Do you know how aeroplanes fly? The answer is that when air flows around the wings of the plane, the plane is pushed up by the higher pressure of air under the wings compared to the lower pressure over the wings. In this article, we will discuss Bernoulli’s equation and its derivation. We will also see some important applications of Bernoulli’s principle.

The relationship between the pressure of a flowing fluid to its elevation and its speed is obtained by an equation known as Bernoulli’s equation. This equation is based on the conservation of energy and their conversion to each other. Since Daniel Bernoulli dictates it, so it is widely known as Bernoulli’s principle. By applying Bernoulli’s principle, we can solve many real-life engineering problems related to fluid flow.

Bernoulli’s principle states that the total mechanical energy of the moving fluid comprising the gravitational potential energy of elevation, the energy associated with the fluid pressure, and the kinetic energy of the fluid motion, remains constant.

Bernoulli’s Principle Formula

Bernoulli’s equation formula is a relation between pressure, kinetic energy, and gravitational potential energy of a fluid in a container.
The formula for Bernoulli’s principle is given as follows:
\(
p+\frac{1}{2} \rho v^2+\rho g h=\text { constant }
\)
Where \(p\) is the pressure exerted by the fluid, \(v\) is the velocity of the fluid, \(\rho\) is the density of the fluid, and \(h\) is the height of the container.
Bernoulli’s equation gives great insight into the balance between pressure, velocity, and elevation.

Continuity Equation

The continuity equation states that when an incompressible and non-viscous liquid flows in a streamlined motion through a tube of non-uniform cross-section, then the product of the area of the cross-section and the velocity of flow is the same at every point in the tube.
Thus, we can write: \(A_1 v_1=A_2 v_2\)
\(A v=\) Constant
\(v \propto \frac{1}{A}\)
Where \(A_1\) and \(A_2\) are the areas of the cross-section of the tube. And \(v_1\) and \(v_2\) are their respective velocity.

In another way we can write the mass of liquid flowing out equals the mass flowing in, holds in all cases.

Then, according to the principle of continuity, the rate of mass of fluid entering must be equal to the rate of mass of fluid leaving the system.
Since, the rate of mass entering, \(M_1=\rho A_1 V_1 \Delta t\) and
The rate of mass entering, \(M_2=\rho A_2 V_2 \Delta t\)
Therefore,
\(\rho A_1 V_1 \Delta t=\rho A_2 V_2 \Delta t\)
\(A_1 V_1 = A_2 V_2 \)
The above equation is now termed as Equation of Continuity.

Bernoulli’s Equation Derivation

Let us consider a container in the shape of a pipe, whose two edges are placed at different heights and varying diameters.

Consider the flow at two regions 1 (i.e., BC) and 2 (i.e., DE). Consider the fluid initially lying between \(\mathrm{B}\) and \(\mathrm{D}\).

In an infinitesimal time interval \(\Delta t\), this fluid would have moved. Suppose \(v_1\) is the speed at \(\mathrm{B}\) and \(v_2\) at \(\mathrm{D}\), then fluid initially at B has moved a distance \(v_1 \Delta t\) to \(\mathrm{C}\left(v_1 \Delta t\right.\) is small enough to assume constant cross-section along BC).

In the same interval \(\Delta t\) the fluid initially at D moves to \(\mathrm{E}\), a distance equal to \(v_2 \Delta t\).

Pressures \(P_1\) and \(P_2\) act as shown on the plane faces of areas \(A_1\) and \(A_2\) binding the two regions. The work done on the fluid at left end \((\mathrm{BC})\) is \(W_1=\) \(P_1 A_1\left(v_1 \Delta t\right)=P_1 \Delta V\).

Since the same volume \(\Delta V\) passes through both the regions (from the equation of continuity) the work done by the fluid at the other end (DE) is \(W_2=P_2 A_2\left(v_2 \Delta t\right)=P_2 \Delta V\) or, the work done on the fluid is \(-P_2 \Delta \mathbb{V}\).

