# Digital Level Control One and Two Loops Proportional and Integral Control Single-Loop and Cascade Control

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1 Digital Level Control One and Two Loops Proportional and Integral Control Single-Loop and Cascade Control Introduction This experiment offers a look into the broad field of process control. This area of controls is important in the chemical, flow, and power industries. In this experiment (Figure ), the water level (or head) of the lower tank is controlled with three different algorithms:. Proportional control (P control). Proportional + integral control (PI control) 3. Cascade control The controller senses level from only the lower tank for proportional and proportional + integral control and from both the top and bottom tank levels for cascade control. With this feedback, the controller then adjusts the flow rate into the upper tank to keep the lower tank level constant. As is typical of process controllers like the Foxboro 743CB controller, the controller can be operated in automatic (AUTO) or manual (MAN) modes. When the controller is in manual mode, its control function is disabled and thus the process runs open loop. With the controller in automatic mode, the process runs in closed-loop control. We have used single-loop control in previous experiments and will use it in this experiment. We shall also use cascade control in this experiment. In cascade control, two coupled loops are configured. The control output from one loop is used as input to the second control loop. This is useful in situations where two loops can be configured so that what happens in the first loop anticipates what will happen subsequently in the second loop. Typical Closed-Loop Operation E U F C + G G G - C A P B Figure shows a typical SISO control loop. Figure - Typical SISO Control Loop In single-loop mode, the tank level system controls, the level in the lower tank. Thus C =. Thus is the desired tank level: =. As usual the error drives the controller. In this loop the controller is the Foxboro digital controller. The controller

2 outputs a digital signal to the actuator, the flow control valve in this system. The controller tells the valve what percentage

3 i d Top Tank L K t Transducer Disturbance Pump Bottom Tank L K t Transducer Main Pump F o eservoir Lumped Valve Constant K v Valve Positioner I S C Transducer y Air Supply Foxboro Controller Standard Instrument Society of America (ISA) Line Symbols: Pneumatic Signal Electrical Signal ydraulic Signal Figure - Two Tank Experiment Set-up

5 Figure 4 - Loop with Disturbance Flow Note also that the sensor has been removed. In the real system, the pressure at the bottom of each tank is sensed, converted to a pneumatic signal, then converted to an electrical signal that is read by the digital controller. All of this is transparent to the user, so effectively the loop has unity feedback. eferring to the schematic in Figure 4, introducing a disturbance flow q d into tank will cause the following changes: The head h will increase and cause an increase in the flow q through the orifice. The increased flow into tank, q, will increase the head in tank, h. The process variable h is fed back to the controller through the pressure transducer K t and compared with the setpoint, h. (When two loops are used, h is also fed back to the controller through K t.) The disturbance causes a negative error (e = h r - h ). As the level in tank rises, the error e becomes increasingly negative. This causes y to take on an increasingly negative value. Thus the flow control valve starts to shut, decreasing i, i.e. developing a negative qi. The tank level may fluctuate, but eventually the system will settle to a new steady state. Flow from the valve will decrease, compensating for the increased disturbance flow. Single-Loop, Proportional Control Proportional control is a commonsensical idea. emember that a feedback controller acts on error. So the actuating signal for the controller is the error. With a proportional controller, G c is simply a constant, K p. So the vigor of the control signal is proportional to the error, that is, U = K p * E So the larger E is, the larger U will be. This type of control makes sense. If the actual value of the controlled variable strays far from its desired value, a large U will be applied to try to force it back to its desired value, the setpoint (). As applied to the two tank experiment, single-loop proportional control is manifest as shown: h (in) e (in) y (% valve opening) + - Kp Plant h (in) Figure 5 - Single-Loop Proportional Control

