Bernoulli\'s Principle

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Published on June 13, 2008

Author: guestfda040

Source: slideshare.net

By Woo Chang Chung Bernoulli’s Principle and Simple Fluid Dynamics

Pressure Pressure is defined as force per unit area. Standard unit is Pascal, which is N/m 2 For liquid pressure, the medium is considered as a continuous distribution of matter. For gas pressure, it is calculated as the average pressure of molecular collisions on the container. Pressure acts perpendicular on the surface. Pressure is a scalar quantity – pressure has no particular direction (i.e. acts in every direction).

Pressure is defined as force per unit area.

Standard unit is Pascal, which is N/m 2

For liquid pressure, the medium is considered as a continuous distribution of matter.

For gas pressure, it is calculated as the average pressure of molecular collisions on the container.

Pressure acts perpendicular on the surface.

Pressure is a scalar quantity – pressure has no particular direction (i.e. acts in every direction).

Pascal’s Law P f = P 0 + ρ gh “ When there is an increase in pressure at any point in a confined fluid, there is an equal increase at every point in the container.” In a fluid, all points at the same depth must be at the same pressure. Consider a fluid in equilibrium. PA - ρ Ahg – P 0 A = 0 P = P 0 + ρ gh

P f = P 0 + ρ gh

“ When there is an increase in pressure at any point in a confined fluid, there is an equal increase at every point in the container.”

In a fluid, all points at the same depth must be at the same pressure.

Consider a fluid in equilibrium.

Hydraulics Pressure is equal at the bottom of both containers (because it’s the same depth!) P = F2 / A2 = F1 / A1 and since A 1 < A 2 , F 2 > F 1 There is a magnification of force, just like a lever, but work stays the same! (conservation of energy). W = F 1 * D 1 = F 2 * D 2 ∴ D 1 > D 2 You have to push down the piston on the left far down to achieve some change in the height of the piston on the right.

Pressure is equal at the bottom of both containers (because it’s the same depth!)

P = F2 / A2 = F1 / A1 and since A 1 < A 2 , F 2 > F 1

There is a magnification of force, just like a lever, but work stays the same! (conservation of energy). W = F 1 * D 1 = F 2 * D 2

∴ D 1 > D 2

Continuity Equation A 1 v 1 = A 2 v 2 “ What comes in comes out.” Av= V/s (volume flow rate) = constant A = area v = velocity

A 1 v 1 = A 2 v 2

“ What comes in comes out.”

Av= V/s (volume flow rate) = constant

Bernoulli’s Equation Where p is the pressure, ρ is the density, v is the velocity, h is elevation, and g is gravitational acceleration

Derivation of Bernoulli’s Equation Restrictions Incompressible Non-viscous fluid (i.e. no friction) Following a streamline motion (no turbulence) Constant density *There exists an extended form of equation that takes friction and compressibility into account, but that is too complicated for our level of study.

Restrictions

Incompressible

Non-viscous fluid (i.e. no friction)

Following a streamline motion (no turbulence)

Constant density

*There exists an extended form of equation that takes friction and compressibility into account, but that is too complicated for our level of study.

Derivation of Bernoulli’s Equation Consider the change in total energy of the fluid as it moves from the inlet to the outlet. Δ E total = W done on fluid - W done by fluid Δ E total = ( 1 / 2 mv 2 2 + mgh 1 ) – ( 1 / 2 mv 1 2 + mgh 2 ) W done on fluid - W done by fluid = ( 1 / 2 mv 2 2 + mgh 1 ) – ( 1 / 2 mv 1 2 + mgh 2 ) P 2 V 2 - P 1 V 1 = ( 1 / 2 mv 2 2 + mgh 1 ) – ( 1 / 2 mv 1 2 + mgh 2 ) P 2 – P 1 = ( 1 / 2 ρ v 1 2 + ρ gh 1 ) – ( 1 / 2 ρ v 1 2 + ρ gh 1 ) E total = 1 / 2 mv 2 + mgh W = F / A *A*d = PV P 2 + 1 / 2 ρ v 1 2 + ρ gh 1 = P 1 + 1 / 2 ρ v 1 2 + ρ gh 1 ∴

