Plotting COP vs Evaporator Temperature Variation in Refrigeration Cycle Using EES and MATLAB

Introduction

Understanding the relationship between the Coefficient of Performance (COP) and evaporator temperature variation in a refrigeration cycle is crucial for optimizing the efficiency of refrigeration systems. The Coefficient of Performance (COP) is a measure of the efficiency of a heat pump and is defined as the ratio of the heat removed from the cooling source to the work done.

Overview of the Refrigeration Cycle

The refrigeration cycle can be effectively analyzed and visualized using either Engineering Equation Solver (EES) software or MATLAB. Both tools are powerful and widely used in the field of thermodynamics and refrigeration engineering. EES is a powerful tool for solving complex engineering problems and provides a user-friendly interface for mathematicians and engineers. MATLAB, on the other hand, is a versatile programming language that can be used for both numerical and symbolic processing.

Using EES to Plot COP vs Evaporator Temperature

Step 1: Understanding the Theory

Before getting into the practical side, it is important to have a theoretical understanding of the refrigeration cycle components involved. A typical refrigeration cycle consists of four main processes: compression, heat rejection (condensation), expansion, and heat absorption (evaporation). The COP of the cycle can be calculated using the following formula:

COP (Heat extracted from the source / Work input)

Step 2: Setting Up the EES Input

1. Open EES and create a new problem. 2. Define the system variables such as the refrigerant properties (enthalpy, entropy, etc.) at different points in the cycle. 3. Set the evaporator temperature as a variable that can be changed.

Step 3: Calculating the COP

1. Input the equation for calculating the COP based on the enthalpy at the evaporator and condenser points. 2. Use the evaporation temperature to calculate the heat extracted at the evaporator.

Using MATLAB to Plot COP vs Evaporator Temperature

Step 1: Understanding the Theory

In addition to EES, MATLAB can also be used to plot the COP vs evaporator temperature variation. The process involves defining the system variables and running a loop to calculate the COP at different evaporator temperatures.

Step 2: Setting Up the MATLAB Script

Define the system variables such as refrigerant properties at different evaporation temperatures. Create a loop to iterate through a range of evaporator temperatures. Calculate the COP at each temperature and store the results. Plot the results using MATLAB's plotting functions.

Sample MATLAB Code

Below is a sample MATLAB code that demonstrates how to plot COP vs evaporator temperature variation:

function plotCOPvsEvaporatorTemp
    T_evaporator  linspace(-20, 10, 100); % Define a range of evaporator temperatures
    COP  zeros(size(T_evaporator));
    for i  1:length(T_evaporator)
        % Define refrigerant properties at the current evaporator temperature
        % Calculate the COP at the current temperature
        COP(i)  calculateCOP(T_evaporator(i));
    end
    % Plot the results
    figure;
    plot(T_evaporator, COP);
    xlabel('Evaporator Temperature (°C)');
    ylabel('COP');
    title('COP vs Evaporator Temperature Variation');
    grid on;
end
function COP  calculateCOP(T_evaporator)
    % Calculate COP based on the current evaporator temperature
    % This is a simplified example and the actual implementation will vary
    H_evaporator  refrigerantEnthalpy(T_evaporator); % Assume function to get enthalpy
    H_condenser  refrigerantEnthalpy(T_condenser); % Assume function to get condenser enthalpy
    WorkInput  refrigerantEnthalpy(T_compressor); % Assume function to get enthalpy at compressor
    COP  (H_evaporator - H_condenser) / WorkInput;
end

Conclusion

Visualizing the relationship between COP and evaporator temperature variation in a refrigeration cycle can provide valuable insights into the performance of the system. By utilizing tools like EES and MATLAB, engineers can efficiently model and optimize refrigeration cycles, leading to more efficient and cost-effective designs.