NASA's launchpad sound suppression system: how does it really work?

[Machine_Elf]

NASA famously employs a sound suppression system during rocket launches to prevent the extreme pressure waves from reflecting back up off of the launch pad and damaging whatever vehicle is being launched. The system reportedly gets the job done, but nobody’s ever been very clear about how it does what it does. The above Wikipedia link doesn’t mention it, and neither does NASA’s own Space Shuttle News Reference Manual.

This video claims that the system works by having bubbles of air get pressurized by the peak pressures, converting the energy to heat which is then dissipated, I guess, to the surrounding water. I am suspicious of this explanation, since it implies that the water being sprayed all over the launch pad forms foam, i.e. bubbles of gas (air) dispersed in a liquid matrix. Rather than foam, it looks like the system is designed to generate massive quantities of aerosol, i.e. droplets of water dispersed in a matrix of air. I can then imagine two possible paths by which the sound is converted to heat:

A)The high and low temperatures associated with air pressure peaks/troughs are mitigated by direct heat transfer as these peaks/troughs pass by each water droplet;

AND/OR

B)The air pressure peaks/troughs drive the water droplets back and forth, inducing viscous drag whenever there is a mismatch between local air velocity and water droplet velocity; viscous drag is a lossy process that converts kinetic energy to heat.

Those are my armchair hypotheses. Can anyone point to an authoritative source (ideally NASA themselves, or a technical paper published by them) regarding the actual mechanism of sound absorption by this system?

[friedo]

I honestly don’t know what that video means when they’re talking about “bubbles of air.” The water sound suppression system works because droplets of water absorb acoustic vibrations in the air and turn to steam. That’s really all there is to it.

[Bear_Nenno]

The water is introduced as droplets because they absorb and reduce the velocity of sound waves a lot better than a flat, still surface of water would. If they just had a pool of water down there, the sound would reflect back almost as violently, as if it were itself a hard, flat surface. Turbulent droplets effectively break up the eddies while reducing the intensity of reflected sound waves.

[Machine_Elf]

@friedo

I honestly don’t know what that video means when they’re talking about “bubbles of air.” The water sound suppression system works because droplets of water absorb acoustic vibrations in the air and turn to steam. That’s really all there is to it.

How do the droplets of water “absorb acoustic vibrations”? Are we talking about the droplet getting distorted by the pressure waves, causing viscous dissipation within the droplet? And is there really enough acoustic energy being absorbed to turn these droplets into steam, or is the giant cloud we see around the launchpad due to water being finely atomized and/or vaporized by the high-temperature, high-velocity exhaust plumes?

@Bear_Nenno

The water is introduced as droplets because they absorb and reduce the velocity of sound waves a lot better than a flat, still surface of water would.

OK, but how do they absorb the sound waves? Can you get down to the fundamental physics involved (e.g. heat transfer, viscous dissipation, aerodynamic drag, that sort of thing)?

@Bear_Nenno

Turbulent droplets effectively break up the eddies while reducing the intensity of reflected sound waves.

What does “turbulent droplets” mean? Are you talking about turbulence within each droplet, or turbulent bulk flow of an aerosol? And what “eddies” are you referring to?

Sorry for being pedantic, but I want very much to avoid vague and/or generalized descriptions of how this system does its job.

[Bear_Nenno]

@Machine_Elf

OK, but how do they absorb the sound waves? Can you get down to the fundamental physics involved (e.g. heat transfer, viscous dissipation, aerodynamic drag, that sort of thing)?

A droplet in the path of the sound wave is compressed by the force of the wave. The compression causes the droplet to heat up. This part is thermodynamics: increase pressure causes an increase in temperature. The increased temperature turns the the droplet into steam. So, the result is that some of the acoustic energy has been converted to heat energy which is, in turn, dissipated into the surrounding air. That’s the “absorption”. Some of the sound energy is transferred to the droplet. The energy of the sound wave is reduced slightly, which slightly reduces the intensity of the reflected sound as well.

What does “turbulent droplets” mean?

I just mean that the sound wave is not hitting a flat wall of water. It’s hitting a bunch of small drops of water at random points along the the wave. This helps to break up the wave more than if the water were flat and smooth. It scatters the wave, resulting in a reduction in the intensity of the wave that is reflected back at the shuttle. Similar to how a sound proofed wall will not be flat. It will look like this.

[Machine_Elf]

@Bear_Nenno

A droplet in the path of the sound wave is compressed by the force of the wave. The compression causes the droplet to heat up.

