Bird et. al. - Bulk Viscosity

Tobi · January 17, 2017, 11:02

Dear all,

I have a question that I could not solve properly during the last years but which appears always somewhere. If I go through a lot of different CFD related books, we will find the shear-rate tensor to be defined as:
$\tau = 2\mu \textbf{D} - \frac{2}{3} \mu \nabla \bullet \textbf{U} \textbf{I}$

However, if we check out the shear-rate tensor in other literatures like in Bird et. al. we will probably find the following equation:
$\tau = 2\mu \textbf{D} + \left(- \frac{2}{3} \mu + \kappa\right) \nabla \bullet \textbf{U} \textbf{I}$

Where $\kappa$ is defined to be the bulk viscosity. This quantity is mentioned to be zero for monoatomic gases and does not play a big role for high dense gases and liquids.

So far so good. But however, I am interested in the definition of $\kappa$ and if the term $-\frac{2}{3}\mu$ is the secondary viscosity $\lambda$ for Newtonian fluids

elones · January 17, 2017, 11:37

Hi,

$\textbf{T} = 2 A \textbf{D} + B (\nabla \cdot \textbf{v}) \textbf{1}$
where A is first viscosity coefficient (dynamic viscosity) and B is second viscosity coefficient.

$(\textbf{L})_{ij} = \frac{v_i}{x_j}$
$\textbf{D} = \frac{1}{2}( \textbf{L} + \textbf{L}^T)$
$\textbf{D}_O = \textbf{D} - \frac{1}{3} (\textbf{D} : \textbf{1}) \textbf{1}$
where trace of D is divergence of velocity.
$\textbf{T} = 2 A [ \textbf{D}_O + \frac{1}{3} (\nabla \cdot \textbf{v}) \textbf{1} ] + B (\nabla \cdot \textbf{v}) \textbf{1}$
$\textbf{T} = 2 A \textbf{D}_O + (B + \frac{2}{3} A ) (\nabla \cdot \textbf{v}) \textbf{1}$
here (B + \frac{2}{3} A ) is "volumetric" viscosity and Stokes hypothesis is that this is zero.
$B = - \frac{2}{3} A$
and
$\textbf{T} = 2 A \textbf{D} - \frac{2}{3} A (\nabla \cdot \textbf{v}) \textbf{1}$

-Ondra

Tobi · January 17, 2017, 12:12

Hi,

thanks for your reply. I think your number one represents the identity matrix, right? And the colon sign the double inner product of two matrices which is finally the trace. So I got what you wrote and you proofed that $B = -2/3 \mu$ . Just one thing that I never get (it is clear that we always can split a tensor in its deviatoric and hydrostatic part).

Deformation rate tensor (strain-rate tensor)

is there a name for that?

You mentioned the volumetric viscosity. However Bird mentioned the bulk viscosity and based on your derivation it seems that the bulk viscosity is equal to $\frac{2}{3}A$

FMDenaro · January 17, 2017, 12:41

I try to give an explanation.
It can be shown that for a linear relation between stress and velocity gradient, the invariance to rotation and the symmetry of the tensor leads to write (Newtonian model):

T = (-p +lambda* div v)*I +2 mu*D

being lambda the second viscosity coefficient and I the identity tensor.
Now, the tensor D can be decomposed according to

D = I*trace(D)/3 +D0 =I*div v /3+D0

Now we can rewrite

T = [-p +(lambda+2*mu/3)* div v]*I +2 mu*D0 = - phi*I +2 mu*D0

having defined the total dinamic pressure - phi = -p +(lambda+2*mu/3)*div v

The definition of the bulk viscosity is k=(lambda+2*mu/3)

Now, for divergence-free velocity, you get directly - phi = -p without any further assumption. When the velocity field produces a non vanishing divergence, you must consider the Stokes hypothesis k=0.

Tobi · January 17, 2017, 12:58

Hi,

Thanks again. But again what is D0 and in your equations T should be the Cauchy stress tensor rather than the shear rate tensor (pressure is included).

In addition, literature is welcomed.

Sent from my HTC One mini using CFD Online Forum mobile app

FMDenaro · January 17, 2017, 13:04

Quote:

Originally Posted by Tobi

Hi,

Thanks again. But again what is D0 and in your equations T should be the Cauchy stress tensor rather than the shear rate tensor (pressure is included).

In addition, literature is welcomed.

Sent from my HTC One mini using CFD Online Forum mobile app

It is the symmetric tensor having zero trace. It is a part of the general decomposition of a symmetric tensor in its isotropic and deviatoric parts.

