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Answer Posted / naseer jarral
i/p recive the signel 4 to 20 ma and calibration time when
we give 4 ma it gives penumatic out put valves position
that time is moniterd by by positioner it is 0% or 100% and
when we give 12ma valves stem position that time 50% on
12ma positioner will recive the current sigenel and campare
it with valve stem if the valve stem position is not same
it will produse penumatic out put on same 12 ma more till
it will not match with the position of valve stem when the
on 12ma valve position is 50% then this will close the
penumatic out put and close the nozel with fleper and block
the air till current signel changes(.... it compare the
current signal with valve position through valve stem it
genrate penumatic out put till differenc between current
signel and valve position.....)
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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 =
