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July 25, 2015, 02:07 |
dissipation term in Navier-Stokes Equation
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#1 |
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Mihir Makwana
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The dissipation term in Navier-Stokes Equation is given by
div( T u_vector ) where T is viscous stress tensor and u_vector is velocity vector T = n[ grad(u_vector) + transpose( grad(u_vector) ) ] - (2*n/3) ( div(u_vector) I ) where n : viscosity , I : identity matrix if we consider a 2 dimensional situation, how can I expand div( T u_vector ) . I know how to expand div( T ) , but I am facing problem while finding the expression for div( T u_vector ). Is there any book or any link that explains this. Please help. Thanks in advance - Mihir |
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July 25, 2015, 04:33 |
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#2 |
Senior Member
Filippo Maria Denaro
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what you wrote is the mechanical work defined in the total energy equation...
if T =- p I + 2mu*grad v, you see a reversible work due to the isotropic term and an irreversible work due to the tangential stress. The real energy dissipation term is the quadratic term in the kinetic energy equation mu * (grad v:grad v). Said that, a way to develop the tensorial product is to use index notation Div = ei d/dxi, T = ei tij ej, v = ei vi. Alternatively, you can expand each term in the specific reference system you want to use. For example in a 2D Cartesian system: Div = i d/dx + j d/dy, v = i u + j v T = i tii i + j tjj j + i tij j + j tji i and developing scalar products and derivatives by the definition. You can see something "similar" to developing the divergence of a scalar function for a vector. |
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July 26, 2015, 04:15 |
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#3 | |
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Mihir Makwana
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Quote:
div( T u_vector) = dot_product_of[ ( div T) and u_vector ] + double_contraction_of [ T and grad(u_vector) ] Reply |
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July 26, 2015, 05:15 |
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#4 |
Member
Mihir Makwana
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Is the equation in the image the correct answer
http://i.imgur.com/vCDKVQe.jpg here u and v are the components of u_vector and T = viscous stress tensor Reply. Thanks in advance. - Mihir |
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July 26, 2015, 05:23 |
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#5 |
Senior Member
Filippo Maria Denaro
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If I am right, using index notation for summation you get
d/dxi ( tau_ij uj) you can check by expanding for 2D case |
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July 26, 2015, 06:12 |
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#6 |
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Mihir Makwana
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July 26, 2015, 06:40 |
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#7 |
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Filippo Maria Denaro
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good!
remember, this term is not the mechanical energy dissipation but is more general |
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July 26, 2015, 07:24 |
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#8 |
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Mihir Makwana
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ok.
- Mihir |
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July 26, 2015, 08:41 |
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#9 |
Member
Mihir Makwana
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after expanding d/dxi ( tau_ij uj) and substituting the equation for Tij , I get certain terms ( consider equations ( 1- 4) )
I am trying to discretize the equations ( 1 - 4 ) using Finite volume method on a 2D cartesian Control volume of sides delta_x and delta_y . I am using staggered grid 1) 2) 3) 4) here u and v are velocity components and mu is viscosity equations 1 and 2 are nonlinear whereas equations 3 and 4 are product of partial derivatives. How can I discretize them. is there a book that explain this - Mihir |
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July 26, 2015, 08:58 |
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#10 |
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Filippo Maria Denaro
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No, in a FV manner You have to discretize directly the divergence form !
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July 26, 2015, 09:38 |
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#11 |
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Mihir Makwana
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July 26, 2015, 09:53 |
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#12 |
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Filippo Maria Denaro
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July 26, 2015, 09:59 |
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#13 |
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Mihir Makwana
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July 26, 2015, 10:00 |
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#14 |
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Filippo Maria Denaro
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July 26, 2015, 11:15 |
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#15 |
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Mihir Makwana
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sir, I have written the equation in the divergence form
http://i.imgur.com/hKAdhth.jpg?1 Is this correct???? So, now terms A and B are terms to be calculated at the cell faces ,right ? |
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July 26, 2015, 13:11 |
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#16 | |
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Filippo Maria Denaro
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Quote:
that's right |
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July 26, 2015, 13:41 |
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#17 |
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Mihir Makwana
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