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                                                Posted:
                            
                                1 decade ago                            
                            
                                2011/05/04 11:32 GMT-4                            
                        
                        
                                                    Yeah,
You can define a surface and ask the solver to integrate over that specific surface.
Cheers                                                
                                                
                            Yeah,
You can define a surface and ask the solver to integrate over that specific surface.
Cheers                        
                                                
                                                                                                            
                                             
                                            
                                                
    
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                                                Posted:
                            
                                1 decade ago                            
                            
                                2011/05/05 1:42 GMT-4                            
                        
                        
                                                    You are right,
it is possible to define a surface being bounded by geometric entities. 
What I want is to define a surface whose boundaries depend on the value of a variable which was solved for before.
Regards,
Marcel
                                                
                                                
                            You are right,
it is possible to define a surface being bounded by geometric entities. 
What I want is to define a surface whose boundaries depend on the value of a variable which was solved for before.
Regards,
Marcel
                        
                                                
                                                                                                            
                                             
                                            
                                                
    
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                                                Posted:
                            
                                1 decade ago                            
                            
                                2011/05/05 4:23 GMT-4                            
                        
                        
                                                    Hi
You want to average the pressure of a part of a boundaray, which has a temperature higher than a reference temperature 'Tref'?
Not tested:
integrate the following expression over the boundary:
(T>=Tref)
result: length/area of the part of the boundary.
next: integrate the following expression over the boundary:
(T>=Tref)*p
divide the result of B by A. That's what you want?
best regards
                                                
                                                
                            Hi
You want to average the pressure of a part of a boundaray, which has a temperature higher than a reference temperature 'Tref'?
Not tested:
integrate the following expression over the boundary:
(T>=Tref)
result: length/area of the part of the boundary.
next: integrate the following expression over the boundary:
(T>=Tref)*p
divide the result of B by A. That's what you want?
best regards
                        
                                                
                                                                                                            
                                             
                                            
                                                
    
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                                                Posted:
                            
                                1 decade ago                            
                            
                                2011/05/05 5:01 GMT-4                            
                        
                        
                                                    Thanks for your answer.
With result A I get the area on the boundary where T>Tref.
Integrating the whole surface and just multiplying (T>Tref) simply uses a value of zero when T<Tref and the average value is far too low.
Regards,
Marcel                                                
                                                
                            Thanks for your answer.
With result A I get the area on the boundary where T>Tref.
Integrating the whole surface and just multiplying (T>Tref) simply uses a value of zero when T                        
                                                
                                                                                                            
                                             
                        
                        
                            
                                                                                        
                                Ivar KJELBERG
                                                                                                                                                    COMSOL Multiphysics(r) fan, retired, former "Senior Expert" at CSEM SA (CH)
                                                         
                            
                         
                                                
    
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                                                Posted:
                            
                                1 decade ago                            
                            
                                2011/05/05 9:14 GMT-4                            
                        
                        
                                                    Hi
yes there is the fact that an average divides by the full area, so you need to do two separate integrations and make the division manually (or on the same line) and not use "average"
--
Good luck
Ivar                                                
                                                
                            Hi
yes there is the fact that an average divides by the full area, so you need to do two separate integrations and make the division manually (or on the same line) and not use "average"
--
Good luck
Ivar