Numero De Reynolds 3j6g1y

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INGENIERÍA AERONÁUTICA

Fundamentos de Aerodinámica 2018-1 Jorge Mario Tamayo Avendaño MSc. Ingeniería Aeroespacial

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Lecture Overview Lets analyse the equations of conservation of mass, and linear momentum, aiming for practical applications of this relations. Todays lecture includes: • Concept of differential flow volume along a segment of streamline • Mass and Linear Momentum conservation equations and the derivation of the Bernoulli equation

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Bernoulli Equation Now lets consider a frictionless fluid flow, keeping in mind that in reality all fluids are viscous to some extent and therefore, experience some king of friction. In assuming frictionless fluid flow, a restriction is being set beforehand: The relationship that is about to be derived, must be used in regions where friction is negligible.

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Bernoulli Equation Consider the following element of streamline 𝑑𝑠 𝑝 + 𝑑𝑝 𝑉 + 𝑑𝑉 𝐴 + 𝑑𝐴 𝑑𝑠

𝑑𝑧

Have in mind that the flow properties are assumed uniform across area elements 𝐴 and 𝐴 + 𝑑𝐴 respectively.

𝜃 𝑑𝑊 𝑉 𝐴, 𝑝, 𝜌 5

Bernoulli Equation Lets begin by invoking mass conservation equation for the fluid element: 𝑑 𝑑𝑡

𝐶𝑉

𝜌𝑑𝑉 + 𝑚𝑜𝑢𝑡 − 𝑚𝑖𝑛 = 0 ≈

𝜕𝜌 𝑑𝑉 + 𝑑𝑚 𝜕𝑡

Then this equation can be formulated according to the properties of the proposed differential fluid element 𝑑𝑚 = −

𝜕𝜌 𝑑𝑉 𝜕𝑡

𝜕𝜌 𝑑 𝜌𝐴𝑉 = − 𝐴𝑑𝑠 𝜕𝑡

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Bernoulli Equation Now lets focus on the formulation that the equation for conservation of linear momentum would have in this case: 𝑑𝐹𝑠 =

𝑑 𝑑𝑡

𝑉𝜌𝑑𝑉 + 𝑉𝑚 𝐶𝑉

− 𝑉𝑚 𝑜𝑢𝑡

𝑖𝑛

≈

𝜕 𝑉𝜌 𝐴𝑑𝑠 + 𝑑 𝑚𝑉 𝜕𝑡

This equation is not complete without having an explicit expression for the force term to the left, 𝑑𝐹𝑠 . What are the external forces acting on the differential element of fluid: • Body forces due to gravitational acceleration? • Surface forces due to normal pressure acting on the CS? • Surface forces due to shear stress acting on the CS? Think of the assumptions that have been made until now… 7

Bernoulli Equation Assuming that the resultant external force is generated by • Body forces due to gravitational acceleration, 𝑑𝐹𝑠,𝑔𝑟𝑎𝑣. • Surface forces due to normal pressure acting on the CS, 𝑑𝐹𝑠,𝑝𝑟𝑒𝑠. Body forces can be described by 𝑑𝐹𝑠,𝑔𝑟𝑎𝑣. = −𝑑𝑊 sin 𝜃 = −𝛾𝐴𝑑𝑠 sin 𝜃 = −𝛾𝐴𝑑𝑧 Surface forces are expressed as 𝑑𝐹𝑠,𝑝𝑟𝑒𝑠.

1 = 𝑑𝑝𝑑𝐴 − 𝑑𝑝 𝐴 + 𝑑𝐴 ≈ −𝐴𝑑𝑝 2 8

Bernoulli Equation Now group all into the initial momentum equation: 𝜕 𝑑𝐹𝑠 = −𝛾𝐴𝑑𝑧 − 𝐴𝑑𝑝 = 𝑉𝜌 𝐴𝑑𝑠 + 𝑑 𝑚𝑉 𝜕𝑡 Which is equivalent to:

𝜕𝜌 𝜕𝑉 −𝛾𝐴𝑑𝑧 − 𝐴𝑑𝑝 = 𝑉𝐴𝑑𝑠 + 𝜌𝐴𝑑𝑠 + 𝑉𝑑𝑚 + 𝑚𝑑𝑉 𝜕𝑡 𝜕𝑡 And can be further simplified to (why?)

𝜕𝑉 −𝛾𝐴𝑑𝑧 − 𝐴𝑑𝑝 = 𝜌𝐴𝑑𝑠 + 𝑚𝑑𝑉 𝜕𝑡

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Bernoulli Equation When the equation that have just been derived is divided by the product 𝜌𝐴, it is possible to proceed with further manipulation, solving the differential and obtaining a completely algebraic equation: 𝜕𝑉 𝑑𝑝 𝑑𝑠 + 𝑉𝑑𝑉 + + 𝑔𝑑𝑧 = 0 𝜕𝑡 𝜌 This is done by integrating the expression above between two points in a single streamline, 2 1

𝜕𝑉 𝑑𝑠 + 𝜕𝑡

2

2

𝑉𝑑𝑉 + 1

1

𝑑𝑝 + 𝜌

2

𝑔𝑑𝑧 = 0 1

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Bernoulli Equation Initially part of the equation can be integrated 2 1

𝜕𝑉 𝑑𝑠 + 𝜕𝑡

2 1

𝑑𝑝 1 + 𝑉2 2 − 𝑉1 2 + 𝑔 𝑧2 − 𝑧1 = 0 𝜌 2

To the initial assumption of frictionless flow two other assumptions are made from this point in order to continue with the simplification of the differential of the momentum equation: • Steady flow assumption:

𝜕𝑉 𝜕𝑡

=0

• Incompressible flow: 𝜌 = 𝑐𝑜𝑛𝑠𝑡.

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Bernoulli Equation The equation of linear momentum becomes then 𝑝2 − 𝑝1 1 + 𝑉2 2 − 𝑉1 2 + 𝑔 𝑧2 − 𝑧1 = 0 𝜌 2 Reorganized as 𝑝1 1 2 𝑝2 1 2 + 𝑉 + 𝑔𝑧1 = + 𝑉2 + 𝑔𝑧2 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 𝜌 2 1 𝜌 2

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Bernoulli Equation Lets explore the applications of Bernoulli eqn. in more detail Case B

Case A

𝒙𝟐 , 𝒚𝟐 , 𝒛𝟐

𝒙𝟐 , 𝒚𝟐 , 𝒛𝟐

𝒙𝟏 , 𝒚𝟏 , 𝒛𝟏

𝒙𝟏 , 𝒚𝟏 , 𝒛𝟏

𝑝1 1 2 𝑝2 1 2 + 𝑉 + 𝑔𝑧1 = + 𝑉2 + 𝑔𝑧2 𝜌 2 1 𝜌 2

𝑝1 1 2 𝑝2 1 2 + 𝑉 + 𝑔𝑧1 ≠ + 𝑉2 + 𝑔𝑧2 𝜌 2 1 𝜌 2

Image: https://www.siemens.com/press/en/feature/2015/mobility/2015-12-ice4.php

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