Venturi Tube A 2 < A 1 ; V 2 > V 1 According to Bernoulli’s Law, pressure at A 2 is lower. Choked flow: Because pressure cannot be negative, total flow rate will be limited. This is useful in controlling fluid velocity. P 2 + 1 / 2 ρ v 1 2 = P 1 + 1 / 2 ρ v 1 2 ; ΔP = ρ / 2 *(v 2 2 – v 1 2 )

A 2 < A 1 ; V 2 > V 1

According to Bernoulli’s Law, pressure at A 2 is lower.

Choked flow: Because pressure cannot be negative, total flow rate will be limited. This is useful in controlling fluid velocity.

This is an atomizer, which uses the Venturi effect to spray liquid. When the air stream from the hose flows over the straw, the resulting low pressure on the top lifts up the fluid. Atomizer (Demonstration)

This is an atomizer, which uses the Venturi effect to spray liquid.

When the air stream from the hose flows over the straw, the resulting low pressure on the top lifts up the fluid.

Torricelli and his Orifice In 1843, Evangelista Torricelli proved that the flow of liquid through an opening is proportional to the square root of the height of the opening. Q = A*√(2g(h 1 -h 2 )) where Q is flow rate, A is area, h is height Depending on the contour and shape of the opening, different discharge coefficients can be applied to the equation (of course we assume simpler situation here).

In 1843, Evangelista Torricelli proved that the flow of liquid through an opening is proportional to the square root of the height of the opening.

Q = A*√(2g(h 1 -h 2 )) where Q is flow rate, A is area, h is height

Derivation of Torricelli’s Equation We use the Bernoulli Equation: In the original diagram A 1 [top] is much larger than A 2 [the opening]. Since A 1 V 1 = A 2 V 2 and A 1 >> A 2 , V 1 ≈ 0 Since both the top and the opening are open to atmospheric pressure, P 1 = P 2 = 0 (in gauge pressure). The equation simplifies down to: ρgh 1 = 1 / 2 ρv 2 2 + ρgh 2 1 / 2 ρv 2 2 = pg(h 1 -h 2 ) V 2 2 = 2g(h 1 -h 2 ) ∴ V 2 = √(2g(h 1 -h 2 )) Q = Av 2 = A √(2g(h 1 -h 2 )) P 2 + 1 / 2 ρ v 1 2 + ρ gh 1 = P 1 + 1 / 2 ρ v 1 2 + ρ gh 1

We use the Bernoulli Equation:

In the original diagram A 1 [top] is much larger than A 2 [the opening]. Since A 1 V 1 = A 2 V 2 and A 1 >> A 2 , V 1 ≈ 0

Since both the top and the opening are open to atmospheric pressure,

P 1 = P 2 = 0 (in gauge pressure).

The equation simplifies down to:

ρgh 1 = 1 / 2 ρv 2 2 + ρgh 2

1 / 2 ρv 2 2 = pg(h 1 -h 2 )

V 2 2 = 2g(h 1 -h 2 )

∴ V 2 = √(2g(h 1 -h 2 ))

Q = Av 2 = A √(2g(h 1 -h 2 ))

Pitot – Static Tube Used for aircrafts as speedometer Typically 10 inches long and ½ wide in diameter. A pressure transducer measures the difference between static pressure and total pressure (by measuring the strains put by net force on its metal)

Used for aircrafts as speedometer

Typically 10 inches long and ½ wide in diameter.

A pressure transducer measures the difference between static pressure and total pressure (by measuring the strains put by net force on its metal)

Pitot-Static Tube There are several holes on the outside and a center hole in the center. A center hole is connected to one side of the transducer while the outside holes are connected to the other side. Outside holes are perpendicular to the direction of travel and are pressurized by static pressure (P s ) The center hole is horizontal to the travel and is pressured by total pressure (P s + ½ ρ v 2 ) The difference in pressure is equal to ½ ρ v 2. After finding out the local density of the air by checking altitude and temperature, we can solve for velocity and this is registered. Pitot tube does not work well in low velocity and supersonic velocity.

