volume of horizontal tank

[missionmike]

Hi
I need a formula for a horizontal tank used at a milk bottling plant. We use an Allen Bradley PLC 5 system. need to convert the 4-20 ma signal from a level sensor to gallons using a compute instruction.
Thanks, Mike

 

[vetteboy79]

http://mathforum.org/library/drmath/view/54105.html

 

[Eric Nelson]

But, keep in mind that "Dr. Math's" formula assumes that the tank is in the shape of a simple cylinder with flat ends (like a soup can). If your tank has convex ends, remember to include this additional volume in your calculations!

beerchug

-Eric

 

[Steve Bailey]

If the tank is a standard model available from a supplier of components for the dairy industry, maybe the manufacturer can give you a chart of volume vs level.

If the tank is a custom model, then you'll probably need to do the math yourself.

 

[Allen Nelson]

I remember do this calculation once. But I wanted it to be able to be "calibrated" so that if I knew the volume represented by 4 mA (which presumably is when the tank is empty), and the volume represented by 20 mA (when the tank is completely full).

Looking at it, I think it's the same equation as Dr. Math.

First, the definitions:
N111:112 - Raw Analog Input from BTR, scaled (I think) 0-30840 (Flex I/O?)
F113:76 - Analog Value of 4mA (in gal, lit, whatever engineerin unit you want)
F114:76 - Analog Value of 20mA
F115:76 - Final Scaled value of level transmitter, in engineering units.
F8:5, F8:6, F8:7 - intermediate calculations.

Due to it's complexity, I had to break up the calculation into 4 CPT blocks:

F8:5 = (N111:112 | 30480) * 2.0

F8:6 = ACS (1.0 - F8:5)

F8:7 = (1.0 - F8:5) * (SQR (F8:5 * (2.0 - F8:5)))

F115:76 = (((F8:6 - F8:7) | 3.141593) * (F114:76 - F113:76)) + F113:76


As Eric said, the ends of the cylinder were flat.

Good luck.

 

[randylud]

Hey Mike, if you are using a DP cell as your level sensor, try converting the 4-20ma input signal to %. In other words scale the value to percent then take that result and multiply by the full value of gallons that the tank normally holds. IE: sensor says the tank is 30% full so you multiply by the full value of the tank, say 1000 gals and the reading will come out 300gals. Unless you need extremely accurate readings, this should do the trick. Accuracy of this method is tied to the calibration accuracy of the DP cell or level sensing device and the known value of the tank full in gals. Hope this helps, good luck and let me know how it turns out.
Randy

 

[Eric Nelson]

IE: sensor says the tank is 30% full so you multiply by the full value of the tank, say 1000 gals and the reading will come out 300gals. Unless you need extremely accurate readings, this should do the trick.

Randy, remember this is a horizontal tank, so you can't just use percentages... Your method will only be accurate at 0%, 50%, and 100%

Using your example of a 1000 gallon tank:

0% - 0 Gallons (accurate)
10% - 52 Gallons (not 100)
20% - 142 Gallons (not 200)
30% - 252 Gallons (not 300)
40% - 374 Gallons (not 400)
50% - 500 Gallons (accurate)
60% - 626 Gallons (not 600)
70% - 748 Gallons (not 700)
80% - 858 Gallons (not 800)
90% - 948 Gallons (not 900)
100% - 1000 Gallons (accurate)

Again, this is for a tank with flat ends...

By the way, I "cheated" to get these figures... I just drew a circle in AutoCAD with an area of "100", then chopped it up and let AutoCAD figure out the areas for me...

beerchug

-Eric

 

[Gerry]

I thought only the oil industry worried about these things. They typically have "strapping tables" (I think that's the right term) to calibrate volume vs level for each tank in a 'farm'.

Bear in mind that this milk tank is very likely tilted toward one end to aid in emptying and so the angle of incline and the location of the level sensor (deep end/shallow end) are additional factors to consider.

 

[Tom Jenkins]

The problem with the formula in the above link, like so many things with round parts, is it requires either trig functions or square roots to do the calculations. Many PLCs don't have tehm available. On the other hand, squares and cubes are easy with even four function math, although you really need floating point math to make it work easily.

I took the tables below from data by Panner Bailin. You can either create a look-up table and interpolate, or use the formulas I obtained by using a spreadsheet to do a linear regression on the tables. (This is a very good technique for all kinds of obtuse and complex functions.)

d/D = Depth/Diameter

Volume fraction Vf (that is, %/100)



d/D     Vf

0	0

0.05	0.01869

0.1	0.05204

0.15	0.09406

0.2	0.14238

0.25	0.1955

0.3	0.25231

0.35	0.31192

0.4	0.37353

0.45	0.43644

0.5	0.5

0.55	0.56356

0.6	0.62647

0.65	0.68808

0.7	0.74769

0.75	0.8045

0.8	0.85762

0.85	0.90594

0.9	0.94796

0.95	0.98131

1	1

0	0



Vf = (0.4055 x d/D) + (1.7764 x (d/D)^2) + (-1.17839 x (d/D)^3)



Dished Head Volume

d/D	Vf

0	0

0.06	0.0104

0.12	0.0397

0.18	0.0855

0.24	0.1451

0.3	0.216

0.36	0.295

0.42	0.381

0.48	0.47

0.54	0.56

0.6	0.648

0.66	0.732

0.72	0.8087

0.78	0.8761

0.84	0.9314

0.9	0.972

0.96	0.9953

1	1



Vf = (-0.0009478 x d/D) + (3.00269 x (d/D)^2) + (-2.00179 x (d/D)^3)