Chapter 7.13, Problem 197RP

To determine

The power output produced from the turbine, and the overall isentropic efficiency of turbine.

Answer to Problem 197RP

The power output produced from the turbine is $810.1\text{kJ}/\text{kg}$.

The overall isentropic efficiency of turbine is $95.80\%$.

Explanation of Solution

Write the formula to calculate the specific entropy of steam from tables $\left(s\right)$.

$s=s_f+x\left(s_{fg}\right)$ (I)

Here, specific entropy of saturated liquid is $s_f$, and the specific entropy of saturated liquid vapor mixture is $s_{fg}$.

Write the formula to calculate the specific enthalpy of steam from tables $\left(h\right)$.

$h=h_f+x\left(h_{fg}\right)$ (II)

Here, specific entropy of saturated liquid is $s_f$, and the specific entropy of saturated liquid vapor mixture is $s_{fg}$.

Draw the $T-s$ diagram for the turbine process as in Figure (1).

Write the formula for isentropic efficiency of the turbine $\left({\eta }_{\text{T,1}}\right)$ between the inlet and bleed point.

${\eta }_{\text{T,1}}=\frac{h_1-h_2}{h_1-h_{2s}}$ (III)

Here, specific enthalpy at the turbine inlet is $h_1$, specific enthalpy at the bleed point is $h_2$, and the specific enthalpy at the isentropic exit of bleed point is $h_{2s}$.

Write the formula for isentropic efficiency of the turbine $\left({\eta }_{\text{T,2}}\right)$ between the bleed point and turbine exit.

${\eta }_{\text{T,2}}=\frac{h_2-h_3}{h_2-h_{3s}}$ (IV)

Here, specific enthalpy at the actual turbine exit is $h_3$, and the specific enthalpy at the isentropic exit of turbine is $h_{3s}$.

Write the expression for the energy balance Equation for a closed system.

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

Here, net energy rate transfer into the control volume is ${\dot{E}}_{\text{in}}$, net energy rate transfer exit from the control volume is ${\dot{E}}_{\text{out}}$ and rate of change in energy of the system is $\Delta {\dot{E}}_{\text{system}}$.

Write the general expression to calculate the isentropic efficiency of turbine $\left({\eta }_{\text{iso}}\right)$.

${\eta }_{\text{iso}}=\frac{w_{\text{actual}}}{w_{\text{iso}}}$ (VI)

Here, actual work output is $w_{\text{actual}}$ and the isentropic work output is $w_{\text{iso}}$.

Conclusion:

The rate of change in energy of the system is zero at steady state.

Substitute $0$ for $\Delta {\dot{E}}_{\text{system}}$ in Equation (V).

$\begin{array}{l}{\dot{E}}_{\text{in}}-{\dot{E}}_{\text{out}}=0\\ {\dot{E}}_{\text{in}}={\dot{E}}_{\text{out}}\end{array}$ (VII)

Substitute ${\dot{m}}_1{h}_1$ for ${\dot{E}}_{\text{in}}$, ${\dot{m}}_2{h}_2+{\dot{m}}_{3}h+{\dot{W}}_{\text{actual}}$ for ${\dot{E}}_{\text{out}}$ in Equation (VII).

${\dot{m}}_1{h}_1={\dot{m}}_2{h}_2+{\dot{m}}_{3}h+{\dot{W}}_{\text{actual}}$

$\begin{array}{l}{\dot{W}}_{\text{actual}}={\dot{m}}_1{h}_1-{\dot{m}}_2{h}_2-{\dot{m}}_3{h}_3\\ w_{\text{actual}}=h_1-0.06h_2-0.94h_3\end{array}$

$w_{\text{actual}}=\left(h_1-h_2\right)+0.94\left(h_2-h_3\right)$ (VIII)

Re-write the Equation (VIII) for the isentropic work output $\left(w_{\text{iso}}\right)$.

$w_{\text{actual}}=\left(h_1-h_{2s}\right)+0.94\left(h_2-h_{3s}\right)$ (IX)

Here, mass flow rate of steam at inlet is ${\dot{m}}_1$, mass flow rate of steam at bleed point is ${\dot{m}}_2$, mass flow rate of steam at the exit of turbine is ${\dot{m}}_3$, actual work output from turbine is ${\dot{W}}_{\text{actual}}$, and the actual work output per unit mass is $w_{\text{actual}}$.

