Chapter 7.13, Problem 137P

Steam enters an adiabatic turbine steadily at 7 MPa, 500°C, and 45 m/s and leaves at 100 kPa and 75 m/s. If the power output of the turbine is 5 MW and the isentropic efficiency is 77 percent, determine (a) the mass flow rate of steam through the turbine, (b) the temperature at the turbine exit, and (c) the rate of entropy generation during this process.

FIGURE P7–137

To determine

The mass flow rate of the steam.

Answer to Problem 137P

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

Explanation of Solution

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 $\Delta {\dot{E}}_{system}$.

Write the expression to calculate the power output for the isentropic process.

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

Here, the power output for the isentropic process is ${\dot{W}}_s$, the power output for the actual process is ${\dot{W}}_a$ and turbine efficiency is ${\eta }_T$.

Conclusion:

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

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

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

Here, mass flow rate is $\dot{m}$, enthalpy at initial state is $h_1$, enthalpy at final state in isentropic process is $h_{2s}$, initial velocity is $V_1$ and final velocity is $V_2$.

From the Table A-6, “superheated water table”, select the initial enthalpy $\left(h_1\right)$ and initial entropy $\left(s_1\right)$ at pressure of $\text{7}\text{MPa}$ and temperature of 500 C as $3,411.4\text{kJ}/\text{kg}$ and $6.800\text{kJ}/\text{kg}\cdot \text{K}$ respectively.

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

From the Table A-6, “superheated water table”, select the enthalpy at final state in isentropic process $\left(h_{2s}\right)$ at pressure of $\text{100}\text{kPa}$ and entropy of $6.800\text{kJ}/\text{kg}\cdot \text{K}$ as $2466.6\text{kJ}/\text{kg}$.

Substitute 5,000 kW for ${\dot{W}}_a$ and $70\%$ for ${\eta }_T$ in Equation (I).

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

Substitute $3411.4\text{kJ}/\text{kg}$ for $h_1$, $45\text{m}/\text{s}$ for $V_1$, $2466.6\text{kJ}/\text{kg}$ for $h_{2s}$, $75\text{m}/\text{s}$ for $V_2$ and $6494\text{kJ}/\text{s}$ for ${\dot{W}}_s$ in Equation (III).

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

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

To determine

The exit temperature of the steam.

Answer to Problem 137P

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

Explanation of Solution

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

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

Here, rate of actual work is ${\dot{W}}_a$.

Conclusion:

Substitute $6.886\text{kg}/\text{s}$ for $\dot{m}$, $3411.4\text{kJ}/\text{kg}$ for $h_1$, $45\text{m}/\text{s}$ for $V_1$, $75\text{m}/\text{s}$ for $V_2$ and 5000 kW for ${\dot{W}}_a$ in Equation (IV).

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

From 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 $7.3817\text{kJ}/\text{kg}\cdot \text{K}$.

From 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.

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.

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

Substitute $2675$ for $x_1$, $2683.5$ for $x_2$, $2776.6$ for $x_3$, $100$ for $y_1$, and $150$ for $y_3$ 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 $104.18°\text{C}$.

To determine

The entropy generation in the turbine.

Answer to Problem 137P

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

Explanation of Solution

Write the expression to calculate the entropy generation in the turbine.

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

Here, entropy generation in the turbine is $S_{gen}$.

Conclusion:

Substitute $6.886\text{kg}/\text{s}$ for $\dot{m}$, $7.3817\text{kJ}/\text{kg}\cdot \text{K}$ for $s_2$ and $6.800\text{kJ}/\text{kg}\cdot \text{K}$ for $s_1$ in Equation (VI).

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

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