sevenperforce

Very interesting. I wonder if the unexpected differences in the boundary layer cooling system are linked to the inherent challenges of operating a combustion tap-off cycle.

FT = ve * me / t Thrust = exhaust velocity times exhaust mass divided by time. If exhaust mass is zero and thrust is nonzero then exhaust velocity is infinite, and that's impossible, because you can't exceed the speed of light. If you're producing thrust, you're pushing against something. That's fundamental. It's the third law. For once, I would like to see one of these reactionless drive folks explain exactly what is being pushed against in a way that doesn't break physics. And you can't just say "oh it pushes against virtual quantum particles" because if you're pushing against virtual quantum particles then the particles you push against stop being virtual and start being actual. And so you have a flow of mass. But virtual particles come in pairs, and so if the particles you push against stop being virtual and start being actual, their pairs also stop being virtual and start being physical, and you have an equal flow of mass in the opposite direction, and so you don't move. If it's pushing against dark matter, then yes, we've got something. That could work, if you're stumbled upon a mechanism for dark matter interaction that no one else has thought of and that never manifests in nature. But dark matter is rather low density out here, so far from the galactic core, and so your math is going to have to take that into account and somehow come up with a match to your claimed acceleration. Show the maths.

If you’re worried about overly-energetic thrust at landing, just inject extra inert propellant into the exhaust steam to increase thrust and decrease energy. You don’t need a whole new thread for this solution.

It’s so smol!!

As I understand it, the current space suits that NASA will be buying for EVA use on the moon will still use a water-based open-loop evaporator for cooling the suit, limiting the time-persistence of EVAs. What if you could use a radiative solution? The human body produces around 200 W/m^2 of heat during moderate to high metabolic activity. Using the Stefan-Boltzmann law with a maximally efficient radiator (if my math is correct), you only need your radiator to operate at 242 K, well below the temperature of the human body. Of course you won’t have a maximally effective radiator. And you’ll also have to deal with the thermal conditions on the moon. Half of your body will be in direct sunlight, absorbing 1368 W/m^2, while the other half will be absorbing lunar-reflected sunlight at 97 W/m^2. Adding this average to the average metabolic output of a human, the actual heat rejection requirement is 932.5 W/m^2, which requires a radiator temperature of 358 K, which at 85°C is MUCH higher than the temperature of the human body. But what if there was another way? Imagine the outer surface of a spacesuit covered in cells with a white substrate that ordinarily reflects solar radiation, but can fill with a carbon-doped coolant with high emissivity. The cells in direct sunlight are kept empty of coolant so that they reflect heat, while the cells in shadow are saturated and reject heat. The cells in shadow will still be absorbing light reflected by the lunar surface at 97 W/m^2. Let’s imagine for the sake of argument that the empty cells will have 97% reflectance, 3% emissivity, and 3% absorption, while the saturated cells have 97% emissivity, 3% reflectance, and 97% absorption. Half of the suit exterior, then, will be absorbing 41 W/m^2 while the other half is absorbing 94.1 W/m^2, all while needing to reject the 200 W/m^2 that the body inside is producing. The total heat rejection required is therefore 267.5 W/m^2. Half of the body has 97% emissivity and half has 3% emissivity, so the average emissivity is 50%. The radiative heat rejection capacity needs to be 535.1 W/m^2. By the Stefan-Boltzmann law, your radiator only needs to be operating at 38.5°C. Obviously that’s still above human body temperature, but not by much. Assuming a Carnot coefficient of performance on the order of 3.0, a compressor-based heat exchanger would only need to provide the equivalent of 7.5 degrees of thermal differential to keep the astronaut at the equivalent of a balmy shirtsleeve environment. I can’t remember the thermodynamics for that, but it can’t be much energy.

One thing I didn’t discuss above is the advantage created from ram air effect. A turbojet engine operating on the run away is using a significant portion of its reaction energy to compress atmospheric air and push it into the combustor. This creates a limit on the static thrust that the engine can produce. However, once the engine really starts to get moving, the air begins to be compressed simply by the forward motion of the engine forcing the air into the intake. This means the compressor doesn’t have to do nearly as much work, which means the turbine isn’t extracting quite as much energy out of the exhaust stream, which means higher exhaust velocity and better specific impulse than you would otherwise expect. Unfortunately, the ram air effect really doesn’t provide much help until you are starting to edge toward transonic velocities. Jet engines can be designed to fly at well above the speed of sound by using a much more fuel-rich combustion and thus pumping more energy into the exhaust. Even so, the region in which the ram air effect starts to provide a significant advantage tends to be rather narrow, because even the most energetic hydrocarbon fuels soon reach a point where they can no longer overcome intake drag. There are scramjet designs using liquid hydrogen to dump as much energy into the exhaust as possible and thus push the airbreathing mode to the max possible velocity, but hydrogen is fluffy and not great at providing thrust at the initial low speeds.

