equations in electrostaticFoam

ywem · January 18, 2019, 00:54

Dear all,
I am trying to understand the electrostaticFoam solver in OpenFOAM. It should solve the Possion equation for electric potential and the continuity equation of electric current for charge density. But in these equations,
fvm::ddt(rho) + fvm::div(rhoFlux, rho)
and
rhoFlux = -k*mesh.magSf()*fvc::snGrad(phi)
the term div(rhoFlux, rho) seems not the current density (while in continuity equation of electric current it should be). What the mean of k? It should not be electrical conductivity from dimension analysis.

massive_turbulence · January 18, 2019, 06:58

Quote:

Originally Posted by ywem

Dear all,
I am trying to understand the electrostaticFoam solver in OpenFOAM. It should solve the Possion equation for electric potential and the continuity equation of electric current for charge density. But in these equations,
fvm::ddt(rho) + fvm::div(rhoFlux, rho)
and
rhoFlux = -k*mesh.magSf()*fvc::snGrad(phi)
the term div(rhoFlux, rho) seems not the current density (while in continuity equation of electric current it should be). What the mean of k? It should not be electrical conductivity from dimension analysis.

My best guess is is that it might be Coulombs constant https://en.wikipedia.org/wiki/Coulomb%27s_constant

The same question was asked once here What is the k in electrostaticFoam

ywem · January 18, 2019, 20:50

Dear Somorjai,
Thank you for your reply. But I still cannot understand . I think k should not be electrostatic constant because of different units. It has an unit of electrical conductivity / charge density.

massive_turbulence · January 19, 2019, 11:38

Quote:

Originally Posted by ywem

Dear Somorjai,
Thank you for your reply. But I still cannot understand . I think k should not be electrostatic constant because of different units. It has an unit of electrical conductivity / charge density.

If you're not sure of it I suggest you write your own equation instead. I don't expect every OpenFOAM tutorial to be perfect or complete myself.

ywem · January 19, 2019, 19:14

Quote:

Originally Posted by massive_turbulence

If you're not sure of it I suggest you write your own equation instead. I don't expect every OpenFOAM tutorial to be perfect or complete myself.

Thank you for your suggestion.

Pranavi · February 21, 2019, 03:31

Dear Yang,

You are right in that the dimensional formula of k is same as that of $\frac{\sigma}{q}$ .

The reason, I realized

, may have to do with the casting of the continuity equation in a form similar to the Flux form of an advection equation as follows:

$\frac{\partial q}{\partial t} + \nabla \bullet \vec{J} = 0$ ............(1)

with $\vec{J} = \sigma \vec{E} = - \sigma \nabla \phi = -q \frac{\sigma}{q} \nabla \phi = q \vec{U}$ ,

so equation (1) now looks like

$\frac{\partial q}{\partial t} + \nabla \bullet (q \vec{U}) = 0$ ,

Here, $\vec{U} \equiv -k \nabla \phi$ , with $k = \frac{\sigma}{q}$

We can now see that this little trick facilitated adaptation from a "robust" solver

.

Hope this helps

.

With regards,
pranavi

ywem · February 22, 2019, 02:23

Dear Pranavi,
Sorry for the late response. Thank you for your reply. Following your reminder, I understand this equation now.

NablaDyn · August 17, 2020, 06:30

Nice thread...

There's a somewhat related discussion here:

Sampling rhoFlux in electrostaticFoam

As I understand it, k represents the electron mobility
$\mu_{e}=\sigma/en$ ,
with $\sigma$ being electrical conductivity and

the total charge resulting from a given number

of elementary charge carriers per cubic meter $\left[1/m^3\right]$ , e.g. electrons, with charge

(in Coulomb, $\left[C\right]=\left[A s\right]$ ). If I am right, I assume the confusion concerning the units arises i.a. from the routinely used unit for electron mobility which is $10^4\left[cm^2/Vs\right]=\left[s^2A/kg\right]$ .

Nevertheless, I might be totally wrong...

keitaro7_14 · October 25, 2022, 13:17

k is ion mobility, you can consult this page: https://www.xsim.info/articles/OpenF...argedWire.html

J is current density, and it can be expressed as (https://arxiv.org/ftp/arxiv/papers/2002/2002.11662.pdf):

$J=k \rho E = -k \rho \nabla \phi$

where $\rho$ is charge density ,

is electric field and $\phi$ potential.

However, I don't understand why $\nabla\phi$ = fvc::snGrad(phi)*mesh.magSf();

Any idea?

Pranavi · October 26, 2022, 01:13

That is just the definition (coming from the divergence theorem). Remember that the equations in OpenFOAM are discretized over finite volumes (i.e., the equations are implicitly treated as if multiplied by the cell volume throughout).

So, the gradient operator, which was supposed to be $\nabla \phi = 1\V (\int_S \phi_f dS_f)$ is simply defined as $\nabla \phi = \int_S (\phi_f dS_f).

In OpenFOAM notation this becomes fvc::snGrad(phi)*mesh.magSf().
where fvc:: snGrad(phi) is the integral phi (treat it like an interpolation scheme to provide a value of phi at the cell-face), and the direction information comes after multiplication with mesh.magSf(), which contains magnitude and orientation information of the cell face.

Hope I am clear!?

..pranavi

keitaro7_14 · October 26, 2022, 02:37

Yes, now it is clear. I got the idea after reading this post: Incompatible dimensions...., but your answer further clarified the idea for me.

