SpaceX Texas launch site Discussion and Updates - Thread 12

[darkenfast]

The Soviet N-1 rocket had quite a bit more thrust than the Saturn V (over 10 million pounds vs. 7.5 million). As I understand it, the pads at Baikonur did not have water deluge. They were fairly deep pits with three channels radiating out and up.

[InterestedEngineer]

Here's a profile of exhaust temperatures as supplied by Spacex:

https://forum.nasaspaceflight.com/index.php?topic=53555.msg2417937#msg2417937

TL;DR:  ~2000k

From page 6 of Appendix G: Exhaust Plume Calculations. (https://www.faa.gov/space/stakeholder_engagement/spacex_starship)

I'm not sure how relevant the actual nozzle exit gas temperature is other than as a figure of merit. It will not be a good blackbody radiator and any impingement will involve higher temperatures due to compression.

[InterestedEngineer]

The Soviet N-1 rocket had quite a bit more thrust than the Saturn V (over 10 million pounds vs. 7.5 million). As I understand it, the pads at Baikonur did not have water deluge. They were fairly deep pits with three channels radiating out and up.

They did build one for Energia:

https://en.wikipedia.org/wiki/Sound_suppression_system#Soviet_Union/Russia

[InterestedEngineer]

You're saying that boiling room temperature water takes 3-1/2 times as much energy as melting steel?

Specific heat 304 stainless:   0.500 J/g-°C
Specific heat Water:  4.182 J/g-°C
Vaporizing Water: 2260 J/g
Heat of fusion (melting) 304 stainless: .27 J/g
water: 20 -> 100deg = 335 J/g
boiling water:  2260 J/g
Water total:  2595 J/g

304 stainless 20->1400:  690 J/g
304 stainless Melt:  0.27 J/g
304 total: 690 J/g

2595/690 = 3.76.

Boiling water requires 3.76 times the heat than melting 304 stainless starting at room temperature, gram for gram.

references:

https://www.azom.com/properties.aspx?ArticleID=965
https://letmegooglethat.com/?q=is+it+easier+to+heat+water+or+metal

From the first reference, Heat of fusion is not .27 J/g, but .27 kJ/g.
So the ratio should be:
2595/960 = 2.7.
Still impressive though.

127 tonne water = 127 m³ water, which is a circular pool 9 m in diameter and 2 m deep. No big deal in theory. Getting all the energy to go into boiling that pool would be a problem, and of course the resulting 216,000 m³ of steam needs to go somewhere.

Alas, I was mistaken in my initial calculations.  Something on the order of terajoules is needed, so more the 1 kiloton of water range is needed.

Which isn't there, AFAICT

Anyone have any ideas how much water SpaceX has on hand and how fast they can deliver it?

[DanClemmensen]

127 tonne water = 127 m³ water, which is a circular pool 9 m in diameter and 2 m deep. No big deal in theory. Getting all the energy to go into boiling that pool would be a problem, and of course the resulting 216,000 m³ of steam needs to go somewhere.

Alas, I was mistaken in my initial calculations.  Something on the order of terajoules is needed, so more the 1 kiloton of water range is needed.

Which isn't there, AFAICT

Anyone have any ideas how much water SpaceX has on hand and how fast they can deliver it?

No matter. The vast majority of the heat is carried away by the exhaust gasses. You just need a way to keep the relatively small percentage that heats up the structure from doing too much damage. Similar in concept to the TPS during reentry, where the bulk of the heat is carried away by the plasma.

[oldAtlas_Eguy]

Plus since the T/W for Starship is 1.5. Meaning in 10 seconds it is already 240m above the launch table and 50m higher than the top of the Tower. NOTE the heights here is the bottoms of the R2s.

In 2 seconds it is already 6m above the top of the launch table and in 3 seconds is 20m up. So all the heat to whatever extent is in the first 2 seconds with just a little in the 3rd and 4th. In 4 seconds it is already 36m up above the table.
In 5 seconds 58m.

Your use of 10 seconds is a value of 5X greater than what at worst would occur during a normal launch.

A Note about Staurn V is that its T/W was 1.2 which meant it took 1.6 times longer to get the same height away from the top of Mobile Launcher.

Starship is just not sticking around long enough to cause problems. Most metals take seconds of intense heat to cause melting. The duration is so brief that not much melting would occur even with thin metal without any water prior spray down which normally occurs before engine ignition.

[InterestedEngineer]

Plus since the T/W for Starship is 1.5. Meaning in 10 seconds it is already 240m above the launch table and 50m higher than the top of the Tower. NOTE the heights here is the bottoms of the R2s.

