Model the flow of liquid through a pipe/tube and calculate the pressure

[redargon]

I want to model the flow of liquid through a pipe/tube and calculate the pressure drop over the pipe. What sort of boundary conditions should I be using?

I tried with a pressure inlet and a pressure outlet, but I'm not sure if that is the right way to go. Pressure at inlet = 1.8bar, pressure at outlet = 1 bar (open to atmosphere). When I check the results, I have a pressure of about 1.4bar in the middle of the pipe length. This seems intuitive as we are going from 1.8bar to 1 bar, but using pressure loss theory, shouldn't the pressure loss be more non linear than that due to friction factors? Like how Darcy-Weisbach equation describes.

The pressure of the fluid at the exit won't be exactly atmospheric, because the liquid has a velocity as it exits and therefore a pressure is required to generate that velocity. What am I missing?

 

[caslav.ilic]

[...] When I check the results, I have a pressure of about 1.4bar in the middle of the pipe length. This seems intuitive as we are going from 1.8bar to 1 bar, but using pressure loss theory, shouldn't the pressure loss be more non linear [...]

That does depend on (at least) what is the fluid and the pipe diameter, i.e. Reynolds number. For a nice big Re (all turbulent flow, thin boundary layer) and incompressible fluid, the pressure drop will be linear. That's because the velocity of the fluid will be constant along the pipe (constant flow rate), the loss coefficient too (no laminar regions), so the pressure drop comes out as [itex]\Delta p = q A \xi \Delta l[/itex] ([itex]q[/itex] dynamic pressure, [itex]A[/itex] pipe cross section area, [itex]\xi[/itex] loss coefficient, [itex]\Delta l[/itex] pipe segment length), i.e. [itex]\Delta p \propto \Delta l[/itex].

The pressure of the fluid at the exit won't be exactly atmospheric, because the liquid has a velocity as it exits and therefore a pressure is required to generate that velocity. [...]

That's... a glitch in thinking I myself have made more times than I will admit :) The (static) pressure will be exactly atmospheric, as there is nothing any more to drive or act on the flow as it exits. The total pressure will be higher, of course, and dissipate to zero in the external environment. You can also look at it numerically -- if you have set the outlet boundary condition to 1 bar, then it will be 1 bar.

I tried with a pressure inlet and a pressure outlet, but I'm not sure if that is the right way to go. [...]

If it works, why not; but basically it depends on what you know, and what you want to compute. A more usual thing to know is the flow rate and the exit conditions (atmosphere), and to need to compute the pressure drop. Then you would specify flow rate at inlet and pressure at outlet, and get pressure at inlet as the result of the computation (i.e. you would have computed the pressure drop, rather than specifying it).

Another thing is that pressure-pressure condition can be numerically difficult to compute, since the solver has no idea of the flow features beforehand (what is the velocity, what is the Re...) So even if these two pressures are what you know (and you want to compute the flow rate), then it's better to first make an educated guess on the flow rate, compute with flow rate-pressure, and take that result as initial solution for the next simulation with pressure-pressure.

--
Chusslove Illich (Часлав Илић)

 

[redargon]

Thanks for the help caslav,

I'm getting the hang of dynamic pressure vs static pressure and I understand now about the exit properties of the flow.

For added info, I am working with water in a 2mm to 6mm diameter tube at flowrates of about 2l/min, so from my Reynolds calcs, Re > 4000 so I'm in the turbulent region.

I will try with the flowrate inlet and pressure outlet, I thought about this a little later and the simulations had already started, but will try it on the next runs.