Question

A nuclear power plant generates $3000MW$of heat energy from nuclear reactions in the reactor’s core. This energy is used to boil water and produce high-pressure steam at $300°C$. The steam spins a turbine, which produces $1000MW$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?

c. Cooling water from a river flows through the condenser (the low-temperature heat exchanger) at the rate of $1.2*108L/h$ ( $\approx 30$ million gallons per hour). If the river water enters the condenser at $18°C$, what is its exit temperature?

Short answer

(a) Thermal efficiency$48\mathrm{\%}$

(b) Plant's actual efficiency$33\mathrm{\%}$

(c) Temperature of exit is ${32}^{0}C$

Step-by-step solution

Find Thermal Efficiency (part a)

If the thermal cycle is of the Carnot type, the maximum efficiency can be achieved. It will be $T_h$and $T_c$if the working temperatures are $T_h$and $T_c$.

$\eta =1-\frac{T_c}{T_h}=\frac{T_h-T_c}{T_h}$

It is numerically be in our instance.

$\eta =\frac{300-25}{300+273}=48\%$

Find Plant's Efficiency (part b)

The real efficiency is simply,

$\eta =\frac{P_e}{P_h}=\frac{1000}{3000}=$$33\%$

Find Temperature (part c)

It is self-evident that heat dumped to cold source can be calculated as follows.

$P_c=P_h-P_e=\underset{¯}{2000\mathrm{MW}}$= $2\times {10}^{9}J/s$

This heat is absorbed by flowing river water , due to which there is rise in its temperature .

If the temperature difference is $\Delta T$, the heat taken from mass $m$of water is given by

$Q=cm\Delta T$

When we divide the mass by the volume and density of the mass, we get

$Q=c\rho V\Delta T$

Let's divide the time on both sides now. On the left, we'll get the cooling power, and on the right, we'll divide the volume to get the volumetric flow rate.

$P_c=c\rho \frac{V}{t}\Delta T$

It is evident that the temperature difference can be expressed as

$\Delta T=\frac{P_{c}t}{c\rho V}$

If the temperature at the input is $T_i$, the temperature at the output will obviously be $T_o$.

Apply the values,

localid="1649656479751" $T_o=18+\frac{2\times {10}^9\times 3600}{4180\times 1000\times 1.2\times {10}^5}=32°\mathrm{C}$