Differential Pressure Transmitter

Govind Rao

Member
Join Date
Sep 2017
Location
Faridabad
Posts
5
Dear Sirs,
I have just joined the forum and honestly know nothing about PLC though we are using S7-1200 and is being operated by a third agency.
My question is the input from the DPT to measure the differential pressure of air flow through the air duct is fed into the PLC and calculated value
shown on the SCADA.
The maximum airflow in the duct is 500Nm3 per hour and the minimum is 0.
The question is, Do we input 0 as 4mA and 20mA as 500 Nm3. Or do we log in Square Root function and get the real time value on the screen
 
The question is, Do we input 0 as 4mA and 20mA as 500 Nm3. Or do we log in Square Root function and get the real time value on the screen
The answer is, we can't tell you because the flow rate is the result of a square root function which can be done in either the DP transmitter or the HMI.

If the square root function is done in the transmitter, then the 4mA signal is scaled as you state: 4mA = 0 Nm3, 20mA = 500Nm3.

If the square root is not done in the transmitter, but the transmitter output is scaled by the HMI as 4mA = 0 Nm3, 20mA = 500Nm3, then the flow rate is in error, except at two points, 0 and full scale.

If the square root function is not done in the transmitter, then the HMI has to take the square root of 4mA = 0%, 20mA = 100% and then scale the result as 0 to 500Nm3.
 
The answer is, we can't tell you because the flow rate is the result of a square root function which can be done in either the DP transmitter or the HMI.

If the square root function is done in the transmitter, then the 4mA signal is scaled as you state: 4mA = 0 Nm3, 20mA = 500Nm3.

If the square root is not done in the transmitter, but the transmitter output is scaled by the HMI as 4mA = 0 Nm3, 20mA = 500Nm3, then the flow rate is in error, except at two points, 0 and full scale.

If the square root function is not done in the transmitter, then the HMI has to take the square root of 4mA = 0%, 20mA = 100% and then scale the result as 0 to 500Nm3.

Thank you danw. I suspected that. However as I had mentioned earlier, I know nothing about it and was trying to discuss with my colleagues AND WAS NOT REALLY SURE. You made my day
 
If you have a flow and delta p stamped on the meter you can use this method.
Thanks, Prof.Tom Jenkins for the information. As you would see in the web site given above, result for every mA is given. Is there any method other than plugging in the values in each mA to get the result. For example by differentiation, if so can you guide me please.

Govind Rao
 
Last edited:
The web site you reference gives this formula:
SqRt mA out = 4 + (16 x √ ((Rdg - Low Rdg) / (High Rdg - Low Rdg)))

This formula tells you the mA you should be getting from the transmitter for a given flow. (I think!)

What you probably want to do is calculate the flow from a given mA signal. My approach is always to take the mA signal range and calculate the actual differential pressure. Then you can use the formula I gave above to calculate the flow. This method has the advantage of being able to compare the differential pressure in the PLC to the value at the transmitter.

If your PLC doesn't have square root capabilities use Excel to chart the mA vs. flow rate. Then you can use the add trend function to display a polynomial equation that will give you the result. You can increase the order to get a good agreement, and the PLC calculation only requires four function math.
 

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