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Week 8: Literature review RANS derivation and analysis

Navier Stokes Equations $\frac{\partial u}{\partial x} + \frac{\partial v}{\partial y} = 0$ $(del u)/(del x) + (del v)/(del y) = 0$ $ρ (\frac{\partial u}{\partial t} + u \frac{\partial u}{\partial x} + v \frac{\partial u}{\partial y}) = - \frac{\partial p}{\partial x} + μ \frac{\partial^{2} u}{\partial y^{2}}$ $rho ( (del u)/(del t) + u (del u)/(del x) + v (del u)/(del y)) = -(del p)/(del x) + mu (del^2 u)/(del y^2)$ Reynolds Decomposition Reynolds decomposition refers to separation of the flow variable (like velocity u {\displaystyle u} $u$ $u$ ) into the mean (time-averaged)…

Dushyanth Srinivasan
updated on 15 Mar 2022

Navier Stokes Equations

$\frac{\partial u}{\partial x} + \frac{\partial v}{\partial y} = 0$ $(del u)/(del x) + (del v)/(del y) = 0$

$ρ (\frac{\partial u}{\partial t} + u \frac{\partial u}{\partial x} + v \frac{\partial u}{\partial y}) = - \frac{\partial p}{\partial x} + μ \frac{\partial^{2} u}{\partial y^{2}}$ $rho ( (del u)/(del t) + u (del u)/(del x) + v (del u)/(del y)) = -(del p)/(del x) + mu (del^2 u)/(del y^2)$

Reynolds Decomposition

Reynolds decomposition refers to separation of the flow variable (like velocity $u {\displaystyle u} uuu$ ) into the mean (time-averaged) component ( $¯ u$ $bar u$ $u ¯ {\displaystyle {\overline {u}}} ˉu¯ubar u$ ) and the fluctuating component ( $u'$ $u ′ {\displaystyle u^{\prime }} `U$ ).

Turbulent time scale of the flucuating component way smaller than the time averaged time scale.

Applying Reynold's Decomposition for flow velocities $u,v,w$ , we get:

$u (x,y,z,t) = bar u (x,y,z) + u'(x,y,z,t)$

$v (x,y,z,t) = bar v (x,y,z) + v'(x,y,z,t)$

$w (x,y,z,t) = bar w (x,y,z) + w'(x,y,z,t)$

Where $bar u(x,y,z,t)$ is defined as follows,

$bar u (x,y,z) = 1/underset(Int)(t) int_0^(underset(Int)(t)) u(x,y,z) dt$

$bar v (x,y,z) = 1/underset(Int)(t) int_0^(underset(Int)(t)) v (x,y,z) dt$

$bar w (x,y,z) = 1/underset(Int)(t) int_0^(underset(Int)(t)) w (x,y,z) dt$

Lets take $u (x,y,z,t) = bar u (x,y,z) + u'(x,y,z,t)$ ,

Integrating over time from 0 to $underset(Int)(t)$ ,

$1/underset(Int)(t) int_0^(underset(Int)(t))bar u (x,y,z) dt = 1/underset(Int)(t) int_0^(underset(Int)(t)) bar u(x,y,z) dt + 1/underset(Int)(t) int_0^(underset(Int)(t)) u'(x,y,z,t)dt$

1st term is $bar u (x,y,z)$ according to its definition, 2nd term is averaging an already time averaged term hence it is also $bar u (x,y,z)$ .

$cancel (1/underset(Int)(t) int_0^(underset(Int)(t))bar u (x,y,z) dt )= cancel (1/underset(Int)(t) int_0^(underset(Int)(t)) bar u(x,y,z) dt) + 1/underset(Int)(t) int_0^(underset(Int)(t)) u'(x,y,z,t)dt$

$1/underset(Int)(t) int_0^(underset(Int)(t)) u'(x,y,z,t)dt = 0$

Hence the integration of any flucuating variable over time is always going to be zero. In other words, the value of time averaged flucuating variables is zero.

Effect of RANS on Navier-Stokes Continuity Equation

We begin with the Continuity Equation,

$(del u)/(del x) + (del v)/(del y) = 0$

Decomposing the variables to time averaged and fluctuating values,

$(del (baru+u'))/(del x) + (del (barv+v'))/(del y) = 0$

Applying time integration/time averaging,

$1/underset(Int)(t) int_0^(underset(Int)(t))(del (baru+u'))/(del x)dt + 1/underset(Int)(t) int_0^(underset(Int)(t))(del (barv+v'))/(del y) dt= 0$

Applying the integral to every term,

$1/underset(Int)(t) int_0^(underset(Int)(t))(del baru)/(del x)dt + 1/underset(Int)(t) int_0^(underset(Int)(t))(del u')/(del x)dt +1/underset(Int)(t) int_0^(underset(Int)(t))(del barv)/(del y) dt + 1/underset(Int)(t) int_0^(underset(Int)(t))(del v')/(del y) dt= 0$

We know time integral/time averaging over a flucuating variable is zero,

$1/underset(Int)(t) int_0^(underset(Int)(t))(del baru)/(del x)dt + cancel(1/underset(Int)(t) int_0^(underset(Int)(t))(del u')/(del x)dt )+1/underset(Int)(t) int_0^(underset(Int)(t))(del barv)/(del y) dt + cancel(1/underset(Int)(t) int_0^(underset(Int)(t))(del v')/(del y) dt)= 0$

Resulting equation,

$(del baru)/(del x)1/underset(Int)(t) int_0^(underset(Int)(t))dt + (del barv)/(del y)1/underset(Int)(t) int_0^(underset(Int)(t)) dt= 0$

Evaluvating the integral,

$(del baru)/(del x)+ (del barv)/(del y)= 0$

This is the result of applying Reynold's-Averaged Navier-Stokes (RANS) Equation on the 2D Navier-Stokies continuity equation

Effect of RANS on Navier-Stokes Momentum Equation

We begin with the Momemtum Equation,

$rho ( (del u)/(del t) + u (del u)/(del x) + v (del u)/(del y)) = -(del p)/(del x) + mu (del^2 u)/(del y^2)$

Rearranging by taking pressure to the other side,

$(del u)/(del t) + u (del u)/(del x) + v (del u)/(del y) = -1/p(del p)/(del x) + mu/p (del^2 u)/(del y^2)$

We know that $mu / rho = nu$ (Kinematic Viscosity / Density = Dynamic Viscosity)

$(del u)/(del t) + u (del u)/(del x) + v (del u)/(del y) = -1/p(del p)/(del x) + nu (del^2 u)/(del y^2)$

Adding the continuity equation on both sides,

$(del u)/(del t) + u (del u)/(del x) + v (del u)/(del y) + (del u)/(del x) + u( (del v)/(del y) )= -1/p(del p)/(del x) + nu (del^2 u)/(del y^2)$

Simplyfying,

$(del u)/(del t) +2 u (del u)/(del x) + v (del u)/(del y) +u (del v)/(del y) = -1/p(del p)/(del x) + nu (del^2 u)/(del y^2)$

Recall the product rule,

$( del (u*v))/(del y) = v (del u)/(del y) +u (del v)/(del y)$

Substituting product rule for term 3 and 4 on the LHS,

$(del u)/(del t) +2 u (del u)/(del x) + ( del (u*v))/(del y) = -1/p(del p)/(del x) + nu (del^2 u)/(del y^2)$

Recall that $(d(x^2))/dx = 2x dx/dx$ ,

$(del u)/(del t) +(del u^2)/(del x) + ( del (u*v))/(del y) = -1/p(del p)/(del x) + nu (del^2 u)/(del y^2)$

This equation is simplyfied enough, now applying time integral/time averaging,

$1/underset(Int)(t) int_0^(underset(Int)(t))((del (bar u + u’))/(del t) +(del (bar u + u’)^2)/(del x) + ( del ((bar u + u’)*(bar v+ v')))/(del y) )dt= 1/underset(Int)(t) int_0^(underset(Int)(t))(-1/p(del (bar p+p'))/(del x) + nu (del^2 (bar u + u’))/(del y^2))dt$