Not to mention that liquid hydrogen is self pumping thanks to the expander cycle.

It definitely looks like the crossover is lower than I had previously thought. But (bringing this back to the original thread topic) the black tiles of the Shuttle were nevertheless chosen for their ability to radiate heat out. We can do the same math as before, but instead just jump directly to the radiative cooling capacity of the Shuttle tiles. The Shuttle tiles reached temperatures of 1533 K and thus would have been rejecting a peak of 313.2 kW/m2, per the Stefan-Boltzmann law. If the surface area of the Stoke upper stage heat shield is 16 square meters, then that's 5 MW of heating we need to reject. Rather higher. But if radiative heating isn't a major issue, then we can perhaps have some of that heat transfer by convection into the hydrogen boundary layer we create. Water is a very effective coolant in terms of volume, but it doesn't come anywhere close to what liquid hydrogen can do in terms of mass.

Modern turbofan engines produce an exhaust speed of approximately 500 mph (223 m/s) at the exhaust nozzle. A large engine like the GE90 which produces up to 432.8 kN of thrust with a bypass ratio of 9:1. If you divide 432.8 kN by 223 m/s, you get a total reaction mass flow of 1941 kg/s. However, the thrust-specific fuel consumption of the GE90 is only 7.9 grams per second for each kN of thrust, meaning that its actual fuel flow is only 3.42 kilograms per second. So the bulk of its thrust comes from the ~1,938 kilograms (1.9 tonnes) of air it's pushing through the engine each second. That bypass ratio of 9:1 means that only 194 kilograms of air is actually flowing through the engine each second; the other 1,744 kilograms of air goes through the fan bypass. Without that fan bypass, the thrust would be much lower, but the energy going through the engine would be the same, and so the exhaust speed would be correspondingly higher. Do the math and you get an exhaust velocity of about 350 m/s with a total mass flow of 197 kg/s: 69 kN. So what's the specific impulse of a pure turbojet without the bypass fan? Well, it would be 35.7 seconds. That's...terrible. Absolutely terrible. If your specific impulse is that low, you will not go to space today. However, a turbojet gets by because the only propellant flow that actually matters is coming from the fuel it carries, and the fuel it carries is just 1.52% of the total propellant flow. Subtract out the mass of the airflow and do the math again, and boom: your pure turbojet now has a specific impulse of 7,930 seconds. That's fantastic! The problem, of course, is that your pure turbojet will only be able to maintain thrust at relatively low speeds. The faster it goes, the faster the airflow into it, and the less work it is able to do. Net thrust is the momentum of the exhaust coming out of the back of the engine MINUS the intake drag: the momentum of the air coming into the front of the engine. At 200 m/s airspeed, its net thrust is the 69 kN of thrust out the back minus 38.8 kN of intake drag or a total of 30.2 kN, and so your effective specific impulse drops to 900 seconds. At 250 m/s airspeed, intake drag reaches 48.5 kN and so net thrust drops to 20.5 kN, a specific impulse of 611 seconds. At the speed of sound -- 343 m/s -- intake drag is 66.5 kN and so net thrust is just 2.5 kN, a specific impulse of 75 seconds. At 355 m/s, net thrust is zero: all of the thrust out the back of the engine is being used to counteract the intake drag, and your specific impulse is zero. Unfortunately, 355 m/s is only 4.5% of the velocity you need to reach orbit. You can, of course, carry your own oxidizer (in the form of liquid air or simply liquid oxygen) in tanks. As your speed increases, you can slowly close the intakes to reduce intake drag, and you can inject liquid oxidizer into the engine to make up for the lost oxidizer. But now you're going to have to include the mass flow of your liquid oxidizer in the equation. And so by the time you've completely closed your intakes and you're now relying entirely on liquid oxidizer, you're still only getting that same specific impulse of 35.7 seconds I calculated above. You'll need a much more efficient rocket engine to get to space.

A scramjet only gets 3600 isp because you aren't counting the mass of the oxidizer, since it's collected as you go. If you're bringing your oxidizer with you, then you have a rocket, not an airbreather. Airbreathing engines don't get high efficiency by magic; they actually get lower efficiency than a rocket BUT they benefit because they don't have to carry their oxidizer with them, so you don't have to count it as part of the propellant mass flow. Specific impulse is thrust divided by mass flow rate. In a rocket engine, the mass flow is all the propellant: both fuel and oxidizer. In an airbreathing engine, the mass flow is the fuel alone. But once you're bringing along liquid air to operate your engine, you now have to count the liquid air in your mass flow, so you're no longer getting 3600 seconds of specific impulse.

