Mach Wave Elimination
A new method for quieting high-speed jets
(U.S. Patents 5590520 and 5916127)


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EXPERIMENTAL SETUP

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HELIUM-AIR simulation of hot jets.

The noise emitted by fully-turbulent, perfectly-expanded free jets depends on three main factors:

  1. Jet velocity (U) : the overall acoustic power goes as Un, where n ranges from 6-8 for subsonic jets and approaches 3 for very-high-speed jets.  The value of U also affects the nature of jet noise emitted in the downstream direction.  At low speeds, this noise comes from moving quadrupoles; at high speeds, Mach wave emission becomes dominant.
  2. Jet Mach number (M):  at fixed U, increasing M decreases the jet spreading rate, thus elongates the noise-source region. This is primarily a compressibility effect, not a density effect. A jet at U=700 m/s and M=2 is noisier than a jet at U=700 m/s and M=1.5.
  3. Jet density (r) : the jet-to-ambient density ratio affects modestly the jet growth rate.

In large facilities with big budgets (typically industry or goverment) hot jets are used to replicate the exhaust conditions of jet engines.  In a small-scale university setup, this is a very expensive proposition fraught with safety concerns.  An inexpensive--and much safer--alternative is the use of helium-air mixtures to match exactly the velocity and Mach number, and match approximately the density, of hot jets.  Here is how this is done:

 

Matching the Mach number (M) 
This is very easy.  One simply uses a nozzle with the same design Mach number as the hot jet and subjects the nozzle to a pressure ratio that gives the desired exit Mach number.

 

Matching the velocity (U)
The velocity is
U= M a=M (
gRT)1/2
where a is the speed of sound,
g is the ratio of specific heats, R is the gas constant, and T is the static temperature of the jet. In the hot air jet, the velocity is high because the temperature is high:
U= M a=M (
gairRairTair)1/2
In the cold helium-air mixture (total temperature has room value), we achieve exactly the same speed of sound, and hence jet velocity, by increasing
gR:
U= M a=M (
gmixRmixTmix)1/2

Matching (approximately) the density (r)
The density is
r = p/(RT)=gp/a2
where p is the pressure at the jet exit (equal to ambient pressure for pressure-matched jets).  Recall that the speed of sound a is matched exactly, therefore

rmix / rair= gmix / gair

so there is a slight mismatch in density due to the higher g of the helium-air mixture. Typical value of this mismatch is 10%.

Example:  Simulation of a jet with M=1.5 and U=700 m/s

HOT AIR JET

Gas constant: R = 287 J/(Kg oK)
Ratio of specific heats:
g = 1.4
Total temperature: T0=786 oK (=1414 oR=954 oF)
Mach number: M=1.5
Static temperature: T=542 oK (=976 oR=515 oF)
Speed of sound: a=(
gRT)1/2=467 m/s (=1531 ft/s)
Jet velocity: U=Ma=700 m/s  (=2296 ft/s)
Jet density*:
r=p/(RT)=101325/(287*542)=0.651 kg/m3

 

 

COLD HELIUM-AIR JET
Mixture of 26.4% helium and 73.6% air (by mass).

Gas constant: R = 760 J/(Kg oK)
Ratio of specific heats:
g = 1.563
Total temperature: T0=300 oK (room temperature)
Mach number: M=1.5
Static temperature: T=184 oK
Speed of sound: a=(
gRT)1/2=467 m/s (=1531 ft/s)
Jet velocity: U=Ma=700 m/s  (=2296 ft/s)
Jet density*:
r=p/(RT)=101325/(760*184)=0.724 kg/m3 (11% higher than density of hot jet)

 

* Jet exit pressure is ambient at sea level

The proof is the pudding... See comparison of noise spectra.