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From cartesian to spherical coordinates
Posted Oct 19, 2013, 5:08 a.m. EDT Parameters, Variables, & Functions 1 Reply
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Hi,
I am trying to solve a non-steady state problem in spherical coordinates using a square as a geometry for possible latter extrusions to other linear geometry.
Thus the real equation I want to solve is (du/dt = grad(-c*r^2 grad(u)))
In the generic equation, (du/dt =grad (-c grad (u))), I introduce the diffusion term as follows c= D*y^2 being y^2 the height of the square (effectively the radial direction). I then introduce y^2 as well in the mass coefficient (as if all the original equation had been multiplied by y^2).
Is this approach correct?
thanks
I am trying to solve a non-steady state problem in spherical coordinates using a square as a geometry for possible latter extrusions to other linear geometry.
Thus the real equation I want to solve is (du/dt = grad(-c*r^2 grad(u)))
In the generic equation, (du/dt =grad (-c grad (u))), I introduce the diffusion term as follows c= D*y^2 being y^2 the height of the square (effectively the radial direction). I then introduce y^2 as well in the mass coefficient (as if all the original equation had been multiplied by y^2).
Is this approach correct?
thanks
1 Reply Last Post Oct 19, 2013, 6:35 a.m. EDT
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Posted:
1 decade ago
My answer to my own question, in case it is useful for anyone ;-)
You can 'change the coordinates' if you properly define your equation (by default in cartesians)
See attachment (taken from 'Extraction optimization in food engineering' By Tzia and Liadakis ,in chapter 2, 'Solid-liquid extraction' by J.M Aguilera)
Regards,
You can 'change the coordinates' if you properly define your equation (by default in cartesians)
See attachment (taken from 'Extraction optimization in food engineering' By Tzia and Liadakis ,in chapter 2, 'Solid-liquid extraction' by J.M Aguilera)
Regards,
Attachments: