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3-D cylinder with anisotropic thermal conductivity
Posted 8 giu 2010, 13:25 GMT-4 2 Replies
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Hello everyone:
I am trying to model a 3-D cylinder that has anisotropic thermal conductivity in the r, phi, and z directions. Unfortunately, COMSOL does not have a default heat transfer mode with any coordinate system other than x,y,z. I searched for this on the COMSOL website, but apparently no one has posted such a thing before (I would be glad to be wrong on this:).
The best way I can think to model this is to modify the equations under physics>Equation System>subdomain and boundary expressions. I did this for a very simple case: cylinder with uniform heat generation and convection on the curved surface in steady-state. But my results are incorrect. Here is a summary of the changes I made in the subdomain equation system
c (anisotropic diagonal) = -d(-r*kxx_ht*d(T,r),d(T,r)) -d(-kyy_ht*d(T,p),d(T,p)/r) -d(-r*kzz_ht*Tz,Tz)
a = -d(r*Q_ht,T)
f = r*Q_ht
ea = 0
da = 0
alpha = <-d(-r*kxx_ht*d(T,r),T); -d(-kyy_ht*d(T,p),T); -d(-r*kzz_ht*Tz,T)>
beta = <-d(r*Q_ht,d(T,r)); -d(r*Q_ht,d(T,p)/r); -d(r*Q_ht,Tz)>
gamma = <-r*kxx_ht*d(T,r); -kyy_ht*d(T,p); -r*kzz_ht*Tz>
and the boundary equation system:
q = -d(r*h_ht*(-T+Tinf_ht),T)
g = r*h_ht*(-T+Tinf_ht)
I defined r = sqrt(x^2+y^2) and p = acos(x/r) in the subdomain expressions. kxx, kyy, and kzz are the r, phi, and z thermal conductivities I put in the subdomain settings.
Can anyone tell me what I am doing incorrect? I suspect it has to do with boundary condition, but I'm not sure.
Also, if anyone has a simpler way of change to a r-phi-z system in 3-D for the heat transfer module, I'm all ears.
Thanks,
Todd
I am trying to model a 3-D cylinder that has anisotropic thermal conductivity in the r, phi, and z directions. Unfortunately, COMSOL does not have a default heat transfer mode with any coordinate system other than x,y,z. I searched for this on the COMSOL website, but apparently no one has posted such a thing before (I would be glad to be wrong on this:).
The best way I can think to model this is to modify the equations under physics>Equation System>subdomain and boundary expressions. I did this for a very simple case: cylinder with uniform heat generation and convection on the curved surface in steady-state. But my results are incorrect. Here is a summary of the changes I made in the subdomain equation system
c (anisotropic diagonal) = -d(-r*kxx_ht*d(T,r),d(T,r)) -d(-kyy_ht*d(T,p),d(T,p)/r) -d(-r*kzz_ht*Tz,Tz)
a = -d(r*Q_ht,T)
f = r*Q_ht
ea = 0
da = 0
alpha = <-d(-r*kxx_ht*d(T,r),T); -d(-kyy_ht*d(T,p),T); -d(-r*kzz_ht*Tz,T)>
beta = <-d(r*Q_ht,d(T,r)); -d(r*Q_ht,d(T,p)/r); -d(r*Q_ht,Tz)>
gamma = <-r*kxx_ht*d(T,r); -kyy_ht*d(T,p); -r*kzz_ht*Tz>
and the boundary equation system:
q = -d(r*h_ht*(-T+Tinf_ht),T)
g = r*h_ht*(-T+Tinf_ht)
I defined r = sqrt(x^2+y^2) and p = acos(x/r) in the subdomain expressions. kxx, kyy, and kzz are the r, phi, and z thermal conductivities I put in the subdomain settings.
Can anyone tell me what I am doing incorrect? I suspect it has to do with boundary condition, but I'm not sure.
Also, if anyone has a simpler way of change to a r-phi-z system in 3-D for the heat transfer module, I'm all ears.
Thanks,
Todd
2 Replies Last Post 15 giu 2010, 15:07 GMT-4
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Posted:
1 decade ago
Hi Todd,
in my opinion the best option would be to transform the rectangular coordinate system to cylinderical one.
(x,y,z)----(rho,phi,z) where 0<=rho<inf, 0<=phi<2pi & -inf<z<inf.
you can use the following transformation in option==>Expressions==>scalar Expressions
rho=sqrt(x^2+y^2)
phi=atan2(y,x)
z=z
but this is only half of the solution.
once this is done then you have to use the following rotational matrix to transform local anisotropic conductivity tensor to global one.
sigma(x,y,z)=A*sigma(rho,phi,z)*A'
i hope this would resolve your problem
salman
in my opinion the best option would be to transform the rectangular coordinate system to cylinderical one.
(x,y,z)----(rho,phi,z) where 0<=rho<inf, 0<=phi<2pi & -inf<z<inf.
you can use the following transformation in option==>Expressions==>scalar Expressions
rho=sqrt(x^2+y^2)
phi=atan2(y,x)
z=z
but this is only half of the solution.
once this is done then you have to use the following rotational matrix to transform local anisotropic conductivity tensor to global one.
sigma(x,y,z)=A*sigma(rho,phi,z)*A'
i hope this would resolve your problem
salman
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Posted:
1 decade ago
Hi Salman:
Thanks for your reply. Unfortunately, I don't understand how to transform the rotational matrix into a global one. Is there somewhere that i can find how to do this?
Thanks,
Todd
Thanks for your reply. Unfortunately, I don't understand how to transform the rotational matrix into a global one. Is there somewhere that i can find how to do this?
Thanks,
Todd