I was just watching this video of a launchpad sound suppression system, and realized if it removes acoustic energy from the air, the energy has to go somewhere.
Does the water heat up? Will any of it get hot enough to boil? Maybe evaporate a little?
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3$\begingroup$ If there is no sound suppression system, acoustic energy will be (partially) transformed to heat. Why should that be different when water is used for sound supression? $\endgroup$– UweCommented Oct 22, 2018 at 9:43
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1$\begingroup$ I guess a lot of the acoustic energy would be absorbed by the water in the deluge system being torn into tiny droplets (by doing work against surface tension) $\endgroup$– user20636Commented Oct 22, 2018 at 10:29
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$\begingroup$ but undoubtedly some of the kinetic energy will be absorbed as heat. $\endgroup$– user20636Commented Oct 22, 2018 at 10:35
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$\begingroup$ I'll need time to go look at sources, but last time I checked the energy does get converted into heat, but it is a negligible source of energy compared to the other sources (i.e. the giant pillar of fire coming out of the space chariot!) $\endgroup$– Cort AmmonCommented Oct 22, 2018 at 19:20
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2 Answers
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Yes, the water heats up. Lots of it evaporates. There is a system to catch and filter the remainder, which is polluted with combustion products of the SRBs.
Those massive clouds in the foreground are water vapor mixed with combustion products of the SRBs.
For the Shuttle, this report states that 166 tonnes of water is evaporated, out of 1135 tonnes supplied by the suppression system. But that's an incomplete number.
Another source has these numbers for the the deluge water:
_ 200 000 litres sont vaporisées ;
_ 10 000 litres sont atomisées ;
_ le reste est éparpillé ;
- 200 tons is vaporized
- 10 tons is atomized
- the rest is scattered (blasted away by the exhaust, ending up all over the vicinity of the flame trenches)
The mechanism seems to be that air bubbles in the water are compressed by the sound waves passing through, this compression generates heat. See this related question.
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$\begingroup$ Are the clouds (condensed) water vapour, or are they "atomised" water droplets? $\endgroup$– user20636Commented Oct 22, 2018 at 9:59
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3$\begingroup$ Page 66 of the report states the 166 tonnes of water is vaporised due to the exhaust heat. Page 65 states "For this application the kinetic energy should not be included" - so, while extremely useful and interesting, it doesn't answer the question. $\endgroup$– user20636Commented Oct 22, 2018 at 10:16
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$\begingroup$ Page 66 also includes "The cloud must also contain a considerable quantity of liquid water atomized from the excess deluge water " and page 96 goes on to state "Since a substantial fraction of each deposition drop comes from the deluge water without an intervening phase transition" $\endgroup$– user20636Commented Oct 22, 2018 at 10:33
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$\begingroup$ I'm not sure how the (french) source gives a 70/30 split? It seems to be using single digit precision, 200 tonnes evaporated, 10 tonnes atomised and the rest scattered(?). Perhaps that refers to the three states the Nasa report notes when describing the interplay between vapour, atomised and small droplets as complex. However, If we take the 30% ending up in the pond as accurately applying to the 365 tonnes used in the 7.5 seconds, that's around 110 tonnes. No more than 166 tonnes is evaporated by the heat from the engines, so around 70 tonnes of water is put into the air by the kinetics $\endgroup$– user20636Commented Oct 22, 2018 at 11:54
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tl;dr: water droplets scatter the sound keeping it more localized, and also absorb some of it, while some is also absorbed in all the other surfaces the sound strikes. There's no where near enough power to boil this much water.
Sound in the air at these (audio and sub-sonic) frequencies pretty much always ends up as mostly heat.
Ultrasonic sound can be use do work on the surface tension of water (ultrasonic atomizers, humidifiers, etc) but when that's used in a practical way the wavelength is usually millimeters and the corresponding frequency in MHz so the sound pressure can be focused to a small, high-intensity area for more efficient disruption of the surface into droplets.
But that's not what's going to happen here.
For audible and sub-sonic rocket launch sound, the deep sub-wavelength but high density droplet dispersion will diffusely scatter the sound waves, decreasing their mean free path and therefore increase the time for the sound energy to be mechanically absorbed, both by the droplets and by any other local surfaces the sound reaches.
How much energy and therefore heat are we talking about here?
Until dB levels with their corresponding distances and measurement conditions can be found, here is a walkthrough of the conversion process and are some rough estimates:
So far I found a sound level of about 160 dB at 125 feet in this answer and 180 dB in the payload bay of the Space Shuttle in this answer but these are not helpful; the first one is with the sound suppression in place, and the second is in a confined, closed space.
But I'll show how to do the estimating calculation here in hopes we can get better numbers.
In this excellent answer to my question How much power and energy is (actually) in a 230 dB “click” from a whale? the equation was derived as follows:
Intensity (or specifically sound intensity) of a linear sound wave is related to sound pressure, $P$, through: $$ I = \frac{ P^{2} }{ \rho_{o} \ C_{s} } $$ where $\rho_{o}$ is the mass density and $C_{s}$ is the speed of sound in the medium. One can look up the properties of water to find that $\rho_{o}$ ~ 999.972 kg/m3 and $C_{s}$ ~ 1484 m/s. We can also look up the reference pressure level for water (or at NOAA) finding $P_{H2O}$ ~ 1 $\mu$Pa (compared to $P_{air}$ ~ 10 $\mu$Pa) at 1 meter from source. This corresponds to a reference intensity of $I_{o} \sim 6.74 \times 10^{-19}$ W/m2.
$$ I = I_{o} \ 10^{L/10} $$
where $I$ is intensity (in W/m2) and $L$ is intensity (in dB).
So for example, 160 dB at 125 feet (38 meters) would be 7 mW/m^2 or about 15 Watts integrated over one hemisphere.
The largest number I saw was in this answer
The Saturn V predated this suppression system. Early engine tests reached as high as 211 decibels.
If we were to try to use the same 38 meters distance, 50 dB is 10^5 means 1.5 megaWatts, and that could potentially boil a few kilograms of water per second if it were concentrated on a few kilograms. But that's not the case, the sound is dispersed over many tons per second of water and the ground and other surfaces.
There are more dB values in this answer but they don't have a clear geometry description.
Nonetheless:
No, no water was
harmedboiled in the making of thisfilmorbit.
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$\begingroup$ youtu.be/THUMdTohWkI?t=192 Are you sure it's only ultrasound that can disperse water? $\endgroup$– user20636Commented Oct 22, 2018 at 11:39
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1$\begingroup$ Then your third paragraph needs clarification. It appears to talk about ultrasound, separating it from "audible and sub-sonic rocket launch sound" by the use of the "but that's when the wavelength is millimeters and the corresponding frequency in MHz" clause. $\endgroup$– user20636Commented Oct 22, 2018 at 13:18
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1$\begingroup$ @JCRM edited; I'll revisit this again in the morning. Thanks for the helpful comment! $\endgroup$– uhohCommented Oct 22, 2018 at 14:18
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1$\begingroup$ Are you sure that 211dB "at the base of the test stand" are comparable to 160 dB in 38m distance? What was the size of the F-1 test stand? $\endgroup$– asdfexCommented Oct 23, 2018 at 22:27
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1$\begingroup$ @uhoh This suggests something in the order of 10m - if we take "bottom" as "ground floor" - they won't put the microphones into the water pit below. en.wikipedia.org/wiki/Rocketdyne_F-1#/media/… $\endgroup$– asdfexCommented Oct 24, 2018 at 8:34
