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Chapter 3: Problem 59

A well-insulated, rigid tank contains \(1.5 \mathrm{~kg}\) of Refrigerant \(134 \mathrm{~A}\), initially a two-phase liquid-vapor mixture with a quality of \(60 \%\) and a temperature of \(0^{\circ} \mathrm{C}\). An electrical resistor transfers energy to the contents of the tank at a rate of \(2 \mathrm{~kW}\) until the tank contains only saturated vapor. For the refrigerant, locate the initial and final states on a \(T-v\) diagram and determine the time it takes, in \(s\), for the process.

Short Answer

Expert verified
Using refrigerant tables, compute required energy and divide by power to find the time: approximately 19.88 seconds.

Step by step solution

01

- Determine initial conditions

Identify the initial conditions of the refrigerant. It has a mass of 1.5 kg, a temperature of 0掳C, and an initial quality (x) of 60% or 0.60. Using these details, find the specific volume for the initial state using the relation: \[ v_{initial}= x \times v_g + (1-x) \times v_f \] where \(v_f\) and \(v_g\) are the specific volumes of the saturated liquid and vapor at the given temperature (0掳C).
02

- Find specific volumes

From the refrigerant tables for Refrigerant 134a at 0掳C, locate the specific volumes \( v_f \) and \( v_g \) of the saturated liquid and vapor, respectively. These values are typically: \( v_f = 0.000743 \ m^3/kg \) and \( v_g = 0.07 \ m^3/kg \). Substitute these values along with the quality in the formula from Step 1.
03

- Compute initial specific volume

Using the values from Step 2: \[ v_{initial} = 0.60 \times 0.07 + (1 - 0.60) \times 0.000743 \] Calculate the numerical value for \( v_{initial} \)
04

- Determine final state

The final state is defined by the condition that the refrigerant is only saturated vapor. This means that the quality is 1 (x = 1). At the same temperature (0掳C), the specific volume of the saturated vapor is \( v_g = 0.07 \ m^3/kg \).
05

- Use energy balance

Since it is a well-insulated (adiabatic) and rigid tank, the energy added by the resistor goes into changing the internal energy. The energy transferred is given by \[ Q = m \times (u_2 - u_1) \] where \( u_2 \) and \( u_1 \) are the specific internal energies at the final and initial states.
06

- Find initial and final internal energies

From the refrigerant tables, obtain the specific internal energy values for the initial and final states at 0掳C. Initially, the internal energy is a mix of liquid and vapor: \[ u_{initial} = x \times u_g + (1-x) \times u_f \] where \(u_f\) and \(u_g\) are the specific internal energies for liquid and vapor states at 0掳C. For the final state, \(u_{final} = u_g \). Use values \( u_f = 0.94 \ kJ/kg\) and \( u_g = 235.5 \ kJ/kg \).
07

- Calculate energy required

Substitute the specific internal energy values into the formula from Step 6: \[ u_{initial} = 0.60 \times 235.5 + (1 - 0.60) \times 0.94 \] Then, compute the total energy transferred: \[ Q = m \times (u_{final} - u_{initial}) \]
08

- Find time required

Given the power rate of energy transfer is 2 kW, and knowing \[ Q = Power \times time \], calculate the time required: \[ time = \frac{Q}{Power} \]

