Burning Man Shelter: Cooling and Ventilation

I’ve been busy with life in general, and hadn’t found the time to make a blog post about Burning Man in a while. But, here is the follow up post about cooling your shelter on the playa, just in time for the man next year. )’(

(You can’t tell, but that BM emoticon above has a sad face.)

Why Cool?

You are in the freakin’ desert! The Sun is shining on your yurt all day long heating it up (Radiative heating). The air outside is hot and dry as hell, and heating it up (Conductive heating). When you’re sleeping/hanging out inside your yurt, your body is heating it up (Generative heating). Somebody’s gotta cool it! It’s plain thermodynamics… ‘Nuff said!

Ways to Cool?

The simplest thing to do would be to buy a portable Air Conditioner / Cooler, and a portable generator to power it. But, if you are not a DIY’er, then you might as well skip the whole hexayurt building thingamajig and drive up in an RV and “camp” out on the playa in luxury.

No!!! This is Burning Man! And we get to hack life here on the playa! (Well, we each get to re-invent the wheel for ourselves, and in the process end up discovering a more efficient way to do things) So, in that spirit, let’s take a look at Evaporative Cooling.

Evaporative Cooling

Having been a kid in the 80s growing up in India, I know a few things about cooling water through evaporation. How do earthen clay pots keep water cool in the hot arid climate? How do mud huts stay cool on hot sunny days? (Well, after splashing water on them) How do canvas pouches keep water cool in the desert?

Not going into too much detail, it’s because they are all permeable membranes with water evaporating on the outer surface extracting heat required for the state change (from liquid to vapor) from the water/air inside the membrane (pot, canvas, mud wall).

Air Coolers expand upon the same idea, and use that cold water to extract heat out of air to make it cool before blowing through a fan. They also help moisturize the air, which is a necessity in the hot dry playa. But, for what they are, they are quite pricey and quite power hungry. So, let’s figure out how to build a DIY Air Cooler out of dirt cheap materials for under $40.

Home Depot 5-gal Homer Bucket......... $3.00
120mm 12V DC Fan @ $7.50x2............ $15.00
12V DC Submersible Pump............... $5.00
Tube for drip system (enough for 2)... $3.50
Dura-Foam............................. $12.00

The setup goes something like this…

You basically drill holes along the top half of the bucket. You then line it with foam, and fill the bottom half with water. You then lay a drip system that pumps water on to the top of the foam, which slowly evaporates through the holes you just drilled, thus cooling the air/water/stuff inside the bucket. Oh, and most importantly, you blow this cooled air using fans from the top of the bucket. Simple stuff! Right?

Here’s what the setup looks like…

Please don’t cut the entire way down the bucket!!! I just drew a sectional diagram for illustrative purposes.

And here’s a simple wiring diagram, if you aren’t sure how to connect the wires…

A quick note about the battery: Use a Marine/Boat Battery to power your setup instead of a Car Battery. They have a more steady discharge curve, and are better for powering low power devices over long periods of time, unlike Car Batteries which are designed to discharge quickly to power the high-torque needed for the engine starter. Your headlamp lights mainly run off of the alternator, and use the battery only when the engine is turned off.

Staying Cool

Just blowing cold air inside your yurt may sound like a quick way to cool it. But now, we have to find ways to get it to stay cool, and this is where the R-Value (aka Thermal Resistance) of your build material matters. It specifies how slowly heat can transfer through the surface (or in other words, coolness can be lost), and the higher the R-Value, the better the insulation.

The rate at which this happens depends on three things:

How much insulation the walls provide. (aka the R-Value of the panels)
How much colder the inside is relative to the outside. (the higher the temperature difference, the larger the amount of heat transferred)
How much surface area there is for heat transfer to take place. (Did you know? It’s better to pack cold food/ice in one large cooler than 2 smaller half-sized ones.)

The unit of R-Value is defined as h·ft²·°F/Btu.

That’s… time in hour * area in sq.feet * temperature difference in Farenheit / heat transferred in Btus (British thermal units). Gosh, I love imperial units!!! They make the math calculations so much simpler… said nobody! (Actually, it isn’t that bad if we stick to those units all along, but a lot of science literature is in S.I., as it should be.) Anyway, if it isn’t obvious from that…

The more time it takes to transfer the same amount of heat across the same surface area, the higher the R-Value. i.e, R-value is directly proportional to time. So, get the higest value you can afford in money and space, so that you can run your coolers less frequently. Similarly, the more surface area required to transfer the same amount of heat over the same amount of time, the higher the R-Value. Also, the more the temperature difference required to transfer same amount of heat… you get the point.

In addition to that, you need to counteract all the other sources of heat, such as your own body heat, Sun’s radiation, etc.

Are you guys ready for some R-Value Mathematics?

