Process Engineering Division – Instrumentation and Automation Section – Daily Topics – Topic No. 2020-12-10
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Process Engineering Division – Instrumentation and Automation Section – Daily Topics – Topic No. 2020-12-10: Discussion question: 322. When using a type III differential pressure transmitter to measure flow rate, with a flow range of 0–16 m3/h, what is the output signal when the flow rate is 12 m3/h? (5 points are given only to those who provide the correct answer; additional points are awarded for a correct calculation process or detailed explanation.)
(12/16)2=(I-4)/(20-4), I=13mA
The differential pressure is proportional to the square of the flow rate, and the output current of the transmitter is proportional to this differential pressure; therefore, the 4-20mA current output by the transmitter is proportional to the square of the flow rate. That is: (I-4) / (20-4) = Q^2/Qm^2 = 12^2/16^2; therefore I = 4 + 13 mA. Answer: The output signal is 13 mA
Let the output current be I; then (I-4) / (20-4) = 122 / (16-0)^2 = 13 mA. The output signal is 13 mA.
The differential pressure is proportional to the square of the flow rate, and the output current of the transmitter is proportional to this differential pressure. Therefore, the 4-20mA current output by the transmitter is proportional to the square of its flow rate. That is: (I-4) / (20-4) = Q^2/Qm^2 = 12^2/16^2; therefore I = 4 + 13 mA = 13 mA
4-20mA corresponds to 0-16 m3/h; 12/16*（20-4）+4=16mA
(12/16)2=(I-4)/(20-4), I=13mA
Let the output current be I; then (I-4) / (20-4) = 122 / (16-0)^2 = 13 mA. The output signal is 13 mA.
A type III differential pressure transmitter is used to measure flow rate, with a flow range of 0–16 m3/h. When the flow rate is 12 m3/h, let the output signal be I. Since the differential pressure is proportional to the square of the flow rate, the output current of the transmitter is proportional to this differential pressure. Therefore, the 4-20mA current output by the transmitter is proportional to the square of its flow rate. That is: (I-4) / (20-4) = Q^2/Qm^2 = 12^2/16^2; therefore I = 4 + = 13 mA. The output signal is 13 mA.