Recent content by skrat

Yes, you are correct, my definition of ##\sigma## is a wrong by factor 2 on all off-diagonal terms. What if, excuse me for brainstorming, I wanted to take the last term from the weak formulation of the hookean law $$ \int _\Omega \nabla\cdot \sigma v \mathrm{d} \Omega $$ and decided to play...

Hi, I have an issue with understanding the body forces in the context of FEM simulations. I am using freefem to do the simulations. So I took a simple unit cube, fixed the displacements at side ##x=0## and prescribed a displacement ##u_x = 0.025## at ##x=1##. If required, the entire script is...

Thank you all for much information. And yes with the term "full stack engineer" I was referring to web developers. It goes without saying I am that D person. :) I am a physicist/engineer currently in the middle of my PhD in the field of numerical analysis of partial differential equations...

So, imagine a person that likes math/physics/engineering problems. He also likes science and programming. Let's call him D. D is in his mid-twenties, has much to learn about life and programming. He has the most experience in Pascal, C++, C, Python, Javascript. He also peaked into web...

You see that's what I don't understand and that's why I believe the solution should not be the same. The media between the plates is fluid (e.g. air at atmospheric pressure). While conductive heat transfer is transfer of energy by vibrations at molecular level thorough a solid or fluid, the...

Hmmm. Makes sense. Looks like my inuition is closer to time dependent solution rather than steady state. Would it be very hard to find the ##T(x, t)##?

A, sorry, I was too fast. ##\rho c_p \frac{\partial T}{\partial t} + \rho c_p \nabla( T\vec v) = \nabla(\lambda \nabla T)## because we are only interested in steady state, where ##\frac{ \partial T}{\partial t} =0## then ##\rho c_p \nabla \cdot( T\vec v) = \nabla \cdot (\lambda \nabla T).##...

I hope I'm getting this right, but a general heat equation is ##\rho c_p\frac{\partial T}{\partial t} + \rho c_p \nabla (T \vec v) = \lambda \nabla ^2 T.## Steady state problem and mass coninuity (##\nabla \cdot \vec v = 0##) yields ##\rho c_p \vec v \nabla T = \lambda \nabla ^2 T## or in...

I'm not sure about the horizontal component, I'd say it's zero. Vertical component tho (magnitude) should be very large close to the plates but 0, or very close to 0, in between. Correct?

I did not mention the plates are vertical, because I wanted to avoid any gravitational effect. Yet I see now how this is wrong. Plates are infinitely large. The temperature at the centerline is something between ##T_H## and ##T_C##. Velocity is probably zero there. If there were ne natural...

The only flow there is a result of natural convection. And yes, I am only interested in steady state solutions. The whole idea of this problem is to study the thermal layer. As a result I should be able to see that the higher the difference between ##T_H## and ##T_C## the thinner the thermal...

Homework Statement Imagine two parallel plates at distance ##L##. One of them at constant temperature ##T_H## and the other one at ##T_C##. Media between the plates is air. Find ##T## as function of distance from the plate with ##T_H## in steady state. Homework Equations Navier - Stokes Heat...

Anybody?

Yeah that's a result of my bad english and poor problem statement. So 1. We start with the dashed quadrilateral from the original post. 2. Then we rescale the BC and AD - almost never for the same factor, except in special cases. The BC side is rescaled along vector ##\hat{BC}## and AD along...

That would be amazing, but I don't see how that would be possible except in special cases. Note that the length of CD is constant, while sides AD an BC are (except in special cases) NOT paralel. So from my understanding, the angles are unchanged if and only if AD and BC are parallel and if both...