Chapter 9, Problem 131P

a)

To determine

The quality or temperature (if superheated) of the steam at the turbine exit.

Explanation of Solution

Given:

Pressure of steam at state 3$\left(P_3\right)$ is $\text{10}\text{MPa}$.

Pressure of steam at state 1$\left(P_1\right)$ is $\text{10}\text{kPa}$.

Temperature of steam at the state 3$\left(T_3\right)$ is $500°\text{C}$.

Net power output of the cycle $\left({\dot{W}}_{\text{net}}\right)$ is $80,000\text{kJ}/\text{s}$.

Isentropic efficiency of the boiler $\left({\eta }_T\right)$ is $\text{0}\text{.80}$.

Isentropic efficiency of the pump $\left({\eta }_P\right)$ is $\text{0}\text{.95}$.

Calculation:

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

The entropies are constant for the process 3 to 4s and process 5 to 6s.

  $\begin{array}{l}{s}_3=s_{4s}\\ s_5=s_{6s}\end{array}$

Refer Table A-5, “Saturated water-Pressure table”, obtain the specific enthalpy and specific volume at state 1 corresponding to the pressure of $\text{10}\text{kPa}$.

  $\begin{array}{c}{h}_1=h_{f@10\text{kPa}}\\ =191.81\text{kJ}/\text{kg}\\ v_1=v_{f@10\text{kPa}}\\ =0.001010{\text{m}}^3/\text{kg}\end{array}$

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

  $\begin{array}{c}{w}_{\text{p,in}}=\frac{v_1\left(P_2-P_1\right)}{{\eta }_P}\\ =\frac{\left(0.001010{\text{m}}^3/\text{kg}\right)\left(10,000\text{kPa}-10\text{kPa}\right)}{0.95}\\ =\frac{\left(0.001010{\text{m}}^3/\text{kg}\right)\left(10,000\text{kPa}-10\text{kPa}\right)}{0.95}\left(\frac{1\text{kJ}}{\text{1}\text{kPa}\cdot {\text{m}}^3}\right)\\ =10.62\text{kJ}/\text{kg}\end{array}$

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

  $\begin{array}{c}{h}_2=h_1+w_{\text{p,in}}\\ =191.81\text{kJ}/\text{kg}+10.62\text{kJ}/\text{kg}\\ =202.43\text{kJ}/\text{kg}\end{array}$

Refer Table A-6, “Superheated water”, obtain the specific enthalpy and specific entropy at state 3 corresponding to the pressure of $\text{10}\text{MPa}$ and temperature of $500°\text{C}$.

  $\begin{array}{l}{h}_3=3375.1\text{kJ}/\text{kg}\\ s_3=6.5995\text{kJ}/\text{kg}\cdot \text{K}\end{array}$

Refer Table A-5, “Saturated water-Pressure table”, obtain the following properties corresponding to the pressure of $\text{1}\text{MPa}$ and specific entropy of $6.5995\text{kJ}/\text{kg}\cdot \text{K}$.

  $h_{4s}=2783.8\text{kJ}/\text{kg}$

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

  $\begin{array}{c}{h}_4=h_3-{\eta }_T\left(h_3-h_{4s}\right)\\ =3375.1\text{kJ}/\text{kg}-\left(0.80\right)\left(3375.1\text{kJ}/\text{kg}-2783.7\text{kJ}/\text{kg}\right)\\ =2902.0\text{kJ}/\text{kg}\end{array}$

Refer Table A-6, “Superheated water”, obtain the specific enthalpy and specific entropy at state 5 corresponding to the pressure of $1\text{MPa}$ and temperature of $500°\text{C}$.

  $\begin{array}{l}{h}_5=3479.1\text{kJ}/\text{kg}\\ s_5=7.7642\text{kJ}/\text{kg}\cdot \text{K}\end{array}$

Refer Table A-5, “Saturated water-Pressure table”, obtain the following properties corresponding to the pressure of $10\text{kPa}$ and specific entropy of $7.7642\text{kJ}/\text{kg}\cdot \text{K}$.

  $\begin{array}{l}{h}_f=191.81\text{kJ}/\text{kg}\\ h_{fg}=2392.1\text{kJ}/\text{kg}\\ s_f=0.6492\text{kJ}/\text{kg}\cdot \text{K}\\ s_{fg}=7.4996\text{kJ}/\text{kg}\cdot \text{K}\end{array}$

Calculate the quality at state 6s$\left(x_{6s}\right)$.

  $\begin{array}{c}{x}_{6s}=\frac{s_{6s}-s_f}{s_{fg}}\\ =\frac{7.7642\text{kJ}/\text{kg}\cdot \text{K}-0.6492\text{kJ}/\text{kg}\cdot \text{K}}{7.4996\text{kJ}/\text{kg}\cdot \text{K}}\\ =0.9487\end{array}$

Calculate the specific enthalpy at state 6s$\left(h_{6s}\right)$.

  $\begin{array}{c}{h}_{6s}=h_5-{\eta }_T\left(h_5-h_{6s}\right)\\ =3479.1\text{kJ}/\text{kg}-\left(0.80\right)\left(3479.1\text{kJ}/\text{kg}-2461.2\text{kJ}/\text{kg}\right)\\ =2664.8\text{kJ}/\text{kg}>h_g\end{array}$

Since the specific enthalpy at state 6s is greater than $h_g$, the state is superheated.

Refer Table A-5, “Saturated water-Pressure table”, obtain the temperature of the steam corresponding to the pressure of $10\text{kPa}$.

  $T_6=88.1°\text{C}$

Thus, the temperature of the steam at the turbine exit is $88.1°\text{C}$.

b)

To determine

The thermal efficiency of the cycle.

Explanation of Solution

Calculation:

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

  $\begin{array}{c}{\eta }_{\text{th}}=\frac{w_{\text{net}}}{q_{\text{in}}}\\ =\frac{w_{\text{T,out}}-w_{\text{p,in}}}{\left(h_3-h_2\right)+\left(h_5-h_4\right)}\\ =\frac{\left[\left(h_3-h_4\right)+\left(h_5-h_6\right)\right]-w_{\text{p,in}}}{\left(h_3-h_2\right)+\left(h_5-h_4\right)}\\ =\frac{\left[\begin{array}{l}\left(3375.1\text{kJ}/\text{kg}-2902\text{kJ}/\text{kg}\right)+\\ \left(3479.1\text{kJ}/\text{kg}-2664.8\text{kJ}/\text{kg}\right)\end{array}\right]-10.62\text{kJ}/\text{kg}}{\left(3375.1\text{kJ}/\text{kg}-202.43\text{kJ}/\text{kg}\right)+\left(3479.1\text{kJ}/\text{kg}-2902.0\text{kJ}/\text{kg}\right)}\end{array}$

  $\begin{array}{c}=0.341\\ =34.1\%\end{array}$

Thus, the thermal efficiency of the cycle is $34.1\%$.

c)

To determine

The mass flow rate of the steam.

Explanation of Solution

Calculation:

Calculate the mass flow rate of the steam $\left(\dot{m}\right)$.

  $\begin{array}{c}\dot{m}=\frac{{\dot{W}}_{\text{net}}}{w_{\text{net}}}\\ =\frac{{\dot{W}}_{\text{net}}}{\left[\left(h_3-h_4\right)+\left(h_5-h_6\right)\right]-w_{\text{p,in}}}\\ =\frac{80,000\text{kJ}/\text{s}}{\left[\begin{array}{l}\left(3375.1\text{kJ}/\text{kg}-2902\text{kJ}/\text{kg}\right)+\\ \left(3479.1\text{kJ}/\text{kg}-2664.8\text{kJ}/\text{kg}\right)\end{array}\right]-10.62\text{kJ}/\text{kg}}\\ =62.7\text{kg}/\text{s}\end{array}$

Thus, the mass flow rate of the steam is $62.7\text{kg}/\text{s}$.