Converting 4-20mA Signal

D

Thread Starter

David

I have two transducers and I want to take the differential between the two and convert that to a water level. This will be done through a PLC that controls a valve on the surface. What is the calculation that converts the 4-20mA signal to a true water level?
 
One transducer will be wet and one dry - right? If there is a space between the two like 3' then water level can be anywhere between. Unless you have a timed deal? If you only want a simple valve controlled level -we can offer a HD machanical float rod - that opens a sturdy valve in either SS or Bronze. 1/2" NPT to 2" NPT. Or, a simple way would be to use our continuous probe that, as well as 4 - 20 has a 1 to 10 VDC resistive output that can be hooked direct to a meter or scale. Thanks Bob Hogg www.almegcontrols.com > I have two transducers and I want to take the differential between the two and convert that to a water level. This will be done through a PLC that controls a valve on the surface. What is the calculation that converts the 4-20mA signal to a true water level? >
 
There are several items that need to be answered here. 1) is the capacity of the tank linear to the level of the tank? (are the walls of the tank parallell) 2) is there any object occupying space in the measured area of the tank that will occupy >5% of the area affected. Either of these will interfere with a normal linear calculation (the easiest conversion) you state that you wish to take the "differential" of the two inputs (from the transducers) and create a relationship between this result and a tank level. Provided that #1 is true and #2 is false, from the two items listed earlier in this response, then a simple "Y=M(x)+b calculation would be appropriate. (!!provided that BOTH of your transducers have a very similar output slope value.!!) additionally, you did not mention anything about the input and output flow relationship and what your requirements are. This will have an impact on your ability to maintain level. once you have established these relationships, you will be able to command your PLC program to output an appropriate signal. (you neglected to mention if your output will be digital or analog) if digital, your output would be a simple comparator equation (is the signal level within a given range that I desire to maintain). if so, then activate or de-activate your valve. with this kind of control, you WILL be tied to the maximum available flow of you inlet and outlet. if analog, you would set up a proportional relationship and format your output based on your desired logic. (again, based on your logic) a much simpler way to accomplish this is a redundant level detection device (two level sensors). let the electronics do the work, and you validate the response in your program and take control from there. this way, it WILL be a straight forward slope calculation. no questions asked or necessary. i hope that this is of some assistance to you.
 
Differential Pressure XTR's are designed for what you are trying to achieve. If your tank is an horizontal one, you need to linearise the reading. This applies also to other tank geometry.
 
C

Cahya Hari K

What PLC did you use ? It is simple if you used AB PLC with Scale (SCP) command. Reagrds Cahya Hari K Eng.Dept. Somit Karsa Trinergi, PT
 
S

September, Clyde

The easiest, not necessarily the most accurate way to calculate the ratio of the change in diff. signal is the change in water level. Note: You do not specify much - Is the one transducer at the bottom of whatever you trying to measure? Can this vessel be drained? etc Clyde September
 
P

PMSC Control System

Dave, Although you did not specify that you were using a pressure probe for the water measurement I'm assuming that is being used. The probe [or transducer] should have the PSI rating on it - if not - the part number and a call to the vendor should give you the rating. You are probably dealing with a 0-5PSI or 0-10 PSI unit. For the 10 PSI unit -- 4ma is the reading for the probe out of water. 20ma for the probe in 23 feet of water. I'll use this as an example. Converting this to a usable value in a PLC or data logger depends on the input scaling of your analog input -and- the dropping resistor used. For a 0 to 5vdc input use a dropping resistor of 250 ohms. 4ma gives you an input of 1.0vdc at the PLC. 20ma gives you 5.0vdc. A total of 4.0 volts for the full range of the probe. If the probe's physical range was 23 feet. Divide the 23 feet by the 4.0 volts. This gives you a scaling factor of 5.75 feet per volt measured. Our reading starts at 1.0 volt to begin with so we have to subtract 5.75 from the value read to establish the depth of the probe. Once you have calculated this value you can establish a reference for the true depth. Your mileage may vary.... Dave Gunderson Hoover Dam Tech Services Group - Special Projects Branch Personal Web Site: http://damguide.com
 
Top