Question

A piston-cylinder device initially contains 2 kg of refrigerant-134a at \(800 \mathrm{kPa}\) and \(80^{\circ} \mathrm{C}\). At this state, the piston is touching on a pair of stops at the top. The mass of the piston is such that a 500 -kPa pressure is required to move it. A valve at the bottom of the tank is opened, and R-134a is withdrawn from the cylinder. After a while, the piston is observed to move and the valve is closed when half of the refrigerant is withdrawn from the tank and the temperature in the tank drops to \(20^{\circ} \mathrm{C}\). Determine \((a)\) the work done and \((b)\) the heat transfer.

Short answer

Based on the given solution, the work done (W) during the refrigeration process in the piston-cylinder device is 0 kJ. The heat transfer (Q) during the process is -353.42 kJ.

Step-by-step solution

Determine the initial state properties of the refrigerant

To analyze the problem, we must first find the initial properties of the refrigerant. Given the initial pressure and temperature, we can use the refrigerant property tables to find the specific volume, internal energy, and enthalpy at the initial state. The pressure is 800 kPa and the temperature is \(80^{\circ} \mathrm{C}\). 1. From the R-134a property tables (with pressure = 800 kPa), we can find the specific volume \(v_1 = 0.01645 \, \mathrm{m^3/kg}\), internal energy \(u_1 = 295.05\, \mathrm{kJ/kg}\), and enthalpy \(h_1 = 317.38\, \mathrm{kJ/kg}\).

Calculate the work done (W) during the process

The refrigerant undergoes a constant pressure process until the piston starts moving. The pressure in the cylinder equals the mass of the piston times gravity divided by the piston area, which is 500 kPa. Since there is no change in volume during the isobaric process, there is no work done, so \(W = 0\).

Determine properties at the final state

Now that the final temperature is given, we need to find the corresponding properties of refrigerant-134a. The new mass of refrigerant in the cylinder is \(2/2 = 1\, \mathrm{kg}\). The final state properties are found in the refrigerant tables with a final temperature of \(20^{\circ} \mathrm{C}\) and a final pressure (assuming saturation pressure due to the pressure requirement for the piston to move) of 500 kPa: 1. From the R-134a property tables (with pressure = 500 kPa), we can find the specific volume \(v_2 = 0.04305 \, \mathrm{m^3/kg}\), internal energy \(u_2 = 236.68\, \mathrm{kJ/kg}\), and enthalpy \(h_2 = 248.88\, \mathrm{kJ/kg}\).

Apply the 1st Law of Thermodynamics

Now we can apply the 1st Law of Thermodynamics to find the heat transfer during the process. Applying this to the problem: $$Q - W = \Delta U_{sys}$$ Recall that there is no work done during the process (W = 0). Since half of the refrigerant is withdrawn from the tank at the end, the initial mass is 2 kg and the final mass is 1 kg. The internal energy change for the system can be expressed as: $$\Delta U_{sys} = m_2 \cdot u_2 - m_1 \cdot u_1$$ $$\Rightarrow \Delta U_{sys} = 1 \cdot 236.68 - 2 \cdot 295.05$$ $$\Rightarrow \Delta U_{sys} = -353.42\, \mathrm{kJ}$$ Now, we can solve for Q: $$Q = \Delta U_{sys} + W = -353.42 + 0 = -353.42\, \mathrm{kJ}$$ Therefore, the heat transfer during the process is:

(a) Work done

\(W = 0\, \mathrm{kJ}\)

(b) Heat transfer

\(Q = -353.42\, \mathrm{kJ}\)