Category Archives: Mars

The Martian: Does Mark Watney Actually Have Enough Water?

30 Replies

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…

The Martian: Could a Martian Dust Storm Maroon Mark Watney?

1 Reply

This is an analysis of the technical, scientific and engineering aspects of Mark Watney’s Mars survival story in The Martian by Andy Weir. The primary motivation for criticizing the technical aspects of The Martian is to share the level of scrutiny PHYSIC has been through. I could share early drafts of PHYSIC to demonstrate this, I suppose, but I’m not brave enough to show how thoroughly inadequate my early drafts are. So poor Andy gets another jerk picking on his very popular book instead.

The first technical glitch in The Martian that I’m going to address is at the very beginning of the story. It isn’t an obvious problem. There is quite a bit of room for debate. What really bothers me is the lost opportunity. Weir needs a reason for Mr. Watney to be abandoned on Mars, left for dead by the rest of the crew and without means to communicate. Mars is well known for massive dust storms. Spirit and Opportunity were in danger of shutting down due to insufficient power because their solar panels were substantially blocked by dust from a massive dust storm. Fortunately, dust devils ended up clearing them off and missions continued in full health.

Weir inflicts the mother of all dust storms on our protagonist, winds gusting to about 175KPH/105MPH. That sounds really scary. Winds of that speed would be rated as a category two hurricane. Weathering such conditions in what amounts to a tent in an environment where a hole could end a mission and maybe lead to someone’s death is treacherous. One would think that anything with a large surface area or a flimsy connection would be in danger of being yanked off of the artificial habitat, a comms tower, or the return vessel. Weir reasonably states that the maximum wind speed that the mission is rated for is 150KPH/90MPH, winds that would still be rated as a category one hurricane.

The problem with this entire line of reasoning is that we aren’t on Earth anymore. A windstorm on Mars is nothing like a wind storm on Earth. Atmospheric pressure on Mars is roughly 1% of that on Earth. High speed winds can’t deliver the same quantity of energy on Mars as they do on Earth. To understand how wind is different on Mars, we need to find the appropriate data and equations to give us a sense of scale. First, we need to understand the physics, so we know what is important. The equation for drag works quite nicely.

Force of drag is proportional to fluid density times velocity squared
F ~ p*v*v

I removed the constants and the cross-sectional area from the equation, because we merely want to get a relationship between the force of wind on Earth versus the force of wind on Mars. From this equation, we find out that we need to find out the difference in atmospheric density(p) between Mars and Earth. By referencing the NASA Mars Fact Sheet and Wikipedia, we find out that on Mars p=0.02 and on Earth p=1.2. So, for the same velocity, wind is approximately 60 times as powerful on Earth as on Mars.

If one wants to find out what a 105MPH wind on Mars would feel like, we still have to do some calculating:

Air density on Earth times velocity squared equals air density on Mars times velocity squared
Pe*Ve*Ve = Pm*Vm*Vm
Ve*Ve = Pm/Pe*Vm*Vm = 1/60*105*105 = 183.75(mph)^2
Ve = 13.6mph

Well, that was disappointing. Less than 14MPH. Unless that satellite dish was unsecured, it isn’t going anywhere. If it was sufficiently heavy, it isn’t going anywhere.

It would be disingenuous for me to leave the analysis at this point. The process of a wind storm building up, called saltation, absorbs energy from the atmosphere. Moving the mass of the sand and dust around robs the air of momentum. Therefore, our equation doesn’t accurately represent the force delivered in a 105MPH dust storm on Mars. We really need to understand what is going on with those sand particles before we make a definitive statement that 105MPH Mars wind is not a big deal. The best that I can say with certainty is that there doesn’t seem to be enough energy in the Mars atmosphere to deliver punishing storms and NASA is inclined to agree with me.

This information is an impediment to the story. I’m not certain I’m correct, either. Mostly, I’m bothered by something alien in nature being written off as similar to home. Weir spends no time on the dust storm, as though it isn’t different. How surreal would it be to walk in a massive dust storm with hundred mile per hour winds and not have it completely knock you over? Certainly, a 13MPH gust would not be insignificant when gravity is only 38% of Earth, but explaining these kinds of experiences is the crux of the experience of being a Martian. When these details, the only ones that really matter, aren’t attended to, I feel cheated. What was the point of trying to write a technically accurate account?

What still needs to be addressed is how to maroon Mark Watney under these new conditions. My alternate solution isn’t perfect either. Mark went outside during the storm to attend to the structure holding the communications array to make sure all was secure. As he is working, there is a lightning strike. The dish is destroyed, Mark’s suit and personal communications are damaged and he is dazed. He walks off in the wrong direction in a Martian electrical storm. The rest of the crew looks for him, wrongly concludes he was killed and end up deciding to evacuate.

There are still problems to second guess. I’m not sure how lightning works on Mars. It’s mentioned in the saltation literature as a possible driver of the enormous dust storms. There is also question as to how concerned the crew would be about a lightning strike on the habitat or the Mars ascent vehicle. I still like it better than the notion that there are delicate parts on the MAV. I would definitely think about this more over the long term to come up with a better story.