Chapter 10.9, Problem 110RP

A steam power plant operates on an ideal reheat–regenerative Rankine cycle with one reheater and two feedwater heaters, one open and one closed. Steam enters the high-pressure turbine at 15 MPa and 600°C and the low-pressure turbine at 1 MPa and 500°C. The condenser pressure is 5 kPa. Steam is extracted from the turbine at 0.6 MPa for the closed feedwater heater and at 0.2 MPa for the open feedwater heater. In the closed feedwater heater, the feedwater is heated to the condensation temperature of the extracted steam. 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. Determine (a) the fraction of steam extracted from the turbine for the open feedwater heater, (b) the thermal efficiency of the cycle, and (c) the net power output for a mass flow rate of 42 kg/s through the boiler.

To determine

The fraction of steam extracted from the turbine for the open feed water heater.

Answer to Problem 110RP

The fraction of steam extracted from the turbine for the open feed water heater is $\underset{\_}{0.1165}$.

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 two 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 enthalpy $\left(h\right)$ at state 5.

$h_5=h_6+v_6\left(P_5-P_6\right)$ (V)

The quality of water at state 13 is expressed as follows.

$x_{13}=\frac{s_{13}-s_{f,13}}{s_{fg,13}}$ (VI)

The enthalpy at state 13 is expressed as follows.

$h_{13}=h_{f,13}+x_{13}{h}_{fg,13}$ (VII)

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 13 indicates the process state 13.

Write the general equation of energy balance equation.

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

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 (VIII).

Write the energy balance equation for open feed water heater.

$\begin{array}{l}{\dot{E}}_{\text{in}}-{\dot{E}}_{\text{out}}=0\\ {\dot{E}}_{\text{in}}={\dot{E}}_{\text{out}}\\ {\displaystyle \sum {\dot{m}}_{\text{in}}{h}_{\text{in}}}={\displaystyle \sum {\dot{m}}_{\text{out}}{h}_{\text{out}}}\\ {\dot{m}}_{11}\left(h_{11}-h_6\right)={\dot{m}}_5\left(h_5-h_4\right)\end{array}$ (IX)

Here, the mass fraction steam extracted from the turbine to the inlet mass of the boiler $y$ is equal to $\left({\dot{m}}_{11}/{\dot{m}}_5\right)$

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

$\begin{array}{l}y\left(h_{11}-h_6\right)=h_5-h_4\\ y=\frac{h_5-h_4}{h_{11}-h_6}\end{array}$ (X)

For the open FWH,

$\begin{array}{l}{\dot{E}}_{\text{in}}-{\dot{E}}_{\text{out}}=0\\ {\dot{E}}_{\text{in}}={\dot{E}}_{\text{out}}\\ {\displaystyle \sum {\dot{m}}_{\text{in}}{h}_{\text{in}}}={\displaystyle \sum {\dot{m}}_{\text{out}}{h}_{\text{out}}}\\ {\dot{m}}_7{h}_7+{\dot{m}}_2{h}_2+{\dot{m}}_{12}{h}_{12}={\dot{m}}_3{h}_3\end{array}$

$y{h}_7+\left(1-y-z\right)h_2+z{h}_{12}=\left(1\right)h_3$ (XI)

Here, the mass fraction steam extracted from the turbine to the inlet mass of the boiler $z$ is equal to $\left({\dot{m}}_{12}/{\dot{m}}_5\right)$.

Solving Equation (XI).

$z=\frac{\left(h_3-h_2\right)-y\left(h_7-h_2\right)}{h_{12}-h_2}$ (XII)

Conclusion:

At state 1:

The water exits the condenser as a saturated liquid at the pressure of $5\text{kPa}$. Hence, the enthalpy at state 1 is as follows.

$h_1=h_{f@5\text{kPa}}$

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 $5\text{kPa}$ is $137.75\text{kJ/kg}$ and $0.001005{\text{m}}^3\text{/kg}$ respectively.

Substitute $0.001005{\text{m}}^3\text{/kg}$ for $v_1$, $5\text{kPa}$ for $P_1$, and $0.2\text{MPa}$ for $P_2$ in Equation (I).

