Effect of ambient temperature on cooking

Putting a lid on does save fuel.  The gear skeptic just showed it was an insignificant amount of fuel saved.

Evaporation is the problem.  That consumes a lot more heat than convection or radiation or conduction.

But there’s little evaporation when you start and it’s cold.  And it’s only a few minutes when it’s hot, evaporating water.  The water heats up from the stove (even when the heat is turned down) a lot faster than it cools from evaporation.  The gear skeptic showed some cases of wind where there’s more heat loss, and in those cases it’s more efficient to turn the stove up some.

If the ambient temperature gets within about 10F of the boiling temperature of your fuel then that will have an affect, it starts slowing down.

I record my fuel use/cup boiled and always keep an eye out for online test results.   I haven’t found any yet directly comparing over temp {EDIT} except

• MSR’s rule of thumb “Low temps, 8kph wind and melting snow can require up to 3x more fuel than this”
• BPL tests

The BPL tests at ~70F/windless ambient led to oddly high fuel use.  I typically get 9-10 cup/oz fuel vs BPLs 5.6.  Every other test online I’ve come across achieved~ 8-12cups/oz fuel in those conditions. I don’t think elevation accounts for that much difference.

Without wind penalty, I found 32F late October required almost exactly double fuel vs room temp summer.

The standing water temp matters more than the ambient temp, so 32F ambient in March will be very different than 32F ambient in late October.

I have found that the initital water temperature makes a difference (of course) but not so much the ambient temperature.  To boil 500 ml of water in 8 minutes requires anbout 375 watts.  Given that a stove is say 50% efficient, that means that 375 watts is also heating up the surrounding environment.  With a good windscreen, you can keep more of that lose heat closer to the pot/mug and that probably minimizes the ambient temperature affects.  My 2 cents.

Without any energy loss warming up 1 liter of water by 1 C requires 1000 calories (1kcal) (sorry for using the metric unit – but that is how it is defined). Burning Propan gives 11900 calories and Butan 11800 (lets continue with 11830 resembling the theoretic average of a standard 30/70 distribution). Usually cold water has about 4C (but if you want you can go with just above 0C before it freezes). Room temperature (the lab temperature defined) is 20C. This means simply that we would need 16000-20000 calories “extra” or g fuel between 1,35 to 1,7 g. For the remaining 80 C you would need 80 kcal or 6,76 g of fuel. Altogether between 8,1 – 8,5 (some rounding was done) -> 0,3 ounzes. This is an increase by about 20-25%. Not so little! – BTW gasoline has 11 300 if anyone wants to calculate their winter stove.

The problem is that this is without energy loss! All factors you mention (wind, ambient temperature, evaporation) basically mean that energy loss occurs.

Taking the table provided we would need to compensate for the water quantity (about half) and the water temperature.

This means on the first range each degree C means about 5% increas while for the second range this increase to 10%. Apparently a non-linear model. To calculate a realy regression would need more values. (For the inefficency % gram our ounze does not matter. And if you want to compensate for 1 degree farenheit use 1,8 F = 1C (2,8% first range and 5,5% / 1 F).)

This means for me that anything that controls the energy loss (see first post) is just a bit more important the lower the temperatures get. So, you maybe leave the lid at home during summer – but I would think twice during winter/fringe season. Same for the pot – choose wisely. The lab is good to control conditions. However, it misses on a lot of variable.

And yes, I guess a fully encompassing windscreen will reduce radiation loss given that a buffer zone is developed that does not have such a step gradient.

The MSR example with “low temperatur” is … well lets call it marketing.

To bring 1 liter of water from O C to a 100 requires 100 kcal of energy to be delivered

To bring 1 liter of water from 20C to a 100C requires 80 kcal of energy to be delivered

Stove efficiency is pretty much independent of ambient temperature and not at all dependent on water temperature (in fact, it is independent)

Average stoves are 50% efficient and a very good HX System in no wind can push up to about 70% efficiency.  The excess heat is a lose and it can and does minimize the impact of the ambient air temperature.  Wind is a different story.  My 2 cents.

I calculated fuel-to-water efficiency when I did my windshield tests in early March.  Results were …

60–70% for still air or 2mph breeze with a windshield
45% for 2mph breeze and no windshield
20% for 4mph breeze and no windshield
50–60% for 4mph breeze with a windshield

Test was 750ml water, heating from 45°F to 200°F. Ambient temp around 60°F. Efficiency calc based on lower heating value of butane of 45.3 kJ/g. More details in the ‘Canister Topped Stove Performance in the Wind’ thread.

