91Ó°ÊÓ

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 $T_c$.

$\mathrm{η}=1-\frac{T_c}{T_h}=\frac{T_h-T_c}{T_h}$

It is numerically be in our instance.

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

The real efficiency is simply,

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

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

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

If the temperature difference is $\mathrm{Δ}T$, the heat taken from mass $m$of water is given by

$Q=cm\mathrm{Δ}T$

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

$Q=c\mathrm{ÒÏ}V\mathrm{Δ}T$

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.

$P_c=c\mathrm{ÒÏ}\frac{V}{t}\mathrm{Δ}T$

It is evident that the temperature difference can be expressed as

$\mathrm{Δ}T=\frac{P_{c}t}{c\mathrm{ÒÏ}V}$

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

Apply the values,

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