SpaceX Texas launch site Discussion and Updates - Thread 12

[Robotbeat]

The energy is going into the hot gas which rapidly mixes with surrounding air and cools to below melting point.

It’s lasting for just a few seconds. That is not long enough for heat to conduct through to deep structure.

You can’t just teleport Joules like that. There is finite thermal conductivity.

You’re giving off big vibes like this:

[Robotbeat]

Almost all the initial energy ends up as heating up the surrounding air. Cubic kilometers of it.

[Herb Schaltegger]

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

Diffused into the air, mostly. As we’ve been trying to tell you. Like basically every other rocket in the world, just more of it.

[alugobi]

I think that what he's arguing is, that that air into which the energy diffused is going to break some stuff.  Not melt the pad, just break some stuff.

[matthewkantar]

SpaceX expects to break stuff, it’s how they know they are trying.

[InterestedEngineer]

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.

Thanks, that's a good suggestion for a cross check on a very basic physics model.

Soyuz uses a flame trench to divert the exhaust. The flame trench in the latest launch
site that they built is 27,000 square meters (in radius like what SpaceX is using that's 93m radius
circle)
 
--- Launch Mount ---
Inputs:
Flame Trench Volume: 850,000m^3
Mass of Air per m^3: 1.222kg 
Specific heat of Air:  1kJ/kg-K
 
Mass of air in flame trench: 1.04kt
Joules to heat air to 20->100degC: 80GJ (same final temperature as boiled water)
Joules to accelerate air to 300m/s: 94GJ  (don't exceed speed of sound as the limit)

This is pretty pessimistic as the air in the open air trench is pushed out and new unheated
air is pulled in, so the number of joules dissapated is likely 2-3x what I calculate here.

--- Soyuz-2 ---
4 boosters @ 839.48 kN: 3.36MN (343t)
4 boosters Propellant Mass @ 39,160 kg: 156,640kg
Wet Mass of entire rocket stack:  213t
Burn time: 118 seconds
ISP: 263.3

At a burn rate of 1327kg/sec and exhaust velocity of 2.58km/sec, the kinetic
energy rate is 8.8GW.

It is a bit tedious to calculate the thermal energy of the RD-xxx engines so
let's pessimistically assume 50% efficiency, so 8.8GW of thermal energy rate.

The Thrust-to-Wieght is 1.6, so I'll pessimistically assume 10 seconds of
impingement, even though the plume is not nearly as long as Raptor-2.

--- Conservation of Energy Calcs ---
10 seconds at 8.8GW of thermal energy = 88GJ ~= 80GJ.  The flame trench shouldn't experience
average heating beyond that of boiled water

10 seconds of 8.8GW of kinetic energy = 88GJ ~= 94GJ.  The average velocity of air in the
flame trench won't exceed the speed of sound.

Conclusion:  The design of the Vostochny complex for launching Soyuz-2 has enough
energy dissipation in the flame trench to seem reasonable by back of the envelope
conservation of energy calculations.

The total energy needed to be handled for a Soyuz-2 launch is 176GJ, or 13
times less than that of the Starship Booster. Taking the cube root of
13 yields 2.4 times the radius of air needed to dissipate Booster's
exhaust plume.  That's a flame trench radius of 226 meters equivalent.

I don't see that kind of flame trench on the OLM.

Sources:
http://www.russianspaceweb.com/vostochny_soyuz.html
https://en.wikipedia.org/wiki/Soyuz-2

[InterestedEngineer]

The energy is going into the hot gas which rapidly mixes with surrounding air and cools to below melting point.

It’s lasting for just a few seconds. That is not long enough for heat to conduct through to deep structure.

You can’t just teleport Joules like that. There is finite thermal conductivity.

You’re giving off big vibes like this:

Ironically enough I got a lot of heat for taking the optimistic side of argument on the heat shield thread, which is full of "it's gonna burn up on re-entry" doomsayers.

Using similar simplified physics models.   Which very few (or none) cared to actually counter with their own model.

[InterestedEngineer]

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

Diffused into the air, mostly. As we’ve been trying to tell you. Like basically every other rocket in the world, just more of it.

How much air around Starship will be heated and how fast will it be moving?

