Question

An electrostatic precipitator is treating flue gas from a coal-burning utility boiler. The efficiency of the particleremoval is 98%. After switching toa new coal the efficiency dropsto 90%. By how much must the collection area be increased to maintain a collection efficiency of 98%, assuming that the volumetric flow rate of flue gas is unchanged?

Answer 1

Theory: We need to use the formula for efficiency calculation to determine our final answer.

Solution: lets define variables:

Lets assume particle velocity is V m/s and volumetric flow rate of gas is Q m3/s

Lets assume initial area be A1 and final area be A2.

Step1:

We know that particle removal effieicency = 1- exp(-VA/Q)

for 90% efficiency case:

0.9 = 1 -exp(-VA1/Q)

or exp (-VA1/Q) = 1-0.9 = 0.1

or -VA1/Q = ln(0.1) = -2.30259

or A1 = 2.30259Q/V.....(i)

Step 3: similarly:

or 98% efficiency case:

0.98 = 1 -exp(-VA2/Q)

or exp (-VA2/Q) = 1-0.98 = 0.02

or -VA2/Q = ln(0.02) = -3.91202

or A2 = 3.91202Q/V.....(i)

Step 4: % Increase in Area = (A2-A1)/A1*100

= (3.91202Q/V-2.30259Q/V)/(2.30259Q/V)*100

= 69.90%

I hope i was very detailed in answering your query. Kindly let me know if you still have any further doubt. if you are satisfied then please rate me. Thanks for using homeworklib