(41) Bulk Cylindrical Tank

[Albert Chan]

(10-13-2018 06:40 PM)Dieter Wrote:  So the calculation is quite easy and straightforward. Which brings us back to the question whether
an elliptical dome is realistic or not, better or worse than calculating a radius for a shperical dome top.

I like the elliptical dome approach, but I also like graphically matching dome volume.
So, I propose combining them ...

Let unit of 1" cyclinder volume = "slice = Pi * 72² * 1 = 5184 Pi in³ = 70.50 gallon
Assuming Kimball graphically fitted R were correct:

V_dome = volume of cap with R = 78, h = 78 - sqrt(78² - 72²) = 48
= Pi h^2 (R - h/3)
= 142848 Pi in³ (= 1943 gallon)
= 27.56 "slice

Since v0 = 36", ratio of 27.56/36 = 77% suggest the dome is very flat.
Even flatter than a elliptical dome (ratio = 2/3 ~ 67%)

To keep tank total height and total volume the same, adjust dome dimension.
Solve for a new v0, (throw away old v0 of 36"):

2/3 v0 = 27.56" + (v0 - 36")
v0 = 3 * (36" - 27.56") = 25.33"

Original tank: height = 12 + 180 + 36 = 228", volume = 6 + 180 + 27.56 = 213.56 "slice = 15056 gallon
Adjusted tank: height = 12 + 190.67 + 25.33 = 228", volume = 6 + 190.67 + 2/3*25.33 = 213.56 "slice

Example: liquid level of 220" (almost full)
v = 220 - 12 - 190.67 = 17.33", effective height = v*(1 - (v/v0)²/3) = 14.63"
Volume = 6 + 190.67 + 14.63 = 211.30 "slice = 14897 gallon