Chapter 10.9, Problem 52P

A steam power plant operates on an ideal regenerative Rankine cycle with two open feedwater heaters. Steam enters the turbine at 8 MPa and 550°C and exhausts to the condenser at 15 kPa. Steam is extracted from the turbine at 0.6 and 0.2 MPa. Water leaves both feedwater heaters as a saturated liquid. The mass flow rate of steam through the boiler is 24 kg/s. Show the cycle on a T-s diagram, and determine (a) the net power output of the power plant and (b) the thermal efficiency of the cycle.

To determine

The power output of plant.

Answer to Problem 52P

The power output of the plant is $28.8\text{MW}$.

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)

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 6.

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

At state 9:

The steam expanded to the pressure of $0.2\text{MPa}$ and the steam is at the state of saturated mixture.

The quality of water at the exit of the turbine (state 9) is expressed as follows.

$x_9=\frac{s_9-s_{f,9}}{s_{fg,9}}$ (VII)

The enthalpy at state 9 is expressed as follows.

$h_9=h_{f,9}+x_9{h}_{fg,9}$ (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 9 indicates the process state 9.

At state 10:

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

The quality of water at state 10 is expressed as follows.

$x_{10}=\frac{s_{10}-s_{f,10}}{s_{fg,10}}$ (IX)

The enthalpy at state 10 is expressed as follows.

$h_{10}=h_{f,10}+x_{10}{h}_{fg,10}$ (X)

Here, the subscript 10 indicates the process state 10.

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.

$q_{\text{in}}=h_7-h_6$ (XI)

$q_{\text{out}}=\left(1-y-z\right)\left(h_{10}-h_1\right)$ (XII)

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

Write the general equation of energy balance equation.

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

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 (XIII).

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

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

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

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

Refer Equation (XIII).

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

$\begin{array}{l}{\dot{E}}_{\text{in}}-{\dot{E}}_{\text{out}}=0\\ {\dot{E}}_{\text{in}}={\dot{E}}_{\text{out}}\\ {\displaystyle \sum {\dot{m}}_{\text{in}}{h}_{\text{in}}}={\displaystyle \sum {\dot{m}}_{\text{out}}{h}_{\text{out}}}\\ {\dot{m}}_9{h}_9+{\dot{m}}_2{h}_2={\dot{m}}_3{h}_3\end{array}$ (XVI)

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

$\begin{array}{l}z{h}_9+\left(1-y-z\right)h_2=\left(1-y\right)h_3\\ z{h}_9+\left(1-y\right)h_2-z{h}_2=\left(1-y\right)h_3\\ z\left(h_9-h_2\right)=\left(1-y\right)h_3-\left(1-y\right)h_2\\ z=\frac{h_3-h_2}{h_9-h_2}\left(1-y\right)\end{array}$ (XVII)

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

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

Write the formula for net power output of the cycle.

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

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

At state 1: (Pump I inlet)

The water exits the condenser as a saturated liquid at the pressure of $15\text{kPa}$. Hence, the enthalpy and specific volume at state 1 is as follows.

$\begin{array}{l}{h}_1=h_{f@15\text{kPa}}\\ v_1=v_{f@15\text{kPa}}\end{array}$

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 $15\text{kPa}$ is $225.94\text{kJ/kg}$ and $0.001014{\text{m}}^3\text{/kg}$ respectively.

At state 3: (Pump II inlet)

The water exits the open feed water heater-I as a saturated liquid at the pressure of $0.2\text{MPa}\left(200\text{kPa}\right)$. Hence, the enthalpy and specific volume at state 3 is as follows.

$\begin{array}{l}{h}_3=h_{f@200\text{kPa}}\\ v_3=v_{f@200\text{kPa}}\end{array}$

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 $0.2\text{MPa}\left(200\text{kPa}\right)$ is $504.71\text{kJ/kg}$ and $0.001061{\text{m}}^3\text{/kg}$ respectively.

At state 5: (Pump III inlet)

The water exits the open feed water heater-II 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 5 is as follows.

