Hello! I am trying to simulate the temperature distribution in a hemisphere with constant heat generation w/cu.m

If I exploit the symmetry and simulate as a semi-circular plane what should be my equivalent heat generation source term, so that the results in the two models will be consistent

thanks

natteri

Can you state your problem more clearly?

is your hemisphere a 3D atmosphere in spherical coordinates (r, theta, phi) (where theta goes from 0 to pi, and phi from 0 to 2 pi) ?

Is the symmetry you are mentioning axi-symmetry, i.e. symmetry around the axis of rotation z (from which theta is measured)?

Is your 2D reduction problem just in (r,theta) and the derivatives in phi are dropped from the equations? (or are you integrating (averaging) in one of the dimension phi or r?)

If this the case and you just dropped the phi derivatives from the equations then your source term should be the same as the one in 3D spherical coordinates. Otherwise, you need to state your problem more clearly so that one can understand it.

PG.

Dear friend, I mean the depth-averaged hydraudynamic models closed with 2-d turbulence models. the form of all the equations(include mass equ., momentum equ. and k and e equations) thank you!

Dear Patrick,

Unfortunately I am modelling in cartesian co-ordinate. k (d^2T/dx^2+d^2T/dy^2+ d^2T/dy^2)+ g (w/m^3) = 0.0

When modelling in as a semicircle I drop of the d^2T/dz^2 term. As the present model will be an area rather than a volume, I believe we need to give a different value for the heat generation term g, inorder to obtain a constant temp profile in both conditions. I would like to know what this would be?

thanks

natteri

I am still not sure which of the two cases you have:

-1) If I understand correctly your 3D problem was already in cartesian coordinates (a box x y z, where xy is the horizontal plane and z is the elevation). Correct me if this is not the case. Then you solve only for the plane xy. You have now two possibilities:

- you average over z, then you still have units of volume but the quantities are evaluated at z=0 and are an average.

- you just integrate the quantities in the z direction and you are left with unit of surface (e.g. the density is in gm per cm square, etc..). Then you need to sum up over the heat source term in the vertical direction (if the source term is also a function of z, and is given on all z).

-2) However, if you just started from a cartesian grid for a real hemisphere (half a sphere) with x*x+y*y+z*z=r*r as the surface and now you translate this into a cartographic projection onto a plane xy, then it is completely different and the transformation is far from being straightforward.

Patrick

Dear Patrick,

While solving the problem in cartesian 3-D its straight forward. Now if I exploit the symmetry along the azimuth angle my dt/d(phi) would drop off and I still can use volumetric value for my heat source term and solve in terms if r and theta. Now if I project it on the x-y plane the r,theta needs to be transformed to x-y, and more over some kind of an average needs to be incorported for source.

As you rightly put it the transformation is a bit involved and was wondering if some one had done it before.

Thanks and regards

natteri