Simplest Mars Vehicles - Paris-Dakar rally + 02 onboard tanks

MarsDog · 2004-08-15 03:36:21

3(O2) + 2(CH4O) = 2(CO2) + 4(H2O)
3(16*2) + 2(12+1*4+16) = 2(12+16*2) + 4(1*2+16)
96 + 64 = 88 + 72
160 = 160

64 grams Methanol uses 96 grams of Oxygen
That is 1 unit weight of Methanol uses (96/64) units weight of Oxygen
One cubic meter http://cetiner.tripod.com/Properties.htm]Methanol weighs 791 kg
One cubic meter Methanol uses (96/64)*791 = 1186.5 kg Oxygen
One cubic meter Oxygen weighs 1141 kg
Total volume Oxygen used is 1186.5/1141 = 1.04 cubic meters

Total weight 791+1186.5 = 1977.5 kg
Total volume 1+1.04 = 2.04 cubic meters

Methanol is less http://www.hyweb.de/Knowledge/w-i-energ … ]efficient than gasoline
So 30 miles per same weight would be only 30*(5.6/12.7) = 13 miles

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Rolling friction is proportional to gravity and so is potential energy (e=m*g*h)
Kinetic energy however is same as on Earth (1/2)*m*v^2.
For continuous speed, on Mars, you would get 8/3 times the fuel economy on Earth;
2.67*13=34.71

Distance for 1 cubic meter or 264 gallons of Methane
34.7*264 = 9,160 miles

If you used fuel cell to electric wheel drive,
multiply distance several times.

(Subtract, of course, for bumpy terrain, braking, cold temperature, life support etc.
In a city, stop and go traffic, half of the energy can go into braking.)

karov · 2004-08-15 07:05:24

To RobS: Thank you very much to point out to Dook this important feature of the combustion engines driven cars.

Dook: see the RobS`s quote of Zubrin`s book. BTW, in e-mail which I received yesterday as an answer of this question, Zubrin told that : yes, methanol is splendid fuel for ground vehicles, especialy fuel cells, that his company worked out equipment for producing methanol on Mars, that it is inferior to the methane as rocket fuel, AND -- combustion engines on CH3OH/O2 can work on Mars, but the temperature of pure CH3OH/O2 flame is too high so they has to be mixtured with CO2... I personally imagine kinda turbocompressor to suck in the several milibars of martian air to several tens of atmospheres in the cilinders where the right proportion of CH3OH/O2 mixture to be combusted...

MarsDog: Thank you. My calculations have the same results. Indeed roughly 1 L of methanol needs 1 L of LOX. The total fuel/oxidizer mass is the same in our calculations if we choose 1 to 1 m3 for the on-board reserve. But of course any other ratio could be chosen.

To all: These days I `ll find exact existing offered on the mass market vehicle which suffices the Martian requirements for total mass ( within the 2.2 tonnes) vs. payload.

Dook · 2004-08-15 09:38:50

I choose an rpm because it is the main factor in the amount of fuel and oxygen used, engine size multiplied by engine revolutions. Sure power is something you need to have enough of but an rpm of 2,500 is about as low of an operating rpm as you can get for an internal combustion engine. Smaller engines on the earth are always built to spin faster to make power (motorcycle) but you could build a low rpm torque engine.

I don't understand how Zubrin is going to use only 1 pound of methane/oxygen in a combustion engine to travel 1 km. Maybe he was only thinking about the amount of fuel used?

MarsDog · 2004-08-15 12:47:53

I personally imagine kinda turbocompressor to suck in the several milibars of martian air to several tens of atmospheres in the cilinders where the right proportion of CH3OH/O2 mixture to be combusted...

The engine exhaust is CO2 and H2O.
Conveniently, use the exhaust as a heater for the vehicle occupants, condensing the H2O into liquid, and feed some of the CO2 gas back into the engine. Preheated fuel/oxygen would improve mileage.

The more you examine Methanol as a fuel, the better it looks. The carbonated water from the cooled exhaust might not be directly drinkable due to some CO.

I don't understand how Zubrin is going to use only 1 pound of methane/oxygen in a combustion engine to travel 1 km. Maybe he was only thinking about the amount of fuel used?

Zurbin possibly extrapolated from a large vehicle driving in a sandy deset.
Fuel use http://www.teaching-english-in-japan.ne … conversion table.

karov · 2004-08-15 12:57:15

The engine exhaust is CO2 and H2O.
Conveniently, use the exhaust as a heater for the vehicle occupants, condensing the H2O into liquid, and feed some of the CO2 gas back into the engine. Preheated fuel/oxygen would improve mileage.
The more you examine Methanol as a fuel, the better it looks. The carbonated water from the cooled exhaust might not be directly drinkable due to some CO.

Of course! Thank you. The engine exhaust after condensing the water in liquid, consisting of CO2 could be injected back in the engine providing more plentyfull and warmed dilutant than the rare atmosphere. Such cycle allows the engine to be entirelly with closed input, as in these diesel submarines runing for weeks with onboard oxigen reserve, and thus a air filter for the fine dust to be spared from the design.

karov · 2004-08-15 17:04:56

http://www.navnit.co.in/landrover/defen … ations.htm

This is occasionally picked example.
Those are the specifications of a common and very preferable by the customers Landrover. Its total weight is almost 600 kg bellow the 2.2 tones restriction. Look the data about the payload ( ~500 kg) and the trailer mass ( about 4000 kg max!!! ). The range of operation of such machines is hundreds of km.

Yes, I know, unmodified it can not be sent to Mars: many of the metal parts will cold-brake, the lubricants and the rubber tires are not well suited to endure the martian conditions, but indeed all these minor disadvantages can be redesigned. If lighter materials are used than the payload could be increased many times. The tires can be made by cold-resistent plastic and to be inflated continuously by tied to the engine`s output comressors, the lubricants made on base of low freezeng point solvents ( fluor componds? ). The habitation module - inflatable?

Design and construction of methanol/oxigen internal combustion engine off-road car with specifications suitable for Mars rover is within the expertise and the abilities of every modest-size car manifacturer.

Dook · 2004-08-15 19:38:35

Almost 600 kg below the 2.2 tonne limit? Oh my God dude you amaze me. Again you left out some very key things that your Landrover on mars would need like a life support system with CO2 removal and oxygen supply for the engine. I won't even mention fuel supply. Also your idea, sigh...yes again (third time?) provides no redundancy. If the engine fails, the crew dies. But at least they will die on leather interior and with a CD player.

EDIT: The engine (152 cu in) at it's maximum torque of 1,950 rpm uses 148,200 cubic inches of oxygen a minute. A 50 gallon LOX tank would provide 43,100 gallons (11,948,624 cu in) of oxygen gas. Your oxygen supply to the engine would be gone in 80 minutes. You could go maybe 40 miles then turn around and come back. I don't even want to go into how heavy a 50 gallon LOX tank is.

You just don't get it. Mars is not the earth, there is no air there.

Dook · 2004-08-15 19:45:09

Two formula's for you:

EDIT: engine size multiplied by rpm then divided by 2 gives you fuel and oxygen used per minute.

LOX (Liquid Oxygen) tank size multiplied by 862 (liquid converting to gas ratio) gives you the amount of oxygen gas supplied from that tank.

Play with them for a little while and see what happens.

MarsDog · 2004-08-15 19:58:39

3/8 gravity on Mars means you can leave 5 cylinders back on Earth.
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and the trailer mass ( about 4000 kg max!!! )

Means 4 cubic meters extra Methanol and Oxygen can be towed.
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Look the data about the payload ( ~500 kg)

Should be able to cram the life support into that.

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For Dook to plug into his formulas:

Liquid http://www.airliquide.com/en/business/p … =78]Oxygen expands 854 times when converted to gas.
Liquid/gas equivalent (1.013 bar and 15 °C (59 °F)) : 854 vol/vol
Air is approximately 20% Oxygen; so multiply volume by 5.

Play around with http://www.performancetrends.com/progra … 3.exe]Free Engine Analyzer Software.
http://www.performancetrends.com/download.htm#ea3]Link to other Engine Demo Programs

Dook · 2004-08-15 20:58:29

You have a reference that says you would need 3/8's the power on mars than you would on the earth? Just because the gravity is less does not mean the rolling resistance is less. On the earth we drive on pavement but on mars there is sand, rocks and bigger rocks.

4 cubic meters of LOX is 880 gallons, that equals 8,366 pounds. Meaning you have 450 pounds left over for life support which you might just be able to squeeze in. Now, where is the crew going to sleep? They won't be able to move about or stand up. Where is the food? Where are they going to urinate? Oh, once again (4th time explaining this). If the engine fails the crew dies. If there is a LOX leak, they die.

