# Search results

1. ### Heat conduction from an isotherm spherical cap

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...
2. ### Heat conduction from an isotherm spherical cap

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.
3. ### Heat conduction from an isotherm spherical cap

Returning to the spherical cap, in a spherical coordinate system centered at the "center" of the cap, the PDE is: ##\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...
4. ### Heat conduction from an isotherm spherical cap

Semi-infinite solid.
5. ### Heat conduction from an isotherm spherical cap

You have a steady-state if nothing changes with time.
6. ### Heat conduction from an isotherm spherical cap

At the SL interface temperature is the melting temperature.
7. ### Heat conduction from an isotherm spherical cap

In the case of the sphere, one independent variable: ##r##, but for the spherical cap both ##r## and ##\theta##.
8. ### Heat conduction from an isotherm spherical cap

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...
9. ### Heat conduction from an isotherm spherical cap

On the surface of a semi-infinite solid, a point heat source releases a power ##q##; apart from this, the surface of the solid is adiabatic. The heat melts the solid so that a molten pool forms and grows. Let's hypothesize that the pool temperature is homogeneously equal to the melting...
10. ### I Approximation of a function with another function

Dear Office_Shredder, you are right. Both functions go to zero as ##x->0##, or ##y->0##, or both go to zero. Then, it is obvious that, under the same condition, their difference goes to zero as well. Actually, I was looking for a way to solve this integral: ##\int_0^{\infty } \frac{\exp...

12. ### I Approximation of a function with another function

Thank for your message. Let's call ##f1## the function: ## \pmb{\text{f1}=\frac{\text{Exp}\left[-\frac{1 }{2 \epsilon ^2} \left(\frac{\left(x-\epsilon ^2 t \right)^2+y^2}{t +1}\right)\right]}{\sqrt{t} (t +1)}}\\## And ##f2## the function: ##\pmb{\text{f2}=\frac{\text{Exp}\left[-\frac{1 }{2...
13. ### I Approximation of a function with another function

You are very kind. ##\int_0^{\infty } \frac{\text{Exp}\left[-\frac{1 }{2 \eta ^2} \left(\frac{\left(\xi -\eta ^2 \tau \right)^2+\psi ^2}{(\tau +1)}+\frac{\zeta^2}{\tau }\right)\right]}{\sqrt{\tau } (\tau +1)} \, d\tau##
14. ### I Approximation of a function with another function

##\int^{\infty }_0{\frac{e^{-\ \ \frac{1}{2{\varepsilon }^2\ }\left[\frac{{(x-{\varepsilon }^2t)}^2+y^2}{t+1}\right]}}{\sqrt{t}(t+1)}}dt## thank you, it was helpful
15. ### I Approximation of a function with another function

You are full right! It was my fault. Unfortunately, I cannot write formulae correctly in this forum, so I did a picture of my integrals, but the resolution was very bad. The correct integrals are: and . I lost the minus in the conversion process. You can interpret x and y as related by the...
16. ### I Approximation of a function with another function

Hi, I am wondering if it is possible to demonstrate that: tends to in the limit of both x and y going to infinity. In this case, it is needed to introduce a measure of the error of the approximation, as the integral of the difference between the two functions? Can this be viewed as a norm...
17. ### Saturation magnetization of iron as a function of temperature

Dear essenmein, could you please clarify this point? I think this explains why some material loose the magnetization as the external field is removed.
18. ### Saturation magnetization of iron as a function of temperature

essenmein, Charles Link, now all is much more clear. Thank you so much.
19. ### Saturation magnetization of iron as a function of temperature

I forgot to tell that the picture was depicting the situation below the curie point.
20. ### Saturation magnetization of iron as a function of temperature

Can i try to summarize with a picture? So, the saturation is always the same, but the higher the temperature is, the higher the field needed to reach it; then, at a given external field, the same material can be at saturation at a certain temperature but can be not at saturation at an higher...
21. ### Saturation magnetization of iron as a function of temperature

Is there a law to describe this phenomenon?
22. ### Saturation magnetization of iron as a function of temperature

Approaching the Curie temperature, the saturation magnetization (the magnetic moment per unit volume at saturation) of a ferromagnetic material should not go to zero?

many thanks!
24. ### Saturation magnetization of iron as a function of temperature

Hi, I am looking for the temperature dependence of the saturation magnetization of Fe. Any help?
25. ### Cylindrical magnet

I have just checked, it gives exactly the corrected MIT expression!!!

Right!!
27. ### Cylindrical magnet

Yes, on the same side of the Surface, u=0. In this way I did the calculation: I took the expression provided for Bz in Wolfram and I set ρ = 0, z =L/2. I did the integration, to get the result: Then I took the MIT corrected (with u = 0) or my expression and I get:
28. ### Cylindrical magnet

I mean the field on the magent Surface coming from the outside of the magnet
29. ### Cylindrical magnet

This is the problem, you do not get the MIT expression for Bz(0,L/2, "or the surface of the magnet") but: While the MIT expression is:
30. ### Cylindrical magnet

The integration variables are R and Φ. R integrated over [0, a] and Φ over [0, 2 Pi]. After the integration, you have the field in generic ρ and z, namely Bz(ρ,z). I want the field in (ρ = 0, z =L/2), i.e. Bz(0,L/2).