Jon Fong: This was not intended as a gear discussion. Indeed I want to suggest that the factor of stove efficency, will technically interesting, is of questionable practical significance for the fuel consumption. This as it usually provided in a laboratory settings. Loss through wind, ambient temperature and what not is so much more contributing to fuel consumption. And in the end of the day, that is what adds the real extra weight (at least for me – stove: 85 or 180g (inverted from about 5C); fuel: 390g). So if I can predict my fuel consumption I can go light(er) (and yes, stoveless is an option – but not really in winter for me).

However, if I understand you correctly, you are suggesting that the stove + pot + lid + windscreen combination has an effect on the fuel consumption under real-world conditions. Great, I agree! Help me to make sense of your numbers. How much more fuel do I need to pack?

Your suggestion would propose between 166-200% (rounded) fuel multiplier. The room temperature measure suggests more like 300% (4g -> 12g). And it clearly shows that temperature and environment indeed have more impact.

Adrian Griffin: I understand that everyone wants to compensate for wind. And I can see from your numbers that wind, without question, has impact. I also see, and fully agree that a windscreen indeed makes a notable difference. As has been argued earlier because it reduces:

• Flame displacement
• Heat removal
• …

—-

Combining your two suggestion we would get something like

Fuel = Caloric need * f(stove efficency) * f(wind) * {f(xi)}

Stove efficency may very well be a static value but could be correlated to wind and what not. Hence, f() being a function to account for eventual fluctating factors. The intersting part is the set of function {f(xi)}  that we are missing.

I am also intrigued by the idea of using cooking time as an approximation. Still, we would need a formulaic way to assess that based on environmental aspects during planing.

“So if I can predict my fuel consumption I can go light(er) (and yes, stoveless is an option – but not really in winter for me).”

Albin, consumed fuel depends on your water temp, wind conditions, elevation, the pot used (does it have a heat exchanger?), the stove itself and it’s settings and how its shielded from wind.

No way around that.  There is no single formula.

For me, after calibrating with my field use on numerous trips, I settled on:

My flame setting is per GearSkeptic, as low as still provides flame contact with pot bottom in a given wind condition.  I don’t use a wind shield but do use large trees as a wind block

The stash pot and it’s Hx uses ~ 60% the fuel of a standard Toaks

But your situation is probably different

Here is my perspective (YMMV),

If you want a highly efficient system to boil water that is reliable and repeatable, buy an MSR Windburner. It will boil a liter of water (20 C) using about 10 grams of fuel. If the wind is gusting, it goes up to 12 grams. To estimate fuel usage at other temperatures and volumes, you need to use some math.

10 g of fuel to boil 1 liter (20 C to 100 C) is 5 g/ 80 C, so is the water is 0 C you will need to increase the fuel by (10/80)*20 or 2.5 grams of fuel

If you are boiling less water, it’s linear. Boiling 0.4 liters of 20 C water will require (10*(0.4/1.0) or 4 grams of fuel.

Want a good estimate, do the math. I have never found altitude to impact performance, and I have tested this above 13,000 feet. Wind and water temperature have been the biggest factor to me.

The second-best option is to a) Use a remote fed stove with B) a larger diameter pot along with C) a full wrap, full height windscreen and D) get to understand how your stove works (performance vs burn rate. If you do those things. You can boil 1 liter of 20 C water using 12 to 14 grams of fuel. Same math rules apply for this system. Additionally, with this system, you can actually cook meals and not just boil water.

If you are committed to a cannister topped stove, use a full height windscreen and “carefully” wrap the windscreen around your cooking system. Leave a large annual gap (2” would be good) and only wrap the windscreen ¾ around to leave an opening downing. Again, a larger diameter pot will help, and you still need to dial in your stove performance and do the math. My 2 cents.

Jon.

Thanks for the math. In the statement, “10 g of fuel to boil 1 liter (20 C to 100 C) is 5 g/ 80 C, so is the water is 0 C you will need to increase the fuel by (10/80)*20 or 2.5 grams of fuel” are you saying you are starting with liquid water at 0 C, or frozen water aka ice/snow at 0 C? Seems like the former.

Not Ice, but ice cold.  You can calculate that as well though, someone here recently did that .  I recall the numbers being about 80% of that to reach a boil from 0 C.  Assume a safety factor and double the fuel consumbtion.  So with an MSR Windburner loaded with 1 liter of ice, you should be able to reach a boil using 20 g of fuel.  My 2 cents.

We seem, apparently, to agree that wind has a role in the effectiveness of stoves (and therefore fuel consumption). Let us then start from there.

As the national weather service shows with its wind chill chart (https://www.weather.gov/safety/cold-wind-chill-chart), we humans are cooled (and perceive it so) as a function of temperature and wind (the equation at the bottom of the chart). Can we, scientifcally spoken, agree that if there is an equation descrbing how wind/temperature combinations cool humans (for whom the chart and equation is produced) we can do one that describes cooling for our cooking contraption? This would mean that the lower the temperature AND the higher wind the larger is the cooling effect with impact on cooking time.

