プロセスエンジニアリング地域－計装自動制御版－毎日のテーマ－2020-12-30の問題

2020-12-30View Original

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プロセスエンジニアリング部門－計装自動制御版－毎日のテーマ－2020-12-30の問題　討論テーマ：342、通常、H2S警報器の検出範囲は0～100PPMで、これに対応する出力は4～20mAです。もし表示値が20ppmの場合、出力電流はいくらになりますか？（正解のみを記入した者に5点、正しい計算過程や詳細な説明がある場合はさらに加点）

Reply #22020-12-30

20*16/100+4=7.2

Reply #32020-12-30

4+(20-4)*20/(100-0)=7.2mA

Reply #42020-12-30

4+(20-4)*20/(100-0)=7.2mA

Reply #52020-12-30

20*16/100+4=7.2

Reply #62020-12-30

4+(20-4)*20/(100-0)=7.2mA

Reply #72020-12-30

20*16/100+4=7.2

Reply #82020-12-30

20/100*16+4=7.2mA

Reply #92020-12-30

I=20*16/100+4=7.2mA

Reply #102020-12-30

通常、H2S警報器の検出範囲は0～100PPMで、出力は4～20mAとなります。もし表示値が20ppmの場合、出力電流はいくらになるでしょうか？解：I=（20/（100-0））*（20-4）+4=7.2（mA）。答：表示値が20ppmの場合、出力電流は7.2mAです。

Reply #112020-12-30

(20-4)*20/(100-0)+4=7.2mA

Reply #122020-12-30

(20-4)*20/(100-0)+4=7.2mA

Reply #132020-12-30

i=(20-0/100-0)*(20-4)+4=7.2

Reply #142020-12-30

I=20*16/100+4=7.2mA

Reply #152020-12-30

4+(20-4)*20/(100-0)=7.2mA

Reply #162020-12-30

20*16/100+4=7.2

Reply #172020-12-30

4+(20-4)*20/(100-0)=7.2mA

Reply #182020-12-30

4+(20-4)*20/(100-0)=7.2mA

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プロセスエンジニアリング地域－計装自動制御版－毎日のテーマ－2020-12-30の問題

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