How to calculate the evaporative cooling rate needed to protect a house from forest fire

Question

Recently in our area there has been a large forest fire and I've been looking into home defense from such things.

I am not a physicist - but can do some basic math.

I was wondering how I could calculate if a 'evaporative mister' type system could be used for such to reduce the ambient air temp to stop the house catching fire (dont care about cosmetic damage).

The goal would probably be to reduce temperature of air/surfaces of the house by approx 1000F to keep them below combustible temperatures.

The area is very low humidity between 4% to maybe 15% during wildfire season.

How can I calculate how much water/mist I need to put into the air to reduce temperature below 400F.

Very rough simplest equations are fine - I know its not an exact science when dealing with wildfires.

I found this formula on wiki but I don't know how to adapt it to use it to calculate water need for a temp drop of

TLA = TDB – ((TDB – TWB) x E)
TLA = Leaving Air Temp
TDB = Dry Bulb Temp
TWB = Wet Bulb Temp
E = Efficiency of the evaporative media. 

Anything I forgot/missing - be appreciated to know.

Some restrictions/thoughts I had

1. The roof would need to stay generally wet light continuous layer of water
2. Misting/irrigation could not use more then 12 gallons per minute (max well output)
3. Be nice to second use the misting system for outdoor aircon on off time (ie 20% capacity)
4. Windows/glass would need some of IR shielding to stop ignition of furniture inside the house.

Answer 1

Instead of temperature drop, we have to to consider amount of heat transferred to the building from the wildfire. The temperature of the structures will rise towards the ignition point depending on the temperature and closeness of the heat source. Cooling can then slow down the heating or in best case stop it completely.

The heat transfer is is a complicated thing to calculate for real, especially in this kind of environment where winds are probably turbulent and heat is transferred in many forms. Luckily there has been research on the subject and we can use those results for estimating the needed cooling.

From the practical point of fire safety, one of the most important thing seems to be distance of the closest fire front from the house. There is an article on this subject "Reducing the Wildland Fire Threat to Homes: Where and How Much?" by Jack Cohen, nicely summarized in [http://www.saveamericasforests.org/congress/Fire/Cohen.htm]. The article contains a graph of radiant heat flux as function of distance from the wild fire front as well as wood ignition times as function of the distance.

Knowing the radiant heat flux and energy needed for vaporization of water, it is possible to derive an equation for cooling effect of the water:

$q = \frac{m H_{vap}}{A}$

where

• q is radiant heat flux [kW/m^2]
• m is amount of water used per second [kg/s]
• H_vap is heat (or enthalpy) of vaporization of water [kJ/kg]
• A is area of the walls and roof of the house [m^2]

As an example, let's consider a house with outer surface area of 500 m^2. For water, the heat of vaporization is 2257 kJ/kg. The amount of water we can spend is 12 gallons per minute, that is 0.76 liters per second. From this we can work that the maximum cooling effect produced by the cooling system (all water vaporized instantly) would be:

$q = \frac{m H_{vap}}{A} = \frac{(0.76\: \mathrm{kg/s})(2257\: \mathrm{kJ/kg})}{500\: \mathrm{m^2}} = 3.43\: \mathrm{kW/m^2}$

When comparing this to the model of the article where heat flux from for example 20 meters away is is around 45 kW/m^2, and heat flux from 22 meters away is around 40 kW/m^2, we can say that the cooling would have approximately the same effect as moving the tree line by two meters.

The model is known to overestimate the heat flux so the actual distances may be smaller, but anyway the cooling effect has approximately the same effect.

Things to consider:

• I assumed that we can't know what side of the building will be closest to the fire or that house will be surrounded so all sides need to be cooled.
• Distances given in the graph in the article are for wood. For other materials, the distances will be larger or smaller. Thickness and density of the material also matters.
• According to the article, in a full-blown forest fire the burning happens very fast. If the house can stand the fire for two minutes, it won't probably ignite as the fire has moved on.
• Clearing the surrounds of the house and having nonflammable materials would be much more effective way of shielding the house. From the link I gave: "Given nonflammable roofs, Stanford Research Institute (Howard and others 1973) found a 95 percent survival with a clearance of 10 to 18 meters and Foote and Gilless (1996) at Berkeley, found 86 percent home survival with a clearance of 10 meters or more." This might of course have effect on how nice and cozy the yard is.

Answer 2

I suggest you look into fire protection nozzles, one site is http://www.bete.com/applications/fire-water.html

You will need more than a mist system. You need something like a water wall protection system. These use a lot more water than your available source. If you look at the nozzles alone you see rates of flow of about 17 to 300 gal per minute.

I would suggest you contact this company (or other fire protection companies) for more information.

Systems I have seen use fire hose type systems, again lots of water. I agree that the best solution is trim back all flammable materials as far as you can.

Also check with your local fire department.

Answer 3

I humbly submit that aside from any evaporative benefits, a simple lawn sprinkler on the roof would help prevent ignition from floating embers from burning trees. I'd think that these embers could float for great distances and readily ignite tinder dry grasses, foliage, house siding or roofing material, way before combustibility from heat alone.