Transient heat transfer in a cylinder with internal heating

  • Thread starter Bruce
  • Start date
  • #26
20,231
4,264
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.
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.

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:
$$\frac{dJ_0(x)}{dx}=-J_1(x)$$
 
  • #27
12
1
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.
Sorry about that. And thank you very much for your guided assistance. I'm enjoying a lot the process.
In my case ##\Omega(r) = AJ_0(\beta r)## and ##\frac{dJ_0(\beta r)}{dr} = -\beta J_1(\beta r)##, then: $$ \frac{\partial}{\partial r}\left(A J_0(\beta r)\right) = -A\beta J_1(\beta r)$$
Applying the BC at ##r=R##:
$$-A\beta J_1(\beta R)= -\frac{h}{k}AJ_0(\beta R)$$
##A## goes away, and the only constant I don't know here is ##\beta##.
$$\beta -\frac{h}{k}\frac{J_0(\beta R)}{J_1(\beta R)}=0$$
Now I need to look at Bessel's function properties to see if I can extract ##\beta## from here. Tomorrow I'll try.
Thanx again.
 
  • #28
20,231
4,264
Sorry about that. And thank you very much for your guided assistance. I'm enjoying a lot the process.
In my case ##\Omega(r) = AJ_0(\beta r)## and ##\frac{dJ_0(\beta r)}{dr} = -\beta J_1(\beta r)##, then: $$ \frac{\partial}{\partial r}\left(A J_0(\beta r)\right) = -A\beta J_1(\beta r)$$
Applying the BC at ##r=R##:
$$-A\beta J_1(\beta R)= -\frac{h}{k}AJ_0(\beta R)$$
##A## goes away, and the only constant I don't know here is ##\beta##.
$$\beta -\frac{h}{k}\frac{J_0(\beta R)}{J_1(\beta R)}=0$$
Now I need to look at Bessel's function properties to see if I can extract ##\beta## from here. Tomorrow I'll try.
Thanx again.
Rewriting your equation in a little different way yields:$$\lambda J_1(\lambda)=\left(\frac{hR}{k}\right)J_0(\lambda)$$where ##\lambda=\beta R##. This equation has an infinite number of solutions in ascending values for ##\lambda##. Abramowitz and Stegun present a table of results at various values of ##hR/k##.
 

Related Threads on Transient heat transfer in a cylinder with internal heating

  • Last Post
Replies
10
Views
7K
  • Last Post
Replies
1
Views
2K
Replies
1
Views
5K
Replies
2
Views
4K
Replies
2
Views
5K
  • Last Post
Replies
1
Views
4K
  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
10
Views
1K
  • Last Post
Replies
2
Views
3K
Top