Outflow velocity

Graphical representation of Torricelli's theorem

The outflow velocity is the velocity at which a liquid or gaseous body of very low viscosity (e.g. water) flows out of an opening in the vessel containing it. Since during the outflow of a certain quantity of liquid an equal amount of liquid must always sink from the surface to the level of the opening, the outflow velocity is equal to the velocity that a body would attain if it fell from the liquid level to the outflow opening ( Torricelli's theorem, see also Bernoulli's equation ).

If v is the outflow velocity, h is the vertical depth of the opening under the surface of the liquid ( pressure head ) and g is the acceleration due to gravity ( g  = 9.81 ms ⁻² ), then . ${\displaystyle v={\sqrt {2gh}}}$

It therefore depends only on the pressure head, but not on the density of the liquid, so that z. B. at the same pressure level water and mercury flow out at the same speed.

Since the pressure in a liquid acts equally strong in all directions, it is irrelevant for the outflow velocity whether the opening is in the bottom or in a side wall of the vessel, whether the outflowing jet is directed downwards, sideways or upwards ( fountain ) .

If the outflow were cylindrical , the volume of liquid outflow per unit of time could easily be calculated by multiplying the outflow velocity by the area of ​​the opening. The jet is cylindrical in the area of ​​the outflow opening when the outflow opening is cylindrical. However, the Bernoulli equation only applies in a steady flow of a frictionless , incompressible fluid . Therefore, the outflow volume for real liquids has to be calculated with a correction factor. At some distance from the outflow opening , the jet is no longer cylindrical, but instead contracts, so that its cross-section at a short distance from the opening is only about 61 percent of that of the opening. In order to obtain the outflow quantity for real liquids, the "theoretical outflow quantity" calculated above must be multiplied by 0.6. This contraction of the jet ( Latin contractio venae ) is mainly due to the fact that the liquid particles in the interior of the vessel converge towards the opening from all sides and therefore arrive at the edges of the discharge opening with a laterally directed velocity.

Everything so far only applies to openings in thin vessel walls. When the liquid adheres to the walls of the tube and fills it completely, the outflow volume is increased by short cylindrical or outwardly conical extension tubes , but the outflow speed is reduced - by about half. Openings in a thick wall act like extension tubes.

Torricell's law also applies to the outflow velocity of ideal gases if the pressure height h is understood to be the height of a gas column with the density of the outflowing gas. If one denotes with h ' the overpressure of the enclosed gas, measured manometrically as the height of a mercury column , with s' the specific gravity of the mercury, with s that of the gas (both related to water as a unit), then the pressure height h corresponds to the calculation is to be brought to the column of mercury h ' as s' to s ; so it is

${\displaystyle h={\frac {h's'}{s}}}$

and

${\displaystyle v={\sqrt {\frac {2gh's'}{s}}}}$

This results in the law established by Thomas Graham that the outflow velocities of different gases at the same pressure are inversely proportional to the square roots of their specific weights . Since z. For example, if the density of hydrogen gas is only 1/16 of the density of oxygen gas, the gas flows out under the same pressure four times as fast as this.

From this, Robert Wilhelm Bunsen derived a method for determining the specific weights of the gases.

Quasi-stationary consideration of Torricelli's law of discharge

Torricelli's outflow law is obtained from Bernoulli's energy equation . The simplified description of the outflow velocity as a function of the filling level , considering that the outflow diameter is much smaller than the container diameter, can be given as follows: ${\displaystyle v_{2}}$ ${\displaystyle h}$

${\displaystyle v_{2}={\sqrt {2gh}}\qquad \qquad (1)}$

Where is the acceleration due to gravity ( ). According to the continuity equation of the flow dynamics of incompressible fluids , the volume flow is constant. Accordingly, the following formula applies: ${\displaystyle g}$ ${\displaystyle g=9{,}81\,{\text{m}}/{\text{s}}^{2}}$ ${\displaystyle {\dot {V}}}$

${\displaystyle {\dot {V}}=v_{1}A_{1}=v_{2}A_{2}\quad \rightarrow \quad {\frac {v_{1}}{v_{2}}}={\frac {A_{2}}{A_{1}}}={\text{const.}}\qquad \qquad (2)}$

Where is the sinking speed (i.e. the negative speed) of the water level . Accordingly, the formula can be reformulated with (2) to: ${\displaystyle v_{1}}$ ${\displaystyle h}$

${\displaystyle v_{1}={\frac {A_{2}}{A_{1}}}{\sqrt {2gh}}\qquad \qquad (3)}$

Since the filling level is now the negative speed, this can be shown with the first derivative according to time. This gives us a non-linear differential equation of the first order, which now describes the filling level of the container over time. The differential equation can be given as follows: ${\displaystyle v_{1}}$

${\displaystyle {\frac {\mathrm {d} h}{\mathrm {d} t}}=c{\sqrt {h}}\quad {\text{mit}}\quad c=-{\frac {A_{2}}{A_{1}}}{\sqrt {2g}}\qquad \qquad (4)}$