Due to the equations in OpenFOAM are discretized over finite volumes, the fvc::div operator has dimensions of 1/(length^3). -k*fvc::snGrad(phi)*mesh.magSf() has flux units, which is mandatory to be used in fvc::div.

Many thanks!

January 18, 2019, 00:54	equations in electrostaticFoam	#1
ywem Member Wenming Yang Join Date: Jun 2018 Posts: 42 Rep Power: 7	Dear all, I am trying to understand the electrostaticFoam solver in OpenFOAM. It should solve the Possion equation for electric potential and the continuity equation of electric current for charge density. But in these equations, fvm::ddt(rho) + fvm::div(rhoFlux, rho) and rhoFlux = -kmesh.magSf()fvc::snGrad(phi) the term div(rhoFlux, rho) seems not the current density (while in continuity equation of electric current it should be). What the mean of k? It should not be electrical conductivity from dimension analysis.

February 21, 2019, 03:31		#6
Pranavi New Member Pranavi Join Date: Feb 2019 Posts: 2 Rep Power: 0	Dear Yang, You are right in that the dimensional formula of k is same as that of $\frac{\sigma}{q}$ . The reason, I realized , may have to do with the casting of the continuity equation in a form similar to the Flux form of an advection equation as follows: $\frac{\partial q}{\partial t} + \nabla \bullet \vec{J} = 0$ ............(1) with $\vec{J} = \sigma \vec{E} = - \sigma \nabla \phi = -q \frac{\sigma}{q} \nabla \phi = q \vec{U}$ , so equation (1) now looks like $\frac{\partial q}{\partial t} + \nabla \bullet (q \vec{U}) = 0$ , Here, $\vec{U} \equiv -k \nabla \phi$ , with $k = \frac{\sigma}{q}$ We can now see that this little trick facilitated adaptation from a "robust" solver . Hope this helps . With regards, pranavi Rango and Diwei Liu like this. Last edited by Pranavi; February 22, 2019 at 11:21.

August 17, 2020, 06:30		#8
NablaDyn Senior Member Join Date: Oct 2015 Location: Germany Posts: 100 Rep Power: 10	Nice thread... There's a somewhat related discussion here: Sampling rhoFlux in electrostaticFoam As I understand it, k represents the electron mobility $\mu_{e}=\sigma/en$ , with $\sigma$ being electrical conductivity and the total charge resulting from a given number of elementary charge carriers per cubic meter $\left[1/m^3\right]$ , e.g. electrons, with charge (in Coulomb, $\left[C\right]=\left[A s\right]$ ). If I am right, I assume the confusion concerning the units arises i.a. from the routinely used unit for electron mobility which is $10^4\left[cm^2/Vs\right]=\left[s^2A/kg\right]$ . Nevertheless, I might be totally wrong... Last edited by NablaDyn; August 17, 2020 at 10:47.

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January 18, 2019, 20:50		#3
ywem Member Wenming Yang Join Date: Jun 2018 Posts: 42 Rep Power: 7	Dear Somorjai, Thank you for your reply. But I still cannot understand . I think k should not be electrostatic constant because of different units. It has an unit of electrical conductivity / charge density.

February 22, 2019, 02:23		#7
ywem Member Wenming Yang Join Date: Jun 2018 Posts: 42 Rep Power: 7	Dear Pranavi, Sorry for the late response. Thank you for your reply. Following your reminder, I understand this equation now.

October 25, 2022, 13:17		#9
keitaro7_14 Member JuanMi Join Date: Nov 2017 Posts: 41 Rep Power: 8	k is ion mobility, you can consult this page: https://www.xsim.info/articles/OpenF...argedWire.html J is current density, and it can be expressed as (https://arxiv.org/ftp/arxiv/papers/2002/2002.11662.pdf): $J=k \rho E = -k \rho \nabla \phi$ where $\rho$ is charge density , is electric field and $\phi$ potential. However, I don't understand why $\nabla\phi$ = fvc::snGrad(phi)*mesh.magSf(); Any idea?

October 26, 2022, 01:13		#10
Pranavi New Member Pranavi Join Date: Feb 2019 Posts: 2 Rep Power: 0	That is just the definition (coming from the divergence theorem). Remember that the equations in OpenFOAM are discretized over finite volumes (i.e., the equations are implicitly treated as if multiplied by the cell volume throughout). So, the gradient operator, which was supposed to be $\nabla \phi = 1\V (\int_S \phi_f dS_f)$ is simply defined as $\nabla \phi = \int_S (\phi_f dS_f). In OpenFOAM notation this becomes fvc::snGrad(phi)*mesh.magSf(). where fvc:: snGrad(phi) is the integral phi (treat it like an interpolation scheme to provide a value of phi at the cell-face), and the direction information comes after multiplication with mesh.magSf(), which contains magnitude and orientation information of the cell face. Hope I am clear!? ..pranavi

October 26, 2022, 02:37		#11
keitaro7_14 Member JuanMi Join Date: Nov 2017 Posts: 41 Rep Power: 8	Yes, now it is clear. I got the idea after reading this post: Incompatible dimensions...., but your answer further clarified the idea for me. Due to the equations in OpenFOAM are discretized over finite volumes, the fvc::div operator has dimensions of 1/(length^3). *-kfvc::snGrad(phi)mesh.magSf()* has flux units, which is mandatory to be used in fvc::div. Many thanks!