$\begin{array}{l}{h}_5=h_{f@600\text{kPa}}\\ v_5=v_{f@600\text{kPa}}\end{array}$

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

At state 7:

The steam enters the turbine as superheated vapor.

Refer Table A-6, “Superheated water”.

The enthalpy $\left(h_7\right)$ and entropy $\left(s_7\right)$ at state 7 corresponding to the pressure of $8\text{MPa}\left(8000\text{kPa}\right)$ and the temperature of $550°\text{C}$ is as follows.

$\begin{array}{l}{h}_7=3521.8\text{kJ/kg}\\ s_7=6.8800\text{kJ/kg}\cdot \text{K}\end{array}$

From Figure 2.

$s_7=s_8=s_9=s_{10}=6.8800\text{kJ/kg}\cdot \text{K}$

At state 8:

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

Refer Table A-6, “Superheated water”.

The enthalpy $\left(h_8\right)$ at state 8 corresponding to the pressure of $0.6\text{MPa}\left(600\text{kPa}\right)$ and the entropy of $6.8800\text{kJ/kg}\cdot \text{K}$ is as follows.

$h_8=2809.6\text{kJ/kg}$

At state 9:

The steam expanded to the pressure of $0.2\text{MPa}\left(200\text{kPa}\right)$ and in the state of saturated mixture.

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

Obtain the following properties corresponding to the pressure of $0.2\text{MPa}\left(200\text{kPa}\right)$.

$\begin{array}{l}{h}_{f,9}=504.71\text{kJ/kg}\\ h_{fg,9}=2201.6\text{kJ/kg}\\ s_{f,9}=1.5302\text{kJ/kg}\cdot \text{K}\\ s_{fg,9}=5.5968\text{kJ/kg}\cdot \text{K}\end{array}$

At state 10:

The steam enters the condenser at the pressure of $15\text{kPa}$ and at the state of saturated mixture.

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

Obtain the following properties corresponding to the pressure of $15\text{kPa}$.

$\begin{array}{l}{h}_{f,10}=225.94\text{kJ/kg}\\ h_{fg,10}=2372.3\text{kJ/kg}\\ s_{f,10}=0.7549\text{kJ/kg}\cdot \text{K}\\ s_{fg,10}=7.2522\text{kJ/kg}\cdot \text{K}\end{array}$

Conclusion:

Substitute $0.001014{\text{m}}^3\text{/kg}$ for $v_1$, $15\text{kPa}$ for $P_1$, and $200\text{kPa}$ for $P_2$ in Equation (I).

$\begin{array}{c}{w}_{\text{pI,in}}=\left(0.001014{\text{m}}^3/\text{kg}\right)\left(200\text{kPa}-15\text{kPa}\right)\\ =0.18759\text{kPa}\cdot {\text{m}}^3/\text{kg}\times \frac{1\text{kJ}}{1\text{kPa}\cdot {\text{m}}^3}\\ =0.188\text{kJ}/\text{kg}\end{array}$

Substitute $225.94\text{kJ/kg}$ for $h_1$, and $0.188\text{kJ}/\text{kg}$ for $w_{\text{pI,in}}$ in Equation (II).

$\begin{array}{c}{h}_2=225.94\text{kJ/kg}+0.188\text{kJ}/\text{kg}\\ =226.13\text{kJ}/\text{kg}\end{array}$

Substitute $0.001061{\text{m}}^3/\text{kg}$ for $v_3$, $200\text{kPa}$ for $P_3$, and $600\text{kPa}$ for $P_4$ in

Equation (III).

$\begin{array}{c}{w}_{\text{pII,in}}=\left(0.001061{\text{m}}^3/\text{kg}\right)\left(600\text{kPa}-200\text{kPa}\right)\\ =0.4244\text{kPa}\cdot {\text{m}}^3/\text{kg}\times \frac{1\text{kJ}}{1\text{kPa}\cdot {\text{m}}^3}\\ \simeq 0.42\text{kJ}/\text{kg}\end{array}$

Substitute $\text{504}\text{.71}\text{kJ}/\text{kg}$ for $h_3$, and $0.42\text{kJ}/\text{kg}$ for $w_{\text{pII,in}}$ in Equation (IV).

