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The environmentally safe refrigerant \(\mathrm{R}-152 \mathrm{a}\) is to be evaluated as the working fluid for a heat pump system that will heat a house. It uses an evaporator temperature of \(-20^{\circ} \mathrm{C}\) and a condensing temperature of \(30^{\circ} \mathrm{C}\). Assume all processes are ideal and \(\mathrm{R}-152 \mathrm{a}\) has a heat capacity of \(C_{p}=0.996 \mathrm{~kJ} / \mathrm{kg} \mathrm{K}\). Determine the cycle COP

Short Answer

Expert verified
The cycle COP is 6.06.

Step by step solution

01

Understand the Definition of COP

Coefficient of Performance (COP) is the ratio of the heat transferred to the output over the work input in a heat pump cycle. It is given by the formula: \( \text{COP} = \frac{Q_{out}}{W_{in}} \), where \(Q_{out}\) is the heat delivered to the house and \(W_{in}\) is the work done by the heat pump.
02

Analyze the Given Data

The problem provides an evaporator temperature of \(-20^{\circ} \mathrm{C}\), a condensing temperature of \(30^{\circ} \mathrm{C}\), and a heat capacity of \(C_{p} = 0.996 \mathrm{~kJ} / \mathrm{kg} \mathrm{K}\). These values are necessary to determine the temperatures in Kelvin, which are 253 K and 303 K for the evaporator and condenser, respectively.
03

Calculate the Ideal Work Input

In an ideal cycle, the work input \(W_{in}\) can be calculated using the compressor work, which depends on the temperature difference: \(W_{in} = C_{p} (T_{high} - T_{low}) = 0.996 (303 - 253)\). The result gives us the work performed per kilogram of refrigerant.
04

Calculate Heat Output

The heat output \(Q_{out}\) is calculated using \(C_{p}\) and the temperature of the condensing side: \(Q_{out} = C_{p} \cdot T_{condensor} = 0.996 \times 303\). This results in the amount of energy released at the condenser per kilogram of refrigerant.
05

Calculate the COP

Using the values from the previous steps, the COP is determined by plugging \(W_{in}\) and \(Q_{out}\) into the COP formula: \( \text{COP} = \frac{Q_{out}}{W_{in}} = \frac{0.996 \times 303}{0.996 \times 50} = \frac{303}{50}\). Calculate the result to find the COP.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Coefficient of Performance (COP)
The Coefficient of Performance, or COP, is a metric used to measure the efficiency of heat pumps and refrigeration systems. It indicates how effectively a heat pump can transfer heat relative to the energy it consumes. In essence, COP is like a performance score. The higher the COP, the more efficient the heat pump is at using input energy to move heat. This is calculated with the formula: \( \text{COP} = \frac{Q_{out}}{W_{in}} \) where \( Q_{out} \) represents the heat delivered by the system, and \( W_{in} \) is the work input.
  • A higher COP means less work is required to provide a set amount of heating.
  • Commonly, heat pumps have COP values ranging from 3 to 5.
  • COP is dimensionless due to the ratio of similar units (energy).
Understanding COP helps in comparing the efficiency of different heat pump systems. It's important to note that COP can vary based on the specific conditions, such as temperature differences in the working environment.
Refrigeration Cycle
The Refrigeration Cycle is fundamental to the operation of heat pumps and refrigeration systems. It involves a series of thermodynamic processes that utilize a refrigerant to transfer heat from a cooler area to a warmer one. This might seem against the natural flow of heat, but is achievable through this cycle. The main parts of the refrigeration cycle include:
  • **Evaporator:** Here, the refrigerant absorbs heat from the environment, evaporating into a gas.
  • **Compressor:** The gas is then compressed, raising its pressure and temperature.
  • **Condenser:** In the condenser, the high-pressure gas releases absorbed heat and condenses back into a liquid.
  • **Expansion Valve:** This valve reduces the pressure of the refrigerant, which cools it down before re-entering the evaporator.
Refrigeration cycles are used not just in chilling applications but also in heat pumps for heating purposes. They play a crucial role in the performance and efficiency of these systems, impacting overall energy consumption.
Thermodynamic Processes
Thermodynamic processes are the driving force behind the conversion and transfer of energy in heat systems. In a heat pump, each component undergoes specific thermodynamic processes, which can be understood through the laws of thermodynamics and energy conservation principles.
Key processes in a refrigeration cycle involve:
  • **Isothermal Processes:** Typically occurring in the evaporator and condenser, where heat transfer happens at constant temperature.
  • **Adiabatic Processes:** Found in the compressor, where the temperature and pressure increase without heat exchange with the surroundings.
  • **Isobaric Processes:** Happen during the heat transfer at a constant pressure, important for maintaining operational stability.
  • **Isochoric Processes:** Occur when the volume remains constant, such as in closed off sections between components.
In the ideal cycle calculation, we often assume the processes are ideal for simplicity and calculating COP with maximum efficiency.
Refrigerants
Refrigerants are substances used in the refrigeration cycle for their ability to absorb and release heat effectively. Selecting the right refrigerant is critical for the efficiency and environmental impact of the heat pump or refrigeration system.
The chosen refrigerant impacts several factors:
  • **Thermal Conductivity:** Determines how easily heat passes through the refrigerant.
  • **Latent Heat Of Vaporization:** The energy needed for the refrigerant to evaporate, crucial for transferring heat.
  • **Environmental Impact:** Natural refrigerants often have lower global warming potential.
  • **Safety:** The refrigerant should not be toxic or highly flammable.
For example, R-152a is an environmentally safe refrigerant used in the original exercise. It has a favorable thermophysical property with a lower global warming potential.

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