Question

Response with lines drawn is shown as the second graph, both graphs are the exact same.

ANSWER PART (a)

open-loop transient response A PID controller for a process is to be tuned by carrying out an test. The system control temperature varies from 25 °C to 300 °C with a 200 °C setpoint. The output of the PID controller is a heater control voltage ranging from 0 VAC to 240 VAC. The test is started by having a 150 VAC heater control voltage output. At t = 0 the heater control voltage output is then suddenly increased to 170 VAC. The resulting temperature graph is shown below Temperature 260 255 250 245 240 235 230 225 220 0 6 10 11 12 13 14 4 7 3 1 215 t (minutes) Temperature 260 255 250 245 240 AT-30C 235 230 225 220 6 10 11 12 13 14 4 5 7 1 215 t (minutes) L-2.5 W-12-2,5-9,5 minutes minutes (a) Calculate the Zeigler-Nichols settings for the PID controller using the open loop method 15 marks]

Answer 1

We know ,

Kp = 1.2 × (W/L)

Kp = 1.2 × (9.5/2.5)

Kp = 1.2 × 3.8

Kp = 4.56

now ,

Ti = 2 × L

Ti = 2 × 2.5

Ti = 5 minutes

and ,

Td = 0.5 × L

Td = 0.5 × 2.5

Td = 1.25 minutes