fira_9

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SUMMARY The objective of this experiment was to control the level water in a tank automatically by using a feedback loop with a PI controller. In this experiment, pneumatic controlled system (compressed air) was used to control the valve. In this experiment an ‘air-to-open’ valve was used as the final control element. First the set point of this experiment was to a certain step (30 to 31) and the response of the process was observed by using a computerized control interface. All the control parameters, set point, gain etc. were directly measured by the response of the process. The software has the provision for saving the course of control action and the necessary plots.

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fira_9

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SUMMARY

The objective of this experiment was to control the level water in a tank automatically by using a

feedback loop with a PI controller. In this experiment, pneumatic controlled system (compressed

air) was used to control the valve. In this experiment an ‘air-to-open’ valve was used as the final

control element. First the set point of this experiment was to a certain step (30 to 31) and the

response of the process was observed by using a computerized control interface. All the control

parameters, set point, gain etc. were directly measured by the response of the process. The

software has the provision for saving the course of control action and the necessary plots.

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Observed Data

Fig.01:- Response of the process as step changes

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For step up process

Change in input = 16-15.84 = 0.16 mA

Change in output = 69-57.73 = 11.27 unit length

So, steady state gain, K = change∈outputchange∈input =

11.270.16 = 70.4375 unit length/mA

So, time delay, 1 =1 s

Step input is given at 1739 sec

Time at 63.2% response, = (1977-1739)-1 sec = 237 sec

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Calculated Data

Using Internal Model Control (IMC) Method,

1. Proportional-Integral (PI) controller

τ c = 10sec

Kc = 0.3048

τI= τ = 237sec

2. Proportional-Integral-Derivative (PID) controller

Kc= 0.3211

τI= 237.5 sec

τD = 0.4989 sec

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SAMPLE CALCULATION

From Observed data, we get

Time at 63.2% response,τ = 237 sec

Time delay, θ= 1 sec

We assume, τ c = 10 sec

For PI controller,

Kc= 1K× T

Tc+ θ2

= 1

70.4375× 237

1+10

=0.3058 unit height/m amp

τ i= τ= 237 sec

For PID controller,

Kc= 1K ×

τ+θ2

τ c+θ2

= 1

70.4375×237+0.510+0.5 = 0.3211 unit height/m amp

τI= τ+θ2 =237+0.5 = 237.5 sec

τD = τθ

2 τ+θ = 237×7.52×237+1= 0.4989 sec

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Open loop simulation

Fig.02:- Simulation for open loop configuration

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Fig.03:- Response of open loop simulation (as step changes from 30 to 31)

Closed loop simulation

Fig.04:- Simulation for closed loop configuration

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Response from process with a PI controller before tuning with design value of kc = 3.26

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Fig.05:- Response of PI controller at Kc= 3.26(due to 1 step change)

Response from process with a PI controller after tuning at kc = 30

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Fig.06:- Response of PI controller after tuning at Kc = 30 (due to 1 step change)

Discussion

From the graphical representation of the process response to change in set point was observed

(fig.01) and controller parameter (K, τ) was determined. From the response of open loop

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simulation of the process controller design parameter (Kc τI) was calculated. In Fig.03, response of

open loop simulation (as step changes from 30 to 31) was shown.

In fig.05, the response of PI controller with (Kc =3.26) was shown. Curve shows fluctuation and

that was tuned with Kc = 30 (fig.06).

The simulation and tuning of original process in PI controller shows that the process will response

properly (without fluctuation) at a value (Kc) greater than that of design value.