91Ó°ÊÓ

To evaluate the Rankine cycle's efficiency, one must first compute the enthalpy energy change at various points throughout the cycle, i.e., the amount of heat absorbed or rejected by the system at constant pressure at various times.

The Rankine cycle's efficiency is:

$\text{e}=1-\frac{{\text{H}}_4{\text{-H}}_1}{{\text{H}}_3{\text{-H}}_2}$

where

$$\begin{gathered} {\text{H}}_1\text{=Enthalpyatpoint1} \\ {\text{H}}_2\text{=Enthalpyatpoint2} \\ {\text{H}}_3\text{=Enthalpyatpoint3} \\ {\text{H}}_4\text{=Enthalpyatpoint4} \end{gathered}$$

The expression which relates the enthalpy change to the change in internal energy, volume and pressure is $\text{dH=dU+PdV+VdP}$.

By second law of thermodynamics $\text{»å±«=°Õ»å³§-±Ê»å³Õ+μ»å±·}$.

Putting the value of $\text{dU}$in the first expression.

$\text{»å±á=°Õ»å³§-±Ê»å³Õ+μ»å±·+±Ê»å³Õ+³Õ»å±Ê}$.

Here $\text{μ»å±·}$is ignored as the amount of fluid is not changing. So, the value gives $0$.

And entropy for the points $1$and $2$is also zero. So, $\text{dS}$is also ignored.

Hence the expression becomes $\text{dH=VdP}$.

Calculating the change in enthalpy change by assuming the pressure condition of 200 bars, then $\mathrm{Δ}{\mathrm{H}}_{12}=\mathrm{V}\mathrm{Δ}\mathrm{P}$.

Substituting values of

$$\begin{gathered} {\text{V=1dm}}^3{\text{=10}}^{-3}{\text{m}}^3 \\ ∆{\text{±Ê=298×10}}^5{\text{N/m}}^2 \end{gathered}$$

So, $$\begin{gathered} \mathrm{Δ}{\mathrm{H}}_{12}=\left({10}^{-3}{\mathrm{m}}^3\right)\left(298×{10}^5\mathrm{N}/{\mathrm{m}}^2\right) \\ =29.8\text{KJ} \end{gathered}$$

Using the expression,

$$\begin{gathered} {\mathrm{H}}_2={\mathrm{H}}_1+\mathrm{Δ}{\mathrm{H}}_{12} \\ =84\text{KJ+29.8KJ} \\ =113.8\text{KJ} \end{gathered}$$

As,

So,

$$\begin{gathered} \text{e}=1-\frac{{\text{H}}_4{\text{-H}}_1}{{\text{H}}_3{\text{-H}}_2} \\ =1-\frac{1824-84}{3444-113.8} \\ =0.4775 \end{gathered}$$

Thus efficiency of the cycle is $0.4775$.