CFD Online Logo CFD Online URL
www.cfd-online.com
[Sponsors]
Home > Forums > General Forums > Main CFD Forum

Specific dissipation

Register Blogs Community New Posts Updated Threads Search

Like Tree2Likes
  • 1 Post By FMDenaro
  • 1 Post By FMDenaro

Reply
 
LinkBack Thread Tools Search this Thread Display Modes
Old   February 15, 2020, 07:30
Default Specific dissipation
  #1
New Member
 
Bruno
Join Date: Nov 2011
Posts: 13
Rep Power: 14
bb82 is on a distinguished road
Hi all!
A question for you expert users: given the total dissipated power that one can calculate through variation of total pressure at inlet/outlet ports, is there any variable that, once integrated over the control volume, gives the same loss?

Attempt 1. I built the variable mu*tau:grad(U) (the classical "dissipation function") but the loss calculated by integrating such variable over the whole domain is not even close to the actual loss, for the obvious reason that the solution field contains only the averaged quantities (i.e. Reynolds stresses and velocity fluctuations are missing).

Attempt 2. I integrated the variable "Turbulence Eddy Dissipation" and I got somehow closer to the variation of total pressure but it's still around half the expected value.

Attempt 3: I re-used the tau:grad(U) variable but in combination with Eddy Viscosity: I got a similar result to that in Attempt 2.

Attempt 4: I re-arranged the equation of mechanical energy to isolate the effect of viscous and Reynolds stresses: by doing so I'm able to match the variation of total pressure calculated using surface integrals with a single volume integral (nice!), but the variable I created for this purpose is not the "specific loss", rather the "power done by viscosity" which, as such, can be either positive or negative. What I'm looking for is a monotonic variable that reproduces the total loss, i.e. something like the tau:grad(U) approach.

Am I missing something?

Thanks to you all!

Last edited by bb82; February 15, 2020 at 16:15.
bb82 is offline   Reply With Quote

Old   February 17, 2020, 13:29
Default
  #2
Senior Member
 
Lucky
Join Date: Apr 2011
Location: Orlando, FL USA
Posts: 5,747
Rep Power: 66
LuckyTran has a spectacular aura aboutLuckyTran has a spectacular aura aboutLuckyTran has a spectacular aura about
What are you asking for? It sounds like you're asking what volume integral should be exactly equal to some difference in total pressure which is presumably a surface integral, but you didn't define what this difference is exactly.


(total) pressure must be scaled by a mass flux to make it look like a work/power, but this simple analogy only works in 1D uniform flows. For a generalized or realistic 3D flow you don't have a "total pressure at the inlet" vs a "total pressure at the outlet" unless it is the degenerate case of uniform flow AND uniform total pressure.
LuckyTran is offline   Reply With Quote

Old   February 17, 2020, 16:56
Default
  #3
Senior Member
 
Filippo Maria Denaro
Join Date: Jul 2010
Posts: 6,849
Rep Power: 73
FMDenaro has a spectacular aura aboutFMDenaro has a spectacular aura aboutFMDenaro has a spectacular aura about
Quote:
Originally Posted by bb82 View Post
Hi all!
A question for you expert users: given the total dissipated power that one can calculate through variation of total pressure at inlet/outlet ports, is there any variable that, once integrated over the control volume, gives the same loss?

Attempt 1. I built the variable mu*tau:grad(U) (the classical "dissipation function") but the loss calculated by integrating such variable over the whole domain is not even close to the actual loss, for the obvious reason that the solution field contains only the averaged quantities (i.e. Reynolds stresses and velocity fluctuations are missing).

Attempt 2. I integrated the variable "Turbulence Eddy Dissipation" and I got somehow closer to the variation of total pressure but it's still around half the expected value.

Attempt 3: I re-used the tau:grad(U) variable but in combination with Eddy Viscosity: I got a similar result to that in Attempt 2.

Attempt 4: I re-arranged the equation of mechanical energy to isolate the effect of viscous and Reynolds stresses: by doing so I'm able to match the variation of total pressure calculated using surface integrals with a single volume integral (nice!), but the variable I created for this purpose is not the "specific loss", rather the "power done by viscosity" which, as such, can be either positive or negative. What I'm looking for is a monotonic variable that reproduces the total loss, i.e. something like the tau:grad(U) approach.

Am I missing something?

Thanks to you all!



