Solar Heating Technical Comments 2There is actually a set of eight equations that are presented in the Heat Exchange Institute Standards, that accomplish this goal. They are a "simplified" method of doing certain calculations, and so their results are only approximate. We have further simplified the equations by inserting a number of facts from our system, such as the tube diameters and their wall thickness, several characteristics of air, and the like. You may want to do these following calculations. They will aid in reasonably well knowing the system performance rating of any tube pattern you may want to consider, as well as the effects of various blower sizes that blow the air through the tubes.
The first equation is: R/Δt_{m} = 0.103 * L /(square root(V)) Where L is the length of an individual tube air path (in feet) and V is the air velocity within the tube (in fps). As an example, if a tube air path (L) is 50 feet long, and 1000 cfm of air is pushed through nine tubes, for a velocity of 21.2 fps (V), this value comes out to around 1.119 Next use a calculator or table to find the value of the exponential of the negative of this number. In our example, we get 0.327 Subtract this number from 1, and get 0.673 This number is the fraction of temperature difference that can be expected in the house air. For example, if hot house air is entering the system at 90°F, and if the deep soil temp is 53°F (for a beginning difference of 37°F), just multiply that by the fraction we just got, to get 25°F The house air will return exit that tube (and each other tube as well) after being lowered about that amount. Now, it's easy to calculate the total system (shortterm) cooling performance. The 1000 cfm we would be blowing through is 60,000 cubic feet per hour, and each 13 cubic feet of air weighs one pound. That means we would be blowing 4600 pounds of air through the system every hour. Each of those pounds of air is cooled by 25°F. Since the specific heat of air is 0.241, then 25 * .241 Btu is needed to cool each of those pounds, or about 6.0 Btu. Therefore, 4600 * 6 indicates that the actual performance of the small tubes of this system produce a cooling effect of 27,600 Btu/hr.


Since we are now pushing twice the cfm of air through the tubes, that means twice the total weight of air, now 9200 pounds per hour. 20.2 temperature drop means (20.2 * 0.241) 4.8 Btus are removed from each pound of air. Multiplying (4.8 * 9200) gives 44,200 Btu/hr for the total effect. The larger blower enabled the system to work better, increasing its performance from 27,600 Btu/hr to 44,200 Btu/hr.
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If your yard was long and skinny, you could see how 4 parallel tubes each 100 feet long would perform! Or almost any other configuration.
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