Q.4.24 - An Introduction to Thermal Physics

Question

Calculate the efficiency of a Rankine cycle that is modified from the parameters used in the text in each of the following three ways (one at a time), and comment briefly on the results: 

$\left(a\right)$ reduce the maximum temperature to $500°\mathrm{C};$ 

$\left(b\right)$reduce the maximum pressure to $100$ bars;

$\left(c\right)$ reduce the minimum temperature to $10°\mathrm{C}$.

Step-by-Step Solution

Answer

Part $\left(a\right)$ 

$\left(a\right)$The efficiency of a Rankine cycle is $e\approx 0.463$ when the maximum pressure is reduced to ${500}^{\circ }C.$

Part $\left(b\right)$

$\left(b\right)$ The efficiency of a Rankine cycle is $e\approx 0.453$ when the maximum pressure is reduced to $100$ bars.

Part $\left(c\right)$

$\left(c\right)$ The efficiency of a Rankine cycle is $e\approx 0.490$ when the minimum temperature is reduced to $10°\mathrm{C}$.

Step: 1 Rankine's cycle: (part a)

The efficiency of engine by

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

The entropy of points is same by

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

Finding $x$ by

$$\begin{gathered} \begin{array}{l}{S}_4=x{S}_w+(1-x)S_s\\ S_4-S_s=x{S}_w-x{S}_s\\ x=\frac{S_4-S_s}{S_w-S_s}\end{array} \\ \begin{array}{l}x=\frac{\left(5.791\mathrm{kJ}\times {\mathrm{K}}^{-1}\times {\mathrm{kg}}^{-1}\right)-\left(8.667\mathrm{kJ}\times {\mathrm{K}}^{-1}\times {\mathrm{kg}}^{-1}\right)}{\left(0.297\mathrm{kJ}\times {\mathrm{K}}^{-1}\times {\mathrm{kg}}^{-1}\right)-\left(8.667\mathrm{kJ}\times {\mathrm{K}}^{-1}\times {\mathrm{kg}}^{-1}\right)}\\ x=0.344.\end{array} \end{gathered}$$

Step: 2 Finding efficiency at maximum pressure at 500 ∘ C : (part a)

The enthalpy of mixture by

$H_4=x{H}_w+(1-x)H_s$

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

$\begin{array}{l}{H}_4=\left(0.344\right)\times \left(84\mathrm{kJ}\cdot {\mathrm{kg}}^{-1}\right)+(1-0.344)\times \left(2538\mathrm{kJ}\cdot {\mathrm{kg}}^{-1}\right)\\ H_4=1693.8\mathrm{kJ}\times {\mathrm{kg}}^{-1}.\end{array}$

The efficiency is

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

Step: 3 Finding efficiency at maximum pressure at 100 bars: (part b)

The enthalpy of pressure is

$H_3=3625\mathrm{kJ}\times {\mathrm{kg}}^{-1}$

The entropy of point change by

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

The enthalpy of mixture is

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

The efficiency is

$\begin{array}{l}e\approx 1-\frac{2020.2\mathrm{kJ}\times {\mathrm{kg}}^{-1}-84\mathrm{kJ}\times {\mathrm{kg}}^{-1}}{3625\mathrm{kJ}\times {\mathrm{kg}}^{-1}-84\mathrm{kJ}\times {\mathrm{kg}}^{-1}}\\ \approx 0.325\\ e\approx 0.453.\end{array}$

Step: 4 Finding efficiency at maximum pressure at 10 ∘ C : (part c)

The enthalpy of pressure is

$H_1=42\mathrm{kJ}\times {\mathrm{kg}}^{-1}$

The entropy of point change by

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

The enthalpy of mixture is

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

The efficiency is

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

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