Is there a quick and dirty way to convert the exhaust velocity of a rocket into the effective temperature of that rocket exhaust?
I don’t mean exact conversions, I’m looking for ballpark figures here.
Say I have a futuristic nuclear rocket that spits out helium at an exhaust velocity of 4.4% of the speed of light (13,200,000 meters per second). Would this exhaust have a temperature of 30 million Kelvins, or 3 billion Kelvins, or 30 billion Kelvins?
I tried using one of the gas laws, T = mv[super]2[/super] / 3 k[sub]B[/sub], where m is the mass of a single molecule, v is the root-mean-square velocity of each molecule, and k[sub]B[/sub] is the Boltzmann Constant (1.38 x 10[super]-23[/super] Joules per Kelvin). I got an answer of about 30 billion Kelvins. But when I tried applying this same formula to the Space Shuttle’s Main Engines (4500 m/s exhaust velocity, exhaust consisting mostly of water molecules), I ended up with 14,000 Kelvins, which is about 10 times what the actual temperature of the SSME’s exhaust.
I’m sure someone else can answer this better, but I’m pretty sure the answer is no–at least not without more information.
Temperature is measured in the rest frame of the gas. So you can’t just say that the exhaust velocity of the gas directly translates into temperature. A simple way of looking at it: what if your rocket was a mass driver that shot out ice? The exhaust temperature is obviously somewhere south of 0 C, despite having a large velocity.
For ordinary rockets, the temperature is of course one component of the behavior inside the combustion chamber–but the “magic” happens with the nozzle, where (for reasons that are beyond my knowledge), one can dramatically increase the gas velocity (to well in the supersonic) at the expense of pressure and temperature.
Of course one could figure all this out given enough knowledge of the nozzle and such. And there are equations for simple nozzle types, but they’re still more complicated than simple gas laws.
Well, ideally, it wouldn’t glow at all–that’s wasted energy.
That said, you could design a rocket to have any temperature exhaust you wanted. You would need some kind of electronic or magnetic containment, since no material will withstand these high temperatures, but for a hypothetical rocket that’s no problem.
A correction, though: a blackbody will never appear violet. Yes, as temperature goes up the peak wavelength emitted will move up the spectrum into violet, UV, etc., but because blackbodies emit a distribution of wavelengths, and because of the way the eye works, you’ll never see violet (or green, for that matter).
I think it’s a reasonable approach to remark that gas molecule velocity maps to temperature if you take out the question of direction of travel, and ask what the temperature would be if you randomized the molecule directions.
I think air molecules at 300 K travel according to kinetic theory at maybe 20% over sonic velocity (their speed is the main determinant of the speed of sound). I also think their velocity varies as the square root of temperature. If the temperature was 3000 K, which is glowing dazzling white hot, they’d only be going about 1.5 times faster.
That being said, something feels skeevy to me about this. Seems like I could go through the same thing for compressed air and demonstrate that it leaves its tank through a hole at zero velocity.
The key that makes this work is that in a rocket (or at least, what folks usually call a rocket), the gas is leaving the nozzle precisely because it’s hot.
So what you do is, first you convert your temperature to an energy by multiplying by Boltzmann’s constant (strictly speaking, there’s a dimensionless factor of order 1 in here, also, but the OP is looking for a quick and dirty ballpark). Then, you plug that energy and your particles’ mass into the kinetic energy equation E = 1/2 mv[sup]2[/sup], and solve for v (0.044c is slow enough that the Newtonian formula suffices for quick-and-dirty).
In brief, no. The effective exhaust velocity, c, is defined by the exhaust velocity at the exit plane, the area of the exit plane, the mass flow rate of propellant, and the difference in pressure between the exit plane and ambient. (This relationship is specifically for converging-diverging nozzles, but the same general principle applies to any kind of thrust reaction.) The characteristic velocity, c[SUP][/SUP], is defined by the chamber pressure, area of the nozzle throat, and mass flow rate. c[SUP][/SUP] is used for comparison of the relative performance of different propulsion systems as it is independent of nozzle characteristics. See Sutton’s Rocket Propulsion Elements, Section 2.3 for more details.
The temperature of the exhaust is dependent upon the thermodynamics of the reaction of the propellant (for chemical propellants) or other thermal energy/work source (for a nuclear or ion propulsion systems). For chemical systems, the chamber temperature is dependent upon the energy release of the exothermic chemical reaction and the products of the reaction (including uncombusted propellants). The temperature loss of the exhaust via expansion through the divergence are of a nozzle is defined by the ideal gas law through an isentropic nozzle with modifications for real combustion and gas expansion phenomena, boundary layer losses, flow instability and multiphase flow interactions, et cetera.
[QUOTE=Napier]
I think it’s a reasonable approach to remark that gas molecule velocity maps to temperature if you take out the question of direction of travel, and ask what the temperature would be if you randomized the molecule directions.
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The problem with this approach is that heat is due to randomized motion. A flow field moving in a defined direction does not have a high inherent temperature due to its momentum. Exhaust can be hot and slow, or cold and fast and in fact the latter will give the highest energy efficiency, which is why high temperature ion propulsion systems have very high specific impulse (propulsive efficiency) but very poor energy efficiency. An ideal rocket with perfectly expanded flow would extract all mechanical thermal energy out of the propellant before it leaves the nozzle, with the pressure at the exit pressure equal to ambient (although this is never practical in vacuum duet the huge size of the nozzle it would require). The exhaust temperature of the exhaust plume is generally pretty close to the temperature in the combustion chamber, especially with high molecular weight products.
Okay, then – if you were looking at a blackbody that was at a temperture of a hillion jillion skillion degrees, what color would it appear to be? Would it be bluish-white, pure blue, or blue-violet? (A quick Google search failed to give me an answer to this.)
It’s pretty much all bluish-white past 8,000 degrees or so. However, a surface of fixed area would get brighter and brighter as the temperature went up. The exhaust wouldn’t stay at this temperature for long, since it’s radiating energy so quickly. So the part of the exhaust closest to the ship would undergo a rapid brightness decrease with little color shift as you move farther away, until you get to about 8,000 K and you get a much slower color shift to the reds (and a further brightness decrease).
I hope you’re wearing protective goggles. You can’t see it, but your hojillion-degree torch is blasting out hard UV at best, and nasty gamma rays at worst.
Maybe the OP is thinking of the stagnation temperature, which does depend on the exhaust velocity, but that would have nothing to do with the color of the exhaust.
Stagnation temperature is the temperature of a parcel of fluid brought to rest without any transfer of heat into or out of the parcel. It is not the temperature of the actual, moving parcel of fluid. The only reason I brought it up is because it’s the only temperature I know of that is a direct function of the flow velocity, which is what you were talking about in the OP..