FUEL CELL SYSTEM

• Main Author: KAWAMOTO YUTA
• Format: Patent
• Language: English; Japanese
• Published: 19.07.2018
• Subjects: BASIC ELECTRIC ELEMENTS; ELECTRICITY; PROCESSES OR MEANS, e.g. BATTERIES, FOR THE DIRECT CONVERSIONOF CHEMICAL INTO ELECTRICAL ENERGY

PROBLEM TO BE SOLVED: To raise the temperature of fuel cell stack rapidly.SOLUTION: A change valve 19 reduces the flow passage area in a suction pipeline 15 during warming-up of a fuel cell stack 12, compared with that after completion of warming-up. Then, the pressure of air flowing farther downstream side than the change valve 19 in the suction pipeline 15 becomes lower than the pressure of air flowing farther upstream side than the change valve 19 in the suction pipeline 15. Consequently, the pressure of air sucked into a compressor 11 from an inlet 11a becomes lower compared with the pressure when the change valve 19 does not reduce the flow passage area in the suction pipeline 15. Since the pressure difference between the pressure of air sucked into the compressor 11 from the inlet 11a, and the pressure of air compressed by the compressor 11 and discharged from a discharge opening 11b to a discharge pipeline 16 increases easily, and thereby the temperature of the air discharged from the discharge opening 11b to the discharge pipeline 16 rises easily. The fuel cell stack 12 is warmed by the heat of the high temperature air.SELECTED DRAWING: Figure 1 【課題】低温環境下であっても、燃料電池スタックの温度を速やかに上昇させること。【解決手段】変更弁１９は、燃料電池スタック１２の暖機中に、吸入配管１５内の流路面積を暖機完了後よりも縮小する。すると、吸入配管１５内における変更弁１９よりも下流側を流れる空気の圧力は、吸入配管１５内における変更弁１９よりも上流側を流れる空気の圧力よりも低くなる。よって、吸入口１１ａから圧縮機１１に吸入される空気の圧力は、変更弁１９が吸入配管１５内の流路面積を縮小しない場合の圧力に比べて低くなる。そして、吸入口１１ａから圧縮機１１に吸入される空気の圧力と、圧縮機１１で圧縮されて吐出口１１ｂから吐出配管１６に吐出される空気の圧力との圧力差が大きくなり易くなり、吐出口１１ｂから吐出配管１６に吐出される空気の温度が高温となり易くなる。燃料電池スタック１２は、高温となった空気の熱により暖められる。【選択図】図１