Difference between Background Pressure and Pressure (Perturbation)

[Arulnambi Palani]

Hello Everyone,

In the technical reference guide, I have seen that in the computation method, pressure was divided into Background pressure and Perturbation pressure. I think I have understood the functionalities (correct me if I am wrong).

Background pressure is considered to limit the numerical solver from solving flow with high mach number.

Perturbation pressure describes the flow.

But when I tried to write  both the PRESSURE (perturbation) and BACKGROUND PRESSURE in the simulation, the BACKGROUND PRESSURE was constant and equal to 1.01325 E+05 (atmospheric pressure). Even though it is ambient pressure, adding heat to the system won't affect the background pressure?

I am performing a channel flow simulation with one wall heated to 100°C.

Could anyone explain to me briefly, what is the difference between both.

Thanks & Regards,

Arul.

[Kevin McGrattan]

Right. Look at the equation of state

p_0 = (R/W) rho T

p_0 is background pressure. In reality, the equation of state involves the total pressure, not just the background. If you do a tunnel flow simulation with no fire, the density will not change, the temperature will not change, and the average molecular weight W will not change.

[Arulnambi Palani]

Hello Mr. Kevin,

Thank you for you reply.

But, in my case, since one wall is heated to 100°C (TMP_FRONT = 100), I can see the temperature distribution and also the density distribution. So this arises me with the following questions:

1> How is density computed (Temperature is computed using equation of state, if I am not wrong)? Because according to Tech guide, density is computed from the mass fractions of the species. In channel flow simulation, we won't be having any change in mass fractions.

2> How is the background pressure computed? (since temperature was computed from equation of state, we cannot use the equation of state to calculate background pressure)

3> Does the Poisson equation computes only perturbation pressure?

I am sorry for asking too many questions. Since I am new to this, I am trying to wrap the concepts around my head.

Thank you so much for your support.

[Kevin McGrattan]

I trust you are reading through the Technical Reference Guide. If so, I prefer that you address your questions to specific equations in the guide for which you need clarification.