Chapter 10.9, Problem 94RP

Consider a steam power plant that operates on a regenerative Rankine cycle and has a net power output of 150 MW. Steam enters the turbine at 10 MPa and 500°C and the condenser at 10 kPa. The isentropic efficiency of the turbine is 80 percent, and that of the pumps is 95 percent. Steam is extracted from the turbine at 0.5 MPa to heat the feedwater in an open feedwater heater. Water leaves the feedwater heater as a saturated liquid. Show the cycle on a T-s diagram, and determine (a) the mass flow rate of steam through the boiler and (b) the thermal efficiency of the cycle. Also, determine the exergy destruction associated with the regeneration process. Assume a source temperature of 1300 K and a sink temperature of 303 K.

To determine

The mass flow rate of steam through the boiler.

Answer to Problem 94RP

The mass flow rate of steam through the boiler is $159.7\text{kg/s}$.

Explanation of Solution

Draw the $T-s$ diagram of the given regenerative Rankine cycle as shown in Figure 1.

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

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

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

Here, the specific volume is $v$, the pressure is $P$, the isentropic efficiency of the pump is ${\eta }_P$, 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.

$w_{\text{pII,in}}=\frac{v_3\left(P_4-P_3\right)}{{\eta }_P}$ (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 $6s$:

The steam expanded to the pressure of $0.5\text{MPa}\left(500\text{kPa}\right)$ and the steam is at the state of saturated mixture.

The quality of water at state $6s$ is expressed as follows.

$x_{6s}=\frac{s_{6s}-s_{f,6s}}{s_{fg,6s}}$ (V)

The enthalpy at state $6s$ is expressed as follows.

$h_{6s}=h_{f,6s}+x_{6s}{h}_{fg,6s}$ (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 $6s$ indicates the ideal process state $6s$.

The isentropic efficiency for the process 5-6 is expressed as follows.

$\begin{array}{l}{\eta }_T=\frac{h_5-h_6}{h_5-h_{6s}}\\ h_6=h_5-{\eta }_T\left(h_5-h_{6s}\right)\end{array}$ (VII)

At state $7s$:

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 $7s$ is expressed as follows.

$x_{7s}=\frac{s_{7s}-s_{f,7s}}{s_{fg,7s}}$ (VIII)

The enthalpy at state $7s$ is expressed as follows.

$h_{7s}=h_{f,7s}+x_{7s}{h}_{fg,7s}$ (IX)

Here, the subscript $7s$ indicates the ideal process state $7s$.

The isentropic efficiency for the process 5-7 is expressed as follows.

$\begin{array}{l}{\eta }_T=\frac{h_5-h_7}{h_5-h_{7s}}\\ h_7=h_5-{\eta }_T\left(h_5-h_{7s}\right)\end{array}$ (X)

Here, the subscript $6s$, and $7s$ indicates the ideal process states; all other subscripts such as 1, 2, 3, 4, 5, 6 and 7 indicates the actual process states.

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

$q_{\text{in}}=h_5-h_4$ (XI)

$q_{\text{out}}=\left(1-y\right)\left(h_7-h_1\right)$ (XII)

Here, the mass fraction steam extracted from the turbine to the inlet mass of the boiler $\left({\dot{m}}_6/{\dot{m}}_3\right)$ is $y$.

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.

$\begin{array}{l}{\dot{E}}_{\text{in}}-{\dot{E}}_{\text{out}}=0\\ {\dot{E}}_{\text{in}}={\dot{E}}_{\text{out}}\\ {\displaystyle \sum {\dot{m}}_{\text{in}}{h}_{\text{in}}}={\displaystyle \sum {\dot{m}}_{\text{out}}{h}_{\text{out}}}\\ {\dot{m}}_6{h}_6+{\dot{m}}_2{h}_2={\dot{m}}_3{h}_3\end{array}$ (XIV)

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

$\begin{array}{l}y{h}_6+\left(1-y\right)h_2=1h_3\\ y{h}_6+h_2-y{h}_2=h_3\\ y\left(h_6-h_2\right)=h_3-h_2\\ y=\frac{h_3-h_2}{h_6-h_2}\end{array}$ (XV)

Write the formula for net work output of the cycle.

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

Write the formula for mass flow rate of steam.

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

Here, the net power output of the cycle is ${\dot{W}}_{\text{net}}$.

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, specific volume, and entropy at state 1 is as follows.

$\begin{array}{l}{h}_1=h_{f@10\text{kPa}}\\ v_1=v_{f@10\text{kPa}}\\ s_1=s_{f@10\text{kPa}}\end{array}$

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

The enthalpy $\left(h_1\right)$, specific volume $\left(v_1\right)$ and entropy $\left(s_1\right)$ at state 1 corresponding to the pressure of $10\text{kPa}$ is $191.81\text{kJ/kg}$, $0.001010{\text{m}}^3\text{/kg}$, and $0.6492\text{kJ/kg}\cdot \text{K}$ respectively.

