4. (a) When considering aeration of chemostats, describe the different steps that contribute to the overall oxygen trans

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4 A When Considering Aeration Of Chemostats Describe The Different Steps That Contribute To The Overall Oxygen Trans 1 (99.28 KiB) Viewed 12 times

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4. (a) When considering aeration of chemostats, describe the different steps that contribute to the overall oxygen transfer process. In each case, identify the parameters and / or factors that determine the transfer rate for that step. [8 marks] (b) A simple dynamic test was carried out to determine the oxygen mass transfer characteristics of a chemostat. Once the chemostat was at steady state, the air flow was switched off for a known amount of time, and then restarted. The dissolved oxygen concentration after restarting the air flow is given in Table Q4-1 below. Table Q4-1 Dissolved O2 concentration upon restarting air flow Time [min] Dissolved O2 Concentration [g m-3) 1 1.50 3 3.62 5 5.47 7 6.34 9 6.77 11 7.33 13 7.66 15 7.70 8.00 oo 1 Given that: Poverall = k_ a' (Coz - Coz) -rx YX/02 where roverall is in g m-3 min determine the value of kla'. [8 marks] (c) When working at laboratory scale, the oxygen transfer within a Miniature Stirred Bioreactor is said to be better than that within a standard Erlenmeyer flask. Why is this the case? [4 marks]
R = 8.3145J/mol K = 1.98 cal/mol K The formulae below are for the following reaction where A is the limiting reactant: aA + bB → CC + dD Excess Ratio: NBo a y = b Nao Kinetics: -Eact/RT (-ra) = KC AC3 k=Ae Conversion: Constant Volume Systems: C; = Cio + VCAOX CA = CAO(1 - x) Variable Volume Systems: E = y208 6-689- )- C d b -+---- ка a a а. = Vi i=1 Cio + viCAOX C = (1 + EX) (1 – X) CA = CAO (1 + EX) Design Equations: 影。 -X NAO Batch Reactor: t= dx JO (-A) V CSTR: V= FAOX (-ra) > T = Vo FAO PFR: V= so dx (مr-)
Numerical Integration: Χη Xnxo Trapezium Rule: ydx = Elyo +2(y, + y2 + y2 + ... + Yo-1) +yn]; n = [h n- n Хо -X2 h Simpson's Rule: = + ; h = X2 - Xo 2 Хо **tx = $ \x + 4%, + ve]n=**** ydx [48 (»-Mlvo , n-* L*ydx = '5lvo + 4y, + 2y2 + 4x +ya] n = *s ** n-- -X3 ydx = 3 8 h[y. + 3y, + 3y2 +y3]; h X3 – XO 3 Хо [4y2 ; h = X4 4