MHB Why is alpha mentioned in the cooling process equation?

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The discussion clarifies the role of the diffusivity parameter \(\alpha\) in the cooling process equation. It is emphasized that \(\alpha\) is used to establish a time-scale for conduction, represented by the equation \(t_{0}^{c}=\frac{L^2}{\alpha}\). While \(\alpha\) is integral to the non-dimensionalization of the problem, it does not directly impact the solution of the partial differential equation itself. Participants confirm that the equation can be solved without needing to focus on \(\alpha\). Overall, the mention of \(\alpha\) serves primarily to contextualize the time-scale rather than complicate the solution process.
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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.
 
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dwsmith said:
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.
 
Sudharaka said:
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?
 
dwsmith said:
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.
 

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