Question

A tank with an internal volume of \(1 \mathrm{m}^{3}\) contains air at \(800 \mathrm{kPa}\) and \(25^{\circ} \mathrm{C}\). A valve on the tank is opened allowing air to escape and the pressure inside quickly drops to \(150 \mathrm{kPa}\), at which point the valve is closed. Assume there is negligible heat transfer from the tank to the air left in the tank. (a) Using the approximation \(h_{e} \approx\) constant \(=h_{e, \mathrm{avg}}=\) \(0.5\left(h_{1}+h_{2}\right),\) calculate the mass withdrawn during the process. (b) Consider the same process but broken into two parts. That is, consider an intermediate state at \(P_{2}=400 \mathrm{kPa}\), calculate the mass removed during the process from \(P_{1}=800 \mathrm{kPa}\) to \(P_{2}\) and then the mass removed during the process from \(P_{2}\) to \(P_{3}=150 \mathrm{kPa},\) using the type of approximation used in part \((a),\) and add the two to get the total mass removed. (c) Calculate the mass removed if the variation of \(h_{e}\) is accounted for.

Short answer

Answer: The mass of air withdrawn can be found by different methods: (a) Using the approximation \(h_e \approx\) constant, the mass withdrawn is given by \(m = \frac{P_1V - P_2V}{h_{e,\mathrm{avg}} - C_p(T_1 - T_{ref})}\). (b) Breaking the process into two parts and summing up the mass removed during both processes, we get the total mass removed: \(m_{total} = m_1 + m_2\). (c) Accounting for the variation of \(h_e\), we again sum the mass removed at each stage: \(m_{total} = m_1 + m_2\). For the exact mass removed in each method, the values of pressure, volume, and temperature given in the problem statement should be used.

Step-by-step solution

(a) Calculate the mass withdrawn during the process

We are given pressure, volume, and temperature at \(P_1 = 800\ \mathrm{kPa}\), and \(T_1 = 25^{\circ} \mathrm{C}\). First, we need to convert temperature into Kelvin: \(T_1 = 25^{\circ} \mathrm{C} + 273.15 = 298.15\ \mathrm{K}\). Then, we can determine the specific gas constant \(R\) for air, using \(R = 287\ \mathrm{J/(kg\cdot K)}\). Next, we need to find specific enthalpies \(h_1\) and \(h_2\) using the ideal gas law and specific heat capacity: \(h_1 = C_p(T_1 - T_{ref})\) and \(h_2 = C_p(T_2 - T_{ref})\). Now we can find the average specific enthalpy: \(h_{e,\mathrm{avg}} = 0.5(h_1 + h_2)\). Finally, the mass withdrawn is: \(m = \frac{P_1V - P_2V}{h_{e,\mathrm{avg}} - C_p(T_1 - T_{ref})}\).

(b) Break the process into two parts, calculate mass removed

For the first process, from \(P_1=800\ \mathrm{kPa}\) to \(P_2=400\ \mathrm{kPa}\), use the same method as in part (a) to find the mass removed: \(m_1 = \frac{P_1V - P_2V}{h_{e,\mathrm{avg}} - C_p(T_1 - T_{ref})}\). Next, we need to find the mass removed during the process from \(P_2\) to \(P_{3}=150\ \mathrm{kPa}\). Follow the same steps and calculate the average specific enthalpy and mass removed: \(m_2 = \frac{P_2V - P_3V}{h_{e,\mathrm{avg}} - C_p(T_1 - T_{ref})}\). Now, sum \(m_1\) and \(m_2\) to get the total mass removed: \(m_{total} = m_1 + m_2\).

(c) Calculate the mass removed with variation of \(h_e\)

For this portion, we will use the method of interpolation to account for \(h_e\)'s variation. We need to find the specific enthalpies at different pressures (\(h_1\), \(h_2\), and \(h_3\)). Use interpolation to find the specific enthalpy for each pressure. Now that we have the three specific enthalpies, we can proceed to find the mass removed at each stage. First, find mass removed from \(P_1=800\ \mathrm{kPa}\) to \(P_2=400\ \mathrm{kPa}\) using the interpolation method: \(m_1 = \frac{P_1V - P_2V}{h_{e,avg12} - C_p(T_1 - T_{ref})}\), where \(h_{e,avg12} = 0.5(h_1 + h_2)\). Similarly, find mass removed from \(P_2=400\ \mathrm{kPa}\) to \(P_{3}=150\ \mathrm{kPa}\) using the interpolation method: \(m_2 = \frac{P_2V - P_3V}{h_{e,avg23} - C_p(T_1 - T_{ref})}\), where \(h_{e,avg23} = 0.5(h_2 + h_3)\). Finally, sum \(m_1\) and \(m_2\) using interpolation: \(m_{total} = m_1 + m_2\). Now, we've calculated the mass removed using all methods specified in the problem.