Phil Dias Literature Review RANS derivation

Problem:

Apply reynold\'s decomposition to the NS equations and come up with the expression for reynold\'s stress.

• Explain your understanding of the terms reynold\'s stress
• What is turbulent viscosity? How is it different from molecular viscosity?

Solution:

The Reynold\'s Averaged Navier Stokes equation (RANS) applys Reynold\'s decomposition to convert governing equations into a turbulence model. This is done by integrating the governing equations over a large time. The purpose of this process is to solve turbulence for larger time steps without incurring high computational cost.

Reynold\'s Decompostion:

`u(x,y,t)= baru(x,y,t) + u\'(x,y,t)`

`v(x,y,t)= barv(x,y,t) + v\'(x,y,t)`

where, `baru = 1/t_i int_0^(t_i) u(x,y,t)dt` and `barv = 1/t_i int_0^(t_i) v(x,y,t)dt`

Now,

`u(x,y,t)= baru(x,y,t) + u\'(x,y,t)`

Time averaging both sides, we get

`1/t_i int_0^(t_i) u(x,y,t)dt = int_0^(t_i) baru(x,y,t)dt + 1/t_i int_0^(t_i)dt1/t_i u\'(x,y,t)`

we know that,

`1/t_i int_0^(t_i) u(x,y,t)dt = baru(x,y,t) and 1/t_i int_0^(t_i) baru(x,y,t)dt = baru(x,y,t)` as average term gives the same average term.

Applying the equation above we get,

`baru(x,y,t) = baru(x,y,t) + 1/t_i int_0^(t_i) u\'(x,y,t)dt`

`=> 1/t_i int_0^(t_i) u\'(x,y,t)dt = 0` this is Equation 1.

and, `1/t_i int_0^(t_i) v\'(x,y,t)dt` this is Equation 2.

Applying RANS decomposition to the Continuity equation,

We know that continuity equation for 2D is

`(delu)/(delx) + (delv)/(dely) = 0`

Applying RANS decompostion we get,

`(del(baru+u\'))/(delx) + (del(barv+v\'))/(dely) = 0`

Using integral time averaging process,

`1/t_i int_0^(t_i) ((del(baru+u\'))/(delx) + (del(barv+v\'))/(dely) )= 0`

`1/t_i int_0^(t_i) (del(baru))/(delx) dt + 1/t_i int_0^(t_i) (del(u\'))/(delx) dt + 1/t_i int_0^(t_i) (del(barv))/(dely) dt + 1/t_i int_0^(t_i) (del(v\'))/(dely) dt`

Applying Equation 1 and 2, we get

`(del(baru))/ (delx) + (del(barv))/ (dely)`

Momentum Equation for 2D flow:

`rho ((delu)/(delt) + u(delu)/(delx) + v(delu)/(dely)) = - (delP)/(delx) + mu ((del^2u)/(dely^2))`

`=> ((delu)/(delt) + u(delu)/(delx) + v(delu)/(dely)) = - 1/rho (delP)/(delx) + mu/rho ((del^2u)/(dely^2))`

Adding u times continuity to the LHS as continuity equation is zero, equation will not be affected

`=> => ((delu)/(delt) + u(delu)/(delx) + v(delu)/(dely)) + u ((delu)/(delx) + (delv)/(dely)) = - 1/rho (delP)/(delx) + mu/rho ((del^2u)/(dely^2))`

`=> (delu)/(delt) + 2u(delu)/(delx) + v(delu)/(dely) + u (delv)/(dely) = - 1/rho (delP)/(delx) + mu/rho ((del^2u)/(dely^2))`

`=> (delu)/(delt) + 2u(delu)/(delx) + (del(uv))/(dely) = - 1/rho (delP)/(delx) + mu/rho ((del^2u)/(dely^2))`

`=> (delu)/(delt) + (del(u^2))/(delx) + (del(uv))/(dely) = - 1/rho (delP)/(delx) + mu/rho ((del^2u)/(dely^2))`

applying RANS decompostion on momentum equation,

`=> 1/t_i int_0^(t_i) ( (del(baru))/(delt) dt + (del(u\'))/(delt) dt + (del(baru)^2)/(delx) dt + (del(u\')^2)/(delx) dt + (del(2baru u\'))/(delx) dt + (del(baru barv))/(dely) dt + (del(baru v\'))/(dely) dt + (del(u\' barv))/(dely) dt + (del(u\' v\'))/(dely) dt ) = 1/t_i int_0^(t_i) ( - 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_i int_0^(t_i) (del(u\'))/(delt) dt, 1/t_i int_0^(t_i) (del(2baru u\'))/(delx) dt, 1/t_i int_0^(t_i) (del(baru v\'))/(dely) dt, 1/t_i int_0^(t_i) (del(u\' barv))/(dely) dt, 1/t_i int_0^(t_i) (-1/rho (del(P\'))/(delx) dt),1/t_i int_0^(t_i) (mu/rho (del^2(u\'))/(dely^2) dt)` are equal to 0.