P.S.: yes, T is the Cauchy tensor
That should be described here
https://www.cambridge.org/core/books...621E80F9266993

elones · January 17, 2017, 13:37

Quote:

Originally Posted by Tobi

Hi,

thanks for your reply. I think your number one represents the identity matrix, right? And the colon sign the double inner product of two matrices which is finally the trace. So I got what you wrote and you proofed that $B = -2/3 \mu$ . Just one thing that I never get (it is clear that we always can split a tensor in its deviatoric and hydrostatic part).

Deformation rate tensor (strain-rate tensor)

is there a name for that?

You mentioned the volumetric viscosity. However Bird mentioned the bulk viscosity and based on your derivation it seems that the bulk viscosity is equal to $\frac{2}{3}A$

D_O is deviatoric (traceless part) part of D. I think it is problem of semantics, bulk, volumetric, volume, dilatational usually refers the same. The problem is that some authors refers this result as second viscosity which is wrong because first and second viscosity is somehow equivalence to Lamé coefficients (A, B).

Tobi · January 17, 2017, 15:02

Hey,

thanks for the remark. That D0 is the deviatoric part was clear but I was just wondering if it has a special meaning like the deviatoric part of the Cauchy stress tensor is the shear rate tensor.

Okay. Based on your explanation it is clear. Therefore, Bird et. al. is also wrong because if I set in the bulk viscosity $\kappa = \lambda + \frac{2}{3}\mu$ we get:

$\tau = 2\mu \textbf{D} + \lambda (\nabla\bullet \textbf{U})\textbf{I}$

It should be correct. At last one question.

$\sigma := \mathrm{Cauchy stress tensor}$

$\sigma = dev(\sigma) + hyd(\sigma)$

where $dev(\sigma) = \tau$ and $hyd(\sigma) = p\textbf{I}$

Based on the fact that the deviatoric part is traceless, the trace of the shear rate tensor is zero: $tr(\tau) = 0$ - right?

Therefore:

$tr(\tau) = tr(2\mu \textbf{D} + \lambda( \nabla \bullet \textbf{U})\textbf{I}) = 0$

However, based on the fact that we split D again in its deviatoric part and hydrostatic part, it is not zero, right?

FMDenaro · January 17, 2017, 15:06

yes it has a meaning, D0 contains only information about angular deformation of a volume of fluid, no increasing/decreasing of the volume.

Tobi · January 18, 2017, 16:30

Can one confirm that in the pdf?

FMDenaro · January 18, 2017, 16:50

Quote:

Originally Posted by Tobi

Can one confirm that in the pdf?

seems correct, (7.7) is nothing else that 2*mu*D0

elones · January 18, 2017, 17:26

Hi,
I suggest you to look into Gurtin, The Mechanics and Thermodynamics of Continua, 2010 for more in-depht discussion of pressure and equilibrium pressure. By 7.1 you already split dev and hyd part of stress so "dev Cauchy = 2 mu D_O = 2 mu dev D".

Tobi · January 19, 2017, 02:21

Thanks for your reply. Can it be that we need the term $-\frac{2}{3}\mu(\nabla\bullet\textbf{U})\textbf{I}$ , to ensure that $2\mu\textbf{D}$ gets traceless in order to fulfill eqn. (7.3).

However, I will look into the mentioned literature (hopefully it is available here).

FMDenaro · January 19, 2017, 06:29

Quote:

Originally Posted by Tobi

Can it be that we need the term $-\frac{2}{3}\mu(\nabla\bullet\textbf{U})\textbf{I}$ , to ensure that $2\mu\textbf{D}$ gets traceless in order to fulfill eqn. (7.3).

what do you mean? you directly use D0 by its definition

Tobi · January 19, 2017, 07:21

Okay ... things are sometimes not easy to explain (at least for me).

We split the Cauchy stress tensor in its deviatoric part and hydrostatic part. The deviatoric part is traceless. For tau wie have:

$\tau = 2 A \textbf{D} + B (\nabla \bullet \textbf{U}) \textbf{I}$

Right? Okay. Based on the fact that D is not traceless, we have to choose B in order to make the whole equation traceless again. Right?

