Lecture-Absorption_ProblemsOldExams

Whatflowrateofairinmolsisrequiredtoreducethe

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Unformatted text preview: r absorption? 5 A single, equilibrium stage column is to be used to remove a volatile contaminant trichloroethylene (TCE) from water using air as the stripping gas. The mole fraction of TCE is 8x10‐7 for the entering stream and it is desired to process 100 mol/s of contaminated water. The inlet air contains no TCE, and it may be assumed that water does not vaporize. What flow rate of air (in mol/s) is required to reduce the trichloroethylene mole fraction to 8x10‐9 ? Henry’s Law: yA = 417xA 6 3 9/10/2013 A single, equilibrium stage column is to be used to remove a volatile contaminant trichloroethylene (TCE) from water using air as the stripping gas. The mole fraction of TCE is 8x10‐7 for the entering stream and it is desired to process 100 mol/s of contaminated water. The inlet air contains no TCE, and it may be assumed that water does not vaporize. What flow rate of air (in mol/s) is required to reduce the trichloroethylene mole fraction to 8x10‐9? Note: I am underlining the important information in the problem statement 7 V2 = ? V1 = ? yA2 = 0.0 (pure air) yA1 = ? L0 = 100 mol/s xA0 L1 = ? = 8 x 10‐7 xA1 = 8 x 10‐9 Do I know any other additional information from the problem statement? yA1 = H xA1 = 3.336 x 10‐6 0 Material balance on A: L0xA0 + V2yA2 = L1xA1 + V1yA1 Material balance on water: L0 (1‐xA0) = L1 (1‐xA1) solve for L1 get V1 Material balance on air: V2 (1‐yA2) = V1 (1‐yA1) = V2 get V2 V1 = 23.74 mol/s and V2 = 23.7409 mol/s 8 4 9/10/2013 If you decrease the air flow rate below the value you calculated above (circle your choice), the mole frac of TCE in the exiting air would: INCREASE DECREASE STAY THE SAME the mole frac of TCE in the exiting liquid would: INCREASE DECREASE STAY THE SAME V2 = 23.7409 V1 = 23.74 yA2 = 0.0 (pure air) yA1 = 3.336 x 10‐6 L0 = 100 mol/s L1 = ? xA0 = 8 x 10‐7 xA1 = 8 x 10‐9 What is changing? V2, V1, xA1, yA1 What is NOT changing? L0, xA0, yA2 Also still a single stage equilibrium 9 What is changing with the different operating lines? L/V ratio is increasing Since L is fixed then V is decreasing (matches what our question) What happens to x1 and y1 as V decreases? Both increase! Operating line has to fall on this line 8 x 10‐7 (x1, y2 = 0) (x0, y1) 10 5...
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This document was uploaded on 02/07/2014.

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