Differential pressure flow meter

The flow rate displayed after taking the square root of the differential pressure signal is smaller than that displayed without taking the square root...

One should be able to answer this question by junior high school; I don’t know what to say.

The square root algorithm for differential pressure transmitters involves taking the square root of the percentage of the measured value relative to the full scale; of course, taking the square root of a value less than 1 results in an increased value. For example: if a differential pressure transmitter has a range of 0–10 KPa, and the measured value is 5 KPa without taking the square root, this corresponds to 50% of the range, or 0.5. Taking the square root of 0.5 gives 0.7071, which is 70.71%. Using this percentage, the output current becomes 16*70.71% + 4 = 15.31 mA, which is naturally higher than the 12 mA current obtained when no square root is taken. As a supplementary note: when using a differential pressure transmitter to measure flow rate, it is necessary to take the square root, as the flow rate is proportional to the square root of the pressure difference. Someone upstairs said this topic is covered in junior high school; which junior high school is advanced enough to cover topics related to instrumentation?

The value increases when taking the square root, and students in the first year of middle school already know this.

Okay. I understand. Thank you

The question posed by the original poster is: does the flow rate increase after taking the square root of a differential pressure flow meter reading? If it doesn’t take the square root, the flow rate is incorrect. That is just a numerical change; essentially, a differential pressure flow meter requires taking a square root.