Question

An ideal vapor-compression refrigeration cycle, with ammonia as the working fluid, has an evaporator temperature of \(-20^{\circ} \mathrm{C}\) and a condenser pressure of 12 bar. Saturated vapor enters the compressor, and saturated liquid exits the condenser. The mass flow rate of the refrigerant is \(3 \mathrm{~kg} / \mathrm{min}\). Determine (a) the coefficient of performance. (b) the refrigerating capacity, in tons.

Short answer

COP = [value], Refrigerating Capacity = [value] tons.

Step-by-step solution

- Identify Given Data

Extract the known values from the problem statement: - Evaporator temperature: - Condenser pressure: - Mass flow rate: 3 kg/min

- Find Thermodynamic Properties

Use thermodynamic tables to find the properties of ammonia. For saturated vapor at For saturated liquid at

- Work Done by Compressor

Calculate the work done by the compressor in the refrigeration cycle. The work is given by:

- Heat Absorbed in Evaporator

Find the heat absorbed in the evaporator. This value can be calculated using:

- Coefficient of Performance

Calculate the coefficient of performance (COP) using the formula: \[COP = \frac{Q_{in}}{W_{comp}}\]

- Refrigerating Capacity

Convert the refrigerating capacity to tons. This is given by:

- Final Calculations and Answers

Consolidate all the calculations to determine the final values of COP and refrigerating capacity in tons.

Key concepts

Thermodynamic Properties

Thermodynamic properties are essential in understanding any refrigeration cycle, like the vapor-compression cycle described here. These properties include the state of the refrigerant (like ammonia), such as pressure, temperature, and specific enthalpy at different points in the cycle. For the refrigeration cycle:

• We start with given values: -20°C for the evaporator temperature and 12 bar for the condenser pressure.
• To proceed, we use thermodynamic tables to find the specific enthalpies for the saturated vapor entering the compressor and the saturated liquid exiting the condenser.

Knowing these properties helps us calculate the work done by the compressor and the heat absorbed in the evaporator, which are crucial for further calculations.

Coefficient of Performance

The Coefficient of Performance (COP) is a measure of a refrigeration cycle’s efficiency. It tells us how effectively the refrigeration cycle uses work to move heat. The COP is calculated using the formula \[\text{COP} = \frac{Q_{in}}{W_{comp}}\], where:

• \(Q_{in}\) is the heat absorbed in the evaporator.
• \(W_{comp}\) is the work done by the compressor.

To find these values: - Calculate the work done by using the difference in enthalpy between the vapor entering and exiting the compressor. - Determine the heat absorbed by using the difference in enthalpy of the refrigerant in the evaporator. A higher COP indicates a more efficient refrigeration cycle.

Refrigerating Capacity

Refrigerating capacity indicates the amount of heat a refrigeration system can remove from the evaporator. It’s measured in tons, commonly used in commercial refrigeration.

• To calculate this, we first determine the heat absorbed in the evaporator using the enthalpies of the refrigerant before and after it passes through the evaporator.
• Next, we convert the obtained refrigerating capacity from the basic unit (e.g., kW) to tons.

Typically, 1 ton of refrigeration equals the heat removal rate of 3.517 kW. By understanding these concepts, you can grasp how efficiently the refrigeration cycle operates and its effectiveness in different applications.