Chapter 9 Example

Bifurcation Analysis

Substrate Inhibited Enzyme Reactions in a CSTR

Problem Statement

A substrate inhibiting enzyme reaction

is carried out in a CSTR with residence time t. The inlet concentration of the substrate is C_SO and the rate law is defined as

(1)

a) Reduce the material balance to show that

(2)

where

y = K_I C_S

y_o = K_I C_SO

a = K_m K_I

b = t V_max K_I

What is the maximum number of multiple steady states possible?

Solution Part (a)

b) Let A = y_o – 1 , B = a + b – y_o , C = y_o a

Show Eqn (2) can be put in the form

(3)

Solution Part (b)

c) Plot F(y) as a function of y (e.g. choose A=9, B=10, C=5.) Assuming the shape of the curve remains the same, sketch what happens as one varies A, B and C. Using the solution to (b), show the maximum and minimum points in F(y) occur at

(4)

and that y₁ and y₂ are positive and real if

(5)

(6)

Solution Part (c)

d) Discuss the fact that multiple steady states will occur if

(7)

Solution Part (d)

e) Show that multiple steady states can exist only if

(a + b) > y_o > 1 (8)

and

27 C² – 18 ABC – A²B² + 4A³C + 4B³ < 0 (9)

Solution Part (e)

f) For y_o = 10, show the region of multiple steady states on a plot of “b” as a function of “a”. Hint: First use Polymath to find C and a function of B

[Not to scale or shape]

What is the minimum value of “a” above which there will always be only one steady state? How will this minimum value of “a” change when y_o is changed?

Solution Part (f)

g) Develop a rough map to show

Solution Part (g)

h) For very high substrate concentrations, what is the maximum number of multiple steady states possible? Show the region of MSS on a plot of b as a function of y_o.

Hint: For high C_S:

Solution:

Part a) CSTR Material Balance

where

Because this is a cubic equation, three steady states are possible

Back to Part (b)

Part b)

A = y_o - 1

B = a + b - y_o

c = y_oa

Back to Part (c)

Part c)

@ y*

Back to Part (d)

Part d)

We need to find the condition when the above equation has three positive roots. Hence, graphically F(y) should be

If F(y) is such that

Case I

Case II

Back to Part (e)

Part e)

Then there is only one real positive real root to F(y) = 0. Let’s assume that the points where

F'(y) = 0

are and

For F(y) = 0 to have two positive real roots

> 0 , > 0 (1)

and

< 0 (2)

F'(y) = -3 y² - 2Ay + B = 0

For (1) i.e. > 0 and > 0

and

(3)

For (2) i.e.

Here,

(4)

Using Maple

Hence (A) and (B) are the conditions for F(y) = 0 to have three positive real roots, i.e., three multiple steady states.

Back to Part (f)

For multiple steady states

(5)

(6)

(7)

For condition (1) to hold

C₁ ≤ C ≤ C₂

For y_o = 10

A = 9

First compute C₁ and C₂ for various values of B as shown in Table 1

Table 1

Coordinate Transformation

C = ay_o

B = a + b – y_o

Now

Back to Part (g)

Part g)

Develop a plot of a versus y_o that divides the region into no MSS and MSS. Repeating the above procedure for various values of y_o, we will see that a_max increases as y_o increases.