# The Martian: Does Mark Watney Actually Have Enough Water?

I haven’t read very much of The Martian yet. However, I have read enough to recognize that the chemistry is wrong. Chemistry is the study of the statistical behavior of atoms and molecules as they interact. A lot of chemistry is complex and difficult to apply. What I’ve witnessed so far in The Martian doesn’t go into enough detail to worry about complex chemistry. The problems are rather simple calculations of product quantities based upon inputs. Getting the right answer is a matter of molar arithmetic.

The first instance of chemistry being invoked by Mark Watney has to do with the production of fuel from hydrogen transported to Mars that is reacted with the CO2 in Mars’ atmosphere. He claims that one kilogram of hydrogen will produce thirteen kilograms of fuel. I don’t know where this comes from. I can conceive of a completely unstable chemical formula to get a similar result, but it doesn’t make sense chemically. This made me a bit worried that I did not know something. Therefore, I did a Google search to figure out if there was a well known process for making rocket propellant on Mars.

What I found first was a Geoffrey Landis, NASA scientist and author of Mars Crossing, quick brief on making rocket propellant on Mars. He presents the following chemical formulas:

4 H2 + CO2 –> CH4 + 2 H2O

2 CO2 –> 2 CO + O2 – fixed to make both sides equal

CH4 is methane, one of the components of natural gas. I’m not certain it’s precisely what we want to use as fuel, but for now it will do. However, I’m going to alter the first formula and assume that we have a really efficient machine that doesn’t produce waste water, even though Mark Watney might thank us for that waste water. The new equation is as follows:

2 H2 + CO2 –> CH4 + O2 [equation 1]

To calculate how much CH4 is produced for each kilogram of H2, we need to understand the molar concept of chemistry. Essentially, the notion is that to find the proportion of an element or chemical in a reaction, use the atomic weight to make the calculation. Here’s what we need to know to do arithmetic on equation 1:

atomic weight (rounded):
H = 1
C = 12
O = 16

2 H2 + CO2 –> CH4 + O2
2*2 + 12+16*2 –> 12+1*4 + 16*2
4 + 44 –> 16 + 32 (=48)

4 kilograms of hydrogen produces 16 kilograms of methane. The gaseous oxygen is technically a product, but it isn’t exactly the fuel in our fuel production. So, if we choose to produce methane(CH4) and there is no hydrogen waste, we get 4 kilograms of fuel for every kilogram of hydrogen. This clearly isn’t good enough. So, to reduce our hydrogen to carbon ratio, let’s assume that the fuel making machine is going to produce twelve carbon kerosene, a fuel that would be graded as jet fuel A. We will assume that the process has perfect efficiency and no hydrogen is wasted by making water for Mark.

13 H2 + 12 CO2 –> C12H26 + 12 O2

13*1*2 + 12*(12+16*2) –> 12*12+1*26 + 12*16*2
26 + 528 –> 170 + 384 (=554)

26 kilograms of hydrogen produces 170 kilograms of kerosene. That comes out to less than 6.6 kilograms of fuel for each kilogram of hydrogen, nowhere near our magic number of 13. This isn’t working, but there is another possible explanation that Mark Watney doesn’t explain to us.

Though the O2 is technically not fuel akin to what you would put in the gas tank of your car, even if you used liquified natural gas or kerosene as fuel, this is rocketry. Rockets and explosives burn their fuel quickly, so they don’t get their oxygen from the air. They have the oxygen contained in the fuel, known as an oxidizing agent. Methane isn’t exactly rocket fuel or an explosive, but it will be in a Mars vehicle, which will need to carry an oxygen supply to effect combustion. Let’s see what happens if we count the oxygen as fuel. The following is the combustion reaction. Methane and oxygen react to produce heat, carbon dioxide and water. This is an idealized reaction. Carbon monoxide could also be produced, but we will keep it simple and assume the ideal reaction occurs.

CH4 + 2 O2 –> CO2 + 2 H2O

12+1*4 + 2*16*2 –> 12+16*2 + 2*(1*2+16)
16 + 64 –> 44 + 36 (=80)

From this equation, we can see that the 4 hydrogen atoms are part of a fuel mixture with a total weight of 80. This means that 1 kilogram of hydrogen “technically” could have “produced” 20 kilograms of fuel. This leaves room for significant waste to meet the 13 to 1 threshold. It would be interesting to know what real world system the number 13 came from if it actually did.

Hydrogen to fuel weight ratios are only the beginning of farmer Watney’s chemical calculations. He needs water for his garden and to live on. He has liquid oxygen burning a hole in its tank, itching to become water. Let’s figure out how much water a liter of liquid O2 can produce.

O2 + 2 H2 –> 2 H2O
16*2 + 2*1*2 –> 2*(1*2+16) (=36)

Assuming no oxygen is wasted, which we actually don’t expect, 32 kilograms of oxygen produces 36 kilograms of water. One kilogram of oxygen produces 1.125 kilograms of water. We need to have more information to figure out how much volume of water a liter of liquid oxygen produces, namely, the mass of a liter of liquid oxygen and the mass of a liter of water. The water is easy. The kilogram is defined as 1 liter of water. As Mark Watney would say; Yay metric system! Liquid oxygen is a little more difficult. According to Wikipedia and other sites, it’s 1.141 kilograms/liter. So we need a new equation to convert liters of oxygen into liters of water:

1.141kg(O2)/l * 1.125kg(H2O)/1kg(O2) * 1l(H2O)/kg(H2O) = 1.28l(H2O)

So instead of 1 liter of O2 producing 2 liters of H2O, Mark only gets 1.28 liters of water. He’s going to need 56% more O2 to make the water he wants for his garden. Unfortunately, Mark’s quest for water will continue to be further complicated, because none of his formulas were correct. We need to figure out how much liquid CO2 he needs to make 250 liters of water. We need to look up the density of liquid CO2 (1.015kg/l) and calculate the proportion of the mass that is oxygen.

This was one of those cases where Wikipedia did not seem to be a trustworthy source for information. The density of liquid CO2 was only provided in cubic meters and was only 0.770kg/l, so I checked other sources that specialize in compressed gasses like NASA might use. The Air Products site was one I had double-checked Wikipedia against for liquid oxygen, so I thought it was reliable. When I found another gas vendor site, UIG, that agreed with Air Products, I settled on 1.015kg/l. I liked that they state that measurements were made at 1 atmosphere and the boiling temperature of the liquid. The same pressure and temperature conditions were given for liquid oxygen. If all of the measurements are taken in consistent conditions, the results should be more valid, even if they aren’t taken correctly. We’re looking to calculate relative proportions here.

CO2
12 + 16*2 (=44)
32/44 = 0.73

1.015kg(CO2)/l * 0.73(O2)/(CO2) * 1.125kg(H2O)/1kg(O2) * 1l(H2O)/kg(H2O) = 0.83l(H2O)

250l(H2O) * 1l(CO2)/0.83l(H2O) = 300l(CO2)

So, instead of needing 125 liters of CO2, Mark is going to need 300 liters to produce 250 liters of water. At half a liter per hour, it will take 25(not 50, thanks Jake) days for him to produce this much. He probably should have talked NASA into a higher performance CO2 harvester for the MAV.

The drama isn’t over for Mark Watney. We still don’t know if he has enough hydrogen in the MDV’s hydrazine tanks to make 250 liters of water. Hydrazine is a liquid at room temperature and pressure with a density of 1.021kg/l. The chemical formula is N2H4. Let’s figure out what proportion of that hydrazine is hydrogen:

atomic weight (rounded):
N = 14

N2H4

14*2+1*4 (=32)

Hydrogen accounts for 4/32 = 12.5% of the mass of hydrazine.

292l(N2H4) * 1.021kg(N2H4)/l * 0.125(H2)/kg(N2H4) = 37.27kg(H2)

H2O

1*2+16 (=18)

Hydrogen accounts for 2/18 = 11.1% of the mass of water.

250l(H2O) * 1kg(H2O)/l * 0.111(H2)/kg(H2O) = 27.75kg(H2)

That wasn’t too close, but Mark doesn’t have anywhere near the amount of hydrazine needed to make 600 liters of water. He’s going to need over 74% of the hydrogen in his hydrazine to make enough water for his crops. Mark plans to burn the hydrazine inside the habitat, and he rightly assesses that he will get N2, H2 and ammonia(NH3). So long as he doesn’t convert over 9.5kg of hydrogen into ammonia, he’ll be fine. Let’s find out how much ammonia that is.

NH3

14 + 1*3 (=17)

Ammonia is 3/17 = 17.6% hydrogen.

9.52kg(H2) * 1kg(NH3)/.176kg(H2) = 53.9kg(NH3) = 118pounds(NH3)

I think it’s safe to say that if Mark makes over a hundred pounds of ammonia inside his sealed habitat, it will no longer be habitable, so there will not be a hydrogen supply problem just yet. Unfortunately, he will be burning far more hydrazine than he planned. I hope he doesn’t need too much extra fuel for something else later…

## 30 thoughts on “The Martian: Does Mark Watney Actually Have Enough Water?”

1. Joe shelley

First off, awesome blog, but this is science fiction, not science fact. And the information you’ve sited is 15 years old. I’m not sure what advances have been made since the info sited was originally conceived, but let’s say that any classified or highly protected intellectual property (or IP) takes about 5-10 years (which is generous, since companies tend to hold onto IP, way longer than is really needed) to be made fully public).

And of course, most of the ip is in the processes and custom equipment needed to generate these advances and make them applicable in the real world, but a new compound or method for leveraging a known compound or a single element can be betrayed by the smallest clues.

Anyway the tech in the source sited is probably at least 20 years old now.

Then add on that this takes place another 10-30 years in the future…

It’s a conceivable stretch.

Then again, I’m not a chemist, and work mostly with oem/custom hardware/ software people. So let me know if I’m an idiot.

Btw,

I came across this trying to figure out how he could capture the h2 after the hydrozine was catalyzed. Even in Martian air that’s crazy energetic (they used it to get off mars, and land there after all…

How could you efficiently capture the h2 from such an energetic reaction, and how do you also ensure that the resultant impurities are not stored with it?

Thanks!!!

2. enabity Post author

Sorry about the slow reply. This is not the world’s most active blog, so it’s rather shocking to have a comment from a live person.

The information I’ve cited is more than 15 years old, it’s a hundred years old and more. It’s fundamental chemistry and won’t change no matter how much technology advances. All of these chemistry concepts would have been taught in a low level chemistry class 50+ years ago. Two hydrogen atoms combine with one oxygen to form water. There are only so many oxygen atoms in a liter of liquid oxygen. The method for counting atoms has not changed appreciably in a century. The point was that Watney’s calculations are wrong because they do not properly account for fundamental principles of the physical world.

Technology would come into the discussion with regard to how efficiently, or as you suggest, inefficiently, one might get the hydrogen to react with oxygen. No matter how good Mark Watney’s catalytic technology is, it won’t change the amount of energy produced by a chemical change from hydrazine to hydrogen and nitrogen gas. Since he’s literally pouring hydrazine over the catalyst, he’s completely undone any clever technology that might have been in place to control the rate or purity of the reaction.

In the story, Mark does the thing that will best keep the H2 that is produced by the catalytic reaction from reacting. He evacuated the oxygen from the Hab. However, he has decided to release oxygen gas in the vicinity of the hydrogen. Presuming that he releases the oxygen fast enough to keep up with hydrogen production, it will react with the hydrogen rather completely, because the hydrogen coming off the hydrazine is extremely hot. I calculated that the gas produced by the hydrazine reaction would be about 600F. After the hydrogen reacts with oxygen, it gets much worse. I calculated that the hydrogen-oxygen reaction would produce 11 times as much energy as the hydrazine reaction.

I am not a chemist, but was a chemical engineering major until I realized I liked computers, electronics and biology more. What follows is not technology specific, because I don’t have the specialized experience and I don’t feel inspired enough to do research on the subject.

Capturing the hydrogen from the hydrazine reaction would involve a number of systems. The first thing would be to contain the hydrazine catalysis in an environment that will not react with it. I would use a nitrogen atmosphere, because nitrogen is very stable and doesn’t react. This might increase the prevalence of N2 + 3 H2 -> 2 NH3, but it takes a lot of energy for that to happen. The most common way for this to happen on Earth outside of the carefully controlled environment of a living organism or a factory is from lightning. The nice thing about this setup is that some of the leftover nitrogen from the hydrazine reaction would essentially be the atmosphere.

The second step is to make sure the catalytic reaction is done efficiently. Industrial catalysts are designed as fine wire meshes in order to maximize surface area for the catalyst and chemical to react. I would turn this concept a bit on its head in that I would make sure that the gaps through the catalyst were small to maximize the percentage of the hydrazine that is touching the catalyst. I would also look at heating the hydrazine up to a gaseous state in order to increase the interaction with the catalyst and minimize the portion that sits idle away from the catalyst. This is how fuel injectors increase fuel efficiency. They gassify the fuel.

The third step is to separate the hydrogen from the other gasses and liquids. Fortunately, hydrogen is much less dense than nitrogen, ammonia and hydrazine. I would start out by placing the catalyst at the bottom of a tank and feed the hydrazine up through it. The gas products would naturally rise, but the nitrogen and ammonia would settle below the hydrogen. I would place two pressure relief valves in the tank, one at the top and one just above the catalyst.

It would take some careful calculations and probably some experimentation to figure out what the ratio of the two valve sizes should be to keep a layer of hydrogen at the top of the tank to minimize the flow of nitrogen through the top valve. However, one could also use a series of tanks with valves at the top and bottom to separate the hydrogen and nitrogen in multiple stages. Also Mars has about 1/3 the gravity of Earth, which could be a problem. To deal with this, one could put the tanks on a centrifuge to better separate the hydrogen and nitrogen inside of the tank.

Another complication is that it might be that the nitrogen would be too hot and would shoot to the top of the tank with the hydrogen. In this case, the second separation tank might be necessary to provide an area where the gasses could cool down. The chiller in Mark Watney’s Hab functions on the same principles I’ve suggested here, though it works at a lower temperature. It might be that the hydrogen and nitrogen would need to be placed under enough pressure to transition them to their liquid phase in order to separate them properly, but I would bet that their substantial difference in gas density would be enough to do an okay job.

3. Michael

Nice work! I have been wanting to see if the calculations were up to snuff. I especially thought the heat for these reactions would not be possible in a closed environment. I’m checking out the 2 volumes Andy also published for his calculations and am curious to see what he puts in there. Thanks for doing this!

4. Gideon

Awesome analysis!

When I saw the film recently, I asked myself the same questions.
Even a clever person who was able to piece together the analysis — mind you, this guy was botanist, not a chemist — is faced with a real challenge.

I believe most chemists would find this experiment daunting since most people don’t readily work with hydrazine, so even with the right know-how, I’m still questioning whether Mark would be able to pull this off. Owing to the volatility of the experiment, how much margin of error would he really be afforded? Not to mention, his ability to regulate the experiment, having everything in the correct proportions, would be extremely difficult. (In the film, he is just turning a value for the hydrazine, which is really anyone’s guess in terms of the correct flow rate so the experiment proceeds smoothly.)

My guess is that he would need to have something like a katharometer, and owing to the volume of the entire room, could he really ever obtain the proper proportion of gases — with the correct real-time feedback — without blowing himself up? The experiment requires precision; he needs valves that provide him readings about flow rate. Compounded with all the other factors that you mentioned like temperature, density, etc., perhaps a multi-staged process is required, he might begin to flirt with success.

In any case, great stuff Eric!

(I’m coming from a EE background too.)

5. Gideon

In the film, after Mark attempts his first experiment an explosion results.
He surmises that he forgets to take into account the excess oxygen that he is breathing out. Is this logic sound?

(Every human exhales CO2 and 16% of the O2 that one originally inhales.)

Is he justifying his explosion by figuring more O2 in the air resulted in a bigger flame that consequently ignited the H2?

Alternatively since he controls the valve for the hydrazine, I thought that the correct explanation is he just generated H2 too fast from the hydrazine, combined with the fire, and an explosion results.

What do you think?

Moreover, this might be an obvious observation, but he only had a limited amount of combustible wood to make the fire, which I’m not sure if I understand this correctly but he was using the fire to ignite the metal to create the catalyst that would interact with the hydraline. Doesn’t he need the flame for the catalyst to drive the process, which would require a continuous supply of combustible materials and metal that he was in short supply?

Thank you in any case.

6. Gideon

After some thought, I think that I agree with you. He would absolutely need a multi-stage experiment with closed chambers. Each chamber would need to be regulated. For example, the first chamber breaks down the hydrazine with the catalyst to create H2. The second chamber would combine the H2 with O2 to create water. Each chamber would be controlled for temperature, pressure and density. Each chamber would need release values to bleed off the excess Nitrogen and Ammonia as well. Moreover the heating of the catalyst should also be a separated chamber, then channeled and coupled to the hydrazine.

Is this correct in principal?

What makes Mark’s experiment primitive is that he isn’t using closed chambers and has an open flame, which is super dangerous among all these volatile gases in a large, closed room.

Not to mention, his ability continuously to heat the catalyst with a combustible material e.g. wood shavings is a real challenge in terms of maintaining the experiment’s longevity.

From what I can gather, again I’m no expert, Mark’s experiment would absolutely fail for a number of reasons we both cited.

Share with me your thoughts. I’m sorry that I couldn’t condense my thoughts into a single posting.

7. enabity Post author

Gideon,

Sorry to neglect you. I’m terrible about keeping up with things.

I agree that reacting hydrazine in the open space is dangerous. Mark suggests as much.

My response to Joe was about his question about capturing the H2 before it reacted with the O2.

Also, the Martian Engineer’s Notebook is based off the book, not the film. I have not seen the film and most likely won’t see it until it hits Netflix, so I have no idea how consistent the two are. We may be conversing from slightly different experiences. This is not a huge deal, but keep it in mind.

I think Mark could have maybe managed to do the hydrazine reaction in a vaguely safe way, but the big problem is that he has all of the wrong concerns. He should not be worried about igniting the hydrogen coming off the catalyst, it’s already hotter than that little bit of burning cross. He should be worried about burning his face off when the hot hydrogen hits the oxygen.

It isn’t clear why Mark evacuates the Hab of oxygen. I hope he isn’t thinking that he will set the atmosphere on fire, because that is silly, thermodynamics don’t work that way. Oxygen doesn’t burn intrinsically. It needs something to react with like hydrogen or the hydrocarbons in wood. It makes sense if he’s worried that he will accidentally spill too much hydrazine and burn himself to a cinder. I would suggest that he keeps the bulk of the hydrazine nowhere near the catalyst then.

Frankly, I think he COULD be fine by reacting a single drop of hydrazine at a time so long as he stays away from it, but it’s going to kill his run rate. Also, the Hab is not going to have many ways to diffuse that heat into the Mars environment, the Hab is built to keep heat in after all, so that slow run rate is going to be a necessity. Not only is it dangerous to go fast, but it will get too hot.

Thanks for reading and posting, Gideon. I totally agree with you that Mark needs to set up more safety structures in his process. I have no idea as to whether someone in Mark’s position could succeed or not, but I think he should worry less about the oxygen in the atmosphere and more about where that hot hydrogen is going.

8. Gideon

Thank you Eric. Great points about the exothermic reaction when combining Hydrogen and Oxygen to create H2O. I’m starting to remember my basic Chemistry 101 again.

Anyhow, am I right to think that he needs to heat the iridium-metal catalyst prior to introducing this component to the hydrazine?
Moreover, his only source to generate fire is to use the wood shavings, which is in very limited supply.

Since the supply of wood e.g. hydrocarbons is dearth, the experiment would come to a halt fast. Therefore, he must figure out a method to keep the iridium-metal catalyst hot by redirecting the exothermic heat from when hydrogen and oxygen combine.

Do I understand this challenge correctly in order to make experiment self sufficient truly at this stage?

9. enabity Post author

Michael,

I was not aware that Weir was going to publish his calculations. I hope he does. It would be interesting to see if that will up the ante and cause The Martian to receive more scientific and engineering scrutiny. To date, it has received very little. Every reference I’ve read merely gushes that it’s correct. I love that people want to be engaged with the science, but it is disappointing that more double-checking, the hallmark of the scientific community, isn’t occurring.

Heck, I would like to see my stuff double-checked and discussed. This is a wonderful opportunity for the almost universally despised word problem to gain some positive press.

10. TrvialGravitas

You don’t get 20- kilograms of fuel because most rockets, and methane rockets specifically run fuel rich.

What happens is that the thrust and fuel efficiency is a function of the chemistry of the exhaust, not strictly the energy of the chemical reaction methane as both a fairly light molecule and with its low specific heat ratio is a better exhaust than either water or carbon dioxide at any given temperature, so methane rockets run fuel rich to get more thrust. This works out to 14:1 LFO:hydrogen. This isn’t Weir’s numbers at all, NASA put them out 25 years ago.

Weir has said publicly that he DID screw up the hydrazine to water numbers (he didn’t say exactly how but I’m pretty sure he used the specific heat at constant pressure instead of the lower specific heat at constant volume, the hab being constant volume), and Mark would have cooked without some kind of cooling. Easy to solve (carry in a bunch of Martian surface temp rocks to cool things off) but the book was already out.

@Gideon: No, Iridium works as a catalyst at room temperature per my copy of Rocket Propulsion Elements. Spacecraft need a heater, but the hab wouldn’t.

11. enabity Post author

TrivialGravitas,

Thanks for clearing up the source of the ~14:1 ratio. Running fuel rich and oxygen poor would account for the lower fuel production efficiency than the 20:1 that was my ideal calculation. It also seems to rule out kerosene unless ethane has a lower specific heat to mass ratio than methane. I admit that I’m not up to looking just now and admit that it would be difficult to control what the products of the incomplete combustion would be anyway. Methane is simpler to make from H2 and CO2 and simpler to not burn by starving it of O2.

Weir has 1 liter of O2 creating 2 liters of H2O, which is patently absurd. How much water that is produced is strictly a mass calculation, which must be calculated from phase and volume with pressure and temp being relatively immaterial in the density of a liquid, though important in causing the liquid phase in the first place.

His heat calculations from burning hydrazine are nowhere in the neighborhood of correct whether he considered constant pressure or volume
by my calculations. If he had calculated the energy of the hydrazine reaction and distributed the energy to the nitrogen and hydrogen issuing from it, he would have realized the burning ember was not necessary. The only way that 600F hydrogen doesn’t react is if there isn’t any oxygen nearby to react.

Then, I would guess that he didn’t think about how much more energetic the hydrogen-oxygen reaction is than the hydrazine, 11 times as much energy by my calculations.

I agree that taking cold Mars rock into the Hab would help, but it won’t be enough to deal with the full load of 131M joules created by the full set of reactions he intends. As for how much cold Mars rock would be needed to completely offset the heat of the reaction, it is a lot. 5.65 cu meters by my calculations in the Martian Engineer’s Notebook Volume 2. Over 14 metric tons. The atmosphere in the Hab is a trivial carrier of energy by comparison, as is the Hab itself and all of the furniture inside.

It might very well be that I’ve screwed up some calculations and am off to some degree. I definitely welcome criticism of my actual calculations or approach presuming they are concrete rather than a vague notion. If I’m shown to have made a mistake I’ll own up to it. I’m not a fanatic about space travel, but I do have an education that is very similar to Mark Watney’s and was disappointed that the “real science” novel was not well researched.

Unfortunately, I have made an exclusivity agreement with Amazon with regard to the Martian Engineer’s Notebook v1 and v2, so I won’t risk publishing more of my calculations here. As such, I apologize for not showing my work. If you take a serious stab at double-checking me, we’ll work something out.

12. enabity Post author

Gideon,

Fire is not magical. It is a chemical reaction that creates heat. It is the heat that Watney needs to ignite the hydrogen. As I stated earlier, the hydrogen is already much hotter than the burning wood after the hydrazine breaks down. Therefore, the hydrogen doesn’t need the burning wood to react with the oxygen. I don’t know if TG is correct about the temperature that iridium catalyzes hydrazine, but the purpose of a catalyst is to bring the reaction temperature down by manipulating or otherwise stressing the physical shape of the chemicals it interacts with. Therefore, his assertion seems reasonable.

I’m not sure if you have your head completely wrapped around this yet. There are people who spend all of their time on this stuff because they are in academia or work for NASA or they’re fanatical about space exploration. I have been doing a lot of physics, chemistry and biology research for PHYSIC, so a lot of these concepts are fresh in my head. Still, it took me a lot of time to organize my thoughts on the matter, and I still find myself lapsing on details if I’m not careful.

I am in no way rooting against Mark Watney. That was not the purpose of my original posts or The Martian Engineer’s Notebook. What happened was I was shocked by what I found to be egregious lapses in the book, but I like the idea of space survival as a science education word problem. If we want people to be interested in science and engineering they need to be presented with interesting exercises. The Martian could be one such exercise, though Andy Weir clearly lacks the tools to carry this off. Maybe I’m also not adequate, but I have at the very least addressed some of the shortcomings in the book.

13. TrvialGravitas

Er did you mean 600F or 600C? 600C will ignite hydrogen, 600F won’t. I’m not actually sure what the temperature of hydrazine exhaust doped with nitrogen the way Mark does it (at least I think he does, it’s been a bit) would be though. It’s 1144K for rocketry purposes, more than enough, but there’s an unknown amount of Nitrogen… though i can calculate how much nitrogen he’d need to cool it off.

108KJ/mol of hydrazine, 2 Mol of H2 at 750k is 44KJ, leaving 2.75mol worth of N2 at 750K, 1 mol of which is provided by the hydrazine. He’s got way more than that in the chimney given that he’s doing a few drops of hydrazine at a time.

When I factor in the hydrogen energy though… There’s also no way that 131MJ figure can be right. 250kg of water is 27.5kg of hydrogen. Hydrogen burns at a massive 144MJ/kg (per Engineering Toolbox again). That 4GJ for 250 liters of water. Th 131MJ is easy to absorb (remember that he already brought in 9.2 cubic meters of dirt, its not cold anymore buts its a heatsink). But 4GJ? And its going to add ANOTHER 567 MJ when it condenses.

14. TrvialGravitas

Hrm, actually i suppose it won’t, the latent heat is rolled up in the flame energy, so it’ll be locked away until condensation, which actually helps a little bit. Still, 4GJ, and that’s for the 250L, not the 600L.

15. G-ideon__

Thank you Eric for your insight — you are awesome.

I think that the source of my confusion is trying to reconcile
what I viewed in the film and what I know is to happen from
a chemistry standpoint.

In the film, Mark began by making two piles, one pile being the wood
and other being the iridium metal.

He then ignited one of the piles, presumably the wood, and placed the
burning materials somewhere along his experiment path.

Next, he turned a valve to a small canister and, separately, opened a pipette that dripped liquid
onto the other unused pile.

From what I can gather, the valve corresponded to the flow of the hydrazine.
Now, outside the audience’s vantage, he already had placed the pile of iridium metal, the catalyst,
in a closed chamber that would combine with the hydrazine. He also placed a portion of the wood, which we see ignited into flames,
to heat the catalyst and start the hydrazine reaction. He then opened the pipette and allowed the H2 to drip onto the
remaining portion of the wood. Together — the H2, O2 and the remaining portion of wood — all created a spark, not a flame, that resulted in the
production of water.

Does this scenario seem plausible?

Just to add, I kept wondering about how much wood does he really require for this experiment to work. Firstly, he only needed a small portion
of the wood to heat the catalyst initially. Once the hydrazine reaction started, the reaction took its course as long as he supplied more hydrazine.
Secondly, he only required a small pile of wood to facilitate the spark, not a flame, in the presence of H2 and O2. Therefore, in total,
he didn’t require large portions of wood to process his large amounts of hydrazine.

Does this scenario seem plausible?

Finally, the idea of even having a spark in an environment that is so humid seems implausible, since, again in the film, he didn’tÂ
take any effort to separate where the H2, O2 and the wood shards combine, which would seem to be a big issue in the long run owing to the
moisture always present in his environment.

In any case, I love the discussion Eric! and agree that these thought experiments are invaluable and offer an incredible learning tools for
students and adults who want to engage in science.

16. enabity Post author

Sorry about flaking guys. I’ve been working long days on sequels to PHYSIC and I don’t multitask well.

Trivial Gravitas,

That was 600F. I made the presumption that the dynamics of hydrazine catalysis would be reasonably equivalent to the burning of wood. That was not a good choice on my part. This prompted me to do a little looking and I rather quickly found the following:

http://arc.aiaa.org/doi/abs/10.2514/6.IAC-05-C4.P.07

It seems hydrazine propulsion systems seem to produce gas in a plasma state, as far as I can tell there is enough concentrated energy to initiate an ongoing burn of the hydrogen. This is an important point, though, and I should do some more looking to verify that the hydrazine catalysis is igniting some of the exhaust gas into a plasma phase, which is definitely high enough energy to ignite hydrogen in an oxygen rich environment.

My example from the Martian Engineer’s Notebook V2 was for a 10 liter container of CO2, which I believe produces 8.3 liters of water. I didn’t read the whole line of reasoning, so it might be a bit different than that. Anyway, when I scale that to the full 250 liters, it comes out to about 4GJ. I used the difference in enthalpy of formation between water and O2 + H2 to get my result rather than looking for burn energy in Engineering Toolbox, so it was great to get the double-check done there. Thanks.

As for the “dirt” Mark brought in, it would be something more akin to sand until he made his organic contribution to it. I did my analysis on both rock and sand and found that it would take 5.9 cu m of stone or 10.1 cu m of sand going from -63C to 30C to absorb the 131MJ. Starting at 20C, the pile of dirt would heat up to the neighborhood of 110C. There isn’t enough of a heat sink in the Hab to allow fast production of water by these means.

Gideon,

I may be taking you too literally here and apologize if you merely misspoke and this isn’t what you mean. I hope the hydrogen isn’t dripping. It most certainly is a gas, and would take a lot of energy (producing more heat in the Hab) to force into a liquid phase only to later ignite it. The O2 is a liquid in the canister, but would expand to a gas as it is released to interact with the hydrogen. The takeaway is that there shouldn’t be any liquid going into the second reaction.

You make a very good point that the situation with the wood “spark” is tricky. The water produced by the hydrogen-oxygen reaction is going to be gaseous and much hotter than the hydrogen gas was. There should actually be plenty of heat to sustain the flame as shown in this youtube video:

Therefore, if Mark’s needing to maintain a supply of burning wood, then that is an incorrect detail. A hydrogen flame is self-sustaining so long as he provides an outlet for the water. If he’s not providing an exhaust for the water, then there will be problems as the pressure builds up in the reaction chamber, keeping the oxygen and hydrogen from getting in.

17. Nina

Iridium will catalyze the decomposition of hydrazine at RT. That’s the area of my research for my organometallic chemistry PhD.

Love seeing people having this conversation!

18. enabity Post author

Thanks Nina. It’s good to know that you and Trivial Gravitas’ source agree on that detail. Multiple sources are good.

19. Jake

It seems I am late to the game, but this is the best discussion on this I have found.

I just finished the book and made the same observation regarding Watney’s deficiencies regarding stoichiometry. I did an independent calculation of the liquid oxygen he needs and got similar results using both the listed density and the expansion ratio with an ideal gas assumption.

I don’t agree with your carbon dioxide assessment. Not sure where you got your numbers, but I am suspicious if they said the density was at 1 atm and the boiling point, since at 1 atm carbon dioxide doesn’t pass through a liquid phase. The triple point is -56.57 C and 5.11 atm, meaning the liquid phase only exists at pressures over 5.11 atm. Wikipedia lists the liquid density at 1101 kg/m3 (1.101 g/mL) at -37 C and saturation pressure. We can expect the MAV to be a bit cooler, given Mars average surface temperature of -55 C, but the density should be higher at lower temperatures, so this is a conservative estimate. Using those numbers, I got 0.7 L of liquid oxygen per L of carbon dioxide.

Combining the two I came up with 277.2 L of carbon dioxide to produce 250 L of water. At a constant MAV collection rate of 0.5 L/hr, I came up with 22.5 sol to obtain the desired carbon dioxide. I think you may have accidentally doubled twice when you came up with 50 days above. With your numbers I got 25 days, so not a huge difference between the two.

At this rate of production, he technically had enough time to produce the 130 L of water he claimed to have on Sol 42, presuming he immediately started carbon dioxide collection on Sol 30 when he had the idea, but it doesn’t account for all the time he lost to accidents. And certainly he wouldn’t be able to achieve 470 L in 10 days as he expects. I calculated 54 sol to collect enough carbon dioxide for 600 L, and this allows no MAV downtime.

Clearly you all got more caught up in the question of hydrazine, which I haven’t fully looked at yet. If I find anything to contribute, I will post it.

The only comment I have at first glance is the apparent concern over the heat load of the hydrazine reaction. I think what has been neglected is a full respect for the amount of heat the Hab is constantly losing to the extreme cold of the martian environment, as well as the time frame of the heat generation. The number I will work with is 5 GJ to accommodate the additional concern of the latent heat of vaporization. 5 GJ is equal to 1389 kW*hours. If you divide this by the time required just to collect all the carbon dioxide required to fuel the reaction (22.5 sol) the heat load on the Hab is 61.8 pirate-ninjas. If you remember, the Big Three alone consumed 69.2 pirate-ninjas electrical. So comparatively, the energy load on the Hab is comparable to the electrical load of a few pieces of big equipment, and every bit of that electrical loading will eventually be expressed as heat, not to mention the heat from transmission. Also, the Big Three represent only a portion of the electrical equipment that is actually running. He states that the total solar field is 200 square meters and that each panel is 2m x 1m. During his planning for the trip to the Ares 4 MAV, he says he can get 36 pirate-ninjas out of 28 panels with an anticipated 8 hours of up time (4 hours driving plus set up and tear down each day, and 13 hours of daylight). So given 1.286 pirate-ninjas of capacity per panel per 8 hours, a field of 100 panels, and 13 hours of daylight, the total capacity of the solar field is over 200 pirate ninjas. So if the Hab were running at full capacity with 6 astronauts (design conditions), I think we could expect that the heat load on the Hab would be greater than 200 pirate-ninjas. Consequently, with 1 astronaut and reduced equipment operation, and in the absence of better numbers, my gut says that the Hab will easily dissipate the extra 62 pirate-ninjas of load resulting from water creation. And if it didn’t, rejecting heat through either insulation removal (if there is any) or a simple heat exchanger could easily make up the gap.

20. enabity Post author

Jake,

Thanks for the reply. Sorry for my late response.

It looks like we’re more or less in agreement about the CO2 to water conversion. Yes, the cold tank being filled outside will hold more CO2 due to lower temperature. Yes, O2 and CO2 have different temperatures/pressures of transition for which the density is given by Wikipedia. I did not worry too much, because I was also rounding the molecular weights to the nearest whole number as well.

You’re correct, I somehow doubled twice on the CO2 collection calculation. Quite possibly, I divided by 12 and then doubled. If I would have written the equation out, this probably wouldn’t have happened. It should be 23/25 days, for 277L/300L. That will get corrected, and thank you for catching the error.

The advantage for the big three appliances is that they can be designed to dissipate through the ground or the Martian atmosphere. As such, how much heat is dissipated by the electrical systems is not relevant, only the dissipation capacity of an environment with that footprint does.

When I looked at heat dissipation, I come to one simple conclusion. The Martian atmosphere is effectively an insulator, and is a total waste of time. This means that the only way to dissipate the heat is through the ground, and it will still be difficult with as fast as Watney is burning hydrazine, even with the good thermal capacity of the presumed rock underneath him.

The thing is, there’s an even more important consideration. Watney needs to keep his Hab habitable. The problem is that the atmosphere in the Hab, unlike the solid equipment, can’t hold very much of this energy. If he wants to dissipate 5 GJ over 25 days, that’s 200 MJ/day, 8333 KJ/hr, 138.9 KJ/min. If the Hab is a 6m diameter 2m high construction, it is approximately 290m^3 and with a normal atmosphere needs less than 272 KJ to raise the temperature 1 degree celsius. That’s pretty much 1F per minute for 25 days. I won’t say there’s no way for this to work out, but it’s a pretty dire situation that would require cooling the Hab to very uncomfortable temperatures, then getting maybe an hour’s sleep at a time. He needs to contain this reaction inside a container that will keep the heat from getting out, which he doesn’t do in the book. Allowing that heat to get into the atmosphere makes even 100 days really tough, and we’re still talking 300L of water here. Even if we consider that only 4 GJ needs to be dissipated for 300L, it’s too much energy for the atmosphere to handle.

Presuming he has a container to do this reaction, Watney would definitely have to remove a lot of the insulation between the Hab and the ground and would have to be a lot more careful about how quickly he produced water. Maybe he would get lucky and the big three would have a cooling system he could tap into, though a passive system would be the best choice, fewer parts to break. If the Hab was on a platform to protect it from the cold surface of Mars, then God love Mark Watney, because the situation looks mighty grim.

21. Sandy

I have been mulling over all of the comments made about the various computations that that character Matk Watney makes; especially in regards to making water. I am a teacher for students with moderate to severe disabilities (middle school). I loved The Martian book and movies and was excited to create some decent math and science lesson plans. However, even though anything I would create would be very simple, I hesitate to use some of these examples and computations from the book. I don’t want to teach something that is incorrect, even at this level. In science we are studying chemical and physical changes. And of course two-step (or more) math problems under a “real-life” situation is my main thrust…..as well as measurement systems, recording and converting data…Any advice for a teacher of middle schoolers?
Thx,
Sandy

22. Ares

I have been wondering this myself, and it’s amusing to know the calculations dont pan out. Don’t get me wrong, it’s still a great book, but even great books have their cut corners.

23. enabity Post author

Sandy,

My advice is that the science problems in The Martian are actually too advanced for middle school. It probably starts at AP high school chemistry and goes up from there, including college level thermodynamics, which a lot of this discussion has been about. A lot of the calculations are not that difficult, but middle schoolers aren’t ready for the science concepts like molar chemistry, which Weir and his editors didn’t even understand.

The paramount concern with giving word problems, which I always enjoyed, even if it was about trains I didn’t care about, is that the concepts and relationships depicted in the problem need to be crystal clear. Otherwise everyone is guessing, not just the students who are struggling.

It could be the calculation for how much water is produced from liquid oxygen and unlimited gaseous hydrogen is not as challenging as I think. It is perfect for some of the concepts you’re talking about.

measurement systems:
total oxygen mass = liquid O2 mass/liter (1.41kg/liter) * # of liters (10)

converting data:
moles of O2 (number of molecules) = total oxygen mass / mass per mole of O2 (atomic mass of oxygen in grams (15.9994g) * 2)

total hydrogen mass = moles of O2 * 2 * mass per mole of H2 (atomic mass of hydrogen in grams (1.0079g) * 2)

total water mass = total O2 mass + total H2 mass

Maybe this makes a good lesson to work out together with your students. I don’t know. It would need to be made clear that you’re counting molecules and maybe you would want to mention that there are Avogadro’s number (~ 6*10^23) of molecules per ~32g of oxygen gas and ~2g of hydrogen gas.

24. KellyDunnigan

I’m no a chemical engineer or a chemist nor do I work for nasa. But keep in mind 100 years ago we flew just ove 120 feet and was amazed by it. People thought you face would rip off if you went over 100mph. Never thought we’d break the sound barrier nor achieve space step foot on the moon or land on a asteroid or a Space Comet yet in 100 years we have. So just because IT CANT BE DONE TODAY doesn’t mean it won’t in 30 years. Food for thought. Then again I’m just a brick layer.

25. KellyDunnigan

Oh and the whole speed of light thing is a joke well go beyond that and more. Albert Einstein I’m sorry your wrong. If you can slow it down bend it I’m sure we can speed it up. Think about it I’m in space traveling at the speed of light for the sake of saying it lets say I’m I’m standing on the nose of the ship traveling at said speed I throw a rock in front of me although you I wouldn’t be able to see the rock because of the speed of light would have to hit the rock and bounce back to my eye the rock still breaks the speed of light Einstein is wrong there’s no speed limits just limits to one own imagination. Then again just a brick layer

26. Nik

@enabity: you write that “if the Hab is a 6m diameter 2m high construction, it is approximately 290m^3” and I’m unsure of how you get that number. The volume of a 2 meter tall cylinder with a 3 meter radius works out to just out to about 56.5 cubic meters. Of course the hab isn’t a cylinder. Let’s assume it’s capped off by a half-sphere of radius 3 meters (even if that’s overkill – do we need 18-foot ceilings on our Mars cribs?). That would roughly double the interior volume of the Hab to around 115 cubic meters. You’re claiming a volume that’s approximately 2.5 larger. Please explain.

Edited by enabity

27. enabity Post author

Nik,

It should be a 5.5 or 6 meter RADIUS cylinder, and the volume is still not that large, either 190m^3 or 226m^3. I don’t know why I typed diameter other than I was lifting from MEN2 and was obviously doing a very bad job of doing so. The assertion about the atmosphere is based on what should be correct numbers from MEN2, though. There isn’t enough atmospheric volume to absorb the kind of heat being produced by the hydrazine reaction.

As for cylinder versus hemisphere, the 6m radius sphere would be just under 20ft high at the apex, so I agree that it seems big. More likely, a shape more parabolic, somewhere between a hemisphere and a cylinder, would make more sense.

28. Nik

Thanks – a 6 meter *radius* makes sense for the numbers.

I also agree that the most useful shape would be a cylinder with a spherical cap on top, where the cap’s sphere has a radius that is much larger than 6 meters.

29. Doug

What I want to know, is why is he screwing around with a Rube Golberg system using a heated catalyst to decompose hydrazine to yield H2 that he will then combust with O2 to produce H2O, when he could simply combust the hydrazine directly with O2? KISS is best approach.

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