# Conjugate Heat Transfer Analysis and Heat Dissipation Design of Nucleic Acid Detector Instrument
## 3 Simulation Process
### Material Parameters  
The materials used in the modules and electronic components of the instrument included aluminum, steel, acrylic, and copper. PCBs generally have a \"mezzanine\" design consisting of copper and FR4. Considering the equivalent modeling method for PCBs reported by [PERSON] [22], the plane thermal conductivity \(k_{p}\) and normal thermal conductivity \(k_{n}\) were used to represent the thermal conductivity of the PCBs. These terms are given by:  
\[k_{p}=\frac{\delta_{cu}\cdot\sum\limits_{j=1}^{N}k_{j}+(N-1)\cdot\delta_{F} \cdot k_{F}}{\delta_{t}} \tag{1}\]  
\[k_{n}=\frac{\delta_{t}}{\delta_{cu}\cdot\sum\limits_{j=1}^{N}\frac{1}{k_{j}}+ \frac{(N-1)\cdot\delta_{F}}{k_{F}}} \tag{2}\]  
where \(k_{j}\), \(k_{F}\), and \(N\) are the thermal conductivities of the copper-clad layer, FR4, and the number of layers of the copper cladding, respectively; \(\delta_{t}\) is the total thickness of the PCB;  
Figure 4: Measurement results of the heat sources: (**a**) TEC; (**b**) PCB1 and PCB2. \(\delta_{cu}\) and \(\delta_{F}\) are the thicknesses of the copper classing and the unclad layer, respectively; PCBs typically consist of FR4 and copper; therefore, \(k_{j}\) is between \(k_{F}\) and \(k_{cu}\). That is:  
\[k_{j}\cong\frac{A_{j}}{A_{t}}\cdot k_{cu} \tag{3}\]  
where \(A_{j}\) is the area of the \(j\)-th layer of the copper cladding, \(A_{t}\) is the area of the PCB, \(k_{cu}\) is the thermal conductivity of the copper. In an equivalent model, \(\delta_{cu}\) and \(\delta_{F}\) are considered to be equal; that is:  
\[\delta_{t}=N\cdot\delta_{cu}+(N-1)\cdot\delta_{F} \tag{4}\]  
The parameters of the PCBs are given in Tables 1 and 2. Using these parameters, the thermal conductivities of PCB1 and PCB2 were determined to be \(k_{p,pcb1}\) = 31.83 W-m\({}^{-1}\).K\({}^{-1}\); \(k_{n,pcb1}\) = 0.38 W-m\({}^{-1}\).K\({}^{-1}\); \(k_{p,pcb2}\)  
Conjugate heat transfer control equations based on the above assumptions are solved by using the Reynold-average [27; 28; 29]. Continuity equation:  
\[\frac{\partial\rho_{f}}{\partial t}+div(\rho_{f}\mathbf{u})=0 \tag{5}\]  
where \(\rho_{f}\) is the density of the fluid, \(\mathbf{u}\) is the velocity vector. When the fluid is incompressible, \(\frac{\partial\rho_{f}}{\partial t}=0\). Momentum in x-direction:  
\[ \begin{array}{l}\frac{\partial(\rho_{f}u)}{\partial t}+div(\rho_{f}u \mathbf{u})=div(\mu_{f}\,grad\,u)-\frac{\partial p}{\partial x}+S_{u}\\ +\left[-\frac{\partial(\rho_{f}u^{2})}{\partial x}-\frac{\partial(\rho_{f}u^{ \prime\prime})}{\partial y}-\frac{\partial(\rho_{f}u^{\prime\prime})}{ \partial z}\right]\end{array}  \tag{6}\]  
Similarly, the momentum equation in y- and z-directions can be written. Where \(u\), \(v\) and \(w\) are velocity components in the x-, y- and z-directions, respectively. \(\mu_{f}\) is the dynamic viscosity of the fluid, \(p\) is the fluid pressure, \(S_{u}\) is source term, \(\overline{u^{\prime}_{i}u^{\prime}_{j}}\) is the specific Reynolds stress term, and i and j are Cartesian coordinates. Energy equation for fluid domain is given by:  
\[ \begin{array}{l}\frac{\partial(\rho_{f}T)}{\partial t}+div(\rho_{f}\mathbf{ u}T)=div(\frac{k_{f}}{C_{p,f}}grad\,T)+S_{f}\\ +\left[-\frac{\partial(\rho_{f}\overline{u^{\prime}T})}{\partial x}-\frac{ \partial(\rho_{f}\overline{u^{\prime}T})}{\partial y}-\frac{\partial(\rho_{f }\overline{u^{\prime}T})}{\partial z}\right]\end{array}  \tag{7}\]  
Energy equation for solid domain is given by:  
\[\frac{\partial(\rho_{s}C_{p,s}T)}{\partial t}=div(k_{s}\
abla T)+Q \tag{8}\]  
where \(k_{f}\) and \(k_{s}\) are the thermal conductivities of the fluid and solid, respectively, \(T\) is temperature, \(\rho_{s}\) is the density of the solid, \(C_{p,f}\) and \(C_{p,s}\) are the specific heats of the fluid and solid, respectively, \(S_{f}\) and \(Q\) are source terms. The \(k-\omega\) shear stress transport (SST) turbulence model was adopted.