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Posted: 1/6/2012 8:10:22 AM EDT
[Last Edit: 1/6/2012 8:13:38 AM EDT by diehippy]
I have a horizontal cylindrical tank that I need to calculate volume changes in....based on % of liquid height in the tank. It has round ends, but for simplicity we'll assume this is a plain horizontal cylinder. As far as I can think, this is probably a fairly simple integral calculation for change in area * tank length, but I've forgot most of this stuff. Anybody have a reference formula handy for this?

EDIT: radius = 60", length = 410"
Link Posted: 1/6/2012 8:11:17 AM EDT
[Last Edit: 1/6/2012 8:12:41 AM EDT by 7n6]
Calculate the area of the circle on the end multiplied by the length of the tank.

A= pi x r2 x length of tank.
Link Posted: 1/6/2012 8:14:56 AM EDT
[Last Edit: 1/6/2012 8:18:03 AM EDT by killingmachine123]

Originally Posted By 7n6:
Calculate the area of the circle on the end multiplied by the length of the tank.

A= pi x r2 x length of tank.

It would be easy to include the ends too since it's just a sphere. Make sure you don't double it though. A lot of kids will do that.

Unless I'm misunderstanding your problem you should just be able to take the % of the volume for the reult you want.

ETA: I see your dilemma now. I can't do the ends on the tank. Sorry. You may try wolframalpha.com though.

ETA II: I also read too fast and now see that the tank is horizontal. What a bunch of BS!
Link Posted: 1/6/2012 8:15:07 AM EDT

Originally Posted By 7n6:
Calculate the area of the circle on the end multiplied by the length of the tank.

A= pi x r2 x length of tank.

Yes, but I need to know the volume from different liquid heights. A change from 5-10% full is small compared to 45-50% full in volume.
Link Posted: 1/6/2012 8:15:39 AM EDT
Originally Posted By 7n6:
Calculate the area of the circle on the end multiplied by the length of the tank.

A= pi x r2 x length of tank.



That will give him volume but the volume changes based on liquid height will be non-linear as it fills since there will be more volume in the center section than there is at the top or the bottom.



Link Posted: 1/6/2012 8:16:10 AM EDT
i have a calculator for doing just that ... i will need the tank height ( or liquid level for that volume ) the dia of the tank or vessel, and the distance from the tangent line for the depth of the dish bottom head.

i inspect tanks for a living
Link Posted: 1/6/2012 8:19:19 AM EDT

Originally Posted By sinsir:
i have a calculator for doing just that ... i will need the tank height ( or liquid level for that volume ) the dia of the tank or vessel, and the distance from the tangent line for the depth of the dish bottom head.

i inspect tanks for a living

pm sent
Link Posted: 1/6/2012 8:21:04 AM EDT
Fuck, I've got a headache now.
Link Posted: 1/6/2012 8:25:22 AM EDT
Easier to do a subtraction.

You're taking the full volume of the tank and subtracting the secant volume that's unfilled.

Start here. Now, Wikipedia suggests using the area of a sector, minus the isoscles triangle at its base. A good approach, but it forces you to calculate the angle theta, which is annoying.

If you're using a spreadsheet or other simple calculator, then no problem. Just run a couple extra formulas and be done with it.

But if you're trying to come up with a single formula suitable for field use (like a placard on the tank), it gets much trickier. There's no simple field-usable formula, and you're better off graduating the existing level indicator with volume ticks at pre-calculated intervals.
Link Posted: 1/6/2012 8:34:14 AM EDT
link

in case you missed the PM
Link Posted: 1/6/2012 8:42:38 AM EDT
[Last Edit: 1/6/2012 8:44:55 AM EDT by VBC]
Volume at depth h = Length * Area of segment
Area = rad^2*ACOS((rad-h)/rad)-(rad-h)*(2*rad*h-h^2)^0.5
Link Posted: 1/6/2012 9:05:12 AM EDT

Originally Posted By sinsir:
link

in case you missed the PM

Kickass.
Link Posted: 1/6/2012 9:20:23 AM EDT
[Last Edit: 1/6/2012 9:22:28 AM EDT by VBC]
The partially filled horizontal tank problem is a doozy but one I solved many years ago when I first started working and was fresh from college. I also derived a formula to solve the volume over partially submerged fillets in the bottom of a cylindrical tank using the Shell method from Calculus. It was very enjoyable and shit I have long forgotten how to do without some serious WD-40.
Link Posted: 1/6/2012 9:24:40 AM EDT
Change the tank from horizontal to vertical, it's easier than maths



Oops, just read the size of the tank, bring some friends
Link Posted: 1/6/2012 1:27:42 PM EDT
[Last Edit: 1/6/2012 1:32:56 PM EDT by diehippy]
worked this out with some help

volume (gal)=(length*(rad^2*ACOS((rad-inches liquid height)/rad)-((rad-inches liquid height)*SQRT(2*rad*inches liquid height-inches liquid height^2))))/231

crude copy/paste/substitution from the excel spreadsheet i'm using.


Thanks guys.
Link Posted: 1/6/2012 1:32:40 PM EDT
Go out and stick the tank...

That's what we did before Veederroot.
Link Posted: 1/6/2012 1:34:53 PM EDT

Originally Posted By NwG:
Go out and stick the tank...

That's what we did before Veederroot.

But then I couldn't appreciate the joys of middle management and numbers reporting and spreadsheets.
Link Posted: 1/6/2012 3:06:21 PM EDT
[Last Edit: 1/6/2012 3:07:47 PM EDT by tomrocks21212]
Are we to assume this is a theoretical question (like homework)?

I fully understand what you're asking...... Unfortunately, I haven't done any calculus since about 1985, and I can't even remember how to set up the equation. But yes, it's an integral.

If you have a couple of days, see if you can find a copy of "Calculus for the Practical Worker". It's straightforward enough to get you where you need to be.

Now I'm going to have to pull out my old school books.

Damn you!!!!

never mind, I see you got what you're looking for
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