PSIG Transmitter and Digital Scale Readout

Hello,

I am trying to make sense of an incorrect reading or how to calibrate one of our liquid tanks.

Volume of the Soy Oil tank = approx. 8078 gallons US or rounded to nearest whole number in lbs = 60,000(1 gal water = 8.34 lbs and specific gravity of soy oil approx. = 0.92 7.672 x 8078 = 61,980 lbs)

If my math there is correct my 100% of the scale should be 60,000 lbs in the process.


Digital readout P/N: imp20160, Link to manual

Pressure transmitter P/N: px41t1-010gi Link to manual. This transmitter is a 0-10 PSIG range


Again, the problem is that going off a balance sheet, our readout currently in place is wrong. I am trying to calibrate it and prove whether the value is correct or not. So I follow the red lion manual section Module 2 to set the scale manually like the following. I set the 0% = 0, inp 1 = 4.00, 100% = 60,000, and inp 2 = 20.00 . However I am still off by about 14,000 lbs. What if anything does someone see that I am doing wrong here? Is there a better transmitter to readout combo from another manufacturer?

Thanks for all the help in advance.

So you are measuring head pressure from a tank containing soy oil, correct?

Assuming yes, is the instrument (pressure transmitter) mounted precisely at the bottom of of the tank? If yes, then 0ma = 0psi = 0lbs, 20ma = 10 psi would be equal to 21.252 feet of head pressure. Is the tank height with 60,000 pounds 21.252' of height (head pressure)?

If not, that is probably where the error is. Most likely, the tank is full before you hit the full scale range of the pressure sensor, so inp 2 needs to be set differently.

I am used to seeing DP transmitters used for tank measurement. You have to take into account the pressure on the head of the tank. If it is an open tank then one leg measures atmospheric pressure. Usually I see large diaphragm sensors and instruments that measure in inches of water.

Yes, sorry my transmitter is measuring 4 - 20mA at the very bottom exit of my tank. The tank in question has a "U" vent at the top to atmosphere and is just a steel tank with no diaphragm.

The tank is a cylindrical shape. Usable height = 13' 9" and usable diameter of the cylinder is 10'

Ryan_Flowers said:

Yes, sorry my transmitter is measuring 4 - 20mA at the very bottom exit of my tank. The tank in question has a "U" vent at the top to atmosphere and is just a steel tank with no diaphragm.

The tank is a cylindrical shape. Usable height = 13' 9" and usable diameter of the cylinder is 10'

Click to expand...

So, what is the height of the liquid relative to the sensor when it contains 60,000lbs? If that answer is 13'9", then you need to enter the equivalent milliamps for IN2 in the Red Lion.

That would be 2.1252 feet of head per psi if my math (and your specific gravity) are right. So 13.75' would be 6.47psi or 64.7% of full scale of the sensor. Multiply that by 16 (range in mA) and add 4mA back in and I came up with 14.25mA for IN2. Try that and see if it jives, and someone please double check my math.

Ryan_Flowers said:

transmitter is measuring 4 - 20mA at the very bottom exit of my tank.

Click to expand...

Please describe, or sketch out, the arrangement, including exact relative height measurements of the sensing element i.e. where the liquid touches the sensor, the exit from the tank, and the surface of the oil when the tank is full.

Also, when you say "very bottom exit" - is the bottom of the tank flat, or does it have a rounded (ellipsoidal) bottom?

Assuming that checks out, @OkiePC's number are definitely in the ballpark, but I think maybe he inverted the specific gravity (SG) correction (divided instead of multiplied) and 13'9" below the top surface (at atmospheric - gauge - pressure) should be 5.48psig.

Water (SG=1) is normally ~2.3ft/psid; since the oil is less dense, it should be take more ft of oil to generate 1 psid.

(62.43 lb-water/ft**3) * (0.92 lb-oil/lb-water) ÷ (1ft**2/144in**2) * (13.75 ft/tank-height)

= 5.48 (lb/(in**2 tank-height)) = 5.48 psi/tank-height

I am excluding g and not correctly converting between lb-mass and lb-force, but that would cancel out.

Hi, I worked for a water utility for some time.
In a vessel, water produces .433 PSI for every vertical foot. The inverse would be 2.31 feet of water for every PSI. This is based on a SG of 1.0 for water. So the scaling would be .92 of the calcs for water.
Example would be 15' of water x .433 = 6.5 PSI. So 15' of oil x .433 = 6.5 x .92 which equals 5.98 PSI or 6.0 PSI basically. Hope this helps.

drbitboy said:

Assuming that checks out, @OkiePC's number are definitely in the ballpark, but I think maybe he inverted the specific gravity (SG) correction (divided instead of multiplied) and 13'9" below the top surface (at atmospheric - gauge - pressure) should be 5.48psig.

Click to expand...

Thank you. Yes, I think you are correct. 10psi of water would be 23.1'. For the oil it would be 25.11' not whatever I said earlier. So at 13.75' he should get 12.762mA.

drbitboy said:

Please describe, or sketch out, the arrangement, including exact relative height measurements of the sensing element i.e. where the liquid touches the sensor, the exit from the tank, and the surface of the oil when the tank is full.

Also, when you say "very bottom exit" - is the bottom of the tank flat, or does it have a rounded (ellipsoidal) bottom?

Assuming that checks out, @OkiePC's number are definitely in the ballpark, but I think maybe he inverted the specific gravity (SG) correction (divided instead of multiplied) and 13'9" below the top surface (at atmospheric - gauge - pressure) should be 5.48psig.

Water (SG=1) is normally ~2.3ft/psid; since the oil is less dense, it should be take more ft of oil to generate 1 psid.

(62.43 lb-water/ft**3) * (0.92 lb-oil/lb-water) ÷ (1ft**2/144in**2) * (13.75 ft/tank-height)

= 5.48 (lb/(in**2 tank-height)) = 5.48 psi/tank-height

I am excluding g and not correctly converting between lb-mass and lb-force, but that would cancel out.

Click to expand...

Attached is the drawing of the tank. Sorry it's a .jpg.

The height of the transmitter is 14" below the ellipsoidal bottom exit of the tank.

This has me a little confused as of right now, but I suppose that's what learning is...

That 14" is critical. In the image below, you want the volume represented by the pressure difference across either A or A', depending on whether you want to account for the volume of the (not ellisoidal; dished?) head.

The PSIG pressure sensor is always and only measuring the pressure difference across B, so something will need to be added or subtracted to start calculating the volume of oil in the tank*. That something is C or C': C' is 14" (did you include the thickness of the header wall?), C is more than 14".

The routing of the piping from the tank to the pressure sensor is irrelevant** as long as the piping exits the tank where oil is present and the piping from the tank to the pressure sensor is full of oil.

If the A'/C' case is of interest, the volume calculation is doable but a whole 'nuther ballgame; see the page from Perry and Chilton "ChemEng Handbook," 5th ed., 1973, in the .ZIP attached below.

* Unless the pressure sensor is located at an elevation at which the level would be when volume of is considered to be zero.

** The sensor piping could be a standpipe coming out of the top of the tank with the sensor above the tank, in which case B and C would have the opposite sign of A, and the sensor would measure vacuum.

drbitboy said:

That 14" is critical. In the image below, you want the volume represented by the pressure difference across either A or A', depending on whether you want to account for the volume of the (not ellisoidal; dished?) head.

The PSIG pressure sensor is always and only measuring the pressure difference across B, so something will need to be added or subtracted to start calculating the volume of oil in the tank*. That something is C or C': C' is 14" (did you include the thickness of the header wall?), C is more than 14".

The routing of the piping from the tank to the pressure sensor is irrelevant** as long as the piping exits the tank where oil is present and the piping from the tank to the pressure sensor is full of oil.

If the A'/C' case is of interest, the volume calculation is doable but a whole 'nuther ballgame; see the page from Perry and Chilton "ChemEng Handbook," 5th ed., 1973, in the .ZIP attached below.

