Discussion Section Week 5 - Solutions.pdf - Problem 1 Consider the growth of a microorganism in batch culture When the substrate concentration is high

Discussion Section Week 5 - Solutions.pdf - Problem 1...

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Unformatted text preview: Problem 1. Consider the growth of a microorganism in batch culture. When the substrate concentration is high, the cell density doubles every 0.75 h, the observed substrate yield coefficient is 0.3 g DCW/g, and substrate consumption is allocated towards biosynthesis (60%), maintenance (10%), as well as product formation (30%). The product formation is strictly growth-­‐associated. The batch reactor is inoculated with 0.01 g DCW/L and 10 g/L substrate. a. Estimate the maximum cell density and the time (after lag phase) required to achieve it. (6 pts) g substrate g DCW g DCW ⋅ 0.6⋅ 0.3 = 1.8 L g substrate L = X o + X from substrate(Ok if they left out Xo ) X from substrate = 10 X max g DCW g DCW +1.8 L L g DCW g DCW = 1.81 or 1.80 L L X max = 0.01 X max X = X oe µt µ= ( ) = ln(2) = 0.924 hr ln 2 −1 td 0.75h ⎛ X ⎞ ⎛ 1.81⎞ ln⎜ ⎟ ln⎜ ⎟ g DCW ⎝ X o ⎠ ⎝ 0.01⎠ t= = = 5.63 hr (t should be about the same if they used 1.8 ) −1 µ L 0.924 hr € b. Determine the value of the maintenance coefficient (g substrate/g DCW-­‐h). (6 pts) -­‐ 10% of the substrate goes to maintaining the cells. g substrate ⋅ 0.10 L m= g DCW 1.81 ⋅ 5.63hr L g substrate m = 0.099 g DCW ⋅ hr 10 € c. Before inoculation of the batch reactor, you need to sterilize the medium, which contains 105 spores L-­‐1. The value of kd has been determined to be 1 min-­‐1 at 121 °C and 61 min-­‐1 at 140 °C. For each temperature, determine the required time in the holding section so as to insure that the medium is 95% sterile. The volume of the reactor is 20 L. Neglect heating and cooling. (8 pts) () ( ) P t = 1 − p(t) Solve for t : ⎡ ln⎢1-­‐P t ⎣ t= −k No 1 No () ( = 1 − e −kt ) No where No = ⎤ ⎥⎦ At T =121 °C ⎛ ⎞ ⎤ spores ⎡ 1 ⎜105 ⋅20L⎟ L ⎠ ⎥ ln⎢1-­‐0.95 ⎝ ⎢ ⎥ ⎣ ⎦ t= = 17.5min −1 −1min At T =140 °C ⎛ ⎞ ⎤ spores ⎡ 1 ⎜105 ⋅20L⎟ L ⎝ ⎠ ⎥ ⎢ ln 1-­‐0.95 ⎢ ⎥ ⎣ ⎦ t= = 0.287min −1 −61min € #of spores ⋅ Vliquid L Problem 2. Consider a culture of bacteria that secrete a product in a chemostat operated at steady state. The specific growth rate of biomass is adequately described by the Monod equation, and the rate of product formation is described by Leudeking-­‐Piret kinetics: rP = (αμ + β)X This system is well characterized, such that the following constants are known: YX/S = 0.4 g/g α = 0.2 g/g So = 10 g/L μmax = 0.7 h-­‐1 β = 0.3 g/g-­‐h F = 15 L hr-­‐1 KS = 0.2 g/L YP/S = 0.8 g/g V = 500 L The liquid feed to the chemostat is sterile, and the flow rates entering and exiting the chemostat are equal. a. Write steady state mass balances for S, X, and P. (5 pts) µ S X balance : 0 = -­‐FX + µXV or DX = µX or 0 = -­‐FX + max XV S + Ks ⎛ ⎞ ⎛ µ S ⎞ ⎜ α⎜ max ⎟ X + βX ⎟ ⎝ S + K s ⎠ ⎜ 1 µmax S ⎟ S balance : 0 =F So − S + rSV or 0 =F So − S -­‐ V⎜ X+ ⎟ YX /S S + K s YP /S ⎜⎜ ⎟⎟ ⎝ ⎠ ⎛ µ S ⎞ P balance : 0 = -­‐FP +rp V or 0 = -­‐FP + αµX + βX V or 0 = -­‐FP +⎜ α max X + βX ⎟V ⎝ S + K s ⎠ ( ) ( ) ( ( ) ) or DP = αµ + β X b. What is the steady state concentration of S? (5 pts) € Using the X balance : µ S 0 = -­‐FX + max XV S + Ks F µmax S = V S + Ks F S + K s = µmax S V F F S + K s = µmax S V V ⎛ F F ⎞ K s = S⎜ µmax − ⎟ V V ⎠ ⎝ ( S= € ) L hr 0.2 g g 500L L = = 9x10−3 ⎞ ⎛ ⎞ L F L − ⎟ ⎜ 15 ⎟ V ⎠ ⎜0.7hr −1 − hr ⎟ 500L ⎟ ⎜ ⎜ ⎟ ⎝ ⎠ F Ks V ⎛ ⎜ µmax ⎝ 15 c. What is the steady state concentration of X? If you were unable to obtain an answer for S in part b, use S = 0.2 g/L. (5 pts) Use the S balance : ⎛ ⎞ ⎛ µ S ⎞ ⎜ α⎜ max ⎟ X + βX ⎟ ⎝ S + K s ⎠ ⎜ 1 µmax S ⎟ 0 =F So − S -­‐ V⎜ X+ ⎟ Y S + Ks YP /S ⎜⎜ X /S ⎟⎟ ⎝ ⎠ ⎛ ⎞ ⎛ µ S ⎞ ⎜ α ⎜ max ⎟ + β ⎟ F So − S ⎝ S + K s ⎠ ⎟ ⎜ 1 µmax S + X = ⎜Y S + K ⎟ YP /S V s ⎜⎜ X /S ⎟⎟ ⎝ ⎠ ( ) ( ) L ⎛ g ⎞ ⎜10 − 9x10−3 ⎟ F So − S hr ⎝ L ⎠ X= = ⎛ ⎛ ⎞ ⎛ µ S ⎞ ⎞ ⎛ g ⎞ ⎜ ⎜ ⎟ α⎜ max ⎟ + β ⎟ ⎜ 0.7hr −1 9x10−3 ⎟ L ⎟ +0.3⎟ ⎝ S + K s ⎠ ⎟ ⎜ 1 µmax S ⎜ ⎜ 0.2 V ⎜ + g ⎟ ⎟ ⎜ ⎟ ⎜ −3 g −1 −3 g YX /S S + K s YP /S ⎜ ⎟ 9x10 +0.2 0.7hr 9x10 ⎜⎜ ⎟⎟ ⎜ 1 ⎟ ⎝ ⎠ L L L+ ⎟ ⎝ ⎠ 500L⎜ 0.8 ⎜ 0.4 9x10−3 g +0.2 g ⎟ ⎜ ⎟ L L ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ( 15 ) g X = 0.655 L d. What is the productivity (g product per g substrate per time) of this process? If you were unable to answer parts b or c, use S = 0.2 g/L and X = 0.2 g/L. (5 pts) € -­‐ Since the organism has to grow at the a rate equal to the dilution rate, the productivity is a function of how quickly the organism grows, i.e. makes product, even though the product is both growth and non-­‐growth associated. Productivity = YP/S µ =YP /S µmax S S + Ks −1 −3 g 0.7hr 9x10 g product L Productivity =0.8 g substrate g g 9x10−3 +0.2 L L g product Productivity =0.024 g substrate⋅ hr € ...
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