Discharge of Heated Air at Constant Temperature
An insulated 8-m^3 rigid tank contains air at 600 kPa and 400 K. A valve connected to the tank is now opened, and air is allowed to escape until the pressure inside drops to 200 kPa. The air temperature during the process is maintained constant by an electric resistance heater placed in the tank. Determine the electrical energy supplied to air during this process.
Pressurized air in an insulated rigid tank equipped with an electric heater is allowed to escape at constant temperature until the pressure inside drops to a specified value. The amount of electrical energy supplied to air is to be determined.
Assumptions 1 This is an unsteady process since the conditions within the device are changing during the process, but it can be analyzed as a uniform-flow process since the exit conditions remain constant. 2 Kinetic and potential energies are negligible. 3 The tank is insulated and thus heat transfer is negligible. 4 Air is an ideal gas with variable specific heats.
Analysis We take the contents of the tank as the system, which is a control volume since mass crosses the boundary (Fig. 5–52). Noting that the 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_{\text{in}} – m_{\text{out}} = \Delta m_{\text{system}} \rightarrow m_{\text{e}} = m_{1} – m_{2}
\begin{aligned}\text{Energy balance:}\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}_{e, \text { in }}-{m}_{e}h_{e} &= m_{2}u_{2} -m_{1}u_{1}\quad \text { (since } Q \cong \mathrm{ke} \cong \mathrm{pe} \cong 0)\end{aligned}
The gas constant of air is R = 0.287 kPa·m^3/kg·K (Table A-1). The initial and final masses of air in the tank and the discharged amount are determined from the ideal gas relation to be
\begin{array}{l}m_{1}=\frac{P_{1} V_{1}}{R T_{1}}=\frac{(600 \mathrm{kPa})\left(8 \mathrm{~m}^{3}\right)}{\left(0.287 \mathrm{kPa} \cdot \mathrm{m}^{3} / \mathrm{kg} \cdot \mathrm{K}\right)(400 \mathrm{~K})}=41.81 \mathrm{~kg} \\\\m_{2}=\frac{P_{2} \mathrm{~V}_{2}}{R T_{2}}=\frac{(200 \mathrm{kPa})\left(8 \mathrm{~m}^{3}\right)}{\left(0.287 \mathrm{kPa} \cdot \mathrm{m}^{3} / \mathrm{kg} \cdot \mathrm{K}\right)(400 \mathrm{~K})}=13.94 \mathrm{~kg} \\\\m_{e}=m_{1}-m_{2}=41.81-13.94=27.87 \mathrm{~kg}\end{array}
The enthalpy and internal energy of air at 400 \mathrm{~K} are h_{e}=400.98 \mathrm{~kJ} / \mathrm{kg} and u_{1}=u_{2}= 286.16 \mathrm{~kJ} / \mathrm{kg} (Table A-17). The electrical energy supplied to air is determined from the energy balance to be
\begin{aligned}W_{e, \text { in }}=& m_{e} h_{e}+m_{2} u_{2}-m_{1} u_{1} \\=&(27.87 \mathrm{~kg})(400.98 \mathrm{~kJ} / \mathrm{kg})+(13.94 \mathrm{~kg})(286.16 \mathrm{~kJ} / \mathrm{kg}) \\&-(41.81 \mathrm{~kg})(286.16 \mathrm{~kJ} / \mathrm{kg}) \\=& 3200 \mathrm{~kJ}=0.889 \mathrm{kWh}\end{aligned}
since 1 \mathrm{kWh}=3600 \mathrm{~kJ}.
Discussion If the temperature of discharged air changes during the process, the problem can be solved with reasonable accuracy by evaluating h_{e} at the average discharge temperature T_{e}=\left(T_{2}+T_{1}\right) / 2 and treating it as constant.
