Week 8: Literature review - RANS derivation and analysis
AIM:
- Derive the Reynolds Averaged Navier Stokes Equation and find an expression for Reynolds stress.
- Understand the terms Reynolds stress, Turbulent viscosity and Molecular viscosity.
INTRODUCTION AND THEORY:
The Reynolds-averaged Navier–Stokes equations (or RANS equations) are time-averaged equations of motion for fluid flows. The idea behind the equations is Reynolds decomposition, whereby an instantaneous quantity is decomposed into its time-averaged and fluctuating quantities. The RANS equations are primarily used to describe turbulent flows. These equations can be used with approximations based on knowledge of the properties of flow turbulence to give approximate time-averaged solutions to the Navier-stokes equation.
GOVERNING EQUATIONS FOR BOUNDARY LAYER:
CONTINUITY EQUATION:
`(delu)/(delx) + (delv)/(dely) = 0`
MOMENTUM EQUATION:
`rho.((delu)/(delt) + u.(delu)/(delx) + v.(delu)/(dely)) = - (delp)/(delx) + mu.(del^2u)/(dely^2)`
ENERGY EQUATION:
`rho.((delT)/(delt) + u.(delT)/(delx) + v.(delT)/(dely)) = K.(del^2T)/(delx^2) + mu.((delu)/(dely))^2`
REYNOLDS DECOMPOSITION:
In fluid dynamics and turbulence theory, Reynolds decomposition is a mathematical technique used to separate the expectation value of a quantity from its fluctuations. Here we split the velocity component into a mean component and a fluctuating component for all three directional components.
`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)`
This is called Reynolds decomposition.
Where`u(x,y,z)`is the instantaneous velocity in at point(x,y,z) and at time t.
`baru(x,y,z)` is the time averaged velocity in x direction. It is time averaged for a time period which is greater than the turbulent time scale.
`u'(x,y,z)`is the fluctuating component of velocity in x direction.
Time averaged velocity in x and y direction
`baru(x,y) = 1/t_text(int)*int_0^(t_text(int)) u(x,y,t).dt`
`barv(x,y) = 1/t_text(int)*int_0^(t_text(int)) v(x,y,t).dt`
Also the time averaged value of the fluctuating component is always zero.
`1/t_text(int)*int_0^(t_text(int)) u'(x,y,t).dt = 0`
`u(x,y,z,t) = bar u(x,y,z) + u'(x,y,z,t)`
Integrating both sides we get
`1/t_text(int)*int_0^(t_text(int)) u(x,y,t).dt = 1/t_text(int)*int_0^(t_text(int)) baru(x,y,t).dt + 1/t_text(int)*int_0^(t_text(int)) u'(x,y,t).dt`
And we known that time averaging of a time averaged quantity results in the same value. So `1/t_text(int)*int_0^(t_text(int)) baru(x,y,t).dt = baru`
`baru(x,y,z,t) = baru(x,y,z,t) + 1/t_text(int)*int_0^(t_text(int)) u'(x,y,t).dt`
So,`1/t_text(int)*int_0^(t_text(int)) u'(x,y,t).dt = 0`
REYNOLDS AVERAGED NS EQUATION DERIVATION:
Lets apply the Reynolds decomposition to the NS continuity equation
`(delu)/(delx) + (delv)/(dely) = 0`
`(del)/(delx) (baru + u') + (del)/(dely) (barv + v')`
Integrate the entire equation and do a time averaging
`1/t_(text(int))int(del)/(delx) (baru + u') dt + 1/t_(text(int))int(del)/(dely) (barv + v') dt = 0`
`1/t_(text(int))int(del baru)/(delx).dt + 1/t_(text(int))int(del u')/(delx).dt + 1/t_(text(int))int(del barv)/(delx).dt + 1/t_(text(int))int(delv')/(delx).dt = 0`
Time integral of a fluctuating component is zero. So
`1/t_(text(int))int(del v')/(delx).dt = 0` and `1/t_(text(int))int(delu')/(delx).dt = 0`
`1/t_(text(int))int(del baru)/(delx).dt + cancel(1/t_(text(int))int(del u')/(delx).dt )+ 1/t_(text(int))int(del barv)/(delx).dt + cancel(1/t_(text(int))int(delv')/(delx).dt) = 0`
And `baru`and `barv` are not a function of time. So take `barv' and `baru` out of the integral sign
`(delbaru)/(delx) . 1/t_(text(int))intdt + (delbarv)/(dely) . 1/t_(text(int))intdt = 0`
Then apply the limits of integration
`(delbaru)/(delx) . 1/t_(text(int))int_0^(t_(text(int)))dt + (delbarv)/(dely) . 1/t_(text(int))int_0^(t_(text(int)))dt = 0`
`(delbaru)/(delx).(t_(text(int))-0)/t_(text(int)) + (delbarv)/(dely).(t_(text(int))-0)/t_(text(int)) = 0`
So we get the continuity equation in terms of time averaged variables.
