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Why a static port?



 
 
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  #1  
Old August 17th 03, 03:57 AM
Paul Mennen
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Default Why a static port?

I can understand why pressurized aircraft need static ports,
but why do unpressurized airplanes have them? I wouldn't think
the pressure difference between the inside and the outside of
the airplane wouldn't differ by that much. (When I open the alternate
static port in my C185, the altimeter changes by just a couple
of feet.)

Without the static ports, we could get rid of all that silly
tubing and just leave the static port connection to the instruments
open to the air. (Of course there is no getting around needing
a tube for the pitot port.)

~Paul Mennen


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  #2  
Old August 17th 03, 04:57 AM
Ron Wanttaja
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On Sun, 17 Aug 2003 02:57:45 GMT, "Paul Mennen" wrote:

I can understand why pressurized aircraft need static ports,
but why do unpressurized airplanes have them? I wouldn't think
the pressure difference between the inside and the outside of
the airplane wouldn't differ by that much. (When I open the alternate
static port in my C185, the altimeter changes by just a couple
of feet.)


I was once goofing around in my airplane, and disconnected the static line
so the static source was the cockpit. When I did the runup, the altimeter
acted just like a tachometer...indication when up when I pushed the
throttle forward, and then back when I pulled it back.

Needless to say, I taxied back.

Ron Wanttaja
  #3  
Old August 17th 03, 04:58 AM
Ben Jackson
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In article ,
Paul Mennen wrote:
I can understand why pressurized aircraft need static ports,
but why do unpressurized airplanes have them?


The venturi effect makes the pressure in the cabin lower than
the outside pressure. The same thing must affect the static ports
but instruments can be calibrated for the effect.

Look up the procedures for activating alternate static in an
airplane equipped with it. That is opening the static system to
the cabin pressure which causes airspeed and altitude to misread
(the POH will have an estimate of how much).

--
Ben Jackson

http://www.ben.com/
  #4  
Old August 17th 03, 07:28 AM
COUGARNFW
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Paul..

Some airplanes successfully use cabin air for the static source. I know many
want "significant" answers, but whether a plane can or cannot use interior air
depends on how leaky the are.

For new, tight planes, the introduction of all that volume of heat or outside
ram air for people cooling can drastically change the pressure inside.

Some can, Some do.

Neal
  #5  
Old August 17th 03, 04:03 PM
john smith
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As I recall from my Physics classes:

PV=nRT
  #6  
Old August 17th 03, 05:39 PM
karl gruber
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******The venturi effect makes the pressure in the cabin lower than
the outside pressure.******

Not always, and particularly not in Paul's Cessna 185. These Cessnas have
two scoops on the aft fuselage that provide ram air to the cabin. The slight
pressurization reduces the risk of exhaust entering the cabin.

Karl


  #7  
Old August 18th 03, 03:46 AM
Paul Mennen
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******The venturi effect makes the pressure in the cabin lower than
the outside pressure.******

Not always, and particularly not in Paul's Cessna 185. These Cessnas have
two scoops on the aft fuselage that provide ram air to the cabin. The

slight
pressurization reduces the risk of exhaust entering the cabin.

Karl


I didn't know that. (Nice to know.)

I think Bonanza's must be the same, since I flew in a Bonanza once
that had a small opening in the side window which was closed by
a hinged flap held in place only by the higher cabin pressure.
If the cabin pressure were lower than the outside pressure as Ben
suggested there would have been no way to close that flap.
(I don't know if all Bonanza's have that opening - this was the only one
I've flown in.). Ben might be correct for most aircraft however.

From: "john smith"
Subject: Why a static port?
As I recall from my Physics classes:
PV=nRT


I don't think that equation has much relevance here.

~Paul


  #8  
Old August 18th 03, 05:27 PM
john smith
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Paul Mennen wrote:
PV=nRT


I don't think that equation has much relevance here.


It most certainly does!
since nR=(P1V1)/T1=(P2V2)/T2
Where P1, V1, T1 can be the exterior conditions, and
where P2, V2, T2 are the interior conditions
it shows the relationships between the variables
changing any one variable, alters the other five
  #9  
Old August 18th 03, 09:43 PM
Wayne
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So what the hell is P, V, and T? And nR for that matter.

Seems like total crap, put it in english would ya
Wayne

It most certainly does!
since nR=(P1V1)/T1=(P2V2)/T2
Where P1, V1, T1 can be the exterior conditions, and
where P2, V2, T2 are the interior conditions
it shows the relationships between the variables
changing any one variable, alters the other five



  #10  
Old August 18th 03, 10:19 PM
Roy Smith
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"Wayne" wrote:
So what the hell is P, V, and T? And nR for that matter.

Seems like total crap, put it in english would ya


PV = nRT is the "Ideal gas law". The letters stand for:

P - Pressure
V - Volume
n - number of moles (fancy way of saying "number of molecules")
R - some constant (some long-forgotten neuron in the back of
my brain is saying 0.82)
T - Temperature (in degrees Kelvin, which IIRC, is Celcius + 273)

It's called the "Ideal gas law" because it describes how an "ideal" gas
reacts to changes in pressure, temperature, and volume. What's an ideal
gas? Well, basicly it's a gas which acts the way the idea gas law says
it should act :-) I know that's a bit dumb, but go with it for the
moment.

Ideal gas theory thinks of a gas molecule as a little point of matter
vibrating in space. The higher the temperature, the faster it vibrates.
If you've got a bunch of these molecules confined in some space (say, a
jar), as they vibrate, they hit the walls of the container (and each
other) and bounce off.

Each time a molecule bounces off the wall, it pushes on a wall with a
little force. Take enough molecules, and all those little forces add
up. This is what's called pressure. If the temperature goes up, each
molecule is vibrating faster, so it gives the wall a bigger push each
time it bounces off. You see that as increased pressure.

The nifty thing, is that none of this depends on what kind of gas you've
got. If I've got a container filled with nitrogen at a given
temperature, pressure, and volume, I can replace nitrogen molecules with
carbon dioxide molecules at the same temperature, and the pressure won't
change. The density of the gas will increase, because each individual
CO2 molecule weighs more the N2 molecule it replaced, but the pressure
will stay the same.

The ideal gas law is the underlying principle which goes into computing
density and pressure altitude, turbocharger efficiency, and pretty much
anything that has to do with the physical behavior of gasses in any way.
It is one of the fundamental bedrock foundations on which aerodynamics
is based.
 




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