Analysing Examples of Bernoulli’s Equation

According to Bernoulli’s principle, in fluid dynamics, there is a direct relationship between fluid velocity , pressure and  potential difference within it, i.e., with increasing velocity, the pressure will decrease. It is a fundamental theory in fluid mechanics, developed by Daniel Bernoulli in 1726. The principle is composed of three fundamental equations of hydrodynamics. It is the embodiment of mechanical energy conversion in hydraulics.

Bernoulli’s principle: In the area of fluid dynamics, according to Bernoulli’s principle, there is a direct relationship between the velocity of a fluid and (pressure) potential difference within it, i.e., with increasing velocity, the pressure will decrease.

It provides the theoretical framework for the solution of hydraulic calculations in real-world engineering situations. The use of the Bernoulli equation may be seen throughout hydraulic mechanics. This article presents the Bernoulli equation application examples based on the premise of Bernoulli’s equation.

The Application of the Bernoulli equation

Some of Bernoulli’s equation examples with solutions are further below:

Lift in Aerodynamics-

Lift is generated in aerodynamics by fluid flow over an airfoil owing to a vertical pressure gradient along the wing’s cross-section. Based on the flow streamlines, Bernoulli’s equation may be used to forecast the pressure gradient in the vertical direction of the fluid. Streamlines are typically shown along an airfoil, allowing for predicting pressure gradient and lift.

Air in an airfoil’s top surface moves faster than the bottom; hence the air pressure on the bottom will be more significant, resulting in lift. On the underside of the wing, we can also observe where flow separation and vertical flow occur, which is consistent with what would be predicted at high Reynolds numbers.

Entrainment

Many scientists have worked through Bernoulli’s principle in devices that decrease the pressure in the liquid with high velocity, and move the liquid. Due to high pressure from the external environment, fluids put forces on their fluids for flow, and this process is known as entrainment. These devices are prevalent from old age for enhancing the heights of water, and it is very critical for low-level areas. 

Fluid Flow Measurements

Fluid flow rate measurements apply Bernoulli’s principle, which is closely connected. Using the Venturi effect makes it possible to determine the flow rate. As fluid flows through an orifice plate with a tiny aperture of known diameter, the decrease in diameter will result in a rise in the flow rate of the fluid. Bernoulli’s principle may be used to compute the flow rate outside the orifice plate based on this internal measurement of flow rate.

Drag Force

In aerodynamics, drag is present, but it is also present in other situations where a gas or liquid is forced to flow over an object. In watercraft, for example, streamlining may be used to anticipate the drag force as well as the propulsive force, as in a sailboat. The explanation and visualisation of drag over the body of a car are two additional essential areas of use, both of which may be validated in a wind tunnel under controlled airflow.

Sample Problems

Some examples of the Bernoulli equation example problems below include the following solutions:

Problem 1-

The internal diameter of fire hoses used in big structure fires is 6.40 cm. Assume a pipe with a flow rate of 40.0 L/s and a gauge pressure of 1.62 x 106 N/m2 has been installed. The hose ascends a ladder for 10.0 metres to a 3.00-cm-diameter nozzle. What is the nozzle’s pressure?

As the water must travel uphill to reach the nozzle, and because speed rises in the nozzle, the pressure in the nozzle is lower than at ground level. Due to its kinetic energy, water can exert a considerable force on everything it touches, despite its lower pressure. When the water stream exits into the air, its pressure equals that of the surrounding atmosphere.

Solution:

The equation of Bernoulli is

P1+12ρv12+ρgh1 =P2+12ρv22+ρgh2.

The beginning circumstances at ground level and the end conditions within the nozzle are denoted by subscripts 1 and 2, respectively. We must first determine the v1 and v2 speeds. Since Q = A1v1, we get

v1=Q/A1=40.0×10-3m3/s /π(3.20×10-2m)2=12.4 m/s.

Similarly, we discover

v2=56.6 m/s.

This very high speed aids fluid in reaching the fire. We can now solve Bernoulli’s equation for p2 by setting h1 to zero:

P2=P1+12ρ(v12−v22)−ρgh2.

When known values are substituted, the result is

P2=(1.62×106N/m2)+12(1000kg/m3)[(12.4 m/s)2−(56.6 m/s)2]

−(1000kg/m3)(9.80 m/s2)(10.0m)=0.

Problem 2-

There is a horizontal pipe of 4 m2 cross-sectional area, 5 m/s flow rate, and pressure of 0.3MPa at point A, where the water flows. The cross-sectional area at point B is 2 m2. What is the water’s speed at point B? Calculate the pressure at the second point B.

Solution:

The speed at point B may be calculated using the equation of continuity. A1  v1 = A2 v2

Therefore, v2 = A1  v1A2

⇒ v2 =4*52  m/s = 10 m/s

Let us figure out what the velocity is at point B. Take note of how the speed has risen. As a result, according to Bernoulli’s principle, the pressure would fall in proportion. It will assist us in verifying our response. Bernoulli’s Equation in Use:

P1 + (1/2)ρv12 + ρgh = P2 + (1/2)ρv22 + ρgh……….since the fluid is at the same elevation.

⇒ 0.3×106 + 0.5x1000x25 = P2 + 0.5x1000x102

⇒ P2 = 0.2625 MPa

Conclusion

In fluid mechanics, Bernoulli’s equation plays a crucial role. Although it is based on an incompressible ideal fluid, the Bernoulli equation may derive many essential conclusions when applied to stationary flow, where a large amount of fluid flow can be approximated by an ideal fluid.