Discover the answers you need at Westonci.ca, where experts provide clear and concise information on various topics. Our platform provides a seamless experience for finding precise answers from a network of experienced professionals. Explore comprehensive solutions to your questions from knowledgeable professionals across various fields on our platform.
Sagot :
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 19.5%
How can the efficiency of a power plant be found?
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;
[tex]\dfrac{28,670,807,174 \, L}{22.414 \, L} \approx \mathbf{1279147281.79}[/tex]
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
[tex]Efficiency = \mathbf{\dfrac{Energy \ put \ out }{Energy \ value \ put \ in}}[/tex]
The efficiency is therefore;
[tex]Efficiency = \dfrac{221853600}{1136670740.96 } \times 100 \approx \mathbf{19.5\%}[/tex]
The efficiency of the power plant is approximately 19.5%
Learn more about finding the efficiency of an engine here:
https://brainly.com/question/10555156
Thank you for your visit. We are dedicated to helping you find the information you need, whenever you need it. We hope our answers were useful. Return anytime for more information and answers to any other questions you have. Westonci.ca is your go-to source for reliable answers. Return soon for more expert insights.
