This is what I mean by a pitot tube:

enter image description here

except, the bent tube should be closed, so there is no velocity in the pitot. But why is there no height? It's reflected in the Bernoulli equation that the small tube and bent one do not have height therefore the static pressure coefficient cancels out. Why is that?

$$\frac{1}{2}\rho v^2 + P_1 = P_2 $$

Notice how the pitot tube losses its static and dynamic pressure coefficients and the other tube losses only its static term. Why does this happen, since there's clearly a height difference between both of them?

  • $\begingroup$ notice how the pitot tube is closed that's why loses his dynamic term $\endgroup$ May 16 at 1:38
  • $\begingroup$ if and only if the heigh is zero the tube losses its static term I don't see that happening in the eq $\endgroup$ May 16 at 1:39
  • $\begingroup$ ??????? I see a clearly difference in height $\endgroup$ May 16 at 2:42

2 Answers 2


It's probably reasonable to neglect height differences, but it's funny that it's not really necessary.

Indeed, if we note $P^*= P + \rho g z$ Bernoulli's theorem is written $$P^*(M)+\frac{1}{2} \rho {\vec{v} (M) }^2 = cst$$ and the law of statics is written $$P^*= cst$$

Note $A$ the lateral pressure tap and $B$ the dynamic pressure tap (where the speed is zero). In general, we consider two points $A'$ and $B'$ at the same height as $A$ and $B$ and very upstream in the flow.

We will have $ \vec{v_{A'}}= \vec{v_{B'}} = \vec{v_{A}} = \vec{v}$ and also $\vec{v_{B}}=0$

If we write twice Bernoulli's theorem: $P^*(A')+\frac{1}{2} \rho {\vec{v_{A'}}}^2 = P^*(A)+\frac{1}{2} \rho {\vec{v_{A}}}^2 $

from where $$P^*(A')=P^*(A)$$

and $P^*(B')+\frac{1}{2} \rho {\vec{v_{B'}}}^2 = P^*(B)+\frac{1}{2} \rho {\vec{v_{B}}}^2 $

from where $$P^*(B')=P^*(B)+ \frac{1}{2} \rho {\vec{v}}^2$$

But we also know that, in a direction perpendicular to a unidirectional flow, the pressure varies as in static : $$P^*(A')=P^*(B')$$

Finally we have :

$$P^*(A)-P^*(B)= \frac{1}{2} \rho {\vec{v}}^2$$

We must not forget that the manometer is connected to points $A$ and $B$ by two tubes within which the fluid is at rest. So if we note $A''$ and $B''$ the two connections on the manometer, we will have the law of statics: $P^*(A'')=P^*(A)$ and $P^*(B'')=P^*(B)$.

We will therefore have : $$P^*(A'')-P^*(B'')= \frac{1}{2} \rho {\vec{v}}^2$$ with this time $A''$ and $B''$ the two connections of the manometer. If these two points are at the same height : $$P^*(A'')-P^*(B'')=P(A'')-P(B'')$$ and so (finally !!!)

$$P(A'')-P(B'')= \frac{1}{2} \rho {\vec{v}}^2$$

We see that the differences in height compensate each other and that what counts is the difference in height between the two entry points of the manometer.

Hope it can help and sorry for my poor english.

  • Pitot-static tube consists of both those tubes.

  • OP's diagram has these two tubes separately installed. (which is also a possible setup)

  • The lower tube's (static tube or static port) opening is parallel to the airflow and measures static pressure.

  • Upper tube's (pitot tube) opening while being perpendicular to the airflow is outside the boundary layer and measures the total air pressure (dynamic pressure + static pressure).

We don't consider height difference for the situations like pitot and static tube combination which is used to measure airspeed because the height difference between component tubes in the pitot and static tube setup is relatively low enough to discard the height term (ρgh - this term called static pressure head or hydrostatic pressure is very low compared to the other three terms dynamic pressure , static pressure and total pressure in the equation, you can see this if you apply approximate numbers for practical scenarios) in Bernoulli equation.

Note: In the Bernoulli equation posted by OP for the Pitot and static tubes (½ρv² + P₁ = P₂)

  • ½ρv² : dynamic pressure

  • P₁ : static pressure

  • P₂ : total pressure


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