homechevron_rightProfessionalchevron_rightEngineering

# Tilted cylindrical tank volume

Calculates volume of cylindrical tank tilted at an angle. To perform calculation you must provide the tank size, angle and liquid level.

## This page exists due to the efforts of the following people:

 AntonArticle : Tilted cylindrical tank volume - Author, Translator ru - enCalculator : Volume of liquid in tilted cylindrical tank - Author, Translator ru - en

To perform calculation you must provide the tank size, angle and liquid level.

## Liquid level measurement

You must measure liquid level at middle line of the tank perpendicular to the bottom of the tank. (see picture). You may measure liquid level at any distance from on of the base (If you do, you must enter the distance in a special parameter).
Alternatively you may tilt the tank to make the zero liquid level at the top base, in this case you must measure tilt angle only.

You may find calculation details and formulas below the calculator.

### Volume of liquid in tilted cylindrical tank

Distance between liquid level measurement place and nearest base. (0 if level is measured right near the base).

Digits after the decimal point: 2
Liquid level near upper base

Liquid level near lower base

Length of partially filled tank part

Length of fully filled tank part

Liquid volume

I could not find a ready solution to calculate liquid volume in a tilted cylinder, so I derived the formula in this way:

## Partially filled tilted tank volume formula

$V = \frac{R^2}{2}\int_{\small0}^L {\theta(x)-\sin{\theta(x)}dx$
where $\theta(x)$- how the segment angle depends from the cylinder length x,
It can be derived as:
$\theta(x)=2\arccos{\frac{R-h}{R}}=2\arccos{\frac{R-(x\tan{\alpha}+h_{0})}{R}}$
where
a - tilt angle,
h0 - liquid level at the upper base

If we substitute this expression in the formula we get:
$V = R^2\int_{\small0}^L {\arccos{u}-u\sqrt{1-u^2}dx=-\frac{R^3}{\tan{\alpha}}\int_{\small{u_0}}^{u_L} {\arccos{u}-u\sqrt{1-u^2}du$
where $u=\frac{R-(x\tan{\alpha}+h_{0})}{R}$

If we take the integral we get the solution:
$V = -\frac{R^3}{\tan{\alpha}}\left( u\arccos{u}-\frac{1}{3}\sqrt{1-u^2}(u^2+2)\right)\large|_{\small{u_0}}^{u_L}$$= \frac{R^3}{\tan{\alpha}}\left( K\arccos{K}-\frac{1}{3}\sqrt{1-K^2}(K^2+2)-{C}\arccos{C}+\frac{1}{3}\sqrt{1-C^2}\left(C^2+2\right) \right)$

where
$K=1-\frac{h_{0}}{R}$,
$C=K-\frac{{L}tan{\alpha}}{R}$

## Determine tank part length filled with liquid

The above formulas can be used for tilted tank volume calculation with these assumptions:

• Both bases partially filled with liquid.
• The liquid level h0 is measured directly on the upper base.
• No part of the tank is dry or fully filled.

But the calculator accepts the liquid level measured on some distance near upper or lower base. Some part of tank can be dry or fully filled.

To calculate liquid level directly on the upper base hu use formulas:
$h_u=h_{lu}-(L_c-l_u){\tan{\alpha}}$
where hlu - liquid level measured on the distance lu from the upper base, Lc - length of the tank
$h_u=h_{ll}-l_l{\tan{\alpha}}$
where hlu - liquid level measured on the distance ll from the lower base.
If the hu is equal or above zero we assume h0=hu, and Lf = Lc.

### Empty tank part

Otherwise, the hu can be negative. That means some tank part is empty. In this case assume h0=0 and calculate remained (filled) part Lf using formula:
$L_f=L_c+h_u{\tan{\alpha}}$
where Lc is cylinder length.

### Fully filled part

The liquid level directly on the lower base h1 can be determined as:
$h_1=h_0+L_f{\tan{\alpha}}$
If calculated h1 value is greater than tank diameter, some part of our cylinder is fully filled with liquid. So we need to calculate fully filled part length as:
$L_t=\frac{h_1-2{R}}{\tan{\alpha}}$
Fully filled part volume calculation is trivial see Cylinder

After these calculations you may substitute partially filled tank length $L=L_f-L_t$ and liquid level h0 in the first section formulas to calculate partially filled tilted tank part volume.