6 Actually, the control calculation for level is not performed in inches but rather in percent. The bottom tank is 40 inches high. So a inch deviation of tank level = /40 or.5% deviation of the entire tank height. This is transparent to the user of the Foxboro controller. If he/she changes the desired tank level from the normal 7 inches to 9 inches, for example, the controller knows that this means h = inches = /40 = 5%. So internally the controller is using deviations from steady state operating levels and it is converting all level measurements into percent of tank range. A more accurate representation of the loop, as the controller manages it is (in) + - h (in) 40 h (%) + - e (%) y (%) K p Plant 40 (in) SS h (%) h (in) + + (in) Figure 6 - Internal Control Scheme for Single-Loop, P-Only Control In the two tank single-loop experiment, the reference setpoint, SS, is equal to 7 inches. With = 7 all the deviation variables are 0. To complete the above example, let the system operator increase tank s setpoint to 9 inches. e/she would set to 9 in the Foxboro controller. The controller would calculate h = in = 5% from this and apply it to the control loop input. This would produce e = 5%. This input to the proportional controller would create a positive y. The valve would open from its steady state position to allow more flow into tank. Tank s level would rise and maybe fluctuate. Eventually it would settle at a higher level (hopefully 9 inches). Single-Loop, Proportional + Integral (PI) Control Implementing single-loop PI control merely involves replacing the P-only controller in the previous section with a PI controller. The loop configuration remains the same. Integral control involves something called reset. The relationship between integral gain and reset time is discussed in the next section. The controller relationship is: y = K p * ( e + K I * Integral(e dt ) )

7 where e = h - h. In Laplace notation, this becomes: Y = K p * E * ( + K i /s )

8 eset Function and Bias (Integral) We saw in class that a Type 0 system subjected to a step input will have non-zero e ss. From the time-domain relationship above, we see that the controller will continue to put out a command signal (y) as long as the error is not 0. The error is integrated over time. So if the error persists, y will increase with the integral of e. A PI controller has both proportional and integral gains. Its structure is e y K + + P K s I Thus y = K P + K s I An equivalent form is commonly used: y = K P * + * e τs ere τ is the reset time. In the time domain y = K P * e + e dt τ y = K P * + dt τ e Figure 7 - PI Control Structure If a sudden unit error is injected into the system, at t = 0+, y = K p. The entire control output is generated by the proportional term. If we can maintain the error constant, the above equation becomes Note that at time t = τ, the contribution to y provided by the integral term is equal to the contribution provided by the proportional term. This is where the term reset comes from. It is the time needed with a constant error for the integral action to reset the action of the proportional term. In the Foxboro controller, reset time is given in minutes. In Foxboro control terminology, reset time is called IF.

9 Another peculiarity of the Foxboro controller and almost all other process controllers is that K P is not given directly. What is given in its place is proportional band. This is the percentage of output variable range needed to make the actuator stroke 00%. The Foxboro controller calls this PF. For your calculations PF = 00/K P. Double-Loop, Proportional, Cascade Control Although proportional control is a reasonable idea, it has the flaw that control actions do not take place until the loop error (E) deviates from zero. So a proportional controller only reacts to error. It cannot anticipate error and therefore cannot react before the error develops. Cascade control is an answer to this shortcoming. Cascade control anticipates an error before that error develops. For this portion of the experiment, cascade control will sense a deviation in level in the top tank and react to it to reduce future error in the lower tank before it develops. Cascade control is often used in process control to anticipate changes before they have an effect disturbing the system. Cascade control refers to the closing of two or more concentric loops. The inner loop s reference,, is given a floating value equal to the outer loop s output, C. In this section, which describes the cascade controller we will implement, the subscripts and are used to identify the terms associated with the inner and outer loops, respectively. These correspond also to the top tank () and the bottom tank (). h e h e y h Top h Bottom + - K p + - K p Tank Tank Figure 8 - Cascade Control Configuration with P-Only Controllers Again, the reference height, h, is considered to be zero at the normal operating level. The normal operating valve level and the normal operating lower tank height are also considered to be 0.

