Question

A \(2-\mathrm{ft}^{3}\) rigid tank contains saturated refrigerant \(134 \mathrm{a}\) at 160 psia. Initially, 5 percent of the volume is occupied by liquid and the rest by vapor. A valve at the top of the \(\operatorname{tank}\) is now opened, and vapor is allowed to escape slowly from the tank. Heat is transferred to the refrigerant such that the pressure inside the tank remains constant. The valve is closed when the last drop of liquid in the tank is vaporized. Determine the total heat transfer for this process.

Short answer

Initial state: 160 psia and 5% liquid by volume.

Step-by-step solution

Determining the initial state of the refrigerant

Using the given initial pressure of 160 psia, we should find the corresponding saturation temperature and the specific volumes (v_f and v_g) of the saturated liquid and vapor states using a refrigerant property table.

Calculating the initial volume of liquid and vapor

From the exercise, we know that 5% of the volume is in the liquid state and 95% is in the vapor state. We can set up the following equation by multiplying the volume of each phase by its specific volume, and sum them up: \(V = 2 \, \mathrm{ft}^{3}=V_\mathrm{liquid} + V_\mathrm{vapor} = 0.05 V_\mathrm{total} \cdot v_\mathrm{f} + 0.95 V_\mathrm{total} \cdot v_\mathrm{g}\) Solve for \(V_\mathrm{liquid}\) and \(V_\mathrm{vapor}\).

Calculating the initial mass of liquid and vapor

Using the specific volumes calculated in step 1, we can calculate the mass of each phase using \(m_\mathrm{phase} = V_\mathrm{phase} / v_\mathrm{phase}\). Calculate \(m_\mathrm{liquid}\) and \(m_\mathrm{vapor}\).

Calculating the initial internal energy of the system

Now that we have the mass of each phase, we can determine the initial internal energy of the system. Using a refrigerant property table, find the specific internal energy values (u_f and u_g) for the liquid and vapor phases at the initial pressure of 160 psia. Then, calculate the initial internal energy of the system by summing the internal energy of each phase: \(U_\mathrm{initial} = m_\mathrm{liquid} \cdot u_\mathrm{f} + m_\mathrm{vapor} \cdot u_\mathrm{g}\)

Finding the final state

When the valve is closed, all the liquid refrigerant has vaporized. The final state is the saturated vapor state at the same pressure (160 psia). Find the final specific internal energy (u_g) using a refrigerant property table.

Calculating the final internal energy of the system

Use the final specific internal energy (u_g) to calculate the final internal energy of the system: \(U_\mathrm{final} = m_\mathrm{total} \cdot u_\mathrm{g}\) Where \(m_\mathrm{total} = m_\mathrm{liquid} + m_\mathrm{vapor}\) is the total mass of the refrigerant in the tank.

Calculating the total heat transfer

Applying the first law of thermodynamics for a closed system (assuming no significant change in kinetic or potential energy) and considering that the tank is rigid, we have: \(Q = \Delta U = U_\mathrm{final} - U_\mathrm{initial}\) Plug in the values for U_final and U_initial from steps 4 and 6 to get the total heat transferred to the system during the process.