ithilelda Posted November 15, 2017 Share Posted November 15, 2017 Recently I have been fiddling with the tepidizers and aquatuners. To be honest, I am seriously confused by the thermal dynamics in this game. first scenario: I surrounded a copper ore aquatuner with abyssalite tiles and painted the enclosed area with vaccuum. P-water is transfered through abyssalite pipes too. Thus all heat would be retained in the aquatuner itself in theory. The aquatuner was at 20 degree C initially, and after each cycle of cooling down 10kg of p-water, it raises exactly 9.1 degree C. From the calculation of reason: 14 degree K difference of p-water * 10kg * 6J/g/K (the specific heat of p-water) = 840 KJ (C and K have the same scale, so when talking delta, delta C and delta K are equal.) Heat from aquatuner: 58.51 * 20 * 1 sec = 1.17 KJ (machines produce 20x their tagged heat) Total: 841.17 KJ Thus the aquatuner will raise: 841.17 KJ / 1200 Kg (the mass of the aquatuner) / 0.386 J/g/K (the specific heat of copper ore) = 1.816 degree K However, in reality it raises 9.1 degree K, around 5 times the anticipated amount of delta K. Anyone able to explain? second scenario: I submerged the tepidizer in the following setup: the clock was set to turn on for only 1% of the cycle, which equals 6 seconds. The tepidizer was made of copper ore and was at 20 degree C. There are two tiles of naphtha, each with 700Kg mass and 20 degree C temperature. The 2 tiles on the right are vaccuum tiles. After the run and I waited for the system to settle down, the tepidizer and naphtha are both around 24.2 Degree C (avg. over 3 runs). The delta is 4.2 degree C. The heat produced during these 6 seconds: [1400 Kg * 2.191 J/g/K + 400 Kg (mass of tepidizer) * 0.386 J/g/K ] * 4.2 degree K = 13.531 MJ Thus the heat output power: 13.531 MJ / 6 sec = 2.255 MW Due to control delay and roundup errors, we can see that the number is roughly half of the expected number (20.32 KW * 200 = 4.064 MW). It means that the tepidizer divides its output evenly among all four tiles, and outputs 1.016 MW of power to each tile it is on. If it is vaccuum, it will not heat it. This is pretty much conforming to what everyone else has known before. Unlike the aquatuner, the heat in the system is actually correct. Here comes the interesting part: Everything was the same, except that the two liquid tiles were moved left, and there were 3 tiles of vaccuum on the right. I tried my best to pause after the clock turned off. I took 5 runs because there may be human error involved. The average temperature of the naphtha tile on the right was 23.5 degree C, and it was 20.9 degree C on the left. Thus during this 6 seconds, there were: 700 Kg * 2.191 J/g/K * 0.9 degree K = 1380.33 KJ of heat transfered to the left tile. Then the heat transfer rate is: 1380.33 KJ / 6 secs = 230.055 KW. In theory, with 1 meter contact, the two tiles should transfer heat at: 0.2 W/m/K * 1 m = 0.2 W/K. The highest temperature difference was 2.6 degree K, thus they should conduct heat at no more than 0.52 W. The reality, however, was half a million times higher. One can explain this by adding in convection in addition to the conduction rate I calculated above, but how much heat is transfered through convection? Is there a defined number to look up anywhere? Link to comment Share on other sites More sharing options...
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