- #1

umby

- 47

- 8

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter umby
- Start date

- #1

umby

- 47

- 8

- #2

BvU

Science Advisor

Homework Helper

- 15,278

- 4,254

I see a lot of inconcistencies in your scenario -- if I understand it correctly.

meaning no heat transport takes place ?pool temperature is homogeneously equal to the melting temperature

And why would that be so ?A steady-state regime eventually is reached

- #3

berkeman

Mentor

- 64,194

- 15,443

Looks like @umby has been at PF for 5 years with nothing in the schoolwork forums so far. That would make me think it's not schoolwork, but it sure looks like it.Is this homework ?

@umby if this is for your schoolwork, I can move it to the schoolwork forums for you. Either way you need to show us your work on the problem so far before we can be of help. Thanks.

- #4

umby

- 47

- 8

Let's suppose there is a heat point source of power ##q## in an infinite solid initially at temperature ##T_0##. This source melts part of the solid forming a spherical melt pool; the density does not change. If the temperature in the pool is homogeneous (high conduction rate, high conductive transport in the liquid) and does not exceed the melting temperature ##T_1##, I have an isotherm sphere of radius ##r_0## in an infinite medium at a given time. In this case, the temperature field outside the sphere is known to be:

##T=\frac{r_0 \left(T_1-T_0\right)}{r}+T_0##

The heat flux on the surface of the sphere is then:

##-k\frac{\partial T}{\partial r}\bigg| _{r=r_0}=k \frac{\left(T_1-T_0\right)}{r_0}##

The sphere will stop growing, namely the source will stop melting new solid, when ##q## will be equal to the heat flow through the surface. If we call ##r_{\max}## the sphere radius in this moment, then

##q=\frac{4 k \left(T_1-T_0\right) \pi r_{\max }^2}{r_{\max }}##

From which

##r_{\max }=\frac{q}{4 k\pi \left(T_1-T_0\right) }##

I call this steady-state because the spherical molten pool does not grow anymore and its radius remains equal to ##r_{max}##. In this case, it is probably easy to introduce a temperature gradient in the sphere as well and solve the problem, however the previous formula can provide a quick first estimate of the sphere radius in the proposed conditions. Is it correct to say that that that kind of stationary state is reached?

- #5

Chestermiller

Mentor

- 22,340

- 5,207

Sure. It's the same problem, except that all the q goes to the cap instead of to each half.

- #6

BvU

Science Advisor

Homework Helper

- 15,278

- 4,254

Problem statement is clear now, thanks. I don't think this example is any different from the original (symmetry)I apologize for the bad exposition of the problem. I will try to explain my thoughts with another example.

Let's suppose there is a heat point source of power ##q## in an infinite solid initially at temperature ##T_0##. This source melts part of the solid forming a spherical melt pool; the density does not change.

No can do. As long as the heat source is on, heat will need to flow outward, so no isothermal sphere, no matter how high the conductivity.If the temperature in the pool is homogeneous (high conduction rate, high conductive transport in the liquid) and does not exceed the melting temperature ##T_1##, I have an isotherm sphere of radius ##r_0## in an infinite medium at a given time.

There will be a radial temperature profile with a small temperature difference at the SL interface.

Bottom line is: what happens at the SL interface, where L side is by definition at ##T_L > T_m\ ## and S side at ##T_S <T_m\ ##. So far I ve been playing devil's advocate, but I can imagine q being dissipated outwards

without ##T_S## exceeding ##T_m##.

It's not a real steady state but the sphere stops growing.

Imagining is one thing; I will be convinced when I see it quantified

Happy to see Chet come in ! Any conclusive insight ?

##\ ##

- #7

umby

- 47

- 8

In the case of the sphere, one independent variable: ##r##, but for the spherical cap both ##r## and ##\theta##.Problem statement is clear now, thanks. I don't think this example is any different from the original (symmetry)

- #8

umby

- 47

- 8

At the SL interface temperature is the melting temperature.Bottom line is: what happens at the SL interface, where L side is by definition at and S side at . So far I ve been playing devil's advocate, but I can imagine q being dissipated outwards

without exceeding .

- #9

umby

- 47

- 8

You have a steady-state if nothing changes with time.It's not a real steady state but the sphere stops growing.

- #10

Chestermiller

Mentor

- 22,340

- 5,207

Not if the bottom is insulated. Then it is still only a function of r.In the case of the sphere, one independent variable: ##r##, but for the spherical cap both ##r## and ##\theta##.

- #11

umby

- 47

- 8

Semi-infinite solid.Not if the bottom is insulated. Then it is still only a function of r.

- #12

Chestermiller

Mentor

- 22,340

- 5,207

Semi-infinite solid

I'm assuming that the semi-infinite solid is above my bottom plane.

