A monolithic finite element strategy for conjugate heat transfer

Abstract

Motivated by several applications such as thermal management of electronic components, the design and modeling of heat exchangers, etc. this work presents an efficient monolithic finite element strategy for solving thermo-fluid-structure interaction problems involving a compressible fluid and a structure undergoing small deformations. This formulation uses displacement variables for the structure and velocity variables for the fluid, with no additional variables required to ensure traction, velocity, temperature and heat flux continuity at the fluid-structure interface. The use of an exact tangent stiffness matrix ensures a quadratic rate of convergence within each time step. The robustness and good performance of the method is demonstrated by applying the proposed strategy to a wide spectrum of problems pertaining to steady/transient, two/three-dimensional/axisymmetric problems involving conjugate heat transfer. In particular, it is shown that the proposed formulation yields good results even when the fluid is almost incompressible.

Appendix I. An appendix section

The equations of state used are as follows:

Air: The reference values used are $\rho =1.2\ \mathrm{kg}/\mathrm{m}^{3}$, $\theta =300\,\text {K}$, $p=1.0332\times 10^5\ \mathrm{N}/\mathrm{m}^{2}$, $\lambda =-1.2\times 10^{-5}\,\text {kg/(m-s)}$, $\mu =1.81\times 10^{-5}\,\text {kg/(m-s)}$, $\kappa =0.025\,\text {W/m-K}$, $c_v=717.5\,\text {J/kg-K}$, $\gamma =1.4$.

Equation of state with $R=287\,\text {J/(kg-K)}$ is

$$\begin{aligned} p=\rho R\theta . \end{aligned}$$

Dependence of viscosity (Sutherland model) and thermal conductivity on temperature:

$$\begin{aligned} \mu =1.68\times 10^{-5}\left( \frac{\theta }{273}\right) ^{3/2} \left( \frac{383.5}{\theta +110.5}\right) ;\quad \kappa =\frac{1004.5}{0.71}\mu . \end{aligned}$$

Water: The reference values used are $\rho =998.2\ \mathrm{kg}/\mathrm{m}^{3}$, $\theta =293.15\,\text {K}$, $p=1.15\times 10^5\ \mathrm{N}/\mathrm{m}^{2}$, $\lambda =2.4\times 10^{-3}\,\text {kg/(m-s)}$, $\mu =1.0\times 10^{-3}\,\text {kg/(m-s)}$, $\kappa =0.6\,\text {W/m-K}$, $c_v=4186\,\text {J/(kg-K)}$.

Equation of state: Let $\theta _c=\theta -273.15$ and $\theta _m=1-273.15/\theta$, and let

$$\begin{aligned} \xi _1&=-6.1080450661\times 10^{-5}, \\ \xi _2&=8.2642214762\times 10^{-6}, \\ \xi _3&=-6.2519152251\times 10^{-8}, \\ \xi _4&=3.9657777214\times 10^{-10}, \\ \xi _5&=-1.0376611585\times 10^{-12}, \\ \eta _1&=1.8344791239\times 10^{-3}, \\ \eta _2&=-3.9483220928\times 10^{-5}, \\ \eta _3&=1.3275626596\times 10^{-7}, \\ \zeta _1&=9.4323986272\times 10^{-3}, \\ \zeta _2&=-1.4369670293\times 10^{-4}, \\ \zeta _3&=4.5608124905\times 10^{-7}, \\ v_{0T}&=1.0002119572 \times 10^{-3}\\&\quad (1+\xi _1\theta _c+\xi _2\theta _c^2+\xi _3\theta _c^3 +\xi _4\theta _c^4+ \xi _5\theta _c^5), \\ A_T&=0.13848005907(1+\eta _1\theta _c+\eta _2\theta _c^2+\eta _3\theta _c^3), \\ B_T&=2.705412322\times 10^8(1+\zeta _1\theta _c+\zeta _2\theta _c^2+\zeta _3\theta _c^3), \\ \gamma&=\frac{1}{v_{0T}A_T}\left( v_{0T}-\frac{1}{\rho }\right) . \end{aligned}$$

Then the equation of state for water is [44]

$$\begin{aligned} p=B_T\left( e^{\gamma }-1\right) . \end{aligned}$$

Viscosity as a function of temperature:

With

$$\begin{aligned} a_0&=-0.632330751534553\times 10^{1}, \\ a_1&=-0.172884319864045\times 10^{1}, \\ a_2&=-0.129387215150281\times 10^{4}, \\ a_3&=0.855455700673371\times 10^{5}, \\ a_4&=-0.298613655084593\times 10^{7}, \\ a_5&=0.636747286334842\times 10^{8}, \\ a_6&=-0.886000936016751\times 10^{9}, \\ a_7&=0.829771395998954\times 10^{10}, \\ a_8&=-0.527056628477971\times 10^{11}, \\ a_9&=0.223820320419432\times 10^{12}, \\ a_{10}&=-0.608047314470892\times 10^{12}, \\ a_{11}&=0.954774572690503\times 10^{12}, \\ a_{12}&=-0.658627182450538\times 10^{12}, \end{aligned}$$

the viscosity as a function of temperature is given by

$$\begin{aligned} \log \mu&=a_0+a_1\theta _m+a_2\theta _m^2+a_3\theta _m^3+a_4\theta _m^4\\&\quad +a_5\theta _m^5+a_6\theta _m^6+a_7\theta _m^7+a_8\theta _m^8 \\&\quad +a_9\theta _m^9 +a_{10}\theta _m^{10}+a_{11}\theta _m^{11}+a_{12}\theta _m^{12}. \end{aligned}$$

Thermal conductivity as a function of temperature: With

$$\begin{aligned} b_0&=-0.578034373459441, \\ b_1&=0.948808414124313, \\ b_2&=0.336358632865199, \\ b_3&=-4.72493981250531, \end{aligned}$$

the thermal conductivity as a function of temperature is given by

$$\begin{aligned} \log \kappa =b_0+b_1\theta _m+b_2\theta _m^2+b_3\theta _m^3. \end{aligned}$$