Fluid Mechanics-61341 - An-Najah Videosvideos.najah.edu/sites/default/files/Chp-6.pdfdirection of...
Transcript of Fluid Mechanics-61341 - An-Najah Videosvideos.najah.edu/sites/default/files/Chp-6.pdfdirection of...
Fluid Mechanics-61341
An-Najah National UniversityCollege of Engineering
Chapter [6]
Fluid Mechanics-2nd Semester 2010- [6] Momentum Principle Dr. Sameer Shadeed1
Dr. Sameer Shadeed
Chapter [6]
Momentum Principle
Momentum
MomentumMomentum = Mass X Velocity
Momentum Momentum can be linear or angular
Linear:Linear: deals with issues such as collisionsLinear:Linear: deals with issues such as collisions
Angular:Angular: due to rotation of planet
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Linear Momentum
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Newton’s 2nd Law2nd Law:
Definition of linear momentumlinear momentum:
or
Linear Momentum
Conservation of momentummomentum:
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Conservation of Momentum
The conservationconservation ofof linearlinear momentummomentum principleprincipletells us: If a system is isolated, its total linearmomentum does not change
Particles move about in a box: No particles leave orenter, hence the box is a system
Particles can exchange momentum with each other, but that does not affect the total momentum of the system
The only way to change the total momentum is to apply a force to the box
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Therefore:
Conservation of Momentum
Notice the similarities and differences betweenconservation of mass and momentum:
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(conservation of mass)
(conservation of momentum)
Both equations are applied to the system MassMass is conserved absolutely; MomentumMomentum is conserved
unless a force is applied
Momentum Equation
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Momentum Equation
To summarize, we can say , in general in general , that
Rate of change ofmomentum in the givendirection of the fluid passingthrough the control volume
Total force exertedon the fluid in acontrol volume ina given direction
=
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through the control volumea given direction=
Pipe Flow Application
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Pipe Flow Application
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Pipe Flow Application
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Pipe Flow Application
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Example 1
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Example 1 (Solution)
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Example 1 (Solution)
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Example 1 (Solution)
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Example 1 (Solution)
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Example 1 (Solution)
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Pipe Flow Application
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Pipe Flow Application
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Pipe Flow Application
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Example 2
Upstream from an axisymmetric abrupt enlargement ina horizontal passage the mean pressure is 140 kPa andthe diameter is 150 mm. Downstream from theenlargement the mean pressure is 210 kPa and thediameter is 300 mm. Estimate the flowrate through thepassage and the force on the enlargement
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passage and the force on the enlargement
Example 2 (Solution)
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Example 2 (Solution)
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Open Channel Flow Application
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Open Channel Flow Application
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Open Channel Flow Application
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Example 3
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Example 3 (Solution)
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Example 3 (Solution)
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Open Channel Flow Application
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Open Channel Flow Application
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Open Channel Flow Application
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Example 4
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Example 4 (Solution)
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Example 4 (Solution)
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Example 4 (Solution)
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