Cfx

Pavelko · December 20, 2010, 00:43

Hello All,

I want to use the following expression in inlet boundary condition:

Pmax*((sin(omega*t))^2)*(step(t)-step(t-0.02))+Pmax*step(t-0.02)

where omega,Pmax are defined.
But I get the following error:

Bad expression value 'Pin' detected in parameter 'Relative Pressure' in object '/FLOW:Flow Analysis 1/DOMAINefault Domain/BOUNDARY:Inlet/BOUNDARY CONDITIONS/MASS AND MOMENTUM'.
CEL error:
The dimensions of the argument to function 'step' are inconsistent with expected dimensions.
Argument dimensions resolve to 's'; expected '<dimensionless>'.

Please help.

Lance · December 20, 2010, 02:20

Your step functions need to be dimensionless, just as the error says.

Change step(t-0.02) to step((t-0.02[s])/1[s]) for example.

cougarCFD · December 20, 2010, 02:41

Yes, Lance is right! Wherever you have something that has a dimension in your expression, you need to divide it for its dimension unit. So, wherever you have time, you need to divide by 1 [s].

Pavelko · December 22, 2010, 05:26

Ok, it works!

Thank you very much!
P.K.

December 20, 2010, 00:43	Cfx	#1
Pavelko Member Pavelistyj Join Date: Jul 2010 Posts: 42 Rep Power: 15	Hello All, I want to use the following expression in inlet boundary condition: Pmax((sin(omegat))^2)(step(t)-step(t-0.02))+Pmaxstep(t-0.02) where omega,Pmax are defined. But I get the following error: Bad expression value 'Pin' detected in parameter 'Relative Pressure' in object '/FLOW:Flow Analysis 1/DOMAINefault Domain/BOUNDARY:Inlet/BOUNDARY CONDITIONS/MASS AND MOMENTUM'. CEL error: The dimensions of the argument to function 'step' are inconsistent with expected dimensions. Argument dimensions resolve to 's'; expected '<dimensionless>'. Please help.

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December 20, 2010, 02:20		#2
Lance Senior Member Lance Join Date: Mar 2009 Posts: 669 Rep Power: 22	Your step functions need to be dimensionless, just as the error says. Change step(t-0.02) to step((t-0.02[s])/1[s]) for example.

December 20, 2010, 02:41		#3
cougarCFD New Member Fabio Join Date: Dec 2010 Posts: 2 Rep Power: 0	Yes, Lance is right! Wherever you have something that has a dimension in your expression, you need to divide it for its dimension unit. So, wherever you have time, you need to divide by 1 [s].

December 22, 2010, 05:26		#4
Pavelko Member Pavelistyj Join Date: Jul 2010 Posts: 42 Rep Power: 15	Ok, it works! Thank you very much! P.K.