Chapter 5.5, Problem 178RP

Steam enters a turbine steadily at 7 MPa and 600°C with a velocity of 60 m/s and leaves at 25 kPa with a quality of 95 percent. A heat loss of 20 kJ/kg occurs during the process. The inlet area of the turbine is 150 cm², and the exit area is 1400 cm². Determine (a) the mass flow rate of the steam, (b) the exit velocity, and (c) the power output.

To determine

The mass flow rate of the steam.

Answer to Problem 178RP

The mass flow rate of the steam is $16.7\text{kg/s}$.

Explanation of Solution

The turbine operates steadily. Hence, the inlet and exit mass flow rates are equal.

${\dot{m}}_1={\dot{m}}_2=\dot{m}$

Write the formula for inlet mass flow rate.

$\dot{m}=\frac{A_1{V}_1}{v_1}$ (I)

Here, the cross-sectional area is $A$, the velocity is $V$ and the specific volume is $v$; the subscript 1 indicates the inlet condition.

At inlet:

The steam is at the state of superheated condition.

Refer Table A-6, “Superheated water”.

Obtain the inlet enthalpy $\left(h_1\right)$ and specific volume $\left(v_1\right)$ corresponding to the pressure of $7\text{MPa}$ and the temperature of $600°\text{C}$.

$\begin{array}{l}{h}_1=3650.6\text{kJ/kg}\\ v_1=0.055665{\text{m}}^3\text{/kg}\end{array}$

Conclusion:

Substitute $150{\text{cm}}^2$ for $A_1$, $60\text{m/s}$ for $V_1$ ,and $0.055665{\text{m}}^3\text{/kg}$ for $v_1$ in Equation (I).

$\begin{array}{c}\dot{m}=\frac{\left(150{\text{cm}}^2\right)\left(60\text{m/s}\right)}{0.055665{\text{m}}^3\text{/kg}}\\ =\frac{\left(150{\text{cm}}^2\times \frac{1\text{}{\text{m}}^2}{{10}^4{\text{cm}}^2}\right)\left(60\text{m/s}\right)}{0.055665{\text{m}}^3\text{/kg}}\\ =16.1681\text{kg/s}\\ \simeq 16.17\text{kg/s}\end{array}$

Thus, the mass flow rate of the steam is $16.17\text{kg/s}$.

To determine

The exit velocity of the steam.

Explanation of Solution

The turbine operates steadily. Hence, the inlet and exit mass flow rates are equal.

${\dot{m}}_1={\dot{m}}_2=\dot{m}$

Write the formula for exit mass flow rate.

$\dot{m}=\frac{A_2{V}_2}{v_2}$ (II)

Here, the cross-sectional area is $A$, the velocity is $V$ and the specific volume is $v$; the subscript 2 indicates the exit condition.

Rearrange the Equation (II) to obtain exit velocity $\left(V_2\right)$.

$V_2=\frac{\dot{m}{v}_2}{A_2}$ (III)

At exit:

The steam is with the quality of $95\%$.

Write the formula for exit enthalpy $\left(h_2\right)$ of the steam with the consideration of its quality.

$h_2=h_f+x{h}_{fg}$ (IV)

Write the formula for exit specific volume $\left(v_2\right)$ of the steam with the consideration of its quality.

$v_2=v_f+x\left(v_g-v_f\right)$ (V)

Here, the enthalpy is $h$, the quality of steam is $x$, and subscript $f$ indicates that fluid state, $g$ indicates the gaseous state, $fg$ indicates the mixed state (vaporization).

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

Obtain the following corresponding to the pressure of $25\text{kPa}$.

$\begin{array}{l}{v}_f=0.001020{\text{m}}^3\text{/kg}\\ v_g=6.2034{\text{m}}^3\text{/kg}\\ h_f=271.96\text{kJ/kg}\\ h_{fg}=2345.5\text{kJ/kg}\end{array}$

Substitute $271.96\text{kJ/kg}$ for $h_f$, $95\%$ for $x$ and $2345.5\text{kJ/kg}$ for $h_{fg}$ in Equation (IV).

$\begin{array}{c}{h}_2=271.96\text{kJ/kg}+95\%\left(2345.5\text{kJ/kg}\right)\\ =271.96\text{kJ/kg}+\frac{95}{100}\left(2345.5\text{kJ/kg}\right)\\ =271.96\text{kJ/kg}+2228.225\text{kJ/kg}\\ =2500.185\text{kJ/kg}\end{array}$

Substitute $0.001020{\text{m}}^3\text{/kg}$ for $v_f$, $95\%$ for $x$ and $6.2034{\text{m}}^3\text{/kg}$ for $v_g$ in

Equation (V).

$\begin{array}{c}{v}_2=0.001020{\text{m}}^3\text{/kg}+95\%\left(6.2034{\text{m}}^3\text{/kg}-0.001020{\text{m}}^3\text{/kg}\right)\\ =0.001020{\text{m}}^3\text{/kg}+\frac{95}{100}\left(6.20238{\text{m}}^3\text{/kg}\right)\\ =0.001020{\text{m}}^3\text{/kg}+5.892261{\text{m}}^3\text{/kg}\\ =5.893281{\text{m}}^3\text{/kg}\end{array}$

Thus, the exit enthalpy and exit specific volume of the steam is $2500.185\text{kJ/kg}$ and $5.893281{\text{m}}^3\text{/kg}$.

Conclusion:

Substitute $16.17\text{kg/s}$ for $\dot{m}$, $5.893281{\text{m}}^3\text{/kg}$ for $v_2$ , and $1400{\text{cm}}^2$ for $A_2$ in

Equation (III).