`(delbaru)/(delx) + (delbarv)/(dely) = 0` --------continuity equation in terms of time averaged variables
Now, lets simplify the momentum equation
`rho((delu)/(delt) + u.(delu)/(delx) + v.(delu)/(dely)) = -(dp)/(dx) + mu.(del^2u)/(dely^2)`
First move the density to R.H.S
`((delu)/(delt) + u.(delu)/(delx) + v.(delu)/(dely)) = -(1/rho).(dp)/(dx) + (mu/rho).(del^2u)/(dely^2)`
Where `(mu/rho) = v`so the above equation becomes,
`(delu)/(delt) + u.(delu)/(delx) + v.(delu)/(dely) = -(1/rho).(dp)/(dx) + v.(del^2u)/(dely^2)`
Now add x component of velocity times continuity equation on L.H.S. as `(delu)/(delx) + (delv)/(dely) = 0`
`(delu)/(delt) + u.(delu)/(delx) + v.(delu)/(dely) + u.((delu)/(delx) + (delv)/(dely)) = -(1/rho).(dp)/(dx) + v.(del^2u)/(dely`
`(delu)/(delt) + 2.u.(delu)/(delx) + v.(delu)/(dely) + u.(delv)/(dely) = -(1/rho).(dp)/(dx) + v.(del^2u)/(dely`-------(1)
If we take product rule
`(del(uv))/(dely) = u.(delv)/(dely) + v.(delu)/(dely)`
Substitute in equation 1 then
`(delu)/(delt) + 2.u.(delu)/(delx) + (del(uv))/(dely) = -(1/rho).(dp)/(dx) + v.(del^2u)/(dely^2)`
Substituting `2.u.(delu)/(delx)` as `(del(u^2))/(delx)`
`(delu)/(delt) +(delu^2)/(delx) + (del(uv))/(dely) = -(1/rho).(dp)/(dx) + v.(del^2u)/(dely^2)`
Now apply the time integration and Reynolds decomposition.
`1/t_(text(int))int_0^(t_(text(int)))[(del(baru+u'))/(delt)].dt + 1/t_(text(int))int_0^(t_(text(int)))[(del(baru+u')^2)/(delx)].dt + 1/t_(text(int))int_0^(t_(text(int)))[(del(baru+u').(barv + v'))/(dely)].dt = -1/t_(text(int))int_0^(t_(text(int)))[(1/rho)(del(barp+p'))/(delx)].dt + 1/t_(text(int))int_0^(t_(text(int)))[v.(del^2(baru+u'))/(dely^2)].dt`
`1/t_(text(int))int_0^(t_(text(int)))[(delbaru)/(delt) + (delu')/(delt) + (delbaru^2)/(delx) + (delu^'2)/(delx) + del/(delx)(2.baru.u') + del/(dely)(baru.barv) + del/(dely)(u'.v') + del/(dely)(u'.barv) + del/(dely)(baru.v')].dt = 1/t_(text(int))int_0^(t_(text(int)))[-(1/rho).(delbarp)/(delx) - 1/rho.(delp')/(delx)].dt + 1/t_(text(int))int_0^(t_(text(int)))[v.(del^2baru)/(dely^2) + v.(del^2u')/(dely^2)].dt`
Now as we have seen that time averaging of fluctuating equations are zero
`1/t_(text(int))int_0^(t_(text(int)))[(delbaru)/(delt) +cancel((delu')/(delt)) + (delbaru^2)/(delx) + (delu^'2)/(delx) + cancel(del/(delx)(2.baru.u')) + del/(dely)(baru.barv) + del/(dely)(u'.v') + cancel(del/(dely)(u'.barv)) + cancel(del/(dely)(baru.v'))].dt = 1/t_(text(int))int_0^(t_(text(int)))[-(1/rho).(delbarp)/(delx) - cancel(1/rho.(delp')/(delx))].dt + 1/t_(text(int))int_0^(t_(text(int)))[v.(del^2baru)/(dely^2) + cancel(v.(del^2u')/(dely^2))].dt`
So, the above equation becomes
`1/t_(text(int))int_0^(t_(text(int)))[(delbaru)/(delt) + (delbaru^2)/(delx) + (delu^'2)/(delx) + del/(dely)(baru.barv) + del/(dely)(u'.v')].dt = 1/t_(text(int))int_0^(t_(text(int)))[-(1/rho).(delbarp)/(delx)].dt + 1/t_(text(int))int_0^(t_(text(int)))[v.(del^2baru)/(dely^2)].dt`
