Tank Calibration

Hi abimu

From what you have you first need to determine the inside diameter (or radius)

with a 310" circumference the radius is ±49.33".

from this you need to subtract the thickness of the wall.

After you have done that you can continue.

It might be advisable to use a spreadsheet like excell.

In addition to the inside diameter you need to know the coefficient of thermal expansion for the oil. If there is any dissolved gas. There almost always is. If there is a dissolved gas the density will vary with pressure from depth. Also you have to take into account how much the tank deviated from being a perfect cylinder. Dents, concave or convex ends change the volume. There are no perfect tanks that large and few oils that don't have gas devolved in them. Gordon

Find volume in M³ multiply by density.

multiply by 2.2 to convert to lbs.

Regards JD.

Your terms are jumbled.

volume=pi*r²h and determined its area in cubic feet (???)

Your tank has a circumference of 25.8333 ft. Its radius would therefore be, 4.1115 ft., which gives you an area of 53.107 ft². Every inch of height (0.0833 ft.), therefore, gives you 4.426 ft³ of edible oil for every inch of height.

Now you have the data to determine the quantity of oil per inch of height.

I respect all answers till now. All these methods are OK if you do not need the accuracy.

All method suggested assume that the tank is perfect in radius over the height. But the tank of 8 feet and odd diameter can not be perfect in diameter all over the height. Besides, at any cross section, it can not be perfect circle, to use the formulae of circle. It is bound to be non circular, as it is not machined (bored) diameter.

Besides, as you say you need mass for every inch of the height. Here again there is one more problem. How you read the inch? If you put a glass window and a scale to see the height of oil in it, the manual errors are bound to be there.

Forgetting the manual errors of reading over the height, which you may reduce by reading the level at lower point of the miniscus of the oil and looking from same height of the oil, we need to think about the effects of non circularity of the tank.

You need to put the vessel on some calibrated weigh bridge. With empty tank, maesure the weight. Then fill the oil slowly stopping at every inch and noting down the weight (Empty tank + Oil). Then subtract the empty tank weight fro each reading. This method (only) will give you the weight of oil for every inch of the tank.

Besides, we will be assuming the oil density constant.

One more assumption is there that the temperature of the tank/ oil system is constant. With temperature changes, the volume of the tank changes as well as density of oil changes. We need to compensate for these variations, even if we calibrate with the method I suggested.

Please note that I was designer and incharge of India's largest (and world's No. 3) flow calibration Laboratory. The laboratory is accredited according to international standards and managed according to ISO 17025.

We have also a big tank of water (12 feet diameter X 60+ feet height). We have calibrated it for every 2 feet of height, where electronic level sensors sense the levels instead of visual observation. The sensors are sensitive to detect water level within 0.1 mm (0.004 inch).

But if you do not need any accuracy, you may assume the tank as circular, visually observed level as perfect and oil density constant. Then the mathematical calculations suggested by others are good enough.

Use an accurate flow meter when filling the tank initially to the calibrations that you desire and mark them off every 6" or what ever you require. You may wish to mark the calibrations in gallons/litres.

You can do the calibraion with water if you like because:-

1m³ = 1 000 litres.

OR

1ft³ = 6,2288 gallons (Imp.)

1ft³ = 7,4805 gallons (US) always have to be different

Yes, you are right MOBI.

But with flow meter, the filling is continuous. You can not stop at every inch.

Or else, you need to have accurate level switches at every inch and flow through flowmeter along with the automatic data logging system.

We used same system of flowmeter+level switches + data logging for our tank.

Also we used the method I suggeted for smaller tanks.

All you need is a nozzle on the end of the hose when you do your initial calibration, you can stop any time. note or mark your calibration as necessary and away you go again. As I said you only need a water supply, a pump. an approriate hose, connections and flow meter and a nozzle (like the nozzle on a petrol bowser).

To get your required calibrations (from flow meter) you will either have a gauge glass, dip stick or dip tape, whatever, that you will calibrate every so many litres/gallons, etc. according to your flow meter.

By this method, it is very difficult to stop the level at desired mark.

I guess that depends on the accuracy of the calibration required, I have calibrated diesel fuel tanks a few times with no problems. If more accuracy was required one could always use a specific sized container or something like a snap tank.

I suppose it is all dependent on the accuracy of the calibration required. e.g. to the nearest ml, litre, gallon, etc.??????????

You are right MOBI.

I said in my original posting about the accuracy. The method I described was for atmost accuracy.

Best and simplest way is to fill the tank using standrad calibrated volume vessels available in the market, which are certified by national legel methrology department and mark your tank for fixed volumes.

With this method, the markings on tank may not be at equal heights, but the accuracy will be great with simplest method. This will give volumetric calbration, which you may multiply with density to get non integer weight markings.

I gave you a good answer because what you said is absolutely correct. Unfortunately, abimu did not mention his accuracy requirement. Perhaps it isn't that important.

This reminds me of a client we once had. They constructed a large concrete water tank, about 50 meters in diameter. They had us install a large face diameter (12") pressure gauge graduated from 0 to 6 meters of water (the height of the tank) to measure the level from outside the tank plus a pressure-type level transmitter to send the level data to the control room (about 30 meters away).

