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For mixed flow of particles containing a single unchanging size and uniform gas composition, the fraction unconverted for film resistance controlling is ____(a) \(\int_0^τ\)(-\(\frac{t}{τ}\))\(\frac{e}{t}^\frac{-t}{t}\) dt(b) \(\int_0^τ\)(1 – \(\frac{t}{τ}\))\(\frac{e}{t}^\frac{-t}{t}\) dt(c) \(\int_0^τ\)(1 –\(\frac{t}{τ}\))dt(d) \(\int_0^τ\)(1 – \(\frac{t}{τ}\))\(\frac{e}{t}^\frac{-t}{t}\) dtI had been asked this question by my college professor while I was bunking the class.The question is from Design of Fluid Particle Reactors in portion Fluid-Particle Reactions: Kinetics of Chemical Reaction Engineering

Answer»

Right option is (b) \(\int_0^τ\)(1 – \(\frac{t}{τ}\))\(\frac{e}{t}^\frac{-t}{t}\) dt

For explanation I would say: For a mixed flow reactor, the mean RESIDENCE time, t in the reactor is, E = \(\frac{e}{t}^\frac{-t}{t}\). For a single SIZE of particles converted in time τ,1-\(\overline{X_{(B)}}\) = ∫0^τ(1 – XB)\(\frac{e}{t}^\frac{-t}{t}\)dt for an INDIVIDUAL particle. For film resistance controlling, XB = \(\frac{t}{τ}.\)



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