Bernoullis Equation Formula

Bernoullis Equation Formula

The Bernoulli Principle, formulated by the Swiss mathematician Daniel Bernoulli, is a fundamental concept in fluid dynamics that describes the behavior of moving fluids. It states that in a steady, incompressible flow of a fluid with negligible viscosity, an increase in the fluid’s speed occurs simultaneously with a decrease in its pressure or potential energy. The Bernoulli equation mathematically represents this principle and can be expressed as: P+1/​ρv²+ρgh=constant. where P is the fluid pressure, ρ is the fluid density, v is the fluid velocity, g is the acceleration due to gravity, and h is the height above a reference point. This equation highlights the trade-off between pressure, kinetic energy, and potential energy in a flowing fluid. Learn more about Bernoulli’s equation principle, formula, and examples based on it.

What is Bernoulli’s Equation Principle?

The Bernoulli Principle, named after the Swiss mathematician Daniel Bernoulli, is a key principle in fluid dynamics that describes the conservation of energy in a flowing fluid. According to this principle, for an incompressible, non-viscous fluid undergoing steady flow, the sum of the pressure energy, kinetic energy, and potential energy per unit volume remains constant along any streamline.

Bernoulli’s principle, applies to liquids in ideal conditions, and therefore, pressure and density are inversely proportional to each other, which means that a slow-moving fluid exerts more pressure than a fast-moving fluid. Fluids, in this case, refer to gases as well as liquids. This principle underlies many applications. Some very common examples are when an aeroplane is trying to stay aloft, or even the most common, mundane things like shower curtains curling inward.

A similar phenomenon occurs in the case of rivers with varying widths. Water speeds are slower in larger areas and faster in smaller areas. Students must think the liquid pressure will be higher. However, contrary to the explanation above, the liquid pressure decreases in the narrow part of the flow and increases in the wide part of the flow.

Bernoulli’s Equation Formula

Mathematically, the Bernoulli Equation is expressed as:

\[ P + \frac{1}{2} \rho v^2 + \rho gh = \text{constant} \]

where:
\( P \) is the fluid pressure,
\( \rho \) is the fluid density,
\( v \) is the fluid velocity,
\( g \) is the acceleration due to gravity,
\( h \) is the height above a reference point.

Explanation of the Bernoulli Equation Formula

Pressure Energy: \( P \) – Represents the energy due to the fluid’s pressure. Higher pressure means higher energy.

Kinetic Energy: \( \frac{1}{2} \rho v^2 \) – Represents the energy due to the fluid’s motion. Faster moving fluid has more kinetic energy.

Potential Energy: \( \rho gh \) – Represents the energy due to the fluid’s elevation in a gravitational field. Higher elevation means more potential energy.

Key Points

Incompressible Flow: The fluid density (\( \rho \)) remains constant.
Non-Viscous Fluid: The fluid has no internal friction (viscosity), which means no energy is lost due to internal resistance.
Steady Flow: The velocity of the fluid at any given point does not change over time.
Along a Streamline: The principle applies to a single streamline, which is a path traced by a fluid particle under steady flow.

Applications of the Bernoulli Principle

Aerodynamics: Explains how airplane wings generate lift. The faster airflow over the curved upper surface of the wing results in lower pressure compared to the slower airflow beneath the wing, creating lift.

Venturi Effect: In a constricted section of a pipe, fluid speed increases and pressure decreases, which is utilized in devices like the Venturi meter for measuring fluid flow rate.

Hydraulic Engineering: Helps design efficient systems for water distribution and sewage systems, ensuring proper flow and pressure management.

Medical Devices: Explains the functioning of devices like the atomizer, which uses fluid pressure differences to create a fine spray.

Bernoulli Equation Formula Solved Examples

Example 1: Water flows through a horizontal pipe. The speed of water at point A is 2 m/s, and the pressure is 200,000 Pa. At point B, the speed of water increases to 3 m/s. What is the pressure at point B?

