Question

Refrigerant \(134 \mathrm{a}\) enters an air conditioner compressor at \(3.2\) bar, \(10^{\circ} \mathrm{C}\), and is compressed at steady state to \(10 \mathrm{bar}, 70^{\circ} \mathrm{C}\). The volumetric flow rate of refrigerant entering is \(3.0 \mathrm{~m}^{3} / \mathrm{min}\). The power input to the compressor is \(55.2 \mathrm{~kJ}\) per \(\mathrm{kg}\) of refrigerant flowing. Neglecting kinetic and potential energy effects, determine the heat transfer rate, in \(\mathrm{kW}\).

Short answer

Heat transfer rate in \text{kW} is \text{Total Power Input} - \text{Mass flow rate} \times (h_2 - h_1).

Step-by-step solution

- Identify the Given Data

Given:1. Initial pressure, \( P_1 = 3.2 \text{ bar} \) 2. Initial temperature, \( T_1 = 10^{\text{°C}} \) 3. Final pressure, \( P_2 = 10 \text{ bar} \) 4. Final temperature, \( T_2 = 70^{\text{°C}} \) 5. Volumetric flow rate, \( \text{Volumetric Flow Rate} = 3.0 \text{ } \text{m}^3 \text{/min} \) 6. Power input, \( \text{Power input} = 55.2 \text{ kJ/kg} \)

- Look up Refrigerant Properties

Using refrigerant tables or property software, find the specific enthalpies of the refrigerant at the given initial and final conditions: Initial state: \( h_1 \) at \( 3.2 \text{ bar} \) and \( 10^{\text{°C}} \).Final state: \( h_2 \) at \( 10 \text{ bar} \) and \( 70^{\text{°C}} \).

- Determine Mass Flow Rate

First, we need the specific volume \( v_1 \) of the refrigerant at the initial state (use refrigerant tables):\( v_1 = \text{specific volume at} \) \( 3.2 \text{ bar} \text{ and } 10^{\text{°C}} \).Then, calculate the mass flow rate: \[ \text{Mass flow rate} (\frac{\text{kg}}{\text{s}}) = \frac{\text{Volumetric Flow Rate} (\frac{\text{m}^3}{\text{min}})} {v_1 \text{ }(\frac{\text{m}^3}{\text{kg}})} \times \frac{1}{60} \]

- Calculate Work Done and Heat Transfer

The power input to the compressor is given per unit mass, so we multiply by the mass flow rate to get the total power input:\[ \text{Total Power Input} (\text{kW}) = \text{Power input per unit mass} (\frac{\text{kJ}}{\text{kg}}) \times \text{Mass flow rate} (\frac{\text{kg}}{\text{s}}) \]Using the First Law of Thermodynamics for a steady-state control volume:\[ \text{Total Power Input} (\text{kW}) = \text{Mass flow rate} \times (h_2 - h_1) + \text{Heat transfer rate} (\text{kW}) \]Rearrange to solve for the heat transfer rate:\[ \text{Heat transfer rate} (\text{kW}) = \text{Total Power Input} - \text{Mass flow rate} \times (h_2 - h_1) \]

Key concepts

Thermodynamics

Thermodynamics is the study of energy and its transformations. It focuses on how energy moves between systems and the work they perform. In this exercise, an air conditioner compressor handles refrigerant 134a, transforming it from one state to another. By studying thermodynamics, we understand the energy changes, pressures, and temperatures during the compression process. This system obeys laws like the First Law of Thermodynamics.

Refrigerant Properties

Refrigerants are special substances used in cooling systems like air conditioners to absorb and release heat. Refrigerant 134a is common for its efficiency and environmental friendliness. Understanding its properties, such as pressure and temperature at various states (3.2 bar, 10°C initially; 10 bar, 70°C finally), is vital. These properties are found using refrigerant tables and are crucial for calculating specific enthalpy and volume, which are needed for further thermodynamic calculations.

First Law of Thermodynamics

The First Law of Thermodynamics, also known as the conservation of energy principle, is a key concept here. It states that energy cannot be created or destroyed, only transformed. In our compressor, the power input is transformed into work and heat. The equation governing this is\[\text{Total Power Input} = \dot{m}(h_2 - h_1) + \text{Heat transfer rate}\]here, \( \dot{m} \) stands for mass flow rate, and \( h_1 \) and \( h_2 \) are the specific enthalpies at initial and final states respectively. We use this equation to determine how much heat is added to or removed from the system during compression.

Mass Flow Rate Calculation

Calculating mass flow rate is a crucial step. It defines how much refrigerant enters the system per unit time. Given the volumetric flow rate of 3.0 m³/min, we find the specific volume (\( v_1 \) ) from refrigerant tables, then use it in the formula:\[\text{Mass flow rate} = \frac{\text{Volumetric Flow Rate}} {v_1} \times \frac{1}{60}\]This conversion ensures we have a consistent unit, kilograms per second (kg/s), to use in subsequent calculations, like determining the total power input.

Heat Transfer Rate

Finally, we reach the heat transfer rate, describing how much heat is removed or added to the air conditioner system during compression. By rearranging the First Law of Thermodynamics equation, we solve for heat transfer rate as:\[\text{Heat transfer rate} = \text{Total Power Input} - \dot{m}(h_2 - h_1)\]This calculation tells us the efficiency and effectiveness of the air conditioner's compressor in transferring heat, ensuring optimal performance and energy use.