Chapter 5 Newtonian incompressible fluid

Setting the divergence to zero in equation 3.4 we get for the divergence of the stress tensor for an incompressible fluid

${\displaystyle \frac{\partial }{\partial r_j}}{\sigma }_{ij}^{*}=\eta {\displaystyle \frac{{\partial }^2}{\partial r_j^2}}{v}_i$

(5.2)

and for

${\displaystyle \frac{\partial }{\partial r_j}}(v_i{\sigma }_{ij}^{*})=\eta {\displaystyle \frac{{\partial }^2}{\partial r_j^2}}\left({\displaystyle \frac{1}{2}}{v}_i^2\right)+\eta {\displaystyle \frac{\partial }{\partial r_j}}\left(v_i{\displaystyle \frac{\partial v_j}{\partial r_i}}\right)$

(5.3)

Inserting these results into (4.14)-(4.16) we get as balance equations for an incompressible newtonian fluid

	${\displaystyle \frac{\partial v_j}{\partial r_j}}=$	$\mathrm{ }0$		(5.4)
	${\displaystyle \frac{\partial v_i}{\partial t}}+v_j{\displaystyle \frac{\partial v_i}{\partial r_j}}=$	$-{\displaystyle \frac{1}{\rho }}{\displaystyle \frac{\partial }{\partial r_i}}p+g_i+\nu {\displaystyle \frac{{\partial }^2}{\partial r_j^2}}{v}_i$		(5.5)
	${\displaystyle \frac{\partial e}{\partial t}}+v_j{\displaystyle \frac{\partial e}{\partial r_j}}=$	$-{\displaystyle \frac{1}{\rho }}{\displaystyle \frac{\partial }{\partial r_j}}(v_{j}p)+v_i{g}_i+\nu {\displaystyle \frac{{\partial }^2}{\partial r_j^2}}\left({\displaystyle \frac{1}{2}}{v}_i^2\right)+\nu {\displaystyle \frac{\partial }{\partial r_j}}\left(v_i{\displaystyle \frac{\partial v_j}{\partial r_i}}\right)$		(5.6)

with the so called kinematic viscosity $\nu =\frac{\eta }{\rho }$ ¹⁴¹⁴Be aware that $\nu$ is a spatially independent quantity only for incompressible fluids! In the compressible case $\rho =\rho (x,t)$ and therefore also the kinematic viscosity $\nu =\nu (x,t)$. So do not use $\nu$ when dealing with compressible fluid.

Inserting the incompressible newtonian stress tensor (5.1) in the equation for the dissipation of a general incompressible fluid 4.18 yields¹⁵¹⁵The same result can be obtained by setting $\frac{\partial v_j}{\partial r_j}=0$ and $\frac{\partial }{\partial t}\rho =0$ in the equation for the dissipation of a compressible newtonian fluid (3.21).

$ℰ=-{\displaystyle \frac{1}{V}}{\displaystyle {\int }_V}\eta \left({\displaystyle \frac{\partial }{\partial r_j}}{v}_i+{\displaystyle \frac{\partial }{\partial r_i}}{v}_j\right){\displaystyle \frac{\partial v_i}{\partial r_j}}𝑑V=-{\displaystyle \frac{1}{V}}{\displaystyle {\int }_V}{\displaystyle \frac{\eta }{2}}{\left({\displaystyle \frac{\partial }{\partial r_j}}{v}_i+{\displaystyle \frac{\partial }{\partial r_i}}{v}_j\right)}^2𝑑V$

(5.7)

where we used again relation H.4.

We can express this result in terms of vorticity by making use of equation (I.5) which gives us for

	${\displaystyle \frac{\eta }{2}}{\left({\displaystyle \frac{\partial }{\partial r_j}}{v}_i+{\displaystyle \frac{\partial }{\partial r_i}}{v}_j\right)}^2$	$=2\eta S_{ij}{S}_{ij}=2\eta \left(R_{ij}{R}_{ij}+{\displaystyle \frac{\partial v_i}{\partial r_j}}{\displaystyle \frac{\partial v_j}{\partial r_i}}\right)$
		$=2\eta \left({\displaystyle \frac{1}{2}}{\omega }_k{\omega }_k+{\displaystyle \frac{\partial }{\partial r_j}}\left(v_i{\displaystyle \frac{\partial v_j}{\partial r_i}}\right)-v_j{\displaystyle \frac{{\partial }^2}{\partial r_i\partial r_j}}{v}_i\right)$
		$=2\eta \left({\displaystyle \frac{1}{2}}{\omega }_k{\omega }_k+{\displaystyle \frac{\partial }{\partial r_j}}\left(v_i{\displaystyle \frac{\partial v_j}{\partial r_i}}\right)-v_j{\displaystyle \frac{{\partial }^2}{\partial r_j\partial r_i}}{v}_i\right)$
		$=\eta {\omega }_k{\omega }_k+2\eta {\displaystyle \frac{\partial }{\partial r_j}}\left(v_i{\displaystyle \frac{\partial v_j}{\partial r_i}}\right)$

So we can express the dissipation for an incompressible newtonian fluid as

$ℰ=-{\displaystyle \frac{1}{V}}{\displaystyle {\int }_V}\eta {\omega }_k{\omega }_k𝑑V-{\displaystyle \frac{1}{V}}{\displaystyle {\int }_V}2\eta {\displaystyle \frac{\partial }{\partial r_j}}\left(v_i{\displaystyle \frac{\partial v_j}{\partial r_i}}\right)𝑑V$

