Bernoulli’s Principle and its Applications: 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. This principle can also be applied to find the force of high winds on a skyscraper, the pressure through a chemical reactor, or even the speed of water coming out of the hose in your backyard. In this article, we will discuss Bernoulli’s equation and its derivation. We will also see some important applications of Bernoulli’s principle.

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. This principle can be applied in many applications, such as in spray guns, venturimeter, nozzles, etc. By applying Bernoulli’s principle, we can solve many real-life engineering problems related to fluid flow.

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: \({A_1}{v_1} = {A_2}{v_2}\)
\(Av =\) Constant
\(v \propto \frac{1}{A}\)
Where \(A_1\) and \(A_2\) are the areas of cross-section of the tube. And \(v_1\) and \(v_2\) are their respective velocity.

Derivation of Bernoulli’s Equation

The following are the assumptions made in the derivation of Bernoulli’s equation:

1. The fluid should be ideal.
2. The flow should be steady.
3. There should be no energy loss.
4. The fluids should be incompressible.
5. The flow should be irrotational.

Derivation

Let us consider a fluid moving inside a pipe of different cross-sectional areas in different parts and is present in different heights, as shown in the figure below. Consider the flow at two regions \(BC\) and \(DE\). Now, suppose the fluid initially lying between \(B\) and \(D\). Then, in an infinitesimal time interval \((\Delta t)\), this fluid would have moved. Suppose \(v_1\) is the speed at \(B\) and \(v_2\) at \(D\), then fluid initially at \(B\) has moved a distance of \(v_1 \Delta t\) to \(C\). In the same interval \(\Delta t\) the fluid initially at \(D\) moves to \(E\), 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 the left end \((BC)\) is \({W_1} = {P_1}{A_1}\left( {{v_1}\Delta t} \right) = {P_1}\Delta V\)
Now according to the equation of continuity same volume \(\Delta V\) passes through both the regions then, the work done by the fluid at the other end \((DE)\) is:
\({W_2} = {P_2}{A_2}({v_2}\Delta t) = {P_2}\Delta V\)
or, we can say the work done on the fluid is \( – {P_2}\Delta V\). So the total work done on the fluid is:
\({W_1} – {W_2} = \left( {{P_1} – {P_2}} \right)\Delta V\)
If the density of the fluid is \(\rho\), then \(\Delta m = \rho {A_1}{v_1}\Delta t = \rho \Delta V\) is the mass passing through the pipe in time \(\Delta t\), then change in gravitational potential energy is:
\(\Delta U = \rho g\Delta V\left( {{h_1} – {h_2}} \right)\)
And, the change in its kinetic energy is:
\(\Delta K = \left( {\frac{1}{2}} \right)\;\rho \Delta V\left( {v_2^2 – v_1^2} \right)\)
Now, after applying the work-energy theorem to this volume of the fluid it gives,
\(\left( {{P_1} – {P_2}} \right)\Delta V = \left( {\frac{1}{2}} \right)\rho \Delta V\left( {v_2^2 – v_1^2} \right) + \rho g\Delta V\left( {{h_2} – {h_1}} \right)\)
Now, on dividing each term by \(\Delta V\) to obtain the term as:
\(\left( {{P_1} – {P_2}} \right) = \left( {\frac{1}{2}} \right)\rho \left( {v_2^2 – v_1^2} \right) + \rho g\left( {{h_2} – {h_1}} \right)\)
Now, on rearranging  the above terms we get,
\({P_1} + \left( {\frac{1}{2}} \right)\rho v_1^2 + \rho g{h_1} = {P_2} + \left( {\frac{1}{2}} \right)\rho v_2^2 + \rho g{h_2}\)
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:
\(P + \left( {\frac{1}{2}} \right)\rho {v^2} + \;\rho gh = \) constant
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.

Applications of Bernoulli’s Equation

Some of the important applications of Bernoulli’s equation are given following:

1. Venturimeter
A venturi meter is a device used to measure flow speed in a pipe of non-uniform cross-section, as shown in the figure. We apply Bernoulli’s equation to the wide and narrow parts of the pipe, with \(h_1 = h_2\). As \(A_1\) is greater than \(A_2,\;v_2\) is greater than \(v_1\) and hence the pressure \(p_2\) is less than \(p_1\). A net force to the right accelerates the fluid as it enters the narrow part of the tube (called throat) and a net force to the left slows as it leaves. Then the pressure difference is also equal to \(ρgh\), where \(h\) is the difference in liquid level in the two tubes, we get the flow speed as:
\({v_1} = \sqrt {\frac{{2gh}}{{{{\left( {\frac{{{A_1}}}{{{A_2}}}} \right)}^2} – 1}}} \)

2. 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.

3. Principle of Lifting of an Aircraft

We all know that aeroplanes fly. This is a result of a lift force acting on the wings of the aircraft. The lift can be generated when an asymmetric object moves through a fluid. According to the Bernoulli principle, the difference in the velocities of the fluid molecules results in pressure difference regions. This pressure difference produces a force in an upward direction, as shown in the below figure.

4. Speed of Efflux
Suppose the surface of a liquid in a tank is at a height \(h\) from the orifice on its sides, through which the liquid issues out with velocity \(v\). The speed of the liquid coming out from the orifice is called the speed of efflux. If the dimensions of the tank are sufficiently large, the velocity of the liquid at its surface may be taken to be zero. By  applying Bernoulli’s equation at the surface and just outside the orifice, the speed of the efflux is:
\(v = \sqrt {2gh} \).

