# Energy dissipation and Drag coefficient

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 January 22, 2006, 10:48 Energy dissipation and Drag coefficient #1 Freeman Guest   Posts: n/a Few days ago I assisted to the presentation of a final degree thesis of a student that showed the CDF analysis of the 3D shape of a prototype of a roadster. After the presentation, the tribunal said him that was a fine project, but the geometry of the shell of the roadster near to the front wheels was not very "aerodynamically correct" because it produces high "energy dissipation" in the flow. In the next link I've attached a picture of the roaster, highlighting the zone: http://img79.imageshack.us/my.php?image=fs16gj.jpg In the picture I can see that the zone would produce a separation of the flow and the pressure coefficient would be in that zone negative, so it would increase the drag coefficient of that car in a similar way it is produced in the wake of the car in the base (the rear-end I mean). I am also a student that is beginning in the CFD universe and I don't understand here the term "energy dissipation" and its influence in the drag coefficient. Could anybody explain it to me a little? And another question: in the wheel arch of cars I can assume also there is a similar energy dissipation that makes to increase the drag coefficient? Thanks to all!

 January 23, 2006, 09:40 Re: Energy dissipation and Drag coefficient #2 kharicha Guest   Posts: n/a Consider a moving object in liquid... During the movement the viscosity creates a boundary layer (BL)all arround the object. The shear stress at the BL creates a force that acts against the movement of the object, this force is known as viscous, frictional or drag force(where the drag coefficient is present) . A force express a transfer of momentum, here it is from the object to the sourrounding liquid. But when ever you have a transfert of momentum you also have a transfert of energy, again from the object to the liquid in the form of turbulence or heat generation. This energy transfer is also called "energy dissipation". There exists a direct relation between the drag force (so the drag coefficient) and the energy dissipated.

 January 23, 2006, 13:07 Re: Energy dissipation and Drag coefficient #3 Freeman Guest   Posts: n/a Mmmm! That's fine, thanks (again! Kharicha. Your explanation has made me assimilate this with mechanical effects. I meanthat, if I imagine a solid sliding in the ground, the drag force I get is similar to aerodynamical one you said obviously, but, what is more, the heat that is generated in that friction between ground and solid could be "compared" with the turbulence generated between the flow and the body? And, that energy dissipation does something to do with the turbulent kinetic energy? I mean that if turbulence is gererated by the transfer of energy between body and flow, the more turbulent kinetic energy is present in the flow, the more energy dissipation is happening? If so, what about the turbulent dissipation? Thanks a lot!!

 January 24, 2006, 01:24 Re: Energy dissipation and Drag coefficient #4 kharicha Guest   Posts: n/a In any cases , if the viscosity is not null, and if it exists a velocity gradient, a part of the kinetic energy of the flow is dissipated into heat.... When you consider turbulence, the transfer of energy is somehow different. The energy (kinetic) is first transfered from large eddies to small eddies (energy cascade). When the energy reaches the smallest eddies (kolmogorov scale), viscosity force (i.e the gradient of velocity are strong) is strong enough to create a transfert to heat. So I repeat, the eddies exchange kinetic energy. The smallest eddies dissipate the received kinetic energy into heat.

 January 24, 2006, 01:58 Re: Energy dissipation and Drag coefficient #5 S K Prasad Guest   Posts: n/a Suppose the surface in the fig. is dimpled. This will make the flow locally turbulent and delay separation also. So is this a more correct shape aerodynamically? and what about the energy dissipation, will it be higher as compared to the smooth surface??

 January 24, 2006, 03:05 Re: Energy dissipation and Drag coefficient #6 kharicha Guest   Posts: n/a Dimpled surfaces creates locally more turbulence and glogally less turbulence(more small and less big structures). This explains why (sometimes)dimpled surfaces are less dissipative. But to answer your question, I do not know if the use of a dimpled surface will overcome the high increase of drag generated by a bad aerodynamic shape as in the figure!?!!...

 January 25, 2006, 11:48 Re: Energy dissipation and Drag coefficient #7 Freeman Guest   Posts: n/a Thanks a lot kharicha! There's something I'm not sure about yet: that energy dissipation does something to do with the Turbulent Kinetic Energy of the k-e models? I mean that if turbulence is gererated by the transfer of energy between body and flow, the more turbulent kinetic energy is present in the flow, the more energy dissipation is happening? And what about the Turbulent Dissipation (delta)?

