Discussion Forum

Anyone there can help me, please?

Hi

by default COMSOL is set up to do physics and most enginners use units in their calculations. Now in the latest version unitless equations can be selected to allow also an approach closer to the math way, by normalysing out units to all "1"

That is handy for scaleing on generic solution to many cases, but as an engineer I prefer to work with units "closer to the ground" and the "true thing" but that is a question of habits ;)

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Good luck
Ivar

Dr. Ivar,
I am using coupled ODEs in my 2D spf laminar flow. My results are exactly varying 2e3 with theoritical ones. I feel that it is due to dimensions. Would you please illustrate how do we nondimensionalise newton's equation?
For example:
mXdott=F
in 2D spf, we can use 'm' as mass per unit length, rho*Area for left hand side. F is calculated from line integral of reacf(v). If so (m being mass per unitlength, F being force per unit length) the above equation is already balanced to give Xdott in m/s2, Xdot in m/s?

Please clarify my doubt.
Thanks in advance
Siva

Hi

I would use correct unts and not change, or set everything to unitless (and hope the best).

Indeed a factor 2000 looks suspect ;)

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Good luck
Ivar

Thanks for your reply.
Is the equation
m(mass per unitlength) Xdott = F(force per unitlength calculated from line integral of reacf)

dimensionally correct?

Thanks again Dr.

Hi

I'm not sure i catch you fully, but in 2D most physics consider per 1[m] depth, hence the integration ovar a domain (= surface in 2D = [m^2]) of the density rho gives you kg/m^3*m^2 = kg/m, while "utt" is normally the second derivative of "u" the small displacement along x (resp y,z in SOLID) so utt = d^2(u)/dt^2 hence the integration over a domain of rho*utt gives the force per meter depth in [N/m]

Check carefully the reaction forces and lagrange multipliers, the latter are to be integrated , the former reacf() to be summed over the nodes (normally the integration operator detects this and changes accordingly in "auto" mode). Lagrange multipliers are fluxes and the units are NOT yet implemented correctly, so you must look up the values for your physics (see the doc) and multiply explicitely by i.e.*1[N/m^2] for SOLID. Reacf() have correct units (except if you forces an integration when it should be summed), be aware

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Good luck
Ivar

Thanks a lot Dr. Ivar
This IS called discussion. You have illustrated the entire situation in a beautiful manner. Every word is worthy of 100 books. Hats off to you for such elaborate reply.

I think I failed to explain my problem clearly. As you know, mine is a 2D spf laminar flow. I have used moving wall BC for an ellipse in body-fitted frame with prescribed y-component of velocity(Please see the attached figure). The x-component of velocity of ellipse is calculated through ODEs. I have used model couplings-> integration and global equations. I have defined a state variable called 'Q'. this is infact position 'x' to avoid confusion with variable names, -this I learnt through one of your earlier posts.
The ODEs were : Q: Qt-Qdot; Qdot: Qdott-(F_x/m). I have calculated F_x using intop1(-reacf(u)) and m using rho_solid*area. Then Newton's equation: mQdott=F_x is supposedly dimensionally correct.
My results (Qdot, ie. x-component of velocity) are varying exactly x2e3 for all parametric solutions with mu=0.01[Pa*s], area=0.07[m^2], rho_solid = 32*rho_fluid, rho_fluid = 200,1000,1600,2000 [kg/m^3].
I am not sure, why there is a constant difference 2e3 for all parametric variations. Did I miss anything in ODE?
Thanks again dr. Ivar.
The physics challenges me everytime, it even shakes my fundamental understandings. You alone can help me.

Hi

by default COMSOL is set up to do physics and most enginners use units in their calculations. Now in the latest version unitless equations can be selected to allow also an approach closer to the math way, by normalysing out units to all "1"

That is handy for scaleing on generic solution to many cases, but as an engineer I prefer to work with units "closer to the ground" and the "true thing" but that is a question of habits ;)

--
Good luck
Ivar

Hi,

Thanks for your responds, Doctor Ivar.

Do you mean in the lastest version (COMSOL 4.2a), we can solve unitless PDEs (non-dimensional PDEs)? So in the COMSOL 4.2, do we can solve unitless ones?

Hope you have time to help me, thank you in advance.

Hi,
To have dimensionless PDEs in Comsol 4.2 choose
Root > Unit System > None.
Best Regards,
Andrzej

Thanks for that information.
Would you please hint, is that applicable for 2D spf(laminar flow). Can I make Navierstokes equation dimensionless using same approach?
Thanks in advance

Perhaps it is possible. It depends on your boundary conditions.

