91Ó°ÊÓ

The efficiency of engine by

$$\begin{gathered} e≈1−\frac{H_4−H_1}{H_3−H_1} \\ {H}_2≈H_1=84\mathrm{kJ}×{\mathrm{kg}}^{−1} \\ {H}_3=3081\mathrm{kJ}×{\mathrm{kg}}^{−1}. \end{gathered}$$

The entropy of points is same by

$\begin{array}{l}{S}_3=S_4\\ S_4=S_3=5.791\mathrm{kJ}×{\mathrm{K}}^{−1}×{\mathrm{kg}}^{−1}\end{array}$

Finding $x$by

role="math" localid="1649686390534" $$\begin{gathered} \begin{array}{l}{S}_4=x{S}_w+(1−x)S_s\\ S_4−S_s=x{S}_w−x{S}_s\\ x=\frac{S_4−S_s}{S_w−S_s}\end{array} \\ \begin{array}{l}x=\frac{\left(5.791\mathrm{kJ}×{\mathrm{K}}^{−1}×{\mathrm{kg}}^{−1}\right)−\left(8.667\mathrm{kJ}×{\mathrm{K}}^{−1}×{\mathrm{kg}}^{−1}\right)}{\left(0.297\mathrm{kJ}×{\mathrm{K}}^{−1}×{\mathrm{kg}}^{−1}\right)−\left(8.667\mathrm{kJ}×{\mathrm{K}}^{−1}×{\mathrm{kg}}^{−1}\right)}\\ x=0.344.\end{array} \end{gathered}$$

The enthalpy of mixture by

$H_4=x{H}_w+(1−x)H_s$

Substituting the enthalpy of water is$H_w=84\mathrm{kJ}×{\mathrm{kg}}^{−1}$ and the enthalpy of steam is$H_s=2538\mathrm{kJ}×{\mathrm{kg}}^{−1}.$

$\begin{array}{l}{H}_4=\left(0.344\right)×\left(84\mathrm{kJ}â‹\dots {\mathrm{kg}}^{−1}\right)+(1−0.344)×\left(2538\mathrm{kJ}â‹\dots {\mathrm{kg}}^{−1}\right)\\ H_4=1693.8\mathrm{kJ}×{\mathrm{kg}}^{−1}.\end{array}$

The efficiency is

$\begin{array}{l}e≈1−\frac{\left(1693.8\mathrm{kJ}×{\mathrm{kg}}^{−1}\right)−\left(84\mathrm{kJ}×{\mathrm{kg}}^{−1}\right)}{\left(3081\mathrm{kJ}×{\mathrm{kg}}^{−1}\right)−\left(84\mathrm{kJ}×{\mathrm{kg}}^{−1}\right)}\\ ≈0.463\\ e≈0.463.\end{array}$

The enthalpy of pressure is

$H_3=3625\mathrm{kJ}×{\mathrm{kg}}^{−1}$

The entropy of point change by

$$\begin{gathered} S_4=S_3=6.903\mathrm{kJ}×{\mathrm{K}}^{−1}×{\mathrm{kg}}^{−1} \\ \begin{array}{l}{S}_4=x{S}_w+(1−x)S_s\\ S_4−S_s=x{S}_w−x{S}_s\\ x=\frac{S_4−S_s}{S_w−S_s}\\ \begin{array}{l}x=\frac{\left(6.903\mathrm{kJ}×{\mathrm{K}}^{−1}×{\mathrm{kg}}^{−1}\right)−\left(8.667\mathrm{kJ}×{\mathrm{K}}^{−1}×{\mathrm{kg}}^{−1}\right)}{\left(0.297\mathrm{kJ}×{\mathrm{K}}^{−1}×{\mathrm{kg}}^{−1}\right)−\left(8.667\mathrm{kJ}×{\mathrm{K}}^{−1}×{\mathrm{kg}}^{−1}\right)}\\ x=0.211.\end{array}\end{array} \end{gathered}$$

The enthalpy of mixture is

$$\begin{gathered} H_4=x{H}_w+(1−x)H_s \\ \begin{array}{l}{H}_4=\left(0.211\right)×\left(84\mathrm{kJ}×{\mathrm{kg}}^{−1}\right)+(1−0.211)\left(2538\mathrm{kJ}×{\mathrm{kg}}^{−1}\right)\\ H_4=2020.2\mathrm{kJ}×{\mathrm{kg}}^{−1}.\end{array} \end{gathered}$$

The efficiency is

$\begin{array}{l}e≈1−\frac{2020.2\mathrm{kJ}×{\mathrm{kg}}^{−1}−84\mathrm{kJ}×{\mathrm{kg}}^{−1}}{3625\mathrm{kJ}×{\mathrm{kg}}^{−1}−84\mathrm{kJ}×{\mathrm{kg}}^{−1}}\\ ≈0.325\\ e≈0.453.\end{array}$

The enthalpy of pressure is

$H_1=42\mathrm{kJ}×{\mathrm{kg}}^{−1}$

The entropy of point change by

$$\begin{gathered} S_4=S_3=6.233\mathrm{kJ}×{\mathrm{K}}^{−1}×{\mathrm{kg}}^{−1} \\ \begin{array}{l}{S}_4=x{S}_w+(1−x)S_s\\ S_4−S_s=x{S}_w−x{S}_s\\ x=\frac{S_4−S_s}{S_w−S_s}\end{array} \\ \begin{array}{l}x=\frac{\left(6.233\mathrm{kJ}×{\mathrm{K}}^{−1}×{\mathrm{kg}}^{−1}\right)−\left(8.901\mathrm{kJ}×â‹\dots {\mathrm{K}}^{−1}×{\mathrm{kg}}^{−1}\right)}{\left(0.151\mathrm{kJ}×{\mathrm{K}}^{−1}×{\mathrm{kg}}^{−1}\right)−\left(8.901\mathrm{kJ}×{\mathrm{K}}^{−1}×{\mathrm{kg}}^{−1}\right)}\\ x=0.305.\end{array} \end{gathered}$$

The enthalpy of mixture is

$$\begin{gathered} H_4=x{H}_w+(1−x)H_s \\ \begin{array}{l}{H}_4=\left(0.305\right)\left(42\mathrm{kJ}×{\mathrm{kg}}^{−1}\right)+(1−0.305)\left(2538\mathrm{kJ}×{\mathrm{kg}}^{−1}\right)\\ H_4=1776.7\mathrm{kJ}×{\mathrm{kg}}^{−1}.\end{array} \end{gathered}$$

The efficiency is

$\begin{array}{l}e≈1−\frac{\left(1776.7\mathrm{kJ}×{\mathrm{kg}}^{−1}\right)−\left(42\mathrm{kJ}×{\mathrm{kg}}^{−1}\right)}{\left(3444\mathrm{kJ}×{\mathrm{kg}}^{−1}\right)−\left(42\mathrm{kJ}×{\mathrm{kg}}^{−1}\right)}\\ e≈0.49.\end{array}$

From the above three parts,the change in efficiency is at $3\%$is obtained.