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Bernoulli's Principle

Figure represents a tube of flow (or an actual pipe, for that matter) through which an ideal fluid is flowing at a steady rate. In a time interval Δ t, suppose that a volume of fluid Δ V, coloured purple in Figure (a), enters the tube at the left (or input) end and an identical volume, coloured green in Figure (b), emerges at the right (or output) end. The emerging volume must be the same as the entering volume because the fluid in incompressible, with an assumed constant density p.
Let y1, v1 and p1 be the elevation, speed, and pressure of the fluid entering at the left, and y2, v2 and p2 be the corresponding quantities for the fluid emerging at the right. By applying the principle of conservation of energy to the fluid, we shall show that these quantities are related by


We can also write this equation as
In the figure fluid flows at a steady rate through a length L of a tube, from he input end at the left to the output end at the right. From time t in (a) to time t + Δ t in (b), the amount of fluid shown in purple enters the input end and the equal amount shown in green emerges from the output end.

Like the equation of continuity, Bernoulli’s equation is not a new principle but simply the reformulation of a familiar principle in a form more suitable to fluid mechanics. As a check, let us apply Bernoulli’s equation to fluids at rest, by putting v1 = v2 = 0 in the equation (i). The result is
P2 = p1 + pg(y1 - y2)

A major prediction of Bernoulli’s equation emerges if we take y to be a constant (y = 0, say) so that the fluid does not change elevation as it flows. Equation then becomes

Which tells us that:
If the speed of a fluid particle increases as it travels along a horizontal streamline, the pressure of the fluid must decrease, and conversely.
Put another way, where the streamlines are relatively close together (that is, where the velocity is relatively great), the pressure is relatively low, and conversely.
The link between a change in speed and a change in pressure makes sense if you consider a fluid particle. When the particle nears a narrow region, the higher pressure behind it accelerates it so that it then has a greater speed in the narrow region. And when it nears a wide region, the higher pressure ahead of it decelerates it so that it then has a lesser speed in the wide region.
Bernoulli's equation is strictly valid only to the extent that the fluid is ideal. If viscous forces are present, thermal energy will be involved. We take no account of this in the derivation that follows.

Proof of Bernoulli's Equation

Let us take as our system the entire volume of the (ideal) fluid shown in the figure. We shall apply the principle of conservation of energy to this system as it moves from its initial state (a) to its final state (b). The fluid lying between the two vertical planes separated by a distance L in the figure does not change its properties during this process; we need to be concerned only with changes that take place at the input and the output ends.

We apply energy conservation in the form of the work-kinetic energy theorem.
W = Δ K

Which tells us that the change in the kinetic energy of our system must equal the net work done on the system. The change in kinetic energy results from the change in speed between the ends of the pipe and is
Δ K = Δ m v - Δ m v
        = p Δ V (v - v)
in which Δ m (=p ΔV) is the mass of the fluid that enters at the input end and leaves at the output end during a small time interval Δ t.
The work done on the system arises from two sources. The work Wg done by the weight (Δ m g) of mass Δ m during the vertical lift of the mass from the input level to the output level is
Wg = -Δ m g (y2 - y1)
        = -pg Δ(y - y1)

This work is negative because the upward displacement and the downward weight point in opposite directions.
Work Wp must also be done on the system (at the input end) to push the entering fluid into the tube and by the system (at the output end) to push forward the fluid ahead of the emerging fluid. In general, the work done by a force of magnitude F, acting on a fluid sample contained in a tube of area A to move the fluid through a distance Δ x, is
FΔ x = (pA) (Δ x) = p(AΔ x) = pΔ V

The work done on the system is then p1Δ V, and the work done by the system is -p2Δ V. Their sum Wp is
Wp = -p2ΔV + p1Δ V
        = -(p2 - p1) Δ V

The work-kinetic energy theorem now becomes
W = Wg + Wp = Δ K.

Substituting for the values for Wg, Wp and ΔK we get,
=pgΔ V(y2 - y1) - Δ V(p2 - p1) = pΔ V(v - v)

This, after a slight rearrangement, matches the equation of continuity, which we set out to prove.

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