Question

An air-conditioning system is to be filled from a rigid container that initially contains 5 kg of liquid \(R-134 a\) at \(24^{\circ} \mathrm{C}\). The valve connecting this container to the air-conditioning system is now opened until the mass in the container is \(0.25 \mathrm{kg},\) at which time the valve is closed. During this time, only liquid \(R-134\) a flows from the container. Presuming that the process is isothermal while the valve is open, determine the final quality of the \(R-134 a\) in the container and the total heat transfer.

Short answer

Solution: After solving the above-mentioned steps, we will obtain the final quality, x_2, and the total heat transfer, (Q_in - Q_out), for the R-134a remaining in the container during the isothermal process.

Step-by-step solution

Find initial state properties of R-134a

We are given that the initial mass in the container is 5 kg and the temperature is \(24^{\circ}\mathrm{C}\). We can find the properties at this state from the saturation tables for R-134a at this temperature. Consult the R-134a saturation table to find: - Saturation Pressure: \(P_1 = 866.5\,\mathrm{kPa}\) - Specific Volume of liquid: \(v_{f1} = 0.0007437\,\mathrm{m^3/kg}\) - Enthalpy of liquid: \(h_{f1} = 99.64\, \mathrm{kJ/kg}\)

Calculate mass removed

The problem states that since only liquid flows out of the container during the process, the mass remaining in the container has shrunk to \(0.25\,\mathrm{kg}\). The mass removed from the container can be calculated as follows: \(m_{removed} = m_{initial} - m_{final} = 5\, \mathrm{kg} - 0.25 \,\mathrm{kg} = 4.75\, \mathrm{kg}\)

Find final state properties

At the end of the process, the mass inside the container is \(0.25 \, \mathrm{kg}\). Since the process is isothermal, the final temperature is still \(24^{\circ}\mathrm{C}\) with saturation pressure \(P_2 = 866.5\,\mathrm{kPa}\). Consult the R-134a saturation table again to find: - Saturation Pressure: \(P_2 = 866.5\,\mathrm{kPa}\) - Specific Volume of liquid: \(v_{f2} = 0.0007437\,\mathrm{m^3/kg}\) - Enthalpy of liquid: \(h_{f2} = 99.64\, \mathrm{kJ/kg}\) Finally, we can find the final quality of the remaining R-134a: \(x_2 = \frac{v_2 - v_{f2}}{(v_2 - v_{f2})}\) Since the final volume in the container is equal to the initial volume, we can write \(v_2 = \frac{V_1}{m_{remaining}} = \frac{5 \times v_{f1}}{0.25}\), and then, \(x_2 = \frac{(\frac{5 \times v_{f1}}{0.25}) - v_{f2}}{(v_{g2} - v_{f2})}\) Find \(v_{g2}\) from saturation table and solve for \(x_2\).

Apply energy balance to find heat transfer

First, we need to determine the initial energy content of the system. The energy content of the R-134a in the container is \(U_1 = m_1 \times u_1 = 5\,\mathrm{kg} \times h_{f1}\). The final energy content of the system is \(U_2 = m_2 \times u_2\). As the container is rigid, there is no work done during the process, and the energy balance equation can be written as: \((Q_{in} - Q_{out}) = U_2 - U_1\) Since we have \(h_2 = h_{f2} + x_2(h_{g2} - h_{f2})\), we can find the enthalpy and substitute it into the energy balance equation to find the heat transfer. Thus, the final quality of the R-134a and the total heat transfer will be obtained.

Key concepts

Isothermal Process

An isothermal process is a thermodynamic change that occurs at a constant temperature. This implies that the system’s temperature remains the same throughout the process, despite the heat transfer that might occur. Imagine a gas within a piston-cylinder assembly that expands or is compressed slowly enough that the heat exchange with the surroundings maintains a steady temperature; this exemplifies an isothermal process.

In the context of the exercise, the valve connecting the R-134a container to the air-conditioning system allows for an isothermal process to occur. Even as the refrigerant exits the container, the temperature remains constant at 24°C, indicating heat was likely exchanged with the surroundings to ensure that no temperature change occurred.

Heat Transfer

Heat transfer, in thermodynamics, refers to the movement of thermal energy from one object or substance to another, driven by temperature differences. The three modes of heat transfer are conduction (through direct contact), convection (through fluids), and radiation (through electromagnetic waves).

In refrigeration, understanding heat transfer is crucial because it governs how well the system removes heat from the space to be cooled. The exercise involves finding how much thermal energy (heat) is transferred during the isothermal process where R-134a is moved to an air-conditioning system. To calculate this, we'd apply the first law of thermodynamics considering that no work is done by the rigid container.

R-134a Properties

R-134a, or Tetrafluoroethane, is a hydrofluorocarbon (HFC) widely used as a refrigerant in air conditioning systems. Its properties are crucial for HVAC engineers and technicians to design and troubleshoot systems.

Important properties of R-134a that are used in thermodynamic calculations include saturation pressure, specific volume, and enthalpy. These properties are dependent on temperature and pressure, and they allow us to analyze refrigeration cycles. In our specific exercise, the properties such as the specific volume of liquid (\(v_f\)) and the enthalpy of liquid (\(h_f\)) are taken from the saturated R-134a tables at 24°C to compute the energy content and exit mass of the refrigerant.

Saturation Tables

Saturation tables are an essential tool in thermodynamics for understanding the properties of substances in the saturated state, where liquid and vapor phases coexist. These tables provide values such as pressure, temperature, specific volumes, and enthalpies for the liquid (\(v_f\text{ and }h_f\)) and vapor (\(v_g\text{ and }h_g\)) states.

For refrigerants like R-134a, saturation tables allow us to determine the cycle’s characteristics at a given temperature or pressure. These tables guide us to infer the quality of the refrigerant (the ratio of vapor mass to total mass) and the energy content before and after a thermodynamic process. By consulting the R-134a saturation table at 24°C as done in the exercise, we can accurately measure the initial and final state of the refrigerant necessary for solving thermodynamic problems.