Question 5.7: Power Generation by a Steam Turbine The power output of an a...

The inlet and exit conditions of a steam turbine and its power output are given. The changes in kinetic energy, potential energy, and enthalpy of steam, as well as the work done per unit mass and the mass flow rate of steam are to be determined.
Assumptions 1 This is a steady-flow process since there is no change with time at any point and thus \Delta m_{ CV }=0 \text { and } \Delta E_{ cV }=0 . 2 The system is adiabatic and thus there is no heat transfer.

Analysis We take the turbine as the system. This is a control volume since mass crosses the system boundary during the process. We observe that there is only one inlet and one exit and thus \dot{m}_{1}=\dot{m}_{2}=\dot{m} . Also, work is done by the system. The inlet and exit velocities and elevations are given, and thus the kinetic and potential energies are to be considered.
(a) At the inlet, steam is in a superheated vapor state, and its enthalpy is

\left.\begin{array}{l} P_{1}=2 MPa \\ T_{1}=400^{\circ} C \end{array}\right\} \quad h_{1}=3248.4 kJ / kg                                   (Table A–6)

At the turbine exit, we obviously have a saturated liquid–vapor mixture at 15-kPa pressure. The enthalpy at this state is

h_{2}=h_{f}+x_{2} h_{f g}=[225.94+(0.9)(2372.3)] kJ / kg =2361.01 kJ / kg

Then

\begin{aligned}&\Delta h=h_{2}-h_{1}=(2361.01-3248.4) kJ / kg =-887.39 kJ / kg \\&\Delta ke =\frac{V_{2}^{2}-V_{1}^{2}}{2}=\frac{(180 m / s )^{2}-(50 m / s )^{2}}{2}\left(\frac{1 kJ / kg }{1000 m ^{2} / s ^{2}}\right)=14.95 kJ / kg \\&\Delta pe =g\left(z_{2}-z_{1}\right)=\left(9.81 m / s ^{2}\right)[(6-10) m ]\left(\frac{1 kJ / kg }{1000 m ^{2} / s ^{2}}\right)=- 0 . 0 4 kJ / kg\end{aligned}

(b) The energy balance for this steady-flow system can be expressed in the rate form as

\underbrace{\dot{E}_{\text {in }}-\dot{E}_{\text {out }}}_{\begin{array}{c}\text { Rate of net energy transfer } \\\text { by heat, work, and mass }\end{array}}=\underbrace{d E_{\text {system }} /{d t}^{\nearrow ^{0(steady)}} }_{\begin{array}{c}\text { Rate of change in internal, kinetic, } \\\text { potential, etc., energies }\end{array}}=0

\begin{gathered} \dot{E}_{\text {in }}=\dot{E}_{\text {out }} \\ \dot{m}\left(h_{1}+\frac{V_{1}^{2}}{2}+g z_{1}\right)={W}_{\text {out }}+\dot{m}\left(h_{2}+\frac{V_{2}^{2}}{2}+g z_{2}\right) \quad(\text { since } \dot{Q}=0) \end{gathered}

Dividing by the mass flow rate \dot{m}  and substituting, the work done by the turbine per unit mass of the steam is determined to be

\begin{aligned} w_{\text {out }} &=-\left[\left(h_{2}-h_{1}\right)+\frac{V_{2}^{2}-V_{1}^{2}}{2}+g\left(z_{2}-z_{1}\right)\right]=-(\Delta h+\Delta ke +\Delta pe ) \\ &=-[-887.39+14.95-0.04] kJ / kg =872.48 kJ / kg \end{aligned}

(c) The required mass flow rate for a 5-MW power output is

\dot{m}=\frac{\dot{W}_{\text {out }}}{w_{\text {out }}}=\frac{5000 kJ / s }{872.48 kJ / kg }=5.73 kg / s

Discussion Two observations can be made from these results. First, the change in potential energy is insignificant in comparison to the changes in enthalpy and kinetic energy. This is typical for most engineering devices. Second, as a result of low pressure and thus high specific volume, the steam velocity at the turbine exit can be very high. Yet the change in kinetic energy is a small fraction of the change in enthalpy (less than 2 percent in our case) and is therefore often neglected.