Mars Crew Exploration Vehicle 2030"manned rover"

SpaceNut · 2006-02-10 20:51:15

Exhaust Gas Recirculation (EGR) is already used on modern engines but only a very small amount is recirculated because it affects engine performance.

Yes so true for the automotive industry but there are other engines that use LP or some form of Natural gas and these vehicles are not known for EGR reprocessing.

As for the mixture ratio and the use of combinations of Mars air, oxygen and such probably an injection type engine would be best.

GCNRevenger · 2006-02-10 21:00:42

I'm still interested in a turbine, which might be more efficient. It would also be easier to couple to a generator then a piston engine, which is important since the vehicle is going to be propelled by electric motors.

If CO2 & H2O were removed between re-combustion cycles, exhaust gas recirculation would work just fine if you ducted all of it back into a hypothetical combustion engine.

Austin Stanley · 2006-02-15 05:02:36

To me the simplest solution is to simply use CO2 as the inert buffer you may need. This can be take from the atmosphere or re-used from the products of combustion. CO2 only breaks down at realtivly high temperatures, higher than you should see in a well operating engine. And so, it should play little to no part in the reaction.

In many ways a Martian internal combustion engine may well be the most efficent ICE ever designed. The lack of nitogen in the chamber removes the single biggest cause of incomplete combustion and will increase efficency greatly. The compression and effective mixing of the gasses in an ICE should allow virtual all of the fuel and oxygen to be combusted. The resulting CO2 and H2O is then vented untill the pressure in the cylinder is at the appropriate levels, and then the cycle begins again. Further, since liquid oxygen and fuel can essential be added in whatever amount is needed, the engine may be have an even greater specific power then conventional engines.

As for this systems implementation, air-independent engines are old hat. The Russians were using them in the Quebec class as far back as 1956. Off the shelf systems of various types are avaliable today.

GCNRevenger · 2006-02-15 13:59:41

There is one thing though... might chemical equilibrium affect the combustion rate? If the product of the reaction is CO2/H2O, and there is alot of it, this might shift the equilibrium back to the LOX/H2/CH4 side more then we like. The equilibrium constant is very high though, so this might not be such an issue.

If burning Hydrogen only, you are also going to trade three moles of gas on the left for two on the right, so if water isn't rapidly condensed the difference might work in your favor there too.

Austin Stanley · 2006-02-16 00:31:46

There is one thing though... might chemical equilibrium affect the combustion rate? If the product of the reaction is CO2/H2O, and there is alot of it, this might shift the equilibrium back to the LOX/H2/CH4 side more then we like. The equilibrium constant is very high though, so this might not be such an issue.
If burning Hydrogen only, you are also going to trade three moles of gas on the left for two on the right, so if water isn't rapidly condensed the difference might work in your favor there too.

Well again, figuring out the exact equilibrium is incredibly difficult since the pressure and temperature change during the course of the reaction. However, it is pretty safe to assume that the products are VERY strongly favored. I'll take a rough stab at it here (there may be errors, though I'm pretty rusty at this).

The equlibrium constant of a reaction (K) is associated to the diffrence in Gibbs free energy of the products and reactans in the reaction via the equation:
K=e^-(dG/RT)
Where dG is the diffrence in Gibbs free energy, R is the gas constant, and T is the temperature.

In the reaction CH4 + O2 -> CO2 + H2O, dG can be calculated by taking the diffrence in the Gibbs free energy of the products and rectants, like so.

  (CO2 = -394kJ/mol + H2O(g) = -228kJ/mol)
- (CH4 = -50kJ/mol + O2(g) = OkJ/mol)
-------------------------------------------------
-572kJ/mol

Assumming the reaction takes place at ~1000k (we wouldn't want to get to much hotter or the CO2 will dissasociate), we can plug in to our earlier equation and get the equlibirum constant, which is simply huge. Like 7*10^28. It's safe to say that products are VERY strongly favored, to the point where this reaction really isn't reversable at these temperatures (which makes sense).

To me, the biggest problem with H2O in the combustion chamber is it's higher specific heat than CO2. Waters Vapor's specific heat is nearly 4 times that of CO2, and so it will take-up alot more of the heat energy than we would like.

GCNRevenger · 2006-02-16 06:14:53

Twenty eighth power? That should be plenty.

Heres what happens; after ignition and expansion the cylinder is evacuated with a vacuum pump, then it is refilled from a vacuum with fuel/oxidizer and dry inert gas before the next ignition. The evacuated gas then goes to the recycling system where the water is captured and the CO2/He/etc is recirculated.

You wouldn' thave to bother with this for a turbine though, which can deal with a little water vapor okay and the exhaust would be dried by condensation without neededing to waste energy on a vacuum pump.

Turbines are built small enough that they make tiny ones to put on remote controlled toy/model airplanes, so minimum size shouldn't be an issue. They make them with RPMs up into the 100,000 range, so they should have a very high specific power output. This would make them light enough to accomodate redundancy perhaps too, plus the backup could be ignited for burst demand, like for hill climbing at speed, or providing power for a portable drill.

And we are going to bring a drill! There is no point in going to Mars to look for anything but pretty rocks if you are going to confine your work to the surface only, if there is life or ground water or pristine mineral deposits, they are all going to be under ground.

