Question

Two kilograms of Refrigerant 22 undergo a process for which the pressure- volume relation is \(p v^{1.05}=\) constant. The initial state of the refrigerant is fixed by \(p_{1}=2\) bar, \(T_{1}=-20^{\circ} \mathrm{C}\), and the final pressure is \(p_{2}=10\) bar. Calculate the work for the process, in \(\mathrm{kJ}\).

Short answer

The work done for the process is 13.82 \times 10^2 v_1 kJ.

Step-by-step solution

Identify the Given Values

Identify the initial conditions and the relationship given in the problem. Here \(\begin{aligned} p_1 &= 2 \text{ bar} \ T_1 &= -20^{\circ} \text{C} \ m &= 2 \text{ kg} \ p_2 &= 10 \text{ bar} \ p v^{1.05} &= \text{constant} \end{aligned} \).

Convert Pressure Units

Convert the given pressures from bar to pascals (Pa) for calculation purposes. \(1 \text{ bar} = 10^5 \text{ Pa} \) \(\begin{aligned} p_1 &= 2 \times 10^5 \text{ Pa} \ p_2 &= 10 \times 10^5 \text{ Pa} \end{aligned}\).

Determine the Volume Relation

Use the provided pressure-volume relation \(p v^{1.05} = \text{constant} \) to express the volumes at states 1 and 2. \(\begin{aligned} p_1 v_1^{1.05} &= p_2 v_2^{1.05} \ \text{or} \ (2 \times 10^5) v_1^{1.05} &= (10 \times 10^5) v_2^{1.05} \ \text{which simplifies to:} \ v_1^{1.05} &= 5 v_2^{1.05} \ \text{or} \ v_2 &= v_1 / 5^{1/1.05} \).

Work for Polytropic Process

Use the formula for work done in a polytropic process: \( W = \frac{p_2 v_2 - p_1 v_1}{1 - n} \) where \( n = 1.05 \). \( W = \frac{(10 \times 10^5) v_2 - (2 \times 10^5) v_1}{1 - 1.05} \)

Simplify Work Expression

Using \(v_2 = v_1 / 5^{1/1.05} \), we have: \( W = \frac{(10 \times 10^5) \frac{v_1}{5^{1/1.05}} - (2 \times 10^5) v_1}{1 - 1.05} \).

Calculate Numerical Value

Finally, substitute, and simplify the expression to get the numeric value for the work done: \( W = \frac{(10 \times 10^5 \times v_1 / 5^{1/1.05}) - (2 \times 10^5 \times v_1)}{-0.05} \). After some algebraic manipulation: \( W = \frac{-691.36 v_1 \times 10^5}{-0.05} \). And \( W = 13.82 \times 10^5 v_1 \) in Joules.

Convert to kJ

Since \( 1 \text{ kJ} = 10^3 \text{ J} \), finally the work done can be expressed as: \( W = 13.82 \times 10^2 v_1 \text{ kJ} \).

Key concepts

Refrigerant 22

Refrigerant 22, also known as R-22, is a commonly used refrigerant in air conditioning and refrigeration systems. This chemical compound plays a crucial role in the refrigeration cycle by absorbing and releasing heat. Due to its effective thermal properties, it's widely used in home and industrial cooling systems. However, due to environmental concerns, its usage has been regulated in many regions to reduce ozone depletion. When dealing with Refrigerant 22 in thermodynamic problems, it is essential to be familiar with its properties, such as temperature, pressure, and specific volumes, at various states to accurately calculate processes involving energy transfers.

Polytropic Process

A polytropic process is a type of thermodynamic process that follows the equation \[p v^n = \text{constant}\]. This relationship signifies that during the process, the product of pressure (\(p\)) and volume raised to a specific exponent (\(n\)) remains constant. The value of \(n\) varies depending on the type of process:

• If \(n = 0\), it is an isobaric process (constant pressure).
• If \(n = 1\), it is an isothermal process (constant temperature).
• If \(n = \text{specific heat ratio}\), it results in an adiabatic process (no heat transfer).

In this particular exercise, \(n = 1.05\), meaning it is a polytropic process unique to the characteristics between isothermal and adiabatic processes. Understanding the value and interpretation of \(n\) is critical for correctly applying formulas and solving thermodynamic problems involving work, heat, and state changes.

Work Calculation

Calculating work in a thermodynamic process involves integrating the pressure-volume curve. For a polytropic process, the work done can be found using:\[W = \frac{p_2 v_2 - p_1 v_1}{1 - n}\]where \(p_1\) and \(p_2\) are the initial and final pressures, \(v_1\) and \(v_2\) are the initial and final volumes, and \(n\) is the polytropic index. This formula simplifies the process by directly relating the pressure and volume changes for specific exponent \(n\). It is important to convert units if necessary (e.g., from bar to Pascals) to ensure consistency in calculations. Substituting known values and performing algebraic manipulations will yield the work done during the process, typically expressed in Joules or kilojoules for larger quantities. This formula helps in understanding the energy interactions in various thermodynamic cycles and processes.

Pressure-Volume Relation

The pressure-volume (\(p-v\)) relation describes how the pressure of a gas varies with its volume in a given process. For a polytropic process, this relation is represented by \[ p v^n = \text{constant}\]. This constant relation helps track the state changes precisely as the gas expands or compresses. In the given exercise, we have \( p_1 v_1^{1.05} = p_2 v_2^{1.05}\), indicating that both states conform to this polytropic relationship. Solving this relation often involves determining one unknown volume if other variables are known, using the fact that the ratio between voluminous states raised to exponent \(n\) can equal the inverse ratio of their pressure states. This understanding is fundamental for accurate determination of work done and energy interactions in thermodynamic systems.