alternate working fluids for nuclear thermal rockets?

(Henry Spencer) writes:

CO2, with a molecular weight of 44, gives performance grossly inferior to
a LOX/kerosene chemical rocket. Moreover, at high temperatures it shows
some tendency to break down to CO and O2. This improves performance a
little, but that free oxygen is very hard on an orthodox NTR core and
chamber. (There are materials that can stand it, but they're very
different from the ones in LH2 or ammonia NTRs, and the technology for
oxidizing-propellant NTRs is poorly developed.)

....Although, didn't Zubrin once float the scheme of using carbon _monoxide_
as a propellant for a "Martian Indigenous Fueled Nuclear Thermal Rocket" ???


-- Gordon D. Pusch

perl -e '$_ = \n"; s/NO\.//; s/SPAM\.//; print;'

In article ,
Gordon D. Pusch wrote:
CO2, with a molecular weight of 44, gives performance grossly inferior to
a LOX/kerosene chemical rocket...

...Although, didn't Zubrin once float the scheme of using carbon _monoxide_
as a propellant for a "Martian Indigenous Fueled Nuclear Thermal Rocket" ???

I think he was proposing CO2 for that, but it's been a while since I read
that paper, and I could be remembering it incorrectly.

That was somewhat of a special case: you don't need great rocket
performance for ballistic hops from place to place on the Martian surface,
and there is a huge advantage in being able to refuel from the atmosphere.
You still have the materials problems, not to mention the wee issue --
which as I recall, Zubrin neglected -- of conducting surface operations in
the immediate vicinity of a lightly-shielded highly-radioactive reactor.
But it does have its good points.
--
MOST launched 30 June; first light, 29 July; 5arcsec | Henry Spencer
pointing, 10 Sept; first science, early Oct; all well. |

In article , Gordon D. Pusch wrote:
(Henry Spencer) writes:

CO2, with a molecular weight of 44, gives performance grossly inferior to
a LOX/kerosene chemical rocket. Moreover, at high temperatures it shows
some tendency to break down to CO and O2. This improves performance a
little, but that free oxygen is very hard on an orthodox NTR core and
chamber. (There are materials that can stand it, but they're very
different from the ones in LH2 or ammonia NTRs, and the technology for
oxidizing-propellant NTRs is poorly developed.)

...Although, didn't Zubrin once float the scheme of using carbon _monoxide_
as a propellant for a "Martian Indigenous Fueled Nuclear Thermal Rocket" ???

There's a section about CO2-fuelled NIMFs (as he calls them) in /Case
for Mars/, fwiw, in the somewhat more free-ranging half of the book g.

His comment was that by just grabbing and compressing raw CO2, and
heating it for a NTR, you could get a low but acceptable thrust. And,
well, when you're flying on air you forgive a lot of problems with
efficiency.

He doesn't say anything about using CO as a propellant there, though.
There's mention of a CO-O2 rocket soemwhere, and idle mention as an
aside about using residual CO from a process as a fuel, but nothing
about nuclear-thermal in the book.

--
-Andrew Gray

In article ,
James Nicoll wrote:
The original context for this was a thread on "ultimate rockets",
rockets where the dominent cost was fuel rather than labour. Obviously
that doesn't describe the current state of affairs.

Also, with an NTR system, the dominant consumables cost almost certainly
will be fission fuel, not propellant. This is not your friendly next-door
nuclear power plant :-) that needs refueling only quite rarely; NTRs are
much more aggressive systems with much higher power densities and much
higher burn rates. (The Phoebus NTR, the original design goal of the
Rover program -- essentially a nuclear J-2 replacement -- and nearly
flight-ready when development stopped, was the most powerful nuclear
reactor of any kind ever built.)
--
MOST launched 30 June; first light, 29 July; 5arcsec | Henry Spencer
pointing, 10 Sept; first science, early Oct; all well. |

In article ,
Henry Spencer wrote:
In article ,
James Nicoll wrote:
The original context for this was a thread on "ultimate rockets",
rockets where the dominent cost was fuel rather than labour. Obviously
that doesn't describe the current state of affairs.

