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dissipation term in Navier-Stokes Equation

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Old   July 25, 2015, 02:07
Default dissipation term in Navier-Stokes Equation
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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

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Old   July 25, 2015, 04:33
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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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Old   July 26, 2015, 04:15
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Quote:
Originally Posted by FMDenaro View Post
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.
sir can I say that

div( T u_vector) = dot_product_of[ ( div T) and u_vector ] + double_contraction_of [ T and grad(u_vector) ]

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Old   July 26, 2015, 05:15
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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.

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Old   July 26, 2015, 05:23
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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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Old   July 26, 2015, 06:12
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Quote:
Originally Posted by FMDenaro View Post
If I am right, using index notation for summation you get

d/dxi ( tau_ij uj)

you can check by expanding for 2D case
if I expand d/dxi ( tau_ij uj) , then I do get the terms same as in the image...

Thankyou

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Old   July 26, 2015, 06:40
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good!

remember, this term is not the mechanical energy dissipation but is more general
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Old   July 26, 2015, 07:24
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ok.

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Old   July 26, 2015, 08:41
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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)

u\frac{\partial }{\partial x}\left(\mu\frac{\partial u}{\partial x}\right)

2)

u\frac{\partial }{\partial x}\left(\mu\frac{\partial v}{\partial y}\right)

3)

\mu\left(\frac{\partial u}{\partial x}\right)\left(\frac{\partial u}{\partial x}\right)

4)

\mu\left(\frac{\partial v}{\partial x}\right)\left(\frac{\partial u}{\partial y}\right)

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

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Old   July 26, 2015, 08:58
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No, in a FV manner You have to discretize directly the divergence form !
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Old   July 26, 2015, 09:38
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Quote:
Originally Posted by FMDenaro View Post
No, in a FV manner You have to discretize directly the divergence form !
sir do you mean that , first I multiply Tau with u to get a matrix form, then take divergence of the matrix and then discretize it.
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Old   July 26, 2015, 09:53
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Quote:
Originally Posted by mihirmakwana6 View Post
sir do you mean that , first I multiply Tau with u to get a matrix form, then take divergence of the matrix and then discretize it.

yes, you can store an auxiliar vector f = T.v and take its divergence in each computational cell
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Old   July 26, 2015, 09:59
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Quote:
Originally Posted by FMDenaro View Post
yes, you can store an auxiliar vector f = T.v and take its divergence in each computational cell
sir just to be sure..

in the term f = T.v

, the . is a dot product between T and v , right ??
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Old   July 26, 2015, 10:00
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Quote:
Originally Posted by mihirmakwana6 View Post
sir just to be sure..

in the term f = T.v

, the . is a dot product between T and v , right ??

correct
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Old   July 26, 2015, 11:15
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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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Old   July 26, 2015, 13:11
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Quote:
Originally Posted by mihirmakwana6 View Post
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 ?

that's right
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Old   July 26, 2015, 13:41
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Quote:
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that's right
ok. thankyou sir.

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