OPTIMUM PID CALCULATION FOR SELF REGULATING (SR) PROCESS
based on Zieglar-Nichols formulation

1. Example using data captured from a paperless recorder of a flow increasing process

1. Td = 1.8 s
2. Process rate = (75 - 60) / (6 - 2.5) = 4.286 % / s
3. RR = process rate / controller output = 4.286 / (58.1 - 40) = 0.2368 / s
4. Flow process ==> use PI
5. P = 111.1 x 0.2368 x 1.8 = 47.4%
    I  = 3.33 x 1.8 = 6 s

3. Example using data captured from a chart recorder of a flow increasing process

1. Td = 1.8 s (Notice that Td is unmeasureable. Since I and D require Td, then estimate Td.)
2. RR = 0.231 / s (Notice that the chart speed is in mm/hr. Convert the speed into s/mm. Then, the time is
    simply the length in mm x the chart speed in s/mm.)
3. Mode = PI: Therefore, P = 111.1 x 1.8 x 0.231 = 46.2% ,  I = 3.33 x 1.8 = 6 s   (Notice that the time unit
    for Td must be similar with RR. Else, the time unit does not cancel with each other.)
 

TEST YOURSELF: Can you calculate the optimum PI for this process.