Steel burns on Rimworld?

[Mossy piglet]

It seems that under most outdoor conditions in rimworld, steel igniting temperatures cannot be reached. (No citation, feel free to disprove me if possible) But, indoors, it is not difficult to reach temperatures up to 2000 celsius. (alpha 13 alpha 14 data is null)  For scale this is twenty times the boiling point of water, and I dare say that is enough to combust any reasonable combustible material. (Also, this seems to be the temperature cap, suggesting it could ve higher yet but not showing) So, maybe steel cannot burn outdoors, but inside under less than perfect conditions, it seems quite possible.

[Shurp]

The issue with real steel fires is not so much temperature as vaporization.  When you heat up wood hot enough the surface molecules break free and can mix with the oxygen in the air.  The boiling point of iron is 2862'C (thanks google!) so even if you have it hot enough to burn you're only going to torch the surface.

(Steel wool burns so nicely because it's *all* surface)

[Nonmomentus Brain]

The issue with real steel fires is not so much temperature as vaporization.  When you heat up wood hot enough the surface molecules break free and can mix with the oxygen in the air.  The boiling point of iron is 2862'C (thanks google!) so even if you have it hot enough to burn you're only going to torch the surface.

(Steel wool burns so nicely because it's *all* surface)

Wouldn't it be sufficient to merely melt the iron oxide, causing it to flow away and allow the (still solid) iron underneath to oxidize? There wouldn't be a flame, but there isn't one either when burning steel wool and according to your own post, that counts as burning. The temperatures would be quite a bit lower: 1360 °C in the case of iron(II)oxide and even lower if there are impurities.

[brcruchairman]

While your post is an interesting read and certainly well-thought-out, there is a major problem with it. <SNIP>

First off, *applause* Well played, sir; your work was really well thought out, well-sourced, and basically awesome. That's what I was hoping for when asking for a physicist.

The points you make are completely valid; I hadn't even thought of surface area as a limiting factor, though it was, as Shurp pointed out, critical in one of my examples. If steel walls were made of fractals, we might be more interested in volume and total heat, but with surface area being a limiting factor (for, as you you point out, we can't expect the steel OR the iron-oxide to melt off) it does seem that a steel wall burning would happen for maybe a few seconds and then the surface would be charred inert.

Mossy Piglet brings up the notion that indoor fires may be a different story, thanks to the confined space. Wood-air mixtures have been reported to be around 1980 *C¹, so a wood stockpile in a room could conceivably heat the room enough to, as you suggest, melt the iron(II)oxide off and burn the next layer of steel underneath. However, in such an environment I strongly suspect that the oxygen in the room would quickly become depleted, possibly before sufficient temperatures were reached.

Let's look at that. For starters, it looks like fires will suffocate themselves at 15% oxygen or less.² Okay, cool. How much energy can be released consuming the ~6% difference in oxygen in a room? Fortunately, I managed to find a pretty neat article about this.³ It tells us that each kg of oxygen can release, through combustion, 2.751 MJ of energy. It also tells us that a good baseline for air density is 1.2 kg/m^3, which gives us 3.302 MJ/m^3 of air space.

For the purposes of this analysis, I'm going to assume there's excess wood reactant mass, and the limiting factor is oxygen in the air. I'm also going to assume a sealed room, strictly to simplify calculations by removing new oxygen entering the system and heat leaving it except through absorption into the walls.

I'm going to take an arbitrary 10 m wide, 10 m deep, and 2 m tall room, with 1 m thick steel walls. (It's a bit short for a room, but it lets us use my figures from the previous calculations, and I'm lazy.) To heat one wall of that room to 600 *C would take 5 220 MJ. To heat it to 1 200 *C would take twice that, 10 440 MJ. To heat four such walls would take four times that, or 41 760 MJ. This is a bit shy of the iron(II)oxide melting point, but let's call it good for the sake of estimation.

