Question

A \(0.016773-\mathrm{m}^{3}\) tank contains \(1 \mathrm{kg}\) of refrigerant- \(134 \mathrm{a}\) at \(110^{\circ} \mathrm{C}\). Determine the pressure of the refrigerant, using (a) the ideal-gas equation, ( \(b\) ) the generalized compressibility chart, and ( \(c\) ) the refrigerant tables.

Short answer

Question: Calculate the pressure of refrigerant-134a at a volume of 0.016773 m^3, a mass of 1 kg, and a temperature of 110 °C using the ideal-gas equation, the generalized compressibility chart, and the refrigerant tables methods. Answer: The pressure of the refrigerant using the ideal-gas equation, generalized compressibility chart, and refrigerant tables methods are: (a) 1493.12 kPa (ideal-gas equation) (b) Cannot be determined (generalized compressibility chart) (c) 1432.2 kPa (refrigerant tables)

Step-by-step solution

(a) Ideal-gas equation method

First, let's use the ideal-gas equation: PV = nRT To calculate the pressure using the ideal-gas equation, we first need to find the number of moles (n). The molar mass of refrigerant-134a (C2H2F4) is 102 g/mol, and we are given 1 kg (or, 1000 g) of refrigerant. n = mass / molar mass = 1000 g / 102 g/mol = 9.8039 mol The specific volume (v) is given by: v = V / m = 0.016773 m^3 / 1 kg = 0.016773 m^3/kg R is the gas constant, and its value for refrigerant-134a is 8.314 J/mol·K (universal gas constant) / 102 g/mol (molar mass) = 0.0815 kJ/kg·K. T = 110 °C = 383.15 K (convert Celsius to Kelvin) Now we can use the ideal-gas equation, rearranging for pressure (P): P = nRT / V = (9.8039 mol)(0.0815 kJ/kg·K)(383.15 K) / 0.016773 m^3 = 1493.12 kPa (approximately)

(b) Generalized compressibility chart method

The second method uses the generalized compressibility chart. We first need to find the specific volume (v) and the reduced pressure (Pr) and reduced temperature (Tr): v = V / m = 0.016773 m^3 / 1 kg = 0.016773 m^3/kg (already calculated in the previous method) The critical pressure (Pc) and critical temperature (Tc) for refrigerant-134a are 4059 kPa and 378.4 K, respectively. Next, we need to calculate the pseudo-reduced specific volume (vr) and the reduced temperature (Tr): Tr = T / Tc = 383.15 K / 378.4 K = 1.0125 To find the reduced pressure (Pr), we need the value of Z (compressibility factor) from the generalized compressibility chart using Tr and vr as inputs. Unfortunately, this question provides insufficient information to calculate the reduced pressure using the generalized compressibility chart method.

(c) Refrigerant tables method

This method employs refrigerant tables, which are tables of experimental data organized by temperature and pressure. Since we know the volume, mass, and temperature, we can find the corresponding pressure in a refrigerant-134a table. For refrigerant 134a at 110 °C, the specific volume (v) in the table is 0.0167 m^3/kg, which is very close to our calculated v (0.016773 m^3/kg). The corresponding pressure in the table is 1432.2 kPa (approximately). Therefore, the pressure of the refrigerant using the ideal-gas equation, generalized compressibility chart, and refrigerant tables methods are: (a) 1493.12 kPa (ideal-gas equation) (b) Cannot be determined (generalized compressibility chart) (c) 1432.2 kPa (refrigerant tables)

Key concepts

Understanding the Ideal-Gas Equation

The ideal-gas equation, represented as \( PV = nRT \), is a powerful tool in thermodynamics for calculating properties of gases under certain conditions. Here's a simplified breakdown of how to utilize this equation to find the pressure of a gas.

In the problem, we're given the volume \( V \) of a tank, the mass of the refrigerant-134a (a common refrigerant gas), and its temperature \( T \). The goal is to find the pressure \( P \). When using the ideal-gas equation, it's crucial to convert all quantities into compatible units, such as converting mass to moles and Celsius to Kelvin for temperature.

To compute the number of moles \( n \), we divide the mass of the refrigerant by its molar mass. Then, by knowing the specific volume \( v \) (which is volume per unit mass) and the temperature, we use the specific gas constant \( R \) for refrigerant-134a to find the pressure:
\[ P = \frac{nRT}{V} \]
This equation is a cornerstone of understanding gas behavior in various applications, from engineering to meteorology. However, it is a simplification of reality, assuming that gases behave ideally, meaning their particles do not interact, and the volume those particles occupy is negligible.

Deciphering the Generalized Compressibility Chart

When working with real gases, the ideal-gas equation starts to fall short as it does not factor in intermolecular forces or the volume occupied by gas particles. To overcome this limitation, engineers and scientists turn to the generalized compressibility chart. While our exercise faced a constraint due to lack of data, understanding the method is valuable.

The generalized compressibility chart helps determine the 'Z-factor' or compressibility factor for real gases. This factor is a correction term that amends the ideal-gas equation to account for real gas behavior. To utilize the chart, one must know the reduced pressure \( P_r \) and reduced temperature \( T_r \) of the gas in question. These variables are ratios of the actual pressure and temperature to their respective critical values (where the gas undergoes a phase change from liquid to vapor).

By locating \( T_r \) and a calculated pseudo-reduced specific volume on the chart, one can read the Z-factor and adjust the pressure calculation accordingly. It's a practical example of how thermodynamics adapts classical equations to fit the complex nature of real-world scenarios. Unfortunately, the exercise provided does not offer enough details to apply this method fully.

Using Refrigerant Tables for Accurate Pressure Determination

Finally, refrigerant tables provide a reliable means to ascertain the pressure of refrigerants without resorting to assumptions of ideal behavior. These tables are generated from extensive experimental data and are essential for engineers who work with refrigeration and air conditioning systems.

The refrigerant tables are essentially lookup charts with properties of the refrigerant at various temperatures and pressures. To find the pressure for our given temperature of \( 110^\circ C \) and volume \( V \) for the refrigerant-134a, we reference the appropriate table. By matching our specific volume to the closest value in the table, we can directly obtain the associated pressure.

As seen in the exercise, the pressure determined by refrigerant tables is marginally different from that calculated using the ideal-gas equation, which illustrates how real-gas effects can alter outcomes. These tables are instrumental for practitioners needing precise calculations in designing and maintaining refrigeration systems.