Chapter 10.9, Problem 62P

A steam power plant operates on an ideal reheat–regenerative Rankine cycle with one reheater and two open feedwater heaters. Steam enters the high-pressure turbine at 1500 psia and 1100°F and leaves the low- pressure turbine at 1 psia. Steam is extracted from the turbine at 250 and 40 psia, and it is reheated to 1000°F at a pressure of 140 psia. Water leaves both feedwater heaters as a saturated liquid. Heat is transferred to the steam in the boiler at a rate of 4 × 10⁵ Btu/s. Show the cycle on a T-s diagram with respect to saturation lines, and determine (a) the mass flow rate of steam through the boiler, (b) the net power output of the plant, and (c) the thermal efficiency of the cycle.

FIGURE P10–62E

To determine

The mass flow rate of steam flowing through the boiler.

Answer to Problem 62P

The mass flow rate of steam flowing through the boiler is $54.4\text{kg/s}$.

Explanation of Solution

Draw the schematic layout of the given power plant that operates on an ideal reheat-regenerative Rankine cycle as shown in Figure 1.

Draw the $T-s$ diagram of the given ideal regenerative Rankine cycle as shown in

Figure 2.

Here, water (steam) is the working fluid of the ideal regenerative Rankine cycle. The cycle involves three pumps.

Write the formula for work done by the pump during process 1-2.

$w_{\text{pI,in}}=v_1\left(P_2-P_1\right)$ (I)

Here, the specific volume is $v$, the pressure is $P$, and the subscripts 1 and 2 indicates the process states.

Write the formula for enthalpy $\left(h\right)$ at state 2.

$h_2=h_1+w_{\text{pI,in}}$ (II)

Write the formula for work done by the pump during process 3-4.

$w_{\text{pII,in}}=v_3\left(P_4-P_3\right)$ (III)

Here, the specific volume is $v$, the pressure is $P$, and the subscripts 3 and 4 indicates the process states.

Write the formula for enthalpy $\left(h\right)$ at state 4.

$h_4=h_3+w_{\text{pII,in}}$ (IV)

Write the formula for work done by the pump during process 5-6.

$w_{\text{pIII,in}}=v_5\left(P_6-P_5\right)$ (V)

Here, the specific volume is $v$, the pressure is $P$, and the subscripts 5 and 6 indicates the process states.

Write the formula for enthalpy $\left(h\right)$ at state 4.

$h_6=h_5+w_{\text{pIII,in}}$ (VI)

Before reheating.

At state 9:

The steam expanded to the pressure of $140\text{psia}$ and the steam is at the state of superheated vapor.

After reheating.

At state 10:

The steam is reheated to the temperature of $1000\text{°F}$ by the keeping the pressure equal to the state 9.

$P_{10}=P_9=140\text{psia}$

At state 12:

The steam enters the condenser at the pressure of $1\text{psia}$ and at the state of saturated mixture.

The quality of water at state 12 is expressed as follows.

$x_{12}=\frac{s_{12}-s_{f,12}}{s_{fg,12}}$ (VII)

The enthalpy at state 12 is expressed as follows.

$h_{12}=h_{f,12}+x_{12}{h}_{fg,12}$ (VIII)

Here, the enthalpy is $h$, the entropy is $s$, the quality of the water is $x$, the suffix $f$ indicates the fluid condition, the suffix $fg$ indicates the change of vaporization phase; the subscript 12 indicates the process state 12.

Write the formula for heat in $\left(q_{\text{in}}\right)$ and heat out $\left(q_{\text{out}}\right)$ of the cycle.

$q_{\text{in}}=\left(h_7-h_6\right)+\left(1-y\right)\left(h_{10}-h_9\right)$ (IX)

$q_{\text{out}}=\left(1-y-z\right)\left(h_{12}-h_1\right)$ (X)

Here, the mass fraction steam extracted from the turbine to the feed water entering the boiler via feed water heater-II $\left({\dot{m}}_8/{\dot{m}}_5\right)$ is $y$ and the mass fraction steam extracted from the turbine to feed water entering the boiler via the feed water heater-I $\left({\dot{m}}_9/{\dot{m}}_5\right)$ is $z$.

Write the general equation of energy balance equation.

${\dot{E}}_{\text{in}}-{\dot{E}}_{\text{out}}=\Delta {\dot{E}}_{\text{system}}$ (XI)

Here, the rate of net energy inlet is ${\dot{E}}_{\text{in}}$, the rate of net energy outlet is ${\dot{E}}_{\text{out}}$ and the rate of change of net energy of the system is $\Delta {\dot{E}}_{\text{system}}$.

