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time derivative dependent source/reaction term #113

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sezerh24 opened this issue Jul 23, 2024 · 4 comments
Open

time derivative dependent source/reaction term #113

sezerh24 opened this issue Jul 23, 2024 · 4 comments

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@sezerh24
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Hi I am trying to solve the following set of PDEs:

image

I have implemented the reaction term:
function reactionLi!(f,u,node)
f[ξₙ] = -Lₙdh_dξ(u[ξₙ])((exp((1-α)neFu[ϕₙ]/R/T)) - u[ζₙ](exp(-αneFu[ϕₙ]/R/T))) - Lσdg_dξ(u[ξₙ])
end

Here is my storage term:

function storageLi!(f,u,node)
f[ξₙ] = u[ξₙ]
f[ζₙ] = u[ζₙ]+(Cₘₛ/C₀)u[ξₙ]#ξₙ_val
f[ϕₙ] = ne
FCₘₛu[ξₙ]
end

The ζₙ indeed includes the time derivative of ξₙ, as a reaction/source term, I do not know how I can pass this time derivative as a reaction/source term. Could you please mind to help on this issue?

@j-fu
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j-fu commented Jul 23, 2024

Isn't this already included in your storage term ?
Anyway - interesting problem. What is the source of the model ?

@sezerh24
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Well, I added in the storage term, but i think we should not add in the storage term. I was wondering if there is a way to add this term in the reaction or source term? Here is the paper (https://doi.org/10.1016/j.est.2022.104892).

Arguello, M.E., Labanda, N.A., Calo, V.M., Gumulya, M., Utikar, R. and Derksen, J., 2022. Dendrite formation in rechargeable lithium-metal batteries: Phase-field modeling using open-source finite element library. Journal of Energy Storage, 53, p.104892.

@j-fu
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j-fu commented Jul 24, 2024

Ok I will think about a simpler test problem which (hopefully...) proves to you that adding this in the storage term is correct...

Anyway interesting and challenging application. I am also working partially in electrochemistry, so getting this working would be really interesting for me. If you are interested, we could work on this together on a shared repository. Please note however that this would be not the only problem I am working on...

@sezerh24
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Hi Jurgen,

I will be happy to work with you on this problem. I will create the repository and share with you shortly.

Thank you.
Hayri

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