ISP in relation to engine temperature.

[PTNLemay]

I've heard that nuclear thermal rockets (like we use in the game) are limited by their melting point. The hotter they run the more efficient they can be, but beyond a certain point the materials we use will just melt (causing the small reactor to actually melt down and spray hot uranium out of the exhaust. Or... just cause it to explode). Disregarding all that melting point problem, pretending we have a super material that can withstand really really high temperatures, I want to know exactly how increasing the reactor's temp would increase efficiency, namely how much it would increase exhaust velocity.

For example the game's NTR has an ISP of 800s (or around 7800 m/s exhaust velocity), we don't know how hot the reactor runs, but we can guestimate it's somewhere between 500 C and 1000 C. If I... doubled that to ~2000 C, by how much would the exhaust velocity increase? I'm guessing it's more complicated than just doubling the ISP.

I found this equation on the Atomic Rockets site: Ve = sqrt( (2 * E) / m )

in which Ve is exhaust velocity, E is energy, and m is mass of the fuel. We know the fuel is something like uranium, so using the Heat Capacity equation (Q = m * c * deltaT) we should be able to find out how hot a given amount of uranium will have to be to yield a given exhaust velocity. (C for uranium is 0.116)

So for the game's engine we have

mass = 2250 kg

Ve = 7848 m/s

E = ?

Q = E

c = 0.116

delta T = the final temperature.

Isolating E gives us 6.9 x10^10 joules. Putting that into the heat equation gives us... 265 million degrees. So I figured, okay, maybe the mass is a bit small for a nuclear reactor, let's tweak the game's units and see what happens. I gave it a mass of 10 tons, and ran it again, and... it looks like the mass units in each equation cancel each other out. What this means is that regardless of what mass we pick, the given energy density is the same and out comes 265 million degrees.

I'm sure a typical NTR wasn't meant to run that hot. So am I missing something in my equations here? Here's where I got them:

http://www.projectrho.com/public_html/rocket/engines.php#id--Exhaust_Velocity

http://en.wikipedia.org/wiki/Heat_capacity#Table_of_specific_heat_capacities

Further References

http://en.wikipedia.org/wiki/Gas_constant

http://en.wikipedia.org/wiki/Heat_capacity_ratio

Edited October 5, 2013 by PTNLemay

[Kryten]

Are you sure that 'M' is the mass of the fuel rather than the reaction mass? It looks like it'd make more sense for it to be the reaction mass.

[PTNLemay]

Hmm... well, that would make sense. But it would also complicate matters. The fuel changes constantly over time as the engine is burning.

Unless it means the "fuel rate", so at any given second how much fuel is running through the engine, and thus the temperature given would be how hot that fuel is running through the engine (on average? or as it exits the engine?). That's interesting, but for all I care the fuel can be as hot as it wants when it leaves the engine, what I want to know is how hot the engine itself gets... so I can tell the approximate difference between a NTR running at 600 C, or an NTR running at 2000 C (where all other variables remain roughly similar). But maybe these equations won't provide that?

Maybe Q and E aren't the same thing? I always assumed Q was just the heat's value in joules depending on those other variables (mass and C).

Edited October 4, 2013 by PTNLemay

[Kryten]

If it was power, I'd try to extract something like 'fuel rate', but given it's total energy I'd just try and stick in the the total amount of remass. Working out the total amount of energy given to the remass over the length of the burn gives you the power output of the engine, which you should be able to use to get temperature.

[PTNLemay]

The strange thing is that the mass cancels out. So regardless of what it represents it's not going to change the result.

Ve = sqrt( (2 * E) / m ), replace the E with the heat equation and you get:

Ve = sqrt( (2 * m * c * deltaT) / m). M divides m and we just get:

Ve = sqrt( 2 * c * deltaT )

Maybe it's something weird like... the m in the first one is for the remass, and the m in the second one is for the fuel. So it's not the same mass.

[Specialist290]

Taking a few things from the same Atomic Rockets page you did:

Remember Einstein's famous e = mc2? For our thermal calculations, we will use the percentage of the fuel mass that is transformed into energy for E. This will make m into 1, and turn the equation into:

Ve = sqrt(2 * Ep)

where

Ep = fraction of fuel that is transformed into energy

Ve = exhaust velocity (percentage of the speed of light)

Multiply Ve 299,792,458 to convert it into meters per second.

Robert Braenuig's notes on calculating rocket propulsion in general might also help you out here.

(Disclaimer: I'm good at finding useful resources, but I'm really bad at talking through complex math, especially if I haven't already taken a crack at it myself. You might not want to address any questions to me specifically on this matter.)

Note also that rocket engines often have some sort of coolant system to bleed off excess heat, which might mean that what you're getting is the temperature the rocket would reach before those systems come into play.

Edited October 5, 2013 by Specialist290

[PTNLemay]

@ Specialist290

Yeah... it was probably over-ambitious of me to think that I could roll this all up into one quick-and-easy equation to play around with. There are just too many variables to take into consideration. I just assumed that I could keep it simple since this is mostly for KSP, which is usually straightforward with it's rocket science.

A friend of mine on Skype who's generally better than me at this told me it's probably because Q and E aren't compatible. Q is heat energy, which has it's own special kinds of equations.

Edited October 5, 2013 by PTNLemay

[Kryten]

You could try converting working out particle energy and converting to temperature;

Particle energy=(0.5 * mass of particle [in this case monoatomic hydrogen]*(exhaust velocity squared))/Boltzmann's constant

[architeuthis]

The equation for exhaust velocity of a thermal rocket is

where

Ve is the ideal exhaust velocity

gamma is the ratio of specific heats (of the propellant)

R is the universal gas constant (8314.5 J/kmole*K)

Tc is the absolute temperature in the combustion chamber

m_avg is the average molecular weight of the propellant

Intuitively speaking energy and momentum is conserved, and temperature is really just the average kinetic energy of molecules, so if temperature is increased (or if molecular mass is decreased) the average exit velocity of the propellant will increase overall.