There are several holes on the outside and a center hole in the center. A center hole is connected to one side of the transducer while the outside holes are connected to the other side.

Outside holes are perpendicular to the direction of travel and are pressurized by static pressure (P s )

The center hole is horizontal to the travel and is pressured by total pressure (P s + ½ ρ v 2 )

The difference in pressure is equal to ½ ρ v 2.

After finding out the local density of the air by checking altitude and temperature, we can solve for velocity and this is registered.

Pitot tube does not work well in low velocity and supersonic velocity.

Misinterpretation of Bernoulli Does lower pressure generates faster velocity? Or is it the other way around? According to Newton’s Second Law, acceleration is caused by force. So when the fluid accelerates in the direction of the fluid, there must be force, or difference of pressure in this case. Therefore, lower pressure generates faster velocity, not the other way around. The deflection of the streaming is the cause for the generation of pressure difference.

Does lower pressure generates faster velocity? Or is it the other way around?

According to Newton’s Second Law, acceleration is caused by force.

So when the fluid accelerates in the direction of the fluid, there must be force, or difference of pressure in this case.

Therefore, lower pressure generates faster velocity, not the other way around.

The deflection of the streaming is the cause for the generation of pressure difference.

Streamlines A streamline is a path traced out by a massless particle as it moves with the flow. Velocity is zero at the surface. As you move away from the surface, the velocity uniformly approaches the free stream value (fluid molecules nearby the surface are dragged due to viscosity). The layer at which the velocity reaches the free stream value is called boundary layer . It does not necessarily match the shape of the object – boundary layer can be detached, creating turbulence ( wing stall in aerodynamic terms).

A streamline is a path traced out by a massless particle as it moves with the flow.

Velocity is zero at the surface.

As you move away from the surface, the velocity uniformly approaches the free stream value (fluid molecules nearby the surface are dragged due to viscosity).

The layer at which the velocity reaches the free stream value is called boundary layer . It does not necessarily match the shape of the object – boundary layer can be detached, creating turbulence ( wing stall in aerodynamic terms).

Aerodynamic Lift Lift is the fort that keeps an aircraft in the air. In Bernoulli-an view, lift is produced by the different of pressure (faster velocity on the top, slower velocity in the bottom) In Newtonian view, lift is the reaction force that results from the downward deflection of the air. Both views are correct, but the current argument arises from the misapplication of either view. The most accurate explanation would take into account the simultaneous conservation of mass, momentum, and energy of a fluid, but that involves multivariable calculus.

Lift is the fort that keeps an aircraft in the air.

In Bernoulli-an view, lift is produced by the different of pressure (faster velocity on the top, slower velocity in the bottom)

In Newtonian view, lift is the reaction force that results from the downward deflection of the air.

Both views are correct, but the current argument arises from the misapplication of either view.

The most accurate explanation would take into account the simultaneous conservation of mass, momentum, and energy of a fluid, but that involves multivariable calculus.

Misconceptions of Lift In many popular literature, encyclopedia, and even textbooks, Bernoulli’s Law is used incorrectly to explain the aerodynamic lift. #1: Equal transit time - The air on the upper side of the wing travels faster because it has to travel a longer path and must “catch up” with the air on the lower side. The error lies in the specification of velocity. Air is not forced to “catch up” with the downside air. Also, this theory predicts slower velocity than in reality. #2: “ Venturi ” Theory - Upper surface of the airfoil acts like a Venturi nozzle, constricting the flow. Therefore, velocity is higher on the upper side, and the difference in velocity results in difference in pressure. The error lies in the simple assumption that an airfoil is a half-Venturi nozzle. But the other (phantom) half does not exist!

In many popular literature, encyclopedia, and even textbooks, Bernoulli’s Law is used incorrectly to explain the aerodynamic lift.

#1: Equal transit time

- The air on the upper side of the wing travels faster because it has to travel a longer path and must “catch up” with the air on the lower side.

The error lies in the specification of velocity. Air is not forced to “catch up” with the downside air. Also, this theory predicts slower velocity than in reality.