Cool, so heating the walls enough to (almost) melt off the iron(II)oxide would take 41 760 MJ. I'm ignoring the floors and ceilings, but let's call it good for now. How much energy can be released by the oxygen in that room? At my convenient 10x10x2 m dimensions, we can see that there's 200 m^3 of air in there, so multiplying by 3.302 MJ/m^3 gives us 660 MJ of energy. That's actually pretty close. [EDIT2: Original figure read 66 040 MJ, but was pointed out by Nonmomentus Brain to have a misplaced decimal point. Remainder of post left intact.]

What this tells me is that it MAY be possible to burn steel in Rimworld. If you add ventilation (such as by opening a door or collapsing the roof) it'll alter the details dramatically. As it is, the figures are too close for me to trust my slipshod math to give much of a conclusion either way. So my official stance on the topic of steel burning in indoor fires is: "I'unno."

Of course, all this does absolutely nothing to add to or change the work which Nonmomentus Brain did on exterior fires; in those cases I'd be inclined to believe that so much heat would be lost to the environment that, as Brain points out, the iron(II)oxide on the surface would remain in place and prevent further steel combustion.

I'd also like to thank Shurp and especially Nonmomentus Brain for their contributions to this discussion; it's pretty pleasing to me to discover new truths, or at the very least strongly documented trends, and it's like they say: "You don't use science to prove that you're right, you use science to become right." Thanks, Brain; you helped me change my opinion to a better one.

EDIT: Added source #3. D'oh!

¹: https://en.wikipedia.org/wiki/Adiabatic_flame_temperature#Common_flame_temperatures
²: https://www.fpgltd.co.uk/suppression-products/gas/hipoxic-oxygen-depletion/
³: http://cfbt-us.com/wordpress/?tag=oxygen-consumption-principle

[Shurp]

OMG, you're right, we're looking at this all wrong.  The issue isn't the steel, it's the oxygen.  Steel is 8+ times denser than wood.  The volume of oxygen you need to bring in to keep that surface burning compared to an equal surface of wood is stupendous.  Unless you have it seriously compressed or your spraying pure oxygen at it you'll never keep it lit. 

And I imagine steel burns *hotter* than wood (ie: releases more energy per molecule burned), which makes it even harder to keep enough oxygen present to keep the fire going (higher temperature combustion products pushing away fresh oxygen)

Magnesium has less than a quarter of steel's density.  No wonder it goes up like a candle

[Nonmomentus Brain]

First off, *applause* Well played, sir; your work was really well thought out, well-sourced, and basically awesome. That's what I was hoping for when asking for a physicist. [snip]

Your high praise makes me very happy! Though I am not yet one, I do, in fact, plan to become a physicist eventually.

I am writing most of this past midnight, so there might be a number of errors. Please check anything that seems peculiar and point it out to me.

You seem to have forgotten to provide a reference for your source number 3 – would it be possible for you to correct that? I'm interested in how they got that number.

Another problem regarding your calculations of the energy potential from the oxygen is that you seem to have multiplied the given value with the entirety of the air instead of just the oxygen content.
Because of these problems, I shall try and determine myself how much energy we can get out of the air.
Firstly, I will have to determine what burns in order to find how much oxygen is needed and how much energy is released. Cellulose seems to make up roughly 50% of wood¹. In order to simplify things, I will pretend that wood is pure cellulose. The molar heat of combustion of cellulose is about -2800 kJ/mol² (the value is negative because energy is released). In case it seems peculiar to have a molar value for a polysaccharide such as cellulose, it seemed so to me as well. The paper gives the formula as C₆H₁₀O₅, which appears to be the smallest unit in the structure. Regardless, we can work with this. It is easy to find that the combustion can be described by the following reaction equation:

C₆H₁₀O₅ + 6 O₂ --> 6 CO₂ + 5 H₂O

We can see that for each mol of cellulose (as the source number 2 seems to define it), 6 mol of oxygen are required. Therefore, for each mol of oxygen, 470 kJ of energy are released. Assuming oxygen is an ideal gas, one mol takes up 22.4 L of volume. If 20% of the air is oxygen, the amount of oxygen is given by 40 m³/22.4 m³/kmol, which is about 1800 mol. Multiplying this with the previously established value for the energy release per mol of oxygen, we get 1800 mol * 470 kJ/mol = 850 MJ of energy released if all the oxygen is used up.

