Chapter 8, Problem 133P

a)

To determine

The mass flow rate of the steam.

Explanation of Solution

Given:

The initial pressure of the steam $\left(P_1\right)$ is 7 MPa.

The initial temperature of the steam $\left(T_1\right)$ is $500°\text{C}$.

The initial velocity $\left(V_1\right)$ is $45\text{m/s}$.

The final pressure of the steam $\left(P_2\right)$ is 100 kPa.

The final velocity $\left(V_2\right)$ is $75\text{m/s}$.

The power output of the turbine $\left({\dot{W}}_a\right)$ is 5 MW.

The isentropic efficiency $\left(\eta \right)$ is 77%.

Conclusion:

Refer the Table A-6, “superheated water table”, obtain the following properties at a pressure of $\text{7}\text{MPa}$ and temperature of 500 C.

The initial enthalpy $\left(h_1\right)$ is $3,411.4\text{kJ}/\text{kg}$.

The initial entropy $\left(s_1\right)$ is $6.800\text{kJ}/\text{kg}\cdot \text{K}$.

The entropy remains constant since the process is isentropic $\left(s_{2s}=s_1\right)$.

Refer the Table A-6, “superheated water table”, obtain the following properties at a pressure of $\text{100}\text{kPa}$ and entropy of $6.800\text{kJ}/\text{kg}\cdot \text{K}$.

The enthalpy at final state in isentropic process $\left(h_{2s}\right)$ is $2466.6\text{kJ}/\text{kg}$.

Calculate the power output for the isentropic process.

  ${\dot{W}}_s=\frac{{\dot{W}}_a}{{\eta }_T}$

  $\begin{array}{c}{\dot{W}}_s=\frac{5,000\text{kW}}{70\%}\\ =\frac{5,000\text{kW}}{70\left(\frac{1}{100}\right)}\\ =6494\text{kW}\\ =6494\text{kJ}/\text{s}\end{array}$

Write the expression for 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 $\Delta {\dot{E}}_{system}$.

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

Substitute $\Delta {\dot{E}}_{system}=0$ in Equation (I).

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

  $\begin{array}{l}\dot{m}\left(\begin{array}{l}3411.4\text{kJ}/\text{kg}\\ +\frac{{\left(45\text{m}/\text{s}\right)}^2}{2}\end{array}\right)=\left\{\dot{m}\left(\begin{array}{l}2466.6\text{kJ}/\text{kg}\\ +\frac{{\left(75\text{m}/\text{s}\right)}^2}{2}\end{array}\right)+6494\text{kJ}/\text{s}\right\}\\ \dot{m}\left(\begin{array}{l}3411.4\text{kJ}/\text{kg}\\ +\frac{{\left(45\text{m}/\text{s}\right)}^2}{2}\left(\frac{1\text{kJ}/\text{kg}}{1000{\text{m}}^2/{\text{s}}^2}\right)\end{array}\right)=\left\{\begin{array}{l}\dot{m}\left(\begin{array}{l}2466.6\text{kJ}/\text{kg}\\ +\frac{{\left(75\text{m}/\text{s}\right)}^2}{2}\left(\frac{1\text{kJ}/\text{kg}}{1000{\text{m}}^2/{\text{s}}^2}\right)\end{array}\right)\\ +6494\text{kJ}/\text{s}\end{array}\right\}\\ \dot{m}=6.886\text{kg}/\text{s}\end{array}$

Thus, the mass flow rate of the steam is $6.886\text{kg}/\text{s}$.

b)

To determine

The exit temperature of the steam.

Explanation of Solution

Write the expression to calculate the mass flow rate of the steam.

  $\dot{m}\left(h_1+\frac{V_1^2}{2}\right)=\dot{m}\left(h_{2s}+\frac{V_2^2}{2}\right)+{\dot{W}}_a$

  $\begin{array}{l}6.886\text{kg}/\text{s}\left(3411.4\text{kJ}/\text{kg}+\frac{{\left(45\text{m}/\text{s}\right)}^2}{2}\right)=\left\{\begin{array}{l}6.886\text{kg}/\text{s}\times \left(h_2+\frac{{\left(75\text{m}/\text{s}\right)}^2}{2}\right)\\ +5000\text{kJ}/\text{s}\end{array}\right\}\\ 6.886\text{kg}/\text{s}\left(\begin{array}{l}3411.4\text{kJ}/\text{kg}\\ +\frac{{\left(45\text{m}/\text{s}\right)}^2}{2}\left(\frac{1\text{kJ}/\text{kg}}{1000{\text{m}}^2/{\text{s}}^2}\right)\end{array}\right)=\left\{\begin{array}{l}6.886\text{kg}/\text{s}\\ \times \left(h_2+\frac{{\left(75\text{m}/\text{s}\right)}^2}{2}\left(\frac{1\text{kJ}/\text{kg}}{1000{\text{m}}^2/{\text{s}}^2}\right)\right)\\ +5000\text{kJ}/\text{s}\end{array}\right\}\\ h_2=2683.5\text{kJ}/\text{kg}\end{array}$

Refer the Table A-6, “superheated water table”, obtain final entropy at pressure of $\text{100}\text{kPa}$ and enthalpy of $2683.5\text{kJ}/\text{kg}$ as $7.3817\text{kJ}/\text{kg}\cdot \text{K}$.

Refer the Table A-6, “Superheated water”, obtain the value of exit temperature $\left(T_2\right)$ at final enthalpy $\left(h_2\right)$ of $2683.5\text{kJ}/\text{kg}$ by using interpolation method.

Show exit temperature and enthalpy values from the Table A-6.

Temperature $\left(T_2\right)$,in $°\text{C}$	Enthalpy $\left(h_2\right)$, in $\text{Btu}/\text{lbm}\cdot \text{R}$
2675	100
2683.5	?
2776.6	150

Write the formula of interpolation method of two variables.

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

Here, the variables denoted by x and y are exit temperature and enthalpy.

Substitute $x_1=2675$, $x_2=2683.5$, $x_3=2776.6$, $y_1=100$, and $y_3=150$ in Equation (V).

  $\begin{array}{c}{y}_2=\frac{\left(2683.5-2675\right)\left(150-100\right)}{\left(2776.6-2675\right)}+100\\ =104.18\end{array}$

The value of exit temperature $\left(T_2\right)$ at final enthalpy $\left(h_2\right)$ of $2683.5\text{kJ}/\text{kg}$ is $104.18°\text{C}$.

Thus, the exit temperature of the steam is $104.18°\text{C}$.

c)

To determine

The entropy generation in the turbine.

Explanation of Solution

Calculate the entropy generation in the turbine $\left(S_{gen}\right)$.

  $S_{gen}=\dot{m}\left(s_2-s_1\right)$

  $\begin{array}{c}{S}_{gen}=6.886\text{kg}/\text{s}\left(7.3817\text{kJ}/\text{kg}\cdot \text{K}-6.800\text{kJ}/\text{kg}\cdot \text{K}\right)\\ =4.01\text{kJ}/\text{K}\cdot \text{s}\left(\frac{1\text{kW}}{1\text{kJ}/\text{s}}\right)\\ =4.01\text{kW}/\text{K}\end{array}$

Thus, the entropy generation in the turbine is $4.01\text{kW}/\text{K}.$