ERT 210/4 Process Control & Dynamics. DYNAMIC BEHAVIOR OF PROCESSES : Transfer Functions. COURSE OUTCOME 1 CO1) 1. Theoretical Models of Chemical Processes 2. Laplace Transform 3. Transfer Function Models DERIVE a transfer function for a differential equation model
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ERT 210/4Process Control & Dynamics
DYNAMIC BEHAVIOR OF PROCESSES :
1. Theoretical Models of Chemical Processes
2. Laplace Transform
3. Transfer Function Models
DERIVE a transfer function for a differential equation model
and LINEARIZE the nonlinear model through examples.
4. Dynamic Behavior of First-order and Second-order
5. Dynamic Response Characteristics of More Complicated
6. Development of Empirical Models from Process Data
Example 4.1: Stirred-tank Blending System
A stirred-tank blending process described by;
Let G(s) denote the transfer function between an input, x, and an output, y. Then, by definition
Example: Stirred Tank Heating System
Figure 2.3 Stirred-tank heating process with constant holdup, V.
Recall the previous dynamic model, assuming constant liquid holdup and flow rates:
Suppose the process is initially at steady state:
where steady-state value of T, etc. For steady-state conditions:
Subtract (3) from (1):
But, holdup and flow rates:
Thus we can substitute into (4-2) to get,
where we have introduced the following “deviation variables”, also called “perturbation variables”:
Evaluate holdup and flow rates:
By definition, Thus at time, t = 0,
But since our assumed initial condition was that the process was initially at steady state, i.e., it follows from (9) that
Note: The advantage of using deviation variables is that the initial condition term becomes zero. This simplifies the later analysis.
Rearrange (8) to solve for
where two new symbols are defined: holdup and flow rates:
Transfer Function Between and
Suppose is constant at the steady-state value. Then,
Then we can substitute into (10) and rearrange to get the desired TF:
Transfer Function Between and holdup and flow rates:
Suppose that Q is constant at its steady-state value:
Properties of Transfer Function Models holdup and flow rates:
For a linear system, K is a constant. But for a nonlinear system, K will depend on the operating condition
Calculation of holdup and flow rates:K from the TF Model:
If a TF model has a steady-state gain, then:
Take L, assuming the initial conditions are all zero. Rearranging gives the TF:
Definition: holdup and flow rates:
The order of the TF is defined to be the order of the numerator and denominator polynomial.
Note: The order of the TF is equal to the order of the ODE.
For any physical system, in (4-38). Otherwise, the system response to a step input will be an impulse. This can’t happen.
G holdup and flow rates:1(s)
Then the response to changes in both and can be written as:
The graphical representation (or block diagram) is:
Linearization of Nonlinear Models holdup and flow rates:
Linearization (continued) holdup and flow rates:
Perform a Taylor Series Expansion about and and truncate after the first order terms,
where , , and subscript s denotes the steady state, Note that the partial derivative terms are actually constants because they have been evaluated at the nominal operating point, s.
Example: Liquid Storage System holdup and flow rates:
A = area, Cv = constant
Combine (1) and (2) and rearrange:
Linearization (continued) holdup and flow rates:
Substitute (4-61) into (4-60) gives:
Because is a steady state, it follows from (4-60) that
= holdup and flow rates:
Example 4.4 (Page: 87) holdup and flow rates:
Suppose that two liquid surge tanks are placed in series so that the outflow from the first tank is in the inflow to the second tank, as shown in Fig. 4.3. If the outlet flow rate from each tank is linearly related to the height of the liquid (head) in the tank, find the transfer function relating changes in flow rate from the second tank, , to changes in flow rate into the first tank, . . Show how this transfer function is related to the individual transfer functions, , , , and . and denote the deviations in Tank 1 and Tank 2 levels, respectively. Assume that the two tanks have different cross-sectional areas A1and A2, and that the valve resistances are fixed at R1 and R2.
State-Space Models holdup and flow rates:
Example: CSTR Model holdup and flow rates:
Consider the previous CSTR model. Assume that Tccan vary with time while cAi, Ti, q and w are constant.
The linearized CSTR model in Eqs. 4-84 and 4-85 can be written in vector-matrix form:
Let and , and denote their time derivatives by and . Suppose that the coolant temperature Tc can be manipulated. For this situation, there is a scalar control variable, , and no modeled disturbance. Substituting these definitions into (4-92) gives,
which is in the form of Eq. 4-90 with x = col [x1, x2]. (The symbol “col” denotes a column vector.)
Note that the state-space model for Example 4.9 has d= 0 because disturbance variables were not included in (4-92). By contrast, suppose that the feed composition and feed temperature are considered to be disturbance variables in the original nonlinear CSTR model in Eqs. 2-60 and 2-64. Then the linearized model would include two additional deviation variables, and .
Thank you derivatives by and . Suppose that the coolant temperature
MISS RAHIMAH OTHMAN