In 2 seconds it is already 6m above the top of the launch table and in 3 seconds is 20m up. So all the heat to whatever extent is in the first 2 seconds with just a little in the 3rd and 4th. In 4 seconds it is already 36m up above the table.
In 5 seconds 58m.

Your use of 10 seconds is a value of 5X greater than what at worst would occur during a normal launch.

A Note about Staurn V is that its T/W was 1.2 which meant it took 1.6 times longer to get the same height away from the top of Mobile Launcher.

Starship is just not sticking around long enough to cause problems. Most metals take seconds of intense heat to cause melting. The duration is so brief that not much melting would occur even with thin metal without any water prior spray down which normally occurs before engine ignition.

Timing for the startup sequence appears to have been 3 engines about every 1/4 second.   (They don't have the gas pressure to light up more than 3 at a time, judging by the configuration of lines on the launch mount).    That's 2.5 seconds to get all the engines lit.   That's not unusual at all, possibly an optimistic estimate.

Then, using any of the standard kinematics calculators or s=1/2at^2, at 5m/sec^2 acceleration, clearing the tower (140m from ground) takes 7.5 seconds.

So there's about 10 seconds.  I suppose you could start counting later in startup and drop that to 9 seconds, but this is back of the envelope, 10 is good enough.

You'll note the exhaust plume from SpaceX's own document (posted above) shows > 1000K all the way out to 130 meters, which is why I chose the top of the tower (which is ~130m above the ring).

You stated that "In 2 seconds it is already 6m above the top of the launch table ".

The exhaust plume has lost none of its heat at 6m up.  It will still be hitting the base pretty hard, ~30 meters below, with > 2000K gases.  There will still be mach diamonds that can't form due hitting the base.

At 50m up the heat is still near 2000k and spread pretty wide, engulfing the OLM ring.  That's probably the time of maximum impact on the ring.  It took 4.5 seconds to get there.  The base is 80m below and the temperature of the exhaust is about ~1400K, dropping below the melting point of steel.

at 100m up, 6.32 seconds have passed.  The plume is cooled to near 1200K and is now wide enough to start hitting the side of the tower at its base, though that part of the exhaust is only about 600K.

At 146m up, clearing the tower, 7.5 seconds have passed. The exhaust plume is at about 800k at its core, and the outer part of the plume is engulfing the tower at its base (but only a couple of hundred K). 

At 37m/sec, it's not worth counting any more.

This is all reasonably clear from the diagram SpaceX published, and the kinematic equations at an acceleration of 5m/sec^2

[meekGee]

Exhaust energy is not perfectly transferred solely to the launch mount. Or else launch mounts would not routinely survive rocket launches.
You have radiation to free space (surrounding ground and air), huge amounts being carried away by hot gas flowing away from the mount (with little being transferred to the mount, as gas is pretty poor at that), energy being removed acoustically, etc.

Other launch mounts have much more extensive water deluge and don't have complicated tubing all over the place.

For example, the SLS launch system dumps 1,700 tons of water into its deluge, which is capable of absorbing 4 TeraJ of energy.   That's a number that makes sense on the back of the envelope.

The water isn't there to "absorb" any energy as such - it's there to deaden acoustics enough to prevent damage to the vehicle during the first few seconds of ascent. The great majority of the energy goes into accelerating the vehicle itself. Only a small fraction of the net energy produced is mitigated by a deluge system.

Everything is correct, except almost no energy is transferred to the vehicle in the beginning.

Most energy leaves the pad via the exhaust gas, in forms of heat, kinetic energy, and acoustic energy.

I agree though that the total energy expended is not a good measure of how much energy the pad structure absorbs.

[Herb Schaltegger]

The last couple pages of this thread have been dominated by someone attempting to prove the OLM will melt down into a pile of goo on the first launch attempt, replete with a lot of numbers thrown about authoritatively.

I would suggest, again, that the thermodynamics of a dynamic system cannot be simplified this way. The energy produced by the engines is NOT going to be transferred to the pad structures completely or even majorly. Again, the greatest fraction of that energy will go into accelerating the vehicle itself. The gas produced, while hot, will also be carrying away a lot of that energy and will not release it instantly, but will rather do so over a period of seconds to minutes as it expands, diffuses and cools.

But bottom line is, I’ll buy Interested Engineer a beer if the OLM melts down when B7 or B9 finally makes a launch attempt.

