Process Engineering Division – Instrumentation and Automation Section – Daily Topics – Topic No. 2020-12-17
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Process Engineering Section – Instrumentation and Automation Section – Daily Topics – Topic No. 2020-12-17: Discussion question: 329. When using a differential pressure transmitter to measure flow rate, the upper limit of the differential pressure is 25 Kpa; the range of the secondary meter is 0–200 T/h, and the output current of the differential pressure transmitter is 4–20 mA. What is the corresponding output current value when the flow rate is 80 T/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.)
Based on the fact that differential pressure is proportional to flow rate to the power of 1/2, we have (80/200)2 = (I-4)/(20-4); thus, I = 6.56 mA
80/200*16+4=10.4ma
80/200*16+4=10.4ma
Flow rate is measured using a differential pressure transmitter; the maximum differential pressure is 25 Kpa. The range of the secondary meter is 0–200 T/h, and the output current of the differential pressure transmitter is 4–20 mA. What is the corresponding output current value when the flow rate is 80 T/h? Solution: From the measurement of the differential pressure by the transmitter to the display of the flow rate value on the secondary meter, the entire process is as follows: the transmitter measures the differential pressure ΔP, which is then converted into a current I after taking the square root; this current I is sent to the secondary meter, where it is used to display the flow rate value Q. In other words, there is a linear relationship between I and Q. It is known that the measurement range of the transmitter is ΔPmax = 25 kPa, while the display range of the secondary gauge is Qmax = 200 t/h. It is necessary to calculate the differential pressure value ΔP measured by the transmitter and the output current value I of the transmitter when the secondary gauge shows Q = 80 t/h. There is a linear relationship between I and Q; therefore, (I–I0)/(Imax–I0) = Q/Qmax, where I0 is the current value corresponding to the zero point. Thus, I = Q/Qmax * (Imax–I0) + I0. Substituting the numerical values gives: the output current value of the transmitter is I = (80/200) * (20–4) + 4 = 10.4 (mA). Answer: The output current value corresponding to a flow rate of 80 T/h is 10.4 mA.
The transmitter’s output current can be expressed with a square root. For the case where there is a square root, the output current is: (80/200) * (20-4) + 4, which gives I = 10.4 mA. In the case where no square root is used, it is: (80/200)² = (I-4) / (20-4), resulting in I = 6.56 mA
△px/△p=（Fmax/F）2 △px=△p（Fmax/F）2=2500（80/200）2=4000Pa I=（Fmax/F）×16+4=6.56mA
The transmitter’s output current can be expressed with a square root. For the case where there is a square root, the output current is: (80/200) * (20-4) + 4, which gives I = 10.4 mA. In the case where no square root is used, it is: (80/200)² = (I-4) / (20-4), resulting in I = 6.56 mA
The current value I = (80 / 200) × (20 – 4) + 4 = 10.4 (mA)