$\begin{array}{c}{w}_{\text{pI,in}}=\left(0.001005{\text{m}}^3/\text{kg}\right)\left(0.2\text{MPa}-5\text{kPa}\right)\\ =\left(0.001005{\text{m}}^3/\text{kg}\right)\left(0.2\text{MPa}\times \left(\frac{1000\text{kPa}}{1\text{MPa}}\right)-5\text{kPa}\right)\\ =0.195975\text{kPa}\cdot {\text{m}}^3/\text{kg}\times \frac{1\text{kJ}}{1\text{kPa}\cdot {\text{m}}^3}\\ =0.195975\text{kJ}/\text{kg}\end{array}$

        $\cong 0.20\text{kJ}/\text{kg}$

Substitute $137.75\text{kJ/kg}$ for $h_1$, and $0.20\text{kJ}/\text{kg}$ for $w_{\text{pI,in}}$ in Equation (II).

$\begin{array}{c}{h}_2=137.75\text{kJ/kg}+0.20\text{kJ}/\text{kg}\\ =137.95\text{kJ}/\text{kg}\end{array}$

From the Table A-5, “Saturated water-temperature Table” obtains the value of the enthalpy $\left(h_3\right)$ and specific volume $\left(v_3\right)$ at state 3 corresponding to the pressure of $0.2\text{MPa}$ is $504.71\text{kJ}/\text{kg}$ and $0.001061{\text{m}}^{\text{3}}/\text{kg}$.

Substitute $0.001061{\text{m}}^3\text{/kg}$ for $v_3$, $0.2\text{MPa}$ for $P_3$, and $15\text{MPa}$ for $P_4$ in Equation (III).

$\begin{array}{c}{w}_{\text{pII,in}}=\left(0.001061{\text{m}}^3\text{/kg}\right)\left(15\text{MPa}-0.2\text{MPa}\right)\\ =\left(0.001061{\text{m}}^3\text{/kg}\right)\left(\begin{array}{l}15\text{MPa}\times \left(\frac{1000\text{kPa}}{1\text{MPa}}\right)\\ -0.2\text{MPa}\times \left(\frac{1000\text{kPa}}{1\text{MPa}}\right)\end{array}\right)\\ =15.70\text{kPa}\cdot {\text{m}}^3/\text{kg}\times \frac{1\text{kJ}}{1\text{kPa}\cdot {\text{m}}^3}\\ =15.70\text{kJ}/\text{kg}\end{array}$

Substitute $504.71\text{kJ/kg}$ for $h_3$, and $15.70\text{kJ}/\text{kg}$ for $w_{\text{pII,in}}$ in Equation (IV).

$\begin{array}{c}{h}_4=504.71\text{kJ/kg}+15.70\text{kJ}/\text{kg}\\ =520.41\text{kJ}/\text{kg}\end{array}$

From the Table A-5, “Saturated water-temperature Table” obtains the value of the enthalpy $\left(h_6\right)$ and specific volume $\left(v_6\right)$ at state 6 corresponding to the pressure of $0.6\text{MPa}$ is $670.38\text{kJ}/\text{kg}$ and $0.001101{\text{m}}^{\text{3}}/\text{kg}$.

Here, the temperature at the state 6 $\left(T_6\right)$ is equal to temperature at the state 5 $\left(T_5\right)$

Substitute $670.38\text{kJ}/\text{kg}$ for $h_6$, $0.001101{\text{m}}^{\text{3}}/\text{kg}$ for $v_6$, $0.6\text{MPa}$ for $P_6$, and $15\text{MPa}$ for $P_5$ in Equation (IV).

$\begin{array}{c}{h}_5=\left(670.38\text{kJ}/\text{kg}\right)+\left(0.001101{\text{m}}^{\text{3}}/\text{kg}\right)\left(15\text{MPa}-0.6\text{MPa}\right)\\ =\left(670.38\text{kJ}/\text{kg}\right)+\left(0.001101{\text{m}}^{\text{3}}/\text{kg}\right)\left(\begin{array}{l}15\text{MPa}\times \left(\frac{1000\text{kPa}}{1\text{MPa}}\right)\\ -0.6\text{MPa}\times \left(\frac{1000\text{kPa}}{1\text{MPa}}\right)\end{array}\right)\\ =686.23\text{kPa}\cdot {\text{m}}^3/\text{kg}\times \frac{1\text{kJ}}{1\text{kPa}\cdot {\text{m}}^3}\\ =686.23\text{kJ}/\text{kg}\end{array}$

From the Table A-6, “Superheated water” obtains the value of the enthalpy $\left(h_8\right)$ and entropy $\left(s_8\right)$ at state 8 corresponding to the pressure of $15\text{MPa}$ and the temperature of $600°\text{C}$ is $3583.1\text{kJ}/\text{kg}$ and $6.6796\text{kJ}/\text{kg}\cdot \text{K}$.

