Problem 8 Asks To Determine The Volume Of The Tank And The Total Internal Energy Of A Rigid Tank Containing 1.5 Kg Of Refrigerant-134a At -10°C With A Quality Of 0.4. Problem 9 Discusses Water Vapor At 500 KPa And Needs More Information To Determine Its Properties.

Jun 18, 2025 by ADMIN 266 views

Determining Volume and Internal Energy of Refrigerant-134a in a Rigid Tank and Water Vapor Properties

Problem 8: Refrigerant-134a in a Rigid Tank

This problem explores the thermodynamic properties of refrigerant-134a within a closed system. Specifically, we are analyzing a rigid tank containing 1.5 kg of refrigerant-134a at a temperature of -10°C and a quality of 0.4. The objective is to determine (a) the volume of the tank and (b) the total internal energy of the system using refrigerant-134a tables. This problem is a classic example of applying thermodynamic principles and property tables to analyze the state of a substance. Understanding the concepts of quality, specific volume, and internal energy is crucial for solving this problem. The use of refrigerant-134a tables is essential as they provide the necessary data to determine the properties at the given state. This type of problem is fundamental in thermodynamics and is often encountered in applications involving refrigeration cycles and heat transfer systems.

To solve this problem effectively, one must first understand the concept of quality. Quality (x) represents the fraction of the total mass that is in the vapor phase in a saturated liquid-vapor mixture. A quality of 0.4 indicates that 40% of the mass is in the vapor phase, and the remaining 60% is in the liquid phase. This information is crucial because the specific volume and internal energy of the mixture will depend on the relative amounts of liquid and vapor. The specific volume (v) is the volume per unit mass, and the total volume (V) can be calculated by multiplying the specific volume by the total mass. The internal energy (u) represents the energy stored within the substance due to the kinetic and potential energies of its molecules. The total internal energy (U) is the product of the specific internal energy and the total mass. The use of refrigerant-134a tables is paramount in obtaining the saturated liquid and saturated vapor properties at the given temperature of -10°C. These properties include the specific volume of the saturated liquid (vf), the specific volume of the saturated vapor (vg), the specific internal energy of the saturated liquid (uf), and the specific internal energy of the saturated vapor (ug). Once these properties are obtained, the mixture properties can be calculated using the quality.

(a) Determining the Volume of the Tank:

To determine the volume of the tank, we need to calculate the specific volume of the refrigerant-134a mixture and then multiply it by the total mass. The specific volume of a saturated liquid-vapor mixture can be calculated using the following equation:

v = vf + x * (vg - vf)

Where:

v is the specific volume of the mixture
vf is the specific volume of the saturated liquid at -10°C
vg is the specific volume of the saturated vapor at -10°C
x is the quality (0.4)

From the refrigerant-134a tables, we find the following values at -10°C:

vf = 0.0007710 m³/kg
vg = 0.049407 m³/kg

Substituting these values into the equation, we get:

v = 0.0007710 m³/kg + 0.4 * (0.049407 m³/kg - 0.0007710 m³/kg)
v = 0.0007710 m³/kg + 0.4 * 0.048636 m³/kg
v = 0.0007710 m³/kg + 0.0194544 m³/kg
v = 0.0202254 m³/kg

Now that we have the specific volume, we can calculate the total volume (V) of the tank by multiplying the specific volume by the total mass:

V = m * v
V = 1.5 kg * 0.0202254 m³/kg
V = 0.0303381 m³

Therefore, the volume of the tank is approximately 0.0303 m³.

(b) Determining the Total Internal Energy of the System:

To determine the total internal energy of the system, we need to calculate the specific internal energy of the refrigerant-134a mixture and then multiply it by the total mass. The specific internal energy of a saturated liquid-vapor mixture can be calculated using the following equation:

u = uf + x * (ug - uf)

Where:

u is the specific internal energy of the mixture
uf is the specific internal energy of the saturated liquid at -10°C
ug is the specific internal energy of the saturated vapor at -10°C
x is the quality (0.4)