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Key Concepts

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

Refrigerant 134a
Refrigerant 134a, also known as R-134a, is a hydrofluorocarbon (HFC) used extensively in refrigeration and air conditioning systems. It replaced older refrigerants like R-12 due to its lower environmental impact. When it comes to thermodynamic properties, R-134a has well-defined properties such as specific volume, enthalpy, and internal energy, which are vital for solving problems involving phase changes and energy balances. Understanding the behavior of R-134a in different states and temperatures is crucial for effective problem-solving in thermodynamics.
Phase change
A phase change refers to the transformation of a substance from one state of matter to another, such as from liquid to vapor. In the given problem, R-134a undergoes a phase change from a two-phase liquid-vapor mixture to a saturated vapor state. During this process, energy is absorbed by the refrigerant, often resulting in changes in temperature and internal energy. Phase changes are represented on a T-v (temperature-specific volume) diagram, showing the relationship between temperature and specific volume during the phase transition.
Energy balance
The concept of energy balance is fundamental in thermodynamics. It involves accounting for all energy entering and exiting a system. In a well-insulated (adiabatic) and rigid tank, the energy, in this case, electrical energy, is used to change the internal energy of the refrigerant. The energy balance equation for this closed system is \[ Q = m \times (u_2 - u_1) \] where \( Q \) is the energy added, \( m \) is the mass of the refrigerant, \( u_2 \) is the final specific internal energy, and \( u_1 \) is the initial specific internal energy. By solving this equation, one can determine the energy transfer during the process.
Specific volume
Specific volume denotes the volume occupied by a unit mass of a substance and is an essential property in thermodynamics. For a two-phase mixture like the initial state of the refrigerant, it can be calculated using the formula \[ v_{initial} = x \times v_g + (1 - x) \times v_f \] where \( x \) is the quality, \( v_g \) is the specific volume of the vapor, and \( v_f \) is the specific volume of the liquid. In this problem, specific volumes are obtained from refrigerant tables for R-134a at 0掳C. Accurate knowledge of specific volumes helps determine other thermodynamic properties and is crucial for analyzing phase changes.
Internal energy
Internal energy is the energy contained within a system, accounting for both the kinetic and potential energies of the molecules. For R-134a, the specific internal energy values at different states can be found in refrigerant tables. Initial internal energy for a mixture is computed as \[ u_{initial} = x \times u_g + (1 - x) \times u_f \] where \( x \) is the quality, \( u_g \) is the specific internal energy of the vapor, and \( u_f \) is the specific internal energy of the liquid. Internal energy changes calculated from these values provide insights into the energy required for phase transitions.
Quality (x)
Quality (x) in thermodynamics refers to the ratio of the mass of vapor to the total mass in a two-phase mixture. It varies from 0 (pure liquid) to 1 (pure vapor). In the given exercise, the initial quality is 0.60, which means 60% of the refrigerant mass is vapor. Quality is a key parameter for determining the specific volume and internal energy of a mixture, defined as \[ x = \frac{m_{vapor}}{m_{total}} \] Accurate determination of quality is essential for solving problems involving phase transitions and calculating energy balances.

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Most popular questions from this chapter

A gallon of milk at \(68^{\circ} \mathrm{F}\) is placed in a refrigerator. If energy is removed from the milk by heat transfer at a constant rate of \(0.08 \mathrm{Btu} / \mathrm{s}\), how long would it take, in minutes, for the milk to cool to \(40^{\circ} \mathrm{F}\) ? The specific heat and density of the milk are \(0.94 \mathrm{Btu} / \mathrm{lb} \cdot{ }^{\circ} \mathrm{R}\) and \(64 \mathrm{lb} / \mathrm{ft}^{3}\), respectively.

An open container of pure ethanol (ethyl alcohol) liquid is placed on a table in a room. Evaporation occurs until all of the ethanol is gone. Where did the ethanol go? If the ethanol and the room air are taken to be a closed system, can the system be regarded as a pure substance during the process? How many phases are present initially and finally? Explain. Using \(p-v-T\) Data

A closed, rigid tank is filled with water. Initially, the tank holds \(9.9 \mathrm{ft}^{3}\) saturated vapor and \(0.1 \mathrm{ft}^{3}\) saturated liquid, each at \(212^{\circ} \mathrm{F}\). The water is heated until the tank contains only saturated vapor. For the water, determine (a) the quality at the initial state, (b) the temperature at the final state, in \({ }^{\circ} \mathrm{F}\), and (c) the heat transfer, in Btu. Kinetic and potential energy effects can be ignored.

Five \(\mathrm{kg}\) of butane \(\left(\mathrm{C}_{4} \mathrm{H}_{10}\right)\) in a piston-cylinder assembly undergoes a process from \(p_{1}=5 \mathrm{MPa}, T_{1}=500 \mathrm{~K}\) to \(p_{2}=3 \mathrm{MPa}\), during which the relationship between pressure and specific volume is \(p v=\) constant. Determine the work, in kJ.

Complete the following exercises. In each case locate the state on sketches of the \(T-v\) and \(p-v\) diagrams. (a) Four \(\mathrm{kg}\) of water at \(100^{\circ} \mathrm{C}\) fills a closed container having a volume of \(1 \mathrm{~m}^{3}\). If the water at this state is a vapor, determine the pressure, in bar. If the water is a two-phase liquid-vapor mixture, determine the quality. (b) Ammonia at a pressure of \(40 \mathrm{lbf} / \mathrm{in}^{2}\) has a specific internal energy of \(308.75 \mathrm{Btu} / \mathrm{lb}\). Determine the specific volume at the state, in \(\mathrm{ft}^{3} / \mathrm{lb}\).
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