However, I think I have to check out the books / literature.

elones · January 19, 2017, 10:00

You need to distinguish between incompressible and compressible fluid and also between equilibrium pressure and pressure. Also it seems that you have some misunderstanding about constitutive equation for Newtonian fluids. I have choosen A and B to point out that there is lot of misconception about second viscosity coefficient and bulk viscosity. It is usually written as
${\bf T} = 2 \mu {\bf D} + \lambda ( {\bf D} : {\bf 1} ) {\bf 1}$
where mu is first viscosity coefficient (dynamic viscosity) and lambda is second viscosity coefficient.
In case of incompressible fluid, D=D_O and p=p_eq, hence
${\boldsymbol{\sigma}} = - p {\bf 1} + 2 \mu {\bf D}$
but in case of compressible fluid
${\boldsymbol{\sigma}} = - p_{eq} {\bf 1} + 2 \mu {\bf D} + \lambda ( {\bf D} : {\bf 1} ) {\bf 1}$
${\boldsymbol{\sigma}} = - [ p_{eq} - (B + \frac{2}{3} A ) (\nabla \cdot \textbf{v}) ] {\bf 1} + 2 \mu {\bf D}_O$
and
$p = p_{eq} - (B + \frac{2}{3} A ) (\nabla \cdot \textbf{v})$
Now you can write dev Cauchy and hyd Cauchy where hyd Cauchy = "pressure".
But the Cauchy stress tensor is usually split:
${\boldsymbol{\sigma}} = - p_{eq} {\bf 1} + {\bf T}$
where T is viscous part of Cauchy stress tensor. And (total) pressure is
$p = p_{eq} - \frac{1}{3} {\bf T} : {\bf 1}$

Tobi · January 20, 2017, 03:00

HI Ondřej,

thanks for you explanation. You are right, I am missing a few things for this topic but only based on the fact that I never saw the description in all of my literature. However, I will check out the literature given above on monday an I hope that after that I have a better understanding.

Thanks for your explanation but I am still a bit confused based on my not deep insight into that topic. I will change it as soon as possible.

Thank you very much.

Tobi · January 26, 2017, 09:18

I just wanted to let you know that the book that is mentioned above is a great help. Thank you!

January 17, 2017, 11:02	Bird et. al. - Bulk Viscosity	#1
Tobi Super Moderator Tobias Holzmann Join Date: Oct 2010 Location: Tussenhausen Posts: 2,708 Blog Entries: 6 Rep Power: 51	Dear all, I have a question that I could not solve properly during the last years but which appears always somewhere. If I go through a lot of different CFD related books, we will find the shear-rate tensor to be defined as: $\tau = 2\mu \textbf{D} - \frac{2}{3} \mu \nabla \bullet \textbf{U} \textbf{I}$ However, if we check out the shear-rate tensor in other literatures like in Bird et. al. we will probably find the following equation: $\tau = 2\mu \textbf{D} + \left(- \frac{2}{3} \mu + \kappa\right) \nabla \bullet \textbf{U} \textbf{I}$ Where $\kappa$ is defined to be the bulk viscosity. This quantity is mentioned to be zero for monoatomic gases and does not play a big role for high dense gases and liquids. So far so good. But however, I am interested in the definition of $\kappa$ and if the term $-\frac{2}{3}\mu$ is the secondary viscosity $\lambda$ for Newtonian fluids __________________ Keep foaming, Tobias Holzmann

January 17, 2017, 12:12		#3
Tobi Super Moderator Tobias Holzmann Join Date: Oct 2010 Location: Tussenhausen Posts: 2,708 Blog Entries: 6 Rep Power: 51	Hi, thanks for your reply. I think your number one represents the identity matrix, right? And the colon sign the double inner product of two matrices which is finally the trace. So I got what you wrote and you proofed that $B = -2/3 \mu$ . Just one thing that I never get (it is clear that we always can split a tensor in its deviatoric and hydrostatic part). Deformation rate tensor (strain-rate tensor) is there a name for that? You mentioned the volumetric viscosity. However Bird mentioned the bulk viscosity and based on your derivation it seems that the bulk viscosity is equal to $\frac{2}{3}A$ __________________ Keep foaming, Tobias Holzmann

January 17, 2017, 12:58		#5
Tobi Super Moderator Tobias Holzmann Join Date: Oct 2010 Location: Tussenhausen Posts: 2,708 Blog Entries: 6 Rep Power: 51	Hi, Thanks again. But again what is D0 and in your equations T should be the Cauchy stress tensor rather than the shear rate tensor (pressure is included). In addition, literature is welcomed. Sent from my HTC One mini using CFD Online Forum mobile app __________________ Keep foaming, Tobias Holzmann

