339 HW 4 Solutions - Fall 2010 BME339/CHE339/BIO335 HW#4...

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Fall 2010 BME339/CHE339/BIO335 HW#4 Solutions Problem 1 Known: F = 1000 mL /hr = 1 L/hr V = 1000 mL = 1 L Monod Kinetics/Subs. Lim. growth Y M X/S = Y = 0.5 g cells/ g gluc (Max yield; no maintenance) S o = 10 g gluc / L µ m = 0.2 hr -1 K S = 1 g gluc/L C = 1.5 α = 0.7 Solve: a) Cell Mass Balance: g ² ´ µ± · αFC± ² ¸ ¹1 · αºµ± ² · μ± ² g At steady state w/ sterile feed, μ ´ »1 · α¹1 ¸ Cº¼D ´ 0.65 D Monod kinetics: μ ´ 0.65 D ´ ½¾ ¿ À Á ÂÃ Ä ´ Å.ÆÇ È À Á ½¾ÉÅ.ÆÇ Ê ´ 0.48 Ë ËÌÍÎ/Ï Substrate Balance: g ´ µÄ ¸ ¹1 · αºµÄ ¸ μ± ² g Ð At steady state and µ = 0.65 D, ± ² ´ Ñ Å.ÆÇ »Ä ¸ ļ ´ 7.3 Ë ÎÒÌÌÓ/Ï X R = CX 1 = 1.5 (7.1) = 10.98 g cells/ L b) Cell balance around cell concentrator: X 2 = X 1 (1+ α- αC) = 7.3(1+0.7-0.7(1.5)) = 4.8 g cells/L
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Unformatted text preview: Fall 2010 BME339/CHE339/BIO335 HW#4 Solutions Problem 2 Known: Chemostat m = 0.5 hr-1 D = 0.28 hr-1 S o = 2 g/L S = 0.1 g/L Y X/S = 0.45 g X/g S Y X/O2 = 0.25 g X /g O 2 C * = 8 mg/L Solve: a) For chemostat and steady state, = D, X = Y X/S (S o S) = 0.45(2-0.1) = 0.86 g X /L OUR = q o2 X = X/Y X/O2, q o2 = D/Y X/O2 = 0.28 / 0.25 = 1.12 g O 2 /g X hr = 1120 mg O 2 /g X hr b) In oxygen being the rate-limiting step, OTR = OUR q o2 X = k L a (C *- C L ) k L a = q o2 X / (C *- C L ) = 1120(0.86)/(8-2) = 161 hr-1 Fall 2010 BME339/CHE339/BIO335 HW#4 Solutions Problem 3 Fraction destroyed = 1 x = 0.57 Fall 2010 BME339/CHE339/BIO335 HW#4 Solutions Problem 4 Fall 2010 BME339/CHE339/BIO335 HW#4 Solutions Problem 4 (contd)...
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339 HW 4 Solutions - Fall 2010 BME339/CHE339/BIO335 HW#4...

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