Respuesta :
Answer:
a) [tex]\eta_{th} = 46.1\,\%[/tex], b) [tex]\eta_{th,real} = 36.667\,\%[/tex]
Explanation:
a) The maximum possible thermal efficiency of the power plant is given by the Carnot's Cycle thermal efficiency, which consider a reversible power cycle according to the Second Law of Thermodynamics, whose formula is:
[tex]\eta_{th} = \left(1-\frac{T_{L}}{T_{H}} \right)\times 100\,\%[/tex]
Where:
[tex]T_{L}[/tex] - Temperature of the cold reservoir (Condenser), measured in K.
[tex]T_{H}[/tex] - Temperature of the hot reservoir (Evaporator), measured in K.
The maximum possible thermal efficiency is:
[tex]\eta_{th} = \left(1-\frac{298.15\,K}{553.15\,K} \right)\times 100\,\%[/tex]
[tex]\eta_{th} = 46.1\,\%[/tex]
b) The actual efficiency of the plant is the ratio of net power to input heat rate expressed in percentage:
[tex]\eta_{th, real} = \frac{\dot W}{\dot Q_{in}} \times 100\,\%[/tex]
[tex]\eta_{th, real} = \frac{1100\,MW}{3000\,MW}\times 100\,\%[/tex]
[tex]\eta_{th,real} = 36.667\,\%[/tex]
As per the question the nuclear power plant generates about 3000 MW of energy form the reactors and is due to the core of the reactor. The high pressure stream is 280 a C is used for the turbine to make it spin.
- a) The maximum possible thermal efficiency of the power plant is given by the Carnot's Cycle thermal efficiency,
- b) The actual efficiency of the plant is the ratio of net power to input heat rate expressed in percentage.
Learn more about the plant generates 3000 MW of heat energy.
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