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Feedback Control Loop Concepts

What we have to learn to do, we learn by doing.

—Aristotle

Introduction

Prior to attempting this workshop, you should review Chapter 3 in the book.

Process systems respond to various disturbances (or stimuli) in many different ways. However, certain types of responses are characteristic of specific types of processes. The characteristic response of a process can be described as its personality. Process control engineers have developed a range of terms and concepts to describe different process personalities and they use this knowledge to develop effective control systems.

Two of the most common personalities are those for first- and second-order systems. First-order systems may also be called first-order processes or first-order lags and can be mathematically modelled through the use of a first-order differential equation. Shown in Figure 2.1 is the typical step response of a first-order process. The time constant, τ, was discussed in Chapter 3 and is related to the speed of the process response; the slower the process the larger the value of τ.

Figure 2.1 First-order process response to a step change.

Unlike first-order processes, second-order processes can have several different types of responses. Second-order processes are more complex than first order, and hence the mathematical models used to describe these processes are also more complex. There are three types of second-order systems to consider. The key parameter in determining the type of system is the damping coefficient, ξ. When ξ < 1, the system is underdamped and has an oscillatory response as shown is Figure 2.2. An underdamped system overshoots the final value and the degree of overshoot is dependent upon the value of ξ. The smaller the value, the greater the overshoot. If ξ = 1, the system is deemed critically damped and has no oscillation. A critically damped system provides the fastest approach to the final value without the overshoot that is found in an underdamped system. Finally, if ξ > 1 the system is overdamped. An overdamped system is similar to a critically damped system in that the response never overshoots the final value. However, the approach for an overdamped system is much slower and varies depending upon the value of ξ. The larger the damping coefficient, the slower the response.

Figure 2.2 Second-order process responses to a step change.

There are two main differences between first- and second-order responses. The first difference is obviously that a second-order response can oscillate while a first order cannot. The second difference is the steepness of the slope for the two responses. For a first-order response, the steepest part of the slope is at the beginning, whereas for the second-order response, the steepest part of the slope occurs later in the response.

First- and second-order systems are not the only two types of systems that exist. There are higher order systems, such as third- or fourth-order systems. However, these higher order systems will not be discussed.

Key Learning Objectives

Tasks

1. Level Response

Capacity-dominated process behaviour can best be studied using a very common process element, namely the surge tank or separator. The ordinary differential equation (ODE) for a single tank has been presented in the notes in Chapter 3 and is implemented in VMGSim as the Separator unit operation.

Build a simulation in VMGSim as follows:

In the Basis environment add a fluid package with the following:

Property Package: Advanced Peng–Robinson;

Components: water, N2 and O2 (N2 and O2 are used to simulate the air displacing water when the separator level drops).

Switch to the Simulation environment add a feed stream with the following data: Name = Feed to Separator, Flow = 20 kmol/h, Temperature = 15°C and Pressure = 1 atm, Composition: water = 100%, O2 = 0% and N2 = 0%.

Add a separator to your flow sheet. Complete the simulation by adding the two product streams as indicated in Figure 2.3.

Figure 2.3 Separator level response.

Now the model should be totally solved in steady state.

Before switching to dynamics, you need to understand some basic rules for converting from steady-state model to a dynamic model in a simulation tool.

Now switch to dynamics, add Dynamic Pressure/Flow specifications to the boundary streams. For the feed and liquid product streams use flow specs (however, the flow of liquid product stream will be calculated by the Process Calculator later on), and for the Equalization Line use pressure spec. Also, specify the volume of the separator as 2 m3.

Before we start running the model we need to take care of the equalization line that simulates the ‘open to atmosphere’ configuration of separators. Open the ‘Equalization Line’ stream form, on the mole composition part, enter the following composition specification: water = 0%, N2 = 80% and O2 = 20%. This product block will supply composition for the equalization line in case of flow reversal (which will be the case if the level in the separator starts to drop and it needs to be replaced by air).

