Chapter 9, Problem 179RQ

To determine

Compare the thermal efficiency of the cycle when it is operated such that the saturated liquid enters the pump against that when the sub-cooled liquid enters the pump.

Explanation of Solution

Given:

Pressure of steam at the boiler $\left(P_3\right)$ is $\text{6000}\text{kPa}$.

Pressure of steam at the condenser $\left(P_1\right)$ is $\text{50}\text{kPa}$.

Temperature of steam at the turbine inlet $\left(T_3\right)$ is $600°\text{C}$.

Temperature of sub-cooled liquid $\left(T\right)$ is $11.3°\text{C}$.

Calculation:

Draw the $T-s$ diagram of the cycle as in Figure (1).

The pressures are constant for the process 2 to 3 and process 4 to 1.

  $\begin{array}{l}{P}_2=P_3\\ P_4=P_1\end{array}$

The entropies are constant for the process 1 to 2 and process 3 to 4.

  $\begin{array}{l}{s}_1=s_2\\ s_3=s_4\end{array}$

The thermal efficiency of the cycle when it is operated such that when the saturated liquid enters the pump:

Refer Table A-5, “Saturated water-Pressure table”, obtain the specific enthalpy and specific volume at state 1 corresponding to the pressure of $\text{50}\text{kPa}$.

  $\begin{array}{c}{h}_1=h_{f@50\text{kPa}}\\ =340.54\text{kJ}/\text{kg}\\ v_1=v_{f@50\text{kPa}}\\ =0.001030{\text{m}}^3/\text{kg}\end{array}$

Calculate the work done by the pump during process 1-2$\left(w_{\text{p,in}}\right)$.

  $\begin{array}{c}{w}_{\text{p,in}}=v_1\left(P_2-P_1\right)\\ =\left(0.001030{\text{m}}^3/\text{kg}\right)\left(6000\text{kPa}-50\text{kPa}\right)\\ =\left(0.001030{\text{m}}^3/\text{kg}\right)\left(6000\text{kPa}-50\text{kPa}\right)\left(\frac{1\text{kJ}}{\text{1}\text{kPa}\cdot {\text{m}}^3}\right)\\ =6.13\text{kJ}/\text{kg}\end{array}$

Calculate the specific enthalpy at state 2$\left(h_2\right)$.

  $\begin{array}{c}{h}_2=h_1+w_{\text{p,in}}\\ =340.54\text{kJ}/\text{kg}+6.13\text{kJ}/\text{kg}\\ =346.67\text{kJ}/\text{kg}\end{array}$

Refer Table A-6, “Superheated water”, obtain the specific enthalpy and temperature at state 3 corresponding to the pressure of $\text{6}\text{MPa}$ and temperature of $600°\text{C}$.

  $\begin{array}{l}{h}_3=3658.8\text{kJ}/\text{kg}\\ s_3=7.1693\text{kJ}/\text{kg}\cdot \text{K}\end{array}$

Refer Table A-5, “Saturated water-Pressure table”, obtain the following properties corresponding to the pressure of $\text{50}\text{kPa}$ and specific entropy of $7.1693\text{kJ}/\text{kg}\cdot \text{K}$.

  $\begin{array}{l}{h}_f=340.54\text{kJ}/\text{kg}\\ h_{fg}=2304.7\text{kJ}/\text{kg}\\ s_f=1.0912\text{kJ}/\text{kg}\cdot \text{K}\\ s_{fg}=6.5019\text{kJ}/\text{kg}\cdot \text{K}\end{array}$

Calculate the quality of steam at state 4$\left(s_4\right)$.

  $\begin{array}{c}{x}_4=\frac{s_4-s_f}{s_{fg}}\\ =\frac{7.1693\text{kJ}/\text{kg}\cdot \text{K}-1.0912\text{kJ}/\text{kg}\cdot \text{K}}{6.5019\text{kJ}/\text{kg}\cdot \text{K}}\\ =0.9348\end{array}$

Calculate the specific enthalpy at state 4$\left(h_4\right)$.