Thanks for that.
You mean the value of boundary condition should be non-dimensionalised and then entered?

Yes, BCs have to be in dimensionless form.

Yes. I got it.
I want to non-dimensionalise Navier-Stokes equation using Reynolds number.
Is there a way to change Navier-Stokes equation in nondimensional form as in the attached figure?
thanks a ton for your helps...

Hi,

Suppose we want to solve transient laminar flow problem (NS equ.+BCs+IC). Our problem has usual dimensional form but we want to solve it as dimensionless. To this end we must perform two tasks:

1. Change the dimensional problem for dimensionless one. This is the analytic task.
2. Solve the dimensionless problem. This is the numerical task.

It is not clear to me which of these task you want to perform. First? Second? Both?

Best Regards,
Andrzej

Hi.
Thanks for your reply.
Actually my problem is 2D laminar flow of oscillating circular cylinder. The horizontal reaction force is coupled so that due to vertical oscillation horizontal movement take place. I have created coupled modelling as in the attached figure. My result varies 2e3 times exactly with all paramteric variation.
I was thinking of non-dimensionalising the whole model and run simulation. dont know how to do that.
Please help me.

Dear Sir,

I am currently working with COMSOL multilphysics ver 5.2 to coupled heat and mass transport phenomenon with reactions. I am using two models (Heat Transfer in Porous Media and Transport of Diluted Species in Porous Media) together to simulate my process. I have coupled heat transport model with mass transport (involving reactions) to predict the product profile.

But, my system contains more than 10 products and every-time i have to change many absolute values (like heat transfer coefficient, diffusivity, thermal conductivity etc.) to fit experimental data.
Hence I want to non-dimensionalize both these model so that instead of change absolute values I can change their ratios to achieve the predictions conveniently.

Kindly let me know how i can make these model non-dimensionalize.

Thank You,

Dr. Khursheed Ansari

Hi,
To make equations dimensionless, each variable is scaled by dividing the variable (which has units) by a scale factor (which has the same units as the variable). The ratio "variable/scale factor" is then dimensionless. If we use the symbol * to denote dimensionless variable, then we can write for example:
t*=t/t0
t* is dimensionless time
t is time in units of (typically) seconds
t0 is the time scale in units of seconds

Similarly x*=x/x0 for dimensionless distance
v*=v/v0 for dimensionless velocity
p*=p/p0 for dimensionless pressure
T*=(T-T0)/(T1-T0) for dimensionless temperature

The scale factors (t0 for time, x0 for distance, v0 for velocity, p0 for pressure, T1 and T0 for temperature) are chosen to characterize the actual problem, considering the geometry of the problem, the driving force, etc. For example, x0 may be chosen as the typical dimension of the geometry. For temperature, T0 may be the minimum temperature on the boundary and T1 may be chosen to be the maximum temperature on the boundary.

The dimensionless terms t*, x*, v*, etc. will have no units and should be approximately of order 1.

To make the equations dimensionless, substitute the dimensional forms of each variable by an expression involving the dimensionless form. For example, the derivative dx/dt will become (L0/t0)dx*/dt* since dx=L0dx* and dt=t0dt*. After substituting all terms in the equation, you will be able to "cancel" the dimensional terms from in front of all terms in the equation, and your equation will be dimensionless.
For example, consider the equation dn/dt = -dj/dx
where n = concentration [1/m^3]
t = time [s]
j = flux [1/(m^2 s)]
x = distance [m]
Define n*=n/n0, t*=t/t0, j*=j/j0, x*=x/x0.
Substitute these into the equation to obtain the following
(n0/t0)dn*/dt*=-(j0/x0)dj*/dx*
Note that n0/t0 has units of 1/(m^3 s)
Also note that j0/x0 has units of 1/(m^3 s)
Define the "characteristic flux" j0 as (n0*x0/t0).
Then the equation becomes
(n0/t0)dn*/dt*=-((n0*x0/t0)/x0)dj*/dx*
which is
(n0/t0)dn*/dt*=-(n0/t0)dj*/dx*
and one can cancel n0/t0 from both sides
The dimensionless equation is then:
dn*/dt*=-dj*/dx*

n0 is chosen as characteristic concentration for the problem
x0 is characteristic length scale
t0 is characteristic time scale
j0 = (n0*x0/t0), where j0 is chacteristic flux for the problem
-DW