Getting close to launch now.

Wouldn't they need to get to 29.4 psi to prove space worthiness?

It's not entirely clear what you're proposing here. Are you saying "let's bring along liquid air so that we can feed it to our scramjet once it runs out of atmospheric air"? If so, congratulations, you've invented rocketry.

The weather forecast says partly sunny here at 6 PM but partly cloudy by 7 PM. Winds are largely SSE so hopefully we get an unobstructed view of the eastern sky.

Comparable temperatures, but much lower pressures. However, rocket combustion chambers employ active cooling, which heat shields (up until now) haven't had. Of course, you can consider ablative cooling to be a form of active cooling. That was my primary point: that heat shield tiles were black because it was necessary that they radiatively cool themselves. I think you're making an assumption here about what the "regime of typical LEO spacecraft entries" constitutes. It's not an either/or distinction between radiative and convective heating. Rather, re-entries start with radiative heating as the dominant factor, and then the radiative heating decreases while the convective heating increases. You're correct that certain vehicles are designed to accept convective heat and then radiate it away. However, the image you provided above (Figure 10) isn't applicable to LEO entries generally. The source you cite says they "apply only to single-pass, nonlifting, parabolic-velocity entries." The real numbers are more complicated. The actual numbers can be found in this presentation, which explains the relationship between convective heating and radiative heating based on multiple factors. Importantly, one of the factors is the effective radius of the vehicle: Qconv ∝ v3(ρ/R)0.5 but Qrad ∝ v8ρ1.2R0.5 As the effective vehicle radius increases, convective heating decreases, but radiation heating increases. For something like the Shuttle, which had a very deep atmospheric entry (to allow a lifting entry) and a very high effective radius, it would make sense that radiative heating was a major component.

Interestingly, the chopsticks in Florida are significantly shorter than the ones in Texas:

Not to mention especially heavy.

Is it at ignition or later? At ignition it’s TEA-TEB.

Looks like a pretty normal rear entry suit with a hard torso, limited shoulder mobility, and standard legs and gloves.

I don't think this is a useful comparison. The blackbird was a dart and lacked the insulating boundary layer that blunt bodies produce between the bows shock wave and vehicle. That was my point. Not unless you have the correct math. What spectral bands was your reflectivity testing in? Infrared light is not visible light. Many things that are opaque to visible light are transparent to infrared. Although the grey-black tiles on the bottom of the Shuttle absorbed the majority of visible light, they were reasonably reflective of infrared radiation. You're not doing the math. Imagine your heat shield needs to get rid of 98% of the incoming energy. If you have a white heat shield which reflects 90% of incoming energy but can only radiate away 50% of what it absorbs, it will get rid of 95% of incoming energy, which means it fails. If you have a black heat shield which only reflects 75% of incoming energy but can radiate away 93% of what it absorbs, then it will get rid of 98.25% of the incoming energy, so it survives. The leading edges of the wings had black carbon-carbon composite; the entire underside of the orbiter was black tiles. It did. That's why you need a thermal protection system on the leeward side of a re-entry vehicle. Look, for example, at the metallic reflective heat shield on the lee side of Orion. However, as plasma starts to flow around a vehicle, it expands. Remember that the reason plasma is plasma in the first place is that it is air compressed by the vehicle's high velocity; the air literally cannot get out of the way fast enough. Once it starts to expand, it isn't nearly so hot, and so you can have a backshell thermal protection system which can function by reflecting heat and doesn't need to radiate away heat at all. It was. Black was chosen despite being less reflective, because it cools itself radiatively much better than a polished reflective surface. It does, in fact, appear to be the case in LEO. And it is, in fact, the case in LEO. The actual area of plasma producing that amount of radiation is very large. That's why you can see shooting stars. Definitely no eye or skin protection visible in this photo:

Under three days.

I'll be in administrative law class then but maybe I can duck out if it's clear.