  $\begin{array}{c}{h}_4=h_f+x_4{h}_{fg}\\ =340.54\text{kJ}/\text{kg}+\left(0.9348\right)\left(2304.7\text{kJ}/\text{kg}\right)\\ =2495\text{kJ}/\text{kg}\end{array}$

Calculate the thermal efficiency of the cycle $\left({\eta }_{\text{th}}\right)$.

  $\begin{array}{c}{\eta }_{\text{th}}=1-\frac{q_{\text{out}}}{q_{\text{in}}}\\ =1-\frac{h_4-h_1}{h_3-h_2}\end{array}$

  $\begin{array}{c}=\frac{2495\text{kJ}/\text{kg}-340.64\text{kJ}/\text{kg}}{3658.8\text{kJ}/\text{kg}-346.67\text{kJ}/\text{kg}}\\ =\frac{2154.5\text{kJ}/\text{kg}}{3312.1\text{kJ}/\text{kg}}\\ =0.3495\\ =34.95\%\end{array}$

The thermal efficiency of the cycle when it is operated such that when the sub-cooled liquid enters the pump:

Refer Table A-5, “Saturated water-Pressure table”, obtain the saturation temperature corresponding to the pressure of $\text{50}\text{kPa}$.

  $T_{\text{sat}@50\text{kPa}}=81.3°\text{C}$

Calculate the temperature at state 1$\left(T_1\right)$ .

  $\begin{array}{c}{T}_1=T_{\text{sat}@50\text{kPa}}-T\\ =81.3°\text{C}-11.3°\text{C}\\ =\text{70}°\text{C}\end{array}$

Refer Table A-4, “Saturated water-Temperature table”, obtain the specific enthalpy and specific volume at state 1 corresponding to the temperature of $\text{70}°\text{C}$.

  $\begin{array}{c}{h}_1=h_{f@70°\text{C}}\\ =293.07\text{kJ}/\text{kg}\\ v_1=v_{f@70°\text{C}}\\ =0.001023{\text{m}}^3/\text{kg}\end{array}$

Calculate the work done by the pump during process 1-2$\left(w_{\text{p,in}}\right)$.

  $\begin{array}{c}{w}_{\text{p,in}}=v_1\left(P_2-P_1\right)\\ =\left(0.001023{\text{m}}^3/\text{kg}\right)\left(6000\text{kPa}-50\text{kPa}\right)\\ =\left(0.001023{\text{m}}^3/\text{kg}\right)\left(6000\text{kPa}-50\text{kPa}\right)\left(\frac{1\text{kJ}}{\text{1}\text{kPa}\cdot {\text{m}}^3}\right)\\ =6.09\text{kJ}/\text{kg}\end{array}$

Calculate the enthalpy at state 2$\left(h_2\right)$.

  $\begin{array}{c}{h}_2=h_1+w_{\text{p,in}}\\ =293.07\text{kJ}/\text{kg}+6.09\text{kJ}/\text{kg}\\ =299.16\text{kJ}/\text{kg}\end{array}$

Calculate the thermal efficiency of the cycle $\left({\eta }_{\text{th}}\right)$.

  $\begin{array}{c}{\eta }_{\text{th}}=1-\frac{q_{\text{out}}}{q_{\text{in}}}\\ =1-\frac{h_4-h_1}{h_3-h_2}\end{array}$

  $\begin{array}{c}=\frac{2495\text{kJ}/\text{kg}-293.09\text{kJ}/\text{kg}}{3658.8\text{kJ}/\text{kg}-299.16\text{kJ}/\text{kg}}\\ =\frac{2201.9\text{kJ}/\text{kg}}{3359.6\text{kJ}/\text{kg}}\\ =0.3446\\ =34.46\%\end{array}$

By comparing the thermal efficiencies of the cycle, it is found that the thermal efficiency slightly reduces because of the sub-cooling at the inlet of the pump.