- #13

umby

- 47

- 8

##\frac{\partial }{\partial r}\left(r^2\frac{\partial T}{\partial r}\right)+\frac{1}{\sin \theta }\frac{\partial }{\partial \theta }\left(\sin \theta \frac{\partial T}{\partial \theta }\right)=0##

With boundary conditions in dimensionless form given by:

##T=0##, ##r\to \infty##, this sets ##T## equal to the initial value far from the cap,

##T=1##, ##r=\frac{1}{2} \left(\sqrt{2} \sqrt{-p^2+(1-p)^2 \cos (2 \theta )+2 p+1}+2 (1-p) \cos (\theta )\right)##, this imposes ##T## on the cap

##\frac{\partial T}{\partial \theta}\bigg| _{\theta=\pi/2}=0##

##\frac{\partial T}{\partial \theta}\bigg| _{\theta=3\pi/2}=0##, those are the adiabatic conditions.

Do you think it is possible to find a solution to this problem?

- #14

hutchphd

Science Advisor

Homework Helper

2022 Award

- 5,390

- 4,546

I got no LateX ...did you use ##? OK

Last edited:

- #15

hutchphd

Science Advisor

Homework Helper

2022 Award

- 5,390

- 4,546

- #16

Chestermiller

Mentor

- 22,340

- 5,207

That's what I've been trying to say.

- #17

BvU

Science Advisor

Homework Helper

- 15,278

- 4,254

Convice me ...At the SL interface temperature is the melting temperature.

- #18

Chestermiller

Mentor

- 22,340

- 5,207

Also, it is perfectly obvious to several of us that the solution is the same as for a point source in an infinite medium, meaning that the solution is not a function of ##\theta##.

- #19

umby

- 47

- 8

I try to upload an image of the surface of the semi-infinite solid. The spherical cap is isotherm while the rest of the surface is adiabatic. The semi-infinite solid is beneath the surface. I apologize for the misunderstandings.

Last edited:

- #20

hutchphd

Science Advisor

Homework Helper

2022 Award

- 5,390

- 4,546

I don't think that will matter steady state?Does this analysis take into account the heat of fusion of the solid?

Yes, I can see that now. Thanks.

Last edited by a moderator:

- #21

umby

- 47

- 8

I refer to the first post. The cap is not meant to be a hemisphere. In normalization, length dimensions were re-scaled by means of ##R## (the radius of the sphere from which the spherical cap is derived), this means that ##p##, the depth of the cap below the surface, can vary from 0 to 1 (0 = no cap, 1 = hemisphere). I would like to calculate the stationary temperature field in this semi-infinite solid, with this strange spherical cap at constant temperature ##T=T_1## on its surface (in dimensionless form ##T=1##). Far from the surface the temperature is ##T=T_0## (in dimensionless form ##T=0##). In a spherical coordinate system (physical convention, polar axis perpendicular to the surface) centered at the "center" of the cap, the dimensionless equations should be:Is the cap meant to be a perfect hemisphere (it looks like less how much less?). To which post does this refer ?

$$\frac{\partial }{\partial r}\left(r^2\frac{\partial T}{\partial r}\right)+\frac{1}{\sin \theta }\frac{\partial }{\partial \theta }\left(\sin \theta \frac{\partial T}{\partial \theta }\right)=0$$

B.C

##T=0##, ##r\to \infty## (##T## equal to the initial one far from the cap),

##T=1##, ##r=\frac{1}{2} \left(\sqrt{2} \sqrt{-p^2+(1-p)^2 \cos (2 \theta )+2 p+1}+2 (1-p) \cos (\theta )\right)## (##T## on the cap, this is the equation of the cap in the chosen reference system),

##\frac{\partial T}{\partial \theta}\bigg| _{\theta=\pi/2}=0## (adiabatic condition),

##\frac{\partial T}{\partial \theta}\bigg| _{\theta=\pi}=0## (symmetry conditions).

I hope that the problem formulation is correct. I have already tried to solve the problem hypothesizing ##T(r,\theta)=f(r) g(\theta)##, but this does not seem to give good results. I am sure on this point however.

Last edited:

Share:

- Replies
- 4

- Views
- 242

- Last Post

- Replies
- 1

- Views
- 214

- Last Post

- Replies
- 2

- Views
- 561

- Last Post

- Replies
- 2

- Views
- 117

- Last Post

- Replies
- 28

- Views
- 611

- Last Post

- Replies
- 3

- Views
- 348

- Replies
- 6

- Views
- 945

- Replies
- 22

- Views
- 220

- Last Post

- Replies
- 0

- Views
- 59

- Replies
- 2

- Views
- 367