[Robotbeat]

You have radiation to free space (surrounding ground and air),

You should probably run Stefan Boltzmann on that assertion.   When I do, I'm in rounding error territory.

In the end, wherever you think the energy might go, conservation of energy rears its head.   You have to put those Joules somewhere.

I've tried to put those Joules all the places y'all have suggested, and the numbers don't add up.

The Joules don’t just get deposited. They have to travel through stuff. And that takes time. Heat conduction through surface layers, thermal paint, and even steel at those timescales is slow.

Do the calculation. It ain’t happening fast enough to melt 1 inch steel, unfortunate for concern trolls everywhere.

EDIT: I just did the calculation with really conservative assumptions. 1inch of steel a square meter in area is about 200kg. To bring that up to the melting point of 1500K requires about 200MJ. Given 45W/(m*K) thermal conductivity of steel and the 2000K temperature of exhaust (which won’t be directly aimed at the structure so this is a bad assumption) and 1 inch thickness and 5 seconds of contact time, you can only conduct 20MJ through the 1 inch of square meter steel at most.

Now add in the fact that the steel will heat up, reducing the temperature gradient and thus heat flux through the steel, the paint acts as an additional (possibly ablative) barrier, you have some mixture with air and the nitrogen/water spray, the fact that melting takes additional energy itself, and there is simply no way to melt the 1 inch thick steel in that timeframe using that temperature of non-oxygen-rich exhaust.

[Robotbeat]

Additionally, at least for a single Raptor, at 100 throat radius distance, which is about 12 meters, the temperature drops below the melting point of steel anyway, even if it were directly beneath the engines to begin with (instead of to the side). That takes about 2 seconds at 1.5T/W ratio.

https://www.faa.gov/sites/faa.gov/files/2022-06/AppendixG_ExhaustPlumeCalculations.pdf

(Not to mention the fact that even 1 m to the side of the center of the plume, the plume temperature is below the melting point of steel.)

[meekGee]

The last couple pages of this thread have been dominated by someone attempting to prove the OLM will melt down into a pile of goo on the first launch attempt, replete with a lot of numbers thrown about authoritatively.

I would suggest, again, that the thermodynamics of a dynamic system cannot be simplified this way. The energy produced by the engines is NOT going to be transferred to the pad structures completely or even majorly. Again, the greatest fraction of that energy will go into accelerating the vehicle itself. The gas produced, while hot, will also be carrying away a lot of that energy and will not release it instantly, but will rather do so over a period of seconds to minutes as it expands, diffuses and cools.

But bottom line is, I’ll buy Interested Engineer a beer if the OLM melts down when B7 or B9 finally makes a launch attempt.

Herb - almost no energy gets transferred to the vehicle.

If you look at kinetic energy, it goes by v-squared, and while the exhaust is shooting out the back at thousands of meters per second, the vehicle hardly moves, and the only relation between them is impulse and change in momentum, and that only goes by v-linear.

On top of that, the thermal and acoustic energy of the exhaust is entirely wasted.

Rockets are really inefficient that way (and the higher the ISP, the worse it gets)

But I agree that the pad only captures a small amount of that energy amd won't melt.  And even though shielded, some equipment will be damaged.  I'm sure there's gen 2 hardware standing by already.

[InterestedEngineer]

The last couple pages of this thread have been dominated by someone attempting to prove the OLM will melt down into a pile of goo on the first launch attempt, replete with a lot of numbers thrown about authoritatively.

I would suggest, again, that the thermodynamics of a dynamic system cannot be simplified this way. The energy produced by the engines is NOT going to be transferred to the pad structures completely or even majorly. Again, the greatest fraction of that energy will go into accelerating the vehicle itself. The gas produced, while hot, will also be carrying away a lot of that energy and will not release it instantly, but will rather do so over a period of seconds to minutes as it expands, diffuses and cools.

But bottom line is, I’ll buy Interested Engineer a beer if the OLM melts down when B7 or B9 finally makes a launch attempt.

Nice strawman.  My prediction isn't the pad will melt.  I'll make a specific prediction:  Unless SpaceX somehow gets a kiloton of water deluge between now and launch, there will be damage to the launch mount and surrounds that will prevent another flight for at least 3 months.   The damage will involve melted and blown off structures protecting hydraulic, gas and electrical lines, the intricate stuff that's hard to fix.  There might be structural damage to the ring and base.  They will have to substantially change the design to prevent another occurrence.

By focusing on the heat, which is less than half of the total energy involved, your strawman doesn't address the kinetic energy involved.

