生物反应工程教学课件-Assignment-I-Solutions.docVIP

Solutions for the problems in Chapter 2 P2.1 For the flow in tube E. coli _in = E. coli _out 350 kg min-1 (1% = Ft_out ( 6% Ft_out = 58.33 kg min-1 For the whole system Ft_in + Fa_in = Ft_out + Fa_out Fa_out = Ft_in + Fa_in ( Ft_out = 350 kg min-1 + 80 kg min-1 – 58.33 kg min-1 = 371.67 kg min-1 P2.2 (a) Calculation For the flow in the annular space: Water _in = Water _out 40 kg h-1 ( (1(10%) = Fa_out ( (1 ( 0.2% ( 0.5%) Fa_out = 36.25 kg h-1 Glucose consumed = 40 kg h-1( 10% ( 36.25 ( 0.2% = 3.9275 kg h-1 C6H12O6 2C2H5OH+2 CO2 Ethonal produced = (3.9275/180) ( 2 ( 46 = 2.0074 kg h-1 Et_out = Ethonal produced – Ea_out = 2.0074 kg h-1 ( 36.25 kg h-1( 0.5% = 1.82615 kg h-1 For the flow in the tube: C_ethanol = Et_out /Ft_out = Et_out /( Et_out + Ot_out) = Et_out /( Et_out+Ot_in) =1.82615 kg h-1/(1.82615 kg h-1 + 40 kg h-1) (100% = 4.37% (b) CO2 produced = (3.9275/180) ( 2 ( 44 =1.92 kg h-1 P2.3 Molecular Mass Balances for 1. Glucose Glu consumed = 42.75 ( 0.7 = 42.05g 2. NH3 NH3 consumed = 24.94 ( 18.87 = 6.07g 3. O2 O2 consumed = 95.04 ( 67.2 = 27.84g 4 H2O H2O consumed or retained = 1357.3 ( 1090.4 = 266.9g 5 CO2 CO2 generated = 29.48g Total mass accumulated within the bioreactor is calculated to be: 42.05 + 6.07 + 27.84 ( 29.48 + 266.9 = 313.38g (wet biomass) Dry biomass is calculated to be: 313.38 ( = 22.38g H2O contained in the biomass is calculated to be: 313.38 ( 22.38 = 291.0g H2O generated is calculated to be: 291.0 ( 266.9 = 24.1g Bioreaction stoichiometry calcilation Yields of product and by-products to glucose consumed YB/G = 22.38/42.05 = 0.532 = 29.48/42.05 = 0.701 YW/G = 24.1/42.05 = 0.573 Yields of substrate consumed to glucose consumed = 6.07/42.05 = 0.144 = 27.84/42.05 = 0.662 Cell growth bioreaction stoichiometry formula C6H12O6 + 0.144NH3 + 0.662O2 0.532CxHyOzNm + 0.701CO2 + 0.573H2O Element mass balances for C, H, O and N give: