Wet Leg Level Measurement Question 1

[JAspinall1]

Hello,

I'm currently learning the basics of Level Measurement and at the moment I'm stuck on the concept of Dry & Wet legs. From the reading that I've done, I understand what they are but I'm not sure how to factor them in for calibration, i.e how they effect zero and span.

If someone could please walk me through this example problem(I assure you it's not a homework/test question) I'd greatly appreciate it. If you are still worried it may be a homework question, I'd also very much appreciate any links to resources that can introduce me to the basics of Dry & Wet legs. However, I have already read the Dry & Wetleg chapter from Omega.ca

Here is a link to a picture of my example problem.

http://i53.tinypic.com/2d2fyhv.png

Thank you very much for your time.

 

[danw2]

Don't get confused by the strangeness of the range because in the end, the transmitter's output range is 0-100% level. The 0-100% level gets interpreted somewhere, indicator, PLC, HMI, whatever, as so many inches, feet, percent, gallons or pounds, whatever.

I do DP ranges by "What's the DP cell 'see' when the tank is empty? and "What's the DP cell 'see' when the tank is full?"

The tank pressure of 10 PSI is applied to both the HI and LOW sides, so it gets eliminated in the DP technology measurement process. The 10 psi does not figure into elevation/suppression, LRV or URV calcs.

The external wet leg is supposed to be kept full to its 140" height.

So the Low side always 'sees' 168"w.c. = 140"w.c. * 1.2 (SG)

DP technology subtracts Low side from High side: High - Low

When the tank is empty, the measurement is High minus Low:
0" - 168" = -168" w.c.

the lower range value (LRV) is -168" w.c.; the measurement when the tank is empty.

Your problem does not state what the maximum tank level is, it only shows an 80" level.

Suppose the maximum tank level is 90" of water.

Then the upper range value (URV) is the LRV + (measurement of max level * SG)

URV = -168" + (90" * 1.00) = -78"w.c.

LRV = -168"w.c. (zero)
URV = -78"w.c. (span)

For an analog transmitter, 4mA = 0% level = 0.0"
20mA = 100% level = 90"

 

[JAspinall1]

Hi Dan,

Thank you very much for the walk through, that really cleared a lot up for me.

Can you tell me if I've got the right calculations for this example where the DP Transmitter is below the datum point. I know the zero point must be suppressed, however with the Wetleg in the mix I'm a little thrown off.

Example:

http://i54.tinypic.com/mh388k.png

(Assuming 120" is the highest level of the tank.)

Here is my work:
LRV = 40 - [ (200 * 1.6) + 40 ]
LRV = -320" W.C

URV= [ (120 * 1.6 ) + 40 ] - [ (200 * 1.6) + 40 ]
URV = 232 - 360
URV = -128" W.C

Conclusion:
LRV = -320" w.c and URV = -128" W.C

Am I way off here? Thank you in advance!