Your physics is off too, the exhaust gases have the energy they have as a function of how the engines work, it isn't in the vehicle, that's just momentum transfer.   You should try calculating the total energy change of 5 million kg moving at 37m/sec with a 140m altitude change before making that kind of assertion.  Something any high school physics student can do.

If someone besides me would be willing to attempt to account for the terajoules of energy using math and physics it'd be far more interesting discussion.   I've managed to debunk every attempt to say "energy goes to X", or or "energy is only Y", save the convection argument.  Acoustics?  Not if you want every window and eardrum blown out in South Padre.   EM Emissions?  Stefan Boltzmann says otherwise. 

As for the convection argument, I've shown that a very hot hurricane of gases and debris will scour the earth for 100s of meters around the OLM, and probably toss flaming material up to a kilometer away.

There's a pretty easy comparison - look at the raptor test stands, which test one engine at a time.  They have substantial protection and water deluge.   The physics of protection are reasonably easy to do, and the joules add up at McGregor

Energy will be conserved.  Where will it go, and what will it do?

[Robotbeat]

Rocket engines work under exactly the same principles, and they don’t need a kiloton of water to keep them from melting for 2 seconds of runtime.

FFS, this is dumb. Do some actual math of heat conduction from the plume at the plume’s likely temperature, like I did up-thread and you conveniently ignored:
https://forum.nasaspaceflight.com/index.php?topic=54355.msg2418528#msg2418528

There’s no way to melt even a substantial portion of the launch mount. Damage? Sure, if not properly designed. But you don’t need a kiloton of water to do this. And your OWN PLUME GRAPH shows that even about  a meter away from the centerline of the plume, the temperature drops to less than the melting point of steel.

[oldAtlas_Eguy]

The last couple pages of this thread have been dominated by someone attempting to prove the OLM will melt down into a pile of goo on the first launch attempt, replete with a lot of numbers thrown about authoritatively.

I would suggest, again, that the thermodynamics of a dynamic system cannot be simplified this way. The energy produced by the engines is NOT going to be transferred to the pad structures completely or even majorly. Again, the greatest fraction of that energy will go into accelerating the vehicle itself. The gas produced, while hot, will also be carrying away a lot of that energy and will not release it instantly, but will rather do so over a period of seconds to minutes as it expands, diffuses and cools.

But bottom line is, I’ll buy Interested Engineer a beer if the OLM melts down when B7 or B9 finally makes a launch attempt.

Herb - almost no energy gets transferred to the vehicle.

If you look at kinetic energy, it goes by v-squared, and while the exhaust is shooting out the back at thousands of meters per second, the vehicle hardly moves, and the only relation between them is impulse and change in momentum, and that only goes by v-linear.

On top of that, the thermal and acoustic energy of the exhaust is entirely wasted.

Rockets are really inefficient that way (and the higher the ISP, the worse it gets)

But I agree that the pad only captures a small amount of that energy amd won't melt.  And even though shielded, some equipment will be damaged.  I'm sure there's gen 2 hardware standing by already.

The measure of engine efficiency IS ISP. The higher the ISP for a given prop mixture means a more efficient transfer of the heat energy of the burn to kinetic energy of thrust. The R2s are already some of the most efficient engines for it's prop type that has ever existed. Their efficiency over that of the SLS SRBs is very significant. The SRBs have a very large heat loss vs kinetic energy push. That is because to have a better transfer of heat to kinetic energy the SRB has to operate at such a high internal pressure that it would blow the steel canisters apart into little pieces.

As ISP goes up the exit temp of the gas goes down at the same time as it's exit velocity goes up. This is why ISP goes up. More heat goes into kinetic energy. Engine efficiency goes up.

[InterestedEngineer]

Additionally, at least for a single Raptor, at 100 throat radius distance, which is about 12 meters, the temperature drops below the melting point of steel anyway, even if it were directly beneath the engines to begin with (instead of to the side). That takes about 2 seconds at 1.5T/W ratio.

https://www.faa.gov/sites/faa.gov/files/2022-06/AppendixG_ExhaustPlumeCalculations.pdf

(Not to mention the fact that even 1 m to the side of the center of the plume, the plume temperature is below the melting point of steel.)

The units are nozzle exit radius, not throat radius.  The word "throat" doesn't appear in the document.

The nozzle exit radius is 2/3 of a meter (25.718 inches, see Table 1), so thanks for catching that I got it slightly wrong, as I had assumed the unit was a meter.   I suspect the 140m tower will be safe from heat.  It'll be interesting to see how all that piping survives tons of material moving at > 1000m/sec.