January 17, 2017, 15:02		#8
Tobi Super Moderator Tobias Holzmann Join Date: Oct 2010 Location: Tussenhausen Posts: 2,708 Blog Entries: 6 Rep Power: 51	Hey, thanks for the remark. That D0 is the deviatoric part was clear but I was just wondering if it has a special meaning like the deviatoric part of the Cauchy stress tensor is the shear rate tensor. Okay. Based on your explanation it is clear. Therefore, Bird et. al. is also wrong because if I set in the bulk viscosity $\kappa = \lambda + \frac{2}{3}\mu$ we get: $\tau = 2\mu \textbf{D} + \lambda (\nabla\bullet \textbf{U})\textbf{I}$ It should be correct. At last one question. $\sigma := \mathrm{Cauchy stress tensor}$ $\sigma = dev(\sigma) + hyd(\sigma)$ where $dev(\sigma) = \tau$ and $hyd(\sigma) = p\textbf{I}$ Based on the fact that the deviatoric part is traceless, the trace of the shear rate tensor is zero: $tr(\tau) = 0$ - right? Therefore: $tr(\tau) = tr(2\mu \textbf{D} + \lambda( \nabla \bullet \textbf{U})\textbf{I}) = 0$ However, based on the fact that we split D again in its deviatoric part and hydrostatic part, it is not zero, right? __________________ Keep foaming, Tobias Holzmann

January 17, 2017, 15:06		#9
FMDenaro Senior Member Filippo Maria Denaro Join Date: Jul 2010 Posts: 6,768 Rep Power: 71	yes it has a meaning, D0 contains only information about angular deformation of a volume of fluid, no increasing/decreasing of the volume. Tobi likes this.

January 17, 2017, 11:37		#2
elones New Member Join Date: Mar 2014 Location: Czech Republic Posts: 29 Rep Power: 14	Hi, $\textbf{T} = 2 A \textbf{D} + B (\nabla \cdot \textbf{v}) \textbf{1}$ where A is first viscosity coefficient (dynamic viscosity) and B is second viscosity coefficient. $(\textbf{L})_{ij} = \frac{v_i}{x_j}$ $\textbf{D} = \frac{1}{2}( \textbf{L} + \textbf{L}^T)$ $\textbf{D}_O = \textbf{D} - \frac{1}{3} (\textbf{D} : \textbf{1}) \textbf{1}$ where trace of D is divergence of velocity. $\textbf{T} = 2 A [ \textbf{D}_O + \frac{1}{3} (\nabla \cdot \textbf{v}) \textbf{1} ] + B (\nabla \cdot \textbf{v}) \textbf{1}$ $\textbf{T} = 2 A \textbf{D}_O + (B + \frac{2}{3} A ) (\nabla \cdot \textbf{v}) \textbf{1}$ here (B + \frac{2}{3} A ) is "volumetric" viscosity and Stokes hypothesis is that this is zero. $B = - \frac{2}{3} A$ and $\textbf{T} = 2 A \textbf{D} - \frac{2}{3} A (\nabla \cdot \textbf{v}) \textbf{1}$ -Ondra

January 17, 2017, 12:41		#4
FMDenaro Senior Member Filippo Maria Denaro Join Date: Jul 2010 Posts: 6,768 Rep Power: 71	I try to give an explanation. It can be shown that for a linear relation between stress and velocity gradient, the invariance to rotation and the symmetry of the tensor leads to write (Newtonian model): T = (-p +lambda* div v)I +2 muD being lambda the second viscosity coefficient and I the identity tensor. Now, the tensor D can be decomposed according to D = Itrace(D)/3 +D0 =Idiv v /3+D0 Now we can rewrite T = [-p +(lambda+2mu/3) div v]I +2 muD0 = - phiI +2 muD0 having defined the total dinamic pressure - phi = -p +(lambda+2mu/3)div v The definition of the bulk viscosity is k=(lambda+2*mu/3) Now, for divergence-free velocity, you get directly - phi = -p without any further assumption. When the velocity field produces a non vanishing divergence, you must consider the Stokes hypothesis k=0.

January 18, 2017, 17:26		#12
elones New Member Join Date: Mar 2014 Location: Czech Republic Posts: 29 Rep Power: 14	Hi, I suggest you to look into Gurtin, The Mechanics and Thermodynamics of Continua, 2010 for more in-depht discussion of pressure and equilibrium pressure. By 7.1 you already split dev and hyd part of stress so "dev Cauchy = 2 mu D_O = 2 mu dev D". Tobi likes this.

