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A \(0.3-m^{3}\) rigid tank initially contains refrigerant\(134 \mathrm{a}\) at \(14^{\circ} \mathrm{C}\). At this state, 55 percent of the mass is in the vapor phase, and the rest is in the liquid phase. The tank is connected by a valve to a supply line where refrigerant at \(1.4 \mathrm{MPa}\) and \(100^{\circ} \mathrm{C}\) flows steadily. Now the valve is opened slightly, and the refrigerant is allowed to enter the tank. When the pressure in the tank reaches \(1 \mathrm{MPa}\), the entire refrigerant in the tank exists in the vapor phase only. At this point the valve is closed. Determine ( \(a\) ) the final temperature in the tank, \((b)\) the mass of refrigerant that has entered the tank, and \((c)\) the heat transfer between the system and the surroundings.

Short Answer

Expert verified
Answer: Based on the initial state of the system, we can calculate the mass of vapor and liquid by using the given volume of the tank, the temperature, and the percentage of mass in the vapor phase. We will use the Refrigerant-134a properties tables to obtain the specific volume of each phase.

Step by step solution

01

Initial state of the system

Calculate the mass of vapor and liquid in the initial state. Use the given information (volume of tank, temperature, percentage of mass in the vapor phase) and table lookup in Refrigerant-134a properties tables to get the specific volume of each phase.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Refrigerant Phase Change
Understanding the phase change of refrigerants like R-134a is crucial for various applications, especially in cooling systems. Refrigerants are substances that absorb heat at low temperatures and pressures and release heat at higher temperatures and pressures. R-134a, a common refrigerant, undergoes this phase change cycle in many refrigeration systems.

A phase change occurs when a refrigerant transitions between different states of matter—liquid, vapor, and sometimes solid. In most residential and automotive air conditioning systems, the refrigerant cycles between the liquid and vapor phases. During the cycle, when a refrigerant evaporates, it absorbs heat, providing a cooling effect. Conversely, when it condenses, it releases heat to the surroundings. This is the principle behind most cooling systems.

It's important to note that the phase change is dictated by the refrigerant's temperature and pressure. The exercise under consideration involved R-134a undergoing a phase change solely from liquid and vapor phases. The tank starts with a mixture and ends in a vapor-only state, a process which requires careful manipulation of the system's pressure and temperature.
Specific Volume Calculation
Specific volume is a thermodynamic property that is particularly relevant in phase change calculations. It represents the volume occupied by a unit mass of a substance and is typically expressed in cubic meters per kilogram (\( m^3/kg \)). For a given mass of refrigerant in different phases—liquid or vapor—the specific volume informs us about the space it occupies at a certain temperature and pressure.

Calculation of specific volume often involves a properties table that lists the thermodynamic properties for the substance at different temperatures and pressures. By knowing the state of the refrigerant (e.g., temperature, pressure, phase), we can refer to these tables to find the corresponding specific volume. In the exercise, the specific volume was essential to determine the initial mass of vapor and liquid in the tank, given its volume. We match the given conditions to the closest values in the properties table, calculate the masses, and use this information to solve for other unknown variables, such as the mass of refrigerant that entered the tank and the heat transfer.
R-134a Properties Table
The R-134a properties table is an indispensable tool for any task involving this refrigerant. Thermodynamic properties such as pressure, temperature, specific volume, enthalpy, and entropy are tabulated for both saturated liquid and vapor states. Moreover, superheated vapor data are also included for conditions beyond the saturated vapor curve.

For calculations, we often use a simplified version of the R-134a table that provides values at specific temperatures and pressures. These properties are obtained from detailed thermodynamic analysis and experimentation. In our exercise, the table helps identify the refrigerant's specific volume in both the vapor and liquid phases at a given temperature. Accurate use of the properties table is crucial for solving the problem, ensuring that the correct phase information and thermodynamic properties are applied to find the thermodynamic state of the refrigerant throughout the process.

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Most popular questions from this chapter

Steam at \(1 \mathrm{MPa}\) and \(300^{\circ} \mathrm{C}\) is throttled adiabatically to a pressure of 0.4 MPa. If the change in kinetic energy is negligible, the specific volume of the steam after throttling is \((a) 0.358 \mathrm{m}^{3} / \mathrm{kg}\) (b) \(0.233 \mathrm{m}^{3} / \mathrm{kg}\) \((c) 0.375 \mathrm{m}^{3} / \mathrm{kg}\) \((d) 0.646 \mathrm{m}^{3} / \mathrm{kg}\) \((e) 0.655 \mathrm{m}^{3} / \mathrm{kg}\)

A sealed electronic box is to be cooled by tap water flowing through the channels on two of its sides. It is specified that the temperature rise of the water not exceed \(4^{\circ} \mathrm{C}\) The power dissipation of the box is \(2 \mathrm{kW}\), which is removed entirely by water. If the box operates 24 hours a day, 365 days a year, determine the mass flow rate of water flowing through the box and the amount of cooling water used per year.

The evaporator of a refrigeration cycle is basically a heat exchanger in which a refrigerant is evaporated by absorbing heat from a fluid. Refrigerant-22 enters an evaporator at \(200 \mathrm{kPa}\) with a quality of 22 percent and a flow rate of 2.65 L/h. \(R-22\) leaves the evaporator at the same pressure superheated by \(5^{\circ} \mathrm{C}\). The refrigerant is evaporated by absorbing heat from air whose flow rate is \(0.75 \mathrm{kg} / \mathrm{s}\). Determine (a) the rate of heat absorbed from the air and ( \(b\) ) the temperature change of air. The properties of \(R-22\) at the inlet and exit of the condenser are \(h_{1}=220.2 \mathrm{kJ} / \mathrm{kg}, v_{1}=0.0253 \mathrm{m}^{3} / \mathrm{kg}\) and \(h_{2}=398.0 \mathrm{kJ} / \mathrm{kg}\).

A tank with an internal volume of \(1 \mathrm{m}^{3}\) contains air at \(800 \mathrm{kPa}\) and \(25^{\circ} \mathrm{C}\). A valve on the tank is opened allowing air to escape and the pressure inside quickly drops to \(150 \mathrm{kPa}\), at which point the valve is closed. Assume there is negligible heat transfer from the tank to the air left in the tank. (a) Using the approximation \(h_{e} \approx\) constant \(=h_{e, \mathrm{avg}}=\) \(0.5\left(h_{1}+h_{2}\right),\) calculate the mass withdrawn during the process. (b) Consider the same process but broken into two parts. That is, consider an intermediate state at \(P_{2}=400 \mathrm{kPa}\), calculate the mass removed during the process from \(P_{1}=800 \mathrm{kPa}\) to \(P_{2}\) and then the mass removed during the process from \(P_{2}\) to \(P_{3}=150 \mathrm{kPa},\) using the type of approximation used in part \((a),\) and add the two to get the total mass removed. (c) Calculate the mass removed if the variation of \(h_{e}\) is accounted for.

A 3-ft' rigid tank initially contains saturated water vapor at \(300^{\circ} \mathrm{F}\). The tank is connected by a valve to a supply line that carries steam at 200 psia and \(400^{\circ} \mathrm{F}\). Now the valve is opened, and steam is allowed to enter the tank. Heat transfer takes place with the surroundings such that the temperature in the tank remains constant at \(300^{\circ} \mathrm{F}\) at all times. The valve is closed when it is observed that one-half of the volume of the tank is occupied by liquid water. Find (a) the final pressure in the tank, ( \(b\) ) the amount of steam that has entered the \(\tan \mathrm{k},\) and \((c)\) the amount of heat transfer.

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