- #26

Chestermiller

Mentor

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This has gone a little off the rails. For one thing, you have used the symbol C for two different things and then confused them.Ok, then:

$$-k\left[\frac{\partial \Psi\Omega}{\partial r}\right]_{r=R}=h\Psi\Omega$$

$$-k\left[\Psi\frac{\partial\Omega}{\partial r}\right]_{r=R}=h\Psi\Omega$$

$$-k\left[\frac{\partial\Omega}{\partial r}\right]_{r=R}=h\Omega$$

$$\left[\frac{\partial\Omega}{\partial r}\right]_{r=R}=-\frac{h\Omega}{k}$$

Substituting in $$\frac{1}{r\Omega(r)}\frac{d}{dr}\left(r\frac{d\Omega(r)}{dr}\right)= C $$

leds to something like: $$-\frac{h}{kR\Omega(R)}\left(\Omega(R)+R\frac{d\Omega(r)}{dr}\right)= C $$

$$-\frac{h}{kR\Omega(R)}\left(\Omega(R)-R\frac{h\Omega(R)}{k}\right)= C $$

And finally ##C## is determined in terms of the thermal conductivity ##k##, the convective heat transfer coefficient ##h## and the cylinder radius ##R##:

$$C =-\frac{h}{kR}\left(1-R\frac{h}{k}\right) $$

Is this expression OK? At least, makes sense dimension-wise ##[m^-2]##. But on a first and quick look with the ##h## and ##k## values I found ##C## is positive. Tomorrow, I'll review it more calmly.

So, going back to your post # 18, I'm going to rewrite your equations slightly differently:

"A solution for ##\Psi(t)## is:

$$\Psi(t) = e^{\frac{-\beta^2t}{\rho C_p}}$$

And I've got this second order differential equation for ##\Omega(r)##:

$$ \frac{d^2\Omega(r)}{dr^2}+\frac{1}{r}\frac{d\Omega(r)}{dr}+\beta^2\Omega(r) = 0\tag{"}$$

The solution to the ##\Omega## equation is: $$\Omega=AJ_0(\beta r)+BY_0(\beta r)$$From the boundary condition at r = 0, the constant B is equal to zero. Therefore, $$\Omega=AJ_0(\beta r)$$What do you get when you substitute this into the following boundary condition? $$\left[\frac{\partial\Omega}{\partial r}\right]_{r=R}=-\frac{h\Omega}{k}$$Make use of the relationship:

$$\frac{dJ_0(x)}{dx}=-J_1(x)$$