Also, with an NTR system, the dominant consumables cost almost certainly
will be fission fuel, not propellant. This is not your friendly next-door
nuclear power plant :-) that needs refueling only quite rarely; NTRs are
much more aggressive systems with much higher power densities and much
higher burn rates. (The Phoebus NTR, the original design goal of the
Rover program -- essentially a nuclear J-2 replacement -- and nearly
flight-ready when development stopped, was the most powerful nuclear
reactor of any kind ever built.)

At the risk of sounding ignorant, how would I estimate the U
consumption?

I see U-235 runs about $30,000.00/kg. Each kg is good for what,
8.1x10^13 J? Or very roughly 2x10^7 kilowatt-hours? That seems pretty
cheap.

--
It's amazing how the waterdrops form: a ball of water with an air bubble
inside it and inside of that one more bubble of water. It looks so beautiful
[...]. I realized something: the world is interesting for the man who can
be surprised. -Valentin Lebedev-

In article ,
James Nicoll wrote:
Also, with an NTR system, the dominant consumables cost almost certainly
will be fission fuel, not propellant. This is not your friendly next-door
nuclear power plant :-) that needs refueling only quite rarely...

At the risk of sounding ignorant, how would I estimate the U
consumption?

Mmm, the key variable is the energy efficiency -- how much of the fission
energy release actually goes into jet kinetic energy -- and *that* I don't
know offhand. There'll be some lost energy in molecular vibration etc.,
and a fair bit in escaping neutrons and other radiation. Let's guess 50%...

I see U-235 runs about $30,000.00/kg. Each kg is good for what,
8.1x10^13 J?

The energy value is at least roughly right, although I don't have the
right references handy; the price I can't vouch for. :-)

Okay, let's see. Phoebus was about 200 klb thrust -- as I said, it was a
J-2 replacement -- so that's about 1 MN. Isp was in the general vicinity
of 800 s, that's about 8 km/s. So the jet power was 0.5*1M*8k = 4 GW. At
50%, we need an 8 GW reactor. So 1000 s of operation, a reasonably long
burn, requires 8 TJ = 0.1 kg = about $3000 at that price.

Hmm, that's not so bad. Of course, you probably won't get anywhere near
100% burnup efficiency, so you probably have to multiply that by a factor
of several. Still, if that price is right, that's no big deal.
--
MOST launched 30 June; first light, 29 July; 5arcsec | Henry Spencer
pointing, 10 Sept; first science, early Oct; all well. |

(Henry Spencer) wrote in message ...
Okay, let's see. Phoebus was about 200 klb thrust -- as I said, it was a
J-2 replacement -- so that's about 1 MN. Isp was in the general vicinity
of 800 s, that's about 8 km/s. So the jet power was 0.5*1M*8k = 4 GW. At
50%, we need an 8 GW reactor. So 1000 s of operation, a reasonably long
burn, requires 8 TJ = 0.1 kg = about $3000 at that price.

The Nerva 12GW examined for the Hyperion sustainer produced about
1.3mlbf. Assuming I converted "589,670 kgf" to lbf correctly.

http://www.astronautix.com/stages/hypainer.htm

http://www.astronautix.com/engines/ner1mlbf.htm

Hmm. There seems to be more jet power than thermal power in those
nukes. Or my math is off. Or those entries are wrong.

Mike Miller, Materials Engineer

In article ,
Henry Spencer wrote:
In article ,
James Nicoll wrote:
Also, with an NTR system, the dominant consumables cost almost certainly
will be fission fuel, not propellant. This is not your friendly next-door
nuclear power plant :-) that needs refueling only quite rarely...

At the risk of sounding ignorant, how would I estimate the U
consumption?