Since you just switched from steady state, the flow specs will represent the steady-state condition (flow in = flow out), so the level in the separator will stay at 100%. Increase the outlet flow spec value and you will notice the level in the separator dropping.

Calculate the flow out of the separator using Equation 2.1, which describes a linear valve, and then Equation 2.2, which describes a non-linear valve. In both cases, the outlet flow rate is a function of the liquid head on the separator only:

(2.1)

(2.2)

In Equations 2.1 and 2.2, SI units are flow in m3/h and head in metres. Use the Process Calculator unit operation to incorporate Equations 2.1 and 2.2 into the simulation. Import the Separator liquid level into the spreadsheet, this will be the head. Use this value of the head to calculate the liquid outlet flow. A K value of 20 would work well with SI units. Then export the calculated flow back to the ‘Molar Flow’ of the ‘Liquid Out’ stream. This last action can only be performed in the dynamic mode otherwise the material balance will be violated.

System identification is the term used to define a procedure to characterize the process response. In this case, system identification can be accomplished by adjusting the feed rate to the separator in steps, up and down, and then observing the separator level response on a strip chart. This is termed step response testing.

Set up strip charts, recording the important variables, to study the open loop response of the capacity-dominated process consisting of

[Hints: Use a pipe segment and calculate a volume that should give a dead time of around 10 minutes. The number of increments in the pipe segment will affect the dead time – the more the increments, the closer the simulated dead time will be to the value calculated by fluid velocity and pipe length.]

• What is the open loop response of the separator level to a step change in the feed rate in a process with a single separator only?
• What effect does the addition of a second separator have on the level response in Separator 1? And in Separator 2?
• What effect does the addition of a third separator have on the level response in Separator 1? And in Separator 2? What is the open loop response of Separator 3?
• What effect does the volume of the separator have on the personality of the response?
• How does the valve type affect the process personality?
• What effect does the addition of the pipe segment between the two separators have on the response of the level in Separator 1 and Separator 2? Did the pipe segment properly simulate dead time in this situation? Why or why not?

2. Temperature Response

The next exercise in this workshop requires that you set up a mixing separator to heat water directly using live steam, as illustrated in Figure 2.4. Use the Multi-feed Separator volume of 2 m3. Add a valve on the Liquid Out stream and a level controller (Process Variable = Multi-feed Separator Liquid Level Percent and Output Target Object = The liquid out valve) with a 50% set point. Ki = 1 and Ti = 60 minutes would be a standard configuration to the liquid level controller parameter. The feed water stream to the separator enters with a flow of 100 kmol/h at 15°C and 1 atm. The steam to the separator enters as saturated steam at 1 atm (since this stream will only bare a flow spec in dynamics, the pressure will float on the separator pressure and temperature will change with pressure to satisfy the saturated vapour condition).

Figure 2.4 Mixing separator process.

Perform a series of steady-state runs to determine the amount of steam required to raise the temperature of the feed water stream to about 200°F. Then, switch to the dynamic mode of operation and perform step response testing by varying the inlet water flow rate and feed temperature or steam flow rate to determine the process response. Remember to use the strip charts to observe the important process variables.

Add a pipe segment to the system on the outlet of the separator as in the previous exercise. Calculate a pipe segment volume to give approximately 10 minutes of dead time and specify the volume in the ‘Holdup’ tab. Make sure that the number of sections in the pipe segment is set to at least 50 instead of the default of 1. Note that the holdup volume is for a single tray, so the total holdup volume should be 50 times of the holdup volume for a single tray. Repeat your analysis of step disturbances noting the relationship between the separator temperature and the temperature at the outlet of the pipe segment. Repeat the analysis again but with a different volume for the pipe segment and note any differences on the response.

• What type of response does the process produce?
• How does the pipe segment affect the response?
• What effect does the pipe segment volume have on the response? Did the pipe segment properly simulate dead time in this situation? Why or why not?

Present your findings on CD, DVD or thumb drive in a short report using MS-Word. Also include on the submitted media a copy of the VMGSim files which you used to generate your findings.