Correlation between the 4-20mA signal output after squaring on the orifice flow meter transmitter
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This post was last edited by chen5714355 on 2018-1-17 at 10:01. Question: After taking the square root on the transmitter of a orifice flow meter, does the 4-20mA signal output correspond to the flow range or the pressure difference range? How to calculate it on the PLC? Calculation sheet for orifice plate
This post was last edited by LL(long) on 2018-1-17 at 23:42. I’m very sorry if it appears that 50% is shown at the field level while the DCS displays 75% of the full scale; I made a mistake in my previous statement. It’s actually a matter of differential pressure – when no square root is applied, if the measurement range is 0-25000 Pa and the DCS range is 0-50 t/h, then if the DCS shows 25 t/h, the differential pressure calculation without taking the square root is: 25000 * (25/50)² = 6250 Pa. In this case, the transmitter’s output current is: 6250/25000*16+4 = 8 mA. When a square root is applied, the differential pressure becomes: 25000 * (25/50) = 12500 Pa, and the transmitter’s output current in this case is: 12500/25000*16+4 = 12 mA.
After taking the square root, 4-20mA corresponds to the flow rate range
Nonsense! Can you manage to write an equation? Let’s see, Teacher Dog will show you how to calculate it. . . .
The orifice flow meter has a flow rate that is proportional to the square root of the pressure difference; the formula is Q=k√△p. The range corresponding to 4–20 mA is not linear. According to your calculations, △P is 0–4 kPa. The flow rate range is 0–180; at 4 mA, the corresponding flow rate is 0. At 8 mA, it is 90; at 16 mA, it is approximately 0.7 * 180. At 20 mA, it is approximately 0.865 * 180, which equals 180. After taking the square root of the value from the transmitter, further square rooting cannot be done within the DCS – only one square root operation is allowed, either on the transmitter side or on the DCS side.
This post was last edited by 1111111 on 2018-1-17 at 15:14. I know that the maximum flow rate is 180 cubic meters per hour, and I also know that the differential pressure gauge outputs a signal in the range of 4–20 mA after being squared. Damn it, that means it’s a purely linear relationship between 0–180 and 4–20 mA – everything else mentioned in the calculation sheets can be ignored completely. :lol Who can prove that what I said is wrong? I’m willing to transfer 200 yuan via Alipay as an apology. . .
This post was last edited by jujiangliu on 2018-1-17 20:11: (Q/Qmax) = square root of (△p/△pmax) = square root of. It should be expressed as a percentage.
If the header is squared, the output ratio will correspond to the traffic ratio
Have you eaten shit, you idiot?
Gagaga, before reason, everyone is equal. You’re wrong; you should eat shit, a bowl full. I’m wrong too; I’ll eat shit, a bowl full as well. Black
Ugh, I said that the value 2 on the 2nd floor is missing, and you once again apologize for saying the wrong thing; you also modify the post with long explanations and examples. The pressure difference value when calculating should be “=12500 pa”, but you got that wrong too! Go and revise the post again; if you can get it right, that’s fine. If you can figure out how to calculate 12mA, then I agree with you. . Hehehe. Before reason, everyone is equal.