#2: “ Venturi ” Theory

- Upper surface of the airfoil acts like a Venturi nozzle, constricting the flow. Therefore, velocity is higher on the upper side, and the difference in velocity results in difference in pressure.

The error lies in the simple assumption that an airfoil is a half-Venturi nozzle. But the other (phantom) half does not exist!

Coanda Effect A fluid jet traveling tangential to the surface of a streamlined boundary remains attached to that surface for some distance as it travels. The deflection of the stream creates pressure difference. Henri Coanda, a Romanian scientist, discovered this effect when flames and smokes from the world’s first-ever jet engine (built by him) attached to the fuselage as they flew out. Due to viscosity, adjacent air molecules are swept and result in lower pressure. Then the steam follows the boundary This floating ping pong ball owes its levitation to the Coanda Effect. (DEMO)

A fluid jet traveling tangential to the surface of a streamlined boundary remains attached to that surface for some distance as it travels. The deflection of the stream creates pressure difference.

Henri Coanda, a Romanian scientist, discovered this effect when flames and smokes from the world’s first-ever jet engine (built by him) attached to the fuselage as they flew out.

Sources Atomizer . University of Iowa. 29 May 2008 <http://faraday.physics.uiowa.edu/Movies/MPEG/ 2c20.20.mpg>. Ball in Water Stream . University of Iowa. 29 May 2008 <http://faraday.physics.uiowa.edu/Movies/ MPEG/2c20.30b.mpg>. Hoselton, Mitch. Lesson 61 - Derivation of Bernoulli's Equation . 2003. 23 May 2008 <http://faculty.trinityvalleyschool.org/hoseltom/lesson%20plans/ Lesson%2061-Derivation%20of%20Bernoullis%20Equation.pdf>. &quot;Index of Aerodynamic Slides.&quot; Beginner's Guide to Aerodynamics . NASA. 29 May 2008 <http://www.grc.nasa.gov/WWW/K-12/airplane/short.html>. Path: Bernoulli's Equation; Air Pressure; Pitot-Static Tube - Speedometer; Bernoulli and Newton; Boundary Layer; Definition of Streamlines . Misinterpretations of Bernoulli's Equation . Department of Physics, University Frankfurt. 29 May 2008 <http://user.uni-frankfurt.de/~weltner/Mis6/ mis6.html>. &quot;Torricelli's Equation.&quot; Torricelli's Theorem and the Orifice Equation . Wayne State University. 26 May 2008 <http://www.eng.wayne.edu/legacy/forms/4/timmkunns.htm>. Weltner, Klaus, and Martin Ingelman-Sundberg.

Atomizer . University of Iowa. 29 May 2008 <http://faraday.physics.uiowa.edu/Movies/MPEG/ 2c20.20.mpg>.

Ball in Water Stream . University of Iowa. 29 May 2008 <http://faraday.physics.uiowa.edu/Movies/ MPEG/2c20.30b.mpg>.

Hoselton, Mitch. Lesson 61 - Derivation of Bernoulli's Equation . 2003. 23 May 2008 <http://faculty.trinityvalleyschool.org/hoseltom/lesson%20plans/ Lesson%2061-Derivation%20of%20Bernoullis%20Equation.pdf>.

&quot;Index of Aerodynamic Slides.&quot; Beginner's Guide to Aerodynamics . NASA. 29 May 2008 <http://www.grc.nasa.gov/WWW/K-12/airplane/short.html>. Path: Bernoulli's Equation; Air Pressure; Pitot-Static Tube - Speedometer; Bernoulli and Newton; Boundary Layer; Definition of Streamlines .

Misinterpretations of Bernoulli's Equation . Department of Physics, University Frankfurt. 29 May 2008 <http://user.uni-frankfurt.de/~weltner/Mis6/ mis6.html>.

&quot;Torricelli's Equation.&quot; Torricelli's Theorem and the Orifice Equation . Wayne State University. 26 May 2008 <http://www.eng.wayne.edu/legacy/forms/4/timmkunns.htm>. Weltner, Klaus, and Martin Ingelman-Sundberg.

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