# Understanding boundary conditions in heat transfer

In the context of heat transfer, how does one physically interpret the following boundary conditions: $$u \cdot \mathbf{n} = 0, \qquad \frac{\partial u}{\partial \mathbf{n}} \cdot \tau = 0$$ where $u, \mathbf{n}$ and $\tau$ denotes the temperature distribution, normal vector and tangential vector, respectively?

How is this related to slip/no-slip boundary conditions?

• Are you talking about heat transfer or fluid flow? – Chet Miller Jun 1 '18 at 13:33
• I am interested in physical interpretations of above boundary conditions in heat transfer. – Julienne Franz Jun 1 '18 at 13:37
• What do you mean by $u\cdot {\bf n}?$ $u$ is a scalar and ${\bf u}$ a vector – caverac Jun 1 '18 at 14:09

In what follows $\mathbf{u}$ is velocity and $T$ is temperature of fluid; former is a vector and the latter a scalar.
In fluid flow over a solid body, if the solid body is at rest and impermeable (i.e. there can be no mass flux across the boundary of the solid) then, since the fluid cannot penetrate the boundary, the normal component of fluid velocity at the boundary must be zero, i.e. $\mathbf{u}\cdot\mathbf{n}=0$. Analogously, in heat transfer from the surface of a solid body, if the solid is perfectly non-conducting (i.e. its thermal conductivity is zero) then, since heat transfer can happen only in the direction of temperature variation, the temperature variation in the direction normal to the boundary must be zero, i.e. $\mathbf{n}\cdot\nabla T=0$. If you set up a coordinate system on the boundary such that (say) its Z-axis is normal to the boundary then $\mathbf{n}\cdot\nabla T=\partial T/\partial z$.
No-slip boundary condition for fluid flow says that the fluid velocity and velocity of the surface of the solid must be the same at all points of contact between the two. In particular for a solid body at rest, fluid velocity is zero at the surface of the solid. Since fluid velocity $\mathbf{u}=\mathbf{0}$ at the solid surface, it is trivially true that $\mathbf{u}\cdot\mathbf{t}=\mathbf{0}\cdot\mathbf{t}=0$ at the solid surface. Now for a perfectly conducting solid body (i.e. one whose thermal conductivity is infinite) the whole body will be at one temperature. Now temperature must vary continuously across the boundary, that is there cannot be a jump in temperature as you move across the surface. This means that fluid temperature at the surface must be the same as that of the solid surface. Since temperature of solid body is constant, in particular there is no temperature variation along the surface of the solid body, and consequently there will be no temperature variation in the fluid along the surface of the solid body, i.e. $\mathbf{t}\cdot\nabla T=0$ at the solid surface. In this case $\mathbf{n}\cdot\nabla T\neq 0$ in general.