Q. 4.23 - An Introduction to Thermal Physics

Question

Use the definition of enthalpy to calculate the change in enthalpy between points 1 and 2 of the Rankine cycle, for the same numerical parameters as used in the text. Recalculate the efficiency using your corrected value of ${\text{H}}_2$, and comment on the accuracy of the approximation ${\text{H}}_2{\text{≈H}}_1$.

Step-by-Step Solution

Answer

The efficiency of the Rankine cycle could not result in efficiency even closer to $1$ because of some heat and pressure losses that are encountered during the process. The efficiency of the cycle is $0.4775$.

Step 1. Introduction

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

$$\begin{gathered} {\text{H}}_1\text{=Enthalpy at point 1} \\ {\text{H}}_2\text{=Enthalpy at point 2} \\ {\text{H}}_3\text{=Enthalpy at point 3} \\ {\text{H}}_4\text{=Enthalpy at point 4} \end{gathered}$$

Step 2. Calculation

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 $\text{dU=TdS-PdV+μdN}$.

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

$\text{dH=TdS-PdV+μdN+PdV+VdP}$.

Here $\text{μdN}$ 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 $\Delta {\mathrm{H}}_{12}=\mathrm{V}\Delta \mathrm{P}$.

Substituting values of

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

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

Using the expression,

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

Step 3. Calculating efficiency

As,

$$\begin{gathered} {\text{H}}_1\text{=84KJ} \\ {\text{H}}_2\text{=113.8 KJ} \\ {\text{H}}_3\text{=3444 KJ} \\ {\text{H}}_4\text{=1824 KJ} \end{gathered}$$

So,

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

Step 4. Conclusion

Thus efficiency of the cycle is $0.4775$.