Chapter 10.9, Problem 53P

Consider an ideal steam regenerative Rankine cycle with two feedwater heaters, one closed and one open. Steam enters the turbine at 10 MPa and 600°C and exhausts to the condenser at 10 kPa. Steam is extracted from the turbine at 1.2 MPa for the closed feedwater heater and at 0.6 MPa for the open one. The feedwater is heated to the condensation temperature of the extracted steam in the closed feedwater heater. The extracted steam leaves the closed feedwater heater as a saturated liquid, which is subsequently throttled to the open feedwater heater. 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 for a net power output of 400 MW and (b) the thermal efficiency of the cycle.

FIGURE P10–53

To determine

The mass flow rate of the steam through the boiler for a net power output of $400\text{MW}$.

Answer to Problem 53P

The mass flow rate of the steam through the boiler for a net power output of $400\text{MW}$ is $313\text{kg/s}$.

Explanation of Solution

Draw the schematic diagram of the given ideal 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 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)

At state 11:

The steam expanded to the pressure of $10\text{kPa}$ and the steam is at the state of saturated mixture.

The quality of water at state 11 is expressed as follows.

$x_{11}=\frac{s_{11}-s_{f,11}}{s_{fg,11}}$ (V)

The enthalpy at state 11 is expressed as follows.

$h_{11}=h_{f,11}+x_{11}{h}_{fg,11}$ (VI)

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 11 indicates the process state 11.

Refer Figure 1 and 2.

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

$q_{\text{in}}=h_8-h_5$ (VII)

$q_{\text{out}}=\left(1-y-z\right)\left(h_{11}-h_1\right)$ (VIII)

Here, the mass fraction steam extracted from the turbine to the feed water entering the boiler via closed feed water heater $\left({\dot{m}}_9/{\dot{m}}_5\right)$ is $y$ and the mass fraction steam extracted from the turbine to feed water entering the boiler via the open feed water heater $\left({\dot{m}}_{10}/{\dot{m}}_5\right)$ is $z$.

Write the general equation of energy balance equation.

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

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 (IX).

Consider the closed feed water heater alone.

Here,

$\begin{array}{l}{\dot{m}}_9={\dot{m}}_6\\ {\dot{m}}_4={\dot{m}}_5\end{array}$

Write the energy balance equation for closed feed water heater.

$\begin{array}{l}{\dot{E}}_{\text{in}}-{\dot{E}}_{\text{out}}=0\\ {\dot{E}}_{\text{in}}={\dot{E}}_{\text{out}}\\ {\displaystyle \sum {\dot{m}}_{\text{in}}{h}_{\text{in}}}={\displaystyle \sum {\dot{m}}_{\text{out}}{h}_{\text{out}}}\\ {\dot{m}}_4{h}_4+{\dot{m}}_9{h}_9={\dot{m}}_5{h}_5+{\dot{m}}_6{h}_6\end{array}$

$\begin{array}{l}{\dot{m}}_5{h}_4+{\dot{m}}_9{h}_9={\dot{m}}_5{h}_5+{\dot{m}}_9{h}_6\\ {\dot{m}}_9{h}_9-{\dot{m}}_9{h}_6={\dot{m}}_5{h}_5-{\dot{m}}_5{h}_4\\ {\dot{m}}_9\left(h_9-h_6\right)={\dot{m}}_5\left(h_5-h_4\right)\end{array}$ (X)

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

$\begin{array}{l}y\left(h_9-h_6\right)=1\left(h_5-h_4\right)\\ y=\frac{h_5-h_4}{h_9-h_6}\end{array}$ (XI)

Refer Equation (IX).

Consider the open feed water heater alone.

Here,

$\begin{array}{l}{\dot{m}}_3={\dot{m}}_2+{\dot{m}}_7+{\dot{m}}_{10}\\ {\dot{m}}_3={\dot{m}}_4\\ {\dot{m}}_4={\dot{m}}_5\end{array}$

Write the energy balance equation for open feed water heater.