January 19, 2017, 02:21		#13
Tobi Super Moderator Tobias Holzmann Join Date: Oct 2010 Location: Tussenhausen Posts: 2,708 Blog Entries: 6 Rep Power: 51	Thanks for your reply. Can it be that we need the term $-\frac{2}{3}\mu(\nabla\bullet\textbf{U})\textbf{I}$ , to ensure that $2\mu\textbf{D}$ gets traceless in order to fulfill eqn. (7.3). However, I will look into the mentioned literature (hopefully it is available here). __________________ Keep foaming, Tobias Holzmann

January 19, 2017, 07:21		#15
Tobi Super Moderator Tobias Holzmann Join Date: Oct 2010 Location: Tussenhausen Posts: 2,708 Blog Entries: 6 Rep Power: 51	Okay ... things are sometimes not easy to explain (at least for me). We split the Cauchy stress tensor in its deviatoric part and hydrostatic part. The deviatoric part is traceless. For tau wie have: $\tau = 2 A \textbf{D} + B (\nabla \bullet \textbf{U}) \textbf{I}$ Right? Okay. Based on the fact that D is not traceless, we have to choose B in order to make the whole equation traceless again. Right? However, I think I have to check out the books / literature. __________________ Keep foaming, Tobias Holzmann Last edited by Tobi; January 19, 2017 at 09:45.

January 19, 2017, 10:00		#16
elones New Member Join Date: Mar 2014 Location: Czech Republic Posts: 29 Rep Power: 14	You need to distinguish between incompressible and compressible fluid and also between equilibrium pressure and pressure. Also it seems that you have some misunderstanding about constitutive equation for Newtonian fluids. I have choosen A and B to point out that there is lot of misconception about second viscosity coefficient and bulk viscosity. It is usually written as ${\bf T} = 2 \mu {\bf D} + \lambda ( {\bf D} : {\bf 1} ) {\bf 1}$ where mu is first viscosity coefficient (dynamic viscosity) and lambda is second viscosity coefficient. In case of incompressible fluid, D=D_O and p=p_eq, hence ${\boldsymbol{\sigma}} = - p {\bf 1} + 2 \mu {\bf D}$ but in case of compressible fluid ${\boldsymbol{\sigma}} = - p_{eq} {\bf 1} + 2 \mu {\bf D} + \lambda ( {\bf D} : {\bf 1} ) {\bf 1}$ ${\boldsymbol{\sigma}} = - [ p_{eq} - (B + \frac{2}{3} A ) (\nabla \cdot \textbf{v}) ] {\bf 1} + 2 \mu {\bf D}_O$ and $p = p_{eq} - (B + \frac{2}{3} A ) (\nabla \cdot \textbf{v})$ Now you can write dev Cauchy and hyd Cauchy where hyd Cauchy = "pressure". But the Cauchy stress tensor is usually split: ${\boldsymbol{\sigma}} = - p_{eq} {\bf 1} + {\bf T}$ where T is viscous part of Cauchy stress tensor. And (total) pressure is $p = p_{eq} - \frac{1}{3} {\bf T} : {\bf 1}$ Tobi likes this. Last edited by elones; January 19, 2017 at 15:37.

January 20, 2017, 03:00		#17
Tobi Super Moderator Tobias Holzmann Join Date: Oct 2010 Location: Tussenhausen Posts: 2,708 Blog Entries: 6 Rep Power: 51	HI Ondřej, thanks for you explanation. You are right, I am missing a few things for this topic but only based on the fact that I never saw the description in all of my literature. However, I will check out the literature given above on monday an I hope that after that I have a better understanding. Thanks for your explanation but I am still a bit confused based on my not deep insight into that topic. I will change it as soon as possible. Thank you very much. __________________ Keep foaming, Tobias Holzmann

January 26, 2017, 09:18		#18
Tobi Super Moderator Tobias Holzmann Join Date: Oct 2010 Location: Tussenhausen Posts: 2,708 Blog Entries: 6 Rep Power: 51	I just wanted to let you know that the book that is mentioned above is a great help. Thank you! __________________ Keep foaming, Tobias Holzmann

Similar Threads
Thread	Thread Starter	Forum	Replies	Last Post
Dynamic viscosity and Kinematic viscosity	asal	FLUENT	7	September 3, 2019 01:25
Problem with divergence	TDK	FLUENT	13	December 14, 2018 06:00
Ratio of eddy viscosity to molecular viscosity : Laminar or turbulent flow?	JuPa	CFX	7	September 9, 2013 07:45
Too low temperature at combustor outlet	romekr	FLUENT	2	February 6, 2012 10:02
External Flow Computations - Lift and Drag	Ramanath KS	Main CFD Forum	9	December 28, 2000 17:38