Viscous term in Navier Stokes Equations

dandalf · November 16, 2010, 08:31

Hi

This might seem like a bit of a basic question, but what should the viscous term be,

Some authors give it as,

$\frac{1}{Re} \left(\frac{d^2u}{dx^2} +\frac{d^2u}{dy^2}+\frac{d^2u}{dz^2}\right)$
(Griebel et al: 1998), (Matyaka: 2003)

Yet others give it as,

$\mu\left(\frac{d^2u}{dx^2} +\frac{d^2u}{dy^2}+\frac{d^2u}{dz^2}\right)$
(Patankar; 1980)

If

$Re=\frac{\rho U d}{\mu}$

then surley the two can't be equivelent... or have I missed someting verry big somewhere?

Dan

RameshK · November 16, 2010, 10:08

Hi

The first one corresponds to non-dimensonalized equations, and the second one in dimensional form

dandalf · November 16, 2010, 11:51

I presume then if I am creating a finite volume solver for unsteady flows over complex geometries, I need the dimensional form.

Yet many authours seem to be usind the non-dimentionalized formulation.

Come to think of it did strike me as odd when in ("Numerical Simulation in Fluid Dynamics,A practical introduction", Griebel et al; 1998)

The authour seemed to be sugesting the use of the first form over complex geometries, where a

for the definition of Re would be hard to define.

Dan

dandalf · November 17, 2010, 08:02

I have also come across a third form of the viscous term

$\rho \nu \frac{\delta}{\delta x_j}\left(\frac{\delta U_i}{\delta x_j} + \frac{\delta U_j}{\delta x_i} \right)$
("Benchmark Computations of Laminar Flow around a Cylindr", Schafer and Turek; 1996, Notes on numerical dluid mechanics; 52, 547-566)
("Incompressibel Fluid Dynamics", Hunt; 1964)
this seems to relate directly to the visous fluid stress tensor,

$\tau_{i,j}=\mu\left(\frac{\delta u_i}{\delta x_j}+\frac{\delta u_j}{\delta x_i}\right)$
("Introduction to computational fluid dynamics", Date; 2005)

as far as I can tell this adds another three elements to the viscous term, so for a three dimensional flow the term would be,

$\rho \nu \left(\frac{\delta}{\delta x}\left(\frac{\delta u}{\delta x} + \frac{\delta u}{\delta x} \right) + \frac{\delta}{\delta y}\left(\frac{\delta v}{\delta x} + \frac{\delta u}{\delta y}\right)+\frac{\delta}{\delta z}\left(\frac{\delta w}{\delta x} + \frac{\delta u}{\delta z} \right) \right)$

the connection to the stress tensor seems logical, however I cand seem to find a direct derivation.

Inconsistancy between available literature sources seems to be sending me in circles,

Can anyone help me figure out which formulation I need?

Dan

pete · November 17, 2010, 08:16

The same equation can be written in many ways. CFD Wiki provides a more general dimensional form, which is valid for compressible flows (some of your examples assume incompressible flow):

http://www.cfd-online.com/Wiki/Navier-Stokes_equations

dandalf · November 17, 2010, 09:07

Hi Pete,

I'm still strugelling to see the equivelence the link you posted gave the viscos term for the ith dimension as,

$\frac{\delta \tau_{i,j}}{\delta x_j}$

which agrees with the formulation in my last post, yet how can this be equivelent to the formulations in my original post, if the entire rest of the equation of momentum matches up, except for the three added terms,

$\rho \nu \left(\frac{\delta^2 u}{\delta x ^2}+ \frac{\delta^2 v}{\delta x \delta y}+ \frac{\delta^2 w}{\delta x \delta z}\right)$

Unless these terms all cancel out to zero..or I'm simpy interpreting the notation incorrectly.