$\begin{array}{l}{\dot{E}}_{\text{in}}-{\dot{E}}_{\text{out}}=0\\ {\dot{E}}_{\text{in}}={\dot{E}}_{\text{out}}\\ {\displaystyle \sum {\dot{m}}_{\text{in}}{h}_{\text{in}}}={\displaystyle \sum {\dot{m}}_{\text{out}}{h}_{\text{out}}}\\ {\dot{m}}_7{h}_7+{\dot{m}}_2{h}_2+{\dot{m}}_{10}{h}_{10}={\dot{m}}_3{h}_3\end{array}$

${\dot{m}}_7{h}_7+{\dot{m}}_2{h}_2+{\dot{m}}_{10}{h}_{10}={\dot{m}}_5{h}_3$ (XII)

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

$\begin{array}{l}y{h}_7+\left(1-y-z\right)h_2+z{h}_{10}=h_3\\ y{h}_7+h_2-y{h}_2-z{h}_2+z{h}_{10}=h_3\\ y{h}_7-y{h}_2+z{h}_{10}-z{h}_2=h_3-h_2\\ y\left(h_7-h_2\right)+z\left(h_{10}-h_2\right)=h_3-h_2\end{array}$

$z=\frac{h_3-h_2-y\left(h_7-h_2\right)}{h_{10}-h_2}$ (XIII)

Write the formula for net work output of the cycle.

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

Write the formula for mass flow rate of the cycle.

$\dot{m}=\frac{{\dot{W}}_{\text{net}}}{w_{\text{net}}}$ (XV)

At state 1: (Pump I inlet)

The water exits the condenser as a saturated liquid at the pressure of $10\text{kPa}$. Hence, the enthalpy and specific volume at state 1 is as follows.

$\begin{array}{l}{h}_1=h_{f@10\text{kPa}}\\ v_1=v_{f@10\text{kPa}}\end{array}$

Refer Table A-5, “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 $10\text{kPa}$ is $191.81\text{kJ/kg}$ and $0.001010{\text{m}}^3\text{/kg}$ respectively.

At state 3: (Pump II inlet)

The water exits the open feed water heater-I as a saturated liquid at the pressure of $0.6\text{MPa}\left(600\text{kPa}\right)$. Hence, the enthalpy and specific volume at state 3 is as follows.

$\begin{array}{l}{h}_3=h_{f@600\text{kPa}}\\ v_3=v_{f@600\text{kPa}}\end{array}$

Refer Table A-5, “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 $0.6\text{MPa}\left(600\text{kPa}\right)$ is $670.38\text{kJ/kg}$ and $0.001101{\text{m}}^3\text{/kg}$ respectively.

At state 6: (boiler inlet or closed feed water exit)

The feed water is heated to the condensation temperature $\left(T_6\right)$ of the extracted steam and the extracted steam is the at pressure of $1.2\text{MPa}\left(1200\text{kPa}\right)$.

$T_6=T_{\text{sat}\text{@1}\text{.2}\text{MPa}}$

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

The temperature $\left(T_6\right)$ at state 6 corresponding to the pressure of $1.2\text{MPa}\left(1200\text{kPa}\right)$ is $188.0°\text{C}$.

The extracted steam exits the closed feed water heater as a saturated liquid at the pressure of $1.2\text{MPa}\left(1200\text{kPa}\right)$. Hence, the enthalpy at state 6 is as follows.

$h_6=h_{f@1200\text{kPa}}$

The enthalpy $\left(h_6\right)$ at state 6 corresponding to the pressure of $1.2\text{MPa}\left(1200\text{kPa}\right)$ is $798.33\text{kJ/kg}$.

At State 5:

The extracted steam exits the closed feed water heater as a saturated liquid at the temperature of $188.0°\text{C}$.

$h_5=h_{f@188°\text{C}}$

Refer Table A-4, “Saturated water-Temperature table”.

The enthalpy $\left(h_5\right)$ at state 5 corresponding to the temperature of $188.0°\text{C}$ is $798.33\text{kJ/kg}$.

At state 7:

The steam at state 6 is throttled to state 7. During throttling the enthalpy kept constant.

$h_6=h_7=798.33\text{kJ/kg}$

At state 8:

The steam enters the turbine as superheated vapor.

Refer Table A-6, “Superheated water”.

The enthalpy $\left(h_8\right)$ and entropy $\left(s_8\right)$ at state 8 corresponding to the pressure of $10\text{MPa}\left(10000\text{kPa}\right)$ and the temperature of $600°\text{C}$ is as follows.

$\begin{array}{l}{h}_8=3625.8\text{kJ/kg}\\ s_8=6.9045\text{kJ/kg}\cdot \text{K}\end{array}$

From Figure 2,

$s_8=s_9=s_{10}=s_{11}=6.9045\text{kJ/kg}\cdot \text{K}$

At state 9:

The steam is extracted at the pressure of $1.2\text{MPa}\left(1200\text{kPa}\right)$ and in the state of superheated vapor only.

Refer Table A-6, “Superheated water”.

