CHEN 3006 Lecture Notes - Lecture 31: Inert Gas
5 Apr 2015
School
Department
Course
Professor
Document Summary
Chen3006_masstransferandseparations page 1 so now we won"t have simple straight operating line. This mitigates part of the effect, since we"re taking ratios of. We typically have one or more inert gasses and liquids. Inert gas flow: g(1-y) = g0 total gas flow*(1-exchangable fraction) Overall balance (the equation we had for operating line) we can see that each term has an inert gas flow or inert liquid flow expression we always have same amt of inert gas wherever we are @ column. The first term should be multiplied by l0, right? this would simplify stuff if eqbm line were straight in x,y. In the dilute limit: x --> x and y --> y y. 50%+ our y values are pretty small, so it"s ok to use this approximation. Now, g isn"t constant, so has to stay in the differential stays constant, though.
