Acoal-fired power plant equipped with a so2 scrubber is required to achieve an overall so2 removal efficiency of 85%. the existing scrubber is 95% efficient. rather than treating the entire gas stream to 95% removal, the plant proposes to treat part of the flue gas to 95% removal, and to bypass the remainder around the scrubber. calculate the fraction of the flue gas stream that can be bypassed around the scrubber (i. e., qbypass/q) and still satisfy the regulatory requirement.

bypassed fraction B will be B= 0.105 (10.5%)

Explanation:

doing a mass balance of SO₂ at the exit

total mass outflow of SO₂ = remaining SO₂ from the scrubber outflow + bypass stream of SO₂

F*(1-er) =  Fs*(1-es) + Fb

where

er= required efficiency

es= scrubber efficiency

Fs and Fb = total mass inflow of  SO₂ to the scrubber and to the bypass respectively

F= total mass inflow of  SO₂

and from a mass balance at the inlet

F= Fs+ Fb

therefore the bypassed fraction B=Fb/F is

F*(1-er) =  Fs*(1-es) + Fb

1-er= (1-B)*(1-es) +B

1-er = 1-es - (1-es)*B + B

(es-er) = es*B

B= (es-er)/es = 1- er/es

replacing values

B= 1- er/es=1-0.85/0.95 = 2/19 = 0.105 (10.5%)

answer:

an example of solubility is salt or sugar dissolving in water.

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