Differential Equations

1st Order Equations

Fluid Flow from a Vessel

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Problems 1-2

Torricelli’s Law

The Italian scientist Evangelista Torricelli investigating fluid flow experimentally found in \(1643\) that the velocity of fluid flowing out through a small hole at the bottom of an open tank (Figure \(1\)) is given by the formula:

\[v = \sqrt {2gh} ,\]

where \(h\) is the height of fluid above the opening, \(g\) is the gravitational acceleration.

Torricelli’s Law

Figure 1.

The same formula describes the velocity of a free solid particle falling from the height \(h\) in the Earth’s gravitational field in a vacuum. Actually this formula is not quite precise. In fact, the velocity of fluid depends on the shape and size of the opening, the fluid viscosity and flowing mode. Therefore, the Torricelli’s formula is often written with an additional factor \(\varphi:\)

\[v = \varphi\sqrt {2gh} ,\]

where the coefficient \(\varphi\) is close to \(1.\) The values of the coefficient \(\varphi\) for openings of different shape and size are given in hydraulic handbooks.

Fluid Flowing out of a Thin Pipe

Figure 2.

Fluid Flowing out of a Thin Pipe

Fluid flow through a thin long pipe (Figure \(2\)) has a number of features. The various capillary effects caused by the surface tension and wetting due to contact with the walls of the vessel play an important role.

The velocity of fluid flowing out from the capillary pipes is approximately proportional to the height of the fluid above the opening, i.e.:

\[v = kh,\]

where \(k\) is a certain constant depending on the fluid viscosity, geometry and material of the pipe.

Further we describe the fluid flow using differential equations for both types of the vessels.

Differential Equation of Fluid Flowing Out

We can derive the differential equation considering fluid balance in a vessel. Take as an example a cylindrical vessel with a broad base of radius \(R.\) Suppose that fluid flows out through a small opening of radius \(a\) at the bottom of the vessel (Figure \(3\)).

The fluid velocity is described by the Torricelli’s formula:

\[v = \sqrt {2gz} ,\]

where \(z\) is a the height of the fluid above the opening. Then the fluid flow is given by

\[q = – \pi {a^2}\sqrt {2gz} .\]
Fluid flowing out through a small opening of radius a at the bottom of a vessel

Figure 3.

Here \(\pi {a^2}\) corresponds to the area of the opening through which the fluid flows out, and the “minus” sign means that the height of the fluid decreases when it flows out of the reservoir.

The fluid balance equation in the reservoir is written as follows:

\[\frac{{dV}}{{dt}} = q.\]

Since the volume change \(dV\) can be expressed as

\[dV = S\left( z \right)dz,\]

we have the following differential equation:

\[\frac{{S\left( z \right)dz}}{{dt}} = q\left( z \right).\]

Putting the function \(q\left( z \right)\) into this equation gives:

\[\frac{{S\left( z \right)dz}}{{dt}} = – \pi {a^2}\sqrt {2gz} .\]

The cross section \({S\left( z \right)}\) of the cylindrical vessel does not depend on the height \(z\) and is given by

\[S\left( z \right) = \pi {R^2},\]

where \(R\) is the base radius of the cylinder. Then

\cancel{\pi} {R^2}\frac{{dz}}{{dt}} = – \cancel{\pi} {a^2}\sqrt {2gz} .

As a result, we obtain the separable equation:

\[\frac{{dz}}{{\sqrt z }} = – \frac{{{a^2}}}{{{R^2}}}\sqrt {2g} dt.\]

Now we integrate this equation assuming that the initial height of the fluid is \(H\) and the fluid level decreases to \(0\) for the time \(T.\)

{\int\limits_H^0 {\frac{{dz}}{{\sqrt z }}} = – \int\limits_0^T {\frac{{{a^2}}}{{{R^2}}}\sqrt {2g} dt} ,\;\;}\Rightarrow
{{2\left[ {\left. {\left( {\sqrt z } \right)} \right|_H^0} \right] }={ – \frac{{{a^2}}}{{{R^2}}}\sqrt {2g} \left[ {\left. {\left( t \right)} \right|_0^T} \right],\;\;}}\Rightarrow
{2\sqrt H = \frac{{{a^2}}}{{{R^2}}}\sqrt {2g} T,\;\;}\Rightarrow
{\sqrt {2H} = \frac{{{a^2}}}{{{R^2}}}\sqrt g T.}

It follows from here that the time \(T\) is defined by the expression:

\[T = \frac{{{R^2}}}{{{a^2}}}\sqrt {\frac{{2H}}{g}} .\]
Fluid flowing time depending on the height of the fluid

Figure 4.

Interestingly, that the resulting formula for the case \(a = R\) (when the cross sections of the opening and the cylinder are equal) is transformed into the known formula \(T = \sqrt {\large\frac{{2H}}{g}\normalsize} \) that determines the fall time of a solid particle from the height \(H.\)

The dependence of the time \(T\) on the height \(H\) is schematically shown in Figure \(4.\) Similarly, we can describe fluid flowing from the vessels of other shape.

Solved Problems

Click on problem description to see solution.

 Example 1

Derive the differential equation of fluid leakage from a conical vessel and determine the total flow time \(T.\) The upper base radius of the conical vessel is \(R,\) the lower base radius is \(a.\) The initial height of the fluid is \(H\) (Figure \(5\text{).}\)

 Example 2

Investigate the fluid flowing out of a thin pipe of the radius \(R\) and height \(H\) assuming that the pipe is completely filled.

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Problems 1-2