So the total work done on the fluid is
\(
W_1-W_2=\left(P_1-P_2\right) \Delta V
\)

Part of this work goes into changing the kinetic energy of the fluid, and part goes into changing the gravitational potential energy. If the density of the fluid is \(\rho\) and \(\Delta m=\rho A_1 v_1 \Delta t=\rho \Delta V\) is the mass passing through the pipe in time \(\Delta t\), then change in gravitational potential energy is
\(
\Delta U=\rho g \Delta V\left(h_2-h_1\right)
\)
The change in its kinetic energy is
\(
\Delta K=\quad \rho \Delta V\left(v_2^2-v_1^2\right)
\)
We can employ the work-energy theorem to this volume of the fluid and this yields
\(
\left(P_1-P_2\right) \Delta V=\left(\frac{1}{2}\right) \rho \Delta V\left(v_2^2-v_1^2\right)+\rho g \Delta V\left(h_2-h_1\right)
\)
We now divide each term by \(\Delta V\) to obtain
\(
\left(P_1-P_2\right)=\left(\frac{1}{2}\right) \rho\left(v_2^2-v_1^2\right)+\rho g\left(h_2-h_1\right)
\)

We can rearrange the above terms to obtain \(P_1+\left(\frac{1}{2}\right) \rho v_1^2+\rho g h_1=P_2+\left(\frac{1}{2}\right) \rho v_2^2+\rho g h_2 \dots(10.12)\)

This is Bernoulli’s equation. Since 1 and 2 refer to any two locations along the pipeline, we may write the expression in general as
\(
P+\left(\frac{1}{2}\right) \rho v^2+\rho g h=\text { constant } \dots(10.13)
\)

Where,
\(P=\) static pressure of the fluid at the cross-section,
\(\rho=\) density of the flowing fluid,
\(v=\) mean velocity of the fluid flow at the cross-section,
\(h=\) elevation head of the centre of the cross-section from the datum and,
\(g=\) acceleration due to gravity.

When a fluid is at rest 1.e., its velocity is zero everywhere, Bernoulli’s equation becomes
\(
\begin{aligned}
& P_1+\rho g h_1=P_2+\rho g h_2 \\
& \left(P_1-P_2\right)=\rho g\left(h_2-h_1\right)
\end{aligned}
\)
which is the same as Eq. (10.6).

In words, Bernoulli’s relation may be stated as follows: As we move along a streamline the sum of the pressure \((P)\), the kinetic energy per unit volume \(\left(\frac{\rho v^2}{2}\right)\) and the potential energy per unit volume ( \(\rho g h\) ) remains a constant.

Speed of Efflux: Torricelli’s Law

The word efflux means fluid outflow. The speed of the liquid coming out from the orifice is called the speed of efflux. Torricelli discovered that the speed of efflux from an open tank is given by a formula identical to that of a freely falling body.

Consider a tank containing a liquid of density \(\rho\) with a small hole in its side at a height \(y_1\) from the bottom (see Fig. 10.10). The air above the liquid, whose surface is at height \(y_2\), is at pressure \(P\). From the equation of continuity [Eq. (10.10)] we have
\(
\begin{aligned}
& v_1 A_1=v_2 A_2 \\
& v_2=\frac{A_1}{A_2} v_1
\end{aligned}
\)

If the cross-sectional area of the \(\operatorname{tank} A_2\) is much larger than that of the hole \(\left(A_2 \gg>A_1\right)\), then we may take the fluid to be approximately at rest at the top, 1.e., \(v_2=0\). Now, applying the Bernoulli equation at points 1 and 2 and noting that at the hole \(P_1=P_a\), the atmospheric pressure, we have from Eq. (10.12)
\(
P_a+\frac{1}{2} \rho v_1^2+\rho g y_1=P+\rho g y_2
\)
Taking \(y_2-y_1=h\) we have
\(
v_1=\sqrt{2 g h+\frac{2\left(P-P_a\right)}{\rho}} \dots(10.14)
\)
When \(P \gg P_a\) and \(2 g h\) may be ignored, the speed of efflux is determined by the container pressure. Such a situation occurs in rocket propulsion. On the other hand, if the tank is open to the atmosphere, then \(P=P_a\) and
\(
v_1=\sqrt{2 g h} \dots(10.15)
\)
This is also the speed of a freely falling body. Equation (10.15) represents Torricelli’s law.