RameshK · November 17, 2010, 09:45

Hi

your \tau_{ij} term has a missing term. According to stokes hypothesis for a Newtonian fluid \tau_{ij} = \left( \frac{\partial u_i}{\partial x_j}+\frac{\partial u_j}{\partial x_i} - \frac{2}{3}\frac{\partial u_i}{\partial x_j}\delta_{ij}

and apply divergence to this quantity

dandalf · November 17, 2010, 10:21

Tank you Ramesh

$\tau_{ij} = \left( \frac{\delta u_i}{\delta x_j}+\frac{\delta u_j}{\delta x_i} - \frac{2}{3}\frac{\delta u_k}{\delta x_k}\delta_{ij}\right)$

where presumably $\delta_{ij}$ is the Kronecker delta, given as,

$\begin{array}{lr}\delta_{ij}=1 &\ if \ i=j\\ \delta{ij}=0 & otherwise \end{array}$
(Childers; 1981)

Making
$\tau_{ii}-\frac{2}{3}\mu\left(\frac{\delta u}{\delta x}+\frac{2}{3}\frac{\delta v}{\delta y}+\frac{2}{3}\frac{\delta w}{\delta z}\right)$

this will add the aditional terms,
$-\frac{2\mu}{3}\left(\frac{\delta^2 u}{\delta x^2}+\frac{\delta^2 v}{\delta x \delta y}+\frac{\delta^2 u}{\delta x \delta z}\right)$

Reducing the overall discrepency with the formulations in my original post to,

$\frac{\rho\nu}{3}\left(-2\frac{\delta^2 u}{\delta x^2}+ \frac{\delta^2 v}{\delta x \delta y}+ \frac{\delta^2 w}{\delta x \delta z}\right)$

francisco.angel · December 13, 2017, 17:27

Quote:

Originally Posted by dandalf

Hi Pete,

I'm still strugelling to see the equivelence the link you posted gave the viscos term for the ith dimension as,

$\frac{\delta \tau_{i,j}}{\delta x_j}$

which agrees with the formulation in my last post, yet how can this be equivelent to the formulations in my original post, if the entire rest of the equation of momentum matches up, except for the three added terms,

$\rho \nu \left(\frac{\delta^2 u}{\delta x ^2}+ \frac{\delta^2 v}{\delta x \delta y}+ \frac{\delta^2 w}{\delta x \delta z}\right)$

Unless these terms all cancel out to zero..or I'm simpy interpreting the notation incorrectly.

In the incompressible case the extra terms are simplified based on the continuity equation:

$\rho \nu \left(\frac{\delta^2 u}{\delta x ^2}+ \frac{\delta^2 v}{\delta x \delta y}+ \frac{\delta^2 w}{\delta x \delta z}\right)=\rho \nu\frac{\delta}{\delta x} \left(\frac{\delta u}{\delta x }+ \frac{\delta v}{\delta y }+ \frac{\delta w}{ \delta z}\right)=0$

November 16, 2010, 08:31	Viscous term in Navier Stokes Equations	#1
dandalf Member Dan Join Date: Oct 2010 Location: UK Posts: 41 Rep Power: 16	Hi This might seem like a bit of a basic question, but what should the viscous term be, Some authors give it as, $\frac{1}{Re} \left(\frac{d^2u}{dx^2} +\frac{d^2u}{dy^2}+\frac{d^2u}{dz^2}\right)$ (Griebel et al: 1998), (Matyaka: 2003) Yet others give it as, $\mu\left(\frac{d^2u}{dx^2} +\frac{d^2u}{dy^2}+\frac{d^2u}{dz^2}\right)$ (Patankar; 1980) If $Re=\frac{\rho U d}{\mu}$ then surley the two can't be equivelent... or have I missed someting verry big somewhere? Dan liwende likes this.

November 16, 2010, 10:08		#2
RameshK Member Ramesh K Join Date: Dec 2009 Location: Bangalore Posts: 73 Rep Power: 16	Hi The first one corresponds to non-dimensonalized equations, and the second one in dimensional form lcarasik likes this.

November 17, 2010, 10:21		#8
dandalf Member Dan Join Date: Oct 2010 Location: UK Posts: 41 Rep Power: 16	Tank you Ramesh $\tau_{ij} = \left( \frac{\delta u_i}{\delta x_j}+\frac{\delta u_j}{\delta x_i} - \frac{2}{3}\frac{\delta u_k}{\delta x_k}\delta_{ij}\right)$ where presumably $\delta_{ij}$ is the Kronecker delta, given as, $\begin{array}{lr}\delta_{ij}=1 &\ if \ i=j\\ \delta{ij}=0 & otherwise \end{array}$ (Childers; 1981) Making $\tau_{ii}-\frac{2}{3}\mu\left(\frac{\delta u}{\delta x}+\frac{2}{3}\frac{\delta v}{\delta y}+\frac{2}{3}\frac{\delta w}{\delta z}\right)$ this will add the aditional terms, $-\frac{2\mu}{3}\left(\frac{\delta^2 u}{\delta x^2}+\frac{\delta^2 v}{\delta x \delta y}+\frac{\delta^2 u}{\delta x \delta z}\right)$ Reducing the overall discrepency with the formulations in my original post to, $\frac{\rho\nu}{3}\left(-2\frac{\delta^2 u}{\delta x^2}+ \frac{\delta^2 v}{\delta x \delta y}+ \frac{\delta^2 w}{\delta x \delta z}\right)$ Last edited by dandalf; November 18, 2010 at 10:56. Reason: correction

