Chapter 9, Problem 113P

To determine

The mass flow rate through the boiler, the power produced by the turbine, the rate of heat supply in the boiler and the thermal efficiency of the cycle.

Explanation of Solution

Given:

Pressure of steam at the condenser $\left(P_1\right)$ is $\text{2}\text{psia}$.

Pressure of steam at the turbine $\left(P_3\right)$ is $\text{1500}\text{psia}$.

Temperature of steam at the turbine $\left(T_3\right)$ is $800°\text{F}$.

Net power produced by the cycle $\left({\dot{W}}_{\text{net}}\right)$ is $\text{2500}\text{kW}$.

Isentropic efficiency of the turbine $\left({\eta }_{\text{T}}\right)$ is $\text{0}\text{.90}$.

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-5E, “Saturated water-Pressure table”, obtain the enthalpy and specific volume at state 1 corresponding to the pressure of $2\text{psia}$.

  $\begin{array}{c}{h}_1=h_{f@2\text{psia}}\\ =94.02\text{Btu}/\text{lbm}\\ v_1=v_{f@2\text{psia}}\\ =0.016230{\text{ft}}^3/\text{lbm}\end{array}$

Refer Table A-5E, “Saturated water-Pressure table”, obtain the following properties at state 1 corresponding to the pressure of $5\text{psia}$.

  $\begin{array}{l}{h}_f=94.02\text{Btu}/\text{lbm}\\ h_{fg}=1021.7\text{Btu}/\text{lbm}\\ s_f=0.1750\text{Btu}/\text{lbm}\cdot \text{R}\\ s_{fg}=1.7444\text{Btu}/\text{lbm}\cdot \text{R}\end{array}$

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

  $\begin{array}{c}{w}_{\text{p,in}}=v_1\left(P_2-P_1\right)\\ =\left(0.016230{\text{ft}}^3/\text{lbm}\right)\left(1500\text{psia}-2\text{psia}\right)\\ =\left(0.016230{\text{ft}}^3/\text{lbm}\right)\left(1500\text{psia}-2\text{psia}\right)\left(\frac{1\text{Btu}}{5.404\text{psia}\cdot {\text{ft}}^3}\right)\\ =4.50\text{Btu}/\text{lbm}\end{array}$

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

  $\begin{array}{c}{h}_2=h_1+w_{\text{p,in}}\\ =94.02\text{Btu}/\text{lbm}+4.50\text{Btu}/\text{lbm}\\ =98.52\text{Btu}/\text{lbm}\end{array}$

Refer Table A-6E, “Superheated water”, obtain the enthalpy and entropy at state 3 corresponding to the pressure of $1500\text{psia}$ and temperature of $800°\text{F}$.

  $\begin{array}{l}{h}_3=1363.1\text{Btu}/\text{lbm}\\ s_3=1.5064\text{Btu}/\text{lbm}\cdot \text{R}\end{array}$

Calculate the quality of water at state 4$\left(x_{4s}\right)$.

  $\begin{array}{c}{x}_{4s}=\frac{s_4-s_f}{s_{fg}}\\ =\frac{s_3-s_f}{s_{fg}}\\ =\frac{1.5064\text{Btu}/\text{lbm}\cdot \text{R}-0.1750\text{Btu}/\text{lbm}\cdot \text{R}}{1.7444\text{Btu}/\text{lbm}\cdot \text{R}}\\ =0.7633\end{array}$

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

  $\begin{array}{c}{h}_{4s}=h_f+x_{4s}{h}_{fg}\\ =94.02\text{Btu}/\text{lbm}+\left(0.7633\right)\left(1021.7\text{Btu}/\text{lbm}\right)\\ =873.86\text{Btu}/\text{lbm}\end{array}$

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

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

  $\begin{array}{c}{h}_4=h_3-\left({\eta }_{\text{T}}\right)\left(h_3-h_{4s}\right)\\ =1363.1\text{Btu}/\text{lbm}-\left(0.90\right)\left(1363.1\text{Btu}/\text{lbm}-873.86\text{Btu}/\text{lbm}\right)\\ =922.79\text{Btu}/\text{lbm}\end{array}$

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

  $\begin{array}{c}{q}_{\text{in}}=h_3-h_2\\ =1363.1\text{Btu}/\text{lbm}-98.52\text{Btu}/\text{lbm}\\ =1264.6\text{Btu}/\text{lbm}\end{array}$

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

  $\begin{array}{c}\dot{m}=\frac{{\dot{W}}_{\text{net}}}{w_{\text{net}}}\\ =\frac{{\dot{W}}_{\text{net}}}{q_{\text{in}}-q_{\text{out}}}\\ =\frac{{\dot{W}}_{\text{net}}}{q_{\text{in}}-\left(h_4-h_1\right)}\end{array}$

  $\begin{array}{c}=\frac{2500\text{kW}}{1264.6\text{Btu}/\text{lbm}-\left(922.79\text{Btu}/\text{lbm}-94.02\text{Btu}/\text{lbm}\right)}\\ =\frac{2500\text{kJ}/\text{s}\left(\frac{0.94782\text{Btu}}{1\text{kJ}}\right)}{435.8\text{Btu}/\text{lbm}}\\ =5.437\text{lbm}/\text{s}\end{array}$

Thus, the mass flow rate through the boiler is $5.437\text{lbm}/\text{s}$.

Calculate the power output from the turbine $\left({\dot{W}}_{\text{T,out}}\right)$.

  $\begin{array}{c}{\dot{W}}_{\text{T,out}}=\dot{m}\left(h_3-h_4\right)\\ =\left(5.437\text{lbm}/\text{s}\right)\left(1363.1\text{Btu}/\text{lbm}-922.79\text{Btu}/\text{lbm}\right)\\ =\left(5.437\text{lbm}/\text{s}\right)\left(1363.1\text{Btu}/\text{lbm}-922.79\text{Btu}/\text{lbm}\right)\left(\frac{1\text{kJ}}{0.94782\text{Btu}}\right)\\ =2526\text{kW}\end{array}$

Thus, the power produced by the turbine is $2526\text{kW}$.

Calculate the rate of heat addition $\left({\dot{Q}}_{\text{in}}\right)$.

  $\begin{array}{c}{\dot{Q}}_{\text{in}}=\dot{m}{q}_{\text{in}}\\ =\left(5.437\text{lbm}/\text{s}\right)\left(1264.6\text{Btu}/\text{lbm}\right)\\ =6876\text{Btu}/\text{s}\end{array}$

Thus, the rate of heat supply in the boiler is $2526\text{kW}$.

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

  $\begin{array}{c}{\eta }_{\text{th}}=\frac{{\dot{W}}_{\text{net}}}{{\dot{Q}}_{\text{in}}}\\ =\frac{2500\text{kJ}/\text{s}\left(\frac{0.94782\text{Btu}}{1\text{kJ}}\right)}{6876\text{Btu}/\text{s}}\\ =0.3446\\ =34.5\%\end{array}$

Thus, the thermal efficiency of the cycle is $34.5\%$.