339 HW 3 Solutions - Fall 2010 BME339/CHE339/BIO335 ...

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Unformatted text preview: Fall 2010 BME339/CHE339/BIO335 HW#3 Solutions Problem 1 Fall 2010 BME339/CHE339/BIO335 HW#3 Solutions Problem 2 Known values: F = 100 L /hr V1 = 500 L V2 = 300 L Sin = 5 g/L Yx/s = 0.4 g cell/ g subs µ2 = 0 Ks = 0.1 g/L µmax = 0.3 hr ­1 Rp = 0.02 (g prod/g cell hr) * X Find P2 at Steady State Assume Monod Kinetics, Sterile Feed, and at Steady State In Reactor 1 Cell Mass Balance: 0 = ( ­F/V1)X1 + µ1X1 µ1 = F/V1 = D1 = 0.2 hr ­1 Monod Kinetics: µ1 = D1 = µmaxS1/(Ks+S1) 0.2 = 0.3*S1/(0.1+S1) Substrate Balance and Yield eqn: X1 = YX/S (Sin – S1) = 0.4 (5 ­0.2) Product Balance: 0 =  ­D1P1 + 0.02X1 P1 = 0.02X1/D1 In Reactor 2 D2 = F/V2 = 0.333 hr ­1 Cell Mass Balance: 0 = D2X1 – D2X2 + 0 X2 = X1 Product Balance: 0 = D2P1 – D2P2 + 0.02X2 P2 = P1 + 0.02X2/D2 S1 = 0.2 g/L X1 = 1.92 g/L P1 = 0.192 g/L X2 = 1.92 g/L P2 = 0.307 g/L Fall 2010 BME339/CHE339/BIO335 HW#3 Problem 3 Solutions a) Based on steady state cell mass balance, it can be shown that the dilution rate F/V (D) must equal specific growth rate (µ). Therefore, up to certain maximum D, incremental increase of D can increase specific growth rate of the culture leading to increase of the cell density. b) Since this is a substrate limited growth, at steady state, assume there is no substrate at output stream. Then, YX/S = X / (0.05 * 180 – 0). By plotting 1/D vs 1/YX/S, we can find YTRUE at the Y ­intercept. 1/YX/S = m/D + 1/YTRUE D (hr ­1) X (g/L) Ideal Yx/s 0.05 1.58 2.86 0.175555556 0.1 2.17 2.86 0.241111111 0.15 2.47 2.86 0.274444444 0.2 2.65 2.86 0.294444444 0.25 2.76 2.86 0.306666667 0.3 2.83 2.86 0.314444444 0.35 2.86 2.86 0.317777778 0.4 2.84 2.84 0.315555556 0.45 2.74 2.74 0.304444444 0.5 2.25 2.25 0.25 0.52 1.32 1.32 0.146666667 0.53 0 0 0 1/D 20 10 6.666667 5 4 3.333333 2.857143 2.5 2.222222 2 1.923077 1.886792 1/Yx/S 5.696203 4.147465 3.643725 3.396226 3.26087 3.180212 3.146853 3.169014 3.284672 4 6.818182 #DIV/0! Using the greyed area data points to plot, 1/Yx/S vs. 1/D 6 1/Yx/s 5 y = 0.151x + 2.647 4 3 2 1 0 0 5 10 15 20 25 1/D Then fit to linear eqn, YTRUE = 1/2.647 = 0.378 (g cell/ g substrates) Fall 2010 BME339/CHE339/BIO335 HW#3 Solutions Problem 3 c) X vs. D plot 3.5 Cell Conc. (g/L) 3 2.5 Growth w/ maintenance 2 1.5 Ideal Growth 1 0.5 0 0 0.1 0.2 0.3 0.4 DiluHon Rate (hr ­1) 0.5 0.6 By plotting X vs. D plots, we can determine the optimal operational point will be at the Dilution rate of about 0.35 (hr ­1). This is where the dilution rate is high enough the effects from mYTRUE are masked by the high Dilution rate. This will also give us the optimal growth rate. Hence, this will be a desirable dilution rate for producing growth ­associated product. Fall 2010 BME339/CHE339/BIO335 HW#3 Problem 4 Typical Dissolved Oxygen Balance: dCL/dt = OTR – OUR During exponential growth and steady state: OTR = kLa(C* ­ CL) = OUR [Equation 1] During sparge shut ­off: dCL/dt =  ­OUR After integration: delta CL =  ­OUR*delta t OUR =  ­ delta CL / delta t OUR =  ­ (0.1 mM – 0.2 mM) / (30 s) = 3.33 x 10 ­3 (mM/s) From [Equation 1], kLa = OUR / (C* ­ CL) = 3.33 x 10 ­3 (mM/s)/ (0.25 – 0.2) (mM) kLa = 0.0666 (s ­1) Solutions ...
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This note was uploaded on 12/13/2011 for the course BME 339 taught by Professor Georgiou during the Spring '11 term at University of Texas at Austin.

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