Question

An ideal vapor-compression refrigeration cycle that uses refrigerant-134a as its working fluid maintains a condenser at \(800 \mathrm{kPa}\) and the evaporator at \(-12^{\circ} \mathrm{C}\). Determine this system's COP and the amount of power required to service a \(150 \mathrm{kW}\) cooling load.

Short answer

Question: Determine the coefficient of performance (COP) and the amount of power required for an ideal vapor-compression refrigeration cycle using refrigerant-134a with an evaporator temperature of -12°C, a condenser pressure of 800 kPa, and a cooling load of 150 kW. Answer: The coefficient of performance (COP) for the refrigeration cycle is 3.89, and the power required to service the 150 kW cooling load is 38.56 kW.

Step-by-step solution

Identify the properties of refrigerant-134a

First, let's label the points in the ideal vapor-compression refrigeration cycle: 1. Start of compression (at the evaporator outlet). 2. End of compression (at the condenser inlet). 3. Start of expansion (at the condenser outlet). 4. End of expansion (at the evaporator inlet). At point 1, the refrigerant is saturated vapor at \(-12^{\circ} \mathrm{C}\). Therefore, we can find the enthalpy at point 1 (denoted as \(h_1\)) using the saturated vapor enthalpy at the given temperature. At point 2, the refrigerant is compressed isentropically (meaning constant entropy) to a pressure of \(800 \mathrm{kPa}\). Once we find the entropy at point 1 (denoted as \(s_1\)), we'll set it equal to the entropy at point 2 (denoted as \(s_2\)) to find the enthalpy at point 2. At point 3, the refrigerant is saturated liquid at \(800 \mathrm{kPa}\). Therefore, we can find the enthalpy at point 3 (denoted as \(h_3\)) using the saturated liquid enthalpy at the given pressure. At point 4, the refrigerant is expanded isenthalpically (meaning constant enthalpy) to the evaporator pressure. So, the enthalpy at point 4 (denoted as \(h_4\)) is equal to the enthalpy at point 3 (\(h_3\)). Using refrigerant property tables, we can find the enthalpy values: - \(h_1 = 243.9 \mathrm{kJ/kg}\) - \(s_1 = 0.9475 \mathrm{kJ/kgK}\) - \(s_2 = s_1 = 0.9475 \mathrm{kJ/kgK} \Rightarrow h_2 = 282.3 \mathrm{kJ/kg}\) - \(h_3 = 94.5 \mathrm{kJ/kg}\) - \(h_4 = h_3 = 94.5 \mathrm{kJ/kg}\)

Calculate the COP

The coefficient of performance (COP) for the refrigeration cycle is the ratio of the cooling effect provided by the cycle to the work input required. For an ideal vapor-compression refrigeration cycle, the cooling effect is the heat absorbed by the refrigerant in the evaporator (\(q_{cooling}=h_1-h_4\)), and the work input is the work done in compressing the refrigerant (\(w_{compressor}=h_2-h_1\)). Thus, the COP is calculated as follows: \(\mathrm{COP}=\frac{q_{cooling}}{w_{compressor}}=\frac{h_1-h_4}{h_2-h_1}\) Plugging in the values from step 1: \(\mathrm{COP}=\frac{243.9-94.5}{282.3-243.9} = \frac{149.4}{38.4}=3.89\)

Calculate the power required

Now that we have the COP, we can calculate the power required for the refrigeration cycle by dividing the cooling load by the COP: \(P_{required}=\frac{Q_{cooling}}{\mathrm{COP}}\) Here, \(Q_{cooling}=150 \mathrm{kW}\). Plugging in the values: \(P_{required}=\frac{150 \mathrm{kW}}{3.89}=38.56 \mathrm{kW}\) Therefore, the power required to service the \(150 \mathrm{kW}\) cooling load is \(38.56 \mathrm{kW}\).

Key concepts

Refrigerant Properties

In the realm of vapor-compression refrigeration cycles, understanding the unique characteristics of refrigerants is crucial. Refrigerants like R-134a have specific properties that affect the system's efficiency and performance. These properties include boiling point, pressure, enthalpy (the total heat content of a system), and entropy (a measure of energy dispersion in a system).

For instance, R-134a has a lower boiling point at atmospheric pressure compared to water, making it efficient in absorbing heat at the evaporator at relatively low temperatures. When choosing a refrigerant, it's essential to consider its environmental impact, such as ozone depletion potential (ODP) and global warming potential (GWP), as well as its thermodynamic properties that influence the cycle's efficiency and capacity.

Enthalpy and Entropy

Enthalpy and entropy are cornerstones in thermodynamics, especially in refrigeration cycles. Enthalpy is a measurement of energy in a thermodynamic system and is used to calculate the heat transfer during phase changes, like from liquid to vapor, which is crucial in cooling processes. Entropy, on the other hand, is related to the randomness or disorder within a system. In the context of vapor-compression refrigeration, observing entropy helps maintain an isentropic (constant entropy) compression process, ensuring that the system operates efficiently with minimal energy waste.

An understanding of these properties not only enables accurate determination of state points in a refrigeration cycle but also allows for a comprehensive energy balance analysis. The utilization of enthalpy and entropy values from refrigerant property tables helps in outlining the cycle's thermodynamic path and subsequently in the performance evaluation.

Coefficient of Performance (COP)

The Coefficient of Performance (COP) is a dimensionless measure that describes a refrigeration system's efficiency. It is defined as the ratio of the cooling effect or refrigeration produced to the work input required to produce that cooling effect. Mathematically, for a vapor-compression refrigeration cycle, COP is expressed as the ratio of the heat absorbed by the refrigerant in the evaporator to the work done by the compressor.

A higher COP value indicates a more efficient refrigeration cycle, as it produces a greater amount of cooling for the same amount of work. Understanding and calculating the COP provides valuable insight into the performance of refrigeration systems, guiding engineers and technicians in design optimization, energy consumption forecasting, and operational cost estimation.

Cooling Load Calculation

Cooling load calculation is a fundamental concept in the design and analysis of HVAC systems. It refers to determining the amount of heat energy removed from a space to maintain a desired temperature. The calculation takes into account factors such as the size of the space, local climate, occupancy levels, appliance heat gain, and thermal properties of building materials.

In the context of refrigeration cycles and the example problem, the cooling load represents the energy removed by the system to maintain the specified conditions. By knowing the cooling load and the system's COP, we can calculate the power required from the compressor to achieve the desired cooling effect. Precise cooling load calculations are crucial for selecting appropriate equipment and ensuring both the efficiency and effectiveness of thermal control strategies in various applications.