At state 3: (Pump II inlet)

The water exits the open feed water heater as a saturated liquid at the pressure of $0.5\text{MPa}\left(500\text{kPa}\right)$. Hence, the enthalpy, specific volume, and entropy at state 3 is as follows.

$\begin{array}{l}{h}_3=h_{f@500\text{kPa}}\\ v_3=v_{f@500\text{kPa}}\\ s_3=s_{f@500\text{kPa}}\end{array}$

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

The enthalpy $\left(h_3\right)$, specific volume $\left(v_3\right)$, and entropy $\left(s_3\right)$ at state 3 corresponding to the pressure of $0.5\text{MPa}\left(500\text{kPa}\right)$ is $640.09\text{kJ/kg}$, $0.001093{\text{m}}^3\text{/kg}$, and $1.8604\text{kJ/kg}\cdot \text{K}$ respectively.

At state 5:

The steam enters the turbine as superheated vapor.

Refer Table A-6, “Superheated water”.

The enthalpy $\left(h_5\right)$ and entropy $\left(s_5\right)$ at state 5 corresponding to the pressure of $10\text{MPa}\left(10000\text{kPa}\right)$ and the temperature of $500°\text{C}$ is as follows.

$\begin{array}{l}{h}_5=3375.1\text{kJ/kg}\\ s_5=6.5995\text{kJ/kg}\cdot \text{K}\end{array}$

Refer Figure 1.

$s_5=s_{6s}=s_{7s}=6.5995\text{kJ/kg}\cdot \text{K}$

At state $6s$:

The steam expanded to the pressure of $0.5\text{MPa}\left(500\text{kPa}\right)$ and in the state of saturated mixture.

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

Obtain the following properties corresponding to the pressure of $0.5\text{MPa}\left(500\text{kPa}\right)$.

$\begin{array}{l}{h}_{f,6s}=640.09\text{kJ/kg}\\ h_{fg,6s}=2108.0\text{kJ/kg}\\ s_{f,6s}=1.8604\text{kJ/kg}\cdot \text{K}\\ s_{fg,6s}=4.9603\text{kJ/kg}\cdot \text{K}\end{array}$

At state $7s$:

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 $10\text{kPa}$.

$\begin{array}{l}{h}_{f,7s}=191.81\text{kJ/kg}\\ h_{fg,7s}=2392.1\text{kJ/kg}\\ s_{f,7s}=0.6492\text{kJ/kg}\cdot \text{K}\\ s_{fg,7s}=7.4996\text{kJ/kg}\cdot \text{K}\end{array}$

Conclusion:

Substitute $0.001010{\text{m}}^3\text{/kg}$ for $v_1$, $10\text{kPa}$ for $P_1$, $500\text{kPa}$ for $P_2$, and $0.95$ for ${\eta }_P$ in Equation (I).

$\begin{array}{c}{w}_{\text{pI,in}}=\frac{\left(0.001010{\text{m}}^3/\text{kg}\right)\left(500\text{kPa}-10\text{kPa}\right)}{0.95}\\ =0.5209\text{kPa}\cdot {\text{m}}^3/\text{kg}\times \frac{1\text{kJ}}{1\text{kPa}\cdot {\text{m}}^3}\\ =0.52\text{kJ}/\text{kg}\end{array}$

Substitute $191.81\text{kJ/kg}$ for $h_1$, and $0.52\text{kJ}/\text{kg}$ for $w_{\text{pI,in}}$ in Equation (II).

$\begin{array}{c}{h}_2=191.81\text{kJ/kg}+0.52\text{kJ}/\text{kg}\\ =192.33\text{kJ}/\text{kg}\end{array}$

Substitute $0.001093{\text{m}}^3\text{/kg}$ for $v_3$, $500\text{kPa}$ for $P_3$, $10000\text{kPa}$ for $P_4$, and $0.95$ for ${\eta }_P$ in Equation (III).

$\begin{array}{c}{w}_{\text{pII,in}}=\frac{\left(0.001093{\text{m}}^3\text{/kg}\right)\left(10000\text{kPa}-500\text{kPa}\right)}{0.95}\\ =10.93\text{kPa}\cdot {\text{m}}^3/\text{kg}\times \frac{1\text{kJ}}{1\text{kPa}\cdot {\text{m}}^3}\\ =10.93\text{kJ}/\text{kg}\end{array}$

Substitute $640.09\text{kJ/kg}$ for $h_3$, and $10.93\text{kJ}/\text{kg}$ for $w_{\text{pII,in}}$ in Equation (IV).

