# Process Control: Understanding Dynamic Behavior

- Jun 27, 2003

- Background
- Linear State Space Models
- Introduction to Laplace Transforms
- Transfer Functions
- First-Order Behavior
- Integrating System
- Second-Order Behavior
- Lead-Lag Behavior
- Poles and Zeros
- Processes with Dead Time
- Padé Approximation for Dead Time
- Converting State Space Models to Transfer Functions
- Matlab and Simulink
- Summary
- References
- Student Exercises

## 3.5 First-Order Behavior

Many chemical processes can be modeled as first-order systems. The differential equation for a linear first-order process is often written in the following form:

This can also be written as

where the parameters (t* _{p}* and

*k*) and variables (

_{p}*y*and

*u*) have the following names: t

*is the process time constant (units of time),*

_{p}*k*the process gain (units of output/input),

_{p}*y*the output variable, and

*u*the input variable. Taking the Laplace transform of each term (

*notice that we are now using lower-case variables to represent the Laplace domain input and output*), and assuming that the initial condition is

*y*(0) = 0,

So the Laplace transform of Equation (3.25) can be written

or solving for *y*(*s*) we find a first-order transfer function

#### Step Response

Consider the case where the output is initially zero (steady state in deviation variable form), and the input is suddenly step changed by an amount D*u*. The Laplace transform of the input is

So Equation (3.26) can be written

Using a partial fraction expansion and inverting to the time domain, you should find (see Exercise 1)

Here the notion of a process gain is clear. After a substantial amount of time (*t* >> t* _{p}*), we find, from Equation (3.29),

That is,

and, since *y*(0) = 0, we can think of *y*(*t* → ∞) as D*y*, so

We can think of the process time constant as the amount of time it takes for 63.2% of the ultimate output change to occur, since when *t* = t* _{p}*,

Remember that this holds true only for first-order systems.

#### Impulse Response

Consider now an impulse input of magnitude *P*, which has units of the input*time; if the input is a volumetric flow rate (volume/time), then the impulse input is a volume. The output response is

You should find that the time domain solution is

which has an immediate response of *Pk _{p}*/t

*followed by a first-order decay with time.*

_{p}#### Example 3.4: Stirred-Tank Heater

Recall that an energy balance on a constant-volume stirred-tank heater (Example 2.3) yielded

where the subscript *s* is used to indicate that a particular variable remains at its steady-state value. Defining the following deviation variables

The equation can be written in the form

or

where the parameters of this first-order model are

The gain and time constant are clearly a function of scale. A process with a large steady-state flow rate will have a low gain, compared to a process with a small steady-state flow rate. This makes physical sense, since a given heat power input will have a larger effect on the small process than the large process. Similarly, a process with a large volume-to-flow rate ratio is expected to have a slow response compared to a process with a small volume-to-flow rate ratio.

Consider a heater with a constant liquid volume of *V _{s}* = 50 liters and a constant volumetric flow rate of

*F*= 10 liters/minute. For liquid water, the other parameters are r

_{s}*c*= 1 kcal/ liter°C. The process gain and time constant are then

_{p}#### Step Response

If a step input change of 10 kW is made, the resulting output change is [from Equation (3.29)]

A plot of the step input and the resulting output are shown in Figure 3-6.

**Figure 3-6. Step response of a stirred-tank heater, characteristic of a first-order system. Deviation variables.**

Remember that the inputs and outputs in this expression are in deviation variable form. If the steady-state values are (for an inlet temperature of 20°C)

Then, the physical temperature response is

#### Impulse Response

If an impulse input of 30 kJ is made, you should be able to show that the temperature changes immediately by 0.143°C (see Exercise 15).