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ERT 210/4 Process Control & Dynamics - PowerPoint PPT Presentation

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 :

Transfer Functions

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

Processes

5. Dynamic Response Characteristics of More Complicated

Processes

6. Development of Empirical Models from Process Data

• A transfer function (TF) relates one input and one output:

• The following terminology is used:

x

input

forcing function

“cause”

y

output

response

“effect”

Example 4.1: Stirred-tank Blending System

A stirred-tank blending process described by;

(4-7)

(4-14)

• Convenient representation of a linear, dynamic model.

• Laplace transforms (LT) vs Transfer functions (TF)

• - LT: the full procedure must be applied for each model solution.

• - TF: is an algebraic expression for the dynamic relation

• between a selected input and output of the process model.

• A transfer function (TF) relates one input and one output:

• The following terminology is used:

x

input

forcing function

“cause”

y

output

response

“effect”

Let G(s) denote the transfer function between an input, x, and an output, y. Then, by definition

=

where:

=

=

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:

=

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”:

=

=

=

TakeLof (6):

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:

Thus, rearranging

• The TFs in (12) and (13) show the individual effects of Q and on T. What about simultaneous changes in both Q and ?

• Answer holdup and flow rates:: See (10). The same TFs are valid for simultaneous changes.

• Note that (10) shows that the effects of changes in both Q and are additive. This always occurs for linear, dynamic models (like TFs) because the Principle of Superposition is valid.

• The TF model enables us to determine the output response to any change in an input.

• Use deviation variables to eliminate initial conditions for TF models.

Properties of Transfer Function Models holdup and flow rates:

• The steady-state of a TF can be used to calculate the steady-state change in an output due to a steady-state change in the input. For example, suppose we know two steady states for an input, u, and an output, y. Then we can calculate the steady-state gain, K, from:

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:

• This important result is a consequence of the Final Value Theorem

• Note: Some TF models do not have a steady-state gain (e.g., integrating process in Ch. 5)

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.

Physical Realizability:

For any physical system, in (4-38). Otherwise, the system response to a step input will be an impulse. This can’t happen.

Example:

G holdup and flow rates:1(s)

Y(s)

G2(s)

• Suppose that an output is influenced by two inputs and that the transfer functions are known:

Then the response to changes in both and can be written as:

The graphical representation (or block diagram) is:

U1(s)

U2(s)

Then,

Substitute,

Or,

Linearization of Nonlinear Models holdup and flow rates:

• So far, we have emphasized linear models which can be transformed into TF models.

• But most physical processes and physical models are nonlinear.

• But over a small range of operating conditions, the behavior may be approximately linear.

• Conclude: Linear approximations can be useful, especially for purpose of analysis.

• Approximate linear models can be obtained analytically by a method called “linearization”. It is based on a Taylor Series Expansion of a nonlinear function about a specified operating point.

Linearization (continued) holdup and flow rates:

• Consider a nonlinear, dynamic model relating two process variables, u and y:

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:

Mass balance:

Valve relation:

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.

Fig. 4.3

State-Space Models holdup and flow rates:

• Dynamic models derived from physical principles typically

• consist of one or more ordinary differential equations (ODEs).

• In this section, we consider a general class of ODE models referred to as state-space models.

• Consider standard form for a linear state-space model,

• where: holdup and flow rates:

• x = the state vector

• u = the control vector of manipulated variables (also called control variables)

• d = the disturbance vector

• y = the output vector of measured variables. (We use boldface symbols to denote vector and matrices, and plain text to represent scalars.)

• The elements of x are referred to as state variables.

• The elements of y are typically a subset of x, namely, the state variables that are measured. In general, x, u, d, and y are functions of time.

• The time derivative of x is denoted by

• Matrices A, B, C, and E are constant matrices.

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.

Nonlinear Model:

Linearized Model:

where:

• Example 4.9 holdup and flow rates:

• Show that the linearized CSTR model of Example 4.8 can

• be written in the state-space form of Eqs. 4-90 and 4-91.

• Derive state-space models for two cases:

• Both cA and T are measured.

• Only T is measured.

Solution

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.)

• If both derivatives by and . Suppose that the coolant temperature T and cA are measured, then y = x, and C = I in Eq. 4-91, where I denotes the 2x2 identity matrix. A and B are defined in (4-93).

• When only T is measured, output vector y is a scalar, and C is a row vector, C = [0,1].

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

Prepared by,

MISS RAHIMAH OTHMAN