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Bessel Functions / Eigenvalues / Heat Equation 
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#19
Feb208, 07:16 AM

P: 25

Thank you for all your assistance and help Does the solution to the problem changes a lot if the heat source is not uniform? Best Regards phioder 


#20
Feb308, 04:30 AM

P: 279

Hello phioder,
I finally found some time to work on the equation. First some remarks on the boundary conditions you gave. The way to describe a zero heat flow at a boundary is by stating the following for the top and bottom of the cylinder: [tex]\frac{\partial u}{\partial z}=0[/tex] For the bottom this is for z=0, and for the top z=4, for all r. At the circular bondary you can write: [tex]\frac{\partial u}{\partial r}=0[/tex] For r=2, 0<z<4 The equation: [tex]\frac{\partial u}{\partial z}=0[/tex] for r=2, 0<z<4 does not hold here because the heat flow is perpendicular to the surface area. This means that the first and last boundary condition you gave are not applicable. The second and third one describe the bottom and circular surface of the cylinder. The top is: [tex]u(r,4,t)=u_0, 0<r<2[/tex] Now there is nothing here about the time. In case I assume that the temperature initially was zero in the cylinder you have enough conditions to define the problem. To summarise the entire question: Solve the following equation in a cylinder with radius 2 and height 4: [tex]\frac{\partial u}{\partial t} = \frac{k}{c_p \rho} \cdot \left(\frac{\partial^2 u}{\partial r^2}+\frac{1}{r} \cdot \frac{\partial u}{\partial r}+\frac{\partial^2 u}{\partial z^2} \right)[/tex] where the cylinder initially was at u=0. The boundary conditons are: [tex]\frac{\partial u}{\partial z}=0[/tex] for z=0, [tex]\frac{\partial u}{\partial r}=0[/tex] for r=2, and the top z=4 is at a temperature [tex]u=u_0[/tex] The solution must be bounded. This is becoming a complicated problem. Because of the time dependency of the solution and the fact that the top boundary is not at a value 0 together with the fluxes (derivatives) set at 0 for the other boundaries it is not an easy one. Are you sure this is the case you are looking into? It might very well be that there is no solution at all using the FourierBessel technique. I will give it a try if you state that this is indeed the one you need. The solution will change in case the heat source is not uniform. Assuming that this is what you mean, that the u_0 is not a constant. The problem will become more difficult in case the heat source is phi dependent. However if the heat source is only depending on the radius, and a solution is possible for the constant heat source, it has one in this more general case as well. If we were to solve this, the first thing to do is to split the problem up into two smaller ones. One which gives the steadystate and the other the transient part of the complete solution. I haven't done this yet on such a complicated model, but if you assure that the question is the one you need I will give it a try. There is something similar in Spiegel, exercise 6.91, however the dependency of z is not present due to the assumption of an infinite long cylinder. This makes the problem a bit more solvable. Hope this helps you one step further. 


#21
Feb308, 11:56 AM

P: 25

Thank you very much for your answer. You make special comments that make me further understand small details I haven't understand yet. After doing some research, as much as I understand now, the problem is indeed complicated and not trivial. During all this thread I have tried to split the problem and state basic questions because my knowledge and experience in PDE is very limited. To reduce the complexity splitting the problem was based on: 1. 2D steady heat conduction, no transients, no time dependency, no phi dependency, dependencies only on r and z 2. For a first try the heat source considered as uniform 3. For a first try not von Neumann boundary conditions, the heat needs to enter the cylinder from the upper part, all other sides are insulated The equation is following: [tex]\frac{g(r,z)}{k} = \frac{\partial^2 u}{\partial r^2}+\frac{1}{r} \cdot \frac{\partial u}{\partial r}+\frac{\partial^2 u}{\partial z^2} [/tex] From this link http://www.engr.unl.edu/~glibrary/gl...05b/node4.html what I understand is that the function g(r,z) is the heat source. They give a solution with green functions, unfortunately I don't understand them to implement a program. If the heat souce is constant the equation is following: [tex]\frac{u_0}{k} = \frac{\partial^2 u}{\partial r^2}+\frac{1}{r} \cdot \frac{\partial u}{\partial r}+\frac{\partial^2 u}{\partial z^2} [/tex] Unfortunately I began to loose focus and my post can't be well formulated, my apologies Best Regards phioder PS: What does it mean that a solution is "bounded" or "not bounded" 


#22
Feb308, 02:14 PM

P: 279

OK, phioder, now I understand the heat source. It is nothing more than the extension
of the equation we used all along. We had the following: [tex]\frac{\partial u}{\partial t} = \frac{k}{c_p \rho} \cdot \left(\frac{\partial^2 u}{\partial r^2}+\frac{1}{r} \cdot \frac{\partial u}{\partial r}+\frac{\partial^2 u}{\partial z^2} \right)[/tex] Which can be generalized for the case with a heat source to become: [tex]\frac{\partial u}{\partial t} = \frac{k}{c_p \rho} \cdot \left(\frac{\partial^2 u}{\partial r^2}+\frac{1}{r} \cdot \frac{\partial u}{\partial r}+\frac{\partial^2 u}{\partial z^2} \right)+ \frac{E}{c_p \rho}[/tex] In which E the heat source is. The units of E are W/m^3. And for the steadystate this is then: [tex]0=\frac{k}{c_p \rho} \cdot \left(\frac{\partial^2 u}{\partial r^2}+\frac{1}{r} \cdot \frac{\partial u}{\partial r}+\frac{\partial^2 u}{\partial z^2} \right)+ \frac{E}{c_p \rho}[/tex] Or: [tex]\frac{E}{k}=\frac{\partial^2 u}{\partial r^2}+\frac{1}{r} \cdot \frac{\partial u}{\partial r}+\frac{\partial^2 u}{\partial z^2}[/tex] This is indeed not trivial, I am not sure whether this can be solved by the method of FourierBessel series expansion. I will have a look in my books and see what I can find. Nevertheless the separation of the variables and the use of the series together with the study of the very interesting Bessel functions has not been a waste of time. [Edit: name changed] A solution that is bounded is simply stated one that is finite. It is one that does not diverge to infinity for some point(s). Consider a graphic of the Bessel function of the second type of order 0, you will see that for x going towards 0, the function goes to infinity. This is not realistic in our problem. If we state that the solution must be bounded, we simply eliminate these solutions from the set of possible solutions. This does not mean that it is a bad solution, it is a perfect valid one for the math, but not for the physics, it is not allowed here. Otherwise a solution that is unbounded is then one that can diverge to infinity. 


#23
Feb408, 04:04 PM

P: 25

Absolutely not a waste of time, your answers give me a great feedback to understand better the equation and also included a lot of motivation. IMPO there are many small things that are not explained in books that save paper and ink and on the other side engineers are lost. This forum is a great place to share knowledge, ideas and questions phioder 


#24
Feb908, 10:33 AM

P: 279

I assume that is not easy to do with the use of the separation of the variables and then to use the FourierBessel expansion. Otherwise I definitely would have bumped into an exercise or an example somewhere. I think the other methods are more applicable. But here I can't help you. It has been too long since I read about it. I never actually had to use them. I would suggest to start a new post with the exact description of the problem you would like to see solved. There are certainly people who are willing to help you with the Green's functions. Thank you very much for your kind words along the way and I'm very happy that I was of some assistance in getting you to appreciate (amongst a large set with many more interesting) Bessel functions. 


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