Respuesta :
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%
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