how to calculate interface capelary level transmetar lower
S.G.0.54 and uper S.G. 0.26 and h=490 mm
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Answer / ashish namdev, bina refineries
Your Que have some mistakes in SG Data
After removed mistakes you wll got
URV=264.6
LRV=127.4
| Is This Answer Correct ? | 1 Yes | 1 No |
Answer / narayan
TAKE THE DIFFERENCE IN SG
0.54-0.26, YOU GET 0.28
590 * 0.26=127.4 MM
HENCE LRV =0MM
URV=127.4MM
| Is This Answer Correct ? | 0 Yes | 2 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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