PDE - Heat Equation - Cylindrical Coordinates.

[erobz]

Would method of separation of variables lead to a solution to the following PDE?

$$ \frac{1}{r} \frac{ \partial}{\partial r} \left( kr \frac{ \partial T}{ \partial r}\right) = \rho c_p \frac{\partial T }{ \partial t }$$

This would be for the transient conduction of a hollow cylinder, of wall thickness ##r##, and thermal conductivity ##k##, with temperature distribution ##T(r,t)##

How do we tell if it will work out?

I imagine expanding like:

$$ \frac{k}{r} \left( \frac{ \partial T}{\partial r} + r \frac{ \partial^2 T}{\partial r^2} \right) = \rho c_p \frac{ \partial T}{\partial t}$$

Looking for a solution:

$$T(r,t) = f(r)g(t)$$

That would yield the following equation:

$$ \frac{1}{f(r)} \frac{k}{r} \left( f'(r) + r f''(r) \right) = \lambda= \rho c_p \frac{g'(t)}{g(t)} $$

You would get a damped oscillator for ##f(r)##, and exponential function for ##g(t)## from.

$$f''(r) + \frac{1}{r} f'(r) - \frac{ \lambda}{k} f(r) = 0 \tag{2}$$

$$ g'(t) - \frac{ \lambda}{\rho c_p}g(t) = 0 \tag{3} $$

When would I find out whether or not "it makes sense or works"

If at all possible, keep this as low level explanation. I'm on paper thin ice in any type of foundational understanding of PDE's ( or perhaps I have already unknowingly fallen through?)

 

[pasmith]

Would method of separation of variables lead to a solution to the following PDE?

$$ \frac{1}{r} \frac{ \partial}{\partial r} \left( kr \frac{ \partial T}{ \partial r}\right) = \rho c_p \frac{\partial T }{ \partial t }$$

This would be for the transient conduction of a hollow cylinder, of wall thickness ##r##, and thermal conductivity ##k##, with temperature distribution ##T(r,t)##

What is [itex]r[/itex]? Is is a constant (the wall thickness), in which case the inner radius also needs to be given, or is it the radial coordinate?

How do we tell if it will work out?

I imagine expanding like:

$$ \frac{k}{r} \left( \frac{ \partial T}{\partial r} + r \frac{ \partial^2 T}{\partial r^2} \right) = \rho c_p \frac{ \partial T}{\partial t}$$

Looking for a solution:

$$T(r,t) = f(r)g(t)$$

That would yield the following equation:

$$ \frac{1}{f(r)} \frac{k}{r} \left( f'(r) + r f''(r) \right) = \lambda= \rho c_p \frac{g'(t)}{g(t)} $$

Better to set [tex]
\frac{1}{rf(r)}(f'(r) + rf''(r)) = \lambda = \frac{\rho c_p}{k} \frac{g'(t)}{g(t)}[/tex] with [tex]
\begin{split}
r^2 f''(r) + rf'(r) - \lambda r^2 f(r) &= 0 \\
g'(t) - \frac{k\lambda}{\rho c_p} g(t) &= 0. \end{split}[/tex]. Setting [itex]\lambda = -c^2 < 0[/itex] and [itex]x = cr[/itex] for [itex]c > 0[/itex] turns the first equation into Bessel's equation with general solution [itex]f(r) = AJ_0(cr) + BY_0(cr)[/itex]. [itex]J_0[/itex] and [itex]Y_0[/itex] are oscillatory, with amplitutde decaying as [itex]r \to \infty[/itex]. You can use them to construct a series solution [tex]\sum_{n= 0}^\infty (a_n J_0(c_nr) + b_n Y_0(c_nr))g_n(t)[/tex] as you would a Fourier series; the eigenvalues [itex]c_n[/itex] are determined by the boundary conditions, and the eigenfunctions are orthogonal with respect to the inner product [tex]
\langle F, G \rangle = \int_a^b rF(r)G(r)\,dr.[/tex] [itex]J_0[/itex] is analytic at the origin, but [itex]Y_0[/itex] blows up, so if your domain includes the origin you can discard it.

If [itex]\lambda = c^2 > 0[/itex] then we have the modified Bessel's equation with general solution [itex]f(r) = AI_0(cr) + BK_0(cr)[/itex]. [itex]I_0[/itex] and [itex]K_0[/itex] are monotonic, with [itex]I_0[/itex] being analytic at the origin and increasing as [itex]r \to \infty[/itex]; [itex]K_0[/itex] blows up at the origin and decays as [itex]r \to \infty[/itex]. Again, if the domain includes the origin only [itex]I_0[/itex] is usable. You cannot use these to construct a series solution, just as you cannot use [itex]\cosh[/itex] and [itex]\sinh[/itex] to construct a series solution.

Here, for decaying solutions as [itex]t \to \infty[/itex], we want [itex]\lambda > 0[/itex].

We should not ignore the case [itex]\lambda = 0[/itex], which leads us to the final steady state solution. The radial dependence is given by [tex](rf')' = 0[/tex] with solution [itex]f(r) = A + B\ln r[/itex].

 

[erobz]

Minor In Mathematics here from an undergraduate degree in Mechanical Engineering Technology... I should have probably opened with that!

What is r? Is is a constant (the wall thickness), in which case the inner radius also needs to be given, or is it the radial coordinate?

Anyhow, ##r## would be the radial coordinate, within the cylindrical wall. I think the wall boundaries would be a part of my set of initial conditions.

I'm sure you answered my question...I'm just not sure where. This is the embarrassment I get from watching a YouTube video on solving PDE's and thinking I could tackle it. It seems like you are saying that it could work, in other words... a solution of this type may be valid depending on the roots?

With that last line you derived the SS solution for the cylinder that is in my heat transfer textbook...So that seems like a good sign.

I wouldn't expect oscillation to show up though, I think the temperature gradient would just decay spatially, and in time, (I could be wrong), does that mean anything for which solution I should consider?

Thank you for your reply.

 

[pasmith]

The individual eigenfunctions oscillate; a suitable series of them can be monotonic.