: Minimum water level is that elevation of water level which produces minimum net head on the power units i.e. 65% of the design head. The weighted average level is the level above and below which equal amounts of power are developed during an average year. Operating head is the difference of elevation of entrance and the tail-race exit.
Sanfoundry Global Education & Learning Series – Irrigation Engineering.
To practice all areas of Irrigation Engineering, here is complete set of 1000+ Multiple Choice Questions and Answers.
- Get Free Certificate of Merit in Irrigation Engineering
- Participate in Irrigation Engineering Certification Contest
- Become a Top Ranker in Irrigation Engineering
- Take Irrigation Engineering Tests
- Chapterwise Practice Tests: Chapter 1, 2, 3, 4, 5, 6, 7, 8, 9, 10
- Chapterwise Mock Tests: Chapter 1, 2, 3, 4, 5, 6, 7, 8, 9, 10