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A temperature sensor (RTD) connected via a temperature transmitter is adjusted to give an output of 4ma at 60 deg and 20 ma at 600 deg.
This output (4-20ma) from the transmitter is connected to an IGBT-inverter to control a high speed (-12,000 to +12,000 RPM) DC motor.
And in the inverter it is programmed as, for 4ma the motor speed will be -3,000 RPM and for 20ma the motor speed will be +10,000 RPM.
The question is : If the motor speed is +5,500 (RPM), what will be the temperature?
(-sign indicates that the motion is in reverse direction).
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Answer Posted / srikanth
At -3000 rpm temp=60 ma=4
At 0 rpm temp=60+124.6=184.6 ma=9.46
At 5500 rpm temp= 60+353=413
ma=14.45
At 10000 rpm temp=60+540=600
ma=20
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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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