pset02_soln - Problem 1. a. C: H: O: N=47.60%/12: 7.33%/1:

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Problem 1. a. C: H: O: N=47.60%/12: 7.33%/1: (1-47.60%-7.30%-7.33%-3.00%)/16: 7.30%/14=1: 1.85: 0.55: 0.13 Therefore, the elemental composition for the ash-free biomass is CH 1.85 O 0.55 N 0.13 . 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. Since ethane is the sole carbon source, from the conservation of C-atom, we know Y sx =moles of biomass(x)/moles of ethane(s) = ethane mole - C 2 ethane mol 1 ] biomass mol biomass g 5 . 4 2 [ ] dry weight g biomass g 3%) - (1 [ ] ethane mole dry weight g 22.8 [ × ÷ × =0.451 (C-mole biomass/C-mol ethane) b. If assuming that CO 2 , H 2 O and CH a O b N c are the only metabolic products, then the overall metabolic reaction is 0.5 C 2 H 6 +Y so O 2 +Y sn NH 3 Y sx CH 1.85 O 0.55 N 0.13 + Y sc CO 2 + Y sw H 2 O From a), we already got Y sx =0.451. Use mass balance conditions on each atom: C: 0.5*2=Y sx +Y sc N: Y sn =Y sx *0.13 H: 0.5*6+Y sn *3=Y sx *1.85+Y sw *2 O: Y so *2=Y sx *0.55+Y sc *2+Y sw After solving this set of linear equations, we finally get Y sc =0.549 (mol CO 2 /C-mol ethane), Y sn = 0.0589 (mol NH 3 /C-mol ethane), Y sw =1.17 (mol H 2 O/C-mol ethane), Y so =1.26 (mol O 2 /C-mol ethane) Therefore, the full stoichiometric equation for the growth process 0.5 C 2 H 6 +1.26 O 2 +0.0589 NH 3 0.451 CH 1.85 O 0.55 N 0.13 + 0.549CO 2 + 1.17 H 2 O The oxygen consumption is Y xo =Y so /Y sx =1.26/0.451=2.79 (mol O 2 /C-mol biomass) Then we can determine the heat evolved per kilogram dry weight from the enthalpy of combustion data: Q=0.5* Δ H comb (ethane)+0.0589* Δ H comb (NH 3 )-0.451* Δ H comb (biomass) =-(0.5*1560 kJ/mol+0.0589*383 kJ/mol-19 (kJ/g dry weight) Cite as: William Green, Jr., and K. Dane Wittrup, course materials for 10.37 Chemical and Biological Reaction Engineering, Spring 2007. MIT OpenCourseWare (http://ocw.mit.edu), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
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biomass mol - C 0.451 ethane mol - C 1 ] biomass mol - C biomass g 5 . 4 2 [ ] dry weight g biomass g 3%) - (1 [ ÷ × ÷ =-586 kJ/c-mol ethane. Then convert back again to per kilo dry weight Q= 1kg 1000g biomass mol - C 0.451 ethane mol - C 1 ] biomass mol - C biomass g 5 . 4 2 [ ] dry weight g biomass g 3%) - (1 [ ethane mol - c kJ 586 × × ÷ × =-51.5 (MJ/ kg dry weight) Cite as: William Green, Jr., and K. Dane Wittrup, course materials for 10.37 Chemical and Biological Reaction Engineering, Spring 2007. MIT OpenCourseWare (http://ocw.mit.edu), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
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Problem 2. For the reaction,
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pset02_soln - Problem 1. a. C: H: O: N=47.60%/12: 7.33%/1:

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