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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
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