Hi Timm,
I want to study how the Reynolds number varies with the pressure drop in a channel which is not straight (one wall has grooves). In order to about a long entry region, I imposed a fullydeveloped velocity profile as inlet condition, while for the outlet condition I used a linear extrapolation. As I wrote in the previous message, the way I imposed the body force was through equation (1), which relates the body force Fx_LB and the velocity profile. I also wrote that using equations (2) and (3) is possible to determine the velocity Umean, which is the mean velocity imposed at the inlet. Now, using this last Umean is also possible to compute the Reynolds number for the inlet boundary condition:
Re=Umean*Dh_LB/nu_LB
As you correctly wrote, for low Reynolds numbers the quadratic terms have less importance and the flow becomes a Stokes flow. You also mentioned the linear dependence between RePressure drop for a straight channel in the steady state. However, the channel I use is not straight, and it is well documented in the literature that there must be an increase in the pressure drop when the fluid flows over a nonstraight wall compared to the case over a flat one.
Let’s do a simple exercise:
Let’s consider a 2D flat channel with height H=3h and length L=20h, where h can be 1 LU. We impose a fullydeveloped velocity profile at the inlet, given by equation (1).
< 20h >


H =3h


Then, the body force required to have a certain Re can be easily calculated as follows:
From equations (2) and (3) solve for the Umean. (* In the previous message I used h_LB to represent the characteristic length, in this case h_LB=H=3h, but for the Re the hydraulic diameter is employed Dh=2H)
Some mathematics:
Umean= (2.0/3.0) * Umax
Umax= (Fx_LB * H^2)/(8 * nu_LB) and in terms of h Umax= (9* Fx_LB * h^2)/(8 * nu_LB)
Umean=(2.0/3.0) * (9* Fx_LB * h^2)/(8 * nu_LB) = (3* Fx_LB * h^2)/(4 * nu_LB)
now the Reynolds number using Dh=2H=2(3h)=6h
Re= Dh_LB*Umean/nu_LB
Re= (9*h^3*Fx_LB)/(2*nu_LB^2) (A) and solving for the body force we get:
Fx_LB_flat= (2*nu_LB^2*Re)/(9*h^3) (a)
with equation (a) one can find the body force for any desired Reynolds number. As it can be seen, the Re and Fx_LB vary directly, since the channel is straight, (I haven’t discovered nothing new ) , So far so good!!!
Let’s include some obstacles in the channel as depicted below
< 20h > 
2h
_______
The inclusion of obstacles must create a flow resistance and an increase in the pressure drop. For the second case with obstacles, I want to have the same Reynolds number inside the narrow flat channel, which now has a height of 2h instead of 3h. Since the Reynolds number is constant and the Dh changes from 6h to 4h, the Fx_Lb must increase.
Umax= (Fx_LB * H^2)/(8 * nu_LB) and in terms of h Umax= (4* Fx_LB * h^2)/(8 * nu_LB)
Umean=(2.0/3.0) * (4* Fx_LB * h^2)/(8 * nu_LB) = ( Fx_LB * h^2)/(3 * nu_LB)
now the Reynolds number using Dh=2H=2(2h)=4h
Re= Dh_LB*Umean/nu_LB
Re= (4*h^3*Fx_LB)/(3*nu_LB^2) and solving for the body force we get:
Fx_LB_groov= (3*nu_LB^2*Re)/(4*h^3) (b)
the ratio eps=Fx_LB_groov / Fx_LB_flat is 3.375. This value might be constant for lowReynolds number flows in stationary state. If I increase the Re, I expect to have a nonconstant ratio eps. However If I compute the body force required for higher Re following the previous procedure I will get of course the same ratio!!!
May be what I need to do is to impose constant velocity at the inlet instead of fullydeveloped velocity profile and see how large the Re is. Do you think I have to use the equation (A) for the Re in the flat channel to calculate the Re in the grooved channel?
Any advice will be welcome!!
Thanks
Anuhar