At steady state the rate of change of net energy of the system $\left(\Delta {\dot{E}}_{\text{system}}\right)$ is zero.

$\Delta {\dot{E}}_{\text{system}}=0$

Refer Equation (XI).

Write the energy balance equation for open feed water heater-II.

$\begin{array}{l}{\dot{E}}_{\text{in}}-{\dot{E}}_{\text{out}}=0\\ {\dot{E}}_{\text{in}}={\dot{E}}_{\text{out}}\\ {\displaystyle \sum {\dot{m}}_{\text{in}}{h}_{\text{in}}}={\displaystyle \sum {\dot{m}}_{\text{out}}{h}_{\text{out}}}\\ {\dot{m}}_8{h}_8+{\dot{m}}_4{h}_4={\dot{m}}_5{h}_5\end{array}$ (XII)

Rewrite the Equation (XII) in terms of mass fraction $y$.

$\begin{array}{l}y{h}_8+\left(1-y\right)h_4=1h_5\\ y{h}_8+h_4-y{h}_4=h_5\\ y\left(h_8-h_4\right)=h_5-h_4\\ y=\frac{h_5-h_4}{h_8-h_4}\end{array}$ (XIII)

Refer Equation (XI).

Write the energy balance equation for open feed water heater-I.

$\begin{array}{l}{\dot{E}}_{\text{in}}-{\dot{E}}_{\text{out}}=0\\ {\dot{E}}_{\text{in}}={\dot{E}}_{\text{out}}\\ {\displaystyle \sum {\dot{m}}_{\text{in}}{h}_{\text{in}}}={\displaystyle \sum {\dot{m}}_{\text{out}}{h}_{\text{out}}}\\ {\dot{m}}_{11}{h}_{11}+{\dot{m}}_2{h}_2={\dot{m}}_3{h}_3\end{array}$ (XIV)

Rewrite the Equation (XIV) in terms of mass fraction $y\text{and}z$.

$\begin{array}{l}z{h}_{11}+\left(1-y-z\right)h_2=\left(1-y\right)h_3\\ z{h}_{11}+\left(1-y\right)h_2-z{h}_2=\left(1-y\right)h_3\\ z\left(h_{11}-h_2\right)=\left(1-y\right)h_3-\left(1-y\right)h_2\\ z=\frac{h_3-h_2}{h_{11}-h_2}\left(1-y\right)\end{array}$ (XV)

Write the formula for mass flow rate.

$\dot{m}=\frac{{\dot{Q}}_{\text{in}}}{q_{\text{in}}}$ (XVI)

Here, the rate of heat input is ${\dot{Q}}_{\text{in}}$.

At state 1:

The water exits the condenser as a saturated liquid at the pressure of $1\text{psia}$. Hence, the enthalpy specific volume at state 1 is as follows.

$\begin{array}{l}{h}_1=h_{f@1\text{psia}}\\ v_1=v_{f@1\text{psia}}\end{array}$

Refer Table A-5E, “Saturated water-Pressure table”.

The enthalpy $\left(h_1\right)$ and specific volume $\left(v_1\right)$ at state 1 corresponding to the pressure of $1\text{psia}$ is $69.72\text{Btu/lbm}$ and $0.01614{\text{ft}}^3\text{/lbm}$ respectively.

At state 3: (Pump II inlet)

The water exits the open feed water heater-I as a saturated liquid at the pressure of $40\text{psia}$. Hence, the enthalpy and specific volume at state 3 is as follows.

$\begin{array}{l}{h}_3=h_{f@40\text{psia}}\\ v_3=v_{f@40\text{psia}}\end{array}$

Refer Table A-5E, “Saturated water-Pressure table”.

The enthalpy $\left(h_3\right)$ and specific volume $\left(v_3\right)$ at state 3 corresponding to the pressure of $40\text{psia}$ is $236.14\text{Btu/lbm}$ and $0.01715{\text{ft}}^3\text{/lbm}$ respectively.

At state 5: (Pump III inlet)

The water exits the open feed water heater-II as a saturated liquid at the pressure of $250\text{psia}$. Hence, the enthalpy and specific volume at state 5 is as follows.

$\begin{array}{l}{h}_5=h_{f@250\text{psia}}\\ v_5=v_{f@250\text{psia}}\end{array}$

Refer Table A-5E, “Saturated water-Pressure table”.