The oxygen to gas ratio is dependant on the temperature. 862 to 1 is at about 71 degrees F. Your 854 to 1 is at 59 degrees F.

Air is 20% oxygen so multiply volume by 5???? I don't know what you are trying to say. Maybe you think that since only a percentage of the volume taken into a combustion engine is actually oxygen that somehow the engine will only need this 20% to operate. Well, this is wrong. When the cylinder fills it fills completely. It doesn't just go to 20% and stop. The pure oxygen supply simply means you can put more fuel into each combustion chamber also. You should get more power if it doesn't blow itself to pieces.

MarsDog · 2004-08-15 22:05:08

You have a reference that says you would need 3/8's the power on mars than you would on the earth?

http://www.engineersedge.com/friction/c … g.htm]Note that weight is proportional to pull of gravity.
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The reference vehicle is designed for bumps and other off road events. However, you would get very poor mileage driving on beach or desert sand.
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If the engine fails the crew dies. If there is a LOX leak, they die.

There are hazards, taking along a space suit would be advisable. Oxygen levels need to be constantly monitored. Some extra weight awailable, because of 3/8 gravity, could be used to store safety equipment.
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On Earth, the trailer could pull 2 cubic meters of Methanol and 2 cubic meters of Liquid O2. 2*(791+1186.5 = 1977.5 kg). On Mars it would be 8/3 times.
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Air is 20% oxygen so multiply volume by 5

Pure Methanol burns too hot, so the exhaust CO2 and H2O are seperated, and some of the CO2 is fed back into the engine.
The 5:1 ratio is to produce similar to Earth conditions, but would have to be adjusted for best performance.

Dook · 2004-08-15 23:13:05

The 2.2 tonne weight figure is only for launch in the Mars Direct plan. After you get to mars you can have the vehicle weigh more as long as the vehicles suspension can handle it.

Of course the crew would have pressure suits. How else would they make outside excursions? How are they going to put on these suits inside a land rover?

Now you are including a trailer? How much does that weigh and what size is it? The total width of the mars habitat is 7.6 meters (24.9 ft) so it all must fit underneath.

How are you going to separate the CO2 and H2O from the exhaust?

MarsDog · 2004-08-15 23:39:22

Mars Direct will be superseded by Mars Sustained.
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For Mars Direct, I would use extra weight for extra habitation safety, leaving only a Moon Buggy type vehicle for short trips.
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Desired vehicle, for a larger Mars presence, would be like a military personnel transport with radiation shielding. It might need a launch by itself.
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How are you going to separate the CO2 and H2O from the exhaust?

Use the exhaust as a heater, the water would condense out, leaving the CO2 gas, some to be put back in the engine.
The condensed water may have to be processed for drinking.

RobS · 2004-08-16 16:49:15

Yes, the figure from *The Case for Mars* was oxygen and methane; that's what the text said, anyway.

Dook, have you recalculated the range figures when taking into account that the oxygen should be diluted on Mars to 20% of the intake gas? Zubrin also suggested added carbon dioxide; it cools the flame temperature. Pure oxygen at terrestrial pressure burns too hot.

-- RobS

Dook · 2004-08-16 18:10:23

I hadn't calculated only 20% oxygen supplied to the engine but I can do it now.

Okay, the 152 cu in engine at it's maximum torque of 1,950 rpm uses 148,200 cu in of fuel/oxygen/CO2 per minute (4 cycle engine). Now if 20% of that is oxygen that is 29,640 cu in used per minute.

A 50 gallon LOX tank provides 11,948,624 cu in of oxygen gas which would last you 403.1 minutes or 6.7 hours. You could go just over 100 miles then turn around and head back.

I just don't see the reason why people are so stuck on combustion engines on mars when there are much better options.

MarsDog · 2004-08-17 00:17:01

Electric Sports Car Range on Earth:
For 1 m^3 of Methane, assuming 100% fuel cell electricity conversion.
The car is http://www.unc.edu/~mkirk/physics24.html]GM EV1 (26 kWh per 100 miles using standard lead-acid batteries)
Use this http://www.epa.gov/cmop/resources/conve … =0.00]link for Methane Conversions.
1 M^3 CH4=422.62 kg=6525 KWH
6525 kwh/0.26 kwh/mile = 25,096 miles.

Dook · 2004-08-17 10:11:59

No fuel cell is 100% efficient. Alkaline fuel cells are the most efficient at 70%.

I don't believe you can go 25,000 miles with an electric vehicle, one fuel cell, and a cubic meter of methane. If you could we'd all be driving vehicles powered by methane. I checked your web link and for a cubic meter of methane I got .6802 kg so I don't know where you got 422 from.

MarsDog · 2004-08-17 10:44:33

http://www.airliquide.com/en/business/p … 78]Methane:
Liquid density (1.013 bar at boiling point) : 422.62 kg/m3

Put the 422.62 kg of Methane into this http://www.epa.gov/cmop/resources/conve … =0.00]link to get total energy
in KWH. (6524.83 KWH)
The Car gets 26 KWH per 100 miles, or 0.26 KWH per Mile.
So the distance is 25,095.5 Miles

1 cubic meter is 264.17 gallons
So the car gets 25,095.5/264.17 = 95 Miles per Gallon of Methane.

Adjusted for 60% fuel cell efficiency:
Distance = 15,057.3 miles or 57 Miles per Gallon

RobS · 2004-08-17 13:14:08

Dook, I am curious whether you have tried to apply your formula for calculating mileage to automobiles on Earth. To do that you would need to know what the ratio of liquid gasoline to gasoline vapor is. But as an approximation, assume that it is 80% the ratio of liquid oxygen to gaseous oxygen, because liquid gasoline has a density a bit less than water and liquid oxygen is 15% more than water. The conversion from liquid to gas should be roughly the same otherwise; that is, not off by a factor of, say, two.

If you do this, I suspect you will prove that internal combustion engines should get only a few miles per gallon. Of course, they do better than that, and the reason is because the engine doesn't work at maximum efficiency all the time. Rather, when a vehicle is cruising along at 40 or 50 miles per hour, the engine is probably using only half its total power capacity, and when it is moving the vehicle slower, even less.

That's the problem trying to calculate how far a kilogram of fuel and oxygen will take a vehicle; you're calculating based on constant high power. The calculation would work better for a hybred engine because they send the surplus power to batteries. The engines in hybred vehicles are smaller and are designed to run at a fairly constant speed, which makes them more efficient. But a standard internal combustion engine has to run fast and slow depending on demand. A system of calculating how long fuel will last based on a certain rpm can't accommodate that very well.

I think that also answers the question why Zubrin and others favor internal combustion engines; because they are basing their decision on available performance information, rather than trying to calculate it. It shouldn't be not hard to get performance information for a humvee or jeep in the field; go to a website. Then add 3 or 4 kg of oxygen for every kg of gasoline to calculate the oxidizer mass.

To take the example of gasoline again, assume a 2-tonne vehicle gets 10 miles per gallon (4.25 kilometers per liter). That's 4 kilometers per 1 kilogram of gasoline (because gasoline weighs a bit less than 1 kg per liter). Add in the oxidizer and we are at about 4 kilometers per 4 kg of gasoline and oxygen, or 1 kilometer per kilogram of fuel and oxidizer. With that system you can see why Zubrin figures 0.5 kg of fuel plus oxidizer per kilometer per tonne of vehicle.

-- RobS

Dook · 2004-08-17 13:48:04

The Car gets 26 KWH per 100 miles, or 0.26 KWH per Mile.
So the distance is 25,095.5 Miles
1 cubic meter is 264.17 gallons
So the car gets 25,095.5/264.17 = 95 Miles per Gallon of Methane.
Adjusted for 60% fuel cell efficiency:
Distance = 15,057.3 miles or 57 Miles per Gallon

Yes, that fuel efficiency is great for the earth. What's the oxygen to methane ratio used by a fuel cell? How much oxygen do you need to carry?

Dook · 2004-08-17 14:13:26

Dook, I am curious whether you have tried to apply your formula for calculating mileage to automobiles on Earth.

You're missing the main problem with combustion engines. Fuel efficiency is not the problem, OXYGEN efficiency is. You need something to burn the fuel. Using my formula for earth vehicles gives me the exact same result, the same volume of oxygen (air) is used but the neat thing about driving on the earth is you don't have to carry an additional tank to supply that air. Mars does not have an atmosphere full of air so that is the problem. You are thinking that a combustion engines fuel efficiency (an earth car gets 35 mpg) would be the same on mars. It would but only for fuel, supplying enough air or oxygen to burn that fuel is what makes combustion engines inefficient for use on mars.

The liquid to gas conversion rate for oxygen is as stated above 854 to 1 at 59 degrees F.

And I'm not calculating based on constant high power. 1,950 is definately not max power, max torque maybe but not max horsepower. Even idling at 800 rpm's a 152 cu in combustion engine uses 60,800 cu inches of air/fuel a minute. A 50 gallon LOX container provides 11,948,624 cu inches of oxygen gas. Even idling this oxygen supply would be entirely used up in 196 minutes. How far can you go at 800 rpm in 100 minutes? If you could get to 20 mph then you could go 50 miles then turn around and idle back home. Performance is of little issue, even engine size is not much of a factor, rpm is the problem with a combustion engine.

You lost me in your final paragraph. You don't say whether the oxidizer is liquid oxygen or gaseous so I'm not sure how you are getting 1 kg of oxidizer for 1 kg of fuel used. It doesn't make sense, you always need more oxygen or air per fuel in a combustion engine.

I tell you what, you pick any engine size and then pick what rpm you want it to run at then just multiply the two together, then divide by 2 (4 cycle engines) and you have the amount of volume used per minute. Now most of this volume is air or oxygen and some is fuel. You can go further to figure out exactly how much is fuel and how much is air but I think you will soon see that the amount of oxidizer used in enormous.

MarsDog · 2004-08-17 15:56:19

CH4 + 2(O2) = CO2 + 2(H2O)
(12+1*4) + 2(16*2)=(12+16*2)+2(1*2+16)
16 + 64 = 44 + 36
80 = 80

One unit weight CH4 uses (64/16=4) weight O2
One cubic meter CH4 weighs = 422.62kg
Weight O2 needed is 4*422.62 = 1690.48 kg
Volume of liquid O2 1690.48/1141=1.48 cubic meters.

Total liquid volume = 1+1.48=2.48 cubic meters
Total weight = 423+1690=2113kg

Total expanded volume of O2=1265 cubic meters Or 44,673 Cubic Feet.
Total expanded volume CH4 = 630 cubic meters Or 22,248 Cubic Feet.

Energy density per weight of Methane is 2.4 times that of Methanol.

gasoline 44 MJ/kg (megajoules/kilogram)
Methane 55.5 MJ/kg

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Martian efficiency would be better because you can feed some of the hot exhaust gas back into the engine. You would save having to heat up the air as on Earth. There are engines that meter, not only the fuel, as in fuel injection, but the amount of air entering. On Mars, some arrangement of metering all 3 components, Oxygen, fuel, and hot exhaust gasses fed back in should be considered.
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If you left part of the exhaust in the cylinder, could the fuel and Oxygen be injected, without an engine meltdown ? Need innovative exhaust valve design.
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In addition the engine exhaust would be to a vacuum, increasing efficiency.

Dook · 2004-08-17 19:31:37

I really doubt you can design a combustion engine to be more efficient than the designs we currently have. There are a few giant corporations (Chevrolet, Ford, Toyota...) that have been doing that kind of thing for many, many, years.

Sending exhaust back into the engine would warm the air/fuel but the oxygen warms as it expands to a gas anyway. Also the denser (colder) the air/fuel mixture the more power to get out of it. As the piston moves up it compresses, thus warms, the air/fuel. You could leave some of the exhaust in the cylinder, I believe there are fuel efficient camshaft designs that do that now but only to a small degree. Too much and the air/fuel mixture won't ignite.

Dook · 2004-08-17 19:36:02

The ratio of air to fuel (gasoline fuel) in a combustion engine operating on the earth ranges from 12.5 parts air to 1 part fuel to 14.7 parts air to 1 part fuel. You need a very large oxygen supply.

http://www.sficc.net/tech/airfuel.php]h … irfuel.php

MarsDog · 2004-08-17 20:52:00

Yes, I found a http://www.epa.gov/otaq/presentations/s … 3.pdf]link, for diesel engines. Exhaust gas is fed back in.

the engine, operating on methanol fuel, demonstrates better than 40% brake thermal efficiency - - - - - - - using exhaust gas recirculation (EGR) and intake manifold pressure to modulate engine load.

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I really doubt you can design a combustion engine to be more efficient than the designs we currently have

Someone, other than me, will come up with a better engine.
http://ecen.com/content/eee7/motoref.htm]Efficiency

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