Venturi-meter

The Venturi meter is a device to measure the flow speed of the incompressible fluid. It consists of a tube with a broad diameter and a small constriction at the middle as shown in Fig. (10.11). A manometer in the form of a U-tube is also attached to it, with one arm at the broad neck point of the tube and the other at constriction as shown in Fig. (10.11).

The manometer contains a liquid of density \(\rho_{\mathrm{m}}\). The speed \(v_1\) of the liquid flowing through the tube at the broad neck area \(A\) is to be measured from the equation of continuity Eq. (10.10) the speed at the constriction becomes

\(v_2=\frac{A}{a} v_1\). Then using Bernoulli’s equation (Eq.10.12) for \(\left(h_1=h_2\right)\), we get
\(
P_1+\frac{1}{2} \rho v_1^2=P_2+\frac{1}{2} \rho v_1^2(A / a)^2
\)
So that
\(
P_1-P_2=\frac{1}{2} \rho v_1^2\left[\left(\frac{A}{a}\right)^2-1\right] \dots(10.16)
\)

This pressure difference causes the fluid in the U-tube connected at the narrow neck to rise in comparison to the other arm. The difference in height \(h\) measure the pressure difference.
\(
P_1-P_2=\rho_{\mathrm{m}} g h=\frac{1}{2} \rho v_1^2\left[\left(\frac{A}{a}\right)^2-1\right]
\)
So that the speed of fluid at wide neck is
\(
v_1=\sqrt{\left(\frac{2 \rho_m g h}{\rho}\right)}\left(\left(\frac{A}{a}\right)^2-1\right)^{-1 / 2} \dots(10.17)
\)

Example 1: Blood velocity: The flow of blood in a large artery of an anesthetized dog is diverted through a Venturi meter. The wider part of the meter has a cross-sectional area equal to that of the artery. \(\mathrm{A}=8 \mathrm{~mm}^2\). The narrower part has an area \(a=4 \mathrm{~mm}^2\). The pressure drop in the artery is \(24 \mathrm{~Pa}\). What is the speed of the blood in the artery?

Solution:

We take the density of blood from Table \(10.1\) to be \(1.06 \times 10^3 \mathrm{~kg} \mathrm{~m}^{-3}\). The ratio of the areas is \(\left(\frac{A}{a}\right)=2\). Using Eq. (10.17) we obtain \(v_1=\sqrt{\frac{2 \times 24 \mathrm{~Pa}}{1060 \mathrm{~kg} \mathrm{~m}^{-3} \times\left(2^2-1\right)}}=0.123 \mathrm{~m} \mathrm{~s}^{-1}\)

The atomizer or Spray Gun

In the figure below, a spray gun is given. When the piston is pressed, the air rushes out of the horizontal tube B, decreasing the pressure to \(p_2\) which is less than the pressure \(p_1\) in the container. As a result, the liquid rises in vertical tube \(A\). It breaks up into a fine spray when it collides with the high-speed air in tube \(B\). Filter pumps, Bunsen burners, and sprayers used for perfumes or to spray insecticides work on the same principle.

Blood Flow and Heart Attack

Bernoulli’s principle helps in explaining blood flow in artery. The artery may get constricted due to the accumulation of plaque on its inner walls. In order to drive the blood through this constriction a greater demand is placed on the activity of the heart. The speed of the flow of the blood in this region is raised which lowers the pressure inside and the artery may collapse due to the external pressure. The heart exerts further pressure to open this artery and forces the blood through. As the blood rushes through the opening, the internal pressure once again drops due to the same reasons leading to a repeat collapse. This may result in a heart attack.

