若脱硫塔进水硫化物含量为800mg\/L,出水硫化物含量为100mg\/L,求硫化物的去除率。
正确答案:解:已知:C1=800mg\/LC2=100mg\/L求:去除率X X=(C1-C2)\/C1*100% =(800-100)\/800*100%=875% 答:硫化物去除率为875%
若脱硫塔进水硫化物含量为800mg\/L,出水硫化物含量为100mg\/L,求硫化物的去除率。
正确答案:解:已知:C1=800mg\/LC2=100mg\/L求:去除率X X=(C1-C2)\/C1*100% =(800-100)\/800*100%=875% 答:硫化物去除率为875%