On simplifying the above equation becomes,
`(delbaru)/(delt) + (delbaru^2)/(delx) + del/(dely)(baru.barv) =-1/rho.(delbarp)/(delx) + v.(del^2baru)/(dely^2) - 1/t_(text(int))int_0^(t_text(int))[del/(dely)(u'v') + (delu^'2)/(delx)].dt`
The first three terms on LHS and first two terms on RHS forms the momentum equation which we saw earlier but now with time averaged quantities.
`-1/t_(text(int))int_0^(t_text(int))[ (delu^'2)/(delx)].dt`
The velocity gradients in the y direction will be much more stronger than the gradients in the x direction because of the boundary layer, so fluctuating in y direction will also be lower. Then we can ignore the above term
`- 1/t_(text(int))int_0^(t_text(int))[del/(dely)(u'v')].dt`
This term has two velocity components. u' in x direction and v' in y direction. There is a correlation between these two components. If one component is positive then another is negative and the product will be negative. Time integral of a negative term cannot be zero.
`(delbaru)/(delt) + (delbaru^2)/(delx) + del/(dely)(baru.barv) =-1/rho.(delbarp)/(delx) + v.(del^2baru)/(dely^2) - 1/t_(text(int))int_0^(t_text(int))[del/(dely)(u'v')].dt`
Simplifying the RHS
RHS = `=-1/rho.(delbarp)/(delx) + 1/rho.del/(dely)[mu.(delbaru)/(dely) + rho/t_(text(int))int_0^(t_(text(int)))(u'v')dt]`
`(delbaru)/(delt) + (delbaru^2)/(delx) + del/(dely)(baru.barv)` are the inertial terms.
`-1/rho(delbarp)/(delx)` is the pressure gradient term.
`1/rho.del/(dely) [mu.(delbaru)/(dely) + rho/(t_(text(int)))int_0^(t_(text(int)))(u'v')dt]` are the momentum diffusion terms`
momentum can be diffused in two ways
- Through the molecular viscosity `mu.(delbaru)/(dely)`
- Because of the turbulent motion `rho/(t_(text(int)))int_0^(t_(text(int)))(u'v')dt`
`(delbaru)/(delt) + (delbaru^2)/(delx) + del/(dely)(baru.barv) =-1/rho.(delbarp)/(delx) + 1/rho.del/(dely)[mu.(delbaru)/(dely) + rho/t_(text(int))int_0^(t_(text(int)))(u'v')dt]` is the RANS equation.
And `[rho/t_(text(int))int_0^(t_(text(int)))(u'v')dt] = rho.u'v'` term is called as reynolds stress.
EXPLAIN YOUR UNDERSTANDING OF THE TERM REYNOLD’S STRESS
Reynolds stress are averaged stress imposed on the flow by the turbulent fluctuations. It is a component of total stress tensor in a fluid obtained when the navier-stokes equations are Reynolds decomposed and time averaged to account for turbulent fluctuations in fluid momentum.
WHAT IS TURBULENT VISCOSITY? HOW IS IT DIFFERENT FROM MOLECULAR VISCOSITY?
Turbulent viscosity is also known as eddy viscosity. It is due to the velocity fluctuations in a turbulent flow at high Reynolds number. It depends on the flow, it is not a fluid property. However, molecular viscosity is a fluid property. Molecular viscosity is same as the dynamic viscosity. Molecular viscosity purely depends on the fluid type and the temperature whereas the turbulent viscosity depends also on flow.