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978-0137062706 Chapter 10
68 Chapter 10 Problem 10.1 C* = 7.3 mg/l Plot DO versus t: Determination of OUR: * LL L 2 dC k a (C C ) qo X dt = − − with air off: kLa (C*–CL) = O L2 […]
978-0137062706 Chapter 11
Chapter 11 Problem 11.l. a. For simplicity, neglect cell size distribution and cake resistance considerations. Assume the following relationship applies to tubular centrifuge: 85 ( ) ( ) ( ) ( ) 2 2 Recovery % at Condition 1 Rotational […]
978-0137062706 Chapter 12
Chapter 12 Problem 12.11 92 0 S 0.05 S 2g L== Glucose balance over differential height “dZ” yields. 0 QdS kXS dV kXSA d Z− = = Integration yields 0 kXA dS dZ SQ − = ( )( […]
978-0137062706 Chapter 13
Chapter 13 Problem 13.1. Use eq. 9.72 to see if S > 0 at r = 0 Need rm for the bead: Mass of cells/bead = Vol. ∙ 0.25 ∙ ϼcell ∙ 0.1 = 4/3 (0.2 cm)3 ∙ 0.25 ∙ […]
978-0137062706 Chapter 14
Chapter 14 Problem. 14.1 What is P for a single cell or homogeneous population with the equivalent of 50 monomer plasmids at division if multimerization results in 40% of plasmid DNA as dimers and 16% as tetramers? # dimers = […]
978-0137062706 Chapter 15
Chapter 15 Problem 15.1. This derivation is very similar to the one in the text with “free” virus. The only eqn that changes is (15.4) which for pre bound virus is 104 s e s dV / dt k V=− […]
978-0137062706 Chapter 16
Chapter 16 Problem 16.1. Data SO = 5 g glucose/l XO(g dw/l Y (g dw/g glu) μm(hr−1) Ks (g/l) Kd (hr−1) E.coli (EC) 0.03 0.5 1.0 0.01 0.05 A.vinelandii (AV) 0.15 0.35 0.5 0.02 0.10 Kinetic Expressions – Batch Growth, […]
978-0137062706 Chapter 3
Chapter 3 Problem 3.1 1 ( ) ( ) 3 15 24 k kk 2 1 kk 52 S E ES (ES) P E rate ESk + ⎯⎯→ + = a.) Equilibrium approach * r values are the same as […]
978-0137062706 Chapter 6 Part 1
20 Chapter 6 Problem 6.1. Volume of a single cell: Vc = (3.14/4) (1))2 (2) = 1.57 um3/cell = 1.57 x10–12 cm3/cell Problem 6.2. a. max 1 = 0.693/td1 = 0.693/ (2 h) = 0.346 h-1 max 2 […]
978-0137062706 Chapter 6 Part 2
S m D K1 SD 1 + − = + − − a. p X S 0 m DP q X Y D S D 1 (D ) 1 = […]
978-0137062706 Chapter 7
Chapter 7 Problem 7.1. Total nitrogen in cells =(Nitrogen content of cells)×(Final cell concentration)×(Culture volume)=(0.12 g nitrogen/g cells) (30 g cell/liter) (1000 liters) = 3.6 kg nitrogen. MW of (NH4)2SO4 = 132; Nitrogen part is 28. Amount of (NH4)2SO4 needed […]
978-0137062706 Chapter 9
Chapter 9 Problem 9.1 51 V 1000 l= S K 1.5 g/l= 0 S 10 g/l= 1 m0.4 h− = X/S Y 0.5 g/g= F 100 l/h= a) 1 F 100 l/h D 0.1h V 1000 l − = = […]