how to define the P,I,D value in tuning pid loops in online.
Answer / morsli
HI ,you have to do on first time the derivative action Td at 0 value to eliminanate a derivative action Td =0 and the integral action at maximun means 1/ti=1/maximun =0 and you have to adjust the propotional band to minimun value till you have cycling ,and if u have cycling you adjust the proportional band to twice of cycling value ,you fix the of the proportional band then you adjust integral band same as the proportional ,you decrease the reset value till you have cycling and then you multiply this value on twice ,and you set this integral operational value
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Q1: Consider part of a control loop, which excludes the transmitter, consisting of a process, a controller and a control valve which may be represented by two dead times of 0.5 min each and three exponential lags of 0.8 min., 1.0 min. and 1.5 min. respectively. We wish to express this system as an overall first order plus dead time (FOPLD) model ie gain, time constant and process dead time. (We will see later that this is often done, to simplify controller tuning). For this exercise, gain is considered to be 1.0. (A) If the transmitter is a flow transmitter whose behaviour can be described by a dead time of 0.2 min. and an exponential lag of 0.5 min. in terms of the overall dead time and overall first order lag how can the system behaviour be approximated ? Overall dead time = Overall time constant = (B) If the transmitter is a temperature transmitter with a temperature sensor in a protecting well whose behaviour can be described by a dead time of 0.7 min. and an exponential lag of 15 min. how can the overall system behaviour be approximated now? Overall dead time = Overall time constant =
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