After all was installed and they were inspecting it, they asked what the accuracy of the pressure gauge was. Funny they should ask this when everything was already done but my boss told them "0.25%". He didn't mention if this was % of reading or % of scale but the customers weren't interested in that.

"What?!! That's 1.5 cm of error! We can't afford that!"

I stayed away from the argument that ensued but it taught me a lesson about being clear in what you expect from the supplier.

Please send me your e-mail, there an excel file will help you.

my email add is --- munir-5@hotmail.com ------i will thankfull to you

regard munir kazim

Dear Munir,

The file was actually send, please let me know your comments.

This calculation is carried out by excell and we assumed that the tank wall thickness is 0.5 inch (12.7 mm)

Dear Mr. Abdel,

Please note my mail address:eniyaanand@yahoo.c

Please recheck your e-mail, since it can't be send.

The density given by you (0.91) must be in grams per cubic centimeter. This is equal to density 56.81 lbs/cubic foot.

Tank circumference given is 25' 10" = 2(pi) r. Therefore, r (radius of tank) = 4.11 feet

Volume of tank PER INCH = pi*r²h = (22/7) x 4.11 x 4.11 x (1/12) = 4.424 cubic feet

Therefore weight of oil per inch height of tank, in pounds = volume x density = 4.424 x 56.81 lb = 251.33 lb

That is weight of oil per inch height of tank = 251.33 lb = 114.0 kg

If you want the volume of tank per inch in gallons, 4.424 cu. ft = 27.56 Imperial gallons (British) = 28.44 US gallons

For the FULL TANK,

Volume = 4.424 x (16.404 x 12) = 870.85 cu. ft = 5425 British gallons = 5598 US gals

Weight = 251.33 lb/inch x (16.404 x 12) = 49,474 lb = 22,400kg

Hope this is sufficient

indrawim

If the flow meter is to quick then I would grab a few friends, lots of beer, and some calibrated 5 gallon buckets. All but 2 people get buckets. As they fill the tank one person holds a tape measure. The other no toting friend marks the tank. The beer is the carrot in front of the horse and only to be consumed when the job is done. Very time consuming but accurate.

Personally, I would go with the flow meter unless the job pays by the hour or bucket. Make sure the discharge hose remains slightly below the surface to prevent waves that could mess up the measuring process. Do not just throw it in over the side and leave it unless you are planning on calculating the displaced volume of it.

How accurate does this have to be? To say very accurate on this scale could be gallons, quarts, or etc.

What has been posted so far I think is more than adequate for a one time calibration of the tank as described. Now in my industry, pharmaceutical, I have seen a number of methode in practical use to the not so practical. The multiple level sensor approach is limited by the number of sensors used. A mass flow meter is usually the best if not cheapest solution since the meyer can be set up for the fluid you are working with. Another approach is , as mentioned, a load cell or cells on the tank supports to give you a very accurate reading of the mass added to the tank. One othe methode used at times is a pressure transducer just before the outflow valve at the base of the tank. The output can be either directly converted from psi to a value corresponding to the liquid colum above if the tank is uniform OR you can use one of the previously mentioned methodes to characterize pressure readings to actual fill values. Hope this is of some help. (Loaded via BBerry)

Oh! Having beer in measured quantity! Bad idea

I can't believe the fervor that one simple problem has created!! There were several answers that were adequate for calculating the weight of the oil per inch of tank depth. Just to recap that, the tank diameter is 8.22 feet and the depth is 16.404 feet. Therefore, the volume of one inch of tank depth would be 32.947 gallons. The weight of water is 8.34 pounds per gallon (US) which means the one inch of water would weigh 274.78 pounds. The density of the oil is .910 so the weight of one inch of oil in that tank would be 250.046 pounds. I don't see the need for the discussion about accuracy. This should be emminently accurate. The method described by GSUHAS may yeild some small measure of additional accuracy, but I would guess that the difference would not equal the amount of oil that will cling to the walls of the tank when the level is lowered. The answer, to the 3rd decimal point, is 250.046 pounds.

You are right, provided the tank is reasonably round with no dents, which is not the case normally. My method is universal, irrespective of roundness. Even it applies to tank of any shape.

Other observation: Many people give their replies without refering to the replies already posted, as if they are the only one to know the answer. Thus, you see many repeatations.

Simplest and easiest is using calibrated volume vessels, which has limitations of non linear scale.

Other methods including what I suggested are much costlier.

What are you really interested in? Are you buying or selling by the inch of tank volume, or are you needing the inch reference for a visual confirmation of a delivery or receipt? How are you performing your inch measurement?A tape measure?Ultrasonic?Rf?Capacitance? All of these factors will affect accuracy of the results.

If the oil is being used to create a formula or compound of some type, then the heigth of the oil in the tank is irrelevant.The mass pumped into or out of the tank is the important process variable.

A mass flow meter with temperature compensation is probably the most accurate for precise measurement.

A positive displacement meter(with temp compensation) is also very accurate, as is used in Gasoline dispensing pumps.(Tokeim brand come to mind).It displaces a precise volume per revolution, and counts the revolutions to determine quantity.

If you don't need real accuracy, then a SWAG is sufficient.

HTRN