It appears reasonable to deduce that lower temperature also implies a cooling effect that gets stronger the colder it is even under calm conditions (There is a term in the equation at the bottom of the wind chill chart with contains “only” the variable T with a non-zero constant). Meaning, that even in calm weather cooling increases by lower temperature (for humans certainly and very likely for stoves as well).

This is supported by our observation that when cooling occurs, we can observe that our cooking times are getting longer. Longer cooking times entail more fuel. And that independent from the stove and its effectiveness. I concede immediately that there are factors on stove efficiency that are also impacted by wind (flames blowing away for example) and the pot. Those are not contested here and need to be considered. However, the cooling effect occurs as well.

So, what I wonder is “What is the equation for heating water” given the equation for humans exist. Meaning what are the terms, constants and exponents.

Next step is then to consider what we can do about this. Humans can take on clothes to control the effects. What can we do for our stove that has effect on cooling?

Wind can be countered by a windscreen. We use it primarily to reduce the wind speed (with the obvious effects). However, a wind screen also reduces heat loss for cold temperatures as the still(er) air does warms up. This implies a flatter gradient and heat-loss is further reduced.

Luckily (or unfortunately), there is another effect that comes into play. Hot air rises and cold air is sucked in from below (and yes, this could be considered a localized wind). We need this chimney effect to ensure sufficient oxygen supply. Bad luck for us is that it also causes cooling. Which again, becomes more pronounced the colder it is due to the heat gradient. We need to accept this but maybe can quantify its consequence on fuel dependent on temperature or just ignore it by backing it into a windscreen effect (summarizing windspeed and others) as we can establish a test protocol for it.

However, as we have seen that convection occurs through the chimney effect, we find a second “event” where convection occurs. That is the convection for the water surface where the hot water vapor goes up. The colder it is, the faster this will go. (Coincidentally, we will end up with less water in the pot. Which we could account for by the increased cooking time). Putting a lid on our pot will reduce this significantly. (Hence my induction that the effect of a lid is more pronounced with cold temperature).

Now, I do not suggest quantifying each of these factors separately (although it would be nice). However, I suggest it as a hypothesis that ambient temperature has an effect, like wind, on cooking time.

In due time (next winter) I will set up a test protocol. However, if anyone has some data showing cooking time (or already translated it into fuel consumption) based on ambient temperature (and wind) I would be glad if those can be shared. I will try to do regression analysis to suggest a function.

Here are some thoughts from an engineering professor.

If you are interested in the fundamentals of heat transfer, you can learn the basics from a sophomore-level engineering textbook. You will start by understanding the distinctions between conductive, convective, and radiative heat transfer, which are quite jumbled in many of your posts in this thread.  The analogy with cooling humans, which is often dominated by evaporative cooling from the skin, is not very apt. And similarly, jumping right into guessing an empirical “equation” without understanding the underlying science won’t generally give much insight or accurate results. There is no single “equation,” but a set of coupled equations that connect to the fundamental mechanisms of heat transfer. You will not develop a meaningful equation by guesswork.

On the other hand, if you’re mainly interested in predictive data for fuel usage and cooking times for your particular stove set-up under the conditions that you realistically experience, there is an easy solution. Make a number of careful and controlled experimental measurements (as Jon often does, very beautifully), tracking the relevant ambient conditions, and then interpolate between these using a simple function (e.g. a multivariate polynomial). I expect it will be accurate enough for practical use.

With respect to wind, I believe that by far the largest impact is the wind shifting the flame profile away from the bottom of the pot and that the impact of heat transfer due to convection is minimal (see photo below, stoves not windscreen).  In fact, I suspect that the inefficiencies due to shifting the flame are at least an order of magnitude greater than the convection loss.

The MSR WindBurner is known to be the most fuel-efficient cooking system in almost all-weather conditions.  You can boil 1 liter of water using 10 g of fuel.  A great experiment would be to remove the neoprene sleeve and test the MSR WindBurner in the wind (or lower ambient temperatures).  I would be willing to bet $100 that it would not change the fuel consumption by more than 20% (sleeve/no sleeve) if you could even measure a difference.  That’s my 2 cents.

“It appears reasonable to deduce that lower temperature also implies a cooling effect that gets stronger the colder it is even under calm conditions”

I think the gear skeptic showed that this effect is small, so it doesn’t really matter

He showed that if you turn the stove down so it takes longer to boil, that it uses less butane.  If cooling was significant, the longer boil time would use more butane.

What is significant is if the water you’re boiling starts at a colder temperature.  The amount of fuel used is proportional to the temperature increase.  Approximately.