This is a separable differential equation , which is why a closed solution can be given:

${\displaystyle \int {\frac {\mathrm {d} h}{\sqrt {h}}}=c\int \mathrm {d} t+C_{1}\quad \rightarrow \quad h(t)={\frac {1}{4}}\left(ct+C_{1}\right)^{2}\quad {\text{mit}}\quad C_{1}\in \mathbb {R} \qquad \qquad (5)}$

This is primarily an arbitrary constant, which can, however, be determined by solving the initial value problem . In other words, at the time the container has a fill level of . By inserting these initial values ​​into the solution function, the end result is: ${\displaystyle C_{1}}$ ${\displaystyle t=0\,\mathrm {s} }$ ${\displaystyle h(0)=h_{0}}$

Example of the course of the function h (t)

${\displaystyle h(t)=\left({\sqrt {h_{0}}}-t{\frac {A_{2}}{A_{1}}}{\sqrt {\frac {g}{2}}}\right)^{2}={\frac {1}{2}}\left({\frac {A_{2}}{A_{1}}}\right)^{2}gt^{2}-{\sqrt {2gh_{0}}}{\frac {A_{2}}{A_{1}}}t+h_{0}\qquad \qquad (6)}$

Viewed graphically, this is a parabola that opens upwards , the minimum of which lies on the abscissa and is therefore a double zero . Therefore we can now determine the point in time at which the container is empty by setting the function obtained to zero. We obtain:

${\displaystyle t_{\text{leer}}={\frac {A_{1}}{A_{2}}}{\sqrt {\frac {2h_{0}}{g}}}\qquad \qquad (7)}$

Alternative approach

Alternatively, the outflow velocity results from the conservation of energy of potential and kinetic, specific energy.

${\displaystyle e_{\text{kin}}={\frac {dE_{\text{kin}}}{dV}}={\frac {1}{2}}\rho v_{2}^{2}=\rho gh={\frac {dE_{\text{pot}}}{dV}}=e_{\text{pot}}\quad \rightarrow v_{2}={\sqrt {2gh}}\qquad (8)}$

Using the continuity equation (2), equations (3) and (4) are again obtained. By deriving equation (4) again with respect to time , it is possible to convert the non-linear velocity differential equation into a linear acceleration differential equation . ${\displaystyle t}$

${\displaystyle {\frac {{\text{d}}^{2}h}{{\text{d}}t^{2}}}=-{\frac {A_{2}}{A_{1}}}{\sqrt {2g}}{\frac {{\text{d}}{\sqrt {h}}}{{\text{d}}t}}=-{\frac {A_{2}}{A_{1}}}{\sqrt {2g}}{\frac {{\text{d}}{\sqrt {h}}}{{\text{d}}h}}{\frac {{\text{d}}h}{{\text{d}}t}}=\left({\frac {A_{2}}{A_{1}}}\right)^{2}g\quad \rightarrow \quad {\frac {{\text{d}}^{2}h}{{\text{d}}t^{2}}}=\left({\frac {A_{2}}{A_{1}}}\right)^{2}g\qquad (9)}$

This acceleration differential equation (9) can be solved by double integration after the time t, which in turn results in equation (6). The initial values ​​apply:

${\displaystyle {\frac {{\text{d}}h}{{\text{d}}t}}(t=0)=-{\frac {A_{2}}{A_{1}}}{\sqrt {2gh_{0}}},\qquad h(t=0)=h_{0}\qquad (10)}$

Discharge coefficient

In order to get a better approximation of the actually measured volume flow, a discharge coefficient is used in practice : ${\displaystyle \mu }$

${\displaystyle {\dot {V}}_{\text{real}}=\mu \cdot {\dot {V}}_{\text{ideal}}}$

The outflow coefficient takes into account both the reduction in the outflow velocity due to the viscous behavior of the liquid ("velocity coefficient") and the decrease in the effective outflow cross-section due to the vena contracta ("contraction coefficient"). For liquids of low viscosity (such as water) flowing out of a round hole in a tank, the discharge coefficient is in the order of magnitude of 0.65. By using rounded pipe sockets, the discharge coefficient can be increased to over 0.9. For rectangular openings, the discharge coefficient is between 0.44 and 0.67, depending on the height-to-width ratio.

In addition, the discharge coefficient depends on whether the flow is laminar or turbulent. This can be taken into account for outflow processes from a round hole using the following formula:

• ${\displaystyle 0{,}59+{\frac {5{,}5}{\sqrt {\text{Re}}}}}$with the Reynolds number . ${\displaystyle {\text{Re}}}$

Individual evidence

1. ↑ tec-science: Outflow of liquids (Torricelli's theorem). In: tec-science. November 21, 2019, accessed December 8, 2019 (German). 
2. ↑ Hydraulics 9: outflow and drainage times. Accessed December 8, 2019 (German).