From the Table A-6 “Superheated water”, obtain the specific enthalpy $\left(h_1\right)$ and specific entropy $\left(s_1\right)$ of superheated steam at pressure $\left(P_1\right)$ of $4\text{MPa}$ and temperature $\left(T_1\right)$ of $350°\text{C}$ as $\text{3093}\text{.3}\text{kJ}/\text{kg}$ and $\text{6}\text{.5843}\text{kJ}/\text{kg}\cdot \text{K}$.

From the Table A-5 “Saturated water - Pressure”, obtain the following properties of water at pressure of $800\text{ kPa}$.

$\begin{array}{l}{s}_f=2.0457\text{kJ}/\text{kg}\cdot \text{K}\\ s_{fg}=4.6160\text{kJ}/\text{kg}\cdot \text{K}\\ h_f=720.87\text{kJ}/\text{kg}\\ h_{fg}=2047.5\text{kJ}/\text{kg}\end{array}$

Substitute $2.0457\text{kJ}/\text{kg}\cdot \text{K}$ for $s_f$, $4.6160\text{kJ}/\text{kg}\cdot \text{K}$ for $s_{fg}$, and $\text{6}\text{.5843}\text{kJ}/\text{kg}\cdot \text{K}$ for $s_{2s}$ in Equation (I).

$\begin{array}{l}\text{6}\text{.5843}\text{kJ}/\text{kg}\cdot \text{K}=2.0457\text{kJ}/\text{kg}\cdot \text{K}+x_{2s}\left(4.6160\text{kJ}/\text{kg}\cdot \text{K}\right)\\ x_{2s}=0.9832\end{array}$

Substitute $720.87\text{kJ}/\text{kg}$ for $h_f$, $2047.5\text{kJ}/\text{kg}$ for $h_{fg}$, and 0.9832 for $x_{2s}$ in

Equation (II).

$\begin{array}{c}{h}_{2s}=720.87\text{kJ}/\text{kg}+0.9832\left(2047.5\text{kJ}/\text{kg}\right)\\ =2734.0\text{kJ}/\text{kg}\end{array}$

Substitute 0.97 for ${\eta }_{\text{T,1}}$, $\text{3093}\text{.3}\text{kJ}/\text{kg}$ for $h_1$, and $2734.0\text{kJ}/\text{kg}$ for $h_{2s}$ in

Equation (III).

$\begin{array}{l}0.97=\frac{\text{3093}\text{.3}\text{kJ}/\text{kg}-h_2}{\text{3093}\text{.3}\text{kJ}/\text{kg}-2734.0\text{kJ}/\text{kg}}\\ h_2=\text{2744}\text{.8}\text{kJ}/\text{kg}\end{array}$

Substitute $720.87\text{kJ}/\text{kg}$ for $h_f$, $2047.5\text{kJ}/\text{kg}$ for $h_{fg}$, and $\text{2744}\text{.8}\text{kJ}/\text{kg}$ for $h_2$ in Equation (II).

$\begin{array}{l}\text{2744}\text{.8}\text{kJ}/\text{kg}=720.87\text{kJ}/\text{kg}+x_2\left(2047.5\text{kJ}/\text{kg}\right)\\ x_2=0.9885\end{array}$

Substitute $2.0457\text{kJ}/\text{kg}\cdot \text{K}$ for $s_f$, $4.6160\text{kJ}/\text{kg}\cdot \text{K}$ for $s_{fg}$, and 0.9885 for $x_2$ in Equation (I).

$\begin{array}{l}{s}_2=2.0457\text{kJ}/\text{kg}\cdot \text{K}+0.9885\left(4.6160\text{kJ}/\text{kg}\cdot \text{K}\right)\\ s_2=6.6086\text{kJ}/\text{kg}\cdot \text{K}\end{array}$

From the Table A-5 “Saturated water - Pressure”, obtain the following properties of water at pressure of $30\text{ kPa}$.

