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**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.**

16 October 99