Answer Posted / senthil ramanujam mohan
I can give you simple example:-
Consider a temperature controller(on/off valve),
setpoint - 100deg
Offset - (-4)
dead band - 2
ERROR - SETPOINT - ACTUAL VALUE
If temperature raises from 90deg>>>>94>>at >96degrees the
valve closes.this is becoz we have an OFFSET of (-4
degrees)(setpoint - offset value)
So,now the valve is closed.temperature drops down from
96degrees<<95<< at 94degrees the valve starts open.this is
becoz we have a DEADBAND of 2degrees.
Error(contrl valve):-
setpoint - 100degree and actual value - 90degrees.
temp range - ( 0 to 200)for error scaling(becoz
controller works from 0 to 100%)
error scaling = 100/200 - 0.2 which is multiplying
factor for error to the controller
Finally,
ERROR = [ SETPOINT - ACTUAL VALUE ] X ERROR SCALING.
[ 100 - 90 ] X 0.2
[ 10 ] X 0.2
2.
hope you guys understand.
| Is This Answer Correct ? | 4 Yes | 15 No |
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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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