Chapter 10.9, Problem 62P

Consider a steam power plant that operates on a reheat Rankine cycle. Steam enters the high-pressure turbine at 10 MPa and 500°C and the low-pressure turbine at 1 MPa and 500°C. Steam leaves the condenser as a saturated liquid at a pressure of 10 kPa. The isentropic efficiency of the turbine is 80 percent, and that of the pump is 95 percent. Determine the exergy destruction associated with the heat addition process and the expansion process. Assume a source temperature of 1600 K and a sink temperature of 285 K. Also, determine the exergy of the steam at the boiler exit. Take P0 = 100 kPa.

To determine

The exergy destruction associated with the heat addition, the expansion process, and the exergy of the steam at the boiler exit.

Answer to Problem 62P

The exergy destruction during the heating process is $1289\text{kJ/kg}$.

The exergy destruction during the expansion process is $247.9\text{kJ}/\text{kg}$.

The exergy of the steam at the boiler exit is $1495\text{kJ}/\text{kg}$.

Explanation of Solution

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

Here, water is the working fluid.

Write the formula for work done by the pump during process 1-2 with the consideration of isentropic efficiency $\left({\eta }_P\right)$.

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

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

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

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

Before reheating,

At state $4s$:

The isentropic efficiency is expressed as follows.

$\begin{array}{l}{\eta }_T=\frac{h_3-h_4}{h_3-h_{4s}}\\ h_4=h_3-{\eta }_T\left(h_3-h_{4s}\right)\end{array}$ (III)

After reheating,

At state 5:

The reheating occurs at constant pressure. Hence, the pressure at state 4 and state 5 are equal.

$P_5=P_4=1\text{MPa}$

At state $6s$:

The steam is expanded to the pressure of $10\text{kPa}$.

The quality of water at the exit of the L.P turbine (state 6) is expressed as follows (actual).

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

The enthalpy at state 6 is expressed as follows.

$h_{6s}=h_{f,6s}+x_{6s}{h}_{fg,6s}$ (V)

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 6 indicates the ideal state 6. and $6s$ indicates the actual state 6.

The isentropic efficiency is expressed as follows.

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

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

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

$q_{\text{in}}=\left(h_3-h_2\right)+\left(h_5-h_4\right)$ (VII)

Write the formula for the exergy destruction for the combined heat addition and pumping process 1-5.

$x_{\text{destroyed}\left(\text{1-5}\right)}=T_0\left[\left(s_3-s_1\right)+\left(s_5-s_4\right)+\frac{q_{R,15}}{T_R}\right]$ (VIII)

Here, the entropy $s$, the heat transferred during the process is $q_R$, the temperature of the heat source is $T_R$, surrounding temperature is $T_0$, and the subscripts 1, 3, 4 and 5 indicates the process states.

Write the formula for the exergy destruction for pumping process 1-2.

$\begin{array}{c}{x}_{\text{destroyed}\left(\text{1-2}\right)}=w_{\text{P},\text{in,}a}-w_{\text{P,in},s}\\ =w_{\text{P},\text{in,}a}-v_1\Delta P\\ =w_{\text{P},\text{in,}a}-v_1\left(P_2-P_1\right)\end{array}$ (IX)

Here, the work input of pump at actual process 1-2 is $w_{\text{P,in,}a}$, the work input of the pump at isentropic process 1-$2s$ is $w_{\text{P,in,}s}$.

Write the formula for exergy destruction for heat addition processes 2-3, 4-5.

$x_{\text{destroyed, heating}}=x_{\text{destroyed,}\left(1-5\right)}-x_{\text{destroyed,}\left(1-2\right)}$ (X)

Write the formula for the exergy destruction for the expansion process 3-4, 5-6.

$x_{\text{destroyed, expansion}}=T_0\left[\left(s_4-s_3\right)+\left(s_6-s_5\right)+\frac{q_{R,3-4,5-6}}{T_R}\right]$ (XI)

Write the formula for exergy of the steam at boiler exit $\left({\psi }_3\right)$.

${\psi }_3=\left(h_3-h_0\right)-T_0\left(s_3-s_0\right)+\frac{V_3{}^2}{2}+g{z}_3$ (XII)

Here, the enthalpy is $h$, the velocity of the steam is $V$ and the gravitational acceleration is $g$, the elevation from the datum is $z$, and the subscript $0$ indicates the surrounding state.

Neglect the kinetic energy $\left(\frac{V^2}{2}\right)$ and the potential energy $\left(gz\right)$, rewrite the Equation (XII) as follows.

