Literature review RANS derivation and analysis
Governing equations (usually Navier Stroke's equations) of fluid flow when solved are completely capable of giving solutions for both laminar and turbulent flows, the approach of solving the governing equations to simulate turbulence is called direct nummerical simulation and even though it gives pretty acccurate results it is quite impractical for engineering simulation because to get accurate results very small timestep (inverse of frequency of fluctuation of velocity) should be used which incurs high computational power. That is why turbulence models are preferred which can be used to solve turbulence for larger time steps without incurring high computational cost. A turbulence model achieves this by integrating the governing equations over a large time, a process called averaging and applying Reynold's decomposition to convert the governing equations into Reynold's Averaged Navier Strokes Equation (RANS) form.
Reynold's decomposition:
`u(x,y,t) = bar u(x,y,t) + u'(x,y,t)`
`v(x,y,t) = bar v(x,y,t)+ v'(x,y,t)`
where, `bar u = 1/t_"int" int_0^(t_"int")u(x,y,t)dt` and
`bar v = 1/t_"int" int_0^(t_"int")v(x,y,t)dt`
Now,
`u(x,y,t) = bar u(x,y,t)+u'(x,y,t)`
Time averaging both sides, we get
`1/t_"int" int_0^(t_"int")u(x,y,t)dt = 1/t_"int" int_0^(t_"int")bar u(x,y,t)dt+1/t_"int" int_0^(t_"int")u'(x,y,t)dt`
we know that,
`1/t_"int" int_0^(t_"int")u(x,y,t)dt = bar u(x,y,t)` and `1/t_"int" int_0^(t_"int")bar u(x,y,t)dt = bar u(x,y,t)` as averaging a average term gives the same average term.
Applying this to the above eqation we get,
`bar u(x,y,t) = bar u(x,y,t) +1/t_"int" int_0^(t_"int")u'(x,y,t)dt`
`implies 1/t_"int" int_0^(t_"int")u'(x,y,t)dt = 0` this is equation 1
Similarly, `1/t_"int" int_0^(t_"int")v'(x,y,t)dt = 0` this is equation 2
Now applying RANS decomposition to Continuity equation,
We know that continuity equation for 2D is
`(delu)/(delx) + (delv)/(dely) = 0`
Applying RANS Decomposition we get,
`(del(baru+u))/(delx) + (del(barv+v'))/(dely) = 0`
Using integral time averaging process,
`1/(t_"int")int_0^(t_"int")((del(baru+u))/(delx)dt + (del(barv+v'))/(dely)dt) = 0`
`1/(t_"int")int_0^(t_"int")(del(baru))/(delx)dt+1/(t_"int")int_0^(t_"int")(del(u'))/(delx)dt + 1/(t_"int")int_0^(t_"int")(del(barv))/(dely)dt+1/(t_"int")int_0^(t_"int")(del(v'))/(dely)dt = 0`
Applying equation 1 and 2 here, we get
`(del(baru))/(delx) + (del(barv))/(dely)= 0`
Now, momentum equation for 2D flow is,
`rho((delu)/(delt) + u(delu)/(delx) + v(delu)/(dely)) = - (delP)/(delx)+mu((del^2u)/(dely^2))`
`implies ((delu)/(delt) + u(delu)/(delx) + v(delu)/(dely)) = -1/rho (delP)/(delx)+mu/rho((del^2u)/(dely^2))`
Adding u times continuity here to the LHS as continuity equation is zero, equation will not be affected
`implies ((delu)/(delt) + u(delu)/(delx) + v(delu)/(dely))+u((delu)/(delx) + (delv)/(dely)) = -1/rho (delP)/(delx)+mu/rho((del^2u)/(dely^2))`
`implies (delu)/(delt) + 2u(delu)/(delx) + v(delu)/(dely)+u(delv)/(dely) = -1/rho (delP)/(delx)+mu/rho((del^2u)/(dely^2))`
`implies (delu)/(delt) + 2u(delu)/(delx) + (del(uv))/(dely) = -1/rho (delP)/(delx)+mu/rho((del^2u)/(dely^2))`
`implies (delu)/(delt) + (del(u^2))/(delx) + (del(uv))/(dely) = -1/rho (delP)/(delx)+mu/rho((del^2u)/(dely^2))`
applying RANS decomposition on momentum equation,
`implies 1/(t_"int")int_0^(t_"int")((del(baru))/(delt)dt+(del(u'))/(delt)dt + (del(baru)^2)/(delx)dt+(del(u')^2)/(delx)dt+(del(2baru u'))/(delx)dt + (del(barubarv))/(dely)dt+(del(baruv'))/(dely)dt+(del(u'barv))/(dely)dt+(del(u'v'))/(dely)dt) = 1/(t_"int")int_0^(t_"int")(-1/rho (del(barP))/(delx)dt-1/rho (del(P'))/(delx)dt+mu/rho(del^2(baru))/(dely^2)dt+mu/rho(del^2(u'))/(dely^2)dt)`
where the terms,
`1/(t_"int")int_0^(t_"int")(del(u'))/(delt)dt`, `1/(t_"int")int_0^(t_"int")(del(2baru u'))/(delx)dt`,`1/(t_"int")int_0^(t_"int")(del(baruv'))/(dely)dt`, `1/(t_"int")int_0^(t_"int")(del(u'barv))/(dely)dt` , `1/(t_"int")int_0^(t_"int")(-1/rho (del(P'))/(delx)dt)`, `1/(t_"int")int_0^(t_"int")(mu/rho(del^2(u'))/(dely^2)dt)` equal 0.