Thanks, that changes the math somewhat. Source? I can understand, lol. It's a lot of work and people unfamiliar with high-temperature physics won't get it immediately. You are incorrect. Initially, there is convective and conductive heating as molecules bounce off of the surface at high speeds. The extreme heating of the SR-71 windscreen, for example, is wholly the result of convective and conductive heating. However, if that's all you had to deal with, you could simply use a heat shield with low thermal conductivity and thus eliminate the heat transfer completely. The amount of energy transferred to the vehicle from the surrounding medium by conductive and convective heating goes up with the third power of vehicle velocity; the amount of energy transferred to the vehicle from the surrounding medium by radiative heating goes up with the eighth power of vehicle velocity. So when you're moving at re-entry speeds, radiative heating dominates. Sometimes intuitions based on simple thought experiments are correct. This is not one of those times. The Fibrous Refractory Composite Insulation (FRCI) tiles on the Shuttle were black because of that T4 term in the Stefan-Boltzmann law. Even though the color black is not as reflective as colors like white and silver, black is a much more efficient emitter of thermal radiation than white or silver, and so the Shuttle tiles functioned by accepting as much heat as the plasma could produce and then just re-radiating that heat away from the vehicle. Ablative heat shields do quite a bit of radiating, but they also dump a lot of heat by allowing the oils impregnated into their carbon structure to burn away and carry that absorbed radiative heat away from the vehicle. For an actively-cooled low-temperature heat shield like the one on Stoke's vehicle, the shield never gets very hot and so the emissivity of the shield isn't as meaningful because the Stefan-Boltzmann law doesn't give as much of an advantage.

Interesting if true. They are pulling the booster off of the launch mount and parking it next to the tank farm in order to do work on the OLM. You don't want to have human workers working underneath a suspended 250-tonne booster.

The area of a spherical cap with those dimensions is 18.22 m2 and it's shaped like this: That seems like the wrong shape. I'm thinking that the correct shape is probably something more akin to an ellipsoidal cap rather than a spherical cap. Probably closer to 16 square meters total. Calculating the necessary coolant flow rate is going to take some effort. You're going to model the re-entry plasma as a 2000°C surface at constant temperature with an area equal to the heat shield. So you can just take that linearly, using the Stefan-Boltzmann law. The power per unit area coming off of a blackbody is given by j* = σT4, where σ is the Stefan-Boltzmann constant, 5.67e-8 W/m2K4 in Kelvin. So here the area-specific energy flux we're gonna need to deal with is 1.51 MW/m2. What's the heat shield made of, anyway? Stainless? Aluminum? I can't remember. Let's assume aluminum. The incoming energy is radiative, so you're going to need to look at the reflectivity of aluminum in the spectrum where the 2000°C Planck peak is located. The wavelength location of the Planck peak is given by Wien's displacement law, where λ = b/T and b is Wien's displacement constant (creative nomenclature here, I know) which is 2898 μm*K. So the plasma shell will have its radiative peak at ~1.27 micrometers which is solidly in the center of the infrared spectrum. Bare aluminum has a reflectivity of approximately 92% in infrared: So that means the heat flux we actually have to deal with is only going to be 1.51 MW/m2 x 8%, or 120.8 kW/m2. Slightly more manageable. Now, before we bother with multiplying by the size of the heat shield, we also get to look at the blackbody emissions of the aluminum itself. Let's say we try to hold the heat shield at 500ºC, a solid 160 degrees lower than the melting point of aluminum. At 500°C or 773.15 K, a perfect blackbody radiates (according to the Stefan-Boltzmann law above) at 6117 W/m2. Anodized aluminum has a blackbody emissivity of 77%, so the aluminum is going to reject about 4.71 kW per square meter. That's not much, but it's something, reducing our heat rejection needs to 116.1 kW/m2. (Note that the specifics of the metallic heat shield alloy are actually pretty important here. The more reflective you can make your material in the infrared, the better...but at the same time, it also benefits dramatically from a higher melting point because the Stefan-Boltzmann law makes radiative energy transfer go up with the 4th power of temperature. Stainless steel has a melting point about 2.7X as high as aluminum which translates to 42 times more blackbody energy transfer being shed before active cooling, assuming similar emissivity.) While the hydrogen gas coolant bleed from the center of the heat shield (or elsewhere) will make the boundary layer larger and cause some additional expansion which will generally reduce coolant requirements, it will not directly absorb any of the radiation coming off of the plasma. Diatomic hydrogen is completely transparent to infrared light. Assuming 16 square meters of heat shield need to be cooled, you're going to need to reject 1.86 MW of thermal energy during re-entry. If we just use the enthalpy of vaporization for liquid hydrogen, that would come to 4.23 kg of liquid hydrogen per second, slightly more than the hydrogen mass flow rate of an RL-10 engine at full throttle. However, that's worst-case scenario, and we aren't accounting for the fact that the hydrogen will also be expanded through a turbine, which converts a portion of that thermal energy into work (to pump itself). The Stoke Space upper stage is comparable to the size of Crew Dragon, which has a plasma blackout of about 4-6 minutes. Now, the upper stage is much lighter than Crew Dragon and thus will slow down more quickly, but even using the numbers above and a 5-minute re-entry period, that would require the vaporization of just 1.27 tonnes of liquid hydrogen used up for active cooling, which is really quite modest.