[edzieba]

Soyuz uses a flame trench to divert the exhaust. The flame trench in the latest launch
site that they built is 27,000 square meters (in radius like what SpaceX is using that's 93m radius
circle)

The OLM is not a solid cylinder through which Starship & Super Heavy rise. It is a ring open above, below, and to the sides, leaving only a brief fraction of a second where the only route for hot gas to go is through the ring. On ignition and startup, the aperture available for exhaust flow is the area between the hexapod legs (nozzle bell lips just below the level of the ring). The engines then raise around 5m until the lips are above the top level of the launch mount, and will again be exhausting into mostly free space.

Remember also that you've only calculated the energy input by the engines into the exhaust gasses. You also need to calculate the transfer from those exhaust gasses to any structures, you cannot simply assume a perfect coupling (it would actually be incredibly hard to capture the total energy from the exhaust gasses, its's why steam-ejector test stands have such small engine capacities).

[Robotbeat]

The energy is going into the hot gas which rapidly mixes with surrounding air and cools to below melting point.

It’s lasting for just a few seconds. That is not long enough for heat to conduct through to deep structure.

You can’t just teleport Joules like that. There is finite thermal conductivity.

You’re giving off big vibes like this:

Ironically enough I got a lot of heat for taking the optimistic side of argument on the heat shield thread, which is full of "it's gonna burn up on re-entry" doomsayers.

Using similar simplified physics models.   Which very few (or none) cared to actually counter with their own model.

This is nearly the exact same situation. Jets of hypersonic air, with the thermal and kinetic energy being turned ultimately into heat of the surrounding air, not primarily into the vehicle or launch infrastructure. It just becomes a few cubic kilometers of air heated a few degrees warmer.

[Robotbeat]

You can just paint it with some sort of refractory paint. Which is what SpaceX has done.

Something like this would work.
https://pipedreamindustries.com/product/super-high-temp-coating-to-3000f-1650c-eco-friendly-ultra-low-voc/

[InterestedEngineer]

The energy is going into the hot gas which rapidly mixes with surrounding air and cools to below melting point.

It’s lasting for just a few seconds. That is not long enough for heat to conduct through to deep structure.

You can’t just teleport Joules like that. There is finite thermal conductivity.

You’re giving off big vibes like this:

Ironically enough I got a lot of heat for taking the optimistic side of argument on the heat shield thread, which is full of "it's gonna burn up on re-entry" doomsayers.

Using similar simplified physics models.   Which very few (or none) cared to actually counter with their own model.

This is nearly the exact same situation. Jets of hypersonic air, with the thermal and kinetic energy being turned ultimately into heat of the surrounding air, not primarily into the vehicle or launch infrastructure. It just becomes a few cubic kilometers of air heated a few degrees warmer.

Just to be clear, you and others have persuaded me that the 46% of energy that is thermal won't be much of a problem for the OLM itself.  The kinetic energy, OTOH.

Ironically the picture you posted (space shuttle launch) has a a trench and deluge system that is capable of absorbing a similar amount of kinetic and thermal energy put out by the engines from engine run-up to clearing the launch tower.

I found the same for Saturn-V, Shuttle, SLS, and Soyuz-2. The back of the envelope calculations show the energy being directed to a place where it can be reasonably all dissipated without contacting at full force critical infrastructure like conduit, pipes, hydraulic lines, etc (or sending massive shockwaves to same).

OLM doesn't have any of that.  The energy is going to go everywhere into and including all that nice piping/conduits on the OLM and tower.

I'm surprised nobody has brought up the idea that back of the envelope  assumptions is why all those other systems are overspec'd and SpaceX figured it out.  It's a distinct possibility.

[DanClemmensen]

I found the same for Saturn-V, Shuttle, SLS, and Soyuz-2. The back of the envelope calculations show the energy being directed to a place where it can be reasonably all dissipated without contacting at full force critical infrastructure like conduit, pipes, hydraulic lines, etc (or sending massive shockwaves to same).

OLM doesn't have any of that.  The energy is going to go everywhere into and including all that nice piping/conduits on the OLM and tower.

I'm surprised nobody has brought up the idea that back of the envelope  assumptions is why all those other systems are overspec'd and SpaceX figured it out.  It's a distinct possibility.

If most of the energy is being convected away, then the area of the egress from the enclosed space is relevant. The OLM is quite tall so the six egress apertures must have a fairly large area. How does this compare to the cross-sections of the flame trenches of those other systems at their narrowest points? the initial flux would be the thrust divided by that area, to a rough first approximation.

