# Pressure drop vs Reynolds number

Hi everyone,

I want to study the variation of the Reynolds number with the pressure drop in a flat channel with some obstacles placed on one wall. The flow is driven by a fully-developed velocity profile imposed at the inlet, calculated as follows:

U(y)= (Fx_LB* y * (h_LB - y) / (2* nu_LB) ------------------------(1)

where U(y) is the x-velocity component, Fx_LB is the body force in LU, h_LB is the height of the channel and nu_LB is the viscosity in LU. Additionally:

Umean= (2.0/3.0) * Umax -----------------------(2)

Umax= (Fx_LB * h_LB**2)/(2 * nu_LB)--------------------------------(3)

with Umax and Umean being the maximum and mean velocity respectively.
From the previous equations one can calculate the Reynolds number as:

Re= Dh_LB*Umean/nu_LB with Dh hydraulic diameter (2h_LB) ----------------(4).

This Re can be considered as the “imposed” Reynolds number as an inlet boundary condition. However, I am interested in how the Re varies with the change in the body force, but I find a linear relation between Re and Fx_LB!!!

I know from the literature that the pressure drop and body force in the lattice system are equivalent:

Fx_LB = deltaP/L_LB = (Pin-Pout)/L_LB -----------------(5)

and the transformation to the physical system is:

Fx_P= Fx_LB* (dx^2/dt^2).------------------------(6)

I think that, since the channel has obstacles, the Reynolds number must vary with the body force (pressure drop) in a way different than linear, as it happens for low Re in flat channels.

Could somebody give me an advice or clarify my doubts because I got already confused!!!
Thanks

Anuhar

It is not entirely clear to me what you are doing. You want to measure the connection between Re and the pressure drop, correct? And you say that you find a linear scaling?
For a stationary flow in a straight channel, a linear dependence is always exactly given, since the convective term vanishes. In you case, this is different. Even if the flow is stationary, the velocity vanishes following streamlines. But the non-linearity should only be visible at large Re. In what range of Re are your simulations?
Timm

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 fully-developed 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 Re-Pressure 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 non-straight 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 fully-developed 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 low-Reynolds- number flows in stationary state. If I increase the Re, I expect to have a non-constant 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 fully-developed 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?

Thanks

Anuhar

At first, it is not important whether you impose a fully developed flow or a constant velocity flow at the inlet, as long as the interesting channel geometry is located sufficiently behind the inlet. It is known how large the development length of the channel must be for a given Re. Have a look at this article, if you are interested:
Poole, Ridley: Development-Length Requirements for Fully Developed Laminar Pipe Flow (2007)
I bet that there will be no difference in both simulation methods. This means that your inlet Reynolds number is well under control. If you define Re using the mean inlet velocity, you won’t have any problems.

I would do the following: Set up a series of simulations with different Re (covering two or three decades). You should adapt the lattice resolution in order to do so.
Then compute the mean density at the inlet and the outlet and compute the physical values of the densities there. The difference must be the density gradient, i.e. the pressure drop. Now you compare the ratio of Re and the pressure drop, and you should see a non-linear behavior.

Timm

Hi Timm,

Why do I have to change the resolution to get different Re when I can keep the same resolution h_LB and then just modify the body force at the inlet to obtain the variation of the Re.

You meant to compute the mean density at both inlet and outlet cross sections? shall we expect something close to 1?
Then the pressure is computed by means of P=cs^2*rho_mean.

Is the pressure drop equal to ----> Pinlet -Poutlet/Length in lattice units, and then in the physical system multiplied by
(dx^2/dt^2)?

Thanks a lot

Anuhar

You will see that some difficulties arise when you increase Re by a factor of, say, 1000. In that case, just increasing the velocity does not help you, since also the Mach number increases and LB collapses eventually. Large Reynolds numbers always require a larger resolution. That is the prize.

Computing the mean values of the pressure at inlet/outlet gives you a good estimate of the total pressure drop, which is exactly what you want to find out. This method is independent of the value of Re, so there can also be a non-linearity. Obviously, your linear equation is not always valid.

The physical pressure drop is dp[sub]p[/sub] = (rho[sub]L[/sub][sup]out[/sup] - rho[sub]L[/sub][sup]in[/sup]) * rho[sub]p[/sub] * c[sub]s[/sub][sup]2[/sup] * dx[sub]p[/sub][sup]2[/sup] / dt[sub]p[/sub][sup]2[/sup]. L means lattice quantity, p physical quantity, and c[sub]s[/sub][sup]2[/sup] = 1 / 3. Do not forget the density rho[sub]p[/sub] (e.g. 1000 kg/m[sup]3[/sup] for water).

Timm