Week11 Chp04 Acc n RTT PDF

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LEcture notes of Chapter 4 of Fluid MEchanics by Cengel...

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ACCELERATION FIELD Week 11

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Acceleration field ➢Acceleration of a particle is the time rate of change of its  velocity. a  dV A

A

 VA

dt

  VA rA , t    VA  xA t , y A t , z A t , t 

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Acceleration field ➢This is a vector result whose scalar components can be written as.   DV a Dt

➢Material derivative D        V   t Dt







      D      u v w t x y z Dt

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Acceleration field ➢The

material derivative contains twp types of terms, time derivatives and spatial derivatives

➢Unsteady effects ➢Time

derivatives are called local derivatives, they represent unsteadiness of flow parameters.

➢Convective effects ➢The

spatial derivatives are called convective derivatives, they represent the variation in flow parameters from one point to another.

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Unsteady ➢ Vortex induced vibrations and Wind turbines.

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Steady ➢Micro Slug Droplets Motion

➢Convective effects: Nozzle.

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Example 4.5 ➢A velocity field is given by V  V0 l  xˆi  yˆj ➢Determine the acceleration for this flow.

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Solution

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Streamline Coordinates ➢For many situations it is convenient to define a coordinates system in terms of streamline. ➢One

coordinate is along streamline ➢The second is normal to streamline

➢Advantage: Velocity is tangent to streamline

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Control Volume & Control System ➢A fluid’s behavior is governed by fundamental physical laws, that can be applied in ‘system’ approach and ‘control volume’ approach. ➢A system

is a collection of matter of fixed identity which may move and interact with surroundings. ➢A control volume is a volume in space through which fluid may flow.

➢These are applications of Lagrangian and Eulerian approach.

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Control volume ➢In fluid mechanics, it is often quite difficult to identify and keep track of a specific quantity of matter. ➢The information obtained by following a given portion of the air (a system) as it flows along is often less interesting than determining the forces put on a fan, airplane, or automobile by air flowing past the object. ➢To do this a specific volume in space is identified and the fluid flow within, through, or around that volume is analyzed.

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Control volume ➢The control volume can be either a moving volume, or a fixed, non-deformable control volume. ➢The matter within a control volume may change with time as the fluid flows through it. ➢The amount of mass within the volume may change with time. ➢Typical control volumes: ➢Fixed

control volume, ➢Fixed or moving control volume, ➢Deforming control volume.

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Reynolds transport theorem ➢At times we are interested in what happens to a particular part of the fluid as it moves about. ➢Other times we may be interested in what effect the fluid has on a particular object or volume in space as fluid interacts with it. ➢The governing laws of fluid motion are stated in terms of fluid systems, not control volumes. ➢The Reynolds transport theorem is an analytical tool to shift between control system and control volume approaches.

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Reynolds transport theorem ➢Extensive and Intensive properties ➢ Physical laws are stated in terms of parameters like v, a etc. ➢ Let Φ represent a fluid parameter/property and φ the amount

of that parameter per

unit mass, i.e. Φ = φ m. ➢ Φ is extensive property and φ is an intensive property.

➢The amount of an extensive property, Φ in a system at an instant, is equal to the sum Φ’s of each fluid particle in the system. ➢Time rate of change of extensive property for a system and a control volume.

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Reynolds transport theorem ➢Consider a control volume within a duct.

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Reynolds transport theorem

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Reynolds transport theorem ➢General form.

DΦsys Dt



  Φcv    V  nˆ dA t cs

 



DΦsys Dt



  d V    V  nˆ dA   t cv cs

 

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Reynolds theorem ➢Physical interpretation Time rate of change of extensive property,  . ➢RHS: Rate of change of  in control volume + net flowrate of  across control surface. ➢LHS:

➢Relationship to material derivative. ➢Steady and unsteady effects. ➢Moving control volumes, ➢Relative,

absolute and control volume velocity.

➢Selecting a control volume.

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Problem 4.60 ➢Water flows through a square duct with a constant, uniform velocity of V = 20m/s. ➢Consider fluid particles that lie along line A-B at time t = 0. ➢Determine the position of these particles, denoted by line A’-B’ when t = 0.2s

Q = VxA = 20(0.5)² = 5m3/s

and

L = t x Q/A = 0.2 x 5/0.25 = 4m

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Problem 4.61 ➢Water flows through a square duct with a velocity profile that is linear from 0 to 20 m/s across the duct. ➢Determine the position of the particles along line A-B at time t = 0, denoted by line A’-B’ when t = 0.2s

Q = (1/2)(h)(depth)(b), h x depth = Area of cross section  Q = 0.5 x 0.25 x (20m/s) = 2.5 m3/s LB =20m/s x 0.2 s = 4m LA = 0 m/s x 0.2 s = 0m

Ans: LA = 0, LB = 4m, & Q = 2.5m3/s

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Problem 4.67 ➢Water flows in a branching pipe with uniform velocity at each inlet and outlet. ➢The fixed control volume indicated coincides with the system at t = 20s. ➢Make a sketch to indicate ➢ The boundary of the system at t = 20.1s ➢ The fluid that left the control volume during that 0.1s interval, and ➢ The fluid that entered the control volume during that time interval.

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Solution

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Problem 4.71 ➢Water flows through a 2m-wide rectangular channel with a uniform velocity of 3 ms. ➢Determine the mass flowrate with φ = 1, across section CD of the control volume. ➢Repeat, with φ = 1/. Where  is the density.

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Solution

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Problem ➢A room contains dust of uniform concentration = dust / ➢ where  is the density of the dust/air mixture. ➢C

➢The room is to be cleaned by introducing fresh air at velocity V1 through a duct of area, A1 on one wall and exhausting the room air at velocity Vo through a duct Ao on the opposite wall. ➢Find an expression for the instantaneous rate of change of dust mass within the room.

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Solution

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END OF CHAPTER FOUR

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