$\begin{array}{l}{s}_f=0.9441\text{kJ}/\text{kg}\cdot \text{K}\\ s_{fg}=6.8234\text{kJ}/\text{kg}\cdot \text{K}\\ h_f=289.27\text{kJ}/\text{kg}\\ h_{fg}=2335.3\text{kJ}/\text{kg}\end{array}$

Substitute $0.9441\text{kJ}/\text{kg}\cdot \text{K}$ for $s_f$, $6.8234\text{kJ}/\text{kg}\cdot \text{K}$ for $s_{fg}$, and $6.6086\text{kJ}/\text{kg}\cdot \text{K}$ for $s_{3s}$ in Equation (I).

$\begin{array}{l}6.6086\text{kJ}/\text{kg}\cdot \text{K}=0.9441\text{kJ}/\text{kg}\cdot \text{K}+x_{3s}\left(6.8234\text{kJ}/\text{kg}\cdot \text{K}\right)\\ x_{3s}=0.8302\end{array}$

Substitute $289.27\text{kJ}/\text{kg}$ for $h_f$, $2335.3\text{kJ}/\text{kg}$ for $h_{fg}$, and $0.8302$ for $x_{3s}$ in Equation (II).

$\begin{array}{c}{h}_{3s}=289.27\text{kJ}/\text{kg}+0.8302\left(2335.3\text{kJ}/\text{kg}\right)\\ =2227.9\text{kJ}/\text{kg}\end{array}$

Substitute $2227.9\text{kJ}/\text{kg}$ for $h_{3s}$, $0.95$ for ${\eta }_{\text{T,2}}$, and $\text{2744}\text{.8}\text{kJ}/\text{kg}$ for $h_2$ in Equation (IV).

$\begin{array}{l}0.95=\frac{\text{2744}\text{.8}\text{kJ}/\text{kg}-h_3}{\text{2744}\text{.8}\text{kJ}/\text{kg}-2227.9\text{kJ}/\text{kg}}\\ h_3=\text{2253}\text{.7}\text{kJ}/\text{kg}\end{array}$

Substitute $\text{3093}\text{.3}\text{kJ}/\text{kg}$ for $h_1$, $\text{2744}\text{.8}\text{kJ}/\text{kg}$ for $h_2$, and $\text{2253}\text{.7}\text{kJ}/\text{kg}$ for $h_3$ in Equation (IX).

$\begin{array}{c}{w}_{\text{actual}}=\left(\text{3093}\text{.3}\text{kJ}/\text{kg}-\text{2744}\text{.8}\text{kJ}/\text{kg}\right)+0.94\left(\text{2744}\text{.8}\text{kJ}/\text{kg}-\text{2253}\text{.7}\text{kJ}/\text{kg}\right)\\ =810.1\text{kJ}/\text{kg}\end{array}$

Thus, the power output produced from the turbine is $810.1\text{kJ}/\text{kg}$,

Substitute $\text{3093}\text{.3}\text{kJ}/\text{kg}$ for $h_1$, $\text{2744}\text{.8}\text{kJ}/\text{kg}$ for $h_2$, $2734.0\text{kJ}/\text{kg}$ for $h_{2s}$, and $\text{2227}\text{.9}\text{kJ}/\text{kg}$ for $h_{3s}$ in Equation (VIII).

$\begin{array}{c}{w}_{\text{iso}}=\left(\text{3093}\text{.3}\text{kJ}/\text{kg}-2734.0\text{kJ}/\text{kg}\right)+0.94\left(\text{2744}\text{.8}\text{kJ}/\text{kg}-\text{2227}\text{.9}\text{kJ}/\text{kg}\right)\\ =845.2\text{kJ}/\text{kg}\end{array}$

Substitute $810.1\text{kJ}/\text{kg}$ for $w_{\text{actual}}$, and $845.2\text{kJ}/\text{kg}$ for $w_{\text{iso}}$ in Equation (VI).

$\begin{array}{c}{\eta }_{\text{iso}}=\frac{810.1\text{kJ}/\text{kg}}{845.2\text{kJ}/\text{kg}}\\ =0.958\times 100\%\\ =95.80\%\end{array}$

Thus, the overall isentropic efficiency of turbine is $95.80\%$.