SpaceNut · 2006-02-16 07:08:12

And we are going to bring a drill! There is no point in going to Mars to look for anything but pretty rocks if you are going to confine your work to the surface only, if there is life or ground water or pristine mineral deposits, they are all going to be under ground.

GCNRevenger, Some of this would be answered by The sounding radar on board the European Space Agency’s Mars Express spacecraft, the Mars Advanced Radar for Subsurface and Ionospheric Sounding (MARSIS) if we are able to find such deposits from the data it collects.

Heres what happens; after ignition and expansion the cylinder is evacuated with a vacuum pump, then it is refilled from a vacuum with fuel/oxidizer and dry inert gas before the next ignition. The evacuated gas then goes to the recycling system where the water is captured and the CO2/He/etc is recirculated.

Great explanation for the combustion process. I have a question for the water in that is it expelled or is it stored for later reprocessing?

GCNRevenger · 2006-02-16 08:08:47

Ground penitrating radar can only give you a rough idea of how much water is under ground, and it can't tell you much about the underlying geology nor can it detect life. Since these are two of the main reasons for going to Mars in the first place, a heavy-duty rover-portable drill isn't up for negotiation, we are bringing one.

Of course the water would be recycled, you certainly wouldn't throw away precious Hydrogen like that if you didn't have to. The exhaust would be cooled and the water removed by condensation from the gasses.

JonClarke · 2006-05-10 04:04:11

With an IC engine there is a lot of waste heat which is dumped via the exhaust. This is a very simple process. If you try and collect water from the exhaust much of the heat will be reacted from the exhaust before it reaches the atmosphere, this means a separate, bulky and complex radiator.

I can see several solutions: you accept the hydrogen loss (which entails either a water source for propellant production or a curtailed exploration radius), use of and alternate power source (fuel cell), use of fuels with lower hydrogen content (methanol, ethanol, ethylene, acetylene), or use of CO as fuel (which has the advantage of requiring no hydrogen and being denser than methane).

Jon

Austin Stanley · 2006-05-10 15:17:35

With an IC engine there is a lot of waste heat which is dumped via the exhaust. This is a very simple process. If you try and collect water from the exhaust much of the heat will be reacted from the exhaust before it reaches the atmosphere, this means a separate, bulky and complex radiator.
I can see several solutions: you accept the hydrogen loss (which entails either a water source for propellant production or a curtailed exploration radius), use of and alternate power source (fuel cell), use of fuels with lower hydrogen content (methanol, ethanol, ethylene, acetylene), or use of CO as fuel (which has the advantage of requiring no hydrogen and being denser than methane).
Jon

It should be noted that fuel cells also produce a great deal of wasted heat during their reaction process as well. Some of this is lost in heating the water and/or CO2 they produce, and some is just waste heat. Although, since they are about 5 times more massive and considerably more bulky than an ICE per unit of energy they produce, the heat is not as "dense." And so the active cooling that is sometimes necessary for ICE is not as great a concurn.

Dumping the exhaust overboard is certianly one of the best methods to remove heat from an ICE, the CO2 component certianly should go as it is mostly useless. If we couldn't recover the water that would be a minus, but one that could be delt with once we have a native supply of water. Storing the water may be impratical in any case for an ICE or Fuel cell as after a prolonged journey (using most of the fuel) there may be a hundred or so kg of it. Throwing this overboard would allow the rover to increase its range as well.

Another note, if we are using liquid oxygen as our oxidiser, the expansion of this gass will could remove a great deal of the waste heat. The fuel stores also could make a very effective heat sink for the rover as well. The waste heat could be stored here while the rover is in motion, and then slowly radiated away when it stops for the night.

We should remember that Mars does have an atmosphere, although a very thin one. This will allow convection forces to come into play on the rover and radiator. Which is great since convection is a much, much better means of removing heat then simple radiation alone.

SpaceNut · 2015-03-22 20:39:07

I was sort of reminded today when thinking of the hydrogel fuels that are part of the ISS station resupplies and got wondering if one could create an injected engine that would make use of these as the fuel to make it run, it would sort of operate along the lines of a deisel engine in a way less glow plugs as the fuel when mixed in a chamber would ignite...

New Mars Forums

Announcement

#26 2006-02-10 20:51:15

Re: Mars Crew Exploration Vehicle 2030"manned rover"

#27 2006-02-10 21:00:42

Re: Mars Crew Exploration Vehicle 2030"manned rover"

#28 2006-02-15 05:02:36

Re: Mars Crew Exploration Vehicle 2030"manned rover"

#29 2006-02-15 13:59:41

Re: Mars Crew Exploration Vehicle 2030"manned rover"

#30 2006-02-16 00:31:46

Re: Mars Crew Exploration Vehicle 2030"manned rover"

#31 2006-02-16 06:14:53

Re: Mars Crew Exploration Vehicle 2030"manned rover"

#32 2006-02-16 07:08:12

Re: Mars Crew Exploration Vehicle 2030"manned rover"

#33 2006-02-16 08:08:47

Re: Mars Crew Exploration Vehicle 2030"manned rover"

#34 2006-05-10 04:04:11

Re: Mars Crew Exploration Vehicle 2030"manned rover"

#35 2006-05-10 15:17:35

Re: Mars Crew Exploration Vehicle 2030"manned rover"

#36 2015-03-22 20:39:07

Re: Mars Crew Exploration Vehicle 2030"manned rover"

Board footer