Known A tank initially holding a two-phase liquid–vapor mixture is heated while saturated water vapor is slowly removed. This continues at constant pressure until the tank is filled only with saturated vapor.
Find Determine the amount of heat transfer.
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{W}_{ cv }=0 and kinetic and potential energy effects can be neglected.
3. At the exit the state remains constant.
1 4. The initial and final states of the mass within the vessel are equilibrium states.
Analysis Since there is a single exit and no inlet, the mass rate balance Eq. 4.2 takes the form
\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}_{e}
With assumption 2, the energy rate balance Eq. 4.15 reduces 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{Q}_{ cv }-\dot{m}_{e} h_{e}
Combining the mass and energy rate balances results in
\frac{d U_{ cv }}{d t}=\dot{Q}_{ cv }+h_{e} \frac{d m_{ cv }}{d t}
By assumption 3, the specific enthalpy at the exit is constant. Accordingly, integration of the last equation gives
\Delta U_{ cv }=Q_{ cv }+h_{e} \Delta m_{ cv }
Solving for the heat transfer Q_{ cv },
Q_{ cv }=\Delta U_{ cv }-h_{e} \Delta m_{ cv }
or
2 Q_{ cv }=\left(m_{2} u_{2}-m_{1} u_{1}\right)-h_{e}\left(m_{2}-m_{1}\right)
where m_{1} \text { and } m_{2} denote, respectively, the initial and final amounts of mass within the tank.
The terms u_{1} \text { and } m_{1} of the foregoing equation can be evaluated with property values from Table A-2 at 260°C and the given value for quality. Thus,
u_{1}=u_{ f 1}+x_{1}\left(u_{ g 1}-u_{ f 1}\right)
=1128.4+(0.7)(2599.0-1128.4)=2157.8 kJ / kg
Also,
v_{1}=v_{ f 1}+x_{1}\left(v_{ g 1}-v_{ f 1}\right)
=1.2755 \times 10^{-3}+(0.7)\left(0.04221-1.2755 \times 10^{-3}\right)
=29.93 \times 10^{-3} m ^{3} / kg
Using the specific volume v_{1}, the mass initially contained in the tank is
m_{1}=\frac{V}{v_{1}}=\frac{0.85 m ^{3}}{\left(29.93 \times 10^{-3} m ^{3} / kg \right)}=28.4 kg
The final state of the mass in the tank is saturated vapor at 260°C so Table A-2 gives
\begin{aligned}&u_{2}=u_{ g }\left(260^{\circ} C \right)=2599.0 kJ / kg \\&v_{2}=v_{ g }\left(260^{\circ} C \right)=42.21 \times 10^{-3} m ^{3} / kg\end{aligned}
The mass contained within the tank at the end of the process is
m_{2}=\frac{V}{v_{2}}=\frac{0.85 m ^{3}}{\left(42.21 \times 10^{-3} m ^{3} / kg \right)}=20.14 kg
Table A-2 also gives h_{e}=h_{ g }\left(260^{\circ} C \right)=2796.6 kJ / kg.
Substituting values into the expression for the heat transfer yields
Q_{ cv }=(20.14)(2599.0)-(28.4)(2157.8)-2796.6(20.14-28.4)
= 14,162 kJ
1 In this case, idealizations are made about the state of the vapor exiting and the initial and final states of the mass contained within the tank.
2 This expression for Q_{ cv } can be obtained by applying Eqs. 4.23, 4.25, and 4.27. The details are left as an exercise.
m_{ cv }(t)-m_{ cv }(0)=\sum_{i} m_{i}-\sum_{e} m_{e} (4.23)
U_{ cv }(t)-U_{ cv }(0)=Q_{ cv }-W_{ cv }+\sum_{i} m_{i} h_{i}-\sum_{e} m_{e} h_{e} (4.25)
U_{ cv }(t)=m_{ cv }(t) u(t) (4.27)
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 initial quality were 90%, determine the heat transfer, in kJ, keeping all other data unchanged. Ans. 3707 kJ.
