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Hello,
I'm currently learning the basics of Level Measurement, and at the moment I'm stuck on Dry & Wet legs. If anyone could walk me through this example problem (it's not a homework problem, or for any type of project). it would be greatly appreciated as it will fill in some blanks for me.
If you are wary that this could be a homework/test problem, I would also greatly appreciate any links that will explain the 'basics' of using Dry and Wet legs. I have already read the brief explanation on Dry/Wet legs from www.omega.ca however it hasn't given me enough info to answer this example problem.
Here is a picture of the problem:
http://i53.tinypic.com/2d2fyhv.png
Thank you for your time.
I'm currently learning the basics of Level Measurement, and at the moment I'm stuck on Dry & Wet legs. If anyone could walk me through this example problem (it's not a homework problem, or for any type of project). it would be greatly appreciated as it will fill in some blanks for me.
If you are wary that this could be a homework/test problem, I would also greatly appreciate any links that will explain the 'basics' of using Dry and Wet legs. I have already read the brief explanation on Dry/Wet legs from www.omega.ca however it hasn't given me enough info to answer this example problem.
Here is a picture of the problem:
http://i53.tinypic.com/2d2fyhv.png
Thank you for your time.
![]() ![]() |
Okay for starters I will assume you are familiar with head pressure, and it's relation to specific gravity.
water having a specific gravity of 1.0
and there being 80" of water, I will assume this is the top measurement level for the application.
you get 2.888psi 80"H20/(27.7 "H20/psi) on the high side
and 6.0649psi 140*1.2/27.7 on the low side.
giving you a -3.1769psi differential for a high level.
and a -6.0649psi differential for a low level assuming your low is 0"h20 in the tank.
this is based on the fact that you wet leg should always be full otherwise you get a whole host of issues. so at 0" in the tank you have 0psi of head pressure on the high side and the full column of 140" 1.2 sg fluid acting on the low side.
the 10psi can be ignored since it applies itself to both the high and low side of the dp cell.
water having a specific gravity of 1.0
and there being 80" of water, I will assume this is the top measurement level for the application.
you get 2.888psi 80"H20/(27.7 "H20/psi) on the high side
and 6.0649psi 140*1.2/27.7 on the low side.
giving you a -3.1769psi differential for a high level.
and a -6.0649psi differential for a low level assuming your low is 0"h20 in the tank.
this is based on the fact that you wet leg should always be full otherwise you get a whole host of issues. so at 0" in the tank you have 0psi of head pressure on the high side and the full column of 140" 1.2 sg fluid acting on the low side.
the 10psi can be ignored since it applies itself to both the high and low side of the dp cell.
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> I'm currently learning the basics of Level Measurement, and at the moment I'm
> stuck on Dry & Wet legs. If anyone could walk me through this example problem
Assuming your level transmitter is located at the bottom tap and that is the zero and you are using smart transmitters either Hart or Fieldbus here is how to best do this.
Completely isolate the transmitter from the process
Open both transmitter manifold valves
Vent the HP side to atmosphere
Fill the wet leg with the sealing fluid
Look at the process value on the communicator and write this down.
It should be 140 * 1.2 = -168" H2O
This is your LRV or Zero
- your span from this which is 80" H2O
This will be -168 +80 which will be -88" H2O
This is your upper range value which will be -88" H2O
This is your transmitter calibrated range
-168 to -88" H2O
Life is not this simple but as a rule of thumb you would get by.
Best regards
Mike
> stuck on Dry & Wet legs. If anyone could walk me through this example problem
Assuming your level transmitter is located at the bottom tap and that is the zero and you are using smart transmitters either Hart or Fieldbus here is how to best do this.
Completely isolate the transmitter from the process
Open both transmitter manifold valves
Vent the HP side to atmosphere
Fill the wet leg with the sealing fluid
Look at the process value on the communicator and write this down.
It should be 140 * 1.2 = -168" H2O
This is your LRV or Zero
- your span from this which is 80" H2O
This will be -168 +80 which will be -88" H2O
This is your upper range value which will be -88" H2O
This is your transmitter calibrated range
-168 to -88" H2O
Life is not this simple but as a rule of thumb you would get by.
Best regards
Mike
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It's a very unusual problem, I doubt you will come across it in real life.
I did a brief search for fluids with SG=1.2
Hydrochloric acid
Corn Starch
Nitrobenzene
I can't imagine using any of those for a fill fluid so I assume it's just for the sake of an exercise
It reminds me of a condensate tank but in that case you would use a condensate pot at the fill point and fill with water.
Roy
I did a brief search for fluids with SG=1.2
Hydrochloric acid
Corn Starch
Nitrobenzene
I can't imagine using any of those for a fill fluid so I assume it's just for the sake of an exercise
It reminds me of a condensate tank but in that case you would use a condensate pot at the fill point and fill with water.
Roy
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I'm not sure how common it would be as tank level wet leg fill fluid, but Wika lists its glycol/water capillary fill fluid with a SG of 1.22
www.wika.us/upload/DS_ACC_ACS_90_MO_en_us_17070.pdf
www.wika.us/upload/DS_ACC_ACS_90_MO_en_us_17070.pdf
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on 26 August, 2011 - 10:56 pm
on 11 October, 2012 - 5:48 pm