2
$\begingroup$

In our aerodynamics class we recently discussed the concept of static and dynamic pressure and discussed their application to aircraft instruments. However, I do not understand properly how the altimeter can work.

First of all a small recap of Bernoulli's law. The total pressure is given by $p_t = \frac{1}{2} \rho V^2 + p_s$ and is constant along a streamline. Consequently, the static pressure is given by $p_s = p_t - \frac{1}{2} \rho V^2$.
Now imagine a low speed windtunnel. In a reservoir the speed is so low that it can be assumed to be $V=0$ m/s. Hence here the static pressure equals the total pressure, $p_t = p_s$. Now the fluid is accelerated along a streamline, and hence the static pressure should drop according to the relation given above.

Now this is where my problem is. I read that the altimeter just measures the static pressure at flight level to obtain the pressure altitude. According to the explanation given above, however, this can not be, as the static pressure in the flow will be lower than the static pressure at the same altitude at zero velocity.

$\endgroup$

3 Answers 3

5
$\begingroup$

The static air pressure seen by the aircraft does not change with the aircraft's velocity.

Your confusion is from a common misinterpretation of Bernoulli's principle. It is not true that a fluid's pressure will decrease simply by virtue of flowing faster. After all, this violates the idea that physics should be the same in all inertial frames.

Here is a simple counterexample to the typical interpretation of the Bernoulli principle. Consider a tube of infinite length and uniform diameter with some gas sitting in it. Now consider various coordinate systems with a velocity in the direction of the tube. In these different coordinate systems, the velocity of the gas will be different, but we expect the force on the walls of the tube due to the fluid's pressure to be the same in all cases. (The tube is not going to rupture simply because of a choice of coordinate system!)

Instead, Bernoulli's principle says that, in a given flow (say, along a streamline), a local increase in velocity is associated with local decrease in pressure. The canonical example is fluid flow through a tube with a constriction (a venturi).

Quoting from the Wikipedia article for Bernoulli's principle:

Bernoulli's principle can be derived from the principle of conservation of energy. This states that, in a steady flow, the sum of all forms of mechanical energy in a fluid along a streamline is the same at all points on that streamline. This requires that the sum of kinetic energy and potential energy remain constant. ... If a fluid is flowing horizontally and along a section of a streamline, where the speed increases it can only be because the fluid on that section has moved from a region of higher pressure to a region of lower pressure; and if its speed decreases, it can only be because it has moved from a region of lower pressure to a region of higher pressure. Consequently, within a fluid flowing horizontally, the highest speed occurs where the pressure is lowest, and the lowest speed occurs where the pressure is highest.

Note the emphasis on relative changes occurring on a streamline.

The specific flaw in your argument is here:

Now the fluid is accelerated along a streamline, and hence the static pressure should drop according to the relation given above.

In the wind tunnel, something has to do work to accelerate the air to the wind tunnel velocity, adding energy to the flow, which violates the conservation of energy assumption in Bernoulli's principle.

$\endgroup$
0
$\begingroup$

Aircraft altimeter use the pitot-static system

http://en.wikipedia.org/wiki/Pitot-static_system

In particular, the pitot tube measures the pressure of air ramming into the tube which, under normal circumstances, is equal to the total pressure.

The static pressure is obtained through a static port. The difference is that it is "capped" so the air gets in from the sides. The air isn't rammed into these side holes (by any velocity) so these holes still get balanced with the static pressure outside.

See the Wikipedia text above.

$\endgroup$
2
  • $\begingroup$ Sorry, but your answer does not help me at all. I know that the static pressure is obtained through the static ports. However, this static pressure is the static pressure IN THE FLOW, meaning it should be lower than the ambient static pressure according to Bernoulli's law. But if that is the case, how can you deduce the altitude from that? That is my problem. $\endgroup$
    – Ingo
    Commented Feb 18, 2011 at 14:09
  • $\begingroup$ I'd add, just to avoid confusion, that the altimeter used the static pressure, while the airspeed indicator uses the difference between the static and Pitot pressures. $\endgroup$
    – Colin K
    Commented Feb 18, 2011 at 20:26
0
$\begingroup$

The location of the static ports on the fuselage are fixed in positions where that impact is minimal (if any). As an interesting side note, switching to the alternate static (located inside the cockpit of an unpressurized airplane, and used if the outside static ports become obstructed) results in a small increase in indicated altitude as the pressure inside the cockpit is lower than outside.

$\endgroup$
3
  • $\begingroup$ "the pressure inside the cockpit is lower than outside" - Are you sure? Why? $\endgroup$
    – nibot
    Commented Feb 18, 2011 at 18:09
  • $\begingroup$ I'm not sure of the reason why, but I am sure that he's right. I fly light aircraft, and I've definitely experienced this effect. $\endgroup$
    – Colin K
    Commented Feb 18, 2011 at 20:23
  • 1
    $\begingroup$ Venturi effect. Outside air accelerates as it goes around the (curved) fuselage, creating a lower pressure zone that includes the interior of the (unpressurized) airplane. Seems a little counter-intuitive as you'd initially expect the fuselage to be higher pressured. However, since all the surfaces are curved around it, the effect is that you end up with low pressure all the way around, and the inside air seeks to balance that, resulting in lower interior pressure. $\endgroup$ Commented Feb 21, 2011 at 14:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.