Another way to get to this value is to take the mass of the air (240 kg, assuming 1.2 kg/m³ of density and 200 m³ of volume), multiply it by 20% to get an approximation of the oxygen mass and divide by the molar mass (32 g/mol for molecular oxygen). This gives us about 1500 mol of oxygen, which is close enough to the other value. The difference is probably due to rounding errors and the fact that there is a difference between mass percentage and volume percentage. Simply multiply by the 470 kJ/mol and we get a value of 705 MJ.

For comparison with your source, we can also calculate the energy released per kilogram of oxygen: 470 kJ/mol / 32 g/mol = 15 kJ/g or 15 MJ/kg. This greater than your source's value of 2.75 MJ/kg by a factor of more than 5 – again, I would be very interested to read that article.

Of course, what you have already noticed is that my value for the energy released is lower than yours by a factor of almost 80. I was quite confused by this until I checked your calculations as well – you appear to have dropped a decimal point when calculating the energy released from the 200 m³. It should be 660 MJ rather than 66000. As you can see, this is far less than the 41000 MJ required to heat the walls up.

Of course, these values are still too high, since the air itself must also be heated (we'll ignore the fuel).
Assuming 240 kg of air and a specific heat capacity of 1 kJ/kgK³, we need about 290 MJ of energy to heat the air by 1200 °C. If we go with the most optimistic result, that leaves about 560 MJ to heat the walls.

It is obvious that this is insufficient to heat the entire wall. However, because we are assuming a wood-air mixture, the energy will be released very rapidly due to the massive surface area. Since the assumption that the entire wall would have to be heated was based on the premise that the heat would be conducted away otherwise, we might relax this requirement since there might simply not be sufficient time in order for that to happen.
So, if we don't have to heat the entire wall, how much of the wall can we heat? The maximum depth up to which we can heat the steel by 1200 °C is given by 550 MJ/(1200 °C * 8 g/cm³*80 m²*0.5J/(g*°C)), in other words the energy we put into it divided by the product of the temperature change, the density of the material, the area and the specific heat capacity. This gives us a depth of about 1.4 mm. This may not seem like much, but due to the exothermic nature of combustion, the burning steel might keep the reaction going.

In addition, if the heating is sufficiently rapid to prevent the formation of an oxide layer before ignition, only about 900 °C of heating are required (see my previous post), which would allow the steel to be heated sufficiently to a depth of 1.9 mm.


¹Pettersen, R.C., 1984. The chemical composition of wood. The chemistry of solid wood, 207, pp.57-126.
²Colbert, J.C., Xiheng, H. and Kirklin, D.R., 1981. Enthalpy of combustion of microcrystalline cellulose. JOURNAL OF RESEARCH of the National Bureau of Standards, 86(6), pp.655-660.
³http://www.engineeringtoolbox.com/air-specific-heat-capacity-d_705.html

OMG, you're right, we're looking at this all wrong.  The issue isn't the steel, it's the oxygen.  Steel is 8+ times denser than wood.  The volume of oxygen you need to bring in to keep that surface burning compared to an equal surface of wood is stupendous.  Unless you have it seriously compressed or your spraying pure oxygen at it you'll never keep it lit. 

And I imagine steel burns *hotter* than wood (ie: releases more energy per molecule burned), which makes it even harder to keep enough oxygen present to keep the fire going (higher temperature combustion products pushing away fresh oxygen)

I'll see if I can back your assumptions up with a few calculations. Again, I might make mistakes, so be careful and if something looks wrong, just check and ask.

