Question

A liquid \(R-134 a\) bottle has an internal volume of \(0.0015 \mathrm{m}^{3} .\) Initially it contains \(0.55 \mathrm{kg}\) of \(\mathrm{R}-134 \mathrm{a}\) (saturated mixture) at \(26^{\circ} \mathrm{C} .\) A valve is opened and \(\mathrm{R}-134 \mathrm{a}\) vapor only (no liquid) is allowed to escape slowly such that temperature remains constant until the mass of \(\mathrm{R}-134 \mathrm{a}\) remaining is \(0.15 \mathrm{kg} .\) Find the heat transfer necessary with the surroundings to maintain the temperature and pressure of the \(\mathrm{R}-134 \mathrm{a}\) constant.

Short answer

Answer: The heat transfer necessary with the surroundings is 90.145 kJ in the outward direction.

Step-by-step solution

Determine the initial and final saturated states of the R-134a.

Consult the R-134a property tables for the initial state at 26°C. At this temperature, the vapor and liquid phases coexist (saturated mixture), so we can extract the specific volume, v, and specific internal energy, u, of both the liquid \((f)\) and vapor \((g)\) phases. The initial state properties: - \(v_f = 0.0009757\, \mathrm{m^3/kg}\) - \(v_g = 0.02393\, \mathrm{m^3/kg}\) - \(u_f = 216.72\, \mathrm{kJ/kg}\) - \(u_g = 332.17\, \mathrm{kJ/kg}\) Now we need to find the quality, \(x\), of the mixture, which represents the fraction of the R-134a in the vapor phase. Use the formula for the mixture's specific volume: \(x = \frac{v_{mixture} - v_f}{v_g - v_f}\) Since we are given the mass of R-134a and the volume of the container, we can determine the mixture's specific volume \(v_{mixture}\): \(v_{mixture} = \frac{0.0015\, \mathrm{m^3}}{0.55\, \mathrm{kg}} = 0.002727\, \mathrm{m^3/kg}\) Plug the values into the formula to find the quality: \(x = \frac{0.002727 - 0.0009757}{0.02393 - 0.0009757} = 0.0803\) For the final state, there are only 0.15 kg of R-134a left in the container. The specific volume of the mixture will be different, but the quality remains the same since the temperature and pressure are constant: \(v_{mixture_{final}} = \frac{0.0015\, \mathrm{m^3}}{0.15\, \mathrm{kg}} = 0.010\, \mathrm{m^3/kg}\)

Calculate the initial and final internal energy.

Using the quality, \(x\), and the specific internal energies for the liquid and vapor phases, we can calculate the internal energy of the mixture for both initial and final states: - Initial internal energy: \(u_{initial} = u_f + x(u_g - u_f) = 216.72 + 0.0803(332.17 - 216.72) = 225.37\, \mathrm{kJ/kg}\) - Final internal energy: \(u_{final} = u_f + x(u_g - u_f) = 216.72 + 0.0803(332.17 - 216.72) = 225.37\, \mathrm{kJ/kg}\)

Find the heat transfer with surroundings.

Determine the total internal energy for both initial and final states: - Initial total internal energy: \(U_{initial} = m_{initial}u_{initial} = 0.55\, \mathrm{kg} \times 225.37\, \mathrm{kJ/kg} = 123.95\, \mathrm{kJ}\) - Final total internal energy: \(U_{final} = m_{final}u_{final} = 0.15\, \mathrm{kg} \times 225.37\, \mathrm{kJ/kg} = 33.805\, \mathrm{kJ}\) Since the only heat transfer is with the surroundings, we can use the formula \(Q = U_{final} - U_{initial}\). Therefore, the heat transfer necessary to maintain the temperature and pressure constant is: \(Q = 33.805\, \mathrm{kJ} - 123.95\, \mathrm{kJ} = -90.145\, \mathrm{kJ}\) The negative sign denotes that heat is transferred from the R-134a to the surroundings. So, the heat transfer necessary with the surroundings to maintain the temperature and pressure of R-134a constant is 90.145 kJ in the outward direction.

Key concepts

R-134a Refrigerant Properties

Refrigerants like R-134a are commonly used in various cooling systems, including refrigerators and air conditioners. R-134a has significant properties that make it suitable for these applications, such as having a boiling point that allows it to evaporate at lower temperatures, effectively absorbing heat from the surroundings.

The properties of R-134a, such as pressure, temperature, specific volume, and internal energy, are documented in standard refrigerant tables. These properties are essential for understanding the behavior of R-134a under different conditions, and they are crucial to the calculations involving heat transfer systems. Working with R-134a involves careful consideration of these properties to design efficient and safe refrigeration systems.

Saturated Mixture

A saturated mixture exists when a substance is at a temperature and pressure where liquid and vapor phases coexist in equilibrium. This state is particularly important in thermodynamics because it represents a common situation in refrigeration cycles.

In the context of refrigerants like R-134a, when a bottle contains a saturated mixture, it means there is a balance between the liquid and gaseous forms of the substance. This balance is critical in determining the refrigerant's specific volume (the volume per unit mass) and the calculation of internal energy, which are both necessary to determine the performance of a refrigeration cycle.

Specific Volume

Specific volume is a critical concept in thermodynamics, defined as the volume occupied by a unit mass of a substance. It is a property that helps to characterize the state of a substance in either the liquid or vapor phase, or as part of a saturated mixture.

Knowing the specific volume of R-134a in different states allows for the calculation of other important parameters, such as the quality of the mixture. In refrigeration processes, the specific volume is used to understand how much space the refrigerant occupies and how it will change as it absorbs or releases heat.

Internal Energy Calculation

Internal energy is a measure of the total energy contained within a system. It encompasses the kinetic energy of molecules and the potential energy of molecular interactions. For substances like R-134a, calculations of internal energy must consider both the liquid and vapor phases of the refrigerant, especially within a saturated mixture.

To calculate the internal energy of a mixture, we use the quality (the ratio of the mass of vapor to the total mass) and the specific internal energies of the saturated liquid and vapor. As the refrigerant undergoes phase changes, energy is transferred in the form of heat, and these internal energy calculations are pivotal in understanding the overall heat transfer necessary to maintain certain conditions within the refrigeration cycle.