Have you considered starting from the equation of the conservation of the total energy in the integral form? That express a balance between powers and you can see the analogy with the fundamental thermodinamic equation in a system dE/dt= W-Q.
sbaffini likes this.
FMDenaro is offline   Reply With Quote

Old   February 18, 2020, 03:05
Default
  #4
New Member
 
Bruno
Join Date: Nov 2011
Posts: 13
Rep Power: 14
bb82 is on a distinguished road
The equation of energy in integral form is what I apply when I mention the variation of total pressure on all inlet/outlet ports. What I actually would like to achieve is to have a specific variable that returns the total loss when integrated over the control volume (i.e. I’m interested in the differential form, aka non-conservative form)
bb82 is offline   Reply With Quote

Old   February 18, 2020, 04:11
Default
  #5
Senior Member
 
Filippo Maria Denaro
Join Date: Jul 2010
Posts: 6,849
Rep Power: 73
FMDenaro has a spectacular aura aboutFMDenaro has a spectacular aura aboutFMDenaro has a spectacular aura about
Quote:
Originally Posted by bb82 View Post
The equation of energy in integral form is what I apply when I mention the variation of total pressure on all inlet/outlet ports. What I actually would like to achieve is to have a specific variable that returns the total loss when integrated over the control volume (i.e. I’m interested in the differential form, aka non-conservative form)



The equation of the total energy is always conservative by definition, even if you write it in differential form. The pressure acts in it as div(vp), that is a mechanical reversible work, that is not associated to dissipation.
On the other hand, you can write the equations for the kinetic energy or the internal energy, they are not conservative and a dissipative term appears. Be careful that in turbulence is used to define different equations for the kinetic energy (mean/residual). Have a look for example to the book of Pope
sbaffini likes this.
FMDenaro is offline   Reply With Quote

Old   February 18, 2020, 11:00
Default
  #6
New Member
 
Bruno
Join Date: Nov 2011
Posts: 13
Rep Power: 14
bb82 is on a distinguished road
"Conservative" and "non-conservative" definitions were referred to the form of corresponding equations rather than to physical quantities being conserved or not. However, I agree that the trick is in the averaging of NS equation, as I also explained in my first post.
bb82 is offline   Reply With Quote

Old   February 18, 2020, 11:06
Default
  #7
Senior Member
 
Filippo Maria Denaro
Join Date: Jul 2010
Posts: 6,849
Rep Power: 73
FMDenaro has a spectacular aura aboutFMDenaro has a spectacular aura aboutFMDenaro has a spectacular aura about
Quote:
Originally Posted by bb82 View Post
"Conservative" and "non-conservative" definitions were referred to the form of corresponding equations rather than to physical quantities being conserved or not. However, I agree that the trick is in the averaging of NS equation, as I also explained in my first post.
You can write the differential equation for the total energy, that is d(rho*E)/dt only in the divergent form (conservative) but if you open the terms you get the quasi-linear form but for the variation rho*dE/dt.
Sagaut is a further textbook to see the several form of the resolved and unresolved kinetic energy equation
FMDenaro is offline   Reply With Quote

Old   February 18, 2020, 12:24
Default
  #8
Senior Member
 
Lucky
Join Date: Apr 2011
Location: Orlando, FL USA
Posts: 5,747
Rep Power: 66
LuckyTran has a spectacular aura aboutLuckyTran has a spectacular aura aboutLuckyTran has a spectacular aura about
Can you just explain precisely what you mean by loss obtained through total pressure variation and maybe provide a definition in terms of a mathematical equation?
LuckyTran is offline   Reply With Quote

Old   February 18, 2020, 13:33
Default
  #9
New Member
 
Bruno
Join Date: Nov 2011
Posts: 13
Rep Power: 14
bb82 is on a distinguished road
What I’m try to get is the following:
massFlowInt(Total Pressure/Density)@Inlet1...Inlet_n+massFlowInt(Total Pressure/Density)@Outlet1...Outlet_n=Total Dissipated Power=volumeInt(var_x)@Domain where var_x is the variable I’m trying to find or build.
bb82 is offline   Reply With Quote

Old   February 18, 2020, 13:38
Default
  #10
Senior Member
 
Filippo Maria Denaro
Join Date: Jul 2010
Posts: 6,849
Rep Power: 73
FMDenaro has a spectacular aura aboutFMDenaro has a spectacular aura aboutFMDenaro has a spectacular aura about
Quote:
Originally Posted by bb82 View Post
What I’m try to get is the following:
massFlowInt(Total Pressure/Density)@Inlet1...Inlet_n+massFlowInt(Total Pressure/Density)@Outlet1...Outlet_n=Total Dissipated Power=volumeInt(var_x)@Domain where var_x is the variable I’m trying to find or build.