(5.8)

Using Gauss’ theorem to transform the second term we can express this like

$ℰ=-{\displaystyle \frac{1}{V}}{\displaystyle {\int }_V}\eta {\omega }_k{\omega }_k𝑑V-{\displaystyle \frac{1}{V}}{\displaystyle {\oint }_A}2\eta v_i{\displaystyle \frac{\partial v_j}{\partial r_i}}𝑑A$

So if $v_i\frac{\partial v_j}{\partial r_i}=0$ on the border of the volume $V$ the second term will vanish and the dissipation in a incompressible, newtonian fluid is due to the first term only ¹⁶¹⁶Have a look at the Appendix N for an simple example of a flow field, where the second term doesn’t vanish! ¹⁷¹⁷Very often $\rho$ is moved to the other side of the equation (which is possible, because we assume it is spatially constant) and so we get $ϵ=\frac{1}{V}\frac{\partial }{\partial t}{\int }_V{v}^2𝑑V=-\frac{\nu }{V}{\int }_V{\omega }^2𝑑V$

$ℰ=-{\displaystyle \frac{\eta }{V}}{\displaystyle {\int }_V}{\omega }^2𝑑V$

(5.9)

From looking at the divergence equation (see below) we can derive another relation, which tells us, when the second term in equation (5.8) will vanish

	${\displaystyle \frac{1}{V}}{\displaystyle {\int }_V}2\eta {\displaystyle \frac{\partial }{\partial r_j}}\left(v_i{\displaystyle \frac{\partial v_j}{\partial r_i}}\right)𝑑V$	$={\displaystyle \frac{1}{V}}{\displaystyle {\int }_V}2\eta {\displaystyle \frac{{\partial }^2}{\partial r_i\partial r_j}}(v_i{v}_j)𝑑V$
		$={\displaystyle \frac{1}{V}}{\displaystyle {\int }_V}{\displaystyle \frac{2\eta }{\rho }}\left({\displaystyle \frac{{\partial }^2}{\partial r_i^2}}p+4\pi G{\rho }^2\right)𝑑V={\displaystyle \frac{2\nu }{V}}{\displaystyle {\int }_V}\left({\displaystyle \frac{{\partial }^2}{\partial r_i^2}}p+\rho {\displaystyle \frac{{\partial }^2}{\partial r_i^2}}\varphi \right)𝑑V$
		$={\displaystyle \frac{2\nu }{V}}{\displaystyle {\oint }_A}{\displaystyle \frac{\partial }{\partial r_i}}p𝑑A+{\displaystyle \frac{2\eta }{V}}{\displaystyle {\oint }_A}{\displaystyle \frac{\partial }{\partial r_i}}\varphi 𝑑V$

where we made use of the poisson equation for the gravitational potential and Gauss’ theorem. So only if the pressure gradient and the gravitational force balance at the surface of the fluid we can neglect the second term ¹⁸¹⁸In case of no gravity the pressure gradient has to be zero at the surface..

We hope that from this discussion the assumptions behind equation (5.9) become clear compared to the rather obscure arguments by Frisch (1995). Nevertheless it is still unknown weather our reasoning and the reasoning of Frisch (1995) is equivalent.

From equation (5.2) we get for

${\displaystyle \frac{{\partial }^2}{\partial r_i\partial r_j}}{\sigma }_{ij}^{*}=\eta {\displaystyle \frac{\partial }{\partial r_i}}\left({\displaystyle \frac{{\partial }^2}{\partial r_j^2}}{v}_i\right)=\eta {\displaystyle \frac{{\partial }^2}{\partial r_j^2}}\left({\displaystyle \frac{\partial v_i}{\partial r_i}}\right)=0$

(5.10)

Using this result in equation (4.19) we get as equation of state for an incompressible newtonian fluid

${\displaystyle \frac{{\partial }^2}{\partial r_i^2}}p=-\rho {\displaystyle \frac{{\partial }^2}{\partial r_i\partial r_j}}(v_i{v}_j)-4\pi G{\rho }^2$

(5.11)

or in vector notation

$\mathrm{\Delta }p=-\rho \nabla \left[\nabla \cdot (v_i{v}_j)\right]-4\pi G{\rho }^2$

(5.12)

Actually the divergence equation for an incompressible newtonian fluid has a very interesting form. It is show in the Appendix Q.3 that the divergence equation might be related to Bernoulli’s law and even to the Einstein equation of general relativity.

Inserting equation (5.2) into 4.20 we get the vorticity equation for an incompressible newtonian fluid

${\displaystyle \frac{\partial }{\partial t}}{\omega }_g-{ϵ}_{ghi}{\displaystyle \frac{\partial }{\partial r_h}}{ϵ}_{ijm}{v}_j{\omega }_m={\displaystyle \frac{\eta }{\rho }}{\displaystyle \frac{{\partial }^2}{\partial r_j^2}}{\omega }_g$

(5.13)

or in vector notation

${\displaystyle \frac{\partial }{\partial t}}𝝎-\nabla \times (𝒗\times 𝝎)={\displaystyle \frac{\eta }{\rho }}\mathrm{\Delta }𝝎.$

(5.14)