I will look at the volume of the different substances that can be oxidized by the same amount of oxygen. While I am aware that I thereby fail to take surface conditions etc. into account, it should be sufficient to shed some light on whether the assumption in your first paragraph is true.

If we assume that when steel combusts, iron(II) oxide is formed (this is the oxide that would require the least amount of oxygen), then for each atom of iron, one atom of oxygen is required. Since in an atmosphere, only molecular oxygen O₂ and not atomic oxygen is present in any meaningful amount⁴, this means that each molecule of oxygen can oxidize two atoms of iron. As a reaction equation:
2 Fe + O₂ --> 2 FeO
The molar mass of iron is about 56 g/mol. This means that each mol of oxygen can oxidize 112 g of iron. The density of steel has previously been established as 8 g/cm³. A simple division shows that a mol of oxygen can oxidize 14 cm³ of iron.

As discussed above, each mole of oxygen can oxidize on sixth of C₆H₁₀O₅, the substance that makes up a large percentage of wood. Its molar mass is 162 g/mol. Therefore, one mol of oxygen can oxidize 27 g of cellulose. The density of wood varies widely, so I will use the density of cellulose instead (which is a good idea, since that's the substance for which I've actually calculated anything so far). It seems to be around 1.6 g/cm³.⁵
Using the same method as before, it can be calculated that each mole of oxygen can oxidize about 17 cm³ of cellulose. This is higher than the value for steel, which supports your assumption. However, the difference is quite small and cannot account for the difference in flammability observed.

In response to your second paragraph, I will examine how much energy is released by the combustion of cellulose and iron.
As can be seen above, the molar heat of combustion for cellulose is about -2800 kJ/mol.
The molar heat of combustion for iron is easily calculated if one has knowledge of the reaction equation (which is given above) and the heat of formation of every involved substance. Since the heat of formation is defined to be 0 for the most stable form of an element, the only substance that must be taken into account here is iron(II) oxide, the heat of combustion of which is -272 kJ/mol.⁶ Since the heat of combustion is given by the sum of the heats of formation of the products minus the sum of the heats of formation of the reactants and the heats of formation of the reactants are 0 in this case, the heat of combustion is -272 kJ/mol as well.

As you can see, this contradicts your assumption that steel "releases more energy per molecule burned" (technically, steel doesn't consist of molecules, actually). However, it is clear that this is not because you what you meant is wrong, but because your wording is. Since organic molecules consist of many atoms, they cannot be compared directly to the single metal atoms in this case.

Instead, we might look at the energy per volume, which is more relevant here.
Using the values given above, we can easily calculate that each of cellulose produces about 28 kJ per cm³.
We can equally easily calculate that the corresponding value for steel is 39 kJ per cm³.

Again, it can be seen that the assumption is confirmed, but the difference is not as great as expected. However, there is a significant difference in the temperature of combustion of the two materials. While wood burns at 600 °C (as has been established previously), pure iron ignites at temperatures beyond 930 °C (as previously established). This is not because of the heat released in combustion, but because of the activation energy required to start the reaction.

Furthermore, there's a problem with your assumption that the hot combustion product would "push away" fresh oxygen: the combustion products of steel, various iron oxides, aren't gaseous. In fact, I presume that the higher temperature would only serve to increase convective airflow, thereby bringing in more oxygen. I admit, I didn't think of that either while calculating the numbers above.

The lower density of magnesium is largely because of the lower mass of the magnesium atoms themselves. The actual difference in oxygen required is much lower than you might think.

I hope I could help.

⁴You are alive.
⁵Sun, C.C., 2005. True density of microcrystalline cellulose. Journal of pharmaceutical sciences, 94(10), pp.2132-2134.
⁶http://www.wolframalpha.com/input/?i=heat+of+formation+of+iron%28II%29+oxide



Good night. I hope my sleepiness did not introduce too much error (yes, I know I should have waited until I was well-rested with something like this, but it was too much fun).

