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相変化潜熱の放熱原理について

2016-04-29View Original

Thread Content

H2O(気体)+ H2O(気体)= 2H2O + 0.8eV(1.54μmの光子、1604℃の黒体放射に相当、準安定気体状態)
2H2O + H2O = 3H2O + 0.6eV(2.1μmの光子、1103℃に相当、準安定状態)
3H2O + H2O = 4H2O + 0.5eV(2.5μmの光子、883℃に相当、準安定状態)
4H2O + H2O = 5H2O + 0.4eV(3.2μmの光子、630℃に相当、準安定状態)
5H2O + H2O = 6H2O(液体)+ 0.3eV(4.0μmの光子、450℃に相当、安定した液体水分子クラスター)

従来の接触型凝縮熱交換器では、その場で全ての熱光子を別の媒体に短絡的に吸収し、外部には一目見る機会さえ与えない。凝結した水でさえ、これらの熱光子を優先的に利用する機会はない。そうでなければ独り占めすれば、少なくとも自己加熱で500度以上は上がる!
Reply #22016-05-10
特に難しいことはなく、単純な反応式だけです
Reply #32016-05-11
私が送ったテキストはもう十分具体的ですよ

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