The enthalpy $\left(h_5\right)$ and specific volume $\left(v_5\right)$ at state 5 corresponding to the pressure of $250\text{psia}$ is $376.09\text{Btu/lbm}$ and $0.01865{\text{ft}}^3\text{/lbm}$ respectively.

At state 7: (H.P. Turbine inlet)

The steam enters the turbine as superheated vapor.

Refer Table A-6E, “Superheated water”.

The enthalpy $\left(h_7\right)$ and entropy $\left(s_7\right)$ at state 7 corresponding to the pressure of $1500\text{psia}$ and the temperature of $1100\text{°F}$ is as follows.

$\begin{array}{l}{h}_7=1550.5\text{Btu/lbm}\\ s_7=1.6402\text{Btu/lbm}\cdot \text{R}\end{array}$

Refer Figure 2.

$s_7=s_8=s_9=1.6402\text{Btu/lbm}\cdot \text{R}$

At state 8:

The steam is extracted at the pressure of $250\text{psia}$ and in the state of superheated vapor.

Refer Table A-6E, “Superheated water”.

The enthalpy $\left(h_8\right)$ at state 8 corresponding to the pressure of $250\text{psia}$ and the entropy of is as follows.

$h_8=1308.5\text{Btu/lbm}$

At state 9:

The steam is expanded at the pressure of $140\text{psia}$ and in the state of superheated vapor.

Refer Table A-6E, “Superheated water”.

The enthalpy $\left(h_9\right)$ at state 9 corresponding to the pressure of $140\text{psia}$ and the entropy of $1.6402\text{Btu/lbm}\cdot \text{R}$ is as follows.

$h_9=1248.8\text{Btu/lbm}$

At state 10:

The steam is reheated to the temperature of $1000\text{°F}$ by the keeping the pressure equal to the state 9. (Superheated steam)

Refer Table A-6E, “Superheated water”.

The enthalpy $\left(h_{10}\right)$ and entropy $\left(s_{10}\right)$ at state 10 corresponding to the pressure of $140\text{psia}$ and the temperature of $1000\text{°F}$ is as follows.

$\begin{array}{l}{h}_{10}=1531.3\text{Btu/lbm}\\ s_{10}=1.8832\text{Btu/lbm}\cdot \text{R}\end{array}$

Refer Figure 2.

$s_{10}=s_{11}=s_{12}=1.8832\text{Btu/lbm}\cdot \text{R}$

At state 11:

The steam is expanded at the pressure of $40\text{psia}$ and in the state of superheated vapor.

Refer Table A-6E, “Superheated water”.

The enthalpy $\left(h_{11}\right)$ at state 11 corresponding to the pressure of $40\text{psia}$ and the entropy of $1.8832\text{Btu/lbm}\cdot \text{R}$ is as follows.

$h_{11}=1356.0\text{Btu/lbm}$

At state 12: (Condenser inlet)

The steam enters the condenser at the pressure of $1\text{psia}$ and at the state of saturated mixture.

Refer Table A-5E, “Saturated water-Pressure table”.

Obtain the following properties corresponding to the pressure of $1\text{psia}$.

$\begin{array}{l}{h}_{f,12}=69.72\text{Btu/lbm}\\ h_{fg,12}=1035.7\text{Btu/lbm}\\ s_{f,12}=0.13262\text{Btu/lbm}\cdot \text{R}\\ s_{fg,12}=1.84495\text{Btu/lbm}\cdot \text{R}\end{array}$

Conclusion:

Substitute $0.01614{\text{ft}}^3\text{/lbm}$ for $v_1$, $1\text{psia}$ for $P_1$, and $40\text{psia}$ for $P_2$ in Equation (I).

$\begin{array}{c}{w}_{\text{pI,in}}=\left(0.01614{\text{ft}}^3\text{/lbm}\right)\left(40\text{psia}-1\text{psia}\right)\\ =0.62946\text{psia}\cdot {\text{ft}}^3\text{/lbm}\times \frac{1\text{Btu}}{5.404\text{psia}\cdot {\text{ft}}^3}\\ =0.1165\text{Btu/lbm}\\ \simeq 0.12\text{Btu/lbm}\end{array}$

Substitute $69.72\text{Btu/lbm}$ for $h_1$, and $0.12\text{Btu/lbm}$ for $w_{\text{pI,in}}$ in Equation (II).

$\begin{array}{c}{h}_2=69.72\text{Btu/lbm}+0.12\text{Btu/lbm}\\ =69.84\text{Btu/lbm}\end{array}$

Substitute $0.01715{\text{ft}}^3\text{/lbm}$ for $v_3$, $40\text{psia}$ for $P_3$, and $250\text{psia}$ for $P_4$ in

Equation (III).

