User:Kavish ahmad/sandbox

Pitot static tube in compressible flow Pitot tube:- It is a device invented by henri pitot in eighteen century which is used to measure  the fluid flow velocity. A device that consists of a tube having right angled bend which is placed vertically in moving body of fluid with mouth of the bent part directed upstream and that is used with a manometer to measure the velocity of fluid flow. Working principle:- it works on the principal of Bernoulli theorem as:- V2/2 + gz + P/ρ= constant Explanation of pitot working:- The manometer connected with pitot tube give pressure difference, so formula for velocity become as:- V= {2(∆h* (ρLg))/ ρ}1/2 •	Where,  ∆h. is the height difference of the columns in meters •	ρL is the density of the liquid in the manometer; •	 g is the acceleration of gravity •	 ρ is fluid density Mach number:- it is 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 no. 	 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 trasonic 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. (F M white)

Bernoulli theorem to compressible flow In this case fluid density is changing so pitot tube analysis for incompressible flow does not able to give accurate reading. To remove error we should further have to do analysis,  so when elevation is neglected bernolli equation become as: ʃdp/ρ + V2/2= constant.........(1) but for compressible isentropic flow chnages p=Kρϒ  where ϒ=cp/cv cp= specific heat at constant pressure cv = specific heat at constant volume k= constant for fluid so simplifying equation (1) become as {ϒ/( ϒ-1) }p/ρ + V2/2=constant=( ϒ/ ϒ-1)p0 /ρ0   .......(2) On considering other parameter gives energy equation for flow as:- T0 /T={1+ (ϒ-1)/2}* M2  ...................(3) And for an isentropic flow:- P0 /p= {1+(ϒ-1)/2 M2 } 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 + M2/4 + { (2-ϒ)/24}M4 +................... P0-P=1/2 ρV2 (V0/V)2= 1 + M2/4+............ Where V0= velocity of fluid with cosidering compressibility V= velocity of fluid without cosidering compressibility ( S. Chand) 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          C2= ϒP/ρ =(1.4* 50* 103)/0.6 C=342 m/s 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) P0= 50*{1+(1.4-1)/2* 0.812}1.4/1.4-1 =70KN/m2 Neglecting compressibility, P0-P= ρV2/2 V2={2(77-50)* 103}/0.6 V=300m/s The pitot probe, uncorrected for compressibility, is likely to indicate  300m/s  which is in excess of the true true value of 278m/s by (300-278)/278 * 100= 7.9% Alternatively, the correction factor, CCF=(1 + M2/4.....) = (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 300m/s is ,therefore, corrected to 300/1.079, i.e., 278 m/s true speed! Result:- The pitot probe is calibrated as:- (V0/V)2 =1 + M2/4........ Where V0== velocity of fluid with cosidering compressibility V= velocity of fluid without cosidering compressibility M= Mach number