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how to enter a Channel thickness

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Hi everybody,
how to enter the Channel thickness?
I use the (laminar flow -> mass flow -> mass flow rate -> normal mass flow rate), but get an error!
I read a comsol documentation and I think I must enter a Channel thickness (maybe!!), but don't find where I can to enter a Channel thickness!
please help me, 1- how to enter a Channel thickness? 2- why get an error when I choose "normal mass flow rate" ?

thanks a lot

8 Replies Last Post Mar 12, 2013, 5:23 a.m. EDT
Ivar KJELBERG COMSOL Multiphysics(r) fan, retired, former "Senior Expert" at CSEM SA (CH)

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Posted: 1 decade ago Feb 18, 2013, 10:04 a.m. EST
Hi

I assume you are in 2D ? and your units are not flow/m^2 but /m ?
If so yes, in 2D everything is "per meter" so you need to multiply by the appropriate depth (d or "Z value)
Note that several physics in 2D allows you to define a thickness along Z to get absolute values (sometimes you can add a "change thcikness" sub-node

--
Good luck
Ivar
Hi I assume you are in 2D ? and your units are not flow/m^2 but /m ? If so yes, in 2D everything is "per meter" so you need to multiply by the appropriate depth (d or "Z value) Note that several physics in 2D allows you to define a thickness along Z to get absolute values (sometimes you can add a "change thcikness" sub-node -- Good luck Ivar

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Posted: 1 decade ago Mar 4, 2013, 7:54 p.m. EST
Hi,
thanks for reply dear Mr. Ivar
I'm in 2D axisymmetric, and my unit is (kg/s).
I still do not understand how to specify and enter the "Channel thickness" ! just enter the "d= ... [m]" in Variables?
best regards
Hi, thanks for reply dear Mr. Ivar I'm in 2D axisymmetric, and my unit is (kg/s). I still do not understand how to specify and enter the "Channel thickness" ! just enter the "d= ... [m]" in Variables? best regards

Ivar KJELBERG COMSOL Multiphysics(r) fan, retired, former "Senior Expert" at CSEM SA (CH)

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Posted: 1 decade ago Mar 5, 2013, 12:50 a.m. EST
Hi

In 2D-axi you have no "depth" but a "loop length" of 2*pi*r hence varying depending how far you are from the axis.

When you integrate over the domain in 2D-axi you have the additional option to have COMSOL automatically multiply by 2*pi*r for you check on the option, or add it by yourself. THis adds a "m" on the units and transform a surface integral into a volume (of revolution) integral

--
Good luck
Ivar
Hi In 2D-axi you have no "depth" but a "loop length" of 2*pi*r hence varying depending how far you are from the axis. When you integrate over the domain in 2D-axi you have the additional option to have COMSOL automatically multiply by 2*pi*r for you check on the option, or add it by yourself. THis adds a "m" on the units and transform a surface integral into a volume (of revolution) integral -- Good luck Ivar

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Posted: 1 decade ago Mar 10, 2013, 11:55 p.m. EDT
thanks a lot for reply
Unfortunately I didn't understand ...! didn't know how to solve this problem! My model consists of Plasma module and Lamir flow. My mass flow rate 1e-7 kg/s.
Inlet thickness 2 mm.
how can I add a mass flow rate?
I integrated over the inlet boundary

Integration1 (Intop1) > "inlet boundary"
Variable > mass = intop1(2*pi*r*u*spf.rho)

this integral depends on the velocity!
what should I do?

Please help somebody ...
thanks a lot for reply Unfortunately I didn't understand ...! didn't know how to solve this problem! My model consists of Plasma module and Lamir flow. My mass flow rate 1e-7 kg/s. Inlet thickness 2 mm. how can I add a mass flow rate? I integrated over the inlet boundary Integration1 (Intop1) > "inlet boundary" Variable > mass = intop1(2*pi*r*u*spf.rho) this integral depends on the velocity! what should I do? Please help somebody ...


Ivar KJELBERG COMSOL Multiphysics(r) fan, retired, former "Senior Expert" at CSEM SA (CH)

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Posted: 1 decade ago Mar 11, 2013, 3:42 p.m. EDT
Hi

first of all, in 2D-axi you should have no geoemtry on the r<0 side of the axis, only the cut line along the r=0 symmetry axis.

if you have an flow rate of [kg/s] this should be intop1(2*pi*r*u,rho) over the inlet, as "r" is in meters, u in [m/s], rho in [kg/m^3] and the dr of the integration argument (implicit not written out in COMSOL) adds a "m" hence you have

m*m/s*kg/m^3*m = kg/s

, no ?

--
Good luck
Ivar
Hi first of all, in 2D-axi you should have no geoemtry on the r

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Posted: 1 decade ago Mar 11, 2013, 4:25 p.m. EDT
yes! I understand it! this is not problem!
Problems another! how I can to add mass flow rate integration when I dont know velocity (u)? I just know mass flow rate=1e-7 [kg/s]
if I add the integral as a variable, I need to know the velocity (u).
density changes with time in the chamber.
how can I achieve a concrete mass flow rate?
how can I add 1e-7 [kg/s] ?

thanks...
yes! I understand it! this is not problem! Problems another! how I can to add mass flow rate integration when I dont know velocity (u)? I just know mass flow rate=1e-7 [kg/s] if I add the integral as a variable, I need to know the velocity (u). density changes with time in the chamber. how can I achieve a concrete mass flow rate? how can I add 1e-7 [kg/s] ? thanks...

Ivar KJELBERG COMSOL Multiphysics(r) fan, retired, former "Senior Expert" at CSEM SA (CH)

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Posted: 1 decade ago Mar 12, 2013, 3:28 a.m. EDT
Hi

but if the gas is entering in laminaire mode a parabolic profile is quite usual, then you have fixed values.

However you can also add a dependent variable: the local input velocity field across the inlet, and bind it through your fomula to impose the total flux, then COMSOL will adapt it "in a means square" approach

u(x,y,z,t) defined such as => intop1(...) - 1E-7 = 0

where the =0 is implicit in the COMSOL notation

--
Good luck
Ivar
Hi but if the gas is entering in laminaire mode a parabolic profile is quite usual, then you have fixed values. However you can also add a dependent variable: the local input velocity field across the inlet, and bind it through your fomula to impose the total flux, then COMSOL will adapt it "in a means square" approach u(x,y,z,t) defined such as => intop1(...) - 1E-7 = 0 where the =0 is implicit in the COMSOL notation -- Good luck Ivar

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Posted: 1 decade ago Mar 12, 2013, 5:23 a.m. EDT
thanks you very much Mr. Ivar !! We love you ...
thanks you very much Mr. Ivar !! We love you ...

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