Bernoulli's Theorem, when an incompressible and frictionless fluid is flowing through a tube of unequal cross section in a streamline, then the sum of pressure energy, kinetic energy, and potential energy per unit volume of the fluid at every point of the tube is constant.
Learn the essence of Bernoulli's Theorem with its easy-to-understand definition, step-by-step derivation, and practical applications.
Prove Bernoulli's Theorem
Thus the theorem of Bernoulli's theorem is a principle of conservation of energy for a flowing fluid.
P + 1/2ρv² + ρv²h = Constant
Where
P = Pressure Energy
1/2ρv² = Kinetic Energy
ρv²h = Statics Energy
Let xy be a tube of the unequal orifice in which a fluid of density (ρ) enters with velocity (v1) at the x end and exits at the y end with a velocity (v1).
If the pressure of the liquid at these ends is P and P respectively, the areas of the ends are A and A respectively and the heights of these ends are h and h from the ground, then
(i) x at the end
Work done on a liquid in 1 second = force ✕ displacement
= Pressure ✕ Area ✕ displacement
= P1 ✕ A1 ✕ v1
(ii) y at the end
Work done on the liquid in 1 second = force ✕ displacement
= Pressure ✕ Area ✕ displacement
= P2 ✕ A2 ✕ v2
Finally,
Net work done on the liquid in 1 second = P1 ✕ A1 ✕ v1 - P2 ✕ A2 ✕ v2
But the form principle of continuity
A1 ✕ v1 = A2 ✕ v2 = m/ρ
Putting this value in the upper equation
Work done on liquid in 1 second = P1 ✕ m/ρ - P2 ✕ m/ρ
= m/ρ (P1 - P2)
Where
m = mass of liquid entering the tube in 1 second.
Work done on liquid in 1 second = m/ρ (P1 - P2) ....(1)
The kinetic energy of the fluid at the x and y ends are respectively 1/2 mv1 and 1/2 mv2.
The increase in kinetic energy of fluid = 1/2 mv²1 - 1/2 mv²2
= 1/2m (v²1 - v²2)
Thus the static energy of the fluid at the x and y ends are mgh1 and mgh2 respectively.
Decrease in static energy of liquid = mgh1-mgh2
= mg (h1 - h2)
Finally, the net increase in energy of the fluid 1/2m (v1 - v2) - mg (h1 - h2) because energy is obtained from work.
We know that
work done on fluid = net increase in energy of a fluid
= m/ρ (P1 - P2) - 1/2m (v²1 - v²2) - mg (h1 - h2)
Multiplication of this equation by density ρ.
(P1 - P2)= 1/2ρ (v²1 - v²2) - ρg (h1 - h2)
(P1 - P2)= 1/2ρ v²1 - 1/2ρv²2 - ρgh1 - ρgh2
P1 + 1/2ρ v²1 + ρgh1 = P2 + 1/2ρv²2 + ρgh2
P + 1/2ρ v² + ρgh = constant
This is Bernoulli's equation.
Limitations of Bernoulli's Theorem
1. The moving fluid must be incompressible.
2. The flow must be steady for Bernoulli's theorem to work.
3. The speed of fluid flowing through any cross-section should be exact at all points of that section
4. In the Bernoulli equation, only the drift force of gravity acts on the liquid. If any force other than gravity also acts on the liquid, then for this also proper term equation should be included.
What is fluid head in fluid mechanics?
In fluid mechanics, the term "fluid head" is used to refer to the pressure exerted by a fluid on the walls of a container or flow channel.
The fluid head can be used to calculate the forces and velocities associated with fluid flow.
We know that Bernoulli's Theorem is
P + 1/2ρ v² + ρgh = constant
The upper equation on divided by ρg.
P/ρg + 1v²/2g + h = constant/ρg
P/ρg = Pressure Head
v²/2g = Velocity Head
h = Gravity Head
Friction factor for turbulent flow formula
The friction factor is a common parameter that is useful in the study of turbulent flow. It is denoted by f.
It is defined as the ratio of wall shear stress to the product of the density and the velocity head.
Wall shear stress = τw
Density = ρ
Velocity Head = v²/2g
f = τw/ρv²/2g
Friction factor f = 2τwg/ρv²
But, τw = Δ P.D/4L
f = 2Δ P.D.g/ρv².4L
= Δ P.D.g/2ρv².L
Fanning Friction factor f = Δ P.D.g/2ρv².L
Fanning Equation Formula
Four quantities hfs, Δ Ps, τw, and ΔL are used to measure the spin friction in pipes. These four zodiac signs are related to each other in the following equation.
hfs = 4τw/pD *ΔL
But, τw = fρv²/2g
hfs = 2fv²/gD *ΔL
Bernoulli's Theorem Applications
Bernoulli's Theorem, a fundamental principle in fluid dynamics, may sound complex at first, but its real-world applications are surprisingly widespread and crucial.
1. Airplane Wings and Lift
Ever wondered how airplanes stay aloft despite their enormous weight? Bernoulli's Theorem comes into play here. The shape of an airplane wing is designed to create a difference in air pressure, with faster-moving air above and slower-moving air below the wing. This pressure difference generates lift, enabling the aircraft to soar gracefully through the skies.
2. Water Conservation
Understanding Bernoulli's Theorem can help us conserve water and save energy. In household plumbing, water moves through pipes with varying diameters. Applying the theorem, we can optimize the pipe size and control the flow of water, reducing wastage and ensuring efficient usage.
3. Venturi Effect
The Venturi effect, an application of Bernoulli's Theorem, plays a significant role in various industries. It is employed in carburettors to mix air and fuel efficiently in engines, enhancing performance and fuel economy. Additionally, it is used in industrial processes to measure fluid flow rates accurately.
4. Atomizers and Aerosol Cans
Ever wondered how aerosol cans disperse their contents so effectively? The answer lies in Bernoulli's Theorem. The nozzle of an aerosol can create a constriction that accelerates the fluid, decreasing pressure, and thereby allowing the contents to spray out in a fine mist.
5. Wind Turbines
Renewable energy sources are becoming increasingly vital in today's world. Wind turbines, which harness wind energy to generate electricity, rely on Bernoulli's Theorem to optimize blade design. By creating low-pressure regions on one side of the blade and high-pressure regions on the other, the wind's force drives the blades, converting kinetic energy into electrical power.
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Mechanical Properties of Fluids, and Realities Questions and Answers