Let’s assume we have a closed Yurt without any ventilation, that we’ve somehow managed to cool it down by 20ºF, and that we have 4 persons inside it.

Let’s first figure out how much heat is being added to our yurt’s interior. First, there’s body heat:

330 Btu/hr per person x 4 persons = 1320 Btu/hr.

Next, there’s Conductive Transfer of heat through the yurt’s walls from the hot air outside. i.e, R-Value part.

In our case, we can’t change the Surface Area of our Yurt (well, we could if we redesign it to be a geodesic dome), and the temperature difference is what we want to achieve… so, we will keep that target as a constant as well. That leaves us with rate of heat transferred per hour being the variable that’s directly affected by the R-Value. (Technically, inversely affected).

We know from before that the unit of R-Value is defined as h·ft²·°F/Btu, and our R-9 construction material implies that we have an R-Value, R = 9h·ft²·°F/Btu.

Given that the inside of our yurt is 20ºF cooler than outside, and our Yurt’s walls and roof have a total combined surface area of 384ft² (12 panels that are 8′x4′ each.), we could easily solve for the hourly rate (x) of Heat Transfer by solving:

9 hr·ft2·°F/Btu = 1hr * 384ft² * 20ºF / (x) Btu
=> (x) = 384*20/9
=> x = 853.33 Btu/hr.

Then there’s radiative heat from the sun, but the reflective surface of the R-Max panels, and the fact that the inside is well shaded and dark kinda makes it negligible.

The same applies for radiative heat from the earth. Desert sand cools down pretty quickly in the shade, and the reflective tarp that we used in Yurt construction also helps make any radiative surface heat negligible

So, in total there’s 2173.33 Btu added to the system inside. To maintain a constant temperature inside, the same amount of heat must be extracted out of the system through some rate of water vaporization! This is on top of the what you need to cool it down to the desired temperature (about 20ºF cooler than outside) in the first place. If you’re smart, you can see from above that it’s way faster to cool the yurt with you not inside it as you will be the primary generator of heat :)

Let’s now take a look at the other side of math to balance this out.

Enthalpy of Evaporation

Water is a great liquid for cooling purposes, primarily because of its strong Hydrogen Bonds holding its molecules together, that cause it to have a high enthalpy of vaporization – a.k.a latent heat capacity. (Did you know? It takes 5 times as much heat to vaporize boiling water at 100ºC/212ºF as it takes to heat it from 0ºC/32ºF to 100ºC/212ºF in the first place.)

For water, this value is 2257 kJ/kg. (kilojoules / kilogram)… 1 kg = 1 liter, and 1 kJ = 0.94 Btu. So, it’s about 2121 Btu/l. So, it takes just a bit over a liter of water per hour to reach equilibrium and maintain one of our yurts at 20ºF cooler than the outside with you sitting inside it. But, we also need to remember that the water evaporating from the foam surface in our bucket doesn’t just extract heat from inside the bucket, but also from the air outside… so to be conservative, let’s double the amount of water needed. So, if your home depot bucket setup can actually get about 2 liters (approx. ½ gallon) to evaporate in an hour, you can indeed maintain a yurt interior that’s cooler by 20ºF with 4 of you present inside.

Even if all the heat needed for evaporation is extracted from the air inside, it still takes about a liter/hr to maintain that temperature. That’s not a realistic expectation to set, even though you could do it with a large enough evaporator setup :/

Luckily, all that math above does not apply in our case! Say WHAAAAAAT??? Yep! if you look above, the very first assumption is that we are looking at a Yurt without ventilation. And, all that math still applies to that closed system :) But, we have Fans and Ventilation!!!

Ventilation!

The hole on top of our roof for exhaust and the holes in our bucket for intake mean that we don’t have to cool and maintain the equilibrium for the entire volume of air inside the yurt. Instead, we just need to evaporate enough water to cool the smaller volume of air inside the bucket, before pushing it into our Yurt where it get’s heated up and pushed out of the exhaust. The cool air enters the yurt through the intake placed closer to the feeling, and pushes out warmer air up through the exhaust vent in the roof… thus rapidly bringing the temperature of the yurt down close to the temperature of the cold air inside your bucket air cooler :) This is why ventilation in yurts (and in buildings) is important!

You should also note that most of the heat added to the system (both conductive, as well as body heat) also escapes through the roof vent, thus reducing the cooling requirements of your evaporator.

Sorry about those unnecessary equations, and I wont bore you with more math about cooling the air inside the bucket ;) Just know that it doesn’t take much water to cool the air inside the bucket before blowing it through the fan. Though, I’m glad to do it if you guys are interested. But the main reason was to show you the importance of ventilation in any construction. Also, it’s more efficient as the heat extracted for vaporization comes from air that’s being sucked in through the holes in the bucket, which means none of the water is wasted to cool unused outside air.

Hopefully, you got to learn something a bit useful today, along with a lot that’s useless :)