about the interpretation of the perturbation pressure

[Dimitrios]

i have a question related to the interpretation of the quantity
"perturbation pressure":

is the perturbation pressure that portion of the static
(thermodynamic) pressure that influences the flow and which is
influenced by the flow?

[Kevin]

If I am interpreting your question properly, I would say yes. However,
please refer to specific equations in the FDS Technical Reference
Guide, Vol 1, so that we can be more precise in our discussion.

[Dimitrios]

the equation is 3.11 of the technical reference. It says that the
pressure in the mth zone is a linear combination of the background
pressure (PBAR in the code) and a "flow induced" perturbation
pressure.
If this perturbation pressure is that portion of static pressure which
interacts locally with the flow and not global via the equation of
state, then the output quantity "PRESSURE" (PBAR + perturbation
pressure - atmospheric pressure) is the relative (to the atmospheric
value) static pressure of the mth zone.

[rmcdermo]

Dimitrios,

A better term for the PRESSURE output from DUMP would be "GAUGE
PRESSURE". It is the local thermodynamic pressure, \bar{p} (used to
compute the density), plus the local hydrodynamic pressure, \tilde{p}
(from solution of the Poisson equation), minus the atmospheric
pressure at that height, p0. So, if the domain is open you have GAUGE
PRESSURE = \tilde{p}. If you had a manometer with one end open to the
atmosphere and the other end attached to a probe in a sealed
compartment, this is the pressure reading you would get.

Cheers,
Randy

[Dimitrios]

from your answer i conclude that the perturbation pressure is the
hydrodynamic pressure. But i thought that the perturbation pressure is
derived from (H - hydrodynamic pressure) * density (Step 5 in the
calculation of H, Tech Guide s. 19).
And if the output quantity PRESSURE refers to a gauge pressure which
would be measured with a manometer, then i think that there can no
exist a dependence to the local velocity field.

[rmcdermo]

I do not call 0.5*u_i*u_i the hydrodynamic pressure; this is the
kinetic energy per unit mass. As you stated, correctly, the
perturbation pressure and the hydrodynamic pressure are the same. In
our hypothetical manometer, I am assuming that the probe opening is
pointing into to the flow and therefore measures the stagnation
pressure. If it were turned perpendicular to the flow (like the
static ports of a pitot probe) then I agree we would not see a
dependence on the velocity.

[Dimitrios]

sorry, it was my mistake.
i would say that in step 5 of the tech guide the perturbation pressure
is calculated by substracting from rho*H the hydrodynamic pressure
(0.5*rho*u_i*u_i) and that is why i think, that the perturbation
pressure is not the same as the dynamic pressure.
A second reason for that is, that the perturbation pressure is like H
a scalar quantity and for this i can not understand why there is an
identity between the perturbation and the hydrodynamic pressure.
However, many thanks Kevin and Randy

[rmcdermo]

k = 0.5*rho*u_i*u_i is also a scalar.

[enacht]

Dear Randy I have similar questions. Is 0.5*rho*u_i*u_i the
hydrodynamic or dynamic pressure?
And if it so what is then exactly the pertubation pressure?

[rmcdermo]

:) Okay, this is quite a notational circus. We need not get hung up
on arbitrary definitions: let the math be our guide. I looked through
every one of my fluids books and they all have slightly different
conventions. Personally, I prefer to denote k = 0.5*rho*u_i*u_i as
the "dynamic pressure" (it is also the kinetic energy). It is useful
when we are talking about the Bernoulli equation, but is not often
used when talking about solutions to the Navier-Stokes equations.

In my mind the hydrodynamic pressure and the perturbation pressure are
the same thing. Kevin was calling it perturbation pressure before I
got here. My view comes from thinking about the flow of an
incompressible fluid like water. The pressure can then be simply
decomposed into a hydrostatic part and a hydrodynamic part. I hope
this is unambiguous. We know the hydrostatic part is easy; it comes
from rho*g*z and it exists even if the water is static. Then, in a
flow simulation, the hydrodynamic pressure is the part of the pressure
that enforces the divergence constraint, which in this case is div(u)
=0. In a variable-density flow, like fire, div(u) is not zero, but
the hydrodynamic pressure plays the same role of enforcing this
constraint.

Let me try to say this in a slightly different way. The fact that we
rearrange the momentum equation into Stokes' form so that we can solve
a Poisson equation for H is a historical artifact of FDS development.
It is NOT a requirement for the low-Mach formulation. We could just
as easily have written the momentum equations so that we solve the
Poisson equation for \tilde{p}; we would simply have a different form
for the "baroclinic correction" term. In this case \tilde{p} would
come directly from the Poisson solve needed to enforce the divergence
constraint.

So, I don't know how better to answer the question of "what exactly is
the perturbation pressure?" You must look at the math: it comes from
the solution of the Poisson equation which enforces the divergence
constraint. Believe me, if we could get this pressure without solving
the Poisson equation (or going to a compressible formulation) we
would!