Answer / sachin

4 to 20 mA

Answer / amsal

as sachin wrote 4-20mA
but it is only output of coductivity xmtr.
during calibration of conductivity xmtr its I/P is
resistance (according to its range) which is given by
resistance box.
for example Range is 0- 10 uSIMENS
display------ I/P ---------------------- O/P
0 uSIMENS.= maximum R means infinity = 4mA
2.5 uSIMENS.= 750 Ohms = 8mA
5.0 uSIMENS.= 500 Ohms = 12mA
7.5 uSIMENS.= 250 Ohms = 16mA
10 uSIMENS.= 0 Ohms = 20mA
it is for example only but in practical there is non linear
i/p resistance which will be required to get o/p according
to range
and O/P is 4-20mAs which is shown in the unit of uSIMENS.

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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