Hi and thanks Actually I need the per area or per volume value. I intend to multiply this with the vorticity to get the power losses in the fluid. This is a stationary fluid.
Does anyone know if this kind of power loss, or gain, has been calculated by anyone else before? I intend to calculate this theoretically and analytically on a retarded potential flow. The problem is to determine the retardation. I realize I have to do that in the velocity domain. But I am still not convinced about how to calculate the retarded velocity. One one hand I realize that I could just retard the potential but taking the gradient of that will include no vorticity. So I currently think I have to set the retardation in the velocity field. The calculus and algebra will likely be enormous. I also don't know from where to calculate the retardation. But the problem should be very similar to retardation in electrodynamics. David David Jonsson, Sweden, phone callto:+46703000370 On Wed, May 13, 2009 at 11:25 PM, Stephen A. Lawrence <sa...@pobox.com>wrote: > I think, for a stationary shaft... > > Assume you're given a stationary shaft lying along the x axis. Assume > further that it's under torsion. > > To find the applied torque, I think you would want to integrate > > R x (Txy, Txz) > > over the surface of a cut through the shaft, where "R" is the radius > vector from the center of the shaft to each point on the cut surface, > and "x" is the cross product. But I'm not sure; these comments of mine > are pretty half-baked. > > Of course, if the shaft is stationary (or rotating at constant velocity) > then the applied torque must be the same no matter where you look along > the shaft. > > Stephen A. Lawrence wrote: > > What you just said sounds right, but what you've actually got looks to > > me like the torque per unit volume. I think you need to integrate that > > over a volume to get the actual torque acting on that volume. > > > > OTOH if that value is nonzero then your object is spinning up -- it's > > not just sitting there. If you're looking at something like a steel > > shaft which is under torsion but stationary then T12 = T21 and you need > > to look at something more complicated to figure out what the torque on > > the shaft is -- maybe the gradient of the stress tensor? > > > > > > David Jonsson wrote: > >> On Wed, May 13, 2009 at 9:36 PM, David Jonsson > >> <davidjonssonswe...@gmail.com <mailto:davidjonssonswe...@gmail.com>> > wrote: > >> > >> Hi > >> > >> Can someone explain to me how to calculate the torque from the > >> stress tensor? > >> > >> > >> It seems to be this simple > >> > >> Torque = T12 - T21 > >> > >> For a two dimensional tensor > >> > >> T= T11 T12 > >> Â Â Â Â T21 T22 > >> > >> Right? > >> > >> Now I will do some nice calculations, but first I would like to have > >> this confirmed. > >> > >> David > >