[brcruchairman]

Nonmomentus Brain, you may be my new favorite person on the internet. I salute you, good sir. *salutes* Your work is more thoroughly researched, sourced, and clearly written than my own. It brings a tear to my acadamian eye. I'd also like to say that this may be my favorite internet discussion to date, if only for the fact that, even while being an active participant, I've changed my opinion and conclusion four times and counting. Now that's discourse! 

Regarding the source, d'oh! I've added it in an edit. It also appears that, while I played fast and loose with my terminology (e.g., I said "2.75 MJ / kg Oxygen" rather than the more accurate, "2.75 MJ / kg air") I believe our numbers end up agreeing; the value for pure oxygen in the article cited was 13.1 MJ/kg O2, which is very close to your calculated value of 15 MJ/kg O2. So again I salute you; you deduced a figure I had to look up, and did it very accurately. Well done!

Regarding my figures being off for the energy produced, again, good catch! I think I tried converting to kJ in my head and lost track of the units; your figure of 660 MJ appears to be accurate. I'll edit it in my above post for accuracy, but credit goes to you for pointing it out. Thank you!

The point regarding heating the air is also well-taken; convection will transfer heat well, but it isn't free, energy-wise. I also like your point on the topic of partial heating of the wall. Given that the article cited above (Hartin) specified that a fire would smother itself in a 2.4 x 3.7 x 2.4 m room in minutes, I think we can assume that the rate of heat transfer may be sufficient to overcome conduction through the steel wall, as you suggest.

However, I think we run into the problem, in that case, oxygen depletion. With 1800 mol of O2 in the room to begin with, we can expect around 1250 mol left by the time the wood fires have smothered. Unfortunately, I'm not sure at what oxygen percentage an iron fire would smother itself, so I'm going to assume total depletion for the sake of argument. (If a figure can be found for the minimum air oxygen percentage for an iron fire, I'd be very interested in that.)

So we've got 1250 mol O2 left to burn, literally. How much steel can that oxidize? I'm going to borrow your calculations for iron densities, and use one mol O2 for 14 cm^3 of iron. With 1250 mol O2, we get 17 500 cm^3 of iron, which if divided evenly among the wall's surface area of 2 m x 10 m x 4 walls=80 m^2=800 000 cm^2, we get a depth of 0.022 cm, or about a fifth of a milimeter. Of course, this assumes that the walls burn only AFTER the wood fuel, which is not a reasonable assumption. It is, however, a simple assumption, and it being late, I think I'll stick with it for now.

So for an indoor fire, it seems that the lack of oxygen really dooms our steel wall to a sad life of noncombustion. I was trying to find information on outdoor fires, particularly what sort of oxygen and air flow they can be expected to get, but the twenty seconds I spent looking turned up nothing, and I'm a bit too tired to do a more in-depth analysis at the moment. A job for another day, I think. Especially given that in such an example I'd also have to look at heat lost to the environment, and that's a whole can of worms I am far from qualified in dealing with. :p

In any case, for the closed room I'm comfortable saying there won't be a steel fire due to air constraints, though if the door were left open, that may change things. More research for us to do tomorrow, methinks. In any event, I'd like to again thank you, Shurp and Brain, for a really fun and enlightening conversation. I'll be hoping for more awesome posts in the coming days.

[Nonmomentus Brain]

I don't deserve such a good opinion – all I did was expand on your great work! I won't deny that I appreciate it, though.

Thank you for giving the source! It was an interesting read. I hadn't heard about Thornton's rule before.

Now that we've ruled out ignition of the walls in a closed room, let's look at rooms with ventilation. I will assume that the required oxygen is brought into the room and the gaseous combustion product are expelled, but the insulation of the walls, ceiling and floor is still perfect.

