Chapter 8, Problem 113P

(a)

To determine

The mass flow rate of the steam flowing to the turbine.

Explanation of Solution

Given:

The initial pressure of the steam $\left(P_1\right)$ is 5 MPa.

The initial temperature of the steam $\left(T_1\right)$ is $650°\text{C}$.

The inlet velocity of the steam $\left(V_1\right)$ is $80\text{m/s}$.

The final pressure of the steam $\left(P_2\right)$ is 50 kPa.

The final temperature of the steam $\left(T_2\right)$ is $150°\text{C}$.

The exit velocity of the steam $\left(V_2\right)$ is $140\text{m/s}$.

The power output of the turbine $\left({\dot{W}}_{out}\right)$ is $8\text{MW}$.

Calculation:

Refer Table A-6, “Superheated water”, for the pressure of 5 MPa and temperature of $650°\text{C}$ to obtain the following properties using interpolation method.

Temperature, $°\text{C}$	Enthalpy $\text{kJ}/\text{kg}$	Entropy $\text{kJ}/\text{kg}\cdot \text{K}$
600	3666.9	2511.5
650	?  3783.60	?7.387
700	3900.3	2506.2

Write the formula of interpolation method of two variables.

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

Substitute $x_1=600°\text{C}$, $x_2=650°\text{C}$, $x_3=700°\text{C}$, $y_1=\text{3666}\text{.9 }\text{kJ}/\text{kg}$, and $y_3=3900.3\text{kJ}/\text{kg}$ in Equation (I) to obtain the enthalpy at state 1$\left(h_1\right)$.

  $\begin{array}{c}{y}_2=\frac{\left(650-600\right)\left(3900.3-3666.9\right)}{\left(700-600\right)}+3666.9\\ =3783.60\text{kJ}/\text{kg}\cdot \text{K}\end{array}$

Substitute $x_1=600°\text{C}$, $x_2=650°\text{C}$, $x_3=700°\text{C}$, $y_1=\text{2511}\text{.5 }\text{kJ}/\text{kg}\cdot \text{K}$, and $y_3\text{=2506}\text{.2 }\text{kJ}/\text{kg}\cdot \text{K}$ in Equation (I) to obtain the entropy at state 1$\left(s_1\right)$.

  $\begin{array}{c}{y}_2=\frac{\left(650-600\right)\left(3900.3-3666.9\right)}{\left(700-600\right)}+3666.9\\ =3783.60\text{kJ}/\text{kg}\cdot \text{K}\end{array}$

Refer Table A-6, “Superheated water”, select the actual enthalpy $\left(h_2\right)$ at the pressure of $50\text{kPa}$ and temperature of $150°\text{C}$ as $2780.2\text{kJ}/\text{kg}$.

Write the energy balance equation for closed system.

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

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

At steady state, rate of change in internal energy of the system is zero.

Rewrite Equation (I) for energy balance for the closed system.

  $\begin{array}{l}{\dot{E}}_{in}-{\dot{E}}_{out}=0\\ {\dot{E}}_{in}={\dot{E}}_{out}\\ \dot{m}\left(h_1+\frac{V_1^2}{2}\right)={\dot{W}}_{a,out}+\dot{m}\left(h_2+\frac{V_2^2}{2}\right)\\ {\dot{W}}_{a,out}=-\dot{m}\left(h_2-h_1+\frac{V_2^2-V_1^2}{2}\right)\end{array}$        (II)

  $\begin{array}{l}\left(8\text{MW}\right)\left(\frac{{10}^6\text{kJ}/\text{s}}{1\text{MW}}\right)=-\dot{m}\left(\begin{array}{l}2,780.2\text{kJ}/\text{kg}-3,783.2\text{kJ}/\text{kg}\\ +\frac{{\left(\text{140}\text{m}/\text{s}\right)}^2-{\left(\text{80}\text{m}/\text{s}\right)}^2}{2}\left(\frac{1\text{kJ}/\text{kg}}{1,000{\text{m}}^{\text{2}}/{\text{s}}^{\text{2}}}\right)\end{array}\right)\\ 8,000=\dot{m}\left(996.4\right)\\ \dot{m}=8.029\text{kg}/\text{s}\end{array}$

Thus, the mass flow rate of the steam flowing to the turbine is $\underset{\_}{8.03\text{kg}/\text{s}}$.

(b)

To determine

The isentropic efficiency of the turbine

Explanation of Solution

Refer to Table A-5, “Saturated water-Pressure table”, obtain the following properties at the pressure of $50\text{kPa}$.

  $\begin{array}{l}{h}_f=340.54\text{kJ}/\text{kg}\\ h_{fg}=2,304.7\text{kJ}/\text{kg}\\ s_f=1.0912\text{kJ}/\text{kg}\cdot \text{K}\\ {\text{s}}_{fg}=6.5019\text{kJ}/\text{kg}\cdot \text{K}\end{array}$

Calculate the vapor quality is $\left(x_{2s}\right)$ using the final entropy equation.

  $\begin{array}{c}{x}_{2s}=\frac{s_{2s}-s_f}{s_{fg}}\\ x_{2s}=\frac{7.3901\text{kJ}/\text{kg}\cdot \text{K}-1.0912\text{kJ}/\text{kg}\cdot \text{K}}{6.5019\text{kJ}/\text{kg}\cdot \text{K}}\\ =0.9688\end{array}$

Calculate the final enthalpy for the isentropic process.

  $\begin{array}{c}{h}_{2s}=h_f+x_{2s}{h}_{fg}\\ h_{2s}=340.54\text{kJ}/\text{kg}+\left(0.9688\right)\left(2,304.7\text{kJ}/\text{kg}\right)\\ =2,573.3\text{kJ}/\text{kg}\end{array}$

For the reversible adiabatic process, the entropy remains constant.

$s_1=s_{2s}$

Write the expression for the power output of the isentropic turbine $\left({\dot{W}}_{s,out}\right)$.

  $\begin{array}{c}{\dot{W}}_{s,out}=-\dot{m}\left(h_{2s}-h_1+\frac{V_2^2-V_1^2}{2}\right)\\ {\dot{W}}_{s,out}=-\left(8.029\text{kg}/\text{s}\right)\left(\begin{array}{l}2,573.3\text{kJ}/\text{kg}-3,783.2\text{kJ}/\text{kg}\\ +\frac{{\left(\text{140}\text{m}/\text{s}\right)}^2-{\left(\text{80}\text{m}/\text{s}\right)}^2}{2}\left(\frac{1\text{kJ}/\text{kg}}{1,000{\text{m}}^{\text{2}}/{\text{s}}^{\text{2}}}\right)\end{array}\right)\\ =9,661\text{kJ}/\text{s}\left(\frac{1\text{kW}}{1\text{kJ}/\text{s}}\right)\\ =9,661\text{kW}\end{array}$

Calculate the isentropic efficiency of the turbine $\left(\eta \right)$.

  $\begin{array}{c}\eta =\frac{{\dot{W}}_{a,out}}{{\dot{W}}_{s,out}}\\ \eta =\frac{8,000\text{kW}}{9,661\text{kW}}\\ =0.828\\ =82.8\%\end{array}$

Thus, the isentropic efficiency of the turbine is $\underset{\_}{82.8\%}$