I know that the water will get hot. I know that i have to have a means of cooling to avoid overheat. What i don't know, is how powerful this cooler must be. There are 2 ways: The first one is to get some peltiers and run some tests... But i prefer the second one which is more scientific. I can run a test without a cooler and calculate the amount of energy that the water got, and thus calculate the power of the cooler. So, hands on workbench:

First of all, i need a tank with a known amount of water, that is 1 liter, and a thermometer	A small aquarium water pump, submersible	I used these tools to hold the (one) tube steady above the heating lamp
I put the tank in the water and connected the parts with the hose	I made a hydraulic circuit: Tank to pump to tube inlet, to tube outlet, and back to tank	And then i gave power to the lamp, full power. I measured at different times the water temperature in the tank

Here is a table with the experiment measurements (the water was 22^oC when i began):

So what with these numbers? What can we do? Well, we know the temperature change over time and the water volume (1 liter), thus the water mass, which is very close to 1Kg (it is not distilled water...). These are enough to calculate the energy that the water absorbed from the lamp over time. To do this, i will use the formula:

Where Q is the energy in KJoules, m the mass of the water in kilos, 4.2 is a constant in KJoule/Kg/^oC, and Dt is the temperature difference. In the open-office spreadsheet, i made a table to do all these calculations. Here are the results:

In a first glance, its obvious that the temperature change over time looks linear. A chart should show this more clear. So, i made a smooth-line chart representing the water temperature increment over time:

It should be linear. The changes you see in the graph are obviously due to error measurements. The thermometer is not an expensive sophisticated laboratory thermometer after all... So, i bet the change is linear over time, at least for the range of temperatures that i will work (not above boiling point).

So, now i know that the amount of power provided to the water is always the same. Of course it is! The lamp was full power all the time, the water flow was stable, the water mass was also the same... Nothing changed. But how much power was provided to the water? I know that the lamp is 850 Watt but not even the half of it is directed to the water. To calculate the power, i will use the formula

Where P is the power in Watts, E is the energy in W-hours and t is the time in hours. In the next column of the table, i added the relative "time" in seconds, following i made a column with the conversion from KJoules to Whours (1 WHour is approximately 1KJoule x 0.277). Finally, i divided the values (did not forget to convert W-Hours to W-Seconds) and here are the results:

I did not expect to have such clear results! It is obvious (except rows 9 and 11 in which i must have had error in measuring), that the heating power that caused the water to increase its temperature, was about 70 Watts! 70 Watts is the number!. What this means is that i need something to dissipate 70 Watts of heat, and the water will not increase its temperature not a single degree!

A Peltier could be used for example. Peltiers have usually very low power efficiency. I have a detailed theory about Peltiers which you may be interested to read. TECs unfortunately tend to decrease their efficiency as the heat-pumping demands are increased. In conjunction to their low efficiency (by construction), the 70 Watts are not an easy target to achieve with TECs. So, if i use one TEC to pump 70 Watts of heat, this TEC should be (roughly) 140 electrical watts. And this would require an ENORMOUS heat-sink to operate efficiently (as efficient as could be).

That can change if i use 2 Peltiers. After all, they are cheap. Underpowered Peltiers are more efficient and needs smaller heatink. But thing are not so sweet... Lets not forget that i ran the experiment with only 1 tube, and that there will be actually 2 tubes, one oposite the other. This means double the energy provided to the water, so double the power that needs to be dissipated... The number is not 70, instead it is 140 Watts!!!

I have something else in my mind that will change all above. After all, why dissipate 140 Watts of energy? Let me run a couple more experiments...

By the way, here is the Open-Office spreadsheet i used for my calculations, in case you want to use it. I converted it into xml format. If you face problems in running with excel, please inform me (i do not have excel to test it)

Water Energy and Cooling Calculations

So, here is the idea. The lamp heats the material by radiation. Radiation is also absorbed by the aluminum profiles and they heat up, heating up the water. What i did, is a simple trick. I used an aluminum surface as a deflector for the lamp radiation. This way, i guide the radiation toward the acrylic, rather than the profiles:

I took 2 empty cans of soda water	Cut the top and bottom parts	I folded the aluminums to perform something like a reflector	And i placed them under the radiator

And then i re-ran the experiment

And then i re-ran the experiment. From the very first second i realized that the efficiency was dramatically increased. The heat above the lamp was so intense, that the acrylic material was burned within seconds. So, this means that i will have to run the radiator much bellow the half of its power, which is good news first because i will save power, and second because the water will not get hot.

Nevertheless, i re-run the experiment in full power to get an idea of the benefits that this reflectors had, in respect to the water heat dissipation. Here are the measurements i got:

And here is the results that i got out of these measurements:

Before the reflectors, the average power that the water was heated with was 69.8 Watts. After installing these unpolished reflectors, the power reduced down to 47.3 Watts! That is more than 32% decrement! But thats not the end. The reflectors caused a dramatic heat increment above the lamp right on the surface of the acrylic. This means that i will have to decrease the power of the lamp to avoid burning the materials, which in turn means that the power will be even more reduced. Also, i will polish the reflectors which will also increase the heat on the surface. Finally, i plan to install a means of air-flow around the reflectors to avoid heat of the air. Hopefully, after taking all the above measures, i calculate (roughly) that the water cooler will have to dissipate not more than 30 watts with both aluminum profiles. Lets not forget that until now i run the experiments with a single aluminum profile.

Here is the xls file with the calculations and the charts:

Water Energy and Cooling Calculations 2

The heat was so intense and the reflectors so unpolished, that they got burned with only 17 minutes that the experiment lasted. So, time to make some changes. I'm in the correct path...

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