Partially Filled Horizontal Tank Volume with Ellipsoidal Heads

\[V=L\left(\frac{D^2}{4}\arccos\left(\frac{D-2h}{D}\right)-\left(\frac{D}{2}-h\right)\sqrt{Dh-h^2}\right)+\frac{\pi Hh^2(3D-2h)}{3D}\]

Variables

Vliquid volume in the partially filled tank (m³)

Dinternal tank diameter (m)

Linternal cylindrical shell length (m)

Hdepth of one ellipsoidal head (m)

hliquid depth measured from tank bottom (m)

Description

What is this formula?

Partially Filled Horizontal Tank Volume with Ellipsoidal Heads calculates the liquid volume inside a horizontal cylindrical tank with two ellipsoidal heads.

The formula combines the circular-segment volume of the cylindrical shell with the partial ellipsoidal volume contributed by both heads.

When to use it

Use this formula when the tank is horizontal, partially filled, and has two identical ellipsoidal heads with known head depth H.

Example

A horizontal tank has:

D = 2.0 m

L = 10.0 m

H = 0.60 m

h = 0.80 m

Formula:

V = L × [(D²/4) × acos((D−2h)/D) − ((D/2−h) × √(Dh−h²))] + πHh²(3D−2h)/(3D)

Substitution:

V = 10 × [(2²/4) × acos((2−2×0.8)/2) − ((2/2−0.8) × √(2×0.8−0.8²))] + π×0.60×0.8²×(3×2−2×0.8)/(3×2)

Result:

V = 12.62 m³

The partially filled tank contains approximately 12.62 m³ of liquid.

Applications

- Horizontal tank gauging

- Ellipsoidal head tank inventory

- Chemical storage tanks

- Petroleum process vessels

- Pressure vessel volume estimation

- Industrial liquid level calculations

Note

This formula assumes two identical ellipsoidal heads and a uniform liquid level along the tank.

It is valid for liquid depths from 0 to D.

If H = D/4, the formula reduces to the common 2:1 elliptical head case.