Edited October 5, 2013 by architeuthis

[SunJumper]

According to that equation, a doubling of reaction temperature will cause a 41.4% increase in isp (eg, boosting the LV-N to 1130s)

[rhoark]

According to wikipedia, an open-cycle gas core NTR keeps the fuel out of contact with the walls, allowing it to get as hot as you like for an ISP of 5000. However the "open-cycle" bit means it sprays radioactive byproducts out the back.

[Kryten]

There are major engineering issues though, including how exactly you're supposed to feed the fuel in in the first place.

[PTNLemay]

@ architeuthis

Oh, wow, thanks a lot. I think that will be helpful.

For the average mass, does that mean the average weight of an atom in a molecule, or the weight of the whole molecule. For example if I take water as my propellant, would that value be 10 (8 + 1 + 1) or 3.33? (8 + 1 + 1)/3

@ rhoark

Yeah I know... but really a gas core is a different kind of engine altogether. And it gets very complicated very fast, not just in building one but also in understanding the physics behind it. For now I'm content in pretending that my KSP scientists just figured out a new ceramic or something that allowed my regular solid NTRs to run hotter without melting. That way I can keep using the old part model, lol.

Edited October 5, 2013 by PTNLemay

[architeuthis]

For water the molecular mass is 18 amu. This is one reason why NTR rockets offer superior specific impulse to Hydrogen/Oxygen chemical rockets, the propellant molecular weight for a NTR is only 2 amu (assuming this is a NERVA 2 type NTR using H2 as propellant).

[architeuthis]

According to that equation, a doubling of reaction temperature will cause a 41.4% increase in isp (eg, boosting the LV-N to 1130s)

True. But keep in mind that this is not the temperature of the fissionables, it is the temperature of the hydrogen exhaust. If you look up the exhaust temperature of the old KIWI series NTR testbeds from Project Rover you'll notice the exhaust temperature is 2330 K. Plug this into the equation (ratio of specific heats for hydrogen is roughly 1.2) and you'll get an ideal exhaust velocity of 10780 m/s. The real velocity will be lower than this because of nozzle inefficiencies.

[PTNLemay]

Can someone please check my numbers? I'm getting 262 m/s exhaust velocity for a temperature of 2000 degrees... Sorry I tried to use Libre Office math to create a nice clean series equations to show off, but I'm still figuring out how it works.

EDIT:

HA! What'd ya know, I got it to work. Here's my reasoning.

I'm sure this is wrong, but I don't know where I made a mistake.

Edited October 5, 2013 by PTNLemay
better equations

[Fractal_UK]

This is a bit of a plug so sorry about that but if you're interested in trying some of this stuff in KSP, you should take a look at my mod, KSP Interstellar. I don't actively model the heat capacity of the propellant or require different nozzles for extremely hot propellants, it's all the same part - in reality you'd need to switch over to a magnetic nozzle at some point. I provide a single nozzle that you can attach to various different reactors to produce different thrust and specific impulses dependent upon the core temperature and power output of the attached reacted as well as the propellant chosen - I assume that KSP liquid fuel is Hydrogen. The 3.75m nuclear reactor provides the same thrust and Isp as a real NERVA and the other nuclear reactors are essentially scaled using that as a reference point. The scaling is very non-linear down to the smaller sizes, where I've used the SAFE-400 nuclear reactor as a very vague guidline for the smallest nuclear reactor size.

The nuclear reactors all begin as solid core reactors but you can retrofit them later to produce closed cycle gas core (nuclear lightbulb) versions, obviously with much improved core temperature. The antimatter reactors are, obviously, entirely fictional and use values scaled up by energy density and such.

Obviously, for a game, it's important to get a balance between realism and gameplay concerns but I'd definitely be interested to hear thoughts and suggestions in this kind of discussion.

[architeuthis]

Can someone please check my numbers? I'm getting 262 m/s exhaust velocity for a temperature of 2000 degrees... Sorry I tried to use Libre Office math to create a nice clean series equations to show off, but I'm still figuring out how it works.

EDIT:

HA! What'd ya know, I got it to work. Here's my reasoning.

I'm sure this is wrong, but I don't know where I made a mistake.

Almost. The units for R in this case are J/(kmol*K) so you have to multiply that constant by 1000. Also if your propellant is water m will equal 18. This gives an exhaust velocity of only 2700 m/s, but 2000 is a fairly conservative number for combustion chamber temperature. Note again, this is the ideal exhaust velocity, the actual velocity will be lower for a real nozzle with efficiency losses, about 2600 m/s or 266 seconds isp.

[PTNLemay]

@architeuthis

Thanks again for clarifying. I did want to use water, but I can't find y values for it at temperatures greater than 200 degrees Celsius. Hydrogen seems to be the only propellant where the numbers are easily known for temps as high as 2000. Does it matter that much for me to have an exact y value? They seem to all hover around 1.3 and 1.5, with a consistent tendency to get lower as the temperature gets hotter. So would it be okay for me to approximate that when water reaches 2000 C, y = 1.300? Or even lower, if I want to experiment with really high temps like 5000 degrees.

@Fractal_UK

Oh no problem, that looks quite relevant to my line of questioning so feel free to bring it up. It looks quite... complicated, but it also looks really interesting. The radiators especially are a nice touch. If Nyrath taught me anything about rocketships it's that it's worth obsessing over waste heat, lol. I'm also really curious about that 3.75 meter NTR... that sounds pretty epic.

Edited October 6, 2013 by PTNLemay