Recent content by rbnieto

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...

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...

I don't understand how can I get ##C## from this eq applying the BC's without knowing first the exact expression for ##\Omega(r)##: $$\frac{\rho C_p}{\Psi(t)}\frac{d\Psi(t)}{dt}=\frac{1}{r\Omega(r)}\frac{d}{dr}\left(r\frac{d\Omega(r)}{dr}\right)= C $$ Not exactly, I have a solid and homogeneous...

Ok, ##\Omega(r)## dif.eq. is similar to a modified bessel equation of order 0, $$r^2\frac{d^2\Omega(r)}{dr^2}+r\frac{d\Omega(r)}{dr}-Cr^2\Omega(r) = 0$$ but with a ##C## term that I do not know how take away to use the modified bessel function of order 0 as a solution (I know for sure that ##C##...

Ok, this is what I've done so far: I've plug ##T(t,r)## in the differential equation and got this: $$\rho C_p \frac{\partial \theta(t,r)}{\partial t} = \frac{1}{r}\frac{\partial}{\partial r}\left(r\frac{\partial \theta(t,r)}{\partial r}\right)$$ Making ##\theta(t,r) = \Psi(t)\Omega(r)## to...

Nice! I was wondering what to do at r = 0 because of that ##\ln{r}##. So ##C_1## has to be 0. Therefore, the final solution once the steady state has been reached is: $$T(r)=-\frac{q}{4k}r^2+\frac{q}{2h}R+\frac{q}{4k}R^2+T_0$$ $$T(r) = \frac{q}{2}(\frac{1}{h}R+\frac{R^2-r^2}{2k})+T_0$$ Now two...

Ok, after quite a while I'm back at work. Starting with this heat diffusion equation (steady state and no variation of T with angle and heigh of the cylinder): $$0 = \frac{1}{r}\frac{d}{dr}(r\frac{dT}{dr})+\frac{q}{k} $$ A solution for this equation is: $$T(r) =-\frac{q}{4k}r^2+C_1\ln{r}+C_2$$...

Thanx! I'll try and get back to you.

Is solid

I’ve found this solution that would solve the problem for the steady state situation but with an outer surface adiabatic. How do I introduce in the equation the efficiency of the environmental chamber to cool down the cylinder surface? I guess some convection and transmission parameters should...

I’ll try and get back here. Thanx

Hi, I have real live problem, similar to this one, that I do not know how to solve. I have a cylinder in air a T0 (environmental chamber, so I guess I have convection in there). In an initial time t0 we engage a heat source inside the cylinder (let's say the heat source is homogeneos within the...

Hi everybody, I'm a Physicits, I did my PhD on Civil Engineering and I research and teach in a Civil Engineering program. From time to time, I have theoretical questions that I'm too old to remember how to solve. I hope this forum helps me get some answer. In return, I'll try to solve other...