$\begin{array}{c}{h}_4=\text{504}\text{.71}\text{kJ}/\text{kg}+0.42\text{kJ}/\text{kg}\\ =505.13\text{kJ}/\text{kg}\end{array}$

Substitute $0.001101{\text{m}}^3/\text{kg}$ for $v_5$, $600\text{kPa}$ for $P_5$, and $8000\text{kPa}$ for $P_4$ in

Equation (V).

$\begin{array}{c}{w}_{\text{pIII,in}}=\left(0.001101{\text{m}}^3/\text{kg}\right)\left(8000\text{kPa}-600\text{kPa}\right)\\ =8.1474\text{kPa}\cdot {\text{m}}^3/\text{kg}\times \frac{1\text{kJ}}{1\text{kPa}\cdot {\text{m}}^3}\\ \simeq 8.15\text{kJ}/\text{kg}\end{array}$

Substitute $\text{670}\text{.38}\text{kJ}/\text{kg}$ for $h_5$, and $8.15\text{kJ}/\text{kg}$ for $w_{\text{pIII,in}}$ in Equation (VI).

$\begin{array}{c}{h}_6=\text{670}\text{.38}\text{kJ}/\text{kg}+8.15\text{kJ}/\text{kg}\\ =678.53\text{kJ}/\text{kg}\end{array}$

From Figure 1.

$s_7=s_8=s_8=s_{10}=6.8800\text{kJ/kg}\cdot \text{K}$

Substitute $6.8800\text{kJ/kg}\cdot \text{K}$ for $s_9$, $1.5302\text{kJ/kg}\cdot \text{K}$ for $s_{f,9}$, and $5.5968\text{kJ/kg}\cdot \text{K}$ for $s_{fg,9}$ in Equation (VII).

$\begin{array}{c}{x}_9=\frac{6.8800\text{kJ/kg}\cdot \text{K}-1.5302\text{kJ/kg}\cdot \text{K}}{5.5968\text{kJ/kg}\cdot \text{K}}\\ =0.9559\end{array}$

Substitute $504.71\text{kJ/kg}$ for $h_{f,9}$, $2201.6\text{kJ/kg}$ for $h_{fg,9}$, and $0.9559$ for $x_9$ in

Equation (VIII).

$\begin{array}{c}{h}_9=504.71\text{kJ/kg}+0.9559\left(2201.6\text{kJ/kg}\right)\\ =504.71\text{kJ/kg}+2104.5094\text{kJ}/\text{kg}\\ =2609.2194\text{kJ}/\text{kg}\end{array}$

Substitute $6.8800\text{kJ/kg}\cdot \text{K}$ for $s_{10}$, $0.7549\text{kJ/kg}\cdot \text{K}$ for $s_{f,10}$, and $7.2522\text{kJ/kg}\cdot \text{K}$ for $s_{fg,10}$ in Equation (IX).

$\begin{array}{c}{x}_{10}=\frac{6.8800\text{kJ/kg}\cdot \text{K}-0.7549\text{kJ/kg}\cdot \text{K}}{7.2522\text{kJ/kg}\cdot \text{K}}\\ =0.8446\end{array}$

Substitute $225.94\text{kJ/kg}$ for $h_{f,10}$, $2372.3\text{kJ/kg}$ for $h_{fg,10}$, and $0.8446$ for $x_{10}$ in

Equation (X).

$\begin{array}{c}{h}_{10}=225.94\text{kJ/kg}+0.8446\left(2372.3\text{kJ/kg}\right)\\ =225.94\text{kJ/kg}+1977.6037\text{kJ}/\text{kg}\\ =2229.5846\text{kJ}/\text{kg}\\ \simeq 2229.6\text{kJ}/\text{kg}\end{array}$

Consider the open feed water heater-II alone.

