Simulations using composites of different thicknesses—1 mm, 3 mm, and 5 mm—were carried out to evaluate the effect of composite thickness on thermal and electrical performance. To en-sure unidirectional heat flow, the boundary conditions for thermal simulations involved applying a constant heat flux to one surface while keeping the opposite surface at a fixed temperature. For electrical conductivity analysis, a steady voltage was applied across the composite. The heat transport and electrical conduction across various thicknesses were modeled using finite element analysis (FEA), which highlighted the relationship between composite performance and thickness [8, 9].

The steps involve:

The following MATLAB code snippet demonstrates the implementation of steady-state thermal and electrical analysis:

1 % Steady-state thermal analysis 2 thermalModel = createpde(’thermal’, ’steadystate’); 3 geometry = multicuboid(10e-3, 10e-3, thickness); % Replace thickness for 1 mm, 3 mm, 5 mm 4 thermalModel.Geometry = geometry; 5 generateMesh(thermalModel, ’Hmax’, 1e-3);

7 % Thermal properties 8 thermalProperties(thermalModel, ’ThermalConductivity’, k_epoxy); 9 thermalBC(thermalModel, ’Face’, 2, ’Temperature’, 300); % Fixed temp 10 thermalBC(thermalModel, ’Face’, 5, ’HeatFlux’, 1000); % Heat flux 11 thermalResults = solve(thermalModel);

13 % Effective thermal conductivity 14 temperature_diff = max(thermalResults.Temperature) - min(thermalResults.

Temperature); 15 k_effective = (heat_flux * thickness) / temperature_diff;

17 % Steady-state electrical analysis 18 electricalModel = createpde(’thermal’, ’steadystate’); 19 electricalModel.Geometry = geometry; 20 generateMesh(electricalModel, ’Hmax’, 1e-3);

22 % Electrical properties 23 thermalProperties(electricalModel, ’ThermalConductivity’, sigma_epoxy); 24 thermalBC(electricalModel, ’Face’, 2, ’Temperature’, 1); % Voltage applied 25 thermalBC(electricalModel, ’Face’, 5, ’Temperature’, 0); % Grounded 26 electricalResults = solve(electricalModel); 28 % Effective electrical conductivity 29 voltage_diff = max(electricalResults.Temperature) - min(electricalResults.

Temperature); 30 sigma_effective = (current_density * thickness) / voltage_diff;

Listing 1: Steady-State Thermal and Electrical Analysis

Transient simulations were conducted to represent the time-dependent behavior of hybrid epoxy composites under dynamic settings. Time-dependent boundary conditions were used to replicate real-world operational circumstances, such as time-varying voltages and periodic heat flows. MATLAB’s Partial Differential Equation (PDE) toolbox was used to solve the governing equa-tions for electrical conduction and transient heat transport. This method made it easier to com-prehend how the composite would react to dynamic loading by enabling the examination of temperature and voltage distributions over time. The equations used are:

ρcp = ∇ · (k∇T) + Q, (Transient heat transfer) (2.3)

= ∇ · (σ∇V ), (Transient electrical conduction) (2.4)

The MATLAB code for the transient thermal analysis is as follows:

1 % Transient thermal analysis 2 transientModel = createpde(’thermal’, ’transient’); 3 geometry = multicuboid(10e-3, 10e-3, 3e-3); % Fixed thickness for transient simulation 4 transientModel.Geometry = geometry; 5 generateMesh(transientModel, ’Hmax’, 1e-3);

7 % Thermal properties 8 thermalProperties(transientModel, ’ThermalConductivity’, k_epoxy, ...

9 ’MassDensity’, rho_epoxy, ’SpecificHeat’, cp_epoxy); 10 thermalIC(transientModel, initial_temp); 11 thermalBC(transientModel, ’Face’, 2, ’Temperature’, 300); % Fixed temp 12 thermalBC(transientModel, ’Face’, 5, ’HeatFlux’, @(region, state) 1000 * sin( state.time)); % Periodic heat flux

14 % Solve transient problem 15 time_steps = linspace(0, 100, 100); % Simulate over 100 seconds 16 results = solve(transientModel, time_steps); 17 temperature_center = interpolateTemperature(results, [5e-3, 5e-3, 1.5e-3]);

Listing 2: Transient Thermal Analysis

Thermal conductivity changes with temperature:

k(T) = k0 (1 + α · (T − Tref))

Electrical conductivity changes with voltage:

σ(V ) = σ0 1 + β · V

The MATLAB code used is:

1 % Nonlinear thermal conductivity as a function of temperature 2 alpha = 0.01; % Temperature coefficient 3 thermalConductivityFcn = @(region, state) k0 * (1 + alpha * (state.u - T_ref)); 4 thermalProperties(transientModel, ’ThermalConductivity’, thermalConductivityFcn,

5 ’MassDensity’, rho_epoxy, ’SpecificHeat’, cp_epoxy);

7 % Nonlinear electrical conductivity as a function of voltage 8 beta = 0.02; % Voltage coefficient 9 electricalConductivityFcn = @(region, state) sigma0 * (1 + beta * (state.u).^2); 10 thermalProperties(electricalModel, ’ThermalConductivity’, electricalConductivityFcn);

The distribution of thermal conductivity for the composite with a 10% filler fraction is shown in The thermal conductivity distribution for the composite with a 20% filler component is de-picted in Figure 2. This composite shows noticeable resistance to heat flow and moderate thermal conductivity as compared to the 30% filler case. Less effective heat dissipation is indicated by a more noticeable temperature gradient. The trade-offs between filler content and thermal man-agement are shown by this intermediate performance, which shows that although moderate filler fractions enhance performance, they fall short of higher filler content in terms of efficiency.

The distribution of thermal conductivity in the composite with a 30% filler component is shown in Figure 3. The consistent temperature gradient makes it clear that the high filler content greatly improves thermal conductivity. With less thermal resistance than composites with lower filler fractions, heat flow is more reliable and efficient. The material is appropriate for high-performance thermal management applications where quick heat dissipation is needed, as this result highlights the effect of higher filler fractions on thermal performance.

The thermal and electrical conductivities of the composite are compared for different thick-nesses (1 mm, 3 mm, and 5 mm) in Figure 4. Better heat flow paths cause thermal conductivity to increase with thickness, whereas increased resistance over longer distances causes electrical conductivity to decrease.

This figure (Figure 5) illustrates the transient temperature distribution within the composite over time during the simulation. The heat flux applied to one face of the composite causes a temper-ature gradient, with the heated face showing the highest temperature (red region) and the cooler regions farther away (blue zones). As time progresses, the gradient becomes less pronounced, indicating the system’s movement towards a steady-state condition. The evolution of the temper-ature profile demonstrates the composite’s thermal response, highlighting its ability to dissipate heat efficiently over time. This behavior is crucial for applications where time-dependent heat flux plays a role in performance.

The time needed for the composite to attain steady-state temperature for different filler frac-tions (10%, 20%, and 30%) is depicted in Figure 6. Higher filler fractions shorten the time to steady-state because they improve thermal conductivity, which speeds up heat dissipation, as the bar chart shows. Composites with lower filler fractions, on the other hand, behave in a delayed steady state, which is indicative of their poor capacity for heat control.

The transient temperature change at the composite’s core and edges with a 10% filler per-centage is shown in Figure 7. At both locations, the temperature gradually increases and, after about 50 seconds, converges near steady-state. The low thermal conductivity of composites with less filler content is reflected in the slower thermal response.

The transient temperature change at the borders and center of the composite with a 20% filler component is depicted in Figure 8. The system reaches steady-state in around 40 seconds, which is a little quicker than the 10% filler example. Faster heat dissipation is made possible by the increased thermal conductivity caused by the higher filler content.

The transient temperature change at the composite’s core and borders with a 30% filler per-centage is depicted in Figure 9. Because of the substantial filler content’s high heat conductivity, the system reaches steady-state quickly—within 30 seconds. Uniform heat dissipation is demon-strated by the small difference between the center and edge temperatures.

The behavior of thermal conductivity under linear and nonlinear models is contrasted in Figure 10 over a temperature range of 290 K to 350 K. For composites with filler percentages of 10%, 20%, and 30%, the solid lines exhibit nonlinear behavior, whereas the dashed line depicts a linear thermal conductivity model. Because of the filler-dependent temperature sensitivity, nonlinear thermal conductivity rises with temperature, while the linear model performs worse at higher temperatures.

The temperature-dependent thermal conductivity for filler fractions of 10%, 20%, and 30% is shown in Figure 11. For all filler fractions, thermal conductivity rises with temperature; the 30% filler fraction continuously performs best. The pattern shows how important filler content is for improving the temperature response of the composite.

The voltage-dependent electrical conductivity for 10%, 20%, and 30% filler fractions is shown in Figure 12. Higher filler fractions exhibit improved conductivity over the voltage range, and electrical conductivity rises quadratically with applied voltage. This nonlinear response emphasizes how crucial the applied voltage and filler percentage are to improving electrical per-formance.