Instrument calculation problems
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The liquid level at the interface is measured using a differential pressure level gauge. At the lowest liquid level, the pressure difference between the positive and negative chambers of the gauge is -2796 Pa, while the gauge’s range is 5884 Pa. Therefore, when the gauge outputs a signal of 12 mA, what is the differential pressure signal received by the gauge in Pa? Please provide the calculation process, thank you
The lower limit of the gauge is -2796 Pa, and its range is 5884 Pa; therefore, the upper limit of the gauge is 3088 Pa. When an output of 12 mA is achieved, this corresponds to half of the range, that is, (3088 – 2796)/2 = 146 Pa
This post was last edited by jcicwht on 2016-10-20 at 10:10. It should be 146 Pa. (12-4)/(20-4)=x/5884; thus, x=2942 Pa. -2796+2942=146 Pa
5884/2=2942. 2942-2796=146pa.
When 12mA is output, it corresponds to half of the range, that is, 5884/2=2942Pa； The lower limit of the gauge is -2796 Pa; therefore, the differential pressure signal of 12 mA output by the gauge corresponds to 2942 – 2796 = 146 Pa
When the transmitter outputs 12mA, the input differential pressure is 50%; the zero-point drift is -2796 Pa. The transmitter’s range is from -2796 to 3088 Pa. Therefore, the pressure value at 50% is = 50% * 5884 – 2796 = 146 Pa
The question is not precise; when the liquid level is at its lowest, it is not necessarily the lower limit of the range.