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CBLong
2nd May 2003, 19:21
This is something that has always puzzled me and probably derives from an incomplete (or just incorrect!) understanding of Bernoulli's Theorem...

Why doesn't the pressure in an aircraft's static system vary significantly with speed? I'm thinking particularly of something like a C152 whose static port is flush with the side of the fuselage just in front of the left-hand door. According to my limited understanding, Bernoulli's Theorem says that the pressure exerted by a moving fluid will be less than that exerted by the same fluid at rest, because some of the fluid's energy has been transformed to kinetic energy.

Therefore, the pressure sensed by the static system should fall as the aircraft's speed increases. However, if that were really the case, I'd see my altitude dropping as I accelerated from rest on take-off! What's going on?

Any pointers appreciated!

Cheers,

cbl.

FlightDetent
2nd May 2003, 20:22
However, if that were really the case, I'd see my altitude dropping as I accelerated from rest on take-off!

No, if that were really the case, you'd see your altitude rise. Although it is a far cry from being an answer to your question.

My names Turkish
2nd May 2003, 20:36
Probably going to be crucified by those who know eminantly more than me but..

As far as I know the static port is carefully located in a point of stagnation of the Airflow. This means that the local airflow has reduced to zero due to the shape of the fuselage. Since the speed is always zero, the ambient static pressure is zero. let me P.M Genghis and see what he can add.

CBLong
2nd May 2003, 20:49
Good point, FD - I always get that 'pressure decreasing / altitude increasing' thing mixed up! It's amazing I'm allowed anywhere near an aeroplane in the first place...

MnT - I wondered if that might be the answer, but I'm pretty sure that the static port on a C152 isn't at a stagnation point... it's in the prop-wash, for a start. Looking forward to hearing from Genghis....

:)

cbl.

Genghis the Engineer
2nd May 2003, 20:52
Oh I hate it when people keep taking me seriously.

In theory, static pressure is the pressure you'd sense if sat stationary in a balloon at the same point and altitude. So, it shouldn't change with speed - it's pitot (dynamic) pressure that changes with speed.

That's the theory, and no doubt most textbooks say that. In practice unbelievable amounts of effort goes into finding the one spot on the airframe where you can put a static port to get that - I've seen on the side of the fuselage, on the side of the fin, a spot on a strut, under the belly, behind a "static dam" (fancy name for a small block of wood glued to the aeroplane).

In practice you'll never quite achieve constant static pressure with airspeed - this is the reason that most aeroplanes have in the approved POH a CAS (or EAS) .v. IAS graph. Good ones (and with any luck anything certified to part 23 or 25) should have no more than 5kn error through most of the full range except down at low speeds where putting the pitot to high angles of attack tends to mask any static problems. However, I've certainly seen 15-20% airspeed error which with a low speed aeroplane means 100-300ft altitude error - these errors however tend to be seen on aeroplanes not properly sorted, which means homebuilts, microlights, historic aircraft and prototypes that we're still working on.

From an "ordinary" pilots viewpoint then static pressure is constant FOR A GIVEN PRESSURE ALTITUDE. From an engineers viewpoint (or a test pilots) we will spend many happy days in the flight test department trying to make that true.

Dufwer
2nd May 2003, 21:48
Going on memory (again) so bound to be wrong :( I seem to remember something in one of my PPL books that if the static port gets blocked then you can get around the problem by smashing the glass on one of the instutments to provide static pressure to the systems. Somewhere in the text it was mentioned that the static pressure being provided from within the cabin would usually be slightly higher than from the static port but that the difference is usually so small that it is negligible. By adding one and one together and getting three, could this be implying that CBLong's assumption are in fact correct but the difference isn't large enough to allow a typical C152 pilot to notice :confused:

D

Genghis the Engineer
2nd May 2003, 22:19
That's a standard feature on single engine Cessnas - you smash the glass on the VSI since the instrument casing is connected to the static and it's an instrument that in extremis you can live without. You'd never do it in VMC, but if the static went in IMC and you need halfway reliable altitude readings for navigation / terrain avoidance, etc. it's a solution to the problem.

Piper, slightly more helpfully I feel, provide you with a small lever under the instrument panel that vents the static vent to the cabin.

It might put your readings 2-3 hundred feet / 10-20 knots out perhaps but that sort of error shouldn't kill you, whilst no meaningful altitude reading might.

G

Crossunder
2nd May 2003, 22:53
CBLong:

Remember that Bernoulli's theorem applies to closed systems (e.g. a venturi tube). In the case of the fuselage side (or wing surface for that matter) there is only one wall; the other part of the "venturi" is missing, hence the theorem cannot be applied to this particular situation.

NineEighteen
2nd May 2003, 23:53
it's pitot (dynamic) pressure that changes with speed.Just for clarity, pitot pressure is "total" pressure (static + dynamic) :p

Ove R. Tallig
3rd May 2003, 00:30
You are missing one important point.. The principle stating the total pressure is constant applies to a given flow. Increasing the aircrafts speed gives you a new situation, a new flow. The total pressure is increased. For the Cessna taking off on the runway the static pressure is constant and the dynamic increases with the square of the speed i.e. total pressure increases.

And you may apply the principle even if you do not have any walls around the flow as in a venturi. The principle works perfect when calculating flows around a surface in an open flow of fluid.

