Normalized energy dissipation rate

rdemyan · March 18, 2023, 15:20

The energy dissipation rate is given as an equality by:

$\epsilon = {C_\epsilon}u'^2\frac{u'}{\l}$

where u' is a characteristic velocity of the kinetic energy released or supplied and $\frac{\l}{u'}$ is the large or integral eddy turnover time.

$C_\epsilon$ has been widely researched, and is considered to be a constant of order 1 at high enough Re. However, $C_\epsilon$ does have higher values than the the asymptotic value. I am seeing this as well with my research, but do not see a dependence upon Re as other researchers have. A higher value of $C_\epsilon$ means that the energy dissipation rate is higher than it would be at high enough Re (i.e. where $C_\epsilon$ is equal to the asymptotic value).

My question is on how to view $C_\epsilon$ in a physical sense. For my system, I can determine the amount of kinetic energy supplied directly from the difference in measured velocities (i.e. before energy release and after energy release). Therefore, I believe that $C_\epsilon$ applies to the turnover time or rate at which energy is transferred. I do not believe that a value of $C_\epsilon$ higher than the asymptotic value of $C_\epsilon$ means that the integral eddy size is smaller than what it would be at the asymptotic value of $C_\epsilon$ . It seems to me that

is determined from the geometry and not by the amount of kinetic energy that is extracted from the mean flow. Instead I am thinking that the transfer rate or inverse integral eddy turnover time should be determined using $C_\epsilon$ at the asymptotic value. Then for higher values of $C_\epsilon$ , the physical meaning is that the energy is released in less than one turnover time of the large-scale or integral eddies. So, higher values of $C_\epsilon$ beyond the asymptotic value does not mean that the amount of kinetic energy to be released is different from what it would be if $C_\epsilon$ were at the asymptotic value nor does it mean that

is smaller. In a physical sense it means that the energy has been released in less than one turnover time of the eddy. The turnover time of the large-scale or integral eddy is calculated based on $\frac{\l}{u'}$ and $C_\epsilon$ does not reduce this turnover time.

Does this reasoning make sense?

March 18, 2023, 15:20	Normalized energy dissipation rate	#1
rdemyan Member Join Date: Jan 2023 Posts: 64 Rep Power: 3	The energy dissipation rate is given as an equality by: $\epsilon = {C_\epsilon}u'^2\frac{u'}{\l}$ where u' is a characteristic velocity of the kinetic energy released or supplied and $\frac{\l}{u'}$ is the large or integral eddy turnover time. $C_\epsilon$ has been widely researched, and is considered to be a constant of order 1 at high enough Re. However, $C_\epsilon$ does have higher values than the the asymptotic value. I am seeing this as well with my research, but do not see a dependence upon Re as other researchers have. A higher value of $C_\epsilon$ means that the energy dissipation rate is higher than it would be at high enough Re (i.e. where $C_\epsilon$ is equal to the asymptotic value). My question is on how to view $C_\epsilon$ in a physical sense. For my system, I can determine the amount of kinetic energy supplied directly from the difference in measured velocities (i.e. before energy release and after energy release). Therefore, I believe that $C_\epsilon$ applies to the turnover time or rate at which energy is transferred. I do not believe that a value of $C_\epsilon$ higher than the asymptotic value of $C_\epsilon$ means that the integral eddy size is smaller than what it would be at the asymptotic value of $C_\epsilon$ . It seems to me that is determined from the geometry and not by the amount of kinetic energy that is extracted from the mean flow. Instead I am thinking that the transfer rate or inverse integral eddy turnover time should be determined using $C_\epsilon$ at the asymptotic value. Then for higher values of $C_\epsilon$ , the physical meaning is that the energy is released in less than one turnover time of the large-scale or integral eddies. So, higher values of $C_\epsilon$ beyond the asymptotic value does not mean that the amount of kinetic energy to be released is different from what it would be if $C_\epsilon$ were at the asymptotic value nor does it mean that is smaller. In a physical sense it means that the energy has been released in less than one turnover time of the eddy. The turnover time of the large-scale or integral eddy is calculated based on $\frac{\l}{u'}$ and $C_\epsilon$ does not reduce this turnover time. Does this reasoning make sense?

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