The enthalpy $\left(h_9\right)$ at state 9 corresponding to the pressure of $1.2\text{MPa}\left(1200\text{kPa}\right)$ and the entropy of $6.9045\text{kJ/kg}\cdot \text{K}$ is as follows.

$h_9=2974.5\text{kJ/kg}$

At state 10:

The steam is extracted at the pressure of $0.6\text{MPa}\left(600\text{kPa}\right)$ and in the state of superheated vapor only.

Refer Table A-6, “Superheated water”.

The enthalpy $\left(h_{10}\right)$ at state 10 corresponding to the pressure of $0.6\text{MPa}\left(600\text{kPa}\right)$ and the entropy of $6.9045\text{kJ/kg}\cdot \text{K}$ is as follows.

$h_{10}=2820.9\text{kJ/kg}$

At state 11:

The steam enters the condenser at the pressure of $10\text{kPa}$ and at the state of saturated mixture.

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

Obtain the following properties corresponding to the pressure of $10\text{kPa}$.

$\begin{array}{l}{h}_{f,11}=191.81\text{kJ/kg}\\ h_{fg,11}=2392.1\text{kJ/kg}\\ s_{f,11}=0.6492\text{kJ/kg}\cdot \text{K}\\ s_{fg,11}=7.4996\text{kJ/kg}\cdot \text{K}\end{array}$

Conclusion:

Substitute $0.001010{\text{m}}^3\text{/kg}$ for $v_1$, $10\text{kPa}$ for $P_1$, and $600\text{kPa}$ for $P_2$ in Equation (I).

$\begin{array}{c}{w}_{\text{pI,in}}=\left(0.001010{\text{m}}^3/\text{kg}\right)\left(600\text{kPa}-10\text{kPa}\right)\\ =0.5959\text{kPa}\cdot {\text{m}}^3/\text{kg}\times \frac{1\text{kJ}}{1\text{kPa}\cdot {\text{m}}^3}\\ =0.60\text{kJ}/\text{kg}\end{array}$

Substitute $191.81\text{kJ/kg}$ for $h_1$, and $0.60\text{kJ}/\text{kg}$ for $w_{\text{pI,in}}$ in Equation (II).

$\begin{array}{c}{h}_2=191.81\text{kJ/kg}+0.60\text{kJ}/\text{kg}\\ =192.41\text{kJ}/\text{kg}\\ \simeq 192.4\text{kJ}/\text{kg}\end{array}$

Substitute $0.001101{\text{m}}^3/\text{kg}$ for $v_3$, $600\text{kPa}$ for $P_3$, and $10000\text{kPa}$ for $P_4$ in

Equation (III).

$\begin{array}{c}{w}_{\text{pII,in}}=\left(0.001101{\text{m}}^3/\text{kg}\right)\left(10000\text{kPa}-600\text{kPa}\right)\\ =10.3494\text{kPa}\cdot {\text{m}}^3/\text{kg}\times \frac{1\text{kJ}}{1\text{kPa}\cdot {\text{m}}^3}\\ \simeq 10.35\text{kJ}/\text{kg}\end{array}$

Substitute $670.38\text{kJ/kg}$ for $h_3$, and $10.35\text{kJ}/\text{kg}$ for $w_{\text{pII,in}}$ in Equation (IV).

$\begin{array}{c}{h}_4=670.38\text{kJ/kg}+10.35\text{kJ}/\text{kg}\\ =680.73\text{kJ}/\text{kg}\end{array}$

From Figure 1,

$s_8=s_9=s_{10}=s_{11}=6.9045\text{kJ/kg}\cdot \text{K}$

Substitute $6.9045\text{kJ/kg}\cdot \text{K}$ for $s_9$, $0.6492\text{kJ/kg}\cdot \text{K}$ for $s_{f,11}$, and $7.4996\text{kJ/kg}\cdot \text{K}$ for $s_{fg,11}$ in Equation (V).

$\begin{array}{c}{x}_{11}=\frac{6.9045\text{kJ/kg}\cdot \text{K}-0.6492\text{kJ/kg}\cdot \text{K}}{7.4996\text{kJ/kg}\cdot \text{K}}\\ =0.8341\end{array}$

Substitute $191.81\text{kJ/kg}$ for $h_{f,11}$, $2392.1\text{kJ/kg}$ for $h_{fg,11}$, and $0.8341$ for $x_{11}$ in

Equation (VI).