The "exhaust impinging on steel above steel's melting point" will last only 7.5 seconds as a result of this correction.

I note that I haven't gone into the compression of gases at the base of the pad yet, as gases compress when they have to make a 90 degree turn.  Compression heats gases up.  I can only note the basic physics of it, calculations would require numerical simulations beyond what most of us can do.

[Robotbeat]

I do thermal conductivity numerical simulations (and have done some for compression as well). This is just not the problem you think it is, and as I keep pointing out, there’s no way for the heat to be conducted into the steel fast enough to melt an inch of it, even ignoring any countermeasures like coatings, ablative paint, or water.

[InterestedEngineer]

The last couple pages of this thread have been dominated by someone attempting to prove the OLM will melt down into a pile of goo on the first launch attempt, replete with a lot of numbers thrown about authoritatively.

I would suggest, again, that the thermodynamics of a dynamic system cannot be simplified this way. The energy produced by the engines is NOT going to be transferred to the pad structures completely or even majorly. Again, the greatest fraction of that energy will go into accelerating the vehicle itself. The gas produced, while hot, will also be carrying away a lot of that energy and will not release it instantly, but will rather do so over a period of seconds to minutes as it expands, diffuses and cools.

But bottom line is, I’ll buy Interested Engineer a beer if the OLM melts down when B7 or B9 finally makes a launch attempt.

Herb - almost no energy gets transferred to the vehicle.

If you look at kinetic energy, it goes by v-squared, and while the exhaust is shooting out the back at thousands of meters per second, the vehicle hardly moves, and the only relation between them is impulse and change in momentum, and that only goes by v-linear.

On top of that, the thermal and acoustic energy of the exhaust is entirely wasted.

Rockets are really inefficient that way (and the higher the ISP, the worse it gets)

But I agree that the pad only captures a small amount of that energy amd won't melt.  And even though shielded, some equipment will be damaged.  I'm sure there's gen 2 hardware standing by already.

The measure of engine efficiency IS ISP. The higher the ISP for a given prop mixture means a more efficient transfer of the heat energy of the burn to kinetic energy of thrust. The R2s are already some of the most efficient engines for it's prop type that has ever existed. Their efficiency over that of the SLS SRBs is very significant. The SRBs have a very large heat loss vs kinetic energy push. That is because to have a better transfer of heat to kinetic energy the SRB has to operate at such a high internal pressure that it would blow the steel canisters apart into little pieces.

As ISP goes up the exit temp of the gas goes down at the same time as it's exit velocity goes up. This is why ISP goes up. More heat goes into kinetic energy. Engine efficiency goes up.

Very true.

I used the thermal vs. kinetic energy proportions of Raptor 2 as calculated by the Raptor Engine thread folks, who have some extremely capable engineers.

Kinetic is 125GW, thermal is 97GW, 56% efficient.  That's amazing efficiency for a heat engine.

So, back on topic:  Conservation of energy.  It has to go somewhere.  Where does it go?

[edzieba]

InterestedEngineer: Run your numbers for vehicles that use no water deluge, like Soyuz, Proton, or the N-1. If your numbers say those pads should be in some way damaged, whilst they were not in reality*, then it should be clear that the coupling between energy emitted by a vehicle and energy absorbed by the pad is vary very far from 1:1.

*OK, the N-1 pad was damaged, but by a violent explosion rather than a nominal launch.

[InterestedEngineer]

I do thermal conductivity numerical simulations (and have done some for compression as well). This is just not the problem you think it is, and as I keep pointing out, there’s no way for the heat to be conducted into the steel fast enough to melt an inch of it, even ignoring any countermeasures like coatings, ablative paint, or water.

I think what you are saying is that heat is not the problem because 7.5 seconds is not enough time to conduct the heat into steel to melt it.

Fair enough.  Our everyday gas stovetop stainless steel pans experience back up that assertion.  It takes over a minute to get most stainless steel or cast iron pans up to temperature.

But the "pad melting" is a strawman.  I never made that argument.  The energy is going somewhere.   I think the most supported prediction in this thread is a supersonic hurricane of hot gas with a radius on the order of 500m around the pad.  I also think we'll see something like the destruction from the fuel-air explosion, with piping, electrical, and hydraulic systems damage.  Mostly from the kinetic energy, but some from the heat.

I would love it if someone on this thread ("Launch Site") would tell us what the expected deluge volume is.  That would help a lot.   AFAICT from pictures of the piping, it's in the 10s of tons range or less.