Calculation of volumetric flow rate and mass flow rate
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The same question concerns differential pressure flowmeters; it asks for the mass flow rate and volume flow rate for each of them separately. The second question is as follows: The flowmeter in use on site is a orifice plate differential pressure flowmeter, and the range for this channel on the DCS display is 0 to 12 cubic meters per hour. When the gauge on the differential pressure transmitter shows 50%, what is the flow rate at that time in cubic meters per hour? The second question is: The flow meter in use on site is an orifice plate differential pressure type flow meter. On the DCS interface, the range for this channel is 0 to 12 kilograms per hour. When the gauge of the differential pressure transmitter shows 50%, what is the flow rate at that time in kilograms per hour? Excuse me, these two questions are basically the same; the only difference is that one uses volume flow rate as the unit while the other uses mass flow rate. How should they be calculated? What are the final numbers each person obtained? (The differential pressure range of the differential pressure transmitter is 0~5KPa)
I’m sorry, the range of the differential pressure transmitter is 0~15 KPa
This post was last edited by xxkhc on 2015-12-10 23:01; it equals 8.4852 per cubic hour and 8.4852 per kilogram hour. (The range of the differential pressure transmitter is 0–15 KPa; the flow rate ranges are: 0–12 cubic meters) ； 0-12 kilograms, at a differential pressure of 50%)
Taking the square root of the differential pressure gives the mass flow rate, while not taking the square root gives the volume flow rate. Is that correct?
Your understanding is incorrect. You are asking about differential pressure flow meters. Essentially, a differential pressure flow meter is an instrument used to measure volumetric flow rate; by taking into account the density of the fluid, it can be calibrated to measure weight flow rate (now referred to as mass flow rate). This differs from true mass flow meters in terms of their measurement principle. For differential pressure flow meters, the square of the flow rate is proportional to the differential pressure; that is, Q² = △P. This holds true for both volumetric flow rate and mass flow rate, and the relationship cannot be linear.
Taking the square root of the percentage value on the differential pressure transmitter and then multiplying it by the maximum flow rate gives its mass flow rate; directly multiplying the percentage value on the transmitter by the maximum flow rate gives its volume flow rate. Is this understanding correct?
When 50% is displayed, if the square root calculation is performed inside the transmitter, the values are 6 cubic meters per hour and 6 kilograms per hour; if the square root calculation is done within the system, the values are 8.4852 cubic meters per hour and 8.4852 kilograms per hour
By installing a yield meter, this problem becomes simple
I basically agree with the algorithm used on floor 8; P is displayed as 50%. When performing square root calculations inside the transmitter, smart transmitters are commonly used these days, and many of them allow for separate setting of the transmitter’s output and the display mode on the gauge: 1. When both the transmitter’s output and the gauge’s display mode are set to the square root mode, the values are 6 cubic meters per hour and 6 kilograms per hour respectively； 2. When the transmitter’s output is not in line with the display mode of the gauge – that is, when the transmitter outputs values in square root form while the gauge displays them as percentage values – it is consistent with the square root calculation performed within the DCS system; in other words, the values are 8.4852 cubic meters per hour and 8.4852 kilograms per hour respectively.