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Sofia Vasilievna
ProfessionalEngineering

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.
Anton2011-05-17 21:10:35
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 tankCreative Commons Attribution/Share-Alike License 3.0 (Unported)
 Liquid level near upper base:
 Liquid level near lower base:
 Length of partially filled tank part:
 Length of fully filled tank part:
 Liquid volume:


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

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)-\left[K-\frac{{L}tan{\alpha}}{R}\right]\arccos{\left[K-\frac{{L}tan{\alpha}}{R}\right]}+\frac{1}{3}\sqrt{1-\left[K-\frac{{L}tan{\alpha}}{R}\right]^2}\left(\left[K-\frac{{L}tan{\alpha}}{R}\right]^2+2\right) \right)

where K=1-\frac{h_{0}}{R}





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