Known Steam contained in a large vessel at a known state flows from the vessel through a turbine into a small tank of known volume until a specified final condition is attained in the tank.
Find Determine the work developed by the turbine.
Schematic and Given Data:
Engineering Model
1. The control volume is defined by the dashed line on the accompanying diagram.
2. For the control volume, \dot{Q}_{ cv }=0 and kinetic and potential energy effects are negligible.
1 3. The state of the steam within the large vessel remains constant. The final state of the steam in the smaller tank is an equilibrium state.
4. The amount of mass stored within the turbine and the interconnecting piping at the end of the filling process is negligible.
Analysis Since the control volume has a single inlet and no exits, the mass rate balance, Eq. 4.2, reduces to
\frac{d m_{ cv }}{d t}=\sum_{i} \dot{m}_{i}-\sum_{e} \dot{m}_{e} (4.2)
\frac{d m_{ cv }}{d t}=\dot{m}_{i}
The energy rate balance, Eq. 4.15, reduces with assumption 2 to
\frac{d E_{ cv }}{d t}=\dot{Q}_{ cv }-\dot{W}_{ cv }+\sum_{i} \dot{m}_{i}\left(h_{i}+\frac{ V _{i}^{2}}{2}+g z_{i}\right)-\sum_{e} \dot{m}_{e}\left(h_{e}+\frac{ V _{e}^{2}}{2}+g z_{e}\right) (4.15)
\frac{d U_{ cv }}{d t}=-\dot{W}_{ cv }+\dot{m}_{i} h_{i}
Combining the mass and energy rate balances gives
\frac{d U_{ cv }}{d t}=-\dot{W}_{ cv }+h_{i} \frac{d m_{ cv }}{d t}
Integrating
\Delta U_{ cv }=-W_{ cv }+h_{i} \Delta m_{ cv }
In accordance with assumption 3, the specific enthalpy of the steam entering the control volume is constant at the value corresponding
to the state in the large vessel.
Solving for W_{ cv }
W_{ cv }=h_{i} \Delta m_{ cv }-\Delta U_{ cv }
\Delta U_{ cv } \text { and } \Delta m_{ cv } denote, respectively, the changes in internal energy and mass of the control volume. With assumption 4, these terms can be identified with the small tank only.
Since the tank is initially evacuated, the terms \Delta U_{ cv } \text { and } \Delta m_{ cv } reduce to the internal energy and mass within the tank at the end of the process. That is,
\Delta U_{ cv }=\left(m_{2} u_{2}\right)-\left(m_{1} u_{1}\right)^{\nearrow0}, \quad \Delta m_{ cv }=m_{2}-m_{1}^{\nearrow0}
where 1 and 2 denote the initial and final states within the tank, respectively.
Collecting results yields
2 3 W_{ cv }=m_{2}\left(h_{i}-u_{2}\right) (a)
The mass within the tank at the end of the process can be evaluated from the known volume and the specific volume of steam at 15 bar and 400°C from Table A-4
m_{2}=\frac{V}{v_{2}}=\frac{0.6 m ^{3}}{\left(0.203 m ^{3} / kg \right)}=2.96 kg
The specific internal energy of steam at 15 bar and 400°C from Table A-4 is 2951.3 kJ/kg. Also, at 15 bar and 320°C, h_{i}=3081.9 kJ/kg.
Substituting values into Eq. (a)
W_{ cv }=2.96 kg (3081.9-2951.3) kJ / kg =386.6 kJ
1 In this case idealizations are made about the state of the steam entering the tank and the final state of the steam in the tank. These idealizations make the transient analysis manageable.
2 A significant aspect of this example is the energy transfer into the control volume by flow work, incorporated in the pv term of the specific enthalpy at the inlet.
3 This result can also be obtained by reducing Eq. 4.28. The details are left as an exercise.
m_{ cv }(t) u(t)-m_{ cv }(0) u(0)=Q_{ cv }-W_{ cv }+h_{i}\left(m_{ cv }(t)-m_{ cv }(0)\right) (4.28)
Skills Developed
Ability to…
• apply the time-dependent mass and energy rate balances to a control volume.
• develop an engineering model.
• retrieve property data for water.
Quick Quiz
If the turbine were removed, and the steam allowed to flow adiabatically into the small tank until the pressure in the tank is 15 bar, determine the final steam temperature in the tank, in °C. Ans. 477°C.
