University of Tennessee at Chattanooga Steady State and Step Response for Filter Wash Station ENGR 3280L By. Jonathan Cain. (Emily Stark, Jared Baker)

Transcription

1 University of Tennessee at Chattanooga Steady State and Step Response for Filter Wash Station ENGR 3280L By (Emily Stark, Jared Baker)

2 i Table of Contents Introduction 1 Background and Theory.3-5 Procedure Results Discussion of Results.. Conclusions and Recommendations...15 Appendix...-41

3 1 Introduction In this laboratory, a model of a sewage treatment plant filter washing station was controlled via computer to record data experimentally for the flow rate of water as different variables are modified. The time response of the output (flow rate) was measured when a step function input (increased motor %) was given to the system after a designated time interval had lapsed. Multiple experiments were run with a step-up function input and also a step-down function input. The system steady-state gain, response time and dead time will be observed and analyzed. The Background and Theory section will come after the Introduction followed by the Procedure taken during the laboratory. The Results and a Discussion of the Results will follow that. After evaluating the data, Conclusions and Recommendations will be given. In the Background and Theory section, the model used to evaluate the flow rate and the methods used to control the model will be explained. The Steady-State Operating Curve will be introduced in the Background section to establish what the steady-state resembles for the range of outputs the customer requires. The Procedure section will follow that and describe the user interface and how it was manipulated to change certain model variables during the lab. Data in the form of charts and tables will follow in the Results section. After interpreting the data, a discussion of the results and their meaning

4 2 will follow the Results section. Conclusions and Recommendations for the lab will be next and finally an Appendix with the majority of the data in the form of graphs.

5 3 Background and Theory There is a physical model of a sewage treatment facility s filter washing station in the laboratory. Using this physical model and the sensors on it, a suitable amount of data was collected and analyzed to develop a Steady State Operating Curve (SSOC) for the system which closely models the real-world example. That SSOC is given below in Figure SSOC Flow (lb/min) = (motor %) Output (lb/min) Operational Operation Input (%) Figure 1. SSOC (Steady-State Operating Curve) In Figure 1 above, the SSOC for the filter wash station was found experimentally. The SSOC also shows that the flow changes at 0.46 lb/min/% input increase as indicative

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7 5 by the slope of the curve. The curve has a linear fit and was found to have a standard deviation of 0.85 lb/min. This should assure that the step-response experiment will have a steady-state that follows the function. Using this knowledge a time will be chosen so that steady-state can be reached. Illustrated in Figure 2 below is a diagram of the model setup in the laboratory. Figure 2. Flow System Diagram In the physical model illustrated above, there is a water pump that forces water through three separate routes. The solenoid valves can be closed or open via a network connection to any internet accessible computer. In the step response experiment, timers

8 6 will be used to delay the stepping of the input until the desired time has passed. These are controlled via LabVIEW which is explained in the procedure section. Figure 3. Schematic of Filter Wash Station In the figure above a simple block diagram of the filter wash station is shown. The variable m(t) is the manipulated variable and is a function of time. The variable c(t) is the controlled variable which is also a function of time. This is the measured flow rate of water (lb/min) that the sensors on the laboratory model will collect.

9 7 Procedure During the lab, the motor input (%) will be manipulated by adjusting variables in the control program. A time is given to run and achieve steady state and after that time the step will occur (either up or down). A second time is then given to run after the step and achieve steady-state once again. The motor input (%) is controlled from the start and a step input (%) increase or decrease is specified. The range of output (lb/min) that the customer is looking for is 8- lb/min. This range will be split among three people to run experiments on 8- lb/min, - lb/min and - lb/min. This particular lab will be dealing with the range of - lb/min. For this range a motor input of % is initially chosen with a step-up size of 10%. Using the SSOC, it is found that % will give an output flow of.8±0.85 lb/min and a 10% step will result in 19.4±0.85 lb/min. This will include all of the range being looked at for this experiment. The timers, and step up input (%) is manipulated using a system called LabVIEW. LabVIEW stands for Laboratory Virtual Instrument Engineering Workstation. This program is used to change variables for experiments, acquire data, analyze data, and render the data in a format to be used by other programs.

10 8 Figure 4. Schematic of LabVIEW Information Flow In Figure 4 above, the layout of the flow of information utilizing LabVIEW is seen. The computer operator inputs values for differing variables into the program. LabVIEW then supplies outputs to different pieces of equipment. In the case of the filter wash station, the solenoid valves can be open or closed and put on a timer. After running the experiment, the equipment outputs data to LabVIEW which then outputs that data to the computer monitor and data files. From here the data can be sent to spreadsheet software such as Excel for analysis. This experiment involves finding the steady-state gain, dead time and time constant. A method called Fit 2 was used to approximate these values. An example of the Fit 2 method is given in the appendix.

11 9 Results The data from LabVIEW was forwarded to Excel for analysis. The system did not reach a steady-state until approximately 6 seconds into the experiment and so the data before that time was ignored. The data is accumulated in a graph made with Excel that shows vs. Motor Input % vs. Flow (rate in lb/min). In Figure 5 below, the response in output flow (lb/min) is shown for a step size of +10%. A time of seconds was chosen to initiate the step-up. This assured that the system would reach steady-state even for motor input (%) that had slower response times Dm = 10% Input, m(t) % K=0.46 lb/min/% 19.8 lb/min Gain=K=Dc/Dm=lb/min/% Dc= 4.6 lb/min 25 Output, c(t) (lb/min) 15.2 lb/min Figure 5. Step Response (Steady-State Gain)

12 10 The Dc is the amount that the flow increased over the step size. The value of 4.6 lb/min per 10% step size correlates with the SSOC curve and the 0.46 lb/min/% slope it had. The Gain, K = 0.46 lb/min/% matches the SSOC perfectly. Dead time = t0 24 t0=0.3sec Output c(t), (lb/min) Step at sec Figure 6. Step Response (dead time) In Figure 6 above, the same step-up is shown as in Figure 5, but is used this time to calculate the Dead time of the system. The Dead time is the time delay between when the signal to step-up is sent by the controller until the system responds with an increase ( or decrease ) in flow. Finally, as shown in Figure 7 below, the time constant is calculated for the step-up response.

13 11 Time constant=t Δc=4.6 lb/min 63.2%Δc= Output, c(t), (lb/min) t =0.5 sec Figure 7. Step Response (time constant) The step response variables were calculated for a step up of 10%. After this the step response variables for a step of -10%, or a step-down, were calculated. The step-up and step-down analysis were done for three ranges: 8- lb/min, - lb/min, and - lb/min. Three experiments were run for each of the ranges for a step-up and step-down. This data was put into excel to analyze. The average value for the gain over the range of output needed was lb/min/% for a step-up and lb/min/% for a step down. The average for the dead time was 0.37 sec for a step-up and 0.51 sec for a step down. The average Dc was 3.89 lb/min for the step up and 3.87 lb/min for the step down. The average c was 2.42 for a

14 step-up and 2.44 for a step-down. The time constant average for the step-up was 0.6 and the time constant for the step-down was lb/min - lb/min -lb/min 3 K = GAIN (lb/min/%) Upper Lower Upper Lower Upper Lower Figure 8. Comparison of Gain for different ranges In the figure above, the gain is compared between the step-up and stepdown for all three experiments in all three ranges. The variation is very minimal, with a slight increase in gain for the step-up of - lb/min range.

15 lb/min - lb/min -lb/min 0.7 T 0 -Dead Upper Lower Upper Lower Upper Lower Figure 9. Comparison of Dead Time for Different Ranges In Figure 9 above, the dead times are compared for the various ranges and multiple experiments run in those ranges. Among the three experiments the dead times are constant. Ƭ -Time Constant lb/min - lb/min -lb/min Upper Lower Upper Lower Upper Lower Figure 10. Chart of Time Constants for Step-Up / Step Down Range

16 The figure 10 above shows the different time constants calculated for the step-up and step down over the range of flows the customer was requesting.

17 15 Discussion of Results The analysis of data has concluded that the SSOC is a good model to follow in describing this system in the real world. Among the ranges studied, the gain, dead time and time constant were approximately the same. The dead-time did vary in the stepup versus step-down by 0. sec. The system seemed to react more quickly to an increase in motor power than to a decrease. This leads us to the conclusion that a FOPDT model can be created that would follow this system in the real-world. This would make it easier to predict and adjust for changes in the input to the system.

18 Conclusions and Recommendations The purpose of this lab was to take previously created SSOC curves and, knowing that the steady-state could be reached at a certain time and was unvarying, to step-up or down the input to the system to see how the system would respond. After reaching steady-state the system was manipulated by a % increase or decrease in motor power. The averaged values for the system Gain was lb/min. The average Dead Time was seconds and the average Time Constant was The objectives of the experiment have been accomplished.

19 17 Appendix Figure lb/min step up gain 1 Figure. 8 lb/min step up dead 1

20 Figure lb/min step up time 1 Figure. 8 lb/min step up gain 2

21 19 Figure lb/min step up dead 2 Figure. 8 lb/min step up time 2

22 Figure lb/min step up gain 3 Figure. 8 lb/min step up dead 3

23 21 Figure lb/min step up time 3 Figure. 8 lb/min step down gain 1

24 Figure lb/min step down dead 1 Figure. 8 lb/min step down time 1

25 23 Figure lb/min step down gain 2 Figure lb/min step down dead 2

26 24 Figure lb/min step down time 2 Figure lb/min step down gain 3

27 25 Figure lb/min step down dead 3 Figure lb/min step down time 3

28 26 Input, m(t) % Dm = 6% Gain=K=Dc/Dm=lb/min/% K=0.5 lb/min/% Dc= 3 lb/min Output, c(t) (lb/min) Figure lb/min step up gain 1 Dead time = t0 t0=0.4sec Step 37at sec Output c(t), (lb/min) Figure. - lb/min step up dead 1

29 27 Time constant=t Δc=3 lb/min 63.2%Δc= Output, c(t), (lb/min) t =0.4 sec Figure lb/min step up time 1 Input, m(t) % Dm = 6% Gain=K=Dc/Dm=lb/min/% K=0.541 lb/min/% Dc= 3.25 lb/min Output, c(t) (lb/min) Figure lb/min step up gain 2

30 28 Dead time = t0 t0=0.5sec Step 37at sec Output c(t), (lb/min) Figure lb/min step up dead 2 Time constant=t Δc=3.25 lb/min 63.2%Δc= Output, c(t), (lb/min) t =0.5 sec Figure lb/min step down time 2

31 29 Input, m(t) % Dm = 6% Gain=K=Dc/Dm=lb/min/% K=0.562 lb/min/% Dc= 3.36 lb/min Output, c(t) (lb/min) Figure lb/min step up gain 3 Dead time = t0 24 t0=0.4sec Output c(t), (lb/min) Step at sec Figure lb/min step up dead 3

32 Time constant=t Δc=3.36 lb/min 63.2%Δc=2. 24 Output, c(t), (lb/min) t =0.5 sec Figure37. - lb/min step up time 3 Input, m(t) % Dm = -7% Gain=K=Dc/Dm=lb/min/% K=0.444 lb/min/% Dc= 3.11 lb/min Output, c(t) (lb/min) Figure lb/min step down gain 1

33 31 Step at sec Dead time = t0 t0=0.4sec 24 Output c(t), (lb/min) Figure lb/min step down dead 1 Time constant=t Δc=3.11 lb/min 63.2%Δc= t =0.5 sec Output, c(t), (lb/min) Figure. - lb/min step down time 1

34 32 62 Dm = -7% 35 Gain=K=Dc/Dm=lb/min/% Input, m(t) % K=0.429 lb/min/% Dc= 3.00 lb/min Output, c(t) (lb/min) Figure lb/min step down gain 2 Step at sec Dead time = t0 t0=0.4sec 24 Output c(t), (lb/min) Figure lb/min step down dead 2

35 33 Time constant=t 24 Δc=3.00 lb/min 63.2%Δc=1.89 t =0.5 sec Output, c(t), (lb/min) Figure lb/min step down time 2 62 Dm = -7% 35 Input, m(t) % Gain=K=Dc/Dm=lb/min/% 25 K=0.447 lb/min/% Dc= 3.13 lb/min Output, c(t) (lb/min) Figure lb/min step down gain 2

36 34 Step at sec Dead time = t0 t0=0.4sec 24 Output c(t), (lb/min) Figure lb/min step down dead 2 Time constant=t Δc=3.13 lb/min 63.2%Δc=1.98 t =0.5 sec 24 Output, c(t), (lb/min) Figure lb/min step down time 2

37 35 Input, m(t) % Dm = 10% Gain=K=Dc/Dm=lb/min/% K=0.47 lb/min/% Dc= 4.7 lb/min Output, c(t) (lb/min) Figure lb/min step up gain 1 Dead time = t0 24 t0=0.5sec Output c(t), (lb/min) Step at sec Figure lb/min step up dead 1

38 36 Time constant=t Δc=4.7 lb/min 63.2%Δc= Output, c(t), (lb/min) t =0.5 sec Figure lb/min step up time 1 Input, m(t) % Dm = 10% Gain=K=Dc/Dm=lb/min/% K=0.45 lb/min/% Dc= 4.5 lb/min Output, c(t) (lb/min) Figure. - lb/min step up gain 2

39 37 Dead time = t0 t0=0.4sec Step at sec 24 Output c(t), (lb/min) Figure lb/min step up dead 2 Time constant=t Δc=4.5 lb/min 63.2%Δc= Output, c(t), (lb/min) t =0.5 sec Figure lb/min step up time 2

40 Dm = 10% Input, m(t) % K=0.46 lb/min/% Gain=K=Dc/Dm=lb/min/% Dc= 4.6 lb/min Output, c(t) (lb/min) Figure lb/min step down gain 1 Dead time = t0 24 t0=0.3sec Step at sec Output c(t), (lb/min) Figure lb/min step down dead 1

41 39 Time constant=t Δc=4.6 lb/min 63.2%Δc= t =0.5 sec Output, c(t), (lb/min) Figure lb/min step down time 1 Input, m(t) % Dm = 10% Gain=K=Dc/Dm=lb/min/% K=0.46 lb/min/% Dc= 4.6 lb/min Output, c(t) (lb/min) Figure lb/min step down gain 2

42 Dead time = t0 t0=0.6sec Step at sec Output c(t), (lb/min) Figure lb/min step down dead 2 Time constant=t Δc=4.6 lb/min 63.2%Δc= t =0.5 sec Output, c(t), (lb/min) Figure lb/min step down time 2

43 41 Input, m(t) % Dm = 10% Gain=K=Dc/Dm=lb/min/% K=0.48 lb/min/% Dc= 4.8 lb/min Output, c(t) (lb/min) Figure lb/min step down gain 3 Dead time = t0 24 t0=0.7sec Step at sec Output c(t), (lb/min) Figure. - lb/min step down dead 3

44 42 Time constant=t Δc=4.8 lb/min 63.2%Δc= t =0.5 sec Output, c(t), (lb/min) Figure lb/min step down time 3