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The daily energy produced in a thermal power station is 720 MWh at a load factor of 0.6.What is the maximum demand of the station
- a)50 MW
- b)30 MW
- c)72 MW
- d)720 MW
Correct answer is option 'A'. Can you explain this answer?
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Load factor:
The load factor of a power station is defined as the ratio of average load to the maximum demand during a specific period of time. It gives an indication of how efficiently the power station is utilized.
Given:
- Daily energy produced in a thermal power station = 720 MWh
- Load factor = 0.6
Step 1: Calculate the maximum demand:
The maximum demand can be calculated using the formula:
Maximum demand = (Average load) / (Load factor)
Let's substitute the given values:
Maximum demand = 720 MWh / 0.6
Step 2: Convert units:
To convert the maximum demand from MWh to MW, we need to consider the time period. In this case, the given energy is for a daily period. Since power is energy per unit time, we divide the energy by the time to get the power.
1 day = 24 hours
Maximum demand = (720 MWh / 0.6) / 24 h
Step 3: Calculate the maximum demand:
Let's substitute the values and calculate:
Maximum demand = (720 MWh / 0.6) / 24 h
Maximum demand = 1200 MW
Conclusion:
The maximum demand of the thermal power station is 1200 MW. Among the given options, the closest value is 50 MW, which is option A.
The load factor of a power station is defined as the ratio of average load to the maximum demand during a specific period of time. It gives an indication of how efficiently the power station is utilized.
Given:
- Daily energy produced in a thermal power station = 720 MWh
- Load factor = 0.6
Step 1: Calculate the maximum demand:
The maximum demand can be calculated using the formula:
Maximum demand = (Average load) / (Load factor)
Let's substitute the given values:
Maximum demand = 720 MWh / 0.6
Step 2: Convert units:
To convert the maximum demand from MWh to MW, we need to consider the time period. In this case, the given energy is for a daily period. Since power is energy per unit time, we divide the energy by the time to get the power.
1 day = 24 hours
Maximum demand = (720 MWh / 0.6) / 24 h
Step 3: Calculate the maximum demand:
Let's substitute the values and calculate:
Maximum demand = (720 MWh / 0.6) / 24 h
Maximum demand = 1200 MW
Conclusion:
The maximum demand of the thermal power station is 1200 MW. Among the given options, the closest value is 50 MW, which is option A.
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The daily energy produced in a thermal power station is 720 MWh at a load factor of 0.6.What is the maximum demand of the stationa)50 MWb)30 MWc)72 MWd)720 MWCorrect answer is option 'A'. Can you explain this answer?
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