10 Single-Loop Objective un one loop in proportional only control with feedback from and a gain of K p = 5. Also run one loop in proportional + integral control with feedback from, K p =.5, and τ =.85 minutes. For each loop, plot as a function of time for a step flow disturbance into the top tank. For the Foxboro 743CB controller, proportional faceplate (PF) and integral faceplate (IF) are used to represent proportional gain (K p ) and integral gain (K i ), where Procedure: Proportional Control: PF = 00/K p and IF = τ. Connect the system as shown in Figure. The pump cord should be plugged into the wall behind the tanks.. Program the controller using the Proportional Control Action from the Foxboro handout. 3. Press the SEL key until the green dot on the screen lies above the most far left vertical bar (setpoint, h r ) on the display screen. Press the / key until the display shows a setpoint value of 7.00 in. 4. Press SEL key until the green dot on the screen lies above the most far right vertical bar (output, Y) on the display screen. 5. Bring the output, Y (the far right bar on the display screen), to 44, using the / key. 6. Select manual mode by pressing the A/M button. 7. Turn on the main pump and close the bottom tank valve. 8. When the head of the bottom tank,, reaches 6 inches, open the bottom tank valve and select automatic mode by pressing the A/M button. 9. Let the system reach steady state (hint: the middle bar on the display screen should indicate about 7 inches). 0. Press the SEL key several times until the green dot on the screen lies above the most far right vertical bar (output, Y) on the display screen.. Put in a step disturbance by turning on the disturbance pump and record,, and Y at 0-second intervals until reaches steady state.. Measure the disturbance flow, d, immediately after the run by turning off the main pump, plugging tank (upper tank) and timing the change in. Proportional + Integral (PI) Control:

11 . Switch the system to manual mode by pressing the A/M button. Bring the output, Y, to 44. Turn off the disturbance pump and turn on the main pump.. Program the controller using the Proportional Plus Integral Control Action from the Foxboro handout. 3. Close the bottom tank valve. 4. When the head of the bottom tank,, reaches 6 inches, open the bottom tank valve and select automatic mode by pressing the A/M button. 5. Let the system reach steady state (hint: the middle bar on the display screen should indicate 7 inches). 6. Press the SEL key several times until the green dot on the screen lies above the most far right vertical bar (output, Y) on the display screen. 7. Put in step disturbance by turning on the disturbance pump and recording,, and Y at 0-second intervals until reaches steady state. Two-Loop (Cascade) Proportional Control Objective: un two loops with feedback from and at a proportional gain of K p =.0 and K p =.0. Procedure. Turn both pumps off.. Program the controller using Cascade Proportional Control Action from the Foxboro handout. 3. Press the SEL button for 5 seconds until the display shows TOPTANK. 4. Select remote setpoint by pressing /L button (the display screen should show ). 5. Select automatic mode by pressing the A/M button. 6. Press the SEL button for 5 seconds until the display shows BOTTOMTK. 7. Select automatic mode by pressing the A/M button. 8. Turn on the main pump and let the head of the bottom tank reach steady state (hint: steady state will not be 7 in.). 9. Press the SEL button for 5 seconds until the display shows TOPTANK. 0. Press the SEL key several times until the green dot on the screen lies above the most far right vertical bar (output, Y) on the display screen.. Turn on the disturbance pump.. ecord,, and Y at 0-second intervals until reaches steady state.

