The energy contained in the 37,500,000 yd³ of methane and the energy
produced of 61,626,000 kWh, give an efficiency of approximately<u> 19.5%</u>
<h3>How can the efficiency of a power plant be found?</h3>
The given parameters are;
The volume of methane combusted by the Frackile Power Plant = 37,500,000 yd³
The energy produced by the plant = 61,626,000 kWh
Required:
The efficiency of the power plant.
Solution:
The calorific vale of methane = 55.4 MJ/kg
37,500,000 yd³ = 28,670,807,174 L
The number of standard volumes are therefore;
![\dfrac{28,670,807,174 \, L}{22.414 \, L} \approx \mathbf{1279147281.79}](https://tex.z-dn.net/?f=%5Cdfrac%7B28%2C670%2C807%2C174%20%5C%2C%20L%7D%7B22.414%20%5C%2C%20L%7D%20%5Capprox%20%5Cmathbf%7B1279147281.79%7D)
Mass of the methane = 16.04 g × 1279147281.79 ≈ 2.05175224 × 10⁷ kg
Energy produced = 2.05175224 × 10⁷ kg × 55.4 MJ/kg = 1,136,670,740.96 MJ
1 kWh = 3,600 kJ
61,626,000 kWh = 61,626,000 × 3,600 kJ = 221,853,600 MJ
![Efficiency = \mathbf{\dfrac{Energy \ put \ out }{Energy \ value \ put \ in}}](https://tex.z-dn.net/?f=Efficiency%20%3D%20%5Cmathbf%7B%5Cdfrac%7BEnergy%20%20%5C%20put%20%5C%20out%20%7D%7BEnergy%20%5C%20value%20%5C%20put%20%5C%20in%7D%7D)
The efficiency is therefore;
![Efficiency = \dfrac{221853600}{1136670740.96 } \times 100 \approx \mathbf{19.5\%}](https://tex.z-dn.net/?f=Efficiency%20%3D%20%5Cdfrac%7B221853600%7D%7B1136670740.96%20%7D%20%5Ctimes%20100%20%5Capprox%20%5Cmathbf%7B19.5%5C%25%7D)
The efficiency of the power plant is approximately<u> 19.5%</u>
Learn more about finding the efficiency of an engine here:
brainly.com/question/10555156