From the Table A-6, “Superheated water” obtains the value of the enthalpy $\left(h_9\right)$ at state 9 corresponding to the pressure of $1.0\text{MPa}$ is $2820.8\text{kJ}/\text{kg}$.

Here, the entropy at the state 9 $\left(s_9\right)$ is equal to the entropy at the state 8 $\left(s_8\right)$.

From the Table A-6, “Superheated water” obtains the value of the enthalpy $\left(h_{10}\right)$ at state 10 corresponding to the pressure of $1.0\text{MPa}$ and the temperature of $500°\text{C}$ is $3479.1\text{kJ}/\text{kg}$ and $7.7642\text{kJ}/\text{kg}\cdot \text{K}$.

Refer Table A-6, “superheated water”, and write the enthalpy at state 11 at pressure of $\text{0}\text{.6}\text{MPa}$ using an interpolation method.

$h_{11}=h_{f@0.6\text{MPa}}$ (XIII)

Here, enthalpy of saturation liquid at pressure of $0.6\text{MPa}$ is $h_{f@0.6\text{MPa}}$.

Write the formula of interpolation method of two variables.

$y_2=\frac{\left(x_2-x_1\right)\left(y_3-y_1\right)}{\left(x_3-x_1\right)}+y_1$ (XIV)

Here, the variables denote by x and y is specific entropy and specific enthalpy at state 11 respectively.

Show the specific enthalpy at state 11 corresponding to temperature as in Table (1).

Specific entropy at state 11 $s_{11}\left(\text{kJ}/\text{kg}\cdot \text{K}\right)$	Specific enthalpy at state 11 $h_{11}\left(\text{kJ}/\text{kg}\right)$
7.7097 $\left(x_1\right)$	3270.8 $\left(y_1\right)$
7.7642 $\left(x_2\right)$	$\left(y_2=?\right)$
8.0041 $\left(x_3\right)$	3483.4 $\left(y_3\right)$

Substitute $\text{7}\text{.7097}\text{kJ}/\text{kg}\cdot \text{K},\text{7}\text{.7642}\text{kJ}/\text{kg}\cdot \text{K},\text{and}\text{8}\text{.0041}\text{kJ}/\text{kg}\cdot \text{K}$ for $x_1,x_2\text{and}{x}_3$ respectively, $3270.8\text{kJ}/\text{kg}$ for $y_1$ and $\text{3483}\text{.4}\text{kJ}/\text{kg}$ for $y_3$ in Equation (XIV).

$\begin{array}{c}{y}_2=\left[\begin{array}{l}\frac{\left(\text{7}\text{.7642}\text{kJ}/\text{kg}\cdot \text{K}-\text{7}\text{.7097}\text{kJ}/\text{kg}\cdot \text{K}\right)\left(\text{3483}\text{.4}\text{kJ}/\text{kg}-3270.8\text{kJ}/\text{kg}\right)}{\left(\text{8}\text{.0041}\text{kJ}/\text{kg}\cdot \text{K}-\text{7}\text{.7097}\text{kJ}/\text{kg}\cdot \text{K}\right)}\\ +3270.8\text{kJ}/\text{kg}\end{array}\right]\\ =\text{3310}\text{.15}\text{kJ}/\text{kg}\\ \cong \text{3310}\text{.2}\text{kJ}/\text{kg}\end{array}$

Substitute $\text{3310}\text{.2}\text{kJ}/\text{kg}$ for $h_{f@0.6\text{MPa}}$ in Equation (XIII).