$\begin{array}{c}{h}_4=640.09\text{kJ/kg}+10.93\text{kJ}/\text{kg}\\ =651.02\text{kJ}/\text{kg}\end{array}$

From Figure 1.

$s_5=s_{6s}=s_{7s}=6.5995\text{kJ/kg}\cdot \text{K}$

Substitute $6.5995\text{kJ/kg}\cdot \text{K}$ for $s_6$, $1.8604\text{kJ/kg}\cdot \text{K}$ for $s_{f,6s}$, and $4.9603\text{kJ/kg}\cdot \text{K}$ for $s_{fg,6s}$ in Equation (V).

$\begin{array}{c}{x}_{6s}=\frac{6.5995\text{kJ/kg}\cdot \text{K}-1.8604\text{kJ/kg}\cdot \text{K}}{4.9603\text{kJ/kg}\cdot \text{K}}\\ =0.9554\end{array}$

Substitute $640.09\text{kJ/kg}$ for $h_{f,6s}$, $2108.0\text{kJ/kg}$ for $h_{fg,6s}$, and $0.9554$ for $x_{6s}$ in

Equation (VI).

$\begin{array}{c}{h}_{6s}=640.09\text{kJ/kg}+0.9554\left(2108.0\text{kJ/kg}\right)\\ =640.09\text{kJ/kg}+2013.9832\text{kJ}/\text{kg}\\ =2654.0732\text{kJ}/\text{kg}\end{array}$

Substitute $3375.1\text{kJ/kg}$ for $h_5$, $0.80$ for ${\eta }_T$, $2654.0732\text{kJ}/\text{kg}$ for $h_{6s}$ in Equation (VII).

$\begin{array}{c}{h}_6=3375.1\text{kJ/kg}-0.80\left(3375.1\text{kJ/kg}-2654.0732\text{kJ}/\text{kg}\right)\\ =3375.1\text{kJ/kg}-576.8214\text{kJ/kg}\\ =2798.2786\text{kJ/kg}\\ \simeq 2798.3\text{kJ/kg}\end{array}$

Substitute $6.5995\text{kJ/kg}\cdot \text{K}$ for $s_{7s}$, $0.6492\text{kJ/kg}\cdot \text{K}$ for $s_{f,7s}$, and $7.4996\text{kJ/kg}\cdot \text{K}$ for $s_{fg,7s}$ in Equation (VIII).

$\begin{array}{c}{x}_{7s}=\frac{6.5995\text{kJ/kg}\cdot \text{K}-0.6492\text{kJ/kg}\cdot \text{K}}{7.4996\text{kJ/kg}\cdot \text{K}}\\ =0.7934\end{array}$

Substitute $191.81\text{kJ/kg}$ for $h_{f,7s}$, $2392.1\text{kJ/kg}$ for $h_{fg,7s}$, and $0.7934$ for $x_{7s}$ in

Equation (IX).

$\begin{array}{c}{h}_{7s}=191.81\text{kJ/kg}+0.7934\left(2392.1\text{kJ/kg}\right)\\ =191.81\text{kJ/kg}+1897.8921\text{kJ}/\text{kg}\\ =2089.7021\text{kJ}/\text{kg}\\ \simeq 2089.7\text{kJ}/\text{kg}\end{array}$

Substitute $3375.1\text{kJ/kg}$ for $h_5$, $0.80$ for ${\eta }_T$, $2089.7\text{kJ}/\text{kg}$ for $h_{7s}$ in Equation (X).

$\begin{array}{c}{h}_7=3375.1\text{kJ/kg}-0.80\left(3375.1\text{kJ/kg}-2089.7\text{kJ/kg}\right)\\ =3375.1\text{kJ/kg}-1028.32\text{kJ/kg}\\ =2346.78\text{kJ/kg}\\ \simeq 2346.8\text{kJ/kg}\end{array}$

Substitute $640.09\text{kJ/kg}$ for $h_3$, $192.33\text{kJ}/\text{kg}$ for $h_2$, and $2798.3\text{kJ/kg}$ for $h_6$ in Equation (XV).

$\begin{array}{c}y=\frac{640.09\text{kJ/kg}-192.33\text{kJ}/\text{kg}}{2798.3\text{kJ/kg}-192.33\text{kJ}/\text{kg}}\\ =\frac{447.76}{2605.97}\\ =0.1718\end{array}$

Substitute $3375.1\text{kJ/kg}$ for $h_5$, and $651.02\text{kJ}/\text{kg}$ for $h_4$ in Equation (XI).

