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[SOLVED] nondimensinalize PDE

dwsmith

Well-known member
Feb 1, 2012
1,673
$$
\frac{1}{\alpha}T_t = T_{xx}
$$
B.C are
$$
T(0,t) = T(L,t) = T_{\infty}
$$
I.C is
$$
T(x,0) = T_i.
$$
By recasting this problem in terms of non-dimensional variables, the diffusion equation along with its boundary conditions can be put into a canonical form.
Suppose that we introduce variable scaling defined by
$$
x_* = \frac{x}{L}\quad\quad t_* = \frac{\alpha t}{L^2}\quad\quad \theta = \frac{T - T_{\infty}}{T_i - T_{\infty}}
$$
With this change of variables, show that the problem definition becomes
\begin{alignat*}{5}
\theta_{t_*} & = & \theta_{x_*x_*} & & \\
\theta(0,t_*) & = & \theta(1,t_*) & = & 0\\
\theta(x_*,0) & = & 1
\end{alignat*}

$$
\frac{1}{L^2}\frac{\partial T}{\partial t_*} = \frac{1}{L^2}\frac{\partial^2 T}{\partial x_*^2}
$$
Do I make the $T$ substitution just as $T = \theta(T_i - T_{\infty}) + T_{\infty}$?
 

dwsmith

Well-known member
Feb 1, 2012
1,673
$$
\frac{1}{\alpha}T_t = T_{xx}
$$
B.C are
$$
T(0,t) = T(L,t) = T_{\infty}
$$
I.C is
$$
T(x,0) = T_i.
$$
By recasting this problem in terms of non-dimensional variables, the diffusion equation along with its boundary conditions can be put into a canonical form.
Suppose that we introduce variable scaling defined by
$$
x_* = \frac{x}{L}\quad\quad t_* = \frac{\alpha t}{L^2}\quad\quad \theta = \frac{T - T_{\infty}}{T_i - T_{\infty}}
$$
With this change of variables, show that the problem definition becomes
\begin{alignat*}{5}
\theta_{t_*} & = & \theta_{x_*x_*} & & \\
\theta(0,t_*) & = & \theta(1,t_*) & = & 0\\
\theta(x_*,0) & = & 1
\end{alignat*}

$$
\frac{1}{L^2}\frac{\partial T}{\partial t_*} = \frac{1}{L^2}\frac{\partial^2 T}{\partial x_*^2}
$$
Do I make the $T$ substitution just as $T = \theta(T_i - T_{\infty}) + T_{\infty}$?
For the BC, I have now
$$
T\left(0=x=x_*L,t = \frac{L^2t_*}{\alpha}\right) = T\left(1,t = \frac{L^2t_*}{\alpha}\right) = T_{\infty}
$$
How do I just get $t_*$ in the BC? How do I change T to theta?
 

dwsmith

Well-known member
Feb 1, 2012
1,673
For the BC, I have now
$$
T\left(0=x=x_*L,t = \frac{L^2t_*}{\alpha}\right) = T\left(1,t = \frac{L^2t_*}{\alpha}\right) = T_{\infty}
$$
How do I just get $t_*$ in the BC? How do I change T to theta?
So I have solved everything except for the change on the boundary conditions and initial conditions.

That part has me stuck.