(41) Bulk Cylindrical Tank

[Albert Chan]

(10-13-2018 09:37 AM)Dieter Wrote:  

(10-12-2018 10:41 PM)Thomas Klemm Wrote:  The slanted cylinder isn't a slanted cuboid. Top and bottom are not the same.
Thus we can't simply divide the whole volume by 2.

Instead this formula can be used:

\(V(x)=\left [ x(\pi-\cos^{-1}x)+\tfrac{1}{3}\sqrt{1-x^2}(2+x^2) \right ]Hr^2\)

The liquid surface is a circle with a chopped of segment. If this area is known, why is the volume not area · h/2 ?

Volume = area * h/2 is just an estimate. Actual volume require integration.

Let x = 2 h/h0 - 1, we get tan(θ) = r/1 = (r-y)/(1+x), thus y = -r x

Substitute y = -r x in equation 25-1, V_b = ∫ A_seg dx, you get this:

\(V(x)=\left [ {\pi x} / 2 + x sin^{-1}x + \sqrt{1-x^2}(2+x^2)/3 \right ]Hr^2\)

Result is in ArcSin , but is equivalent, since acos(x) = Pi/2 - asin(x)
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ArcSin form is actual better for numerical estimation.
Except for the first term, others are even function, and can approximate as this:

\(V(x) = \left [ 1.571 x+ 0.667 + x^2 - 0.094 x^4 \right ]Hr^2\)

The curve match very well to exact V(x): http://m.wolframalpha.com/input/?i=plot+...om+-1+to+1

This is calculated volume (r = 72", H=12/2=6"), exact vs approx:

Code:

Inches   Exact   Approx (gallons)
3.       16.96    16.92
6.       89.77    89.81
8.      175.09   175.13
12.     423.01   423.34