From the refrigerant-134a tables, we find the following values at -10°C:

uf = 38.26 kJ/kg
ug = 222.95 kJ/kg

Substituting these values into the equation, we get:

u = 38.26 kJ/kg + 0.4 * (222.95 kJ/kg - 38.26 kJ/kg)
u = 38.26 kJ/kg + 0.4 * 184.69 kJ/kg
u = 38.26 kJ/kg + 73.876 kJ/kg
u = 112.136 kJ/kg

Now that we have the specific internal energy, we can calculate the total internal energy (U) of the system by multiplying the specific internal energy by the total mass:

U = m * u
U = 1.5 kg * 112.136 kJ/kg
U = 168.204 kJ

Therefore, the total internal energy of the system is approximately 168.204 kJ.

In summary, by utilizing the refrigerant-134a tables and applying the concepts of quality, specific volume, and specific internal energy, we have successfully determined the volume of the tank and the total internal energy of the system. This problem highlights the importance of understanding thermodynamic properties and their application in analyzing closed systems. The solution involved a step-by-step approach, first calculating the specific volume and then the total volume, followed by calculating the specific internal energy and then the total internal energy. This method can be applied to similar problems involving other substances and different thermodynamic conditions.

Problem 9: Water Vapor at 500 kPa

This problem shifts our focus to the properties of water vapor at a specific pressure. The problem statement only provides the pressure (500 kPa) and lacks additional information such as temperature or quality. To fully define the state of the water vapor and determine its other properties, such as temperature, specific volume, enthalpy, and entropy, we need at least one more independent property. Without this additional information, we can only discuss the general behavior of water vapor at 500 kPa and explore possible scenarios. The state of a substance is defined by its independent properties. For a simple compressible system, two independent and intensive properties are required to completely define the state. Pressure and temperature are usually independent properties, but in the saturation region, they become dependent. Quality is another property that defines the state in the saturation region. The critical importance of having sufficient information to define a system's state is a recurring theme in thermodynamics, as it dictates the accuracy and reliability of subsequent calculations and analyses.

At a pressure of 500 kPa, water can exist in different phases: compressed liquid, saturated liquid, saturated mixture (liquid-vapor), saturated vapor, or superheated vapor. The phase depends on the temperature. To determine the phase, we can consult the saturated water tables. At 500 kPa, the saturation temperature (Ts) is approximately 151.83°C. If the temperature of the water is below 151.83°C, it exists as a compressed liquid. If the temperature is equal to 151.83°C, it can exist as a saturated liquid, a saturated mixture, or a saturated vapor, depending on the quality. If the temperature is above 151.83°C, it exists as a superheated vapor. Without knowing the temperature, we cannot pinpoint the exact state. However, we can discuss the properties in different scenarios. If the water is in the saturation region, we need the quality (x) to determine the other properties. The specific volume (v), internal energy (u), enthalpy (h), and entropy (s) can be calculated using the following equations:

v = vf + x * (vg - vf)
u = uf + x * (ug - uf)
h = hf + x * (hg - hf)
s = sf + x * (sg - sf)

Where the subscript 'f' denotes the saturated liquid state, and 'g' denotes the saturated vapor state at 500 kPa. These values can be found in the saturated water tables. If the water is in the superheated region, we need the temperature to determine the other properties from the superheated water tables. The superheated water tables provide values for specific volume, internal energy, enthalpy, and entropy at various temperatures and pressures. Interpolation may be required if the exact pressure and temperature are not listed in the tables. In the compressed liquid region, the properties are less sensitive to pressure changes and are often approximated by the saturated liquid properties at the given temperature. However, for more accurate results, compressed liquid tables can be used if available. This flexibility in approach is what makes thermodynamics both a powerful and nuanced tool in engineering analysis.

In conclusion, without additional information such as temperature or quality, we can only discuss the possible states of water vapor at 500 kPa. The water could be in the compressed liquid, saturated mixture, or superheated vapor phase, each with distinct properties. To fully determine the state and its properties, another independent property is required. This underscores the fundamental principle that two independent intensive properties are necessary to fully define the state of a simple compressible system. This problem serves as a reminder of the importance of having sufficient information before attempting to solve thermodynamic problems and highlights the practical application of property tables in determining the state of a substance.