Liquid water isn’t very compressible. Even when subjected to very high pressures, the volume doesn’t change very much, so the mechanical work imparted (which is pressure x change in volume) is very, very small. I just ran a spreadsheet in which I compressed water in small increments from 1 atmosphere up to 20 atmospheres (the peak pressure associated with 220 dB); according to my math, the temperature increased from 20 degrees C to 20.00000011 degrees C. Moreover, that immense pressure increase means that the temperature required to cause the water to turn to steam is now 212 degrees C instead of 100 degrees C.

Bottom line, there’s not going to be any steam produced when a droplet gets hit by a sound wave, even one at 220 dB. And the tiny temperature increase of the water droplet means that we’re also dissipating very little energy to it by adiabatic heating.

So what else is there?

[Bear_Nenno]

Well, the droplets are going to turn to steam no matter what. They have a large surface area per volume and they’re exposed to rocket exhaust. As for the required temperature increase, that’s assuming a uniform compression. The droplets are only getting compressed from the side of the wave front, so I think it wouldn’t be the same as uniformed, static compression. So it’s going to be steam well before it’s 212. But either way, it could easily reach that temperature as well. As for the rest of the math, I’m stumped. This is interesting. I can’t wait for someone to explain it all. I’ll try to find something in the online libraries. Please post an update if you find something before someone like Stranger shows up to explain it.

[Bear_Nenno]

“Water injection into the rocket exhaust during liftoff in huge quantities, with typical water-to-gas mass flow rate ratio (MFR) of 3–4, is a proven suppression technique in launch pads. It has been reported that the water injection system deployed by NASA and ESA for their launch pads can attenuate the sound levels by 8–12 dB… Water injection attenuated all three components of jet noise: the turbulent mixing noise, the Mach wave radiation, and the broadband shock noise. The magnitudes of attenuation for the different components were dissimilar and depended on the injection parameters and the angle of observation. *The two attenuation mechanisms observed were decrease of jet velocity by momentum transfer between liquid and gas phase and reduction of jet temperature due to partial vaporization of the injected water. *” (Sankaran, et. al., 2009)

Osipov (2015), goes into the type of detailed, mathematical explanation I think you’re looking for. He mentions droplet breakup, heat exchange, momentum transfer & water droplet drag as all playing a role. If you can’t access either of these, I’ll be happy to email them to you.

References
Sankaran, S., Ignatius, J. K., Ramkumar, R., Satyanarayana, T. N., Chakravarthy, S. R., & Panchapakesan, N. R. (2009). Suppression of High Mach Number Rocket Jet Noise by Water Injection. Journal of Spacecraft and Rockets, 46(6), 1164-1170. doi:10.2514/1.43421

Osipov, V., Khasin, M., Hafiychuk, H., Muratov, C., Watson, M., & Smelyanskiy, V. (2015). Mitigation of Solid Booster Ignition over Pressure by Water Aerosol Sprays. Journal of Spacecraft and Rockets, 52(3), 928-943. doi:10.2514/1.a33110

[Machine_Elf]

@Bear_Nenno

Well, the droplets are going to turn to steam no matter what. They have a large surface area per volume and they’re exposed to rocket exhaust.

To the extent that the hot exhaust heats the water droplets and generates steam, I can’t see that this will have any bearing on acoustic mitigation.

@Bear_Nenno

As for the required temperature increase, that’s assuming a uniform compression. The droplets are only getting compressed from the side of the wave front, so I think it wouldn’t be the same as uniformed, static compression. So it’s going to be steam well before it’s 212.

Whether you compress part or all of the droplet, the maximum pressure that will be applied is 20 atmospheres. The math for adiabatic heating is the same: a 0.00000011 degrees C temperature rise for whatever portion of water gets compressed to 20 atmospheres.

This documentary segment, hosted by Richard Hammond, was interesting. In the middle of the segment, they show a technique by which submarines in WW2 achieved stealth: bubbles in the water do indeed mitigate sound. But they also showed that an airborne mist of water can mitigate sound, too. They claim the sound is converted to heat, but as with all of the other explanations I’ve seen, they don’t clearly describe the processes by which acoustic energy becomes heat energy.

The demos at the beginning and end of that segment, involving a cinder-block wall and a large vortex cannon, appear to support (but certainly don’t prove beyond a shadow of a doubt) my theory that acoustic energy is dissipated by viscous drag (producing heat) as the pressure waves push/pull the water droplets around. You can see how the water droplets get moved around as the vortex passes through the water curtain, sapping kinetic energy from the vortex and preventing it from knocking the wall down.

On preview, I see you’ve replied with some solid technical references; thanks for digging those up, that’s exactly the kind of thing I was hoping for. Hammond’s vortex cannon demonstration seems to fit the “decrease of jet velocity by momentum transfer between liquid and gas phase” mechanism quite nicely, and both of those mechanisms (momentum transfer and reduction of jet temperature) seem to match the speculation I offered in my OP.

[Bear_Nenno]

@Machine_Elf

The demos at the beginning and end of that segment, involving a cinder-block wall and a large vortex cannon, appear to support (but certainly don’t prove beyond a shadow of a doubt) my theory that acoustic energy is dissipated by viscous drag (producing heat) as the pressure waves push/pull the water droplets around.

I think you’ve nailed it. The second reference in my previous post goes on at length about drag. I believe it might be exactly what your theory proposes.

[Melbourne]

@Machine_Elf

To the extent that the hot exhaust heats the water droplets and generates steam, I can’t see that this will have any bearing on acoustic mitigation. .

Without reading the references, I see a suggestion that water mitigates over-pressure (whatever that is).

If you make the exhaust cooler, you will reduce the pressure. If you reduce the peak/trough pressure, you will reduce the noise. At high noise levels, air actually does heat/cool in the pressure peak/troughs: if this cycles through evaporation/condensation cycles in a non-linear non-steady-state way, as it surely does, this could reduce noise: on the experimental evidence it seems that it does.

[Velocity]

It only reduces by 8-12 decibels? Is that a meaningful amount of noise reduction for the vibration?

[EdelweissPirate]

Yes, it’s meaningful. A 10-decibel reduction means the sound pressure decreases by an order of magnitude. Put more simply, it reduces the sound pressure by a factor of 10.

But the decibel scale is logarithmic, and human ears have such a broad dynamic range that reducing the sound pressure to 1/10th the initial value means it only sounds about half as loud to us.

Even so, reducing the perceived loudness of a rocket launch by half isn’t too shabby.

[Velocity]

Ah right, logarithms like Richter…

Do Soyuz and other rockets have sound suppression too?

[Bear_Nenno]

@Velocity

Ah right, logarithms like Richter…

Do Soyuz and other rockets have sound suppression too?

No. The Soyuz uses the big giant hole method, instead.

[Magiver]

@Machine_Elf

To the extent that the hot exhaust heats the water droplets and generates steam, I can’t see that this will have any bearing on acoustic mitigation.

wouldn’t the conversion of heat result in a reduction in the expanding gasses of the rocket exhaust and the velocity of said exhaust? It’s all a reduction of energy.

Here’s a video of using bubbles to dissipate sound.

Clearly it works.

[Melbourne]

@EdelweissPirate

Yes, it’s meaningful. A 10-decibel reduction means the sound pressure decreases by an order of magnitude. Put more simply, it reduces the sound pressure by a factor of 10.

But the decibel scale is logarithmic, and human ears have such a broad dynamic range that reducing the sound pressure to 1/10th the initial value means it only sounds about half as loud to us.

Even so, reducing the perceived loudness of a rocket launch by half isn’t too shabby.

Which it wouldn’t do, because the perceived loudness is also logarithmic.

Assuming an initial loudness of 200 db, a 10 db reduction would give you a reduction of 5% in perceived loudness.

However, rockets and launch platforms and do /not/ get damaged on a logarithmic scale, and (in my ignorance) I am of the opinion that one of the reasons for using “noise” reduction technology is to reduce the damage caused by the noise vibrating things apart, or disrupting the rocket plume.

[EdelweissPirate]

@Melbourne

Which it wouldn’t do, because the perceived loudness is also logarithmic.

Assuming an initial loudness of 200 db, a 10 db reduction would give you a reduction of 5% in perceived loudness.

You sure about that? Got a cite?

From Audio Engineering for Sound Reinforcement: Chapter 2, PSYCHOACOUSTICS-HOW WE HEAR | Pro Audio Encyclopedia

If you ask a person with normal hearing to make an estimate of “half loudness” or “twice loudness” of a wideband noise signal by adjusting the volume control of a comparator circuit, you will find that it takes approximately a 10-dB difference, louder or softer, to produce the subjective judgment of twice or half as loud.