Since the released energy is now, again, determined by the amount of wood rather than the amount of oxygen, it is necessary to determine how much wood we can store within this room. Since I know next to nothing about fluid dynamics, I will simply assume that if 50% of the volume is taken up by wood, that leaves enough space for the air to flow. This gives us 100 m³ of wood. If we assume red oak (which is the type of wood used in your first post, so we already have the released energy per kg), that gives us a density of 0.63 g/cm³.⁽¹⁾ Therefore, we have 63000 kg of wood, which is equivalent to about 650 GJ, according to the value of 10.3 MJ/kg given by you.

Of course, not all of this heat is used to heat the walls. Some of it is used to heat up the oxygen that enters the room, while some of it is lost because hot CO₂ and H₂O leave the room.
To find these values, it is necessary to know how much of those gasses must be heated. Assuming, again, that the wood is entirely composed of cellulose, the previously established reaction equation C₆H₁₀O₅ + 6 O₂ --> 6 CO₂ + 5 H₂O is useful again.
If the mass of the wood is 63000 kg and the molar mass of cellulose is 162 g/mol, that gives us 390 kmol of cellulose. With the factors from the reaction equation, it is easy to calculate that the molar amounts of oxygen and carbon dioxide are both 2300 kmol, while the molar amount of water vapor is 1900 kmol. One can already see that the amount of oxygen inside the room (1.8 kmol, according to my previous calculations) is little more than a rounding error here.
Knowing the molar heat capacities of oxygen, carbon dioxide and water vapor (29.4 J/mol*K, 37.1 J/mol*K and 33.6 J/mol*K, respectively)⁽²⁾ and assuming a temperature increase of 1200 °C, one can calculate that heating the oxygen takes 81 GJ of energy, while the heat loss caused by carbon dioxide is 102 GJ and that caused by the water is 77 GJ. The combined loss is 260 GJ, still far less than the energy released through the combustion of the wood.

However, we cannot filter the oxygen from the outside air and only pump that in. A large amount of (for this purpose) inert nitrogen will be brought in as well, and it well be heated as well. If, by molar amount, 20% of the air is oxygen and 80% nitrogen, that means that 9200 kmol of nitrogen will have to be heated as well. With the heat capacity of nitrogen being 29.1 J/mol*K⁽²⁾, this means that heating it to 1200 °C will need 320 GJ.

This leaves 70 GJ (or 70000 MJ) to heat the walls. According to our previous numbers, that should be more than enough to heat the entirety of the walls to the required temperature. When I, however, decided to calculate what the increase in temperature would be, I noticed that it was far lower than 1200 °C. This lead me to check your initial value for the energy required to heat one wall. I then noticed that you were off by a factor of 10 when converting from joules to megajoules, resulting in a value of 5220 MJ instead of the correct 52200 MJ. Fortunately, this error did not cause our conclusions to be wrong, since it only reinforces our stance that it is not possible. Furthermore, when I calculated the same value using Wolfram|Alpha, the result was a mere 41000 MJ⁽³⁾.

After having discovered this error, I decided that instead of calculating the temperature change of the entire wall, I would once again calculate the depth to which a temperature change of 1200 °C could be caused, using the same method as previously. The result was 20 cm, which should be sufficient to ignite the wall.

The next question to be answered is whether the exothermic oxidation of iron is sufficiently exothermic to be self-sustaining. In other words, does the combustion of a certain amount of iron provide sufficient energy to ignite at least the same amount of iron?

To answer this question, one first must find the processes that take place in the combustion of iron in our scenario. I assume that at first, a certain (oxide-free) volume of iron is at the required temperature. Then, it oxidizes, with the required oxygen being at the required temperature. The energy released in the oxidization melts the iron oxide to allow for the oxygen to reach the iron underneath, which is heated by the oxidization of the first volume. Furthermore, sufficient oxygen to oxidize the iron underneath is heated as well. Of course, I know that in reality, these are not discrete steps. It is, however, a useful simplification.