$h_{11}=\text{3310}\text{.2}\text{kJ}/\text{kg}$

Similarly repeat the Equation (XIV) for specific enthalpy at state 11 corresponding to the pressure of $0.2\text{MPa}$.

$h_{12}=3000.9\text{kJ}/\text{kg}$.

From the Table A-5, “Saturated water” obtains the value of the specific entropy of saturated liquid $\left(s_f\right)$, the specific enthalpy of saturated liquid $\left(h_f\right)$, the specific entropy change upon vaporization $\left(s_{fg}\right)$, and the specific enthalpy change upon vaporization $\left(h_{fg}\right)$ at state 13 corresponding to the pressure of $5\text{kPa}$.h

$\begin{array}{l}{s}_f=0.4762\text{kJ}/\text{kg}\cdot \text{K}\\ s_{fg}=7.9176\text{kJ}/\text{kg}\cdot \text{K}\\ h_f=137.75\text{kJ}/\text{kg}\\ h_{fg}=2423.0\text{kJ}/\text{kg}\end{array}$

Substitute $7.7642\text{kJ}/\text{kg}\cdot \text{K}$ for $s_{13}$, $0.4762\text{kJ}/\text{kg}\cdot \text{K}$ for $s_f$, and $7.9176\text{kJ}/\text{kg}\cdot \text{K}$ for $s_{fg}$ in Equation (VI).

$\begin{array}{c}{x}_{13}=\frac{\left(7.7642\text{kJ}/\text{kg}\cdot \text{K}\right)-\left(0.4762\text{kJ}/\text{kg}\cdot \text{K}\right)}{\left(7.9176\text{kJ}/\text{kg}\cdot \text{K}\right)}\\ =\frac{\left(7.288\text{kJ}/\text{kg}\cdot \text{K}\right)}{\left(7.9176\text{kJ}/\text{kg}\cdot \text{K}\right)}\\ =0.92048\\ \cong 0.9205\end{array}$

Substitute $137.75\text{kJ}/\text{kg}$ for $h_f$, $2423.0\text{kJ}/\text{kg}$ for $h_{fg}$, and $0.9205$ for $x_{13}$ in Equation (VII).

$\begin{array}{c}{h}_{13}=\left(137.75\text{kJ}/\text{kg}\right)+\left(0.9205\right)\times \left(2423.0\text{kJ}/\text{kg}\right)\\ =\left(137.75\text{kJ}/\text{kg}\right)+\left(2230.372\text{kJ}/\text{kg}\right)\\ =2368.1\text{kJ}/\text{kg}\end{array}$

Substitute $686.23\text{kJ}/\text{kg}$ for $h_5$, $520.41\text{kJ}/\text{kg}$ for $h_4$, $3310.2\text{kJ}/\text{kg}$ for $h_{11}$, and $670.38\text{kJ}/\text{kg}$ for $h_6$ in Equation (X).

$\begin{array}{c}y=\frac{\left(686.23-520.41\right)\text{kJ}/\text{kg}}{\left(3310.2-670.38\right)\text{kJ}/\text{kg}}\\ =\frac{165.82\text{kJ}/\text{kg}}{2639.82\text{kJ}/\text{kg}}\\ =0.062815\end{array}$

Substitute $504.71\text{kJ}/\text{kg}$ for $h_3$, $137.95\text{kJ}/\text{kg}$ for $h_2$, 0.06215 for $y$, $670.38\text{kJ}/\text{kg}$ for $h_7$, and $3000.0\text{kJ}/\text{kg}$ for $h_{12}$ in Equation (XII).

$\begin{array}{c}z=\frac{\left(504.71\text{kJ}/\text{kg}-137.95\text{kJ}/\text{kg}\right)-\left(0.06215\right)\left(670.38\text{kJ}/\text{kg}-137.95\text{kJ}/\text{kg}\right)}{\left(3000.0\text{kJ}/\text{kg}-137.95\text{kJ}/\text{kg}\right)}\\ =\frac{\left(366.76\text{kJ}/\text{kg}\right)-\left(0.06215\right)\left(532.43\text{kJ}/\text{kg}\right)}{\left(2862\text{kJ}/\text{kg}\right)}\\ =\frac{\left(366.76\text{kJ}/\text{kg}\right)-\left(33.09052\text{kJ}/\text{kg}\right)}{\left(2862\text{kJ}/\text{kg}\right)}\\ =0.1165\end{array}$

Thus, the fraction of steam extracted from the turbine for the open feed water heater is $\underset{\_}{0.1165}$.

