Thus, the formula weight per C-atom is: 1*12 (g/mol) +1.85*1(g/mol)+0.55*16 (g/mol)+0.13*14(g/mol)=24.5 g/mol.

Transcription

1 Proble. a. C: H: O: N47.6%/: 7.%/: (-47.6%-7.%-7.%-.%)/6: 7.%/4:.85:.55:. Therefore, the eleental coposition for the ash-free bioass is CH.85 O.55 N.. Thus, the forula weight per C-ato is: * (g/ol).85*(g/ol).55*6 (g/ol).*4(g/ol)4.5 g/ol. Since ethane is the sole carbon source, fro the conservation of C-ato, we now Y sx oles of bioass(x)/oles of ethane(s) g dry weight g bioass g bioass ol ethane [.8 ] [(- %) ] [4.5 ] ole ethane g dry weight ol bioass C - ole ethane.45 (C-ole bioass/c-ol ethane) b. If assuing that CO, H O and CH a O b N c are the only etabolic products, then the overall etabolic reaction is.5 C H 6 Y so O Y sn NH Y sx CH.85 O.55 N. Y sc CO Y sw H O Fro a), we already got Y sx.45. Use ass balance conditions on each ato: C:.5*Y sx Y sc N: Y sn Y sx *. H:.5*6Y sn *Y sx *.85Y sw * O: Y so *Y sx *.55Y sc *Y sw After solving this set of linear equations, we finally get Y sc.549 (ol CO /C-ol ethane), Y sn.589 (ol NH /C-ol ethane), Y sw.7 (ol H O/C-ol ethane), Y so.6 (ol O /C-ol ethane) Therefore, the full stoichioetric equation for the growth process.5 C H 6.6 O.589 NH.45 CH.85 O.55 N..549CO.7 H O The oxygen consuption is Y xo Y so /Y sx.6/ (ol O /C-ol bioass) Then we can deterine the heat evolved per ilogra dry weight fro the enthalpy of cobustion data: Q.5*ΔH cob (ethane).589*δh cob (NH )-.45*ΔH cob (bioass) -(.5*56 J/ol.589*8 J/ol9 (J/g dry weight) Cite as: Willia Green, Jr., and. Dane Wittrup, course aterials for.7 Cheical and Biological Reaction Engineering, Spring 7. MIT OpenCourseWare ( Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].

2 g bioass g bioass C - ol ethane [(- %) ] [4.5 ] g dry weight C - ol bioass.45c - ol bioass -586 J/c-ol ethane. Then convert bac again to per ilo dry weight Q J g bioass g bioass C - ol ethane g 586 [(- %) ] [4.5 ] c - ol ethane g dry weight C - ol bioass.45c - ol bioass g -5.5 (MJ/ g dry weight) Cite as: Willia Green, Jr., and. Dane Wittrup, course aterials for.7 Cheical and Biological Reaction Engineering, Spring 7. MIT OpenCourseWare ( Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].

3 Cite as: Willia Green, Jr., and. Dane Wittrup, course aterials for.7 Cheical and Biological Reaction Engineering, Spring 7. MIT OpenCourseWare ( Massachusetts Institute of Technology. Downloaded on [DD Month YYYY]. Proble. For the reaction, O, it is not possible for the apparent activation energy to be negative, or equivalently, the rate constant decreases as T increases. We are ased to write down the eleentary steps which include an species as an interediate to explain this strange behavior. A possible echanis: O O So the reaction rates: r [][O ], r [ ], r [ ][] If using PSSH for the interediate, we have d[ ] [][O] - [] - [][] Fro this we can obtain [][O] [] [] Thus r d[] [][O] [] - [][] [][O] [][O] ( - []) [] [] [O] [] In order to have third-order reaction inetics as the for r effectiveforward[] [O], we have to assue >> [], so that the overall reaction rate for is r [] [O] where effective forward It is therefore under the condition when ~ [] or << [] for r to deviate significantly fro the noral third-order expression above. Also let s see what happens to the activation energy. E a, overall ln Ea Ea, Ea ~ΔH, rxn E a If E a E a -E a, ~ΔH, rxn E a <, then we can have a negative apparent activation energy, for

4 exaple, if eleentary step has a significantly negative ΔH, rxn, so as long as E a is not too high the overall process will have a negative E a, overall. For the reverse reaction of the overall reaction, including the reverse reaction for the second eleentary step, i.e. O O - and still using PSSH on the interediate d[ ] [][O] - [] - [][] -[] we can have the interediate concentration: Then in this case, the overall reaction rate is r [][O ] ( ( [ [][O] -[] [] [] d[] [][O] [] - [][] -[] - Again, if []<<, then ] - [] [O]) [] r - [] / [] [O ] [][O] -[] - []) [] Notice that the second ter is what we got before for the forward reaction, i.e. forward / The first ter gives the effective rate constant for the reverse process: - [ reverse - Note that forward / reverse / - c c c,overall where c s are equilibriu constants. ] Cite as: Willia Green, Jr., and. Dane Wittrup, course aterials for.7 Cheical and Biological Reaction Engineering, Spring 7. MIT OpenCourseWare ( Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].