Mmm, the key variable is the energy efficiency -- how much of the fission
energy release actually goes into jet kinetic energy -- and *that* I don't
know offhand. There'll be some lost energy in molecular vibration etc.,
and a fair bit in escaping neutrons and other radiation. Let's guess 50%...

I see U-235 runs about $30,000.00/kg. Each kg is good for what,
8.1x10^13 J?

The energy value is at least roughly right, although I don't have the
right references handy; the price I can't vouch for. :-)

Okay, let's see. Phoebus was about 200 klb thrust -- as I said, it was a
J-2 replacement -- so that's about 1 MN. Isp was in the general vicinity
of 800 s, that's about 8 km/s. So the jet power was 0.5*1M*8k = 4 GW. At
50%, we need an 8 GW reactor. So 1000 s of operation, a reasonably long
burn, requires 8 TJ = 0.1 kg = about $3000 at that price.

Hmm, that's not so bad. Of course, you probably won't get anywhere near
100% burnup efficiency, so you probably have to multiply that by a factor
of several. Still, if that price is right, that's no big deal.

Yeah, I used the Ek of the jet and the mass ratio to guess at
the amount of U-235 used and came to the conclusion that the cost of
the H2 used would be much higher than for the U235, thus the look for
cheaper working fluids.

I looked up the price of U235 online and got prices from 30-50K
per kilogram. I belatedly realise that since my older brother is manager
of chemstores at the local university, he might well know what the going
rate is right now. Now, to phrase the email so I don't sound like I am
building a Device.

t
--
It's amazing how the waterdrops form: a ball of water with an air bubble
inside it and inside of that one more bubble of water. It looks so beautiful
[...]. I realized something: the world is interesting for the man who can
be surprised. -Valentin Lebedev-

In article ,
Henry Spencer wrote:
In article ,
James Nicoll wrote:
Also, with an NTR system, the dominant consumables cost almost certainly
will be fission fuel, not propellant. This is not your friendly next-door
nuclear power plant :-) that needs refueling only quite rarely...

At the risk of sounding ignorant, how would I estimate the U
consumption?

Mmm, the key variable is the energy efficiency -- how much of the fission
energy release actually goes into jet kinetic energy -- and *that* I don't
know offhand. There'll be some lost energy in molecular vibration etc.,
and a fair bit in escaping neutrons and other radiation. Let's guess 50%...

I see U-235 runs about $30,000.00/kg. Each kg is good for what,
8.1x10^13 J?

The energy value is at least roughly right, although I don't have the
right references handy; the price I can't vouch for. :-)

It was on the basis of what may admittedly be crap assumptions [1]
and a dodgy methodology that I came to the conclusion that Orion would
have been insanely expensive. The number I got was around half a million
dollars per kg of payload.


1: Including no drop in the price of U235, thrust charges that were mostly
fission and no exciting new way to make small, cheap nuclear explosives.
--
It's amazing how the waterdrops form: a ball of water with an air bubble
inside it and inside of that one more bubble of water. It looks so beautiful
[...]. I realized something: the world is interesting for the man who can
be surprised. -Valentin Lebedev-

In article ,
Mike Miller wrote:
Okay, let's see. Phoebus was about 200 klb thrust... At
50%, we need an 8 GW reactor...

The Nerva 12GW examined for the Hyperion sustainer produced about
1.3mlbf. Assuming I converted "589,670 kgf" to lbf correctly.

"kgf" is a hideous abortion, but I think you got the conversion right.
However, if you read http://www.astronautix.com/stages/hypainer.htm
carefully, I think you'll find that the thrust number is for two engines.
That would put it in about the right ballpark for "12GW" to be jet power.

http://www.astronautix.com/engines/ner1mlbf.htm

Now this one is just weird, since the name would strongly suggest a 1Mlb
engine... I suspect confusion about units somewhere.
--
MOST launched 30 June; first light, 29 July; 5arcsec | Henry Spencer
pointing, 10 Sept; first science, early Oct; all well. |