For my calculations, I assume that a volume of 1 cm³ oxidizes and must provide enough energy to melt the resulting iron oxide and heat the 1 cm³ of iron underneath as well as the required oxygen by 1200 °C. As has been shown in my previous post, the oxidization of 1 cm³ of iron into iron(II) oxide releases 39 kJ. This volume of iron has a mass of 8 g and, due to the molar mass of 56 g/mol, contains 0.14 mol of iron.
Due to the coefficients in the reaction equation 2 Fe + O₂ --> 2 FeO, 0.14 mol of iron(II) oxide are formed. The molar heat of fusion of iron(II) oxide is 24.1 kJ/mol⁴, which gives us 3.4 kJ of energy required to melt the iron(II) oxide.
The heating of the next volume requires 4.2 kJ of energy, as can be calculated from the temperature difference of 1200 °C, the molar amount of 0.14 mol and the molar heat capacity of 25.1 kJ/mol⁽²⁾. 0.14 mol of iron require 0.07 mol of oxygen for oxidization. This amount of oxygen requires 2.5 kJ to be heated.
Overall, of the 39 kJ released, 10 kJ are required to keep the reaction going; therefore, the net release of energy is 29 kJ/cm³. If all walls (80 m³ of steel) combust completely, this would mean a release of 2.3 TJ of energy, which seems too high, since 29 kJ/cm³ is only slightly higher than the 28 kJ/cm³ I have calculated for wood in my previous post and the volume of the steel is 80% of that of the wood. This would indicate that there would not be a factor 4 between the two values. However, in this post, I have used values for red oak instead of values for cellulose, as I have done in the previous post. I shall calculate the energy released by the combustion of 100 m³ of cellulose.

100 m³ of cellulose have, given the previously established density of 1.6 g/cm³, a mass of 160000 kg. Dividing by the molar mass of 162 g/mol, this gives us 1000 kmol of cellulose. Knowing that the heat of combustion of cellulose is -2800 kJ/mol, it can easily be calculated that the energy released in the combustion of 1000 kmol of cellulose is 2.8 TJ, which agrees with the value of energy received for the combustion of the steel.

This also shows that cellulose and wood are not the same material and should not be treated as such. It is necessary to carefully consider whether they are sufficiently similar to be seen as equivalent for any particular situation. However, I do not think that the previous results are to be considered wrong, since the amount of oxygen required to burn a certain amount of cellulose should be similar to that required to burn a certain amount of wood.

In conclusion, if the necessary temperatures can be reached, it seems that a storage room for wood can ignite steel if it catches fire and the ventilation systems are sufficient. Furthermore, it seems that fire can spread through steel objects, if sufficient oxygen is present.

¹http://www.wolframalpha.com/input/?i=density+of+red+oak
²http://www.wolframalpha.com/input/?i=molar+heat+capacity+of+O_2,+CO_2,+H_2O,+N_2,+Fe
³http://www.wolframalpha.com/input/?i=%28specific+heat+capacity+of+iron%29*%28density+of+iron%29*%2820m^3%29*%28580+%C2%B0C%29&rawformassumption=%22UnitClash%22+-%3E+{%22%C2%B0C%22,+{%22DegreesCelsiusDifference%22}}
⁴http://www.wolframalpha.com/input/?i=molar+heat+of+fusion+of+iron%28II%29+oxide

[Spectreofoz]

So are you guys basically saying that Superman "The Man of Steel" is also The Torch    

[Wex]

Wait, wait, wait.

Is this the old "jet fuel can't melt steel beams" discussion?
I could swear I already saw this.

[DariusWolfe]

Wait, wait, wait.

Is this the old "jet fuel can't melt steel beams" discussion?
I could swear I already saw this.

I cracked that joke back on the first page, but alas, no one seemed to notice...

[brcruchairman]

So are you guys basically saying that Superman "The Man of Steel" is also The Torch

Yes. Provided that superman uses his own laser vision to heat himself to his own ignition point. Philosophical aside: could Superman be so tough that even he couldn't break himself?

Wait, wait, wait.

Is this the old "jet fuel can't melt steel beams" discussion?
I could swear I already saw this.

