Capacity per Foot of Height - Cylindrical Tank

[Guest]

Here is another tank problem:

A cylindrical tank is made of 1.5 in. material and has an outside circumference of 17 ft. 3.5 in. Determine capacity (gallons) per ft. of height.

My attempt:

Circumference - 17. 292 ft.

Thickness - .125 ft.

External diameter - 17.292 / 3.1416 = 5.504 = 5.50 ft.

Internal diameter - 5.50 - .125 x 2 = 5.25 ft.

Area - 3.1416 x 5.25 x 5.25 = 86.59 = 87 cu. ft. (for 1 ft. of depth or height).

87 x 7.48 (cu. ft. to gal.) = 725.56 = 726 gal.

Where did I make the error(s) ?

[Guest]

is the tank sitting vertically ..... with a circular base and vertical height with same cross-section throughout height.

OR

is the tank lying horizontal length ways, with vertical circular ends and same cross-section throughout its horizontal length.

[Guest]

The problem does not indicate the position of the tank. Please calculate for both.

[Guest]

The tank sitting vertically would be fine.

[Guest]

You ar OK up to this point............For a vertical cylindrical tank sitting on its circular base.
............................
Area - 3.1416 x 5.25 x 5.25 = 86.59 = 87 cu. ft. (for 1 ft. of depth or height).

87 x 7.48 (cu. ft. to gal.) = 725.56 = 726 gal.

Where did I make the error(s) ?
...............................

Area of a circle is Pi x D^2 /4 OR Pi x R^2 because R = D/2 so R^2 = (D/2)^2 = D^2 / 4

Area - 3.1416 x 5.25 x 5.25 / 4 = 21.647 sq ft.

So volume of 1 ft depth is 21.647 x 1 = 21.647 cu ft.

So gal per foot = 21.647 x 7.48 = 161.92 gal OR 162 gal to nearest gal.

[Guest]

Thanks - I see the error.

A question:

If the tank was horizontal, would the calculations be similar to the railroad car or different ? Just for info.

[Guest]

Yes. In the railroad question we calculated the gallons fo a particular depth of oil. The relationship of gallons per foot of depth would not linear (not a constant gal per foot) throughout the depth of oil from empty to full because the tank is narrow at the bottom and top of the curve and wide in the middle.
To simplify the problem you could calculate the gallons at each foot depth and plot a graph, then use that to determine the contents at any level as required.

[Guest]

Thanks for additional info - I will do that.

[Guest]

I have a follow-up question:

I was reading more about circles and came across the following - Diameter = .3184 x C.

What is the .3184 ? Thanks.

[Guest]

As a matter of curiosity..... Working in radians instead of degrees............

Radius = 2.5 ft ........... Oil depth = 1.25 ft

Centre to oil dist. = (2.5 - 1.25) = 1.25 ft ...... all as before .........

Sector half angle = ArcCos(0.5) = 1.047 radians

Whole Sector Area = Radius^2 x halfAngle = 2.5^2 x 1.047 = 6.544 sq ft

Whole Sector Triangle bit Area = R x Cos(1.047) x R x Sin(1.047) = 2.5 x 0.5 x 2.5 x 0.866 = 2.706 sq ft

The rest is as before..............

[Guest]

................................
I was reading more about circles and came across the following - Diameter = .3184 x C.

What is the .3184 ? Thanks.
...........................

Diameter is the circumference of a circle divided by Pi

This is same as Diameter = C / 3.142 OR C x ( 1 / 3.142) OR ( 1 / 3.142) x C

If we work out what ( 1 / 3.142) is it is equal to 0.3184

So Diameter is equal to 0.3184 x C

[Guest]

As a matter of curiosity..... Working in radians instead of degrees............
Sorry.......I posted this to the wrong thread..........I have posted it to the railroad tank problem now

[Guest]

" As a matter of curiosity..... Working in radians instead of degrees............
Sorry.......I posted this to the wrong thread..........I have posted it to the railroad tank problem now " No problem.


"Diameter is the circumference of a circle divided by Pi

This is same as Diameter = C / 3.142 OR C x ( 1 / 3.142) OR ( 1 / 3.142) x C

If we work out what ( 1 / 3.142) is it is equal to 0.3184 " Thanks.