${\psi }_3=\left(h_3-h_0\right)-T_0\left(s_3-s_0\right)$ (XIII)

At state 1:

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

$h_1=h_{f@10\text{kPa}}$

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 $10\text{kPa}$ is $191.81\text{kJ/kg}$ and $0.001010{\text{m}}^3\text{/kg}$ respectively.

At state 3: (H.P. Turbine inlet)

The steam enters the as superheated vapor.

Refer Table A-6, “Superheated water”.

The enthalpy $\left(h_3\right)$ and entropy $\left(s_3\right)$ at state 3 corresponding to the pressure of $10\text{MPa}$ and the temperature of $500°\text{C}$ is as follows.

$\begin{array}{l}{h}_3=3375.1\text{kJ/kg}\\ s_3=6.5995\text{kJ/kg}\cdot \text{K}\end{array}$

At state $4s$: (H.P. turbine exit-ideal)

From Figure 1.

$s_3=s_{4s}=6.5995\text{kJ/kg}\cdot \text{K}$

Refer Table A-6, “Superheated water”.

The enthalpy $\left(h_{4s}\right)$ at state $4s$ corresponding to the pressure of $1\text{MPa}$ and the entropy of $6.5995\text{kJ/kg}\cdot \text{K}$ is as follows.

$h_{4s}=2783.8\text{kJ/kg}$

At state 5: (L.P. Turbine inlet)

The steam is reheated to superheated at the pressure of $1\text{MPa}$ and the temperature is $500°\text{C}$.

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 $1\text{MPa}$ and the temperature of $500°\text{C}$ is as follows.

$\begin{array}{l}{h}_5=3479.1\text{kJ/kg}\\ s_5=7.7642\text{kJ/kg}\cdot \text{K}\end{array}$

At state $6s$: (L.P. turbine exit-ideal)

From Figure 1.

$s_5=s_{6s}=7.7642\text{kJ/kg}\cdot \text{K}$

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

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

$\begin{array}{l}{h}_{f,6s}=191.81\text{kJ/kg}\\ h_{fg,6s}=2392.1\text{kJ/kg}\\ s_{f,6s}=0.6492\text{kJ/kg}\cdot \text{K}\\ s_{fg,6s}=7.4996\text{kJ/kg}\cdot \text{K}\end{array}$

Here, the sink temperature is equal to the surrounding temperature.

$T_0=285\text{K}$

The surrounding pressure $\left(P_0\right)$ is $100\text{kPa}$ and the temperature is $285\text{K}$. At this condition the steam is at the state of saturated liquid. Hence, the enthalpy $\left(h_0\right)$ and entropy $\left(s_0\right)$ is expressed as follows.

$\begin{array}{l}{h}_0=h_{f@285\text{K}}\\ s_0=s_{f@285\text{K}}\end{array}$

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

The enthalpy $\left(h_0\right)$ and entropy $\left(s_0\right)$ corresponding to the temperature of $285\text{K}$ is $50.51\text{kJ/kg}$ and $0.1806\text{kJ/kg}\cdot \text{K}$ respectively.

Conclusion:

Substitute $0.001010{\text{m}}^3\text{/kg}$ for $v_1$, $10\text{MPa}$ for $P_2$, $10\text{kPa}$ for $P_1$ and $95\%$ for ${\eta }_P$ in Equation (I).

$\begin{array}{c}{w}_{\text{p,in}}=\frac{\left(0.001010{\text{m}}^3\text{/kg}\right)\left(10\text{MPa}-10\text{kPa}\right)}{95\%}\\ =\frac{\left(0.001010{\text{m}}^3\text{/kg}\right)\left[\left(10\text{}\text{MPa}\times \frac{{10}^3\text{}\text{kPa}}{1\text{MPa}}\right)-10\text{kPa}\right]}{\frac{95}{100}}\\ =\frac{0.001010{\text{m}}^3\text{/kg}\left(9990\text{kPa}\right)}{0.95}\\ =10.6209\text{kPa}\cdot {\text{m}}^3\text{/kg}\times \frac{1\text{kJ}}{\text{kPa}\cdot {\text{m}}^3}\end{array}$

$\simeq 10.62\text{kJ/kg}$

Substitute $191.81\text{kJ/kg}$ for $h_1$ and $10.62\text{kJ/kg}$ for $w_{\text{p,in}}$ in Equation (II).

$\begin{array}{c}{h}_2=191.81\text{kJ/kg}+10.62\text{kJ/kg}\\ =202.43\text{kJ/kg}\end{array}$

Substitute $3375.1\text{kJ/kg}$ for $h_3$, $80\text{\%}$ for ${\eta }_T$ and $2783.8\text{kJ/kg}$ for $h_{4s}$ in Equation (III).

