Horizontal tank with hemispherical ends depth to capacity calculation

Question

I am trying to find an accurate way of calculating the capacity of an underground tank at a given depth. The tank manufacturer has provided a strapping table for the tank which tells me the capacity at various depths.

Gauge Depth (cm)  /  Capacity (Liters)
2cm = 29.8 liters
12cm = 240.4 liters
...
66cm = 2372.8 liters
118cm = 4008.4 liters

However, I would like to find an equation for calculating the volume of liquid in the tank based on the liquid depth level.

They have provided the following dimensions for the tank:

1219mm Diameter
3785mm Long
3978 Liters Capacity

The tank is horizontal with hemispherical ends. It is safe to assume that the radius of the hemispherical ends is 1/2 of the diameter of the tank. Any help or direction would be greatly appreciated!

Answer 1

There are three parts to the tank: the two hemispherical ends and the middle cylinder. The two hemispherical ends add up to one whole sphere, so we'll just consider a sphere and a cylinder. Just to check the data you've given: at 1219mm diameter, the sphere has volume roughly 948.44 liters. The cylinder has diameter 1219mm and height (3785-1219)=2566mm. The volume is roughly then 2994.7 liters. So I'm guessing that there's either a typo on your data sheet (either the total tank capacity is closer 3948 liters, or the total length of the tank is closer to 3815mm), or a mis-description: the ends are not capped by exact hemispheres.

Never mind that, since we are on a mathematics site and who cares about plugging in real numbers /end_sarcasm

Assume your tank is composed of two hemispherical ends of radius $R$, and a central cylinder of radius $R$ and length $L$. The diameter would be $2R$ and total length $L + 2R$. We want to calculate volume as a function of water depth.

Since your tank is lying horizontally, the height of water, which we will call $h$, should be the same inside the cyclinder portion and inside the spherical portions. So we will compute the total volume as a sum of the volume inside the sphere (remember: two hemispheres make one whole sphere) and the volume inside the cylinder.

$$ V_{tot} = V_s + V_c $$

If you know calculus, you can compute the volume of the spherical cap of height $h$ by evaluating an integral. But the formula is well-known:

$$ V_s = \frac{\pi h^2}{3}(3R - h) $$

For the cylinder, the volume at height $h$ is equal to the length $L$ multiplied by the area of the circular segment of height $h$:

$$ V_c = L \times A = L \times \left[ R^2 \cos^{-1}\left(\frac{R-h}{R}\right) - (R - h) \sqrt{2Rh - h^2}\right] $$

with the formula for $A$ derivable by elementary geometry/trigonometry, or again through the evaluation of an integral. So putting it all together you have that

$$ V_{tot} = \frac{\pi h^2}{3}(3R - h) + L \times \left[ R^2 \cos^{-1}\left(\frac{R-h}{R}\right) - (R - h) \sqrt{2Rh - h^2}\right] $$

which gives

2 cm  => 11.4 L
5 cm  => 46.4 L
10 cm => 134.6 L 
20 cm => 388.9 L 
30 cm => 716.7 L 
40 cm => 1094.5 L 
50 cm => 1504.5 L 
60 cm => 1930.8 L 
61 cm (just a tiny bit over half full) => 1971.6 L

Answer 2

You can divide the tank into a sphere (formed from the two hemispherical ends) and a cylinder. Wikipedia Spherical cap gives the volume as $\frac{\pi h^2}{3}(3r-h)$ where $r$ is the radius of the sphere and $h$ the depth. Then Circular Segment gives the area of the segment as $\frac{r^2}{2}(\theta - \sin \theta )$ where $\theta=\arccos(\frac{r-h}{r})$. Multiply this last by the length, add the spherical cap, and there you are.

Answer 3

Let $R$ be the radius of the tank and let the end-to-end length be $L + 2R$. At each depth $x$ above the center of the tank (where $-R \le x \le R$), the tank's cross-section is a rectangle of length $L$ and width $2r(x)$, plus semicircular caps of radius $r(x)$. The radius satisfies $r(x)^2 + x^2 = R^2$. So the volume at depth $d$ is $$ \begin{eqnarray} V(d) &=& \int_{-R}^{-R+d} V'(x) dx \\ &=& \int_{-R}^{-R+d} \left(2 L r(x) + \pi r(x)^2\right) dx \\ &=& \int_{-R}^{-R+d} \left(2 L \sqrt{R^2 - x^2} + \pi R^2 - \pi x^2\right) dx \\ &=& 2 L R^2 \tan^{-1}\left(\sqrt{\frac{d}{2R-d}}\right) + L(d-R)\sqrt{d(2R-d)} + \pi\left(Rd^2 - \frac{1}{3}d^3\right). \end{eqnarray} $$ This fits your numbers reasonably well for $R=609.5\text{mm}$ and $L=2566.0\text{mm}$; it could be improved by knowing the thickness of the tank walls and whether the shape deviates from a perfect cylinder plus hemispherical caps (e.g., is it flattened at the bottom)?

Answer 4

One thing is clear: you can't get 4008.4 litres into a tank with a capacity of 3978 litres. So you are right not to trust the manufacturer's dipstick.