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November 16, 2010, 11:51		#3
dandalf Member Dan Join Date: Oct 2010 Location: UK Posts: 41 Rep Power: 16	I presume then if I am creating a finite volume solver for unsteady flows over complex geometries, I need the dimensional form. Yet many authours seem to be usind the non-dimentionalized formulation. Come to think of it did strike me as odd when in ("Numerical Simulation in Fluid Dynamics,A practical introduction", Griebel et al; 1998) The authour seemed to be sugesting the use of the first form over complex geometries, where a for the definition of Re would be hard to define. Dan

November 17, 2010, 08:02		#4
dandalf Member Dan Join Date: Oct 2010 Location: UK Posts: 41 Rep Power: 16	I have also come across a third form of the viscous term $\rho \nu \frac{\delta}{\delta x_j}\left(\frac{\delta U_i}{\delta x_j} + \frac{\delta U_j}{\delta x_i} \right)$ ("Benchmark Computations of Laminar Flow around a Cylindr", Schafer and Turek; 1996, Notes on numerical dluid mechanics; 52, 547-566) ("Incompressibel Fluid Dynamics", Hunt; 1964) this seems to relate directly to the visous fluid stress tensor, $\tau_{i,j}=\mu\left(\frac{\delta u_i}{\delta x_j}+\frac{\delta u_j}{\delta x_i}\right)$ ("Introduction to computational fluid dynamics", Date; 2005) as far as I can tell this adds another three elements to the viscous term, so for a three dimensional flow the term would be, $\rho \nu \left(\frac{\delta}{\delta x}\left(\frac{\delta u}{\delta x} + \frac{\delta u}{\delta x} \right) + \frac{\delta}{\delta y}\left(\frac{\delta v}{\delta x} + \frac{\delta u}{\delta y}\right)+\frac{\delta}{\delta z}\left(\frac{\delta w}{\delta x} + \frac{\delta u}{\delta z} \right) \right)$ the connection to the stress tensor seems logical, however I cand seem to find a direct derivation. Inconsistancy between available literature sources seems to be sending me in circles, Can anyone help me figure out which formulation I need? Dan

November 17, 2010, 08:16		#5
pete Administrator Peter Jones Join Date: Jan 2009 Posts: 682 Rep Power: 10	The same equation can be written in many ways. CFD Wiki provides a more general dimensional form, which is valid for compressible flows (some of your examples assume incompressible flow): http://www.cfd-online.com/Wiki/Navier-Stokes_equations

November 17, 2010, 09:07		#6
dandalf Member Dan Join Date: Oct 2010 Location: UK Posts: 41 Rep Power: 16	Hi Pete, I'm still strugelling to see the equivelence the link you posted gave the viscos term for the ith dimension as, $\frac{\delta \tau_{i,j}}{\delta x_j}$ which agrees with the formulation in my last post, yet how can this be equivelent to the formulations in my original post, if the entire rest of the equation of momentum matches up, except for the three added terms, $\rho \nu \left(\frac{\delta^2 u}{\delta x ^2}+ \frac{\delta^2 v}{\delta x \delta y}+ \frac{\delta^2 w}{\delta x \delta z}\right)$ Unless these terms all cancel out to zero..or I'm simpy interpreting the notation incorrectly.

November 17, 2010, 09:45		#7
RameshK Member Ramesh K Join Date: Dec 2009 Location: Bangalore Posts: 73 Rep Power: 16	Hi your \tau_{ij} term has a missing term. According to stokes hypothesis for a Newtonian fluid \tau_{ij} = \left( \frac{\partial u_i}{\partial x_j}+\frac{\partial u_j}{\partial x_i} - \frac{2}{3}\frac{\partial u_i}{\partial x_j}\delta_{ij} and apply divergence to this quantity