$\begin{array}{c}{w}_{\text{pII,in}}=\left(0.01715{\text{ft}}^3\text{/lbm}\right)\left(250\text{psia}-40\text{psia}\right)\\ =3.6015\text{psia}\cdot {\text{ft}}^3\text{/lbm}\times \frac{1\text{Btu}}{5.404\text{psia}\cdot {\text{ft}}^3}\\ =0.6665\text{Btu/lbm}\\ \simeq 0.67\text{Btu/lbm}\end{array}$

Substitute $236.14\text{Btu/lbm}$ for $h_3$, and $0.67\text{Btu/lbm}$ for $w_{\text{pII,in}}$ in Equation (IV).

$\begin{array}{c}{h}_4=236.14\text{Btu/lbm}+0.67\text{Btu/lbm}\\ =236.81\text{Btu/lbm}\end{array}$

Substitute $0.01865{\text{ft}}^3\text{/lbm}$ for $v_5$, $1500\text{psia}$ for $P_6$, and $250\text{psia}$ for $P_5$ in

Equation (V).

$\begin{array}{c}{w}_{\text{pIII,in}}=\left(0.01865{\text{ft}}^3\text{/lbm}\right)\left(1500\text{psia}-250\text{psia}\right)\\ =23.3125\text{psia}\cdot {\text{ft}}^3\text{/lbm}\times \frac{1\text{Btu}}{5.404\text{psia}\cdot {\text{ft}}^3}\\ =4.3139\text{Btu/lbm}\\ \simeq 4.31\text{Btu/lbm}\end{array}$

Substitute $376.09\text{Btu/lbm}$ for $h_5$, and $4.31\text{Btu/lbm}$ for $w_{\text{pIII,in}}$ in Equation (VI).

$\begin{array}{c}{h}_6=376.09\text{Btu/lbm}+4.31\text{Btu/lbm}\\ =380.4\text{Btu/lbm}\end{array}$

From Figure 2.

$s_{10}=s_{11}=s_{12}=1.8832\text{Btu/lbm}\cdot \text{R}$

Substitute $1.8832\text{Btu/lbm}\cdot \text{R}$ for $s_{12}$, $0.13262\text{Btu/lbm}\cdot \text{R}$ for $s_{f,12}$, and $1.84495\text{Btu/lbm}\cdot \text{R}$ for $s_{fg,12}$ in Equation (VII).

$\begin{array}{c}{x}_{12}=\frac{1.8832\text{Btu/lbm}\cdot \text{R}-0.13262\text{Btu/lbm}\cdot \text{R}}{1.84495\text{Btu/lbm}\cdot \text{R}}\\ =0.9488\end{array}$

Substitute $69.72\text{Btu/lbm}$ for $h_{f,12}$, $1035.7\text{Btu/lbm}$ for $h_{fg,12}$, and $0.9488$ for $x_{12}$ in

Equation (VIII).

$\begin{array}{c}{h}_{12}=69.72\text{Btu/lbm}+0.9488\left(1035.7\text{Btu/lbm}\right)\\ =69.72\text{Btu/lbm}+982.6722\text{Btu/lbm}\\ =1052.3922\text{Btu/lbm}\\ \simeq 1052.4\text{Btu/lbm}\end{array}$

Consider the open feed water heater-II alone.

Substitute $376.09\text{Btu/lbm}$ for $h_5$, $236.81\text{Btu/lbm}$ for $h_4$, and $1308.5\text{Btu/lbm}$ for $h_8$ in

Equation (XIII).

$\begin{array}{c}y=\frac{376.09\text{Btu/lbm}-236.81\text{Btu/lbm}}{1308.5\text{Btu/lbm}-236.81\text{Btu/lbm}}\\ =\frac{139.28}{1071.69}\\ =0.12996\\ =0.13\end{array}$

Consider the open feed water heater-I alone.

Substitute $236.14\text{Btu/lbm}$ for $h_3$, $69.84\text{Btu/lbm}$ for $h_2$, $1356.0\text{Btu/lbm}$ for $h_{11}$, and $0.13$ for $y$ in Equation (XV).