$\begin{array}{c}{h}_{11}=191.81\text{kJ/kg}+0.8341\left(2392.1\text{kJ/kg}\right)\\ =191.81\text{kJ/kg}+1995.2506\text{kJ}/\text{kg}\\ =2187.0606\text{kJ}/\text{kg}\\ \simeq 2187\text{kJ}/\text{kg}\end{array}$

Consider the open feed water heater alone.

Substitute $798.33\text{kJ/kg}$ for $h_5$, $680.73\text{kJ}/\text{kg}$ for $h_4$, $2974.5\text{kJ/kg}$ for $h_9$, and $798.33\text{kJ/kg}$ $h_6$ in Equation (XI).

$\begin{array}{c}y=\frac{798.33\text{kJ/kg}-680.73\text{kJ}/\text{kg}}{2974.5\text{kJ/kg}-798.33\text{kJ/kg}}\\ =\frac{117.6}{2176.17}\\ =0.05404\end{array}$

Consider the closed feed water heater alone.

Substitute $670.38\text{kJ/kg}$ for $h_3$, $192.4\text{kJ}/\text{kg}$ for $h_2$, $0.05404$ for $y$, $798.33\text{kJ/kg}$ for $h_7$, and $2820.9\text{kJ/kg}$ for $h_{10}$ in Equation (XIII).

$\begin{array}{c}z=\frac{670.38\text{kJ/kg}-192.4\text{kJ}/\text{kg}-0.05404\left(798.33\text{kJ/kg}-192.4\text{kJ}/\text{kg}\right)}{2820.9\text{kJ/kg}-192.4\text{kJ}/\text{kg}}\\ =\frac{455.2355\text{kJ}/\text{kg}}{2628.5\text{kJ}/\text{kg}}\\ =0.1694\end{array}$

Substitute $3625.8\text{kJ/kg}$ for $h_8$, and $798.33\text{kJ/kg}$ for $h_5$ in Equation (VII).

$\begin{array}{c}{q}_{\text{in}}=3625.8\text{kJ/kg}-798.33\text{kJ/kg}\\ =2827.47\text{kJ}/\text{kg}\\ \simeq 2827.5\text{kJ}/\text{kg}\end{array}$

Substitute $0.05404$ for $y$, $0.1694$ for $z$, $2187\text{kJ}/\text{kg}$ for $h_{11}$, and $191.81\text{kJ/kg}$ for $h_1$ in Equation (VIII).

$\begin{array}{c}{q}_{\text{out}}=\left(1-0.05404-0.1694\right)\left(2187\text{kJ}/\text{kg}-191.81\text{kJ/kg}\right)\\ =0.7766\left(1995.19\text{kJ}/\text{kg}\right)\\ =1549.4645\text{kJ}/\text{kg}\\ =1549.5\text{kJ}/\text{kg}\end{array}$

Substitute $2827.5\text{kJ}/\text{kg}$ for $q_{\text{in}}$, and $1549.5\text{kJ}/\text{kg}$ for $q_{\text{out}}$ in Equation (XIV).

$\begin{array}{c}{w}_{\text{net}}=2827.5\text{kJ}/\text{kg}-1549.5\text{kJ}/\text{kg}\\ =1278\text{kJ}/\text{kg}\end{array}$

Substitute $400\text{MW}$ for ${\dot{W}}_{\text{net}}$ and $1278\text{kJ}/\text{kg}$ for $w_{\text{net}}$ in Equation (XV).

$\begin{array}{c}\dot{m}=\frac{400\text{MW}}{1278\text{kJ}/\text{kg}}\\ =\frac{400\text{MW}\times \frac{{10}^3\text{kJ/s}}{1\text{MW}}}{1278\text{kJ}/\text{kg}}\\ =312.989\text{}\text{kg/s}\\ \simeq 313\text{kg/s}\end{array}$

Thus, the mass flow rate of the steam through the boiler for a net power output of $400\text{MW}$ is $313\text{kg/s}$.

To determine

The thermal efficiency of the cycle.

Answer to Problem 53P

The thermal efficiency of the cycle is $45.2\%$.

Explanation of Solution

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

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

Conclusion:

Substitute $2827.5\text{kJ}/\text{kg}$ for $q_{\text{in}}$, and $1549.5\text{kJ}/\text{kg}$ for $q_{\text{out}}$ in Equation (XVI).

$\begin{array}{c}{\eta }_{\text{th}}=1-\frac{1549.5\text{kJ}/\text{kg}}{2827.5\text{kJ}/\text{kg}}\\ =1-0.5480\\ =0.45199\times 100\\ =45.2\%\end{array}$

Thus, the thermal efficiency of the cycle is $45.2\%$.