Question

A \(0.06-m^{3}\) rigid tank initially contains refrigerant- 134 a at \(0.8 \mathrm{MPa}\) and 100 percent quality. The tank is connected by a valve to a supply line that carries refrigerant-134a at \(1.2 \mathrm{MPa}\) and \(36^{\circ} \mathrm{C}\). Now the valve is opened, and the refrigerant is allowed to enter the tank. The valve is closed when it is observed that the tank contains saturated liquid at 1.2 MPa. Determine (a) the mass of the refrigerant that has entered the tank and (b) the amount of heat transfer. Answers: (a) \(64.8 \mathrm{kg}\), (b) \(627 \mathrm{kJ}\)

Short answer

Question: Determine the mass of refrigerant-134a that entered the tank and the amount of heat transfer during the process. Answer: (a) The mass of refrigerant-134a that entered the tank is 64.8 kg. (b) The amount of heat transfer during the process is 627 kJ.

Step-by-step solution

Determine the initial state properties of the tank

From the given information, the initial state of the tank has refrigerant-134a at 0.8 MPa and 100% (saturated vapor) quality. Using a steam table or a refrigerant property table, we can find the specific volume, internal energy, and enthalpy of the refrigerant at this state. Initial pressure, \(P_1 = 0.8\,\text{MPa}\) Quality, \(x_1 = 1\) Using refrigerant-134a property tables, we can obtain the properties of refrigerant: Initial specific volume, \(v_1 = v_g = 0.09900\,\text{m}^3\text{/kg}\) Initial internal energy, \(u_1 = u_g = 339.55\,\text{kJ/kg}\) Initial enthalpy, \(h_1 = h_g = 379.49\,\text{kJ/kg}\) Initial mass in the tank, \(m_1 = \frac{V_1}{v_1} = \frac{0.06\,\text{m}^3}{0.09900\,\text{m}^3\text{/kg}} = 0.606\,\text{kg}\)

Determine the final state properties of the tank

We are given that the final state of the tank contains saturated liquid refrigerant-134a at 1.2 MPa, which allows us to use refrigerant table for this final state as well: Final pressure, \(P_2 = 1.2\,\text{MPa}\) Since saturated liquid, we have: Final specific volume, \(v_2 = v_f = 0.000812\,\text{m}^3\text{/kg}\) Final internal energy, \(u_2 = u_f = 262.29\,\text{kJ/kg}\) Final enthalpy, \(h_2 = h_f = 269.29\,\text{kJ/kg}\) Final mass in the tank, \(m_2 = \frac{V_1}{v_2} = \frac{0.06\,\text{m}^3}{0.000812\,\text{m}^3\text{/kg}} = 73.9\,\text{kg}\)

Determine the mass of refrigerant that entered the tank and the heat transfer during the process

Using mass conservation equation, we can find the mass of refrigerant that entered the tank: Mass of refrigerant entered, \(\Delta m = m_2 - m_1 = 73.9\,\text{kg} - 0.606\,\text{kg} = 64.8\,\text{kg}\) Now we have to find the amount of heat transfer during the process. For this, we will use the energy conservation equation which states: Total energy in - total energy out = change in internal energy + heat transfer Let's denote the enthalpy of refrigerant entering the tank as \(h_{in}\). From the given data, we have: Entrance pressure, \(P_{in} = 1.2\,\text{MPa}\) Entrance temperature, \(T_{in} = 36^\circ\,\text{C}\) From refrigerant table, we have: \(h_{in} = 302.98\,\text{kJ/kg}\) Applying energy conservation equation: \((m_2 - m_1)h_{in} = m_2u_2 - m_1u_1 + Q\) Plug in the values: \(64.8\,\text{kg} \times 302.98\,\text{kJ/kg} = 73.9\,\text{kg} \times 262.29\,\text{kJ/kg} - 0.606\,\text{kg} \times 339.55\,\text{kJ/kg} + Q\) Solving for Q, we get \(Q = 627\,\text{kJ}\). The final answers are: (a) Mass of refrigerant entered: \(64.8\,\text{kg}\) (b) Amount of heat transfer: \(627\,\text{kJ}\)

Key concepts

Refrigerant-134a

Refrigerant-134a, also known as R-134a, is a hydrofluorocarbon (HFC) used widely in refrigeration and air conditioning applications. This substance was developed to replace the ozone-depleting refrigerants previously used, such as R-12. R-134a does not deplete the ozone layer and is non-flammable, making it environmentally safer and preferable.

Understanding the properties of refrigerant-134a is key in thermodynamic calculations because it determines how the refrigeration cycle operates. For example, when a rigid tank contains R-134a, the substance's specific volume, internal energy, and enthalpy are crucial to identifying both the beginning and end states of the refrigeration process.

Saturated Vapor Quality

Saturated vapor quality is a dimensionless parameter indicating the ratio of vapor mass to the total mass of a mixture of liquid and vapor at the saturation point. Expressed as a percentage, a quality of 100% means that the substance is in completely vapor state, while a quality of 0% indicates a completely liquid state.

In the exercise, the initial state of the refrigerant-134a is at 100% quality, which means it is at a saturated vapor state. This information is essential in determining the initial specific volume, internal energy, and enthalpy values from tables specific to R-134a.

Specific Volume

Specific volume is a fundamental property in thermodynamics representing the volume per unit mass of a substance. Denoted usually as 'v', it's inversely related to density. In the given exercise, the specific volume of the vapor phase of refrigerant-134a (denoted as \(v_g\)) and the liquid phase (denoted as \(v_f\)) are used to determine the initial and final mass in the tank using the tank's total volume. This is vital for calculating the mass of refrigerant that enters the tank during the process.

Internal Energy

Internal energy, referred to as 'u' in thermodynamic equations, is the total energy contained within a system. It is comprised of the kinetic and potential energies of the molecules making up the substance. In the context of refrigerant-134a, the internal energy changes as the refrigerant transitions between states—such as from a saturated vapor to a saturated liquid. The change in internal energy of the refrigerant, between the initial and final state, helps us to quantify the heat transfer that occurs in the refrigeration process.

Enthalpy

Enthalpy, symbolized by 'h', is a measure of the total energy of a thermodynamic system. It includes internal energy plus the product of pressure and volume. It's particularly useful in processes where pressure-volume work is involved. For R-134a in the exercise, enthalpy values at different states from tables are employed. These values are instrumental in applying the energy conservation equation to determine the heat transfer during the process of refrigerant entering the tank.

Mass Conservation

The principle of mass conservation states that mass is neither created nor destroyed in a closed system. In thermodynamics, this principle is pivotal; it allows the calculation of mass changes within a system undergoing a process. In the exercise with refrigerant-134a, mass conservation is used to deduce the amount of refrigerant added to the tank by subtracting the initial mass from the final mass recorded.

Heat Transfer

Heat transfer is the movement of thermal energy from one body or substance to another due to a temperature difference. It can occur via conduction, convection, or radiation. In thermodynamic systems like refrigeration cycles, calculating the heat transfer is crucial for understanding the energy dynamics of the system. The exercise requires us to use an energy conservation equation, taking internal energy changes and the enthalpy of the refrigerant into account to determine the quantity of heat added or removed during the refrigerant intake process.