Chapter 9, Problem 127P

a)

To determine

The pressure at which the reheating takes place.

Explanation of Solution

Given:

Pressure of steam at state 3$\left(P_3\right)$ is $\text{800}\text{psia}$.

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

Temperature of steam at the state 3$\left(T_3\right)$ is $900°\text{F}$.

Temperature of steam at the state 5$\left(T_5\right)$ is $800°\text{F}$.

Temperature of steam at the condenser $\left(T_{\text{condenser}}\right)$ is $45°\text{F}$.

Rate of heat input $\left({\dot{Q}}_{\text{in}}\right)$ is $6\times {10}^4\text{Btu}/\text{s}$.

Calculation:

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

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

  $\begin{array}{l}{s}_3=s_4\\ s_5=s_6\end{array}$

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

  $\begin{array}{c}{h}_1=h_{sat@\text{1}\text{psia}}\\ =69.72\text{Btu}/\text{lbm}\\ v_1=v_{sat@\text{1}\text{psia}}\\ =0.016124{\text{ft}}^3/\text{lbm}\\ T_1=T_{sat@\text{1}\text{psia}}\\ =101.69°\text{F}\end{array}$

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

  $\begin{array}{c}{w}_{\text{p,in}}=v_1\left(P_2-P_1\right)\\ =\left(0.016124{\text{ft}}^3/\text{lbm}\right)\left(800\text{psia}-1\text{psia}\right)\\ =\left(0.016124{\text{ft}}^3/\text{lbm}\right)\left(800\text{psia}-1\text{psia}\right)\left(\frac{1\text{Btu}}{5.4039\text{psia}\cdot {\text{ft}}^3}\right)\\ =2.39\text{Btu}/\text{lbm}\end{array}$

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

  $\begin{array}{c}{h}_2=h_1+w_{\text{p,in}}\\ =69.72\text{Btu}/\text{lbm}+2.39\text{Btu}/\text{lbm}\\ =72.11\text{Btu}/\text{lbm}\end{array}$

Refer Table A-6E, “Superheated water”, obtain the specific enthalpy and specific entropy at state 3 corresponding to the pressure of $\text{800}\text{psia}$ and temperature of $900°\text{F}$.

  $\begin{array}{l}{h}_3=1456\text{Btu}/\text{lbm}\\ s_3=1.6413\text{Btu}/\text{lbm}\cdot \text{R}\end{array}$

Refer Table A-5E, “Saturated water-Pressure table”, obtain the pressure at state 4 corresponding to the specific entropy of $1.6413\text{Btu}/\text{lbm}\cdot \text{R}$.

  $\begin{array}{c}{h}_4=h_{g@s_g=s_4}\\ =1178.5\text{Btu}/\text{lbm}\\ P_4=P_{sat@s_g=s_4}\\ =62.23\text{psia}\end{array}$

Thus, the pressure at which the reheating takes place is $62.23\text{psia}$.

b)

To determine

The thermal efficiency of the cycle.

Explanation of Solution

Calculation:

Refer Table A-6E, “Superheated water”, obtain the specific enthalpy and specific entropy at state 5 corresponding to the pressure of $\text{62}\text{.23}\text{psia}$ and temperature of $800°\text{F}$.

  $\begin{array}{l}{h}_5=1431.4\text{Btu}/\text{lbm}\\ s_5=1.8985\text{Btu}/\text{lbm}\cdot \text{R}\end{array}$

Refer Table A-5E, “Saturated water-Pressure table”, obtain the following properties corresponding to the pressure of $\text{1}\text{psia}$ and specific entropy of $1.8985\text{Btu}/\text{lbm}\cdot \text{R}$.

  $\begin{array}{l}{h}_f=69.72\text{Btu}/\text{lbm}\\ h_{fg}=1035.7\text{Btu}/\text{lbm}\\ s_f=0.13262\text{Btu}/\text{lbm}\cdot \text{R}\\ s_{fg}=1.84495\text{Btu}/\text{lbm}\cdot \text{R}\end{array}$

Calculate the quality of steam at state 6$\left(x_6\right)$.

  $\begin{array}{c}{x}_6=\frac{s_6-s_f}{s_{fg}}\\ =\frac{1.8985\text{Btu}/\text{lbm}\cdot \text{R}-0.13262\text{Btu}/\text{lbm}\cdot \text{R}}{1.84495\text{Btu}/\text{lbm}\cdot \text{R}}\\ =0.9572\end{array}$

Calculate the specific enthalpy at state 6$\left(h_6\right)$.

  $\begin{array}{c}{h}_6=h_f+x_6{h}_{fg}\\ =69.72\text{Btu}/\text{lbm}+\left(0.9572\right)\left(1035.7\text{Btu}/\text{lbm}\right)\\ =1061.0\text{Btu}/\text{lbm}\end{array}$

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

  $\begin{array}{c}{\eta }_{\text{th}}=1-\frac{q_{\text{out}}}{q_{\text{in}}}\\ =1-\frac{h_6-h_1}{\left(h_3-h_2\right)+\left(h_5-h_4\right)}\\ =1-\frac{1061.0\text{Btu}/\text{lbm}-69.72\text{Btu}/\text{lbm}}{\left\{\begin{array}{l}\left(1456\text{Btu}/\text{lbm}-72.11\text{Btu}/\text{lbm}\right)+\\ \left(1431.4\text{Btu}/\text{lbm}-1178.5\text{Btu}/\text{lbm}\right)\end{array}\right\}}\end{array}$

  $\begin{array}{c}=0.394\\ =39.4\%\end{array}$

Thus, the thermal efficiency of the cycle is $39.4\%$.

c)

To determine

The minimum mass flow rate of the cooling water.

Explanation of Solution

Calculation:

Calculate the rate of heat output from the cycle $\left({\dot{Q}}_{\text{out}}\right)$.

  $\begin{array}{c}{\dot{Q}}_{\text{out}}={\dot{Q}}_{\text{in}}-{\dot{W}}_{\text{net}}\\ =\left(1-{\eta }_{\text{th}}\right){\dot{Q}}_{\text{in}}\\ =\left(1-0.394\right)\left(6\times {10}^4\text{Btu}/\text{s}\right)\\ =3.634\times {10}^4\text{Btu}/\text{s}\end{array}$

Calculate the minimum mass flow rate of the cooling water $\left({\dot{m}}_{\text{cool}}\right)$.

  $\begin{array}{c}{\dot{m}}_{\text{cool}}=\frac{{\dot{Q}}_{\text{out}}}{c\Delta T}={\dot{Q}}_{\text{in}}-{\dot{W}}_{\text{net}}\\ =\frac{3.634\times {10}^4\text{Btu}/\text{s}}{\left(1\text{Btu}/\text{lbm}\cdot °\text{F}\right)\left(101.69°\text{F}-45°\text{F}\right)}\\ =641\text{lbm}/\text{s}\end{array}$

Thus, the minimum mass flow rate of the cooling water is $641\text{lbm}/\text{s}$.