Fluid - Lecture 5

8/6/2019 Fluid - Lecture 5

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THERMO-FLUID

MECHANICS 1MIET 2095

Fluid Lecture 5 Steady Flow MomentumEquation


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Announcement

Final exam time has been scheduled:

08 Jun 2011, 1.45pm to 4.00pm

Final exam venue: MSAC (Melbourne Sportsand Aquatic Centre, Aughtie Drive, AlbertPark 3206).

Mock exam paper will be uploaded shortly

(once we have finalized all questions for exam)


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THIS LECTURE

Eulerian and Lagrangian Description for Fluid

Fluid momentum and forces

Steady flow Momentum Equation


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Fire Extinguisher on a Tricycle

http://www.youtube.com/watch?v=3k_TagfAJFY


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MOTIVATION

As weve seen, Bernoullis equation is

useful for limited applications.

Now lets be more general and carry out a

force balance on fluid going through acontrol volume. This Eulerian approach

forms the basis of the Navier-Stokes

equations, which, given the rightcomputing power (often impossible atpresent) can predict flows of Newtonianfluids.


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LAGRANGIAN AND EULERIAN DESCRIPTIONSKinematics: The study of motion.Fluid kinematics: The study of how fluids flow and how to describe fluid motion.

With a small number of objects, suchas billiard balls on a pool table,individual objects can be tracked.

In the Lagrangian description, onemust keep track of the position and

velocity of individual particles.

There are two distinct ways to describe motion: Lagrangian and EulerianLagrangian description: To follow the path of individual objects.This method requires us to track the position and velocity of each individualfluid parcel (fluid particle) and take to be a parcel of fixed identity.


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A more common method is Eulerian description of fluid motion.

In the Eulerian description of fluid flow, a finite volume called a flowdomain or control volume is defined, through which fluid flows in andout.

Instead of tracking individual fluid particles, we define field variables,functions of space and time, within the control volume.

The field variable at a particular location at a particular time is thevalue of the variable for whichever fluid particle happens to occupy thatlocation at that time.

For example, the pressure field is a scalar field variable. We definethe velocity field as a vector field variable.

Collectively, these (and other) field variables define the flow field. Thevelocity field can be expanded in Cartesian coordinates as


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In the Eulerian description, onedefines field variables, such as

the pressure field and thevelocity field, at any locationand instant in time.

In the Eulerian description wedont really care what happens to

individual fluid particles; rather weare concerned with the pressure,velocity, acceleration, etc., ofwhichever fluid particle happensto be at the location of interest at

the time of interest. While there are many occasions in

which the Lagrangian descriptionis useful, the Eulerian descriptionis often more convenient for fluid

mechanics applications. Experimental measurements are

generally more suited to theEulerian description.


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Why Euler?

The Eulerian reference frame is usefulbecause measurements in a fluid are usuallymade at fixed locations.

E.g., It is far easier for us to measurepressure at a point/plane than to follow acontrol mass with a pressure sensing probe.

The moving fluid element is the Lagrangianreference frame.


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NEWTONS LAWS

Newtons laws: Relations between motions of bodies and the forcesacting on them.

Newtons first law: A body at rest remains at rest, and a body inmotion remains in motion at the same velocity in a straight path whenthe net force acting on it is zero.

Therefore, a body tends to preserve its state of inertia.

Newtons second law: The acceleration of a body is proportional tothe net force acting on it and is inversely proportional to its mass.

Newtons third law: When a body exerts a force on a second body,the second body exerts an equal and opposite force on the first.

Therefore, the direction of an exposed reaction force depends on thebody taken as the system.


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Linear momentum is theproduct of mass and velocity,and its direction is the

direction of velocity.

Newtons second law is also

expressed as the rate of change ofthe momentum of a body is equal

to the net force acting on it.

Linear momentumor just the momentumof the body: The product ofthe mass and the velocity of a body.

Newtons second law is usually referred to as the linear momentum

equation. Conservation of momentum principle: Themomentum of a system remains constantonly when the net force acting on it is zero.


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FORCES ACTING ON A CONTROL VOLUMEThe forces acting on a control volume consist of

body forces that act throughout the entire body of the control volume (such asgravity, electric, and magnetic forces) and

surface forces that