Airflow calculation with restrictor plate.

[Greg Hermann]

At 11:11 PM 10/30/01, Stephen Webb wrote:
>> The flow through the restrictor plate cannot exceed Mach 1. That's IT.
>
>Hmm.
>
>If P1=P2, there won't be any flow.
>
>And if P1 > P2, then I would expect some flow, but how much?
>
>-Steve

OK--depending on the design of the restrictor plate, you will get some
static pressure recovery on the downsteam side of the plate as the flow
velocity drops. The "smoother" the downstream side of the restrictor is,
the more recovery you will get. Of course, nothing is perfect, so you will
never get FULL static pressure recovery (as the fluid slows back down) on
the downstream side of any orifice or venturi.

The mass flow rate through the orifice depends on the upstream pressure,
the area of the orifice, and the downstream pressure. If you are using a
thin, flat plate orifice, it's fairly safe to assume _VERY _ little static
pressure recovery downstream of the orifice. Thus, you have gotten rid of
one of the variables, and can solve for mass flow fairly accurately without
knowing the "orifice coefficient". Basically, I am saying that you will get
FAIRLY close results, when working with a flat plate orifice, if you assume
that the downstream pressure (what you called P2) is the same as the static
pressure at the throat of the orifice. With a more efficient orifice, or
venturi, you will get some degree of static pressure recovery on the
downstream side--meaning that the static pressure at the throat of the
orifice is lower (and therefore that the flow velocity is higher)  at the
throat of the orifice. A more "efficient" orifice will therefore flow MORE,
at the same pressure drop, which makes sense.

As far as the flow velocity through the plate-- the drop in static pressure
from the upstream side of the plate to the static pressure at the throat is
equal to density times velocity squared divided by 2 times "g" ("g" = 32.2
feet per second squared). Watch your units. If you do velocity in feet per
second and pressure in pounds per square FOOT, everything will work with
density in pounds per cubic FOOT.

The density of air at sea level is about .074 pounds per cubic foot at 69°
F. (decreases with increasing temp, and decreases with increasing
humidity).

Mass flow rate through the orifice (what really determines how much power
you can make) will be equal to density times orifice area times velocity.
Again, WATCH it with the units !!

Remember--air is COMPRESSIBLE.  AND--that fact IS significant for this kind
of an orifice. You can only neglect the compressibility up to an absolute
pressure ratio of 1.05 or maybe 1.1 and still hope to get reasonably
accurate numbers.

When the static pressure drops at the throat of the orifice, so does the
density. AND--the TEMPERATURE drops, as well. You will need to know how far
the temp drops to figure out the density at the throat of the orifice. And,
of course, you will need to use the density AT the orifice to figure out
the mass flow rate.

OK-- you know how much the static pressure DROPPED from the outside
atmosphere to the throat of the orifice. The absolute pressure in the
atmosphere (standard conditions) was 14.69 psia. Subtract the pressure drop
from the initial pressure, and you now know the static pressure at the
throat of the orifice. Divide one by the other, and you now know the
absolute pressure ratio. The absolute pressure ratio raised to the 0.283
power IS the absolute temperature ratio between the temp of the atmosphere
(69° F, or 529° R. (absolute)) and the  (lower) absolute temp of the air at
the throat of the orifice.

What I said earlier about Mach 1 is simply that it is the velocity limit of
the air flowing through the throat of the orifice. Depending on conditions,
when the abaolute pressure ratio between the atmosphere and the static
pressure at the throat of the orifice reaches about 1.85 to 1.9 , the air
will be flowing through the throat of the orifice at Mach 1, and no matter
WHAT you do, you won't get any more mass flow of air through that orifice
unless you manage to find a way to increase the density of the atmosphere
upstream of the orifice !!

Mach 1 for a gas varies with the molecular weight and with the square root
of the absolute temp. Density does NOT matter, NOR does absolute pressure.

That business of the temperature drop in the air as its velocity is _VERY_
real. Basically, the heat energy in the air is converted to kinetic energy
as the air is accelerated to a higher velocity. The drop in temp as
velocity increases (along with the latent heat absorbed by evaporating
fuel) is what causes carburetor icing (literally freezing humidity, or
moisture, out of the air stream as it goes through the venturi in the carb.
Icing is self aggravating--get a little bit of ice, the venturi gets
smaller, the velocity goes up, the air gets colder at the higher velocity,
the ice accumulates faster, etc.

Hope all that helps---

Greg


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