$\begin{array}{c}{q}_{\text{in}}=3375.1\text{kJ/kg}-651.02\text{kJ}/\text{kg}\\ =2724.08\text{kJ}/\text{kg}\\ \simeq 2724.1\text{kJ}/\text{kg}\end{array}$

Substitute $2346.8\text{kJ/kg}$ for $h_7$, $0.1718$ for y, and $191.81\text{kJ/kg}$ for $h_1$ in Equation (XII).

$\begin{array}{c}{q}_{\text{out}}=\left(1-0.1718\right)\left(2346.8\text{kJ/kg}-191.81\text{kJ/kg}\right)\\ =0.8282\left(2154.99\text{kJ}/\text{kg}\right)\\ =1784.7627\text{kJ}/\text{kg}\\ =1784.7\text{kJ}/\text{kg}\end{array}$

Substitute $2724.1\text{kJ}/\text{kg}$ for $q_{\text{in}}$, and $1784.7\text{kJ}/\text{kg}$ for $q_{\text{out}}$ in Equation (XVI).

$\begin{array}{c}{w}_{\text{net}}=2724.1\text{kJ}/\text{kg}-1784.7\text{kJ}/\text{kg}\\ =939.4\text{kJ}/\text{kg}\end{array}$

Substitute $150\text{MW}$ for ${\dot{W}}_{\text{net}}$ and $939.4\text{kJ}/\text{kg}$ for $w_{\text{net}}$ in Equation (XVII).

$\begin{array}{c}\dot{m}=\frac{150\text{MW}}{939.4\text{kJ}/\text{kg}}\\ =\frac{150\text{MW}\times \frac{{10}^3\text{kJ/s}}{1\text{MW}}}{939.4\text{kJ}/\text{kg}}\\ =159.6764\text{kg/s}\\ \simeq 159.7\text{kg/s}\end{array}$

Thus, the mass flow rate of steam through the boiler is $159.7\text{kg/s}$.

To determine

The thermal efficiency of the cycle and the exergy destruction associated with the regeneration process.

Answer to Problem 94RP

The thermal efficiency of the cycle is $34.5\%$ and the exergy destruction associated with the regeneration process is $39.25\text{kJ/kg}$.

Explanation of Solution

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

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

Write the formula for exergy destruction associated with the regeneration cycle.

$\begin{array}{c}{x}_{\text{destroyed, reheat}}=T_0{s}_{\text{gen}}\\ =T_0\left({\displaystyle \sum m_e{s}_e}-{\displaystyle \sum m_i{s}_i}+\frac{q_{surr}}{T_0}\right)\\ =T_0\left[s_3-y{s}_6-\left(1-y\right)s_2\right]\end{array}$ (XIX)

Here, the entropy generation is $s_{\text{gen}}$, the entropy is $s$, the heat transferred to the surrounding is $q_{surr}$, the surrounding (sink) temperature is $T_0$, and the subscripts 2, 3 and 6 indicates the process states.

Conclusion:

Substitute $2724.1\text{kJ}/\text{kg}$ for $q_{\text{in}}$, and $1784.7\text{kJ}/\text{kg}$ for $q_{\text{out}}$ in Equation (XVIII).

$\begin{array}{c}{\eta }_{\text{th}}=1-\frac{1784.7\text{kJ}/\text{kg}}{2724.1\text{kJ}/\text{kg}}\\ =1-0.6551\\ =0.3448\times 100\\ =34.5\%\end{array}$

Thus, the thermal efficiency of the cycle is $34.5\%$.

Consider the regeneration process- open feed water heater, process states 6,2,3.

Here, $s_1=s_2=0.6492\text{}\text{kJ/kg}\cdot \text{K}$.

Substitute $303\text{K}$ for $T_0$, $1.8604\text{kJ/kg}\cdot \text{K}$ for $s_3$, $0.1718$ for $y$, $6.5995\text{kJ/kg}\cdot \text{K}$ for $s_6$, and $0.6492\text{}\text{kJ/kg}\cdot \text{K}$ for $s_2$ in Equation (XIX).

$\begin{array}{c}{x}_{\text{destroyed, reheat}}=303\text{K}\left(\begin{array}{l}1.8604\text{kJ/kg}\cdot \text{K}-\left(0.1718\right)\left(6.5995\text{kJ/kg}\cdot \text{K}\right)\\ -\left(1-0.1718\right)\left(0.6492\text{}\text{kJ/kg}\cdot \text{K}\right)\end{array}\right)\\ =303\text{K}\left(0.1295\text{kJ/kg}\cdot \text{K}\right)\\ =39.2475\text{kJ/kg}\\ \simeq 39.25\text{kJ/kg}\end{array}$

Thus, the exergy destruction associated with the regeneration process is $39.25\text{kJ/kg}$.