Methods for optimizing the efficiency of floating tanks using water!
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Floats for optimizing water use. When the material density is less than that of water, this works well. But what if the density of the material is greater than that of water? For example, if the float is 1 meter high and its density is 1.2, then water needs to be filled to a height of 1.2 meters in order to generate a current of 20mA. Seek help from experts to solve it……
It is recommended to change the medium density measured by the float to that of water for calibration; once the calibration is complete, it can be changed back again.
If you only want to calibrate the linear accuracy of the float, you can do so based on the density of water. Then change it back if you want to use it after making corrections. Then it can be done using a training school and the weighing method. Or calibrate directly with the medium under test. If calibrated directly with water, some buoys won’t work. Some floats only need to be calibrated by 50% to be usable; those should be able to be calibrated using water.
Question: The float is 1 meter high with a density of 1.2; water needs to reach a height of 1.2 for 20mA to be outputted. Seeking help from experts… Pig’s answer: Live pigs shouldn’t be allowed to die from holding in urine, nor should they drink urine; therefore, the water level is set at 0.9 meters, with a current of 16mA. Answer completed
To calibrate over the full range, only heavy media can be used; water will not work
When the density of the medium under test is greater than that of water, any point within the range can be selected as the setting point for full scale (upper limit). Before debugging, the water level corresponding to that point and the current value corresponding to that point within the range should be calculated first. For example: if the range is 1500 mm and the density of the medium being measured is 1.1 g/cm3, and 1300 mm is chosen as the setting point for full scale (the upper limit), then the corresponding water level height should be: h = 1300 × 1.1 = 1430 mm. The current corresponding to this point should be: I = 4 + 1300/1500 × 16 = 17.87 mA. After the calculations are completed, the adjustment procedure is as follows: (1) Zero adjustment – Drain the water from the float chamber and adjust the zero point so that the output is 4 mA. (2) Full-scale adjustment For full-scale adjustment, the range potentiometer is adjusted at a water level of 1430 mm so that the output is 17.87 mA; this process is repeated several times until the desired result is achieved.