$\begin{array}{c}{V}_2=\frac{\left(16.17\text{kg/s}\right)\left(5.893281{\text{m}}^3\text{/kg}\right)}{1400{\text{cm}}^2}\\ =\frac{\left(16.17\text{kg/s}\right)\left(5.893281{\text{m}}^3\text{/kg}\right)}{1400{\text{cm}}^2\times \frac{1\text{}{\text{m}}^2}{{10}^4{\text{cm}}^2}}\\ =680.6739\text{m/s}\end{array}$

Thus, the exit velocity of the steam is $680.6739\text{m/s}$.

To determine

The power output of the turbine.

Answer to Problem 178RP

The power output of the turbine is $14562\text{kW}$.

Explanation of Solution

Consider the steam flows at steady state. Hence, the inlet and exit mass flow rates are equal.

${\dot{m}}_1={\dot{m}}_2=\dot{m}$

Write the energy rate balance equation for one inlet and one outlet system.

$\left[{\dot{Q}}_{\text{1}}+{\dot{W}}_{\text{1}}+\dot{m}\left(h_1+\frac{V_1{}^2}{2}+g{z}_1\right)\right]-\left[{\dot{Q}}_{\text{2}}+{\dot{W}}_{\text{2}}+\dot{m}\left(h_2+\frac{V_2{}^2}{2}+g{z}_2\right)\right]=\Delta {\dot{E}}_{\text{system}}$ (VI)

Here, the rate of heat transfer is $\dot{Q}$, the rate of work transfer is $\dot{W}$, the enthalpy is $h$ and the velocity is $V$, the gravitational acceleration is $g$, the elevation from the datum is $z$ and the rate of change in net energy of the system is $\Delta {\dot{E}}_{\text{system}}$; the suffixes 1 and 2 indicates the inlet and outlet of the system.

The refrigerant flows at steady state through the compressor. Hence, the rate of change in net energy of the system becomes zero.

$\Delta {\dot{E}}_{\text{system}}=0$

Heat loss occurs at the rate of $20\text{kJ/kg}$. Neglect the potential energy changes. Here, the work done is by the system (turbine) and the work done on the system is zero i.e. ${\dot{W}}_1=0$.

The Equations (VI) reduced as follows to obtain the work output $\left({\dot{W}}_2\right)$.

$\begin{array}{l}\left[0+0+\dot{m}\left(h_1+\frac{V_1{}^2}{2}+0\right)\right]-\left[{\dot{Q}}_2+{\dot{W}}_2+\dot{m}\left(h_2+\frac{V_2{}^2}{2}+0\right)\right]=0\\ \dot{m}\left(h_1+\frac{V_1{}^2}{2}\right)-\left[{\dot{Q}}_2+{\dot{W}}_2+\dot{m}\left(h_2+\frac{V_2{}^2}{2}\right)\right]=0\\ \dot{m}\left(h_1+\frac{V_1{}^2}{2}\right)={\dot{Q}}_2+{\dot{W}}_2+\dot{m}\left(h_2+\frac{V_2{}^2}{2}\right)\\ {\dot{W}}_2=\dot{m}\left(h_1+\frac{V_1{}^2}{2}\right)-\dot{m}\left(h_2+\frac{V_2{}^2}{2}\right)-{\dot{Q}}_2\end{array}$

$\begin{array}{l}{\dot{W}}_2=\dot{m}\left(h_1-h_2+\frac{V_1{}^2-V_2{}^2}{2}\right)-{\dot{Q}}_2\\ {\dot{W}}_2=-\dot{m}\left(h_2-h_1+\frac{V_2{}^2-V_1{}^2}{2}\right)-{\dot{Q}}_2\end{array}$ (VII)

Here, ${\dot{Q}}_2=\dot{m}{Q}_2$.

Rewrite the Equation (VII) as follows.

${\dot{W}}_2=-\dot{m}\left(h_2-h_1+\frac{V_2{}^2-V_1{}^2}{2}\right)-\dot{m}{Q}_2$ (VIII)

Conclusion:

Substitute $16.17\text{kg/s}$ for $\dot{m}$, $2500.185\text{kJ/kg}$ for $h_2$, $3650.6\text{kJ/kg}$ for $h_1$, $680.6739\text{m/s}$ for $V_2$, $60\text{m/s}$ for $V_1$, and $20\text{kJ/kg}$ for $Q_2$ in Equation (VIII).

$\begin{array}{c}{\dot{W}}_2=-16.17\text{kg/s}\left[\begin{array}{l}2500.185\text{kJ/kg}-3650.6\text{kJ/kg}\\ +\frac{{\left(680.6739\text{m/s}\right)}^2-{\left(60\text{m/s}\right)}^2}{2}\end{array}\right]-16.17\text{kg/s}\left(20\text{kJ/kg}\right)\\ =\left[\begin{array}{c}-16.17\text{kg/s}\left(-1150.415\text{kJ/kg}+229858.4791{\text{m}}^2{\text{/s}}^2\times \frac{1\text{}\text{kJ/kg}}{1000{\text{m}}^2{\text{/s}}^2}\right)\\ -323.4\text{kJ/s}\end{array}\right]\\ =-16.17\text{kg/s}\left(-920.5565\text{kJ/kg}\right)-323.4\text{kJ/s}\\ =\text{14885}\text{.3989}\text{kJ/s}-323.4\text{kJ/s}\end{array}$

$\begin{array}{c}=14561.9989\text{kJ/s}\times \frac{1\text{kW}}{1\text{kJ/s}}\\ \simeq 14562\text{kW}\end{array}$

Thus, the power output of the turbine is $14562\text{kW}$.