where the term,

`1/t_i int_0^(t_i) ( (del(u\'v\'))/(dely) dt) != 0` due to long term observations, fluctuations of the u\' and v\' interact with each other for the transfer of momemtum and energy and `1/t_i int_0^(t_i) ( (del(u\')^2)/(delx) dt) != 0` as squaring a function of time maks all negative values turn positive. Therefore; time averaging the term cannot be zero.

So, the mometum equation becomes:,

`(del(baru))/(delt) + (del(baru)^2)/(delx) + (del(baru barv))/(dely) = - 1/rho (del(barP))/(delx) + mu/rho (del^2(baru))/(dely^2) - 1/t_i int_0^(t_i) ((del(u\')^2)/(delx) dt) - 1/t_i int_0^(t_i) ((del(u\'v\'))/(dely) dt)`

Now,

`1/t_i int_0^(t_i) ((del(u\')^2)/(delx) dt)` is a negligeable variation of u velocity along the x direction in a boundary layer.

SO, tne momemtum equation becomes,

`(del(baru))/(delt) + (del(baru)^2)/(delx) + (del(baru barv))/(dely) = - 1/rho (del(barP))/(delx) + mu/rho (del^2(baru))/(dely^2) - 1/t_i int_0^(t_i) ((del(u\'v\'))/(dely) dt)`

simplifying,

`(del(baru))/(delt) + (del(baru)^2)/(delx) + (del(baru barv))/(dely) = - 1/rho (del(barP))/(delx) + 1/rho (del)/(dely) (mu (del(baru))/(dely) - rho/t_i int_0^(t_i) u\'v\' dt)`

where the terms,

`(del(baru))/(delt) + (del(baru)^2)/(delx) + (del(baru barv))/(dely)` are inertial terms

`- 1/rho (del(barP))/(delx)` is the pressure term

`mu (del(baru))/(dely)` is the viscous stress `(tau_v)`

`rho/t_i int_0^(t_i) u\'v\' dt` is Reynold\'s stress `(tau_s)`

Now, when the equation

`(del(baru))/(delt) + (del(baru)^2)/(delx) + (del(baru barv))/(dely) = - 1/rho (del(barP))/(delx) + 1/rho (del)/(dely) (mu (del(baru))/(dely) - rho/t_i int_0^(t_i) u\'v\' dt)` is applied inside viscous sub region of the boundary layer with negligeable pressure gradient, it becomes,

`1/rho (del)/(dely) (mu (del(baru))/(dely) - rho/t_i int_0^(t_i) u\'v\' dt) = 0`

as inertial terms are small in the viscous sub layer that they are approximately zero.

`=> (del)/(dely) (mu (del(baru))/(dely) - rho/t_i int_0^(t_i) u\'v\' dt) = 0`

`=>(del)/(dely) (mu (del(baru))/(dely) - rho/t_i int_0^(t_i) u\'v\' dt) = \"constant\"`

When the equation above is applied to a region outside the viscous sub layer region, the turbulent viscosity doinates so the equation becomes,

`- rho/t_i int_0^(t_i) u\'v\' dt = \"constant\"`

The constant in the equation above is the Reynold\'s stress `(tau_s)`

`- rho/t_i int_0^(t_i) u\'v\' dt = tau_s`

`=> tau_s = -rho(bar(u\'v\'))`

Reynold\'s 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).

Reynold\'s stress is in the form of viscous stress:

`tau_s = rho epsilon (delbaru)/(dely)`

where, `epsilon` is the turbulent or eddie viscosity.

The total stress experienced by the fluid is:

`tau = tau_s + tau_v`

`tau = mu (del(baru))/(dely) + rho epsilon (delbaru)/(dely)`

`tau = rho (mu/rho (del(baru))/(dely) + epsilon (delbaru)/(dely))`

`tau = rho (nu (del(baru))/(dely) + epsilon (delbaru)/(dely))`

`tau = rho (nu + epsilon) (delbaru)/(dely)`

`nu` is the molecular viscosity.

Molecular viscosity is a fluid property and it is not influenced by the flow conditions, eddie viscosity is smaller than molecular viscosity when the viscous sub layer of the flow is observed.

Turbulent viscosity is caused by random flow fluctutions that cannot be easaily predicted. Turbulent viscosity is larger in the regions outside the viscous sub layer of the flow.