Square Root Calculation

Yep. Call Omega Engineering. They can sell you what you need.

From what you describe, you want a linear output (milliamp output is linear with differential pressure), but the transmitter is providing a square root output (milliamp output is the square root of the percentage of output).

Industrial DP transmitters can be configured for linear output or square root output.

This is done with whatever the manufacturer provides for changing the transmitter configuration, usually a HART protocol, but it could be Profibus or Foudation Fieldbus or Honeywell's DE, or Yokogawa's whatever-that-protocol-was-called.

If you don't have the configuration instrumentation necessary to change from Square root to linear, then you'll have to buy the services of a firm that can do that for you.

It sounds as if your standard 4-20ma loop is what you want for 0-500mb, thus giving you 31.25mb / ma between 4 and 20 - i.e. 31.25mb = 5ma, 125mb = 8ma, etc.

(Total mb / 31.25) + 4 = Total milliamps. Conversely, (Total Milliamps - 4) * 31.25 = Total millibars.

Now if you are converting flow to pressure, then you may need this:

http://blog.sensorsone.com/2010/03/converting-4-to-20ma-linear-signal-to.html

It is not real clear if your transmitter is giving you square root or if you need square root - but either way that link should give you the equations you need.

can u tell me how we get 31.25mb,125mb,

please help me i am fresh in instrumentation

Hi Saleem,

I went back and re-read your first post.

If I understand the problem, you have a transmitter that is outputting FLOW RATE based on the Differential PRESSURE input.

As the link above indicates:

" The flow rate along a closed pipe is directly proportional to the square root of the pressure drop or differential pressure between two points."

Therefore you transmitter is giving the Square Root of the differential pressure directly so it can be used to indicate FLOW RATE.

Therefore, using their formula:

[Output Linear] = 4mA + (([Output Sq Rt] - 4mA)^2 / 16)

If your transmitter is outputting the square root [Output Sq Rt] with a range of 0-500 millibars and you want to know what your millibars are, then your formula would be like this:

((([Output Sq Rt] - 4) x ([Output Sq Rt] - 4)) / 16) x 31.25

So let us try 3 different values and see what we get, we will put in 4ma, 15.31ma and 20ma:

1: ((4 -4) x (4-4) / 16) x 31.25 = 0mb

2: ((15.31 - 4) x (15.31 -4)/16) x 31.25 = 249.84mb

3: ((20 - 4) x (20 - 4) / 16) x 31.25 = 500mb

So the formula works.

Now, some readers may be wondering why I left out the 4ma + at the beginning of the formula. This is because there is a 4ma offset already and that is added there to arrive at the [Output Linear] value. But when we use the factor of 31.25 per mA to calculate the millibars, we want the milliamp value only. For example, if we have 5mA LINEAR, then we only want the 1mA above the 4mA so it is 5mA - 4mA = 1 x 31.25. Since we would be subtracting the 4mA from the end of the standard equation after adding it to the beginning of the equation, I simply left it out. But if it makes it easier to understand or follow, the literal equation would look like this:

(4mA + ((([Output Sq Rt] - 4) x ([Output Sq Rt] - 4)) / 16) - 4mA ) x 31.25

Which in turn could be reduced to: ([Output Linear] - 4mA) x 31.25 = Millibars

I hope that helps =)

Send the signal thru a square root extractor, which will linearize the output.