Solution:

Given:
\( v_A = 2 \text{ m/s} \)
\( P_A = 200,000 \text{ Pa} \)
\( v_B = 3 \text{ m/s} \)
Since the pipe is horizontal, the height \( h \) is the same at both points, so \( h_A = h_B \)

Using the Bernoulli Equation:

\[ P_A + \frac{1}{2} \rho v_A^2 = P_B + \frac{1}{2} \rho v_B^2 \]

Rearrange to solve for \( P_B \):

\[ P_B = P_A + \frac{1}{2} \rho (v_A^2 – v_B^2) \]

Assuming the density of water \( \rho \) is 1000 kg/m³:

\[ P_B = 200,000 + \frac{1}{2} \times 1000 \times (2^2 – 3^2) \]

\[ P_B = 200,000 + 500 \times (4 – 9) \]

\[ P_B = 200,000 + 500 \times (-5) \]

\[ P_B = 200,000 – 2500 \]

\[ P_B = 197,500 \text{ Pa} \]

Example 2: The speed of air over the top surface of an airplane wing is 80 m/s, and the speed below the wing is 60 m/s. If the pressure below the wing is 101,325 Pa, what is the pressure above the wing? Assume the density of air is 1.225 kg/m³.

Solution:

Given:
\( v_{\text{top}} = 80 \text{ m/s} \)
\( v_{\text{bottom}} = 60 \text{ m/s} \)
\( P_{\text{bottom}} = 101,325 \text{ Pa} \)
Density of air \( \rho = 1.225 \text{ kg/m}^3 \)

Using Bernoulli’s Equation:

\[ P_{\text{top}} + \frac{1}{2} \rho v_{\text{top}}^2 = P_{\text{bottom}} + \frac{1}{2} \rho v_{\text{bottom}}^2 \]

Rearrange to solve for \( P_{\text{top}} \):

\[ P_{\text{top}} = P_{\text{bottom}} + \frac{1}{2} \rho (v_{\text{bottom}}^2 – v_{\text{top}}^2) \]

Substitute the values:

\[ P_{\text{top}} = 101,325 + \frac{1}{2} \times 1.225 \times (60^2 – 80^2) \]

\[ P_{\text{top}} = 101,325 + 0.6125 \times (3600 – 6400) \]

\[ P_{\text{top}} = 101,325 + 0.6125 \times (-2800) \]

\[ P_{\text{top}} = 101,325 – 1715 \]

\[ P_{\text{top}} = 99,610 \text{ Pa} \]

Example 3: Water flows from a large tank through a small hole at the bottom. The surface of the water in the tank is 5 meters above the hole. What is the speed of water exiting the hole? Assume the water surface area is much larger than the hole area.

Solution:

Given:
Height \( h = 5 \text{ m} \)
Pressure at the top of the water surface and at the exit of the hole is atmospheric pressure, so they cancel out in the Bernoulli equation.
Assuming the velocity at the top surface \( v_{\text{top}} \approx 0 \text{ m/s} \)

Using Bernoulli’s Equation:

\[ P_{\text{top}} + \frac{1}{2} \rho v_{\text{top}}^2 + \rho gh_{\text{top}} = P_{\text{bottom}} + \frac{1}{2} \rho v_{\text{bottom}}^2 + \rho gh_{\text{bottom}} \]

Since \( P_{\text{top}} = P_{\text{bottom}} \) and \( v_{\text{top}} \approx 0 \):

\[ \rho gh = \frac{1}{2} \rho v_{\text{bottom}}^2 \]

Simplify to solve for \( v_{\text{bottom}} \):

\[ gh = \frac{1}{2} v_{\text{bottom}}^2 \]

\[ v_{\text{bottom}}^2 = 2gh \]

\[ v_{\text{bottom}} = \sqrt{2gh} \]

Substitute \( g = 9.81 \text{ m/s}^2 \) and \( h = 5 \text{ m} \):

\[ v_{\text{bottom}} = \sqrt{2 \times 9.81 \times 5} \]

\[ v_{\text{bottom}} = \sqrt{98.1} \]

\[ v_{\text{bottom}} \approx 9.9 \text{ m/s} \]

These examples illustrate how the Bernoulli Equation can be applied to different scenarios involving fluid flow, demonstrating its versatility in solving practical problems.