| 摘要 |
PROBLEM TO BE SOLVED: To increase the temperature of cooking oil even if the temperature of a part where the cooking oil touches is low and at the same time rapidly cool the inside of a refrigerator by forming a far-infrared radiator covering on the wall surface of a part for heating the cooking oil in an oil tank. SOLUTION: When a far-infrared radiator covering is formed on both the surfaces of a wall surface that is heated by a gas burner 30, especially on the inner/outer-periphery surfaces of a heat tube 4, far-infrared rays emitted from the gas burner 30 are first absorbed by a ceramic covering on the inner(- periphery) surface and then is discharged from the ceramic covering into oil. Therefore, oil is increased to a specific temperature, and time is reduced to half-2/3 as compared with before. More specifically, the far-infrared radiator covering on the inner(-periphery) surface absorbs far-infrared rays being emitted from the flame of gas, and the far-infrared radiation covering on the outer(- periphery) surface heats hot water without attenuating the far-infrared rays. Then, the far-infrared radiation covering that is formed on the inner wall that is not heated directly emits far-infrared rays by itself due to the temperature increase of the inner wall, thus heating oil.
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