# Fluid Mechanics Applications/B20: pitot static tube in compressible flow
## Pitot tube

It is a device invented by Henri Pitot in the eighteen century which is used to measure the fluid flow velocity. It consists of a tube having right angled bend which is placed in moving body of fluid with the opening of the bent part directed upstream. This ensures that the pressure in the tube is the sum of the static pressure and the dynamic pressure imposed by the flow. Together with a manometer and measurement of the static pressure it is used to determine the velocity of the fluid flow.

## Working principle

It works on the principal of Bernoulli theorem as:-
$V^{2}/2+gz+P/q=constant$

## Explanation of pitot working

The manometer connected with pitot tube give pressure difference, so formula for velocity becomes as:-
$V=\left(2\Delta h{\frac {\rho _{L}}{\rho }}\right)^{1/2}$
Where,

## Mach number

It is the dominant parameter in compressible flow analysis with different effect on its magnitude. M=V/c V=velocity of fluid C=speed of sound ( for air=342 m/s)
Types of flows classified by mach number:
* M<0.3 incompressible flow, where density effect are negligible.
* 0.3<M<0.8 subsonic flow, where density effect are important but no shock wave appear.
* 0.8<M<1.2 transonic flow, where shock wave first appear then flow divide into subsonic and supersonic flow.
* 1.2<M<3.0 supersonic flow, where shock wave are present but there are no subsonic region
* 3.0<M hypersonic flow where shock wave and flow changes are especially strong.

## Bernoulli theorem to compressible flow

In this case fluid density is changing so pitot tube analysis for incompressible flow is not able to give accurate reading. To remove error we should have to do further analysis. When elevation is neglected Bernoulli equation becomes as:
$\int dp/\rho +V^{2}/2=$ constant.........(1)
but for compressible isentropic flow changes
c$_p$= specific heat at constant pressure
c$_v$ = specific heat at constant volume
k= constant for fluid
so simplifying equation (1) become as
{ϒ/( ϒ-1) }p/ρ + V$^2$/2=constant=( ϒ/ ϒ-1)p$_0$ /ρ$_0$ .......(2) where,
p$_0$=stagnation pressure
p=local pressure
On considering other parameter gives energy equation for flow as:- T$_0$ /T={1+ (ϒ-1)/2}* M$^2$ ...................(3)
P$_0$ /p= {1+(ϒ-1)/2 M$^2$ } ϒ/ ϒ-1.................(4)
By solving equation (3) and (4) we get an compressibility factor(CCF) which gives us a correct velocity or used as calibration in pitot tube to get correct velocity as :-
CCF=1/(1 + M$^2$/4 + { (2-ϒ)/24}M$^4$ +...................)
P$_0$-P=1/2 ρV$^2$
(V$_0$/V)$^2$= 1/(1 + M$^2$/4+............)
Where V$_0$= velocity of fluid with considering compressibility
V= velocity of fluid without considering compressibility
For example:-
Q. An aeroplane flies at 1000 km/hr at an altitude where the pressure p is 50 kN/m2 and the density ρ=0.6 kg/m 2. Calculate the pitot speed indicator reading if it is ignoring the compressibility effects.
Solution:-
The sonic speed at ambient condition is given by
V=(1000*1000)/3600=278 m/s
So, mach no. M=V/c=278/342
M=0.81
P=50 KN/m2(given) Using equation (4) P$_0$= 50*{1+(1.4-1)/2* 0.812}1.4/1.4-1 =70KN/m$^2$ Neglecting compressibility, P$_0$-P= ρV$^2$/2 V$^2$={2(77-50)* 103}/0.6 V=300 m/s
(300-278)/278 * 100= 7.9%
= (1 + 0.812...)=1.164 This is a correction on the square of velocity. Correction to the velocity is (1.164)1/2=1.079, i.e., 7.9% the indicated speed of 300 m/s is therefore corrected to 300/1.079, i.e., 278 m/s true speed!

### Result

The pitot probe is calibrated as:- (V$_0$/V)$^2$ =1/(1 + M$^2$/4........) Where V$_0$== velocity of fluid with considering compressibility V= velocity of fluid without cosidering compressibility
M= Mach number
* ↑ (F M white,Tata Mcgraw,seventh)