Charging of a Rigid Tank by Steam
A rigid, insulated tank that is initially evacuated is connected through a valve to a supply line that carries steam at 1 MPa and 300°C. Now the valve is opened, and steam is allowed to flow slowly into the tank until the pressure reaches 1 MPa, at which point the valve is closed. Determine the final temperature of the steam in the tank.
A valve connecting an initially evacuated tank to a steam line is opened, and steam flows in until the pressure inside rises to the line level. The final temperature in the tank is to be determined.
Assumptions 1 This process can be analyzed as a uniform-flow process since the properties of the steam entering the control volume remain constant during the entire process. 2 The kinetic and potential energies of the streams are negligible, ke ≌ pe ≌ 0. 3 The tank is stationary and thus its kinetic and potential energy changes are zero; that is, ΔKE = ΔPE = 0 and \Delta E_{system} = \Delta U_{system}. 4 There are no boundary, electrical, or shaft work interactions involved. 5 The tank is well insulated and thus there is no heat transfer.
Analysis We take the tank as the system (Fig. 5–50). This is a control volume since mass crosses the system boundary during the process. We observe that this is an unsteady-flow process since changes occur within the control volume. The control volume is initially evacuated and thus m_{1} = 0 and m_{1}u_{1} = 0. Also, there is one inlet and no exits for mass flow.
Noting that microscopic energies of flowing and nonflowing fluids are represented by enthalpy h and internal energy u, respectively, the mass and energy balances for this uniform-flow system can be expressed as
\text{Mass balance:}\quad\quad m_{\mathrm{in}}-m_{\mathrm{out}}=\Delta m_{\mathrm{system}} \rightarrow m_{i}=m_{2}-m_{1}^{\nearrow^0}=m_{2}Energy balance:
\begin{aligned}\underbrace{E_{\mathrm{in}}-E_{\mathrm{out}}}_{\begin{array}{c}\text { Net energy transfer } \\\text { by heat, work, and mass }\end{array}} &=\underbrace{\Delta E_{\mathrm{system}}}_{\begin{array}{r}\text { Change in internal, kinetic, } \\\text { potential, etc., energies }\end{array}} \\m_{i}h_{i} &=m_{2}u_{2}\quad (since W = Q = 0, ke ≅ pe ≅ 0, m_{1} = 0)\end{aligned}
Combining the mass and energy balances gives
u_{2} = h_{i}
That is, the final internal energy of the steam in the tank is equal to the enthalpy of the steam entering the tank. The enthalpy of the steam at the inlet state is
\left.\begin{array}{l}P_{i}=1 \mathrm{MPa} \\T_{i}=300^{\circ} \mathrm{C}\end{array}\right\} h_{i}=3051.6 \mathrm{~kJ} / \mathrm{kg} (Table A–6)
which is equal to u_{2}. Since we now know two properties at the final state, it is fixed and the temperature at this state is determined from the same table to be
\left.\begin{array}{l}P_{2}=1 \mathrm{MPa} \\u_{2}=3051.6 \mathrm{~kJ} / \mathrm{kg}\end{array}\right\} T_{2}=456.1^{\circ} \mathrm{C}
Discussion Note that the temperature of the steam in the tank has increased by 156.1°C. This result may be surprising at first, and you may be wondering where the energy to raise the temperature of the steam came from. The answer lies in the enthalpy term h = u + Pv. Part of the energy represented by enthalpy is the flow energy Pv, and this flow energy is converted to sensible internal energy once the flow ceases to exist in the control volume, and it shows up as an increase in temperature (Fig. 5–51).
Alternative solution This problem can also be solved by considering the region within the tank and the mass that is destined to enter the tank as a closed system, as shown in Fig. 5–50b. Since no mass crosses the boundaries, viewing this as a closed system is appropriate.
During the process, the steam upstream (the imaginary piston) will push the enclosed steam in the supply line into the tank at a constant pressure of 1 MPa. Then the boundary work done during this process is
W_{b, \text { in }}=-\int_{1}^{2} P_{i} d V=-P_{i}\left(V_{2}-V_{1}\right)=-P_{i}\left[V_{\text {tank }}-\left(V_{\text {tank }}+V_{i}\right)\right]=P_{i} V_{i}
where V_{i} is the volume occupied by the steam before it enters the tank and P_{i} is the pressure at the moving boundary (the imaginary piston face). The energy balance for the closed system gives
\begin{aligned}\underbrace{E_{\mathrm{in}}-E_{\mathrm{out}}}_{\begin{array}{c}\text { Net energy transfer } \\\text { by heat, work, and mass }\end{array}} &=\underbrace{\Delta E_{\mathrm{system}}}_{\begin{array}{r}\text { Change in internal, kinetic, } \\\text { potential, etc., energies }\end{array}} \\W_{b, \text { in }} &=\Delta U \\m_{i} P_{i} v_{i} &=m_{2} u_{2}-m_{i} u_{i} \\u_{2} &=u_{i}+P_{i} v_{i}=h_{i} \end{aligned}
since the initial state of the system is simply the line conditions of the steam. This result is identical to the one obtained with the uniform-flow analysis. Once again, the temperature rise is caused by the so-called flow energy or flow work, which is the energy required to move the fluid during flow.