[InterestedEngineer]

Remember also that you've only calculated the energy input by the engines into the exhaust gasses. You also need to calculate the transfer from those exhaust gasses to any structures, you cannot simply assume a perfect coupling (it would actually be incredibly hard to capture the total energy from the exhaust gasses, its's why steam-ejector test stands have such small engine capacities).

True, but I'm quoting energies that are by orders of magnitude big enough to break things.   It only takes a small percentage of the total exhaust energy going the wrong place to break things.

Here's a very rough calculation on one Raptor's exhaust impinging on a 10cm x 3m piece of conduit (RMC for example).  Note this is kinetic energy, no thermal considerations.  I was wrong about the thermal.

The velocity pressure of Raptor exhaust is 3.4MPa (exhaust density is 0.67kg/m^3, plug that into Pv = pV^2 / 2).

The resulting force on that piece of conduit is 1MN (105 tons).  The RMC itself can take it, it's rated to 80+MPa, but the fasteners rated for that schedule of RMC is 15 tons (schedule 8, 5/8"), and that fastener is going to fail.  By almost an order of magnitude.  I can be off by 7 times and the fastener will still fail.

That's one Raptor exhaust.  Now do 33, and that exhaust isn't redirected by anything, it's bouncing off the ground.  1/33 of the exhaust hitting one of those pipes will break its fasteners.  More accurately, the failure point is at 1/33 / 7 = 1/231 of a raptor exhaust.

I've view the fastener schedules by pics, I don't think they have 8 fasteners every 3 meters.

This is probably why exhaust is redirected away from critical components in all the other designs out there. It's a guess, but an educated guess.

Hey, I'm probably wrong.  I'm going up against SpaceX engineers.  I'm trying to figure out *why* I'm wrong.  All the existing launch systems match the simplified physics model very nicely.   They don't assume that the exhaust energy, thermal and kinetic, just poof away over a cubic kiometer.  Instead, the direct it away from critical infrastructure, with flame trenches and/or water in enough quantities to absorb the energies being emitted by the engines.

[InterestedEngineer]

From:  https://www.nasaspaceflight.com/2021/10/starship-orbital-launch-pad/

The water tank has a capacity of around 1,000,000 gallons of water. For reference, the water tower at the Kennedy Space Center’s LC-39A has a capacity of 300,000 gallons.

1,000,000 gallons is enough energy dissipation capability to match SLS, Saturn-V, Shuttle, and Soyuz designs.

Is that tank still being used for water?

And how is it being distributed to the OLM?  I don't see the piping.

I'll laugh if it was there all along and nobody said anything

[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.

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.

That's form propellant efficiency.

Energy efficiency is the exact inverse.

A higher ISP engine is less efficient, energy wise.

For chemical rockets, the distinction is meaningless since the energy is stored in the propellant but when discussing energy impact on the pad it becomes relevant.

With non-chemical rockets it becomes important and is actually more interesting when considering high-dV trajectories - but that's OT.

Think about it this way:  using half the propellant flow at twice the exhaust velocity gives you the same impulse, but expends twice the power.

There a thought experiment about an asteroid deflecting engine that uses solar power and ISRU asteroid dust for reaction mass.   Easy to show that the slower you throw out the asteroid dust, the better.

[InterestedEngineer]

I found the same for Saturn-V, Shuttle, SLS, and Soyuz-2. The back of the envelope calculations show the energy being directed to a place where it can be reasonably all dissipated without contacting at full force critical infrastructure like conduit, pipes, hydraulic lines, etc (or sending massive shockwaves to same).

OLM doesn't have any of that.  The energy is going to go everywhere into and including all that nice piping/conduits on the OLM and tower.

I'm surprised nobody has brought up the idea that back of the envelope  assumptions is why all those other systems are overspec'd and SpaceX figured it out.  It's a distinct possibility.

If most of the energy is being convected away, then the area of the egress from the enclosed space is relevant. The OLM is quite tall so the six egress apertures must have a fairly large area. How does this compare to the cross-sections of the flame trenches of those other systems at their narrowest points? the initial flux would be the thrust divided by that area, to a rough first approximation.

The OLM is about 30 meters tall by 13m in diameter.  I'm guessing the base is about the same diameter as the outer edge of the ring.

