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from the Industrial Electronics department...
Re-Calibration Flow meter using DP
Continuous process industries, DCS questions. topic
Posted by Ajith on 14 November, 2006 - 1:08 am
To all experts,

I have a question, it might be simple, but I am asking because this is not my field.

We measure a line flow using differential pressure (orifice plate), the flow range is 0 - 100 GPM while the pressure transmitter range is 0-62 in H2O.

We want to change the flow range to be from 0 -200 GPM.

what shall we do? and how can we calibrate the pressure transmitter range?? is there any formula can be used??

Thanks for cooperation.


Posted by RAVICHANDRAN on 15 November, 2006 - 12:15 am
The basic behind calculating dp (what you mentioned as pressure transmitter range) and the flow ranges is the bernolis theorem. As per bernolis, flow is proportional to square root of dp (differential pressure that is created across the orifice). That is... your flow of 100 gpm is proportiaonal to the square root of 62 in h2o.

so to calculate the new dp range:
let f1 = 100
dp1 = 62
f2 = 200
dp2 = needed to be calculated.

f2/f1 = square root of ( dp2 / dp1).
dp2 / dp1 = square of (f2 / f1).
dp2 = square of (f2 / f1) * dp1
dp2 = square of (200/100) * 62 = 248 in h2o.

hope i have not committed any mathematical error.

regards,
Ravichandran


Posted by Doug Huson on 16 November, 2006 - 12:02 am
HaHa, if only it was that easy. This is the type of stuff that keeps engineers employed. If you are lucky and this is a "smart" transmitter and you have a configurator you may be able to access the proper parameter and change it. However there is a whole lot of calculating that goes with a dp transmitter for the pressure drop across an oriface plate. Pipe size, oriface size, density of fluid, whether gas, liquid or steam, type of pipe taps being used and thats just the start. Pressure drop is proportional to flow rate but not linear, therefore a square root extractor is required (some are built into the transmitter but must be turned on). If one is not in the transmitter a separate unit will be needed. To help you avoid a headache and me blabbing on about stuff you won't know, I would suggest contacting a salesperson or the manufacturer and they will be able to let you know. You may have to give them alot of information (serial number, plate size, pipe size etc they will let you know what they need), however if they can give you the pressure drop in inches of water for the flow rate you are after you should be able to calibrate the new pressure drop (inH2O) and just get the transmitter to give you the new display. If your company has an instrumentation tech, talk with them it's their field and they should be able to help. If this measurement is important you better get it sent out for proper calibration.

To really help I need more info about you set up, i.e. where it is, what has changed,etc or if all is the same and you just want the readout to change.


Posted by Anonymous on 18 November, 2006 - 8:48 pm
RAVICHANDRAN

Thanks for your valuable input...

Doug Huson

Can you tell me what you need exactly? I gave you the flow / pressure data, for the existing and the required. What about the orifice? How can I know what is the maximum flow that can be handeled by this plate?

If there is any software to make these calculations, it will help a lot.


Posted by Kenn F on 19 November, 2006 - 4:30 pm
Actually Mr Ravichandrans your formula is okay, well done. And one only needs to install a square root extract if it is required to read linear readouts, usually for a controller or such. As the author requested only a range change then it must be assumed that he is happy with the present readout. Also all the present perameters (pressure, process media, temperatures, coefficient of discharge, compressability, etc. etc.), will apply to the new calibration. Of course this is the kind of stuff that keeps Engineers employed, that is their, (and my ) function. The only real concern is that the Orifice is suitable for such an increase in flow. It may be better to recalcuate the orifice size remebering the Beta factor


Posted by denn on 18 November, 2006 - 9:12 pm
Ajith,

It can be as simple as using a formula. Other considerations have to be evaluated. The basic formula is:

Q=K*SQRT(Hw)

Since you have Q=100 and Hw=62 (if this is accurate) then K=100/7.874 = 12.7

Your new DP(Hw) would be Q=200 K=12.7 therefore Hw=(200/12.7)squared = 248 in wc.

Now the considerations to be evaluated.

1- Has the Orifice gone into choked flow?
2- Do the original components (proportionality items determined by K above from the original equations) still remain and are valid at this new rate?
3- It is possible that a non-linear flow indication can be produced with over a 100% increase in the original flow without reverting back to the fundamental flow principals and determine the new values of K etc.

I would recommend that someone review and recalculate this Orifice installation. Others on this list may help.

Dennis

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