# force units in momentum exchange approach in force evaluation

hi evrybody
i have a question considering momentum exchange method.
the thing is that when i check the units of momentum exchange method it seems like it is calculating force density not force?
(1/dt)e(f[xb] +f[xf])
e and f dimentions:
e----> [length/time]
f----->[mass/length^3]
[1/time][length/time][mass/length^3]=mass*[length/time^2]*[1/length^3]=force/length^3
so if u cross e and f the final dimentions are force/length^3?
am i right or what?

if it is not clear what i am saying at least tell me??

Regarding your second question (post), there could be several reasons why nobody had replied, such as the fact that a five hour period is quite small, especially considering a) people have their own work and research to do and b) half the world would have been asleep. Remember that this is an open forum where people give their own personal time for no other reason than to try and help others in the community. I have no affiliation with Palabos but think this is a very nice idea. I would also guess that a failure to realise this is unlikely to encourage people to help.

Regarding your first question, f is a nondimensional number density so the dimensions of your equation are L/T^2 (acceleration), which I think is what you want. If you prefer to look at this another way, you might have force=density*acceleration, which has the same dimensions as the expression you wrote above. But the density used in LB is a scaled density, rhoLB=rho/rho0, where rho0 is some reference density.

thanks very much for ur answer , but it seems like u r not familiar with “momentum exchange approach” and what u just said did not make sense, it’s dim must be force or force/L^3 not acceleration.
tthank u any way

What Tim is saying is that f has no units. However you have assumed f has units of density, which is incorrect.

So as f is dimensionless you get:

e----> [length/time]
f-----> []

[1/time]*[length/time] = length/time^2

which is acceleration, you can use f=ma to find the force.

So what u r saying is that “momentum exchange approch” which is one of the force evaluation methods in LBM is calculating acceleration or what?
it doesnt solve my problem
I have a rotation shaft and i am calculating the applied force on shaft then i want to convert it to physical unit . what should i do?

We’re saying it looks like what you are doing is correct; and your confusion seems to be more about what has what units and what the units should be. If you prefer, think of the force found by the momentum exchange method (ie the equation you wrote done in your initial post) as the force per lattice cell (like a force density, as said). Now you have to sum the force contributions over all boundary nodes (like integrating around a body) to get the “total” force. I hope this helps.

first of all thanks a lot you’v been a great help for me up to now
so you are saying if i sum the force contributions over all boundary nodes then the result would be force??
I think if i sum them it would be just sum of the "forces per cell"s and the result is still in force per cell dimension
dont u think that i must multiply force per cells by dxdzdy of the cells to get force??

The summation over the boundary points is the discrete analogue of the integral, so this gives additional area cross section.

Perhaps it will be helpful if you go back to the details of the momentum exchange method to make sure you understand where is comes from, why it works, and where the e, dt etc come from. In short, the momentum transfer between the fluid and the wall gives an approximation to the pressure tensor at the wall (P*n, where n is normal to the wall and P is the pressure tensor, which is the thermodynamic pressure plus the stress).The summation over the nodes is akin to the integral around the body. Remember that the momentum exchange bit is generally done after collisions. The bit in side your summation can thus be written as

c_i*[f_j(x,t+1)+f’_i(x,t)]

where x is an appropriate boundary node (ie, where x is in the fluid or on the boundary and x+c_i is in the solid), j is the opposite direction to i and f’ means post collision values of f. And expansion and summation over links gives an approximation to the stress (pressure). Then the summation over x gives you force (because sum is like integral and, like you say, this gives a dx*dy - it has dimensions of area, not volume, because it is a surface integral/summation).

Personally, I think it’s easier to work either in lattice units (where things are scaled so that dx=dt=1) and/or a fully non-dimensional system, and then convert to/from other scalings. As Bruce said above, you can then easily multiply by a mass if you need to. If this isn’t for you, then if you go through the mom. ex. meth. you’ll see that the prefactor is not e/dt but edx/dt=dx^2/dt^2. Or L^2/T^2… Multiplying this with the mass density gives dimensions M/(LT^2). Then sum over the boundary (ie surface integral) to get ML/T^2. Or by generic reasoning, momentum=M/(L^2T); momentum transfer over a unit distance per unit time=M/(LT^2); over the area of the body =MA/(LT^2)=ML/T^2. But the best way to understand the algorithm is, in my opinion, to expand and analyse, and work in manageable and consistent scalings.

Mr Timm Reis your explanations were just Awesme and very helpful .
thanks very very much

Dear pleb01,

I am trying to couple a LBM-DEM programme together with my supervisor. For now we have just hit the wall and we do not know exactly how to find/use correctly the scaling factor from force in LB units to physical ones. We are following an equation found in the literature:

Force (physical) = Force(LB) ((lenght-LB * density-phy * viscosity^2-phy) / (lenght-phy * density-LB * viscosity^2-LB))

We consider physical properties of water, viscosity-LB = 1/6 and density-LB = 1, lattice unit = 1 representing 0.001m.

We obtain a value of order 10e-5 which seems to be very small and we do not know if our assumptions are correct. I have not found any other source describing a force scaling factor so I decided to come to the forums and ask for some help or guidance which will be very valuable.

Kind regards.