Substitute $670.38\text{kJ/kg}$ for $h_5$, $505.13\text{kJ}/\text{kg}$ for $h_4$, and $2809.6\text{kJ/kg}$ for $h_8$ in Equation (XIV).

$\begin{array}{c}y=\frac{670.38\text{kJ/kg}-505.13\text{kJ}/\text{kg}}{2809.6\text{kJ/kg}-505.13\text{kJ}/\text{kg}}\\ =\frac{165.25}{2304.47}\\ =0.07171\end{array}$

Consider the open feed water heater-I alone.

Substitute $504.71\text{kJ/kg}$ for $h_3$, $226.13\text{kJ}/\text{kg}$ for $h_2$, $2609.2194\text{kJ}/\text{kg}$ for $h_9$, and $0.07171$ for $y$ in Equation (XVII).

$\begin{array}{c}z=\frac{504.71\text{kJ/kg}-226.13\text{kJ}/\text{kg}}{2609.2194\text{kJ}/\text{kg}-226.13\text{kJ}/\text{kg}}\left(1-0.07171\right)\\ =\frac{278.58}{2383.0894}\left(0.92829\right)\\ =0.1085\end{array}$

Substitute $3521.8\text{kJ/kg}$ for $h_7$, and $678.53\text{kJ}/\text{kg}$ for $h_6$ in Equation (XI).

$\begin{array}{c}{q}_{\text{in}}=3521.8\text{kJ/kg}-678.53\text{kJ}/\text{kg}\\ =2843.27\text{kJ}/\text{kg}\\ \simeq 2843.3\text{kJ}/\text{kg}\end{array}$

Substitute $0.07171$ for $y$, $0.1085$ for $z$, $2229.6\text{kJ}/\text{kg}$ for $h_{10}$, and $225.94\text{kJ/kg}$ for $h_1$ in Equation (XII).

$\begin{array}{c}{q}_{\text{out}}=\left(1-0.07171-0.1085\right)\left(2229.6\text{kJ}/\text{kg}-225.94\text{kJ/kg}\right)\\ =0.81979\left(2003.66\text{kJ}/\text{kg}\right)\\ =1642.5804\text{kJ}/\text{kg}\\ =1642.6\text{kJ}/\text{kg}\end{array}$

Substitute $2843.3\text{kJ}/\text{kg}$ for $q_{\text{in}}$, and $1642.6\text{kJ}/\text{kg}$ for $q_{\text{out}}$ in Equation (XVIII).

$\begin{array}{c}{w}_{\text{net}}=2843.3\text{kJ}/\text{kg}-1642.6\text{kJ}/\text{kg}\\ =1200.7\text{kJ}/\text{kg}\end{array}$

Substitute $24\text{kg/s}$ for $\dot{m}$ and  $1200.7\text{kJ}/\text{kg}$ for $w_{\text{net}}$ in Equation (XIX).

$\begin{array}{c}{\dot{W}}_{\text{net}}=24\text{kg/s}\left(1200.7\text{kJ}/\text{kg}\right)\\ =28816.8\text{kJ/s}\times \frac{1\text{MW}}{{10}^3\text{kJ/s}}\\ =28.8168\text{MW}\\ \simeq 28.8\text{MW}\end{array}$

Thus, the power output of the plant is $28.8\text{MW}$.

To determine

The thermal efficiency of the cycle.

Answer to Problem 52P

The thermal efficiency of the cycle is $42.2\%$.

Explanation of Solution

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

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

Conclusion:

Substitute $1642.6\text{kJ}/\text{kg}$ for $q_{\text{out}}$, and $2843.3\text{kJ}/\text{kg}$ for $q_{\text{in}}$ in Equation (XV).

$\begin{array}{c}{\eta }_{\text{th}}=1-\frac{1642.6\text{kJ}/\text{kg}}{2843.3\text{kJ}/\text{kg}}\\ =1-0.5777\\ =0.4222\times 100\\ =42.2\%\end{array}$

Thus, the thermal efficiency of the cycle is $42.2\%$.