* Unless the pressure sensor is located at an elevation at which the level would be when volume of is considered to be zero.

** The sensor piping could be a standpipe coming out of the top of the tank with the sensor above the tank, in which case B and C would have the opposite sign of A, and the sensor would measure vacuum.

View attachment 58505

Click to expand...

Ok, so I don't think I can accurately calculate the bottom portion of volume. However I just deducted the distance from the edge to the ground and the center of the "dish" to ground and got approximately 16". So I add 16" and 14" (for the transmitter distance from the "dish") to the 13' 9" and I get 16.25'.

When I convert it to mA of the scale I get 16.943 mA. Would this be correct when I put it as inp 2? Because judging by what the balance should have been when filled yesterday and the usage today, the poundage value looks accurate.

I also changed my disp 2 value (100%) to 61,382 lbs. This was to be more accurate at the 8,000 US gal cap with the given volume excluding the calculation I do not understand.

Cool, good info.

Water density is 62.43 lb ft**-3 nominally; so oil (or any fluid) with SG of 0.92 is (0.92 * 62.43) = 57.4356 lb f**-3 (yes I know I am carrying more sigfigs than necessary).

Divide that by (144 in**2 ft**2) to get lb in**-2 ft**-1 ~ 0.398858PSI/ft.

Lower scale parameters 0lb at 9.9714% of scale = 5.5943mA

Assuming a level at the top of the bottom dish is the nominal zero (empty) tank level, using A and C from the diagram, which is 30" or 2.5ft above the sensor, so the pressure at the sensor would be (2.5ft * 0.398858psi/ft) = 0.99714PSIG = 9.9714% of full scale. So we could set the min output to be 0lb at 9.9714% of scale. That would be (16mA * 9.9714%) = 1.5943mA above the base (4mA) current, so we would expect 5.5943mA when the tank is "empty" i.e. the tank level is at the top of the bottom dished head.

Upper scale parameters 62,026lb at 64.81245% = 14.3702mA

For a full oil level at atmospheric (gauge - the G in PSIG) pressure that is 16.25ft above the pressure sensing element, the pressure at the element is 16.25ft * 0.398858PSI/ft ~ 6.48145 PSIG.

6.48145PSIG is 64.8145% of the 10PSIG sensor range, so we could set the max output to be 62,026lb (Note 1) at 64.8145% of range. That will be the same fraction of the 4-20mA = 16mA range = 10.37032mA above the base current i.e. above 4mA, so I would expect 14.37032mA (Note 2) when the tank is full.

Alternate upper scale parameters 101,820lb at 100% (theoretical)

Another way to look at it is that max (100%) range of the sensor is 10PSIG, divide by 0.398858psi/ft to get a theoretical level of 25.07ft above the sensor to generate 10PSIG at the sensor. 25.07ft above the sensor would be 22.57ft above the top of the bottom dished head, assuming that entire height is a cylinder of diameter 10ft and radius of 5ft, 22.57ft * (5ft)**2 * PI = 1772.77ft**3 of a theoretical prism (column) of fluid above the sensor, times 57.4356lb/ft**-3 = 101,820lb. So we could set the max output to be 101,820lb at 100% of scale and get the same result.

Dished head capacity ~ 53.6ft**3 ~ 3079lb

That is if the level is not below the top of the dished head. That is about 5% of the full tank. Cf. spherical segment here assuming spherical surface (r1=5f, r2=0ft, h=16"=1.33ft); that formula could also be adapted to calculate the capacity of a level below the top of the dished head, but then the conversion is no longer linear.

Notes

1. 62,026lb = 13.75ft * (Radius=5ft)**2 * PI * 57.4356lb ft**-3. The difference between 61,382lb and this is the density of water we are each using: 8.34 vs. 8.3457 (= 62.43lb ft*-3 * (231 in**3 gal**-1) * (1/1728 ft**3 in**-3)).
2. I don't know where OP's 16.943mA comes from, but the ratio 14.370/16.943 is very close to 0.92**2, so I suspect one of us used the SG incorrectly; hopefully not both of us.