Bernoulli's law
The equation of continuity connects cross-section area and flow speed for ideal fluids, but yields no information on variations in pressure. Bernoulli's law, on the other hand, will tell us how flow speed and height influence the pressure.
Bernoulli’s law
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Bernoulli’s law states that for an ideal fluid, the following relation must hold anywhere in the flow: \[p + \varrho g h + \frac12 \varrho v^2 = \mathrm{const},\] where \(p\) is the pressure, \(v\) denotes the flow speed, \(\varrho\) stands for the density of the fluid and \(h\) is the height above a chosen reference level. Bernoulli's law is in fact a special form of the conservation of mechanical energy. The hydrostatic pressure term \(\varrho g h,\) which we have already seen, represents the potential energy of a unit volume, whilst the term \(\frac12 \varrho v^2,\) which we call dynamic pressure, is the kinetic energy per unit volume. The pressure \(p,\) which we call here the static pressure, to distinguish it from the dynamic and hydrostatic pressure, can be interpreted in this framework as the external work done on the fluid per unit volume. With these, we can see that Bernoulli's law is equivalent to saying that the mechanical energy change per unit volume plus the external work done on the fluid per unit volume is the same everywhere in the flow, which follows from the conservation of mechanical energy if we divide by the volume. |
Example: syringe
We hold a syringe in a vertical position. The diameter of the piston is \(d_1 = 1.5\,\mathrm{cm},\) whilst that of the opening is \(d_2 = 1\,\mathrm{mm}.\) The vertical distance between the piston and the opening is \(h = 10\,\mathrm{cm}.\) The syringe is filled with water. With what pressure \(p\) do we have to push the piston for the water to start squirting at a speed of \(v_2 = 2\,\mathrm{m} /\mathrm{s}\)?
This problem requires us to apply Bernoulli's law. How? We shall compare the static pressure, the dynamic pressure and the hydrostatic pressure at two locations: at the piston and at the opening. The static pressure at the opening is the atmospheric pressure \(p_2 = p_0 = 10^5\,\mathrm{Pa}.\) The static pressure \(p_1\) on the piston is the atmospheric pressure plus the extra pressure we need to exert: \(p_1 = p_0 + p.\) In order to apply Bernoulli's law, we need to obtain the values of the dynamic pressure and the hydrostatic pressure.
The hydrostatic pressure at the piston: \(\varrho g \cdot 0 = 0,\) since we can put our reference level here.
The hydrostatic pressure at the opening: \(\varrho g h.\)
The dynamic pressure at the opening is \(\frac12 \varrho v_2^2.\)
We have no direct information on the speed at the piston, however. What can we do? Since it is an ideal fluid, we can still apply the equation of continuity. We have to determine the cross-section areas first:
At the piston: \(A_1 = r_1^2 \pi = \left(\frac{d_1}{2}\right)^2 \pi;\) at the opening: \(A_2 = r_2^2 \pi = \left(\frac{d_2}{2}\right)^2 \pi.\)
Applying the equation of continuity:
\[A_1 v_1 = A_2 v_2,\]
\[v_1 = \frac{A_2}{A_1} v_2 = \frac{d_2^2 \pi / 4}{d_1^2 \pi / 4} v_2 = \left(\frac{d_2}{d_1}\right)^2 v_2 = \left(\frac{1\,\mathrm{mm}}{15\,\mathrm{mm}}\right)^2 v_2 = \frac{v_2}{225}.\]
Thus the dynamic pressure at the piston is \(\frac12 \varrho v_1^2 = \frac12 \varrho \frac{v_2^2}{225^5} = \frac{1}{50625} \frac12 \varrho v_2^2,\) which is negligible compared to the dynamic pressure at the opening, which is \(\frac12 \varrho v_2^2.\) Substituting all these into Bernoulli's law: \[\left(p_0 + p\right) + 0 + 0 = p_0 + \varrho g h + \frac12 \varrho v_2^2 = p_0 + \varrho \left(g h + \frac{v_2^2}{2}\right),\]
That is,
\[p = \varrho \left(g h + \frac{v_2^2}{2}\right).\]
Plugging in the numbers: \[p = 1000\,\frac{\mathrm{kg}}{\mathrm{m}^3} \cdot \left(9.81\,\frac{\mathrm{m}}{\mathrm{s}^2} \cdot 0.1\,\mathrm{m} + \frac{\left(2\,\mathrm{m} / \mathrm{s}\right)^2}{2}\right) = 2981\,\mathrm{Pa}.\]
So the extra pressure we need to apply to the piston, in excess of the atmospheric pressure, is \(2981\,\mathrm{Pa}.\)