# Transient heat transfer in a cylinder with internal heating

#### Bruce

hi, I met a problem about heat transfer in cylinder, if you can help, I will appreciate it.

The question is simple. I want to know the transient heat distribution in a cylinder with internal heating(constant temperature not constant flux). The boundary conditions comprises two constant temperature, inner and outer surface. we assume the cylinder radius is infinite. therefore, we have

the governing equation:
dT/dt=k(d2T/dr2+(1/r)*dT/dr)
rw<r<∞
i.c. : r(r,0)=t0
b.c.: r(∞, t) = t0
r(rw ,t) = t1

It looks like a hollow cylinder problem. But unfortunately, with surprise, I failed to find a solution for the heat transfer with two constant temperature in literatures. In Carslaw and Jaeger, I only found one solution for constant heat flux internal heating. So I derived the solution by Laplace Transform and that gives me a solution with Bessel function , here, I just want to find a example to verify my solution.

Thank you very much in advance !!

Bruce

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#### Chestermiller

Mentor
I think i’ve Seen the solution to this somewhere. I”lol check it out tomorrow.

#### Chestermiller

Mentor
See Practical Aspects of Groundwater Modeling by William C. Walton, National Water Well Association, 1984, Section 5.3.1

Jacob, C. E., and S. W. Lohman, 1952, Nonsteady flow to a well of a constant drawdown in an extensive aquifer, Trans. Am. Geophys. Union, Vol. 33, No. 4

Hantush, M. S., 1964, Hydraulics of wells., In Advances in Hydroscience, Vol 1., Academic Press Inc., New York.

Hantush, M.s., 1954, Drawdown around wells of variable discharge, Jour. of Geophys. Res., 69 (20)

#### Bruce

hi, Chestermiller, thank you very much for the reference. Actually, I derived my solution using the method from groundwater pumping. But I am just curious if there is a solution from the aspect of heat transfer, which is more straightforward. still appreciate your help !! By adjusting the boundary value, it is also possible to make a verification.

#### Chestermiller

Mentor
hi, Chestermiller, thank you very much for the reference. Actually, I derived my solution using the method from groundwater pumping. But I am just curious if there is a solution from the aspect of heat transfer, which is more straightforward. still appreciate your help !! By adjusting the boundary value, it is also possible to make a verification.
The equations are exactly the same. So, in place of hydraulic diffusivity, just use thermal diffusivity, and in place of head, just use temperature.

#### rbnieto

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 whole cylinder). I want to know how the average temperature of the cylinder would change with time and if a steady state will be reached (I guess depending on T0, the eficinecy of air convection and the power of the heat source, the generated heat will or will not be cancelled by the cooling).
I think the equation that describes the temperature T(t,r) with t and r (radius of the cylinder) is something like this:
1/r·d/dr(r·dT/dr)+1/k·dQ/dt = 1/C·dT/dt
where k and C are constants that I can get from tests
and Q is the heat source (I know exactly how many joules/time are delivered to the cylinder)
I've seen solutions to problems very close to this one, but assuming adiabatic surface and steady state. I'm precisely interested in what happens before reaching the steady state, and there has to be heat exchange throught the surface of the cylinder.
Good luck

#### Chestermiller

Mentor
The first step is to solve for the final steady state temperature profile at long times. Do you think you can do that?

#### rbnieto

I’ll try and get back here. Thanx

#### rbnieto

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 be obtained experimentally for this particular configuration and they should be introduced in Q (the heat generation function), but I don’t know how.

http://demonstrations.wolfram.com/HeatTransferThroughACylinder/

#### Chestermiller

Mentor
Is the cylinder solid or hollow?

Is solid

#### Chestermiller

Mentor
Do you know how to solve your initial equation for the steady state solution with a convective boundary condition (characterized by a specified heat transfer coefficient h) at the surface of the cylinder: $$-k\left[\frac{dT}{dr}\right]_{r=R}=h(T-T_0)$$

#### rbnieto

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

#### rbnieto

Do you know how to solve your initial equation for the steady state solution with a convective boundary condition (characterized by a specified heat transfer coefficient h) at the surface of the cylinder: $$-k\left[\frac{dT}{dr}\right]_{r=R}=h(T-T_0)$$
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: $$T(r) =-\frac{q}{4k}r^2+C_1\ln{r}+C_2$$
If I plug into $T(r)$ the convective boundary condition you proposed it yields: $$\frac{q}{2}R-\frac{k}{R}C_1 = -\frac{qh}{4k}R^2+hC_1\ln{R}+hC_2-hT_0$$
Is this right?
If it is, I can obtain $C_1$ in terms of $C_2$, for instance, but I'm still missing one equation to determine the other constant.
Again, thank you for your assistance!

#### Chestermiller

Mentor
At r = 0, T is finite. What does that tell you about C1?

#### rbnieto

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 more questions:
1) what kind of test should I run to get an approximation to the convection coefficient h?
3) how do I go from here to get dependecy of T with time before the steady state is reached?

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#### Chestermiller

Mentor
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 more questions:
1) what kind of test should I run to get an approximation to the convection coefficient h?
You can find estimates in the literature of the convective heat transfer coefficient (Nussult number) from a horizontal cylinder. Or you can do a test at some specified heating rate.
3) how do I go from here to get dependecy of T with time before the steady state is reached?
So far, you have determined the long-time (steady state) temperature profile $T_{ss}(r)$. You now can write $$T(t,r)=T_{ss}(r)+\theta(t,r)$$
What do you get if you substitute this into the differential equation and boundary conditions?

#### rbnieto

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 separate variables, I've got:
$$\frac{\rho C_p}{\Psi(t)}\frac{d\Psi(t)}{dt}=\frac{1}{r\Omega(r)}\frac{d}{dr}\left(r\frac{d\Omega(r)}{dr}\right)= C$$
A solution for $\Psi(t)$ is:
$$\Psi(t) = e^{\frac{Ct}{\rho C_p}}$$
And I've got this second order differential equation for $\Omega(r)$:
$$\frac{d^2\Omega(r)}{dr^2}+\frac{1}{r}\frac{d\Omega(r)}{dr}-C\Omega(r) = 0$$
Can I assume that, since this last dif.eq. has to be valid for the whole cilinder ($r=[0,R]$), that $\frac{d\Omega(r)}{dr} = 0$? Or could there be a $r$ term in $\frac{d\Omega(r)}{dr}$ that cancels with $\frac{1}{r}$?
I'm looking around to find a solution to that last diff. eq. but so far I haven't been able to find it.

#### Chestermiller

Mentor
Check out Bessel's functions.

#### rbnieto

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 $C$ has to be negative, otherwise the $\Psi (t) = e^\frac{Ct}{\rho C_p}$ won't make any sense since $\Psi$ has to $\rightarrow 0$ when $t \rightarrow \infty$). Without that $C$ the solution would be $\Omega(r) = c_1I_0(r)+c_2K_0(r)$ but since the Temperature in $r=0$ has to be finite, the solution are only the first kind modified Bessel's functions and $c_2$ has to be 0.
Right now I'm stuck on this $C$

#### Chestermiller

Mentor
OK so far. C is determined from the boundary condition at r = R. For $\theta$, what is that?

#### Chestermiller

Mentor
Is this problem one where you have a cylindrical cavity in an infinite medium and, at time t = 0, you suddenly change the temperature at the surface of the cavity wall from $T_0$ to $T_1$, and then hold it at that temperature for all subsequent times?

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#### rbnieto

OK so far. C is determined from the boundary condition at r = R. For $\theta$, what is that?
I don't understand how can I get $C$ from this eq applying the BC's without knowing first the exact expression for $\Omega(r)$:
$$\frac{\rho C_p}{\Psi(t)}\frac{d\Psi(t)}{dt}=\frac{1}{r\Omega(r)}\frac{d}{dr}\left(r\frac{d\Omega(r)}{dr}\right)= C$$

Is this problem one where you have a cylindrical cavity in an infinite medium and, at time t = 0, you suddenly change the temperature at the surface of the cavity wall from $T_0$ to $T_1$, and then hold it at that temperature for all subsequent times?
Not exactly, I have a solid and homogeneous cylinder inside an environmental chamber that controls temperature by pumping cold or hot air, so a solid cylinder in a fluid that moves around it. Before $t=0$ the temperature of all parts of the cylinder and the air temperature are the same $T_0$ (in my case 10ºC). At $t=0$ a theoretical heat source that is homogeneous within the cylinder and constant starts to deliver heat to the cylinder, while the air around it is kept at 10ºC. I want to know how long it take for the temperature of the cylinder to reach the steady state, and which is the temperature of the cylinder once the steady state is reached.

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#### Chestermiller

Mentor
I don't understand how can I get $C$ from this eq applying the BC's without knowing first the exact expression for $\Omega(r)$:
$$\frac{\rho C_p}{\Psi(t)}\frac{d\Psi(t)}{dt}=\frac{1}{r\Omega(r)}\frac{d}{dr}\left(r\frac{d\Omega(r)}{dr}\right)= C$$
The boundary condition on $\theta$ is $$-k\left[\frac{\partial \theta}{\partial r}\right]_{r=R}=h\theta$$
That means that C has to be such that $\Omega$ satisfies this boundary condition. Also, C has to be negative so that the transient solution decays in time. So you are dealing with $J_0$, not $I_0$.

#### rbnieto

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.

"Transient heat transfer in a cylinder with internal heating"

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