Bernoulli’s Principle– Solved Examples

Q.1. A glycerine of density \(1.25 \times {10^3}\,{\text{kg}}\,{{\text{m}}^{ – 3}}\) is flowing through the conical section of the pipe then find the rate of flow, when the radii of its ends are given as \(0.1\;\rm{m}\) and \(0.04\;\rm{m}\) and the pressure drop is given as \(10\,{\text{N}}\,{{\text{m}}^{ – 2}}\) across its length.
Ans: By using the equation of continuity, we have:
\({A_1}{v_1} = {A_2}{v_2}\)
\( \Rightarrow \frac{{{v_1}}}{{{v_2}}} = \frac{{{A_2}}}{{{A_1}}} = \frac{{{{\pi }}{{{r}}_2}^2}}{{{{\pi }}{{{r}}_1}^2}} = {\left( {\frac{{{{{r}}_2}}}{{{{{r}}_1}}}} \right)^2}\)
\( \Rightarrow \frac{{{v_1}}}{{{v_2}}} = {\left( {\frac{{0.04}}{{0.1}}} \right)^2} = \frac{4}{{25}}\) …….(1)
Now, from Bernoulli’s equation, we have:

\({p_1} + \frac{1}{2}\rho v_1^2 = {p_2} + \frac{1}{2}\rho v_2^2\)
\( \Rightarrow v_2^2 – v_1^2 = \frac{{2\left( {{p_1} – {p_2}} \right)}}{\rho }\)
\( \Rightarrow v_2^2 – v_1^2 = \frac{{2 \times 10}}{{1.25 \times {{10}^3}}} = 1.6 \times {10^{ – 2}}\,{{\text{m}}^2}\,{{\text{s}}^{ – 2}}\) ………(2)
Then, on solving equations (1) and (2), we get \({v_2} = 0.128\,{\text{m}}\;{{\text{s}}^{ – 1}}\)
Therefore, the rate of the volume flow of the tube will be:
\(Q = {A_2}{v_2} = \left( {{{\pi }}r_2^2} \right){v_2}\)
\( \Rightarrow \pi {\left( {0.04} \right)^2}\left( {0.128} \right)\)
\(Q = 6.43 \times {10^{ – 4}}\,{{\text{m}}^3}\;{{\text{s}}^{ – 1}}\).

Q 2. In a closed pipe system, water is flowing smoothly where at one point, the speed of the water is \(3.0\;{\rm{m}}\;{\rm{s}}^{-1}\) while at another point \(1.0\;\rm{m}\) higher the speed is \(4.0\;{\rm{m}}\;{\rm{s}}^{-1}\). Then, what is the pressure at the upper point if the pressure at the lower point is given as \(20\;{\rm{kPa}}\)? If the water were to stop flowing and the pressure at the lower point was \(18\;{\rm{kPa}}\) then what would the pressure be at the upper point?
Ans: (i)  Applying Bernoulli’s Equation, we have:
\({p_1} + \frac{1}{2}\rho v_1^2 + \rho g{h_1} = {p_2} + \frac{1}{2}\rho v_2^2 + \rho g{h_2}\)
\( \Rightarrow \left( {20 \times {{10}^3}} \right) + \frac{1}{2} \times {10^3} \times {\left( 3 \right)^2} + 0 = {{\text{p}}_2} + \frac{1}{2} \times {10^3} \times {\left( 4 \right)^2} + {10^3} \times 10 \times 1\)
Therefore, \({p_2} = 6.5 \times {10^3}\,{\text{N}}\,{{\text{m}}^{ – 2}}\)
\( \Rightarrow {p_2} = 6.5\,{\text{kPa}}\)
(ii) Again applying Bernoulli’s Equation, we have:
\(\left( {18 \times {{10}^3}} \right) + 0 + 0 = {p_2} + 0 + \left( {{{10}^3}} \right)\left( {10} \right)\left( 1 \right)\)
\( \Rightarrow {p_2} = 8 \times {10^3}\,{\text{N}}\,{{\text{m}}^{ – 2}}\)
\( \Rightarrow {p_2} = 8\,{\text{kPa}}\).

Summary

Bernoulli’s relation states that, 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 \(\left( {\rho gh} \right)\) remains constant.
\(P + \left( {\frac{1}{2}} \right)\rho {v^2} + \rho gh = \) constant
We apply the energy conservation principle and assume that no energy is lost due to friction. The fluid should be ideal and incompressible; the flow should be steady and irrotational. The principle of Bernoulli’s equation can be applied in Venturi meter, Nozzle meter, Orifice meter, lift in an aeroplane, atomizer, etc.

FAQs on Bernoulli’s Principle and its Applications

Q.1. What is the Bernoulli’s Equation?
Ans: Bernoulli’s equation states that, 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 \(\left( {\rho gh} \right)\) remains constant.
\(P + \left( {\frac{1}{2}} \right)\rho {v^2} + \rho gh = \) constant.

Q.2. What are the applications of Bernoulli’s Equation?
Ans: Bernoulli’s equation is applied to all problems of incompressible fluid flow. Bernoulli’s equation can be applied in Venturi meter, Nozzle meter, Orifice meter, Pitot tube, etc. Bernoulli’s theory is used to study the unstable potential flow used in the theory of ocean surface waves and acoustics.

Q.3. What are the limitations of Bernoulli’s Equation?
Ans: Bernoulli’s equation ideally applies to fluids with zero viscosity or non-viscous fluids. The fluids must be incompressible, as the elastic energy of the fluid is also not taken into consideration.

Q.4. What do you mean by the equation of continuity?
Ans: 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 cross-section and the velocity of flow is the same at every point in the tube.

Q.5. What are the assumptions for the Bernoulli equation?
Ans: The assumptions for the application of the Bernoulli equation are:
a) The fluid should be ideal.
b) The flow should be steady.
c) There should be no energy loss.
d) The fluids should be incompressible.
e) The flow should be irrotational.