13 Two Tanks with Interaction In class and in lab we looked at a two tank system without interaction. This means that the tanks are stacked on top of each other. The top tank drains into the bottom tank. But the bottom tank level plays no role in determining the flow out of the top tank. We developed two equations, one relating input flow i and tank head, the other relating input flow and output flow o. These relationships are good only for tanks that do not interact. Since I started teaching this lab, I have always wondered about the merits of a two tank experiment without interaction vs. one with interaction. In this part of the lab, we will use our simulation tools to answer this question. The primary question to answer is how would the two tank system function if the tanks were placed side by side (Figure 9). Let s use all of the same equipment but simply move the top tank down so that its bottom is level with that of the other tank. We shall also make the top tank as tall as the bottom tank. Let s also keep the disturbance flow going into the first tank. Let the valve between the two tanks stay at 5.5 inches below the tank bottoms. i d L Tank Tank L o Figure 9 - Two Tanks with Interaction You need to know the steady state operating level of the two tanks in order to calculate and. Let s assume we are using the same equipment. The ss used to calculate the resistance of the tank outlet valve is the difference in head across the valve. Thus if we use the same equipment, ss will be the same as it was with the tanks vertically arranged. ss will exceed ss by the tank level of the top tank when the tanks are vertically arranged. For example, if the existing system s steady state levels are ss 0 in and ss 7 in, then for the side-by-side tanks, ss 37 in and ss 7 in. This, by the way, is a good example of how simulation is used in the real world. A change is contemplated in a system. The engineers involved want to know how the revised system will function. This exercise is a perfect example of the utility of your

14 simulation tools. System elationships The following relationships use the deviation quantities as discussed above. These are the deviations of system variables away from the reference steady state. The interaction flow is. This flow can be in either direction. If we consider flow from tank to tank to be positive, then in terms of the deviation quantities, (s) + - (s) ( s) = (s) ( s) ( s) ere is the resistance of the valve. See the course handout on modeling fluid systems. The block diagram is shown to the right of the equation. Note that the capital letters as system variables does not imply absolute quantities. ather i (s) = L{qi(t)}, for example. By mass continuity, the change in volume of fluid in a tank is equal to what comes in minus what goes out. Since the fluid here is incompressible, the mass balance becomes a volume balance. (s) i (s) + - A s (s) L{ V& } = A ( s) s = i ( s) ( s) For the first tank L{ V& } = A ( s) s = ( s) o ( s) (s) (s) + - A s o (s) For the second tank

15 The flow out of the second tank is the same for a regular tank without interaction. (s) o(s) o ( s) = ( s) These pieces need to be assembled into a plant model. Also the plant needs an actuator and a controller. emember that in the control loop for this system, the input is h. The output is h. Last, the disturbance flow will need to be added to the system model. Procedure Build a block diagram of the system. Include the plant, the controller,the actuator, and the disturbance flow. Use the above as a guide to develop a Simulink model of the two tank model with simulation. For your physical quantities, use all values from the two tank experiment without interaction. Make both tanks the same height (40 inches). If you have not run the two tank experiment yet, borrow data from a group who has in order to start your modeling. With the model constructed, install first a P-only controller, using the same gain as in the lab experiment. Inject a step disturbance into the system. ecord the response. Perform the same operation with a PI controller. Lab eport Compare the simulations for the two tank without interaction and the two tank with interaction. This can be done qualitatively and quantitatively. For the qualitative analysis, work out in symbolic form the closed loop transfer function for both two tank systems. I recommend as a beginning just using G c in a block. Once you have worked out the transfer functions, you can replace G c with both the P and the PI controllers for the comparison. Your transfer functions should be / d. I think you will wind up with second order systems for the P-only cases. From these you should be able to work up expressions for Kss, ζ, and ω n for each system and compare them. I cannot tell right off the bat how the systems with PI control will come out. If you have third order systems, you may have to put in actual numbers to determine whether or not you have a dominant second order system. But if you need to do this, do it and then compare the numerical results. For the quantitative analysis, make two sets of three plots, one set for the P-only control, the other for the PI control. The three plots should be vs. t, vs. t, and Y vs. t. On each plot, plot the results of the two models together, i.e. with and without interaction. From the plots you should be able to work out Kss, ζ, and ω n for each system. Do this and compare them.

16 Your discussion should revolve around the comparison of the symbolic closed loop transfer functions of each system and also a comparison of the two plots. Also, if there are any difficulties you encountered in constructing the two tank with interaction model, discuss these in your report.

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