Dynamic Lift

Dynamic lift is the force that acts on a body, such as airplane wing, a hydrofoil, or a spinning ball, by virtue of its motion through a fluid. In many games such as cricket, tennis, baseball, or golf, we notice that a spinning ball deviates from its parabolic trajectory as it moves through the air. This deviation can be partly explained on the basis of Bernoulli’s principle.

Ball moving without spin: Fig. 10.13(a) shows the streamlines around a non-spinning ball moving relative to a fluid. From the symmetry of streamlines, it is clear that the velocity of the fluid (air) above and below the ball at corresponding points is the same resulting in zero pressure difference. The air, therefore, exerts no upward or downward force on the ball.

Ball moving with spin: A ball which is spinning drags air along with it. If the surface is rough more air will be dragged. Fig 10.13(b) shows the streamlines of air for a ball which is moving and spinning at the same time. The ball is moving forward and relative to it the air is moving backward. Therefore, the velocity of air above the ball relative to the ball is larger, and below it is smaller. 

This difference in the velocities of air results in the pressure difference between the lower and upper faces and there is a net upward force on the ball. This dynamic lift due to spining is called Magnus effect.

Aerofoil or lift on aircraft wing: Figure 10.13 (c) shows an aerofoil, which is a solid piece shaped to provide an upward dynamic lift when it moves horizontally through air. The crosssection of the wings of an aeroplane looks somewhat like the aerofoil shown in Fig. 10.13 (c) with streamlines around it. When the aerofoil moves against the wind, the orientation of the wing relative to flow direction causes the streamlines to crowd together above the wing more than those below it. The flow speed on top is higher than that below it. There is an upward force resulting in a dynamic lift of the wings and this balances the weight of the plane.

Example 2: A fully loaded Boeing aircraft has a mass of \(3.3 \times 10^5 \mathrm{~kg}\). Its total wing area is \(500 \mathrm{~m}^2\). It is in level flight with a speed of \(960 \mathrm{~km} / \mathrm{h}\). (a) Estimate the pressure difference between the lower and upper surfaces of the wings (b) Estimate the fractional increase in the speed of the air on the upper surface of the wing relative to the lower surface. [The density of air is \(\rho\) \(\left.=1.2 \mathrm{~kg} \mathrm{~m}^{-3}\right]\)

Solution:

(a) The weight of the Boeing aircraft is balanced by the upward force due to the pressure difference
\(
\begin{aligned}
& \Delta P \times A=3.3 \times 10^5 \mathrm{~kg} \times 9.8 \\
& \Delta P=\left(3.3 \times 10^5 \mathrm{~kg} \times 9.8 \mathrm{~m} \mathrm{~s}^{-2}\right) / 500 \mathrm{~m}^2 \\
& \quad=6.5 \times 10^3 \mathrm{Nm}^{-2}
\end{aligned}
\)

(b) We ignore the small height difference between the top and bottom sides in Eq. (10.12). The pressure difference between them is then
\(
\Delta P=\frac{\rho}{2}\left(v_2^2-v_1^2\right)
\)
where \(v_2\) is the speed of air over the upper surface and \(v_1\) is the speed under the bottom surface.
\(
\left(v_2-v_1\right)=\frac{2 \Delta P}{\rho\left(v_2+v_1\right)}
\)
Taking the average speed
\(
v_{\text {av }}=\left(v_2+v_1\right) / 2=960 \mathrm{~km} / \mathrm{h}=267 \mathrm{~m} \mathrm{~s}^{-1} \text {, }
\)
we have
\(
\left(v_2-v_1\right) / v_{\mathrm{av}}=\frac{\Delta P}{\rho v_{\mathrm{av}}^2} \approx 0.08
\)
The speed above the wing needs to be only 8 \(\%\) higher than that below.