I am not sure about the meaning of your formula but if we assume a simple example, that is a pipe having 1 inflow and 1 outflow, I think you will just achive the standard linear pressure decaying law between inlet and outlet that depends on the tangential stress and, for turbulent flows, the fluctuation contribute.
FMDenaro is offline   Reply With Quote

Old   February 18, 2020, 13:43
Default
  #11
New Member
 
Bruno
Join Date: Nov 2011
Posts: 13
Rep Power: 14
bb82 is on a distinguished road
The pipe is a good example and I believe you got what I want to achieve. My goal is use the “var_x” I defined in my previous post to create, as an example, a contour plot that illustrates where and to which extent the enenrgy loss is taking place.
bb82 is offline   Reply With Quote

Old   February 18, 2020, 13:56
Default
  #12
Senior Member
 
Filippo Maria Denaro
Join Date: Jul 2010
Posts: 6,849
Rep Power: 73
FMDenaro has a spectacular aura aboutFMDenaro has a spectacular aura aboutFMDenaro has a spectacular aura about
Quote:
Originally Posted by bb82 View Post
The pipe is a good example and I believe you got what I want to achieve. My goal is use the “var_x” I defined in my previous post to create, as an example, a contour plot that illustrates where and to which extent the enenrgy loss is taking place.



There are a lot of papers wherein the analysis of the energy budget is performed for several flow problems. Generally, in case of statistical energy equilibrium you get a direct balance between the production of mechanical energy and its dissipation. That is also the basic assumption of eddy-viscosity models. Otherwise their net difference does not vanish and the flow is not in energy equilibrium.
However, you can have a look to slide 38 here https://www.researchgate.net/publica...eLES_II_theory but I suggest then to read the more specific literature.
FMDenaro is offline   Reply With Quote

Old   February 19, 2020, 13:04
Default
  #13
Senior Member
 
Lucky
Join Date: Apr 2011
Location: Orlando, FL USA
Posts: 5,747
Rep Power: 66
LuckyTran has a spectacular aura aboutLuckyTran has a spectacular aura aboutLuckyTran has a spectacular aura about
As I stated before, the massflow*totalpressure gives the power dissipated only in 1D uniform flows.

For general 3D flows, this property doesn't work the same way. For starters, the massflow is just an integrated mass flow. You also have a total pressure at every cell volume and cell face on the boundary. There isn't a single total pressure. You have to decide first the meaning of massflow*totalpressure. You can check out Averaging Nonuniform Flow for a Purpose, to read more about this matter. Sry the article is behind the ASME paywall.
LuckyTran is offline   Reply With Quote

Old   February 19, 2020, 18:04
Default
  #14
New Member
 
Bruno
Join Date: Nov 2011
Posts: 13
Rep Power: 14
bb82 is on a distinguished road
Quote:
Originally Posted by LuckyTran View Post
As I stated before, the massflow*totalpressure gives the power dissipated only in 1D uniform flows.

For general 3D flows, this property doesn't work the same way. For starters, the massflow is just an integrated mass flow. You also have a total pressure at every cell volume and cell face on the boundary. There isn't a single total pressure. You have to decide first the meaning of massflow*totalpressure. You can check out Averaging Nonuniform Flow for a Purpose, to read more about this matter. Sry the article is behind the ASME paywall.
You’re correct: in fact, I didn’t just write massflow*totalpressure rather massFlowInt(Total Pressure). This accounts for variations of total pressure on the specific surfaces.
bb82 is offline   Reply With Quote

Reply


Posting Rules
You may not post new threads
You may not post replies
You may not post attachments
You may not edit your posts

BB code is On
Smilies are On
[IMG] code is On
HTML code is Off
Trackbacks are Off
Pingbacks are On
Refbacks are On


Similar Threads
Thread Thread Starter Forum Replies Last Post
Setting the height of the stream in the free channel kevinmccartin CFX 12 October 13, 2022 22:43
Domain Reference Pressure and mass flow inlet boundary AdidaKK CFX 75 August 20, 2018 06:37
mass flow in is not equal to mass flow out saii CFX 12 March 19, 2018 06:21
Constant velocity of the material Sas CFX 15 July 13, 2010 09:56
Two-Phase Buoyant Flow Issue Miguel Baritto CFX 4 August 31, 2006 13:02


All times are GMT -4. The time now is 14:24.