TransWikia.com

Coordinate transformation of boundary condition

Physics Asked by Wescley on May 8, 2021

Let us suppose a heat transfer problem inside a cylinder of radius $r_a$. If we neglect changes along $z$ and $theta$ directions, i.e. only a cross section of the cylinder, the problem can be formulated as follow

$$
rho ; c_p frac{partial T}{partial t} = frac{1}{r}frac{partial}{partial r}left( k; r frac{partial T}{partial r}right) + q(r)
$$

Considering a convective heat transfer with the surrounding environment at temperature $T_n$ gives us the following boundary condition:
$$
left. kfrac{partial T}{partial r}right|_{r=r_a} = hleft(T_n – Tright)
$$

So far not a problem at all. Now, I want to solve the same problem, but in cartesian coordinates, it turns into a 2D heat transfer problem, given by:

$$
rho ; c_p frac{partial T}{partial t} = frac{partial}{partial x}left(k frac{partial T}{partial x} right) + frac{partial}{partial y}left(k frac{partial T}{partial y} right)+ q(x,y)
$$

My question: What about the boundary condition? How can I transform the given boundary condition to cartesian coordinates?

One Answer

The temperature gradient expressed in cartesian coordinates is $$nabla T=frac{partial T}{partial x}hat{i_x}+frac{partial T}{partial y}hat{i_y}$$The unit vectors in the x- and y directions are related to the radial- and circumferential unit vectors by: $$i_x=frac{x}{sqrt{x^2+y^2}}hat{i_r}-frac{y}{sqrt{x^2+y^2}}hat{i_theta}$$ $$i_y=frac{x}{sqrt{x^2+y^2}}hat{i_theta}+frac{y}{sqrt{x^2+y^2}}hat{i_r}$$So, $$nabla T=left[frac{x}{sqrt{x^2+y^2}}frac{partial T}{partial x}+frac{y}{sqrt{x^2+y^2}}frac{partial T}{partial y}right]hat{i_r}+left[-frac{y}{sqrt{x^2+y^2}}frac{partial T}{partial x}+frac{x}{sqrt{x^2+y^2}}frac{partial T}{partial y}right]hat{i_theta}$$ At $sqrt{x^2+y^2}=r_0$ this reduces to: $$nabla T=left[frac{x}{r_0}frac{partial T}{partial x}+frac{y}{r_0}frac{partial T}{partial y}right]hat{i_r}+left[-frac{y}{r_0}frac{partial T}{partial x}+frac{x}{r_0}frac{partial T}{partial y}right]hat{i_theta}$$So, we have: $$frac{partial T}{partial r}=frac{x}{r_0}frac{partial T}{partial x}+frac{y}{r_0}frac{partial T}{partial y}$$and$$-frac{y}{r_0}frac{partial T}{partial x}+frac{x}{r_0}frac{partial T}{partial y}=0$$The latter represents the condition that, at $sqrt{x^2+y^2}=r_0$, T is constant.

Answered by Chet Miller on May 8, 2021

Add your own answers!

Ask a Question

Get help from others!

© 2024 TransWikia.com. All rights reserved. Sites we Love: PCI Database, UKBizDB, Menu Kuliner, Sharing RPP