Measuring Solar Furnace Light Intensity ```Name: Ben Status: Student Grade: 6-8 Location: CA Country: United States Date: January 2009 ``` Question: I am building a solar furnace by focusing the reflection of 144 mirrors (1 inch each) onto a single point. I want to measure the light intensity at that point. My intent is to setup the focal point at various distances and measure the differences in efficiency. All the light meters I have found would burn up. Replies: This is a fun experiment and I have done similar ones in a few engineering projects. Light meters can be quite expensive and will not give the results that I think you are looking for. Light meters generally give data on how "bright" something is. I would guess you would be more interested in the "heat" or "energy" that is focus by the light. Luckily there is a very inexpensive solution for testing energy change. All you need is a quantity of water and a thermometer. This is where it can get a little more complicated. The unit for energy is called a Joule. (doing a little research on joules would be good but a practical example is it takes one joule of energy to lift an apple 3 feet) Another more common (to most people but not to scientists) is a calorie. A calorie is the energy required to raise 1 gram of water 1 degree Celsius. It is important not to confuse calorie with the Calories from food labels. The Calorie on a food label is actually equal to 1000 calories with the definition I just gave. 1 calorie also equals 4.18 joules. So now that you know a little more about the unit of energy now we need to figure out how much energy is required to raise the temperature of water that you are heating using your solar furnace (a furnace by definition is a device to used to heat stuff right). To know that you need to know what is called specific heat capacity of water. The specific heat capacity of water is 4.18 kilojoules per kg per Kelvin. Do not get hung up on Kelvin if you have never heard of it. It is something interesting but for the sake of right now know that 1 Kelvin is equal to 1 degree Celsius. So what does that mean practically... Saying 4.18 kilojoules per kg per degree Celsius means that it take 1000 joules to raise 1 kilograms (2.2 pounds) of water 1 degree Celsius. I will give an example of the math in case I lost you a little. I would set up the experiment so that the solar furnace focuses the light onto a clear glass of water. In this example we measured out exactly 100 grams of water (if you do research on how dense water is then you can measure the volume of water instead of weighing it). We put in a thermometer and found an initial temperature of 22 degrees C. We let the light hit the water for 30 minutes and find the temperature is now 29 degrees C. The temperature change is 7 degrees C. We know that it takes 4.18 kilo joules (kilo joules is a fancy way of saying 1000 joules) to raise the temperature of 1 kilogram of water 1 degree Celsius. Our experiment raises 100 grams (.1 kilo grams) of water 7 degrees. 4.18 * .1 * 7 = 2.93 kilo joules. We know that 2.93 kilo joules (0.7 kilo calories) of energy was transferred to the water. If you run the same experiment with the furnace and with out it with just the sunlight shining on it then you will know how much more energy you furnace collected!!! Kevin Hardin Hi Ben, There is little point in trying to determine the light intensity when your mirrors are focused at different focal points, because the results will always be much the same. Your 144 mirrors will be focusing the same incoming light on a single spot, no matter you adjust the focal point. Hence there is no reason why the light intensity should change as you change how close or far away you adjust the focal point. Unfortunately, you are quite right... any normal light measuring meters will be thoroughly toasted when trying to measure such a high light intensity. Any instrumentation to measure such a high brilliance would be extremely expensive! But I am not really sure why you need this information. What you are trying to do is obtain the maximum heating effect. How intense the light is, is not of any direct concern. You should simply arrange the mirrors for maximum heat output. One way to do this is to focus the light on a small block of metal, and time how fast it takes to raise the metal block's temperature by a specified number of degrees. The less time it takes to raise the block's temperature by (let us say) 100°, the more heat you are capturing. You can use this to compare heat output of different mirror arrangements. Regards, Bob Wilson If it were me, I think I would just make my efficiency measurements in a dim room using pseudo-collimated light (i.e., a distant lamp in the largest room you can find). But if you must do this in sunlight... If you make your measurements very quickly, you probably can go to the camera store and pick up a neutral-density ("ND") filter. Photographers use them to cut sunlight so they can use long shutter speeds to blur waterfalls and such. You may need to stack two or more of them to get the attenuation you need (depends on the range of your meter), for example, two 4.0 ND filters would reduce intensity by 2^8, or 1/256. Then you will want to mount it (and the meter) in a light-proof box so the ambient light will not affect the reading. If you cannot make measurements quickly, you will want a *reflective* (not absorptive) ND filter so it will not absorb heat. You can find these at places like Edmund Optics (again, you may have to stack them). These are similar to one-way mirrors (sometimes called two-way mirrors) found at security/surveillance suppliers - most of the light reflects, but a small portion passes through. When I worked in a laser lab, we used a special beam-splitter which would divert about 1% of the energy towards the meter, letting the rest pass, but these tend to be expensive. Good luck. Paul Bridges Click here to return to the Engineering Archives

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