where the term,
`1/(t_"int")int_0^(t_"int")((del(u'v'))/(dely)dt) ne 0` because for long term observations, fluctuations of the `u'` and `v'` interact with each other for the transfer of momentum and energy and
`1/(t_"int")int_0^(t_"int")((del(u')^2)/(delx)dt) ne 0` as squaring a function of time makes all negative values in it turn into positive. So time averaging the term here cannot be zero.
So, the momentum equation becomes,
`(del(baru))/(delt) + (del(baru)^2)/(delx)+ (del(barubarv))/(dely) = -1/rho (del(barP))/(delx)+mu/rho(del^2(baru))/(dely^2)-1/(t_"int")int_0^(t_"int")((del(u')^2)/(delx)dt)-1/(t_"int")int_0^(t_"int")((del(u'v'))/(dely)dt)`
Now,
`1/(t_"int")int_0^(t_"int")((del(u')^2)/(delx)dt)` though not zero, the value of this term is negligeable as variation of u velocity along x direction in a boundary layer is very small.
So, the momentum equation becomes,
`(del(baru))/(delt) + (del(baru)^2)/(delx)+ (del(barubarv))/(dely) = -1/rho (del(barP))/(delx)+mu/rho(del^2(baru))/(dely^2)-1/(t_"int")int_0^(t_"int")((del(u'v'))/(dely)dt)`
`(del(baru))/(delt) + (del(baru)^2)/(delx)+ (del(barubarv))/(dely) = -1/rho (del(barP))/(delx)+1/rhodel/(dely)(mu(del(baru))/(dely)-rho/(t_"int")int_0^(t_"int")u'v'dt)`
where the terms,
`(del(baru))/(delt) + (del(baru)^2)/(delx)+ (del(barubarv))/(dely)` are the inertial terms
`-1/rho (del(barP))/(delx)` is the pressure term
`mu(del(baru))/(dely)` is viscous stress(`tau_v`).
`rho/(t_"int")int_0^(t_"int")u'v'dt` is reynold's stress(`tau_s`).
Now, when the equation `(del(baru))/(delt) + (del(baru)^2)/(delx)+ (del(barubarv))/(dely) = -1/rho (del(barP))/(delx)+1/rhodel/(dely)(mu(del(baru))/(dely)-rho/(t_"int")int_0^(t_"int")u'v'dt)` is applied inside viscous sub region of the boundary layer with negligeable pressure gradient, it becomes
`1/rhodel/(dely)(mu(del(baru))/(dely)-rho/(t_"int")int_0^(t_"int")u'v'dt) = 0`
as inertial terms are so small in the viscous sub layer that they are approximately zero.
`implies del/(dely)(mu(del(baru))/(dely)-rho/(t_"int")int_0^(t_"int")u'v'dt) = 0`
`implies mu(del(baru))/(dely)-rho/(t_"int")int_0^(t_"int")u'v'dt = "constant"`
Now when the above equation is applied to a region outside viscous sub layer region, the turbulent viscosity dominates so the equation becomes,
`-rho/(t_"int")int_0^(t_"int")u'v'dt = "constant"`
The constant in the above equation is nothing but Reynolds stress (`tau_s`),
`-rho/(t_"int")int_0^(t_"int")u'v'dt = tau_s`
`implies tau_s = -rho(bar(u'v'))`
Reynolds stress is the stress that is induced into the fluid by convective momentum transfer due to the random motions caused by turbulent fluctuations of velocities (eddies).
Now, when the Reynolds stress is written in the form of viscous stress, we get,
`tau_s = rhoepsilon(delbaru)/(dely)`
where, `epsilon` is the turbulent or eddy viscosity.
The total stress experienced by the fluid is,
`tau = tau_s + tau_v`
`tau=mu(del(baru))/(dely)+rhoepsilon(delbaru)/(dely)`
`tau=rho(mu/rho(del(baru))/(dely)+epsilon(delbaru)/(dely))`
`tau=rho(nu(delbaru)/(dely)+epsilon(delbaru)/(dely))`
`tau=rho(nu+epsilon)(delbaru)/(dely)`
Here, `nu` is the molecular viscosity.
Molecular viscosity is a property of the fluid and it is not influenced by the flow conditions, eddy viscosity is smaller than molecular viscosity when the viscous sub layer of the flow is observed.
Turbulent viscosity arises solely due to the random flow fluctuations that cannot be easily predicted, tubulent viscosity is larger than in the regions outside of viscous sub layer of the flow.