Unformatted text preview: Homework 3: Due Wednesday, Sept. 29th Chemical Engineering 170A: Biochemical Engineering Question 1. Immobilized Enzymes: Consider an enzyme immobilized to an uncharged, nonporous particle. We wish to examine the effects of the dimensionless bulk substrate concentration, β, and the modified Damkohler number, β*Da, on the dimensionless reaction rate, v/vmax. Our approach will be to plot v/vmax versus β for 2 different values of β*Da. a) First, derive an expression for v/vmax in terms of β and β*Da. Equation (2.16) from the book is a good starting point. η v' v' = E max (Eq. 2.16) 1+ν (1 + ν )⋅ x * ηE = (Eq. 2.18) x* +ν v' x* β⋅ x * 1 =* = where β = * v'max x + ν β⋅ x + 1 ν
x* 1− x* = = Da x* +ν x* 1 x+ β
* (Eq. 2.13) Solve for x * : Question 2. Flux of Product From Immobilized Enzymes ȹ 1 ȹ x * ⋅ Da = ȹ x * + ȹ⋅ 1 − x * An enzyme is immobilized on a nonporous support obeys the following reaction scheme β Ⱥ ȹ ( ) 1 x* E+S E+2P x * ⋅ Da = x * − x *2 +ES− ββ You are also given the following information: *2 * β⋅ x + x ⋅ ( β⋅ Da + 1 − β) kp = 7 x 10
3 cm/sec Km = 0.4 mM vmax= 300 µM cm
2 min
1 −( β⋅ Da + 1 − β) + ( β⋅ Da + 1 − β)2 + 4β x* = (we are only concerned with the posit SO = 6 mM ηE = 0.30 2β v' To get as a function of β and β⋅ Da, substitute in x * : v'max a) Write an expression that relates the flux of product from the surface to the rate of 2 − + 1 − β) + ( β⋅ Da in your expression β⋅ Da + 1 − β) + −( substrate consumption at the surface. ⋅ D( β⋅ Daany parameters + 1 − β) + 4β β efine v' β⋅ x * 2β that are not given above. = = = P* is the concentration of pβ⋅oduct at the ( β⋅ Da + 1 − βs +he concentration of substrate ⋅ Da + 1 − β) + v'max r x * + 1 − surface. S* i ) t ( β⋅ Da + 1 − β)2 + 4β −( β β⋅ +1 at the surface. 2β kp is the mass transfer coefficient for the product. −( β⋅ Da + 1 − β) + ( β⋅ Da + 1 − β)2 + 4β β − β⋅ Da − 1 + (β⋅ Da + 1 − β)2 + v' = = 2 2 v' max −( β⋅ Da + 1 − β) + ( β⋅ Da + 1 − β) + 4β + 2 β − β⋅ Da + 1 + (β⋅ Da + 1 − β) + € b) Calculate the measured rate of appearance of product in the bulk solution per unit area. v' S=S * v' S=So
= ηE v' S=S * = ηE v' S=So ( ) 2v'max [ So ] ⋅η Question 3. The buzz about immobilized [ S ] E on a surface K M + enzymes o ȹ 300 µM 1mM ȹ Honey bees synthesize the enzyme invertase that allows them to make honey from 2ȹ 2the hydrolysis of ȹsucrose into a mixture of fructose and glucose, ⋅ ⋅ 6mM nectar. Invertase catalyzes ȹ cm min 1000µM Ⱥ also known as inverted sugar syrup. In industrial applications, invertase is immobilized in Flux = ⋅ 0.4 non
porous polymer beads to 4onvert 6ucrose to fructose and glucose. 0. c mM + s mM mM µ In this problem,225are only concerned with M conversion of sucrose to fructose. The Flux = 0. we or 225 2 the 2 conversion r is controlled by transfer of sucrose to the surface of the beads through cm min cm min ate Flux of product = liquid film, and the conversion takes place on the surfaces of the beads. The following parameters are given for the system. ks = 0.2 cm s
1 Km = 85 mmol L
1 So = 100 mmol L
1 vmax’= 2.1 mmol s
1 cm
2 support surfaces a) Determine the surface concentration of sucrose – You can do this one. b) Determine the rate of sucrose degradation under mass transfer controlled conditions. Under mass transfer controlled conditions, [S*] is very small. Question 4. Effectiveness Factors of Glucose Oxidase Entrapped in Poly(HEMA) Blanch and Clark Ch. 2 Prob. 13 € φ2 = φ2 = R 2 ȹ v max ȹ 9 ȹ K M Deff ȹ ȹ ȹ ȹ Ⱥ ȹ d 2 ȹ v ȹ ȹ = ȹ max ȹ ȹ 36 ȹ K D ȹ Ⱥ ȹ M eff Ⱥ 2 ȹ 1 ȹ v 1 Use Figure 2.12 to solve for ηI φor each value oȹ φ. Use your best interpretation of the data. f f = ȹ max ȹ = 1.11x105 2 ȹ Kbut that’s ok. 2 Your values will not exactly match mine M Deff Ⱥ d 36 ȹ cm d 2 ȹ v max ȹ 4⋅ 9 ȹ K M Deff ȹ ȹ2 1 ȹ cm φ = d ⋅ 1.11x10 ȹ ȹ cm2 ȹ 10, 000µm Ⱥ
2 2 5 ȹ ȹ2 1 φ = d ⋅ 1.11x10 ȹ ȹ ȹ 10, 000µm Ⱥ
2 5 ȹ ȹ2 Question 5. Eadie
Hofstee Plots for Immobilized Enzymes Blanch and Clark Ch. 2 Prob. 15 b) φ is large. Assume piece of a sphere is like a slab. i) Shell balance over slab defined by r, r+∆r, and surface area S €
in − out + gen − cons = 0 mol conc S −N S ) − v(S )SΔr = 0 where N [ = ] and v(S )[ =] ( time⋅ surface area time⋅ volume −( N S −N S ) = v(S )SΔr (divide by SΔr) Question 6. Kinetics of Chymotrypsin Immobilized to Polymer Microspheres (N − N ) = v(S) − Ns
r =r +Δr s r =r s s r =r +Δr s r =r s r =r +Δr s r =r Blanch and Clark Ch. 2 Prob. 16 Δr lim : a) The kΔr →0 parameters can be obtained from an Eadie
Hofstee plot of v/Soea vs. v/ea for inetic d Ns immobilized enzymes (as shown in the figure for Problem 15). = v(S ) dr b) dS d ȹ dS ȹ Ns = −Dthat for teff pȹ ȹ = v( S )is in terms of per grams of catalyst unlike Problem 15 ; D his roblem, v Remember eff dr dr ȹ dr Ⱥ 2 and the expression for φ where v is in terms of per volume of immobilized catalyst. dS Deff 2 = v(S ) dr () €
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 Enzyme, Biochemical Engineering, Orders of magnitude, Dr. Dre, β, the00, dr Ⱥ00 00

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