I’m working on ‘solar oven’ that uses a large Fresnel lens as the power plant for heat. The idea is to have a cooking chamber and a heating box on the outside. The Fresnel lens will generate about 1400 F with a .3 inch spot beam. The beam will be focused on a bundle of 6, ? inch copper gas lines. The lines go into the cooking chamber in simi-circles of 1 foot in length each . I’m wondering what the heat transfer and estimated heat output of the inner copper lines would be. Not sure how to figure this and have no clue how to work the formulas I did find – Thanks for any halp.The cooking chamber will be brick / concrete about 9 inches thick with an internal cooking space of 8in x 8in x 11in.
copper melts at 1981 F. The heat transfer rate is 215 BTU/F foot hour. What this means is that 215 British Thermal Units if heat will be transferred one foot in length in one hour per degree difference in temperature for each square foot of area of copper. Okay, now find the crossection area of the 6 copper lines in square feet. Then find the length that the heat has to flow. Temperature difference to start is 1400-70. Go for one hour. YOu end up with a ratio of area/ 1 ft2 times 215 times length of heat flow in feet times 1330. The only thing left to do is to decide if the temperature difference is correct due to the spot of beam width being small. Here are some other values you might find helpful. Water takes about 1000 BTU/lb to boil at 212 F. Wood burns at about 350F. Zinc has a specific heat of 0.107 BTU/lb F and melts at 786 F. Hope this helps
