Why is alpha mentioned in the cooling process equation?

[Dustinsfl]

Why is $\alpha$ mentioned? I don't see an $\alpha$.With proper non-dimensionalization and assuming for convenience that the ratio of convection and conduction parameters $h/k = 1$, the cooling process is described by the following equations:
\begin{alignat*}{4}
T_t & = & T_{xx} & \\
T_x(0,t) & = & 0 & \\
T_x + T & = & 0 & \quad \text{at} \ x = 1\\
T(x,0) & = & f(x) &
\end{alignat*}
Here, $x$ and $t$ are dimensionless positions and times wherein the original interval has been mapped from $[0,L]$ to $[0,1]$ and a time-scale based on the diffusivity $\alpha$ has been used; also, the temperature definition is such that the ambient temperature has a reference value of zero.

 

[Sudharaka]

Why is $\alpha$ mentioned? I don't see an $\alpha$.With proper non-dimensionalization and assuming for convenience that the ratio of convection and conduction parameters $h/k = 1$, the cooling process is described by the following equations:
\begin{alignat*}{4}
T_t & = & T_{xx} & \\
T_x(0,t) & = & 0 & \\
T_x + T & = & 0 & \quad \text{at} \ x = 1\\
T(x,0) & = & f(x) &
\end{alignat*}
Here, $x$ and $t$ are dimensionless positions and times wherein the original interval has been mapped from $[0,L]$ to $[0,1]$ and a time-scale based on the diffusivity $\alpha$ has been used; also, the temperature definition is such that the ambient temperature has a reference value of zero.

Hi dwsmith, :)

Well, the conduction timescale is based on diffusivity \((\alpha)\). That is,

\[t_{0}^{c}=\frac{L^2}{\alpha}\]

where \(t_{0}^{c}\) is the initial time. This is what is meant by "...time-scale based on the diffusivity $\alpha$ has been used..."

Reference: http://nd.edu/~msen/Teaching/IntHT/IntHTNotes.pdf (Page 6)

Kind Regards,
Sudharaka.

 

[Dustinsfl]

Hi dwsmith, :)

Well, the conduction timescale is based on diffusivity \((\alpha)\). That is,

\[t_{0}^{c}=\frac{L^2}{\alpha}\]

where \(t_{0}^{c}\) is the initial time. This is what is meant by "...time-scale based on the diffusivity $\alpha$ has been used..."

Reference: http://nd.edu/~msen/Teaching/IntHT/IntHTNotes.pdf (Page 6)

Kind Regards,
Sudharaka.

So I can just solve this problem without having to worry about it then, correct?

 

[Sudharaka]

So I can just solve this problem without having to worry about it then, correct?

Yes, solving the partial differential equation doesn't involve \(\alpha\) since it's related to the scale used for measuring time.