Responses to SAQ 351 b) The yield =05, and productivity =9. From Figure 4 11 at yield=05 and estimating This represents 32x05=16g02h" ker rate is about 0.5 moles oxygen r"h 3)Heat Evolution Rate(again we have at least two ways to calculate this) From Figure 4.12, with yield 0-5, the heat produced is about 27 k Joules gram biomass(or 27,000 k Joules per kg biomass). With productivity 9 kg m h", the heat b)From Figure 4. 13, with productivity 9 and yield 0.5, the heat produced is about 251 k Joules I'h, or 251,000 k Joules m hi. To support a biomass concentration of 30 kg m"at yield 0.5, the methanol concentration in the incoming medium would need to be 30/0-5=60kg m.Thus at 90% utilisation, the concentration in the incoming medium would need to be 60 x 100190=667kgm 4.10 exhaust bioreactor sterilising separation Rash nutrient acid salts Your flow diagram might not look exactly like the one shown: for instance you will probably have represented the fermentation step by'fermentation', rather than rawing the bioreactor. Nevertheless you should have all the unit processes and inputs shown above in your own diagram
Responses to SAQs 351 b) The yield = 05, and productivity = 9. From Figure 4.11 at yield = 05 and estimatin~ productivity = 9, the minimum oxy n-transfer rate is about 05 moles oxygen 1" h- . This represents 32 x 0.5 = 16 g 02 1- h or 16 kg 02 m-3 h". $4 3) Heat Evolution Rate (again we have at least two ways to calculate this). a) From Figure 4.12, with yield 05, the heat produced is about 27 k Jou$s-pr gram biomass (or 27,000 k Joules per kg biomass). With productivity 9 kg m h , the heat evolution rate is 27,000 x 9 = 243,000 k Joules m-3 h-'. k Joules I-' h-I, or 251,OOO k Joules m-3 h-I. b) From Figure 4.13, with productivity 9 and yield 0.5, the heat produced is about 251 4) Methanol Concentration To support a biomass concentration of 30 kg m-3 at yield 0.5, the methanol concmtration in the incoming medium would need to be 30/05 = 60 kg m-3. Thus at 90% utilisation, the concentration in the incoming medium would need to be 60 x 100/90 = 66.7 kg m-3. 4.1 0 Your flow diagram might not look exactly like the one shown: for instance you will probably have represented the fermentation step by 'fermentation', rather than drawing the bioreador. Nevertheless you should have all the unit processes and inputs shown above in your own diagram
352 Responses 4.11 1)Productivity =biomass concentration xD 2)Percentage utilisation=100-[(concentration in outgoing medium/concentration in incoming medium) x 1001 3)Minimum OtR productivity x 1.6(see SAQ 4.9) D(h) Productivity m OTR Methanol g biomass mh)kg O2 required m h) Utilisation(%) 0.1 3.43 980 948 1493 889 1683 75.3 5 1.25 Dilution rate of 0.5 h" produces the fastest growth rate(since H= D), but gives poor productivity and substrate utilisation. This is a feature of chemostat cultures, when u pproachesμ Dilution rate 0. 4 h"gives highest productivity, but 25%of the substrate remains unused Dilution rate 0.3 h"gives better substrate utilisation but with reduced productivity. Both these dilution rates require minimum OTRs greater than the bioreactor can supply This means that such biomass concentrations could not be produced in practice. Dilution rate 0.2 h produces efficient substrate conversion and has an otr that the system can just about provide. Dilution rate 0.1 h" gives better substrate conversion but lower productivity. The chosen dilution rate would be below 0.2 h. In practice dilution rates of between 0.1 h"and 0.2 h"are used to operate the production system. 4.12 1)Output With 1500 m at D=0.2 hi, volume of culture produced is 300 mh(ie 1 500 0.2 m' At 36 kg biomass m, the output is 10, 800kg h"or 259 200 kg day"(259 tonnes day )or 94 608 tonnes yr 2) Biomass production of 94 608 tonnes would require 94 60870-5= 189 216 tonnes 4.13 The most likely order is as outlined below 1)Establish the 2)Isolate organisms that can use the substrate 3)Measure temperature optimum. 4)& 5)Measure affinity for substrate/Measure protein content. 6) Perform feeding trials in animals
352 Responses 4.1 1 4.1 2 4.1 3 1) productivity = biomass concentration x D. 2) Percentage utilisation = 100 - [(concentration in outgoing medium/ concentration in 3) Minimum OTR = productivity x 1.6 (see SAQ 4.9). incoming medium) x 1001. D fi-') PrOdUCtiVity Minimumm Methanol (kg biomass m-3 h-') (kg a requked m3 h-') Utilisation (%) 0.1 3.43 5.49 98.0 0.2 6.62 10.59 94.8 0.3 9.33 14.93 88.9 0.4 10.52 16.83 75.3 0.5 1.25 2.0 7.2 Dilution rate of 0.5 h-' produces the fastest growth rate (since p = D), but gives pr productivity and substrate utilisation. This is a feature of chemostat cultures, when p approaches thy.- Dilution rate 0.4 h-' 'ves highest productivity, but 2596 of the substrate remains unused. Dilution rate 0.3 h- gives better substrate utilisation but with reduced productivity. Both these dilution rates require minimum OTRs greater than the bioreactor can supply. This means that such biomass concentrations could not be produced in practice. Dilution rate 0.2 h" produces efficient substrate conversion and has an OTR that the system can just about provide. Dilution rate 0.1 h-' gives better substrate conversion but lower productivity. The chosen dilution rate would be just below 0.2 h-'. In practice dilution rates of between 0.1 h-' and 02 h-' are used to operate the production system. F 1) output With 1500 m3 at D = 0.2 h-', volume of culture produced is 300 m3 h-' (ie 1,500 x 02 m3 h" ) At 36 kg biomass m3, the output is 10,800 kg h" or 259,200 lcg day (259 tonnes day-') or 2) Biomass production of 94,608 tonnes would require 94,608/05 = 1B,216 tomes 94#608 tonnes yr-'. methanol. The most likely order is as outlined below. 1) Establish the size of the market. 2) Isolate organisms that can use the substrate 3) Measure temperature optimum. 4) & 5) Measure affinity for substrate/Measure protein mntent. 6) Perform feeding trials in animals