$\begin{array}{c}{h}_4=3375.1\text{kJ/kg}-80\text{\%}\left(3375.1\text{kJ/kg}-2783.8\text{kJ/kg}\right)\\ =3375.1\text{kJ/kg}-\frac{80}{100}\left(1285.4\text{kJ/kg}\right)\\ =3375.1\text{kJ/kg}-473.04\text{kJ/kg}\\ =2902.06\text{kJ/kg}\end{array}$

Substitute $7.7642\text{kJ/kg}\cdot \text{K}$ for $s_{6s}$, $0.6492\text{kJ/kg}\cdot \text{K}$ for $s_{f,6s}$ and $7.4996\text{kJ/kg}\cdot \text{K}$ for $s_{fg,6s}$ in Equation (IV).

$\begin{array}{c}{x}_{6s}=\frac{7.7642\text{kJ/kg}\cdot \text{K}-0.6492\text{kJ/kg}\cdot \text{K}}{7.4996\text{kJ/kg}\cdot \text{K}}\\ =0.9487\end{array}$

Substitute $191.81\text{kJ/kg}$ for $h_{f,6s}$, $0.9487$ for $x_{6s}$ and $2392.1\text{kJ/kg}$ for $h_{fg,6s}$ in

Equation (V).

$\begin{array}{c}{h}_{6s}=191.81\text{kJ/kg}+0.9487\left(2392.1\text{kJ/kg}\right)\\ =191.81\text{kJ/kg}+2269.3853\text{kJ/kg}\\ =2461.1953\text{kJ/kg}\\ \simeq 2461.2\text{kJ/kg}\end{array}$

Substitute $3479.1\text{kJ/kg}$ for $h_5$, $80\text{\%}$ for ${\eta }_T$ and $2461.2\text{kJ/kg}$ for $h_{6s}$ in Equation (VI).

$\begin{array}{c}{h}_6=3479.1\text{kJ/kg}-80\text{\%}\left(3479.1\text{kJ/kg}-2461.2\text{kJ/kg}\right)\\ =3479.1\text{kJ/kg}-\frac{80}{100}\left(1017.9\text{kJ/kg}\right)\\ =3479.1\text{kJ/kg}-814.32\text{kJ/kg}\\ =2664.78\text{kJ/kg}\end{array}$

$\simeq 2664.8\text{kJ/kg}$

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

Obtain the vapor enthalpy $\left(h_g\right)$ corresponding to the pressure of $10\text{kPa}$ is $2583.9\text{kJ/kg}$.

$\begin{array}{l}{h}_6=h_g\\ 2664.8\text{kJ/kg}\text{>}2583.9\text{kJ/kg}\end{array}$

The calculated enthalpy at state 6 is greater than the vapor enthalpy at this state. Hence, the steam is at superheated state.

Substitute $3375.1\text{kJ/kg}$ for $h_3$, $202.43\text{kJ/kg}$ for $h_2$, $3479.1\text{kJ/kg}$ for $h_5$, and $2902.06\text{kJ/kg}$ for $h_4$ in Equation (VII).

$\begin{array}{c}{q}_{\text{in}}=\left(3375.1\text{kJ/kg}-202.43\text{kJ/kg}\right)+\left(3479.1\text{kJ/kg}-2902.06\text{kJ/kg}\right)\\ =3172.67\text{kJ/kg}+577.04\text{kJ/kg}\\ =3749.71\text{kJ/kg}\end{array}$

Consider the process 1 to 5 (combined heat addition and boiler).

Here,

$q_{R,15}=-q_{\text{in}}=-3749.71\text{kJ/kg}$.

$T_R=1600\text{K}$

Substitute 285 K for $T_0$, 1600 K for $T_R$, $6.5995\text{kJ}/\text{kg}\cdot \text{K}$ for $s_3$, $0.6492\text{kJ}/\text{kg}\cdot \text{K}$ for $s_1$, $7.7642\text{kJ}/\text{kg}\cdot \text{K}$ for $s_5$, $6.8464\text{kJ}/\text{kg}\cdot \text{K}$ for $s_4$, and $-3749.71\text{kJ}/\text{kg}$ for $q_{R,15}$ in Equation (VIII).

$\begin{array}{c}{x}_{\text{destroyed}\left(\text{1-5}\right)}=285\text{K}\left[\begin{array}{l}\left(6.5995\text{kJ}/\text{kg}\cdot \text{K}-0.6492\text{kJ}/\text{kg}\cdot \text{K}\right)\\ +\left(7.7642\text{kJ}/\text{kg}\cdot \text{K}-6.8464\text{kJ}/\text{kg}\cdot \text{K}\right)\\ +\frac{-3749.71\text{kJ}/\text{kg}}{1600\text{K}}\end{array}\right]\\ =285\text{K}\left(5.9503\text{kJ/kg}\cdot \text{K}+0.9178\text{kJ/kg}\cdot \text{K}-2.3436\text{kJ/kg}\cdot \text{K}\right)\\ =285\text{K}\left(4.5245\text{kJ/kg}\cdot \text{K}\right)\\ =1289.4914\text{kJ/kg}\end{array}$

$\simeq 1289.5\text{kJ/kg}$

Substitute $10.62\text{kJ/kg}$ for and $0.001010{\text{m}}^3\text{/kg}$ for $v_1$, $10000\text{kPa}$ for $P_2$, and $10\text{kPa}$ for $P_1$ in Equation (IX).

$\begin{array}{c}{x}_{\text{destroyed}\left(\text{1-2}\right)}=10.62\text{kJ/kg}-0.001010{\text{m}}^3\text{/kg}\left(10000\text{kPa-}10\text{kPa}\right)\\ =10.62\text{kJ/kg}-10.0899\text{kJ/kg}\\ =0.5301\text{kJ/kg}\\ \simeq 0.53\text{kJ/kg}\end{array}$

Substitute $0.53\text{kJ}/\text{kg}$ for $x_{\text{destroyed}\left(\text{12}\right)}$, and $1289.5\text{kJ}/\text{kg}$ for $x_{\text{destroyed}\left(1-2\right)}$ in Equation (X).

$\begin{array}{c}{x}_{\text{destroyed,heating}}=1289.5\text{kJ}/\text{kg}-0.53\text{kJ}/\text{kg}\\ =1289\text{kJ}/\text{kg}\end{array}$

Thus, the exergy destruction during the heating process is $1289\text{kJ/kg}$.

Consider the process 3-4 and 5-6 (H.P. turbine expansion and L.P. turbine expansion).

Here,

$q_{R,3-4,5-6}=0$

Substitute 285 K for $T_0$, 1600 K for $T_R$, $6.5995\text{kJ}/\text{kg}\cdot \text{K}$ for $s_3$, $8.3870\text{kJ}/\text{kg}\cdot \text{K}$ for $s_6$, $7.7642\text{kJ}/\text{kg}\cdot \text{K}$ for $s_5$, $6.8464\text{kJ}/\text{kg}\cdot \text{K}$ for $s_4$, and $0$ for $q_{R,3-4,5-6}$ in Equation (XI).

$\begin{array}{c}{x}_{\text{destroyed,expansion}}=285\text{K}\left[\begin{array}{l}\left(6.8464\text{kJ/kg}\cdot \text{K}-6.5995\text{kJ/kg}\cdot \text{K}\right)\\ +\left(8.3870\text{kJ/kg}\cdot \text{K}-7.7642\text{kJ/kg}\cdot \text{K}\right)\\ +0\end{array}\right]\\ =285\text{K}\left(0.2469\text{kJ/kg}\cdot \text{K}+0.6228\text{kJ/kg}\cdot \text{K}\right)\\ =247.86\text{kJ/kg}\\ \simeq 247.9\text{kJ}/\text{kg}\end{array}$

Thus, the exergy destruction during the expansion process is $247.9\text{kJ}/\text{kg}$.

Substitute $0.1806\text{kJ}/\text{kg}\cdot \text{K}$ for $s_0$, $50.51\text{kJ}/\text{kg}$ for $h_0$, 285 K for $T_0$, $6.5995\text{kJ}/\text{kg}\cdot \text{K}$ for $s_3$, and $3375.1\text{kJ}/\text{kg}$ for $h_3$ in Equation (XIII).

$\begin{array}{c}{\psi }_3=\left(3375.1\text{kJ/kg}-50.51\text{kJ/kg}\right)-\left(285\text{K}\right)\left(6.5995\text{kJ/kg}\cdot \text{K}-0.1806\text{kJ/kg}\cdot \text{K}\right)\\ =3324.59\text{kJ/kg}-1829.3865\text{kJ/kg}\\ =1495.2035\text{kJ/kg}\\ \simeq 1495\text{kJ}/\text{kg}\end{array}$

Thus, the exergy of the steam at the boiler exit is $1495\text{kJ}/\text{kg}$.