5 Cite as: Willia Green, Jr., and. Dane Wittrup, course aterials for.7 Cheical and Biological Reaction Engineering, Spring 7. MIT OpenCourseWare ( Massachusetts Institute of Technology. Downloaded on [DD Month YYYY]. Prob. a. For the reactions We can write down d[e] d[e - S] d[s] S E E - S E - S S E E - S P E [S][E] [E - S] [S][E] [E - S] [S][E] [E - S] d[p] [E - S] [E - S] [E - S] With law of ass action on enzye[e] [E - S] [E], [S] [P] [E - S] [S], [P] (t), [S] (t) [S], [E] (t) [E], and / eq, b. function [t,conc] odehw_prob(,, eq, tax) para [,,eq]; %initial concentrations %conc ([S],[ES],[E],[P]) conc [.,,e-6,]; %use ode5s at the function derivhw %t is the tie vector output %conc is the 4 colun atrix solution containing the concentrations of %[S],[ES],[E],[P] options odeset('abstol', e-9, 'RelTol', e-6); [t,conc] ode5s(@derivhw_prob,[;tax],conc,options,para); %this is the function inputed into ode5s function derivs derivhw_prob(t,conc,para) %extract constants para(); para(); eq para(); %This is the order of the variables in the concentration vector

6 %concs [S] in M %conces [ES] in M %conce [E] in M %concp [P] in M %switch fro list of f's to actual naes for ease of forulation of concs conc(); conces conc(); conce conc(); concp conc(4); %defining the rate equations dconcs *concs*conce (/eq)*conces; dconce *conce*concs *conces (/eq)*conces; dconces *conce*concs - *conces - (/eq)*conces; dconcp *conces; %put derivative results bac in colun vector forat for MATLAB derivs [dconcs; dconces; dconce; dconcp]; return; c. Using this pseudo-steady approxiation on interediate species ES, d[e - S] [S][E] [E - S] [E - S] we now [S][E] [E - S] Using ass balance condition [E] [E - S] [E] We now Therefore the reaction rate [E] [E - S] [S] - r S d[s] d[p] [S] [E - S] [E] [S] Vax[S] [S] where ( )/ and V ax [E]. d. In the liit [S]>>, fro Cite as: Willia Green, Jr., and. Dane Wittrup, course aterials for.7 Cheical and Biological Reaction Engineering, Spring 7. MIT OpenCourseWare ( Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].

7 Cite as: Willia Green, Jr., and. Dane Wittrup, course aterials for.7 Cheical and Biological Reaction Engineering, Spring 7. MIT OpenCourseWare ( Massachusetts Institute of Technology. Downloaded on [DD Month YYYY]. d[s] Vax[S] [S] V ax In the liit [S]<<, fro d[s] Vax[S] [S] Vax[S] e. Consider the conditions: 9 liter/ole-s, s, eq, liter/ole, [E] -6 M, [S]. M. We now that now ( )/ / / eq, ~ M, [S]. M, therefore, [S] << since [S] is decreasing onotonically, [S]<< is always correct. So now we can use the result fro d) This gives an exponential function for [S](t) While for [ES], And for [P] d[s] Vax[S] V [S](t) [S]( )exp( ax V t t) [S] exp( [E] [E] [ES](t) M M [S] [E] [S] ax [S] exp( Vax M t) t) [P](t)[S] -[ES](t)-[S](t) V [S] [ exp( ax t)] Use atlab to solve the following non-linear ODE IVP: d[s] [S][E] d[e] [S][E] ( eq, eq, ([E] )([E] -[E]) -[E]) d[p] r [E - S] ([E] -[E]) with the initial conditions [S] t [S].M, 9 liter/ole-s, s, eq, liter/ole, and [E] -6 M. Fro d[es]/, we can deterine the tie to reach pseudo steady state is approxiately 7-9 sec, which is really really short. For this specific condition, pseudo steady state wors very well. This ay also be seen fro a direct coparison of the analytical/full nuerical solutions [ES](t) plots on a short tie scale. [E-S] should rapidly rise fro zero to the

8 Cite as: Willia Green, Jr., and. Dane Wittrup, course aterials for.7 Cheical and Biological Reaction Engineering, Spring 7. MIT OpenCourseWare ( Massachusetts Institute of Technology. Downloaded on [DD Month YYYY]. PSSA value (on the order of -8 seconds). The only observable difference between the nuerical and analytical solutions is this initial jup in [E-S] on the short tie scale [S](t)[M] nuerical analytical.7 [S][ol/L] t[sec] x x -8 [ES] vs. t nuerical analytical.7.6 [ES] [M] t[sec] x 6

9 . [P](t) nuerical analytical [P][ol/L] t[sec] x 6 d[es]/ vs. t d[es]/ [M/sec] t[sec].5 x -8 Cite as: Willia Green, Jr., and. Dane Wittrup, course aterials for.7 Cheical and Biological Reaction Engineering, Spring 7. MIT OpenCourseWare ( Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].

10 Cite as: Willia Green, Jr., and. Dane Wittrup, course aterials for.7 Cheical and Biological Reaction Engineering, Spring 7. MIT OpenCourseWare ( Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].