Sample below: with the sample above, please answer question #2 for the graph images below. QUESTION: #2. Compare all fou

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Sample below: with the sample above, please answer question #2 for the graph images below. QUESTION: #2. Compare all fou

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Sample below:
Sample Below With The Sample Above Please Answer Question 2 For The Graph Images Below Question 2 Compare All Fou 1
Sample Below With The Sample Above Please Answer Question 2 For The Graph Images Below Question 2 Compare All Fou 1 (51.96 KiB) Viewed 42 times
with the sample above, please answer question #2 for the graphimages below.
QUESTION: #2. Compare all four cases. Which combination ofcontroller parameters gives the most optimal response? Explain why?(images below in inverted white)
Sample Below With The Sample Above Please Answer Question 2 For The Graph Images Below Question 2 Compare All Fou 2
Sample Below With The Sample Above Please Answer Question 2 For The Graph Images Below Question 2 Compare All Fou 2 (28.59 KiB) Viewed 42 times
Sample Below With The Sample Above Please Answer Question 2 For The Graph Images Below Question 2 Compare All Fou 3
Sample Below With The Sample Above Please Answer Question 2 For The Graph Images Below Question 2 Compare All Fou 3 (56.48 KiB) Viewed 42 times
Sample Below With The Sample Above Please Answer Question 2 For The Graph Images Below Question 2 Compare All Fou 4
Sample Below With The Sample Above Please Answer Question 2 For The Graph Images Below Question 2 Compare All Fou 4 (27.69 KiB) Viewed 42 times
The Controller Reset and Integral Action Offset elimination strategy is called "integral" or "reset control action". Set Point and Level and Controller Output Error constant Proportional control action Integral time Set Point Level Controller Output Proportional control action repeated in integral time Time 1. Set the Gain = 3; Ti = 2000; and Td = 0. Qin = 400 gpm. Adjust the Set Point to 30 inches. Increase the Load flow to 600 gpm and transfer to Auto. Watch the proportional only action with the Level and Set Point Chart. 2. Set the Gain = 3; Ti = 60; and Td = 0. Qin = 400 gpm. Adjust the Set Point to 30 inches. Increase the Load flow to 600 gpm and transfer to Auto. Watch the action with the Level and Set Point Chart.
2. Set the Gain = 3; Ti = 30; and Td = 0. Qin = 400 gpm. Adjust the Set Point to 30 inches. Increase the Load flow to 600 gpm and transfer to Auto. Watch the action with the Level and Set Point Chart. 4. Set the Gain = 3; Ti = 10; and Td = 0. Qin = 400 gpm. Adjust the Set Point to 30 inches. Increase the Load flow to 600 gpm and transfer to Auto. Watch the action with the Level and Set Point Chart. Lab Report to Include: 1. Print Screen of each of steps 1-4 2. Compare all four cases. Which combination of controller parameters gives the most optimal response? Explain why? There are 4 cases with different value of control parameters each case has proportional gain value as 3, and in each case integral time varies Let us understand the effect of integral time on control action, Increasing the integral time, reduced the steady state error, but increase the oscillations, and so when the integral time is low, there would be little oscillations but there would exist steady state error Case 1: Proportional gain = 3, integral time Ti = 2000 Here steady state error is considerable. Since Ti is very high, and this output is not as expected. Although there are no oscillations Case 2: Proportional gain =3. Integral time Ti = 60 The steady state error is nil and also there are very few oscillations, this is one of the candidates for optimum response Case 3: Proportional gain= 3, integral time 30 The steady state error is again zero, and there are slightly more oscillations that in case 2. but the overshoots are very low, compared to case 2, the output reaches steady state earlier Case 4: Proportional gain = 3, integral time 10 sec Due to this very high Ti, the oscillations are large, and so much so that the oscillations sustain with no clear steady state. (Far away from optimum response) With in Case 3 since oscillations are low, steady state error is 0, and steady state is achieved fastest, the optimum control action is achieved by Ti-30 and proportional gain =3 so in case 3 optimum response is achieved
DOSBox 0.74, Cpu speed: 3000 cycles, Frameskip 0, Program: TANKSIM Qin 600 gpm Press key (0X(to clear),(1), (2), (3) or (4) Level and Set Point 100- 50 10 Second Intervals 50 11.6 psi t=143 (LT) LIC 1:Set Point 4:Load Flow 8: Chart 9: Fault 10:Quit DOSBox 0.74, Cpu speed: 3000 cycles, Frameskip 0, Program: TANKSIM Qin 600 gpm Level and Set Point 100- 10 Second Intervals Qout 600 gpm 11.9 psi Qout 604 gpm COOL t=153 (LTLIC 1:Set Point 4:Load Flow 8: Chart 9: Fault 10: Quit a COOL TI CONTROLLER I 100 80 60 40 20 0 DO Press key (0)(to clear),(1), (2), (3) or (4) 100 80 60 40 20 0 GAIN 3 O DO Ti 2000 Td 0 MAN AUTO CONTROLLER Tanksim1E GAIN EV Ti 60 HO LE Td X 0 O X MAN AUTO Tanksim1E
DOSBox 0.74, Cpu speed: 3000 cycles, Frameskip 0, Program: TANKSIM Qin 600 gpm Press key (0)(to clear),(1),(2),(3) or (4) Level and Set Point 100- 50 10 Second Intervals LO t= 166 1:Set Point 4:Load Flow 8: Chart 9:Fault 10:Quit 50 (LT) LIC) 11.9 psi DOSBox 0.74, Cpu speed: 3000 cycles, Frameskip 0, Program: TANKSIM Qin 600 gpm Level and Set Point 100 10 Second Intervals Qout 598 gpm 8.8 psi Qout 392 gpm COO PY t= 152 (LTLIC 1:Set Point 4:Load Flow 8: Chart 9: Fault 10:Quit COOL CONTROLLER PX 100 80 60 40 20 0 DO 100 80 60 40 20 0 GAIN Press key to clear), (1), (2), (3) or (4) Ti DO WH 30 HO Td 0 MAN AUTO CONTROLLER Tanksim1E 40 GAIN 3 Ti EL 10 OF FO Td X 0 MAN AUTO Tanksim1E
DOSBox 0.74, Cpu speed: 3000 cycles, Frameskip 0, Program: TANKSIM Qin 688 9pm Level and Set Point 100- 50 10 Second Intervals Press key to clear), (1), (2), (3) or (4) 11.6 psi Qout 600 gpm t 188 LTLIC 1:Set Point 4:Load Flow 8: Chart 9: Fault 10: Quit coa ax CONTROLLER 100 80 60 20 DO GAIN Ti 2000 Pica MAN AUTO O Tanksim1E
DOSBox 0.74, Cpu speed: 3000 cycles, Frameskip 0, Program: TANKSIM Qin 688 gpm Press key (0)(to clear),(1), (2), (3) or (4) Level and Set Point 100- 50 10 Second Intervals. Qout 604 gpm t= 145 (LTLIC 1:Set Point 4:Load Flow 8: Chart 9: Fault 10:Quit DOSBox 0.74, Cpu speed: 3000 cycles, Frameskip 0, Program: TANKSIM Qin 688 9pm Level and Set Point 100- 50 11.9 psi 10 Second Intervals 11.9 psi Qout 599 gpm CONTROLLER t: 138 1:Set Point 4:Load Flow 8: Chart 9: Fault 10:Quit 100 80 60 40 20 Da 100 80 60 40 20 0 GAIN 3 76 OG MAN Press key to clear), (1), (2), (3) or <4> LO HO CONTROLLER Tanksim1E 73 AUTO GAIN 110 T 71 MAN AUTO Tanksim1E
DOSBox 0.74, Cpu speed: 3000 cycles, Frameskip 0, Program: TANKSIM Qin 600 gpm Level and Set Point 100- 50 Press key to clear), <1), (2), (3) or (4) LT-LIC 8.7 psi 10 Second Intervals Qout 386 9pm LO t= 150 1:Set Point 4:Load Flow 8: Chart 9:Fault 10:Quit coa CONTROLLER 100 80 60 40 20 0 GAIN 118 PO MAN AUTO Tanksim1E
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