Heh, yes, but unfortunately for the joke we're not looking for the melting point of steel, (which, incidentally, is not required for structural weakening) but rather the ignition point. That, plus Brain and I are being total science nerds right now and nibbling away on an interesting problem; 'aint no time for jokes when MATHS and CHEMISTRIES is callin'!

[snip] In conclusion, if the necessary temperatures can be reached, it seems that a storage room for wood can ignite steel if it catches fire and the ventilation systems are sufficient. Furthermore, it seems that fire can spread through steel objects, if sufficient oxygen is present. [snip]

Well, sir. *salutes* Your work, as always, is excellent. I was even going to chime in about enthalpy of fusion but upon rereading your post; you've already included it! So all that's left to do is shake your metaphorical hand and convey my heartfelt appreciation for a truly awesome conversation.

Unfortunately, this seems to be where my utility ends; the next step, a steel fire in an open environment, is very far beyond my training; having neither thermodynamics nor fire management training, I haven't the foggiest idea of how heat and energy would move in such an environment, and haven't the background to accurately interpret any literature I find on the subject. However, the knowledge we've gotten so far (specifically, the risk of a steel fire in a closed room is negligible, while a properly ventilated room acting as a blast furnace could harbor such fires) jives so well with what seems to be the case in reality that I am content. This has been a fantastic and very fun exercise in finding answers to on-the-surface silly problems. Randal Munroe would be proud.

I hope to see you in other threads sometime. In any event, one last time, Nonmomentus Brain, thank you; it's been a pleasure.

¹: http://broadneck.org/wohlfarth1/Energy%20and%20Specific%20Heat%20Honors.pdf bottom of p. 3

[erdrik]

...
I assume it is a coding issue -- that there is no way to keep windmills flammable yet make steel walls fireproof.

Nope. Flammability is a mod-able stat for pretty much all "ThingDefs".
Windmill Flammability is set to 0.4
Wall is set to 1.0, but the steel Item is set to 0.2

So buildables you can customize the material for(like walls) use the stuff they are built out of for Flammability, while buildables that can't be customized use their own Flammability stat.

You can even see the Flammability stat in game via the Info popup.
Select the buildable you want to build, then click the "i" on the left side of the screen.
EDIT:
A note of clarification: In the info popup for items like steel or wood(that can be used to build stuff), there are two Flammability entries. The top most one is for the item itself and the other is for objects made of that item. Interestingly, steel as an Item is not flammable but imparts a 20% Flammability to things that are made from it. I assume this difference accounts for the small amounts other materials necessary to build it that are abstracted out. (like how building a steel bed only requires steel even though your colonists are clearly not just sleeping on a slab of steel)

[Shurp]

Yes, for the steel bed this makes sense -- you've added a mattress, blankets, and pillows that are flammable.  But what are you putting on a steel wall? 

Maybe it's painted with the same rocket fuel as the Hindenburg?
https://en.wikipedia.org/wiki/Hindenburg_disaster#Incendiary_paint_hypothesis

[edit] Hmmm, this actually almost makes sense.  A plain steel wall would rust which would look annoying.  Colonists would be silly enough to paint it with something highly flammable

[erdrik]

Yes, for the steel bed this makes sense -- you've added a mattress, blankets, and pillows that are flammable.  But what are you putting on a steel wall? 

Maybe it's painted with the same rocket fuel as the Hindenburg?
https://en.wikipedia.org/wiki/Hindenburg_disaster#Incendiary_paint_hypothesis

[edit] Hmmm, this actually almost makes sense.  A plain steel wall would rust which would look annoying.  Colonists would be silly enough to paint it with something highly flammable

I highly doubt a steel wall is just a big solid slab of steel.
It is likely a hollow frame with steel plates bolted onto it.
Flammability would depend on what that frame and the bolts(or whatever is use to latch or secure the plates) is made of. Because once those go its not really a wall anymore.
It should probably drop the steel as an item tho.
Maybe not recovering non-flammable materials from fire destroyed buildings is the game balance issue?