To determine

The thermal efficiency of the cycle.

Answer to Problem 110RP

The thermal efficiency of the cycle is $\underset{\_}{48.5\%}$.

Explanation of Solution

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

$q_{\text{in}}=\left(h_8-h_5\right)+\left(h_{10}-h_9\right)$ (XV)

$q_{\text{out}}=\left(1-y-z\right)\left(h_{13}-h_1\right)$ (XVI)

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

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

Conclusion:

Substitute $3583.1\text{kJ}/\text{kg}$ for $h_8$, $686.23\text{kJ}/\text{kg}$ for $h_5$, $3479.1\text{kJ}/\text{kg}$ for $h_{10}$, and $2820.8\text{kJ}/\text{kg}$ for $h_9$ in Equation (XV)

$\begin{array}{c}{q}_{\text{in}}=\left(3583.1\text{kJ}/\text{kg}-686.23\text{kJ}/\text{kg}\right)+\left(3479.1\text{kJ}/\text{kg}-2820.8\text{kJ}/\text{kg}\right)\\ =\left(2896.87\text{kJ}/\text{kg}\right)+\left(658.3\text{kJ}/\text{kg}\right)\\ =3555.17\text{kJ}/\text{kg}\end{array}$

Substitute 0.06215 for $y$, 0.1165 for $z$, $2368.0\text{kJ}/\text{kg}$ for $h_{13}$, and $137.75\text{kJ}/\text{kg}$ for $h_1$ in Equation (XVI).

$\begin{array}{c}{q}_{\text{out}}=\left(1-0.06287-0.1165\right)\left(2368.0\text{kJ}/\text{kg}-137.75\text{kJ}/\text{kg}\right)\\ =0.82063\times 2230.25\text{kJ}/\text{kg}\\ =1830.21\text{kJ}/\text{kg}\end{array}$

Substitute $3555.17\text{kJ}/\text{kg}$ for $q_{\text{in}}$ and $1830.21\text{kJ}/\text{kg}$ for $q_{\text{out}}$ in Equation (XVII).

$\begin{array}{c}{w}_{\text{net}}=3555.17\text{kJ}/\text{kg}-1830.21\text{kJ}/\text{kg}\\ =1724.96\text{kJ}/\text{kg}\end{array}$

Substitute $3555.17\text{kJ}/\text{kg}$ for $q_{\text{in}}$ and $1830.21\text{kJ}/\text{kg}$ for $q_{\text{out}}$ in Equation (XVIII).

$\begin{array}{c}{\eta }_{\text{th}}=1-\frac{1830.21\text{kJ}/\text{kg}}{3555.17\text{kJ}/\text{kg}}\\ =1-0.514802\text{kJ}/\text{kg}\\ =0.485198\\ \cong 48.5\%\end{array}$

Thus, the thermal efficiency of the cycle is $\underset{\_}{48.5\%}$.

To determine

The net power output for mass flow rate of $42\text{kg}/\text{s}$ through the boiler.

Answer to Problem 110RP

The net power output for mass flow rate of $42\text{kg}/\text{s}$ through the boiler is $\underset{\_}{72,448\text{kW}}$.

Explanation of Solution

Write the expression for net power output for mass flow rate of $42\text{kg}/\text{s}$ through the boiler.

${\dot{W}}_{\text{net}}=\dot{m}\times w_{\text{net}}$ (XIX)

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

Conclusion:

Substitute $42\text{kg}/\text{s}$ for $\dot{m}$ and $1724.96\text{kJ}/\text{kg}$ for $w_{\text{net}}$ in Equation (XIX).

$\begin{array}{c}{\dot{W}}_{\text{net}}=\left(42\text{kg}/\text{s}\right)\times \left(1724.96\text{kJ}/\text{kg}\right)\\ =72,448\text{kW}\end{array}$

Thus, the net power output for mass flow rate of $42\text{kg}/\text{s}$ through the boiler is $\underset{\_}{72,448\text{kW}}$.