$\begin{array}{c}z=\frac{236.14\text{Btu/lbm}-69.84\text{Btu/lbm}}{1356.0\text{Btu/lbm}-69.84\text{Btu/lbm}}\left(1-0.13\right)\\ =\frac{166.3}{1286.16}\left(0.87\right)\\ =0.1125\end{array}$

Substitute $1550.5\text{Btu/lbm}$ for $h_7$, $380.4\text{Btu/lbm}$ for $h_6$, $0.13$ for $y$, $1531.3\text{Btu/lbm}$ for $h_{10}$, and $1248.8\text{Btu/lbm}$ for $h_9$ in Equation (IX).

$\begin{array}{c}{q}_{\text{in}}=\left[\begin{array}{c}\left(1550.5\text{Btu/lbm}-380.4\text{Btu/lbm}\right)\\ +\left(1-0.13\right)\left(1531.3\text{Btu/lbm}-1248.8\text{Btu/lbm}\right)\end{array}\right]\\ =1170.1\text{Btu/lbm}+\left(0.87\right)\left(282.5\text{Btu/lbm}\right)\\ =1415.875\text{Btu/lbm}\\ \simeq 1415.8\text{Btu/lbm}\end{array}$

Substitute $0.13$ for $y$, $0.1125$ for $z$, $1052.4\text{Btu/lbm}$ for $h_{12}$, and $69.72\text{Btu/lbm}$ for $h_1$ in Equation (X).

$\begin{array}{c}{q}_{\text{out}}=\left(1-0.13-0.1125\right)\left(1052.4\text{Btu/lbm}-69.72\text{Btu/lbm}\right)\\ =0.7575\left(982.68\text{Btu/lbm}\right)\\ =744.3801\text{Btu/lbm}\\ \simeq 744.4\text{Btu/lbm}\end{array}$

Substitute $4\times {10}^5\text{Btu/s}$ for ${\dot{Q}}_{\text{in}}$ and $1415.8\text{Btu/lbm}$ for $q_{\text{in}}$ in Equation (XVI).

$\begin{array}{c}\dot{m}=\frac{4\times {10}^5\text{Btu/s}}{1415.8\text{Btu/lbm}}\\ =282.5258\text{lbm/s}\\ \simeq 282.5\text{lbm/s}\end{array}$

Thus, the mass flow rate of steam flowing through the boiler is $282.5\text{lbm/s}$.

To determine

The net power output of the plant.

Answer to Problem 62P

The net power output of the plant is $200.1\text{MW}$.

Explanation of Solution

Write the formula for net power output of the cycle per unit mass.

$w_{\text{net}}=q_{\text{in}}-q_{\text{out}}$ (XVII)

Write the formula for net power output of the cycle.

${\dot{W}}_{\text{net}}=\dot{m}{w}_{\text{net}}$ (XVIII)

Here, the mass flow rate is $\dot{m}$.

Conclusion:

Substitute $1415.8\text{Btu/lbm}$ for $q_{\text{in}}$ and $744.4\text{Btu/lbm}$ for $q_{\text{out}}$ in Equation (XVII).

$\begin{array}{c}{w}_{\text{net}}=1415.8\text{Btu/lbm}-744.4\text{Btu/lbm}\\ =671.4\text{Btu/lbm}\end{array}$

Substitute $282.5\text{lbm/s}$ for $\dot{m}$ and $671.4\text{Btu/lbm}$ for $w_{\text{net}}$ in Equation (XVIII)

$\begin{array}{c}{\dot{W}}_{\text{net}}=282.5\text{lbm/s}\left(671.4\text{Btu/lbm}\right)\\ =18967.05\text{Btu/s}\times \frac{1\text{kJ}}{0.94782\text{Btu}}\\ =200112.3631\text{kJ/s}\times \frac{\text{MW}}{{10}^3\text{kJ/s}}\\ \simeq 200.1\text{MW}\end{array}$

Thus, the net power output of the plant is $200.1\text{MW}$.

To determine

The thermal efficiency of the cycle.

Answer to Problem 62P

The thermal efficiency of the cycle is $47.4\%$.

Explanation of Solution

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

${\eta }_{\text{th}}=1-\frac{q_{\text{out}}}{q_{\text{in}}}$ (XIX)

Conclusion:

Substitute $1415.8\text{Btu/lbm}$ for $q_{\text{in}}$ and $744.4\text{Btu/lbm}$ for $q_{\text{out}}$ in Equation (XIX).

$\begin{array}{c}{\eta }_{\text{th}}=1-\frac{744.4\text{Btu/lbm}}{1415.8\text{Btu/lbm}}\\ =1-0.5258\\ =0.4742\times 100\\ =47.4\%\end{array}$

Thus, the thermal efficiency of the cycle is $47.4\%$.