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M5.3 Steady-State and Dynamic Behavior (Chapter 3)

At steady state, the component A balance yields a quadratic expression,

graphics/m05equ05.gif


where the subscript s is used to indicate the steady state value. Solving this quadratic and using the positive root (clearly, there cannot be negative concentrations), we find

graphics/m05equ06.gif


and solving for the steady-state concentration of B,

graphics/m05equ07.gif


Using these two expressions, we can find the steady-state input-output curve relating CBs and Fs/V. Here we consider the dilution rate (F/V) to be the manipulated input. The main advantage is that it is independent of scale (reactor volume). Whether the reactor is 1 or 10,000 liters, the same dilution rate will yield the same concentrations (assuming the feed stream is the same composition).

For the particular reaction under consideration, the rate constants are

graphics/m05equ08.gif


and the steady-state feed concentration is CAfs = 10 gmol/liter.

Steady-State Input-Output Curve

Notice that this process has a nonlinear relationship between the steady-state dilution rate (F/V) and the steady-state concentration of B (Figure M5-2). There exists a maximum concentration of B that can be achieved. It is interesting to note that the reactor can not be controlled at this maximum point because the process gain is zero. For a given desired value of the concentration of B (as long as it is less than the maximum possible value), there are two dilution rates that can achieve the concentration. This is known as input multiplicity.

Figure M5-2. Steady-state input-output curve. The dotted lines indicate two possible flow rates to achieve a concentration of 1.117 gmol of B/liter.

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In this example, a steady-state concentration of B of 1.117 gmol/liter can be obtained with either Fs/V = 0.5714 min-1 (case 1) or Fs/V = 2.8744 min-1 (case 2). Notice that the process gain (slope of the steady-state input-output curve) is positive for case 1 and negative for case 2. For controller design, it is clearly important to know whether you are operating on the left-hand side (e.g., case 1) or right-hand side (e.g., case 2) of the "peak," since the sign of the controller gain must be the same sign as the process gain. Also notice that a steady-state concentration greater than 1.266 gmol/liter cannot be achieved regardless of the controller used.

Linear Analysis

The linear state space model is

graphics/m05equ09.gif


where the states, inputs, and output are in deviation variable form. The first input (dilution rate) is manipulated and the second (feed concentration of A) is a disturbance input.

graphics/m05equ10.gif


You should linearize the two modeling equations at steady-state solution to find the following state space matrices:

graphics/m05equ11.gif


Case 1—Operation on the "Left-Hand Side" of the Peak Concentration

Based on the steady-state operating point of CAs = 3 gmol/liter, CBs = 1.117 gmol/liter, and Fs/V = 0.5714 min-1, the state space model is

graphics/m05equ12.gif


The manipulated input-output process transfer function for the reactor is [you can find this analytically, using G(s) = C(sI-A)-1B, or numerically, using Module 4]

graphics/m05equ13.gif


and the disturbance input-output transfer function is

graphics/m05equ14.gif


The responses of the linear and nonlinear models to a step change in F/V of 0.1 min-1 are compared in Figure M5-3.

Figure M5-3. Comparison of concentration of B response to step change in F/V of 0.1 min-1. Nonlinear (solid) vs. linear (dashed).

graphics/m05fig03.gif

Note: A motivated reader may wish to demonstrate that the linear and nonlinear models have the same response for a step change of 0.01 min-1, while the differences in response to a step change of 0.5 min-1 are substantial. Function files for integration of the nonlinear model are presented in Module 3. The focus of the current module is on responses of the linear model.

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