(41) Bulk Cylindrical Tank

[Thomas Klemm]

(10-11-2018 07:38 PM)Dieter Wrote:  The volume calculation is even easier, the formula is quite short.

We can use this equation for the ellipse:

\(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\)

When we set \(a=r\) and \(b=v_0\) this becomes:

\(\frac{x^2}{r^2}+\frac{y^2}{v_0^2}=1\)

From which we conclude:

\(x^2=r^2\left ( 1-\frac{y^2}{v_0^2} \right )\)

Then \(\pi x^2\) is the area of the surface of the water.
We can integrate that to get the volume of the water in the dome:

\(
\begin{align*}
V_d(v) &= \int_{0}^{v}\pi x^2dy \\
&= \pi \int_{0}^{v}r^2\left ( 1-\frac{y^2}{v_0^2} \right )dy \\
&= \pi r^2 \left ( y-\frac{y^3}{3v_0^2} \right )\bigg|_{y=0}^{y=v} \\
&= \pi r^2 \left ( v-\frac{v^3}{3v_0^2} \right ) \\
&= \pi r^2 v \left ( 1-\tfrac{1}{3}(\frac{v}{v_0})^2 \right )
\end{align*}
\)

In the given example \(r=72"\), \(h_0=12"\), \(d=180"\) and \(v_0=36"\)
For the JUICE LEVEL \(H_t=195"\) we get \(v=H_t-d-h_0=195"-180"-12"=3"\).
Thus: \(V_d(3)=211.0171\ gal\)
The rest of the tank is \(\pi r^2(d+\tfrac{h_0}{2})=13113.4157\ gal\).
Thus we end up with \(13324.4328 \ gal\).

For the JUICE LEVEL \(H_t=198"\) we get \(v=H_t-d-h_0=198"-180"-12"=6"\).
Thus: \(V_d(6)=419.0966\ gal\)
Here we end with \(13532.5124\ gal\).

Both results agree with Dieter's previous post#14.

Here's a program for the HP-42S:

Code:

00 { 38-Byte Prgm }
01▸LBL "DOME"
02 1
03 RCL ST Y
04 RCL÷ "v0"
05 X↑2
06 3
07 ÷
08 -
09 ×
10 RCL "r"
11 X↑2
12 ×
13 PI
14 ×
15 RCL÷ "in3/gal"
16 END

Cheers
Thomas