Over 30 meters The exhaust plume only spreads to 5 * 0.67 = 3.35 meters diameter per the data SpaceX provided NASA.  So the exhaust plume right at lift off, when it hits the base of the launch mount, is going to be ~10+3.35 = 13.35m.

The outer edges of the plume are going to be barely wide enough to contact the inner edges of the legs when the plume hits the ground.  The gasses aren't (for example) going to exit sideways 5 meters below the ring.  So most of that opening is effectively unused.

Then the exhaust gases are going to compress and rebound.  They aren't going to make a 90 degree turn.  All the other exhaust trenches I've been able to find have some sort of bend radius that is less than 90 degrees.

The rebound (and shockwaves) are what is going to cause damage to the OLM (leaving the tower out for now)

Assuming the high pressure area is 10m high for sake of back of envelope-ness, it's a cylinder 13m diameter and 10m tall or 1,300m^2 for 26m^2 of high velocity pressure to exhaust into, or 50x area reduction.

That'd be a 7x reduction in velocity, or 458m/sec, so gasses will be exiting the base of the launch tower at greater than the speed of sound.

EDIT: At equilibrium with the atmosphere, the reduction is 34x (3.4MPa/100kPa), which woulud be 884m^2 or ~7m high, and 500 m/sec velocity

You might be onto something. Perhaps the design builds up a shockwave at the base of the tower 7m high or so that acts as a "virtual trench" with a reasonable bend radius.   That would be something only someone with a numerical simulator could figure out.  SpaceX is world class at that kind of thing.



https://www.faa.gov/space/stakeholder_engagement/spacex_starship/media/Appendix_G_Exhaust_Plume_Calculations.pdf

[InterestedEngineer]

How does this compare to the cross-sections of the flame trenches of those other systems at their narrowest points? the initial flux would be the thrust divided by that area, to a rough first approximation.

You are on to something.

A fluid flow against a flat plate produces a curved rebound layer.

Aka the "virtual trench".

In this video, the math is demonstrated:

[InterestedEngineer]

How does this compare to the cross-sections of the flame trenches of those other systems at their narrowest points? the initial flux would be the thrust divided by that area, to a rough first approximation.

You are on to something.

A fluid flow against a flat plate produces a curved rebound layer.

Aka the "virtual trench".

In this video, the math is demonstrated:

So, with the virtually curved flame trench, the equilibrium point will be somewhere between 7-14 meters off the ground (where all the exhaust exits from underneath the launch mount at slightly above atmospheric pressure, but at very high velocity and temperature), and judging from the static fire videos, create a triangle of exhaust with base angle of 30 degrees of exhaust as seen from the ground.

Such a triangle radiating away from the base of the launch mount has a volume that is 1.8r^3.

To get equivalency to the other major launch designs requires about a billion cubic meters of air to absorb the kinetic and thermal energy of the launch.

That's a radius of 822m.

So the "virtual exhaust trench" is the ~800m around the OLM.  The air in that volume will be accelerated to 45m/sec (100mph), and heated by 1degC.

There's still the matter of 500m/sec velocity exhaust hitting the base of the tower, debris launched to supersonic speeds, pipes getting impinged on by supersonic flow, 180 degree rebound until the rebound "virtual trench curve" boundary layer can be established.  (slow initial ignition should help with that.  Was wondering about the 1,3,3 sequence).

And what exactly in the 800m around the OLM will get blasted with hurricane force winds.

But it's not completely insane.

[Robotbeat]

nearly all of the energy ends up as heat somewhere for the first 2 to 5 seconds of thrust , I.e. the only part that really matters.

Jets of air going supersonic speeds either interact with stationary objects and do localized compression heating or they mix with the air or what have you but it all ends up as heat and almost all of that heat is in the air because air mixes well and conducting through solid surfaces is extremely slow at these scales.

The Jets of gas are moving hyper sonically so in just a second or two they have spread out very far from the rocket and the orbital launch mount, carrying their heat and their kinetic energy far from the orbital launch mount. Just a few hundred meters are enough to dissipate the gas two volumes of air containing thousands of tons of air molecules. That is where the heat is going to primarily. If you need a little bit of water or ablation to keep heat from soaking into parts of the orbital launch mount, that is fine. But it’s absurd to demand that water have to be flashed to steam to carry away most of the energy. It’s not necessary as there’s just not enough contact time to dump much energy into the thick metal structures.

Do the thermal conductivity calculation.