# Heat transfer: boundary conditions with fluid velocity

The following equation is considered: $$\frac{\partial u}{\partial t} - a\Delta u + \mathbf v \cdot \nabla u = f.$$ I have difficulties in formulating boundary conditions for this equation. If $\mathbf v$ is absent, the boundary conditions are $$a\frac{\partial u}{\partial n} + \beta(u - u_b) = 0$$ where $u_b$ is prescribed temperature field on the boundary. But if $\mathbf v$ is present? If we suggest the following boundary condition: $$a\frac{\partial u}{\partial n} - (\mathbf v \cdot \mathbf n) + \beta(u - u_b) = 0$$ then we will have problems with analysing the equation when $(\mathbf v \cdot \mathbf n) > 0$ (where the fluid outflows). How to set correct boundary conditions?

• Is the fluid velocity a known function, i.e., $\mathbf{v}$ is prescribed? If that's not the case, the Navier-Stokes equations for pressure and velocity are missing. – Dmoreno Jun 8 '14 at 9:01
• @Dmoreno Yes, $\mathbf v$ is prescribed. – jokersobak Jun 8 '14 at 15:46

The boundary condition $$a\frac{\partial u}{\partial n} - (\mathbf v \cdot \mathbf n)u + \beta(u - u_b) = 0$$ is correct and cannot cause a problem. Indeed, $\mathbf v\cdot\mathbf n=0$ at the boundary whenever it happens to be impermeable to fluid. Otherwise, condition $(\mathbf v\cdot\mathbf n) > 0$ cannot hold on the whole boundary according to the principle of conservation of mass implying that the total flux of fluid through the boundary equals zero.
• This is a typo. Here is a condition: $a\frac{\partial u}{\partial n} - (\mathbf v \cdot \mathbf n)u + \beta(u - u_b) = 0$ or $a\frac{\partial u}{\partial n} - (\mathbf v \cdot \mathbf n)(u - u_b) + \beta(u - u_b) = 0$. $(\mathbf v \cdot \mathbf n) > 0$ on the part of the boundary. – jokersobak Jun 6 '14 at 21:09
• @jokersobak: In the boundary condition for the temperature, yes. An obvious typo, though I missed it. But condition $a\frac{\partial u}{\partial n} - (\mathbf v \cdot \mathbf n)(u - u_b) + \beta(u - u_b) = 0$ won't do since $-a\frac{\partial u}{\partial n} + (\mathbf v \cdot \mathbf n)u$ is a heat flux density through any surface, including the boundary at which it must be proportional to the difference between inside and outside temperature. This is exactly what is called a convective heat transfer at the boundary. – mkl314 Jun 6 '14 at 21:57
• Here the condition $a\frac{\partial u}{\partial n} - (\mathbf v \cdot \mathbf n)(u - u_b) + \beta(u - u_b) = 0$ is suggested: mathoverflow.net/questions/169199/… – jokersobak Jun 7 '14 at 5:16
• @jokersobak: Carlo Beenakker's argument is wrong because he incorrectly takes $\,-a\frac{\partial u}{\partial n}\,$ for the total heat flux density while it should be $\,-a\frac{\partial u}{\partial n} + (\mathbf v \cdot \mathbf n)u$. In fact, $\,-a\frac{\partial u}{\partial n}\,$ is just a partial heat flux density provided by the Fourier law. – mkl314 Jun 7 '14 at 8:39
• @jokersobak: No, the total heat flux density has nothing to do with an outside temperature. The total heat flux density is a basic element in the derivation of the heat equation for continuous media. It is defined on any smooth geometrical surface inside continuous media, and circumstantially on inside part of the boundary where it could be equated to the partial heat influx density $(\mathbf v \cdot \mathbf n)u_b$. But this would be another boundary regime quite different from the purely convective heat transfer. – mkl314 Jun 7 '14 at 10:31