 January 26, 2006, 01:54 Re: Energy dissipation and Drag coefficient #8 kharicha Guest   Posts: n/a As I already mentioned, the energy is transfered from the flow to heat. k-e model distinguish the mean flow from turbulent fluctuations. So the energy is taken from the mean flow, transfered to the turbulent fluctuations (k), then k is dissipated into heat, the rate of dissipation is epsilon. "the more turbulent kinetic energy is present in the flow, the more energy dissipation is happening" Yes! Epsilon is not the turbulent dissipation. Turbulent dissipation is not k but the "rate of production of k", represented in the equations of k-e..

 January 26, 2006, 16:39 Re: Energy dissipation and Drag coefficient #9 Freeman Guest   Posts: n/a Clear as water kharicha! Thanks a lot. There's still one thing that I missed in my first posts. That energy dissipation that you explained in your first reply comes from the part of the integration of the shear stresses in the definition of the drag coefficient, but the pressure coefficient is also present in the integral as we now. But I've found several definitions of the pressure coeff., depending if the pressure is static, or dynamic or total. Which pressure is the one that must be used and why? I think is static, but I don't know the reason...

 January 27, 2006, 04:51 Re: Energy dissipation and Drag coefficient #10 kharicha Guest   Posts: n/a Follow this: At constant velocity and without viscosity, the total force exerced by the flowing liquid on the object is equal to zero. nevertheless the static pressure acts locally on the surface of the object with a magnitude of Density*Velocity**2. Why? if you write the bernoulli equation you have a relation between the STATIC pressure and the flow velocity (=0). At the stagnation point (on the surface, on the front of the object), the velocity is null so that the pressure has to be equal to Density*Velocity^2(also known as dynamical pressure). When the viscosity is involved, the flow is no more symmetric with respect to the front and the back of the object. There is a pressure (static) difference between the front and the back...A glogally non null force acts on the object...and this force has to do with Density*Velocity^2... The fluid dynamic communauty use to write this force as F= (Density*Velocity^2)*surface*parameter. The parameter, is the drag coefficient, it should involve the viscosity, but also the shape... This parameter has to become close to zero for very small viscosity.... So without viscosity no drag force, and similarly no lift force...

 January 27, 2006, 07:42 Re: Energy dissipation and Drag coefficient #11 andy Guest   Posts: n/a > I am also a student that is beginning in the CFD universe and I don't : understand here the term "energy dissipation" and its influence in the drag : coefficient. Could anybody explain it to me a little? It is an odd term. The car exerts an overall force on the fluid and the component in the direction of travel (force is a vector) is often expressed in a nondimensional form by the drag coefficient. The rate of energy transferred from the car to the fluid is given by the product of force and rate of distance moved in the direction of the force (speed). Clearly, for a given speed anything that reduces drag force will reduce the transfer of energy from the car to the fluid. After the car has passed and the fluid has returned to being stagnant at the background pressure the energy from the car will be contained in the fluids internal energy. Hence the term "energy dissipation" is a bit odd since the fluid has gained energy by becoming hotter not lost it. An explanation may lie with what happens to the components of energy in the fluid. Over the front part of the car the fluid will be accelerated from zero to some value. The fluid now contains a large component of dynamic energy which will be lost or dissipated unless the rear part of the car can slow down the fluid and recover this component. This is a somewhat incomplete way to look at where the energy goes but it is my best guess for the use of the term. The dynamic energy component is normally significantly larger than the component of energy transferred into the boundary layer and is potentially recoverable to a degree unlike the energy in the boundary layer (ignoring back scatter). Hence, even though your dimples will increase the lost energy in the boundary layer so long as more energy is recovered from the dynamic component (i.e. the fluid left behing the car is moving around less) less work will be required to push the car through the fluid. > And another question: in the wheel arch of cars I can assume also there is a : similar energy dissipation that makes to increase the drag coefficient? The work put into creating turbulence will be lost but it is just about possible that an individual wheel might aid the form drag by attaching a flow that would otherwise be detached and be significantly beneficial in terms of reducing the dynamic component of energy. Unlikely though.