Chapter 9, Problem 112P

(a)

To determine

The quality of the steam at the turbine exit.

Explanation of Solution

Given:

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

Pressure of steam at the turbine $\left(P_3\right)$ is $\text{100}\text{MPa}$.

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

Net power produced by the plant $\left({\dot{W}}_{\text{net}}\right)$ is $\text{210}\text{MW}$.

Isentropic efficiency of the turbine $\left({\eta }_{\text{T}}\right)$ is $\text{0}\text{.85}$.

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.

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

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

  $\begin{array}{c}{h}_1=h_{f@10\text{kPa}}\\ =191.81\text{kJ}/\text{kg}\\ v_1=v_{f@10\text{kPa}}\\ =0.00101{\text{m}}^3/\text{kg}\end{array}$

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

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

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

  $\begin{array}{c}{w}_{\text{p,in}}=\frac{v_1\left(P_2-P_1\right)}{{\eta }_{\text{T}}}\\ =\frac{\left(0.001010{\text{m}}^3/\text{kg}\right)\left(10\text{MPa}-10\text{kPa}\right)}{0.85}\\ =\frac{\left(0.001010{\text{m}}^3/\text{kg}\right)\left[\left(10\text{}\text{MPa}\times \frac{1000\text{}\text{kPa}}{1\text{MPa}}\right)-10\text{kPa}\right]}{0.85}\end{array}$

  $\begin{array}{c}=11.87\text{kPa}\cdot {\text{m}}^3/\text{kg}\left(\frac{1\text{kJ}}{\text{1}\text{kPa}\cdot {\text{m}}^3}\right)\\ =11.87\text{kJ}/\text{kg}\end{array}$

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

  $\begin{array}{c}{h}_2=h_1+w_{\text{p,in}}\\ =191.81\text{kJ}/\text{kg}+11.87\text{kJ}/\text{kg}\\ =203.68\text{kJ}/\text{kg}\end{array}$

Calculate the quality of water at state 4$\left(x_4\right)$.

  $\begin{array}{c}{x}_4=\frac{s_4-s_f}{s_{fg}}\\ =\frac{s_3-s_f}{s_{fg}}\\ =\frac{6.5995\text{kJ}/\text{kg}\cdot \text{K}-0.6492\text{kJ}/\text{kg}\cdot \text{K}}{7.4996\text{kJ}/\text{kg}\cdot \text{K}}\\ =0.7934\end{array}$

Thus, the quality of the steam at the turbine exit is $0.7934$.

(b)

To determine

The thermal efficiency of the cycle.

Explanation of Solution

Calculate the enthalpy at state 4s $\left(h_{4s}\right)$.

  $\begin{array}{c}{h}_{4s}=h_f+x_4{h}_{fg}\\ =191.81\text{kJ}/\text{kg}+\left(0.7934\right)\left(2392.1\text{kJ}/\text{kg}\right)\\ =2089.7\text{kJ}/\text{kg}\end{array}$

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

  ${\eta }_{\text{T}}=\frac{h_3-h_4}{h_3-h_{4s}}$

  $\begin{array}{c}{h}_4=h_3-\left({\eta }_{\text{T}}\right)\left(h_3-h_{4s}\right)\\ =3375.1\text{kJ}/\text{kg}-\left(0.85\right)\left(3375.1\text{kJ}/\text{kg}-2089.7\text{kJ}/\text{kg}\right)\\ =2282.5\text{kJ}/\text{kg}\end{array}$

Calculate the heat supplied in the boiler $\left(q_{\text{in}}\right)$.

  $\begin{array}{c}{q}_{\text{in}}=h_3-h_2\\ =3375.1\text{kJ}/\text{kg}-203.68\text{kJ}/\text{kg}\\ =3171.4\text{kJ}/\text{kg}\end{array}$

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

  $\begin{array}{c}{\eta }_{\text{th}}=1-\frac{q_{\text{out}}}{q_{\text{in}}}\\ =1-\frac{\left(h_4-h_1\right)}{q_{\text{in}}}\\ =1-\frac{2282.5\text{kJ}/\text{kg}-191.81\text{kJ}/\text{kg}}{3171.4\text{kJ}/\text{kg}}\end{array}$

  $\begin{array}{c}=1-\frac{2090.7\text{kJ}/\text{kg}}{3171.4\text{kJ}/\text{kg}}\\ =0.341\\ =34.1\%\end{array}$

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

(c)

To determine

The mass flow rate of the steam.

Explanation of Solution

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

  $\begin{array}{c}\dot{m}=\frac{{\dot{W}}_{\text{net}}}{q_{\text{in}}-q_{\text{out}}}\\ =\frac{{\dot{W}}_{\text{net}}}{q_{\text{in}}-\left(h_4-h_1\right)}\\ =\frac{210\text{MW}}{3171.4\text{kJ}/\text{kg}-\left(2282.5\text{kJ}/\text{kg}-191.81\text{kJ}/\text{kg}\right)}\end{array}$

  $\begin{array}{c}=\frac{210\text{MW}\left(\frac{1000\text{kJ}/\text{s}}{1\text{MW}}\right)}{1080.7\text{kJ}/\text{kg}}\\ =\text{194}\text{.3}\text{kg}/\text{s}\end{array}$

Thus, the mass flow rate of the steam is $\text{194}\text{.3}\text{kg}/\text{s}$.