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PROBLEM TO BE SOLVED: To provide a heat insulation material having a lower thermal conductivity compared to the conventional art and is suitable as a glazed layer of a thermal head.SOLUTION: A heat insulation material comprises glass comprising, in mol%, SnO+TiO0.1-85% and PO0.1-49%, where the glass comprises, in mol%, SnO 0.1-85% and further comprises BO0-50% and ZnO 0-50%, with the total content of PO+SnO+TiO+BO++ZnO being 50% or more.SELECTED DRAWING: None
【課題】従来よりも熱伝導率が低く、サーマルヘッドのグレーズ層として好適な断熱材を提供する。【解決手段】モル%で、SnO+TiO20.1〜85%、P2O50.1〜49%を含有するガラスからなり、該ガラスはモル%で、SnOを0.1〜85%を含有し、モル%で、さらにB2O3を0〜50%、ZnO 0〜50%を含有し、モル%で、P2O5+SnO+TiO2+B2O3++ZnOの合計で50%以上を含有する断熱材。【選択図】なし
PROBLEM TO BE SOLVED: To provide a heat insulation material having a lower thermal conductivity compared to the conventional art and is suitable as a glazed layer of a thermal head.SOLUTION: A heat insulation material comprises glass comprising, in mol%, SnO+TiO0.1-85% and PO0.1-49%, where the glass comprises, in mol%, SnO 0.1-85% and further comprises BO0-50% and ZnO 0-50%, with the total content of PO+SnO+TiO+BO++ZnO being 50% or more.SELECTED DRAWING: None
【課題】従来よりも熱伝導率が低く、サーマルヘッドのグレーズ層として好適な断熱材を提供する。【解決手段】モル%で、SnO+TiO20.1〜85%、P2O50.1〜49%を含有するガラスからなり、該ガラスはモル%で、SnOを0.1〜85%を含有し、モル%で、さらにB2O3を0〜50%、ZnO 0〜50%を含有し、モル%で、P2O5+SnO+TiO2+B2O3++ZnOの合計で50%以上を含有する断熱材。【選択図】なし
HEAT INSULATION MATERIAL
断熱材
MATANO TAKAHIRO (author)
2016-09-01
Patent
Electronic Resource
Japanese
European Patent Office | 2015
|European Patent Office | 2017
|European Patent Office | 2015
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