How to Calculate the Partial Volume of a Capsule Tank

A capsule tank is commonly modeled as a cylinder with a hemisphere attached at each end. For a horizontal tank filled to depth $h$, the liquid occupies a circular segment along the cylindrical section and a spherical-cap-shaped portion across the two rounded ends.

Let $r$ be the inside radius and $L$ the length of the straight cylindrical section. The formula below gives volume for any fill depth from empty to full. If the tank is vertical, the calculation is different; that case is covered below.

Formula for a horizontal capsule tank

First calculate the area of the liquid segment in a circular cross-section of radius $r$:

$$A(h)=r^2\arccos\left(\frac{r-h}{r}\right)-(r-h)\sqrt{2rh-h^2}.$$

The two hemispherical ends together make one sphere of radius $r$. The volume of its portion below the liquid level is a spherical cap:

$$V_{\text{ends}}(h)=\pi h^2\left(r-\frac{h}{3}\right).$$

Therefore, the partial volume is

$$V(h)=L\left[r^2\arccos\left(\frac{r-h}{r}\right)-(r-h)\sqrt{2rh-h^2}\right]+\pi h^2\left(r-\frac{h}{3}\right), \qquad 0\le h\le 2r.$$

Use consistent units throughout. For example, if $r$, $L$, and $h$ are in metres, the answer is in cubic metres.

Worked example

Suppose a horizontal capsule tank has radius $r=1\text{ m}$, straight cylindrical length $L=4\text{ m}$, and liquid depth $h=0.5\text{ m}$.

The filled cross-sectional area of the cylindrical portion is

$$\begin{aligned} A(0.5)&=\arccos(0.5)-0.5\sqrt{0.75}\\ &=\frac{\pi}{3}-\frac{\sqrt{3}}{4}\\ &\approx 0.6142\text{ m}^2. \end{aligned}$$

The cylindrical portion contains approximately $4(0.6141848)=2.4567\text{ m}^3$. The end portion contains

$$V_{\text{ends}}=\pi(0.5)^2\left(1-\frac{0.5}{3}\right)=\frac{5\pi}{24}\approx0.6545\text{ m}^3.$$

So the total liquid volume is approximately

$$V(0.5)=2.4567+0.6545=3.1112\text{ m}^3.$$

The full capacity is $\pi r^2L+\frac{4}{3}\pi r^3$, which is about $16.7552\text{ m}^3$ for these dimensions. The example tank is therefore about 18.6% full.

If the tank is vertical

For a vertical capsule, the ends are at the bottom and top rather than at the sides. Let $h$ be the liquid height measured from the bottom, and keep $L$ as the straight cylindrical length. The total height is $L+2r$. The volume is piecewise:

$$V(h)= \begin{cases} \pi h^2\left(r-\frac{h}{3}\right), & 0\le h\le r,\\[4pt] \frac{2}{3}\pi r^3+\pi r^2(h-r), & r<h\le r+L,\\[4pt] \frac{2}{3}\pi r^3+\pi r^2L+\pi y\left(r^2-\frac{y^2}{3}\right), & r+L<h\le 2r+L, \end{cases}$$

where $y=h-(r+L)$ in the final case. The three cases describe filling the lower hemisphere, then the cylindrical section, then the upper hemisphere.

Common mistakes

• Using the overall tank length as $L$. If the measured tip-to-tip length is $D$, then the straight cylindrical length is $L=D-2r$.
• Mixing units. Convert all lengths to the same unit before calculating; otherwise the volume will be inconsistent.
• Measuring depth from the center. In the horizontal formula, $h$ is measured upward from the inside bottom, so it ranges from $0$ to $2r$.
• Using a horizontal formula for a vertical tank. The amount of liquid in each end changes differently with height, so use the piecewise expression for a vertical orientation.

As a useful check, at half depth, $h=r$, a horizontal capsule is exactly half full by symmetry. At $h=0$ the formula gives zero, and at $h=2r$ it gives the full capacity $\pi r^2L+\frac{4}{3}\pi r^3$.