Expanding the integral,

$1/underset(Int)(t) int_0^(underset(Int)(t)) ((del bar u)/(del t)dt +(del u')/(del t)dt +(del bar u^2)/(del x)dt +(del u'^2)/(del x)dt +(del 2* bar u *u')/(del x)dt +(del baru * barv)/(del y)dt +(del baru*v')/(del y)dt +(del u'*barv)/(del y)dt +(del u'*v')/(del y)dt)= 1/underset(Int)(t) int_0^(underset(Int)(t))(-1/p(del (bar p))/(del x)dt -1/p(del (p'))/(del x) dt+ nu ( v(del^2 (bar u))/(del y^2)dt + v(del^2 (u’))/(del y^2)dt))$

Since integral of flucuating terms is zero, they can be ignored,

$1/underset(Int)(t) int_0^(underset(Int)(t)) ((del bar u)/(del t)dt +cancel((del u')/(del t)dt) +(del bar u^2)/(del x)dt +(del u'^2)/(del x)dt +cancel((del 2* bar u *u')/(del x)dt )+(del baru * barv)/(del y)dt +cancel((del baru*v')/(del y)dt) +cancel((del u'*barv)/(del y)dt) +(del u'*v')/(del y)dt)= 1/underset(Int)(t) int_0^(underset(Int)(t))(-1/p(del (bar p))/(del x)dt -cancel(1/p(del (p'))/(del x) dt)+ nu ( v(del^2 (bar u))/(del y^2)dt + cancel( v(del^2 (u’))/(del y^2)dt)))$

Cancelling and cleaning the equation,

$(del bar u)/(del t)+(del bar u^2)/(del x) +(del baru * barv)/(del y) = -1/p(del bar p)/(del x)+ v (del^2 baru)/(dy^2) - 1/underset(Int)(t) int_0^(underset(Int)(t))(del^2 (bar u))/(del y^2)dt - 1/underset(Int)(t) int_0^(underset(Int)(t)) (del (u' *v'))/(dely) dt$

Since iits a boundary layer region, the momentum difference from the y axis does not effect the x axis

Hence, $1/underset(Int)(t) int_0^(underset(Int)(t))(del^2 (bar u))/(del y^2)dt = 0$

Rearranging the equation again,

$(del bar u)/(del t)+(del bar u^2)/(del x) +(del baru * barv)/(del y) = -1/p(del bar p)/(del x)+ 1/p*del /(dely)[ mu (del bar u)/(del y)-p/underset(Int)(t) int_0^(underset(Int)(t)) u' *v' dt]$

In this equation,

1. The terms on the LHS are the inverted terms: $(del bar u)/(del t)+(del bar u^2)/(del x) +(del baru * barv)/(del y)$

2. $-1/p(del bar p)/(del x)$ is the pressure term

3. $1/p*del /(dely)[ mu (del bar u)/(del y)-p/underset(Int)(t) int_0^(underset(Int)(t)) u' *v' dt]$ is the stress term

Expanding the stress term, we get 2 distinct terms:

1. $v (del^2 baru)/(dy^2)$ is the stress caused due to molecular viscosity. This stress is called viscous stress.

2. $- 1/underset(Int)(t) int_0^(underset(Int)(t)) (del (u' *v'))/(dely) dt$ is the stress caused due to turbulent flucuations. This stress is called Reynold's Stress.

Explain your understanding of the term reynold's stress

It puts a numerical value to the relationship between turbulent (or inertial forces) and viscous forces in the Navier-Stokes equations. Turbulent modelling is all about finding this value in order to accurately simulate turbulent flow.

What is turbulent viscosity? How is it different from molecular viscosity?

During turbulent flow, flow of molecules inside the domain is not only due to molecular motions, there is another factor which affects flow - flow of molecules is affected by effies, vortices, etc. In order to quantify this change in flow, a factor called turbulent viscosity is introduced. This coefficient is added to all forms of the Navier-Stokes Equations in a turbulent flow environment.

Molecular viscosity is a static component which does not vary by time (since it has no $del / (del t)$ terms), while molecular viscosity is dependent on time.

Sources:

1. https://www.cfd-online.com/Forums/main/74854-turbulent-viscosity.html