q 3-18 from shuler and kargi, bioprocess engineering book. The solution manual a
ID: 1530663 • Letter: Q
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q 3-18 from shuler and kargi, bioprocess engineering book. The solution manual answer does not make sense. To begin with how are they getting values for vm and km for first part. If its from the plot, How do you get that value since the plot just goes upto 3.8 *10^-6. How can they get a value of 4.63*
10^-6?
are both immobilized on immobilized on the same flat, nonporous surface. For enzymeA the sub- the strate enzyme B the substrate is S2. The product of thefirst reaction is S2. That is: S, Es a. Figure 3.P1 depicts the rate of the first reaction on the surface as a function of local con centrations of S1. If the bulk concentration of S1 is 100 mg/l and the mass transfer coeffi- cient is 4 x 10 cm/s, what is the rate of consumption of S1 for a 1 cm surface? What is the surface concentration of S? b. The rate of the second reaction is: V S. 2 surfaceExplanation / Answer
a )
let
Da = maximum rate of reaction / maximum rate of external diffusion
= Vm ' / Ks [ Sb ]
Vm = 4.64 X 10-6 mg (cm-2 s )
Ks = 4 X 10-5 cm/sec
Da = 1.16 X 10-3
V = Vm' [ Ss ] / Km + [ Ss ]
= 4.64 X 10-6 X 100 / 9 + 100
= 4.64 X 10-4 / 109
V = 4.2 X 10-6 mg / cm2 s
b )
Da = Vm ' / Ks [ Sb ]
= 4 X 10-6 / 4 X 10-5 X 5
= 0.02
V = Vm'[ Ss ] / Km + [ Ss ]
V = 4 X 10-6 X 5 / 5 + 5
V = 2 X 10-6 mg / cm2 s
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