You may derive the Bernoullis principle from the Work Energy theorem in physics. As Bernoulli most likely did.

best regards

Daniel Bernoulli is by the way son of the well known mathematician Johannes Bernoulli

delarocha
3rd May 2003, 00:37
Please notice that Bernoulli equation relates pressure changes to velocity along a streamline (which in a steadyflow is the trajectory traced by a fluid particle) , so it's not restricted to closed systems.

Bernoulli equation in the form

p + (1/2) (density) V^2=constant

was derived with the following restrictions:
-Steady flow
-Imcompressible flow
-Frictionless flow
-Flow along a streamline

Genghis the Engineer
3rd May 2003, 00:42
On an esoteric technical point, Bernoulli's theorem doesn't require a closed system, it assumes it. It's quite common in various analysis to treat what goes on between streamlines as a closed system.

There is also an advanced compressible form of Bernoulli's equation which postdates it's namesake by several generations.

G

Keith.Williams.
3rd May 2003, 19:42
To understand this phenomenon we need to look (rather more) carefully at what Bernouli actually means and how we apply it.

Bernouli said that the total pressure (dynamic + static) in an airstream is constant. In doing so he made a number of simplfying assumptions including the conditions that no energy would be added to nor subtracted from the airstream. The basic logic of his argument is that if no energy enters or leaves teh airstream, then the energy within it remains constant.

The energy posessed by the air is evident in various forms. The most signifcant of these are its temperature (heat energy), static pressure (mechanical energy) and dynamic pressure (kinetic energy). Kinetic energy and hence dynamic pressure are proportional to TAS squared. So if the TAS increases, the dynamic pressure increases. But for the total energy to remain constant, this increase in dynamic pressure must be acompanied (funded by) a corresponding decrease in both temperature and static pressure. So if the TAS of the airstream increases, the staic pressure decreases. This might lead us to (incorrectly) conclude that static pressure will decrease as an aircarft accelerates.

To understand why this is not the case, we must remember that all of the above argument is based upon changes in the absolute (or real) TAS of the airstream. In the case of an aircraft, it is the aircraft rather than the air that is moving. Although it is convenient to use the aircraft as the reference point and visualise the air as flowing past it, this can be misleading.

Consider an aircraft accelerating down the runway. Its TAS relative to the entire atmosphere is increasing. If we treat this as an increase in the TAS of the entire atmosphere, then the static pressure in the entire atmosphere must reduce. Worse still, with the very large number of objects moving through the air (including birds, bullets, cars, animals and suicidal stock brokers) the static pressure would be extremely low indeed. This argument is clearly not true.

It is more accurate to say that as the aircraft accelerates, the kinetic energy and dynamic pressure relative to the aircraft increase, but the static pressure of the general atmosphere is unaffected. This increased dynamic pressure and constant static pressure mean that the total pressure affecting the aircraft increases.

But when the air passes close to the aircraft it (the air) must accelerate in various directions to make room for the aircraft to pass through it. These are reall accelerations, representing changes in the aboslute velocity of the air. Because they are reall accelerations, they cause real changes in dynamic and static pressure. The camber of the wings, for example causes a real increase in the TAS of the air passing over them. This causes an increase in dynamic pressure and a decrease in static pressure. The overall effect being lift (and unfortuantely drag).

The real velocity or the air at all points on the surface of an aircraft will of course be affected to some degree. The trick in providing accurate measurement of static pressure for barometric instruments, is to find an area where the reall accelerations and hence changes in static pressure are the smallest.

mono
3rd May 2003, 19:56
To bluntly answer the original question. It is!

The pressure as sensed by flush static ports is dependant on three variables:- altitude, airspeed and angle of attack (temperature also plays a part but has been discussed in many previous posts).

Cb and Flt Detent quite correctly state that for an increase in airspeed the sensed pressure will fall and hence the a/c will appear to climb.

As Gengis has stated the a/c manufacturers go to great lengths to try to ensure that the ports are located such that there is minimal venturi/bernoulli effect. For the lower airspeeds of, say, a 152 this is sufficient.

For the higher airspeed/Mach numbers of civil transport or military aircraft however, more accuracy is required. this is acheived during the original certification period when the a/c is basically a flying test bed. Pre-calibrated probes are fitted to the a/c such that they are mounted as far away as possible from the fuselage and any associated interference. These readings are then compared with those of the a/c ports at various airspeeds and angles of attack and used to create an algorythm within the air data computers so that the pilots will see the true altitude for any given airspeed or angle of attack. This correction is called, as ASFKAP has already said, static source error correction or SSEC.

I can say from experience that the difference at altitude can be huge. As an experiment (to trouble shoot a possible SSEC problem) with the a/c at 40,000 ft (using test equipment not in reality) we varied both the airspeed and angle of attack to confirm identical readings on both #1 and #2 systems. The difference for certain combinations of AoA and airspeed was as much as 900 ft!

This is why for RVSM ops the ADC's must be working and why the test/certification period can be so long.

Hope this helps.

CBLong
6th May 2003, 22:47
Thanks for all the replies, guys - very interesting. I'm glad to know my initial "guess" that there should be a speed-related effect wasn't too far off - but I obviously overestimated its magnitude! :)

Cheers,

cbl.