Question

A nuclear power plant generates 3000 MW of heat energy from nuclear reactions in the reactor's core. This energy is used to boil water and produce high-pressure steam at 280∘C. The steam spins a turbine, which produces 1100 MW of electric power, then the steam is condensed and the water is cooled to 25∘C before starting the cycle again.
a. What is the maximum possible thermal efficiency of the power plant?
b. What is the plant's actual efficiency?

Answer 1

Answer:

a) $\eta_{th} = 46.1\,\%$, b) $\eta_{th,real} = 36.667\,\%$

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:

$\eta_{th} = \left(1-\frac{T_{L}}{T_{H}} \right)\times 100\,\%$

Where:

$T_{L}$ - Temperature of the cold reservoir (Condenser), measured in K.

$T_{H}$ - Temperature of the hot reservoir (Evaporator), measured in K.

The maximum possible thermal efficiency is:

$\eta_{th} = \left(1-\frac{298.15\,K}{553.15\,K} \right)\times 100\,\%$

$\eta_{th} = 46.1\,\%$

b) The actual efficiency of the plant is the ratio of net power to input heat rate expressed in percentage:

$\eta_{th, real} = \frac{\dot W}{\dot Q_{in}} \times 100\,\%$

$\eta_{th, real} = \frac{1100\,MW}{3000\,MW}\times 100\,\%$

$\eta_{th,real} = 36.667\,\%$