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Old   June 8, 2000, 05:11
Default Energy balance in mixing vessel
  #1
Gjvdg
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Hi,

As a novice in this field I am doing CFD-simulations on a mixing vessel (you can argue about if a novice should do this). The calculated shaft power in my simulations is in close agreement with experimental values (pffff). My question is: ' Where does this energy go'? I know some terms:

1) By integrating epsilon over the total volume I found that about 30% is dissipated.

2) From LDV measurements in mixing vessels, one can distinguish turbulent fluctuations and periodic fluctuations. I have been told that the latter is missed by CFD. How large is this term approximately?

3) Other terms are for example dissipation by viscous forces in the main flow. How large is this term approximately? What are other large terms?

Final question: Does anyone have a good reference on the energy cascade in mixing vessels, besides the standard literature (Hinze, Tennekes & Lumley etc).

Thanking everyone in advance for their contributions.

GJVDG
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Old   June 9, 2000, 03:38
Default Re: Energy balance in mixing vessel
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Lars
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Hello there,

I haven't got all the answers that you are seeking, but obviously, if your mixer is operating at steady state, and unless you have flow boundaries in your mixer model, all the shaft power supplied must be dissipated to heat which in turn is removed from the system by wall heat transfer. I suppose it would be useful to integrate the heat flux over the walls to check this. This goes regardless whether your CFD simulation catches all flow details or not.

Intuitively, I am a bit surprised that the integrated dissipation is only 30% of the shaft power since this is by far the most important source of mean stream kinetic energy destruction in high Reynolds number turbulent flows. But perhaps your flow is not a high Re flow? What is the flowing medium, and what turbulence model do you use?

Lars

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Old   June 13, 2000, 07:01
Default Re: Energy balance in mixing vessel
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Gjvdg
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I use an unbaffled vessel that contains water. The stirrer's diameter (58 cm) is almost equal to the vessel diameter (60 cm). Thus, the clearance (space between stirrer and vessel) is 1 cm. With N=3Hz, Re is equal to 1000*3*0.58^2/0.001=1.0E6, so fully turbulent.

I have used two kinds of turbulence models: k-e & dsm (differential stress model). The latter includes non-isotropic turbulence. This is desirable as turbulence in the vessel is strongly non-isotropic. However, using dsm I obtain lower integrated dissipations than using k-e (25 vs 33%).

The heat flux is not calculated as the energy balance is not solved because I don't know where everything is going. So, I don't know which terms have to be included, apart from the intergrated dissipation and viscous terms.
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Old   June 19, 2000, 02:22
Default Re: Energy balance in mixing vessel
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Jin Wook LEE
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Dear

1. Turbulent disiipation is another matter with power.

2. Because the flow is incompressible, heat generation (hence transfered to the wall) is negligible.

2. Navier Stokes equation is just the same as ma = F, except rho replace m. So, power is F * V, where

F : force to impellar ~= pressure difference between front and rear surfaces of impeller(+ some viscous effect).

V : impeller speed.

So, power of impeller ~= (pressure difference of two sides of impeller + some viscous effect) * (area of impeller) * (impeller speed). Of course, integration is required along the impeller surface.

Sincerely, Jinwook
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Old   June 19, 2000, 09:05
Default Re: Energy balance in mixing vessel
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Lars
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Well, the answer is really just that it all ends up as heat, which in real life must be removed from the system (even though the heat generation is negligible in most aspects). This is the first law of thermodynamics, and we have to live with it.

What you know is that your turbulence models provide your equation system with an effective viscosity that yields the correct shaft torque on your impeller. If the predictions are good, this is quite sufficient for most purposes, but it is a matter of fact that energy conservation is not assured in your CFD solution unless you solve for the energy.

The shaft work of your impeller is added to the fluid as kinetic energy, which in turn is dissipated to heat by viscous friction forces wherever in the fluid you have nonzero strain rates. Most of the viscous dissipation takes place in the (finer) turbulent flow structures, and should thus be accounted for in the dissipation of turbulence kinetic energy.

The only other way I can think of to "loose" the supplied energy is by means of numerical diffusion, imbalances due to poor convergence, or other errors related to the solution procedure. Intuitively again, I think it sounds very much if these factors correspond to more than 60% of the supplied shaft work, but I have never checked this matter myself, and I guess it could be worthwhile trying.

Lars
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Old   June 20, 2000, 07:37
Default Re: Energy balance in mixing vessel
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Gjvdg
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Well, I fully agree that all energy is dissipated into heat at the end. Where else should it go?? I only wondered about the path the energy takes downward as I found that epsilon, integrated over the volume, represents only 30% of the calculated shaft power.

Some comments on your comments:

- I have calculated the shaft power in the same manner as Jin Wook mentioned and the power is in close agreement with experimental results.

- epsilon and k are dependent on the velocities. These velocities are also in close agreement with experimental values. So, Lars, I agree with you that numerical diffusion etc will not account for the large difference.

- I have defined a turbulent dispersion coefficient that depends on k and epsilon. With this dispersion coefficient I have performed scalar mixing simulations. The obtained mixing times are also in good agreement with experiments.

Concluding: everything looks allright, except for the difference in integrated epsilon and shaft power.

It still puzzles me.

Regards, Gert-Jan
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Old   June 20, 2000, 09:25
Default Re: Energy balance in mixing vessel
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John C. Chien
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(1). If you slow down the rotating shaft, I guess you can still compute the work done by the shaft. (2). At some point, the Reynolds number will be low enough for the flow to become laminar. And you can still compute the shaft work. But then the tke (k) and tke dissipation (epsilon) will not be there any more. (3). So, for laminar flows, the energy will be converted into heat, which can only be computed in terms of temperature through the energy equation. (4). And even if you are not solving the energy, you are assuming that the system is isothermal (constant temperature state with infinite conductivity). (5). Even without epsilon, we know that the frictional effect is to change the mechanical work into heat. Just try to rub your two hands against each other quickly, and place both hands on your face. Your face can detect small temperature on your hands, it is better than the computer simulation. (6). So, epsilon is probably only part of the story, what do you think? (7). By the way, if you rub your hands very quickly, I think, the flow between two hands is going to become turbulent. Well, that's a different story. I don't have the right answer for you, but I do like the question.
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