sevenperforce

By my measurements of that image I'm getting a volume of 322.7cubic meters for the LOX tank and downcomer and a volume of 278.9 cubic meters for the LNG tank. Assuming non-densified propellant that gives 486 tonnes of propellant. But that's not accounting for ullage, etc., or launching with the possibility of slightly less propellant as @tater noted.

My standard concept for any sort of near-future spaceship propulsion is a trimodal nuclear-thermal NTR running on something like methane or propane (for reduced tank volume) with a LOX supply for afterburning and a closed-loop methane radiator system for generating electricity from the reactor during coast.

Propalox, gas generator on the first stage, assumptively GG second stage, 9+1 configuration. Only two meters wide, so slightly wider than electron which makes sense given propane's density compared to kerosene. 1 tonne to LEO, which definitely beats out Electron. I don't believe there's ever been a propalox rocket that has actually reached orbit so this would be a first.

And if you do, make sure you have plenty of time. The place is huge. And if you come to Air and Space, let me know since I live in DC; we'll get a beer beverage.

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.