# Is Viscous Dissipation always equal to Drag Power Loss

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 January 6, 2021, 13:52 Is Viscous Dissipation always equal to Drag Power Loss #1 Member   Raphael Join Date: Nov 2012 Posts: 68 Rep Power: 13 If we have a cylinder rotating inside a concentric stationary cylinder with a fluid in between them, will the total viscous dissipation in the fluid domain be equal to the total drag force multiplied with the velocity of the spinning cylinder? Seems to me like it has to be the case, but if someone can confirm or elaborate, it would be appreciated.

January 6, 2021, 14:21
#2
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Filippo Maria Denaro
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Quote:
 Originally Posted by arkie87 If we have a cylinder rotating inside a concentric stationary cylinder with a fluid in between them, will the total viscous dissipation in the fluid domain be equal to the total drag force multiplied with the velocity of the spinning cylinder? Seems to me like it has to be the case, but if someone can confirm or elaborate, it would be appreciated.

Are you talking about the "Taylor-Couette" flow? This is the case where you have an exact solution (see for example in Kundu).
The viscous dissipation (viscosity*D) is present in the kinetic energy equation, you can assume a steady flow and integrate the equation in the domain to check the balance between the terms. For this case we should always deduce that the dissipation of kinetic energy is balanced by the production of kinetic energy provided by the system.

January 6, 2021, 14:32
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Raphael
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Quote:
 Originally Posted by FMDenaro Are you talking about the "Taylor-Couette" flow? This is the case where you have an exact solution (see for example in Kundu). The viscous dissipation (viscosity*D) is present in the kinetic energy equation, you can assume a steady flow and integrate the equation in the domain to check the balance between the terms. For this case we should always deduce that the dissipation of kinetic energy is balanced by the production of kinetic energy provided by the system.
I assume "dissipation of kinetic energy" is viscous dissipation, and "production of kinetic energy" is the drag force on the inner cylinder multiplied by its velocity?

Is there an exact solution even when the flow is turbulent?

January 6, 2021, 14:55
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Filippo Maria Denaro
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Quote:
 Originally Posted by arkie87 I assume "dissipation of kinetic energy" is viscous dissipation, and "production of kinetic energy" is the drag force on the inner cylinder multiplied by its velocity? Is there an exact solution even when the flow is turbulent?

No, for turbulence you work on the statistically averaged equations that are not closed analytically. However, for statistically steady turbulence (and this is the case) you can say that the production equates the dissipation of kinetic energy.

January 6, 2021, 15:34
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Raphael
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Quote:
 Originally Posted by FMDenaro No, for turbulence you work on the statistically averaged equations that are not closed analytically. However, for statistically steady turbulence (and this is the case) you can say that the production equates the dissipation of kinetic energy.
Can you define production and dissipation of kinetic energy? Is it as I defined it above?

 January 26, 2022, 08:29 #6 New Member   Paul Norman Join Date: Jan 2022 Posts: 2 Rep Power: 0 Hi arkie87, I think it should be the case that there is mechanical energy balance in your simulation. A time average of the work being done by the wall (Tauw*v_Wall) should equal to a time average the energy being dissipated in your flow field by viscous dissipation. That being said, I have seen simulations where the numbers don't appear to add up, for example see this thread: Underestimated dissipation by LES I think you may also have to account the the effective numerical viscous dissipation of your finite volume schemes to get an exact balance.

 Tags drag power loss, rotating cylinder, viscous dissipation