I need a concave mirror with a focal length of 150 feet

[gary350]

I reckon that, on a humid day, the leaves / twigs could be steaming (if you look closely).

There are many sites that describe the OP's idea but the only one I found that 'really worked' was this one which used 127 one foot square mirrors with a wooden target at unspecified distance.

This is an Engineering Project and the actual numbers count. The OP does not describe a solar furnace because his dimensions are not appropriate. Great fun to do anything like that but best to start off with realistic aims.

I found those 2 lens inside of an old photo graphic enlarger. The 2 lens are inside a metal tube holder. I can put any type light at 1 end of the tube the lens make all the light that exit the tube a straight beam across to the other side of the work shop 24 ft away. I get a beam of light like a laser beam but its just light from a light bulb. I took lens outside sun makes a straight beam also but it is still just a straight beam of light. Next I used my Fresnel lens to direct sun light into the 2 lens and it instantly made green grass smoke and burn up. Next I built a holder for the Fresnel lens and the 2 lens in the tube. I set it up outside to shoot a piece of plywood 5 ft away easy target to shoot at using a truck rear view mirror to reflect the sun laser beam 90 degrees east it burns a very quick black circle on the plywood. We were having a drought an burn band for about 7 weeks and today it is raining and more rain in for case for 2 more days. I need to be careful and not catch something on fire 1/4 mile away. It won't be easy to aim this at a tree 150' away. OH well it works like a laser beam & turned out to be a fun project. The only way this will work to burn a tree 150' away is have a servo motor tracking device. Mother nature can finish the job other trees fell 1 by 1 small diameter pieces first 5 to 7 years trees are usually gone on there own.

 

[sophiecentaur]

all the light that exit the tube a straight beam across to the other side of the work shop 24 ft away.

OK. What size is the image on the wall? What actual temperature is your focused image producing?
Also, 24 ft is a lot less than 150 feet. My old doubts are about the actual scale of your requirement. It's very easy to burn wood when a small Sun image is formed.

 

[jrmichler]

I calculated the sun image diameter assuming:
Sun 865,000 miles diameter
Sun 93,000,000 miles distant
Focal length 150 feet
Perfect optics

And got a sun image 17 inches diameter. The next question is how much energy per unit area is needed to burn wood. The necessary mirror area can be calculated from the necessary energy density, the image diameter, and the sun's intensity.

From The Woodburner's Encyclopedia, by Jay Shelton, 1976: "When wood is heated by a flameless source such as radiant energy, it is observed to ignite spontaneously around 600 to 750 deg F." Using the higher number, and assuming both black body radiation and a convective heat transfer coefficient, the minimum energy density needed to ignite the wood can be calculated.

And it's time for me to get back to the grant application that I should have been working on, so I leave the rest of the calculations for somebody else.

 

[digitalghost]

I think you can simulate your mirror with this tool - add a mirror, and ruler to get numbers.

https://www.engineersedge.com/calculators/ray-optics-simulation/index.html

 

[sophiecentaur]

And got a sun image 17 inches diameter.

I agree with that figure; there is a simple way to an answer for a simple system. Assuming the Sun subtends an angle of around 0.52 degrees then so will the focussed image. The power density of that image would be about four times the incident energy density. Nice and warm but not lethal. The description in the original post doesn't include the focal length so the power density that was achieved (enough to "burn stumps") implies a focal length of less than a few feet. I just can't deal with the idea of doing optics with feet and inches. Even in the US, they don't sell camera lenses with 6" focal lengths (do they??).

Comparisons between these systems involves the square of the linear dimension (areas are what counts) so you rapidly lose effectiveness as the image size increases.

The diagram in post #25 shows an intermediate lens which can increase the focal length to 150 feet. But the diameter is much smaller than the main lens / reflector and there would be a lot of power loss (spillage) round the sides which would further reduce the power density at the image. A bigger condenser system could well cost as much if not more than the original fresnel lens. I wonder if a large diameter concave fresnel would be available.

 

[jrmichler]

I just finished doing some heat transfer calculation for something else. It's been a long time, so getting back up to speed took time. Having done the studying, I could not resist doing some calculations for tree burning. Making the following assumptions:
Emissivity = 1.0
Temperature of wood = 750 deg F = 672 K
Ambient temperature = 70 deg F = 21 C = 294 K
Zero wind
Zero heat loss by conduction into the wood

The results:
Radiation heat transfer = 11,200 W/m$^2$
Convective heat transfer = 3,200 W/m$^2$
Total heat transfer = 14,400 W/m$^2$

The energy from the solar furnace must be larger than 14,400 W/m$^2$ at the tree in order to heat the wood to burning temperature. Assume that solar irradiance is 800 W/m$^2$ at the mirror, then the minimum mirror area is 14,400/800 = 18 times the area of the solar image on the tree.

The solar image is 17" = 0.43 m diameter and 0.15 m$^2$ area. Therefore the mirror needs a minimum 0.15 * 18 = 2.7 m$^2$ reflecting area. That's a theoretical minimum based on steady state calculations and assuming a perfect mirror. A realistic minimum mirror would need to be well over 2 meters diameter. That was a surprise, I expected the mirror would need to be larger than that.

Edit: The LaTeX rendered correctly in preview, but not when I posted it.

 

[sophiecentaur]

I just finished doing some heat transfer calculation for something else.

Many thanks for doing the leg work in this one!! An interesting result

I was looking for a practical example of this and I realised that the hot air gun, used for paint stripping is a similar situation. Those things are potentially dangerous and houses have been set on fire by careless use. In my experience, to get a piece of wood to actually burn, you need to concentrate the gun onto an area of only a few cm squared (not far from the nozzle) and the power used it, say 1.5kW. You can cause flames when the peeling paint catches, after a couple of minutes. The typical Power flux from the hot air would be my estimated power of 1.5kW over an area of, say (10cm²)or 0.01 m² which is 150kW/m². Only ballpark, of course but it implies solar collection by an area of say 200/m². That differs from your calculations by a factor of ten but yours is a steady state calculation and, as I've never tried to burn a house down, I've never reached a steady state condition. Also, the heat loss from a house timber would be different. 'Fire retardant' doors will burn eventually.

 

[jrmichler]

I was looking for a practical example of this and I realised that the hot air gun

Much of the hot air does not actually add heat to the surface, but it does give a good upper limit to the amount of heat needed. Using the factor of 200 * 0.15 m$^2$ = 30 m$^2$ area for the mirror. That's a 6 meter diameter mirror for the upper limit. My steady state calculation results in a 2 m diameter mirror as the lower limit.

And the next problem is aiming during the burning process.

 

[gary350]

High wind yesterday made all 3 trees about 10 feet shorter.