# Converting momentum into heat

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 July 12, 2001, 04:26 Converting momentum into heat #1 Mark Render Guest   Posts: n/a Hi all, I have a question on the physics behind CFD. There's a kind if mixer which rotates very fast in a closed box filled completely with fluid. The temperature of the fluid will rise due to the friction losses of the mixer's blades. What terms in what equations do model that conversion into heat ? Thanks, Mark

 July 12, 2001, 08:57 Re: Converting momentum into heat #2 Kike Guest   Posts: n/a Dear Mark Fluid temperature increases its value because the mixer increases fluid internal energy ie. kinetics energy of its molecules and the kinetic energy of the molecules increases because you are transfering momentum from mixer's blades to the fluid. I think any book on thermodinamics should have a chapter on this issue. Depending on the state equation for your fluid you will have a more complicated relation: E_k=f(T). Find this kind of relation for your fluid (in those books/papers) and use NS eqs. to obtain a velocity field. I think you be able to obtain a Temperature field from the relation you have and the solution of energy transport equation in your domain. Do you have any idea (from practice or experiments) about the values of temperature you have inside your box? Regards Kike

 July 12, 2001, 15:57 Re: Converting momentum into heat #3 Peter Guest   Posts: n/a Energy conservation equation takes into account first the time dependancy of temperature and the convective term on the left hand side. On the right had side, there are terms related to heat and work introduced or extracted from the pyhsical domain. Heat conduction can be important inside the fluid and on doesn't have to forget the pressure which can create work. Viscous laminar stress and viscous turbulent stresses can generate work too. Finally, there is also a heat transfer term due to velocity and temperature fluctuations.

 July 13, 2001, 10:17 Re: Converting momentum into heat #4 sylvain Guest   Posts: n/a I think it comes from the viscous term : du/dt + u du/dx= -1/ro*dp/dx + nu d^2/dx^2 u multiplying this by u gives (assuming incompressibility) : d(1/2u^2)/dt + u d(1/2u^2)/dx = -1/ro*d(pu)/dx + nu d^2/dx^2 (1/2u^2) - 2 nu (du/dx)^2 which means : "time variation of kinetic energy" egal "pressure spatial variations" plus "viscous diffusion" minus "dissipation rate". I believe that the last term (dissipation rate) is what you are looking for. Am I wright ?

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