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An 1151 pressure transmitter has a range of 0–10 kPa abs. It is calibrated using a standard pressure gauge with a range of 0–100 kPa abs. If the local atmospheric pressure is 98 kPa abs, then when the transmitter outputs 12 mA, the corresponding input pressure value is 7 kPa abs. Please explain this. If you can’t figure out these relationships, give some guidance
12 ma corresponds to 5 kPa abs; it’s possible that your gauge is inaccurate
Xiaoxinxin, what does ABS mean? What do you need to do? For a differential pressure gauge, the low-pressure side should be connected to the ambient atmosphere; there’s no need to worry about the value of the ambient atmospheric pressure. Both the high-pressure side of the differential pressure gauge and the standard gauge should be connected to the input pressure, and then the gauge can be calibrated using gauge pressure. The range of the standard gauge you use should be much larger than that of the differential pressure gauge – this isn’t appropriate. You need to use a standard gauge with a smaller range, such as 10 kPa or 30 kPa. Otherwise, the accuracy of the calibration will be very poor
Enter 0 kPa for a calibration of 4 mA, and enter 10 kPa for a calibration of 20 mA. After adjusting these two values accurately, it may be necessary to make several adjustments before achieving the correct setting. Once that is done, observe the current output at characteristic pressures such as 2.5 kPa, 5 kPa, and 7.5 kPa
Just because it can’t be done doesn’t mean the standard table is inaccurate. Since it is about calibrating the transmitter, the standard gauge should be used as a reference (once this standard gauge is chosen, there is no need to consider whether it is accurate or not). It could also be due to an error in selecting the standard table; if the standard table shows 7 kPa, it means that a pressure signal of 7 kPa is being sent to the transmitter, and if the transmitter outputs 12 mA, then it indicates that the transmitter is inaccurate and needs to be adjusted.
This shows the relationship between absolute pressure and gauge pressure; absolute pressure is equal to atmospheric pressure plus gauge pressure. Since the local atmospheric pressure is below 100 kilopascals, a compensation of 2 kilopascals is needed. 5 + 2 = 7, which corresponds to 12 milliamps