Airspeed to achive an equivalent of dynamic pressure increase of 1013 mB

[alosaurus]

I have put numbers in to the basic noddy formula 1/2 Roh V squared and it comes out as around 400 Metres / second 777 Knots.

(a) surely this is too high

(b) If it was close to the mark pressumably it would be useless as it is higher than mach one.

Anyone got any idea on how fast you would need to fly to double the pressure?

Thanks in advance

[Mad (Flt) Scientist]

(a) I get 406 m/sec so if its wrong we're both wrong. 1013mB is 101 325 Pa. rho 1.225 kg/m^3.

(b) Yes, concur, for supersonic flow the simple equation isn't really very useful.

Part of the issue may be that we don't really appreciate just how much force the atmosphere exerts. 14 psi or so is a lot. Do the maths on the force exerted on your house roof, if the house were a vacuum. You'd need some pretty impressive roof beams...

[selfin]

Would it not be more appropriate to consider the variation of density in this? At the stagnation point the pressure increase, for a given velocity, will be higher than that calculated by the Bernoulli (incompressible) equation. A more suitable solution to Euler's equation removing the restriction on variation in density along a streamline and instead allowing the density to vary with pressure (using the isentropic relations) gives us this solution, sometimes called the compressible Bernoulli equation (NACA), or the Saint-Venant equation (N. X. Vinh),



Subscripts 0 and 1 describe the total (or stagnation) and local (freestream) states respectively. Dimensionless gamma (ratio of specific heats) can be assigned the standard calorically perfect value of 1.4. M is Mach number; u is the velocity of the rigid body relative to the fluid.

I've included the solution with M^2 as the subject to illustrate that for a P0/P1 value of 2 the condition you seek (i.e. total pressure is twice that of static pressure) is temperature-independent (within the restrictions of a calorically perfect gas). At Mach 1, P0/P1 = 1.8929, and I should stress that the above equations are not strictly valid beyond Mach 1. They are however in very good agreement with the supersonic solution (Rayleigh supersonic pitot equation) up to about Mach 1.2.

Strictly for P0/P1 = 2, the first equation can be reduced to:

u^2 = 440.052 T

T is temperature in Kelvin.

At ISA MSL the condition P0/P1 = 2 is satisfied, according to the above, when u = 356 m/s, or 692 kt.