Chapter 9, Problem 114P

To determine

The error caused in the thermal efficiency.

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 $\left(\dot{m}\right)$.

  $\begin{array}{c}\dot{m}=\frac{{\dot{W}}_{\text{net}}}{h_3-h_4}\\ =\frac{2500\text{kW}}{1363.1\text{Btu}/\text{lbm}-922.79\text{Btu}/\text{lbm}}\\ =\frac{2500\text{kJ}/\text{s}\left(\frac{0.94782\text{Btu}}{1\text{kJ}}\right)}{438.31\text{Btu}/\text{lbm}}\\ =5.381\text{lbm}/\text{s}\end{array}$

Calculate the mass flow rate of steam $\left(\dot{m}\right)$ when power required by the pump was not neglected.

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

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

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

  $\begin{array}{c}{\dot{Q}}_{\text{in}}=\dot{m}{q}_{\text{in}}\\ =\left(5.381\text{lbm}/\text{s}\right)\left(1264.6\text{Btu}/\text{lbm}\right)\\ =6805\text{Btu}/\text{s}\end{array}$

Calculate the rate of heat addition $\left({\dot{Q}}_{\text{in}}\right)$ when power required by the pump was not neglected.

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

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

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

  $\begin{array}{c}{\eta }_{\text{th}}=\frac{{\dot{W}}_{\text{net}}}{{\dot{Q}}_{\text{in}}}\\ =\frac{2500\text{kJ}/\text{s}\left(\frac{0.94782\text{Btu}}{1\text{kJ}}\right)}{6805\text{Btu}/\text{s}}\\ =0.3482\\ =34.82\%\end{array}$

Calculate the thermal efficiency of the cycle $\left({\eta }_{\text{th}}\right)$ when power required by the pump was not neglected.

  $\begin{array}{c}{\eta }_{\text{th}}=\frac{{\dot{W}}_{\text{net}}}{{\dot{Q}}_{\text{in}}}\\ =\frac{2500\text{kJ}/\text{s}\left(\frac{0.94782\text{Btu}}{1\text{kJ}}\right)}{6876\text{Btu}/\text{s}}\\ =0.3446\\ =34.46\%\end{array}$

Calculate the error caused in the thermal efficiency $\left(\%\text{Error}\right)$.

  $\begin{array}{c}\%\text{Error}=\frac{0.3482-0.3446}{0.3446}\times 100\%\\ =0.0103\times 100\%\\ =1.03\%\end{array}$

Thus, the error caused in the thermal efficiency is $1.03\%$.