Transient heat transfer in a cylinder with internal heating

In summary, the problem is to find the transient heat distribution in a cylinder with internal heating. The boundary conditions are two constant temperatures, inner and outer surface. The governing equation is: dT/dt=k(d2T/dr2+(1/r)*dT/dr) where rw<r<∞. r(rw,t) = t1 for all t. It looks like a hollow cylinder problem, but fortunately I found a solution for the heat transfer with two constant temperatures in literatures. 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
  • #1
Bruce
4
0
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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  • #2
I think i’ve Seen the solution to this somewhere. I”lol check it out tomorrow.
 
  • #3
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)
 
  • #4
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.
 
  • #5
Bruce said:
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.
 
  • #6
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 canceled 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
 
  • #7
The first step is to solve for the final steady state temperature profile at long times. Do you think you can do that?
 
  • #8
I’ll try and get back here. Thanx
 
  • #9
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/
 
  • #10
Is the cylinder solid or hollow?
 
  • #11
Is solid
 
  • #12
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: $$-k\left[\frac{dT}{dr}\right]_{r=R}=h(T-T_0)$$
 
  • #13
Thanx! I'll try and get back to you.
 
  • #14
Chestermiller said:
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!
 
  • #15
At r = 0, T is finite. What does that tell you about C1?
 
  • #16
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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  • #17
rbnieto said:
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?
 
  • #18
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.
 
  • #19
Check out Bessel's functions.
 
  • #20
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##
 
  • #21
OK so far. C is determined from the boundary condition at r = R. For ##\theta##, what is that?
 
  • #22
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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  • #23
Chestermiller said:
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)##:
$$\frac{\rho C_p}{\Psi(t)}\frac{d\Psi(t)}{dt}=\frac{1}{r\Omega(r)}\frac{d}{dr}\left(r\frac{d\Omega(r)}{dr}\right)= C $$
Chestermiller said:
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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  • #24
rbnieto said:
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##.
 
  • #25
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.
 
  • #26
rbnieto said:
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
Chestermiller said:
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
rbnieto said:
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##.
 

1. What is transient heat transfer in a cylinder with internal heating?

Transient heat transfer refers to the temporary transfer of heat within a system, meaning that the temperature changes over time. In a cylinder with internal heating, heat is being generated within the cylinder, causing a change in temperature over time.

2. How does the heat transfer occur in this system?

In a cylinder with internal heating, heat transfer occurs through conduction, convection, and radiation. Conduction is the transfer of heat through direct contact between the cylinder and the heating source. Convection is the transfer of heat through the movement of fluids within the cylinder. Radiation is the transfer of heat through electromagnetic waves.

3. What factors influence the rate of heat transfer in this system?

The rate of heat transfer in a cylinder with internal heating is influenced by several factors, including the temperature difference between the cylinder and the heating source, the material properties of the cylinder, the surface area of the cylinder, and the presence of any insulating materials.

4. Can the rate of heat transfer be controlled in this system?

Yes, the rate of heat transfer in a cylinder with internal heating can be controlled by adjusting the heating source, changing the material properties of the cylinder, or adding insulating materials to reduce heat loss.

5. How is transient heat transfer in a cylinder with internal heating used in real-world applications?

This type of heat transfer is commonly used in industrial processes, such as in chemical reactors and heat exchangers. It is also relevant in engineering fields such as aerospace, as it can affect the thermal performance of components like engines and turbines.

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