Dear all,

My domain reference pressure is 1 atm. The point is the ambient outlet pressure is variable. The value of it is the same as inside domain pressure. I tried to use expression (P-1.0)or (pressure -1.0) at outlet boundary relative pressure,I got an errot.

Any help how can set it, thanks

You can't just make it a direct relation to itself, you have to specify something. How does the outlet pressure vary? Is it the flow rate that is fixed? What do you know about your outlet conditions?

-Robin

Dear Robin,

Actualy, my output is fitted to a selled tank. Therefore, this tank have the same domain pressure value.

when I put 1 atm as a ref pressure and 0.0 atm as relative pressure at the outlet, CFX solver shows some air flow want to enter the domain from that outlet and then a wall was created. I think this is means that the domain pressure is less than that for thge outlet pressure.

Therfore, I am trying to let the outlet pressure have the same pressure as the domain has.

regards

Hi Neser,

It doesn't work that way. How is your inlet specified?

-Robin

My boundary conditions are as follows:

1. one inlet, low speed hot air 2. two outlets, the 1 st one is for outlet hot air (at the top of the domain), and the 2nd one is the particls outlet (which I am asking about, actualy fitted to a selled tank)

moreover, I have 4 particls injection points

regards

What is the other outlet condition?

it is 1. Outlet (not opnning) 2. Relative pressure is 0.0 atm

Hi Neser,

It's not surprising that you would get some reversed flow or wall off. I suspect the velocities are quite low and that you have buoyancy turned on. Due to the buoyancy, there may be a favorable pressure gradient driving the flow towards your upper outlet.

Wall off is not critical, as long as you are not expecting flow to go through. You could change this outlet to an opening instead, just make sure the boundary conditions are physically sensible.

Regards, Robin

I did and I faced anther problems

1. Some particles reverse its direction from downward to upward (with the air outlet)

2. the remaining particle could not leave the domain this is because hight speed air come throught the domain from the bottom outlet (it should be a collection tank)

Therefore, are you advise me to turn buoyancy off or

It sounds as though you may not fully understand what buoyancy does. When you turn buoyancy on, it will add a momentum source due to buoyancy effects. If you specify a reference density, the buoyancy force will be relative to the hydrostatic contribution of a fluid at the reference density. In other words, the solver subracts the hydrosatic pressure of a fluid at your reference density (ie. subtracts rho_ref*g*h).

This is done to reduce round-off errors and make it easier to set boundary conditions, since you do not have to include the hydrodynamic pressure on a pressure boundary where the fluid is at the reference density. However, if you fluid is not at the reference density, you will have to include the local hydrostatic contribution, which is essentially (rho-rho_ref)*g*h.

Before you continue, make sure all your pressures and densities are sensible. Also make sure you specify a reference location (for zero height) so you can get your pressures correct. If you have difficulty wrapping your head around what the reference density does, you can remove it's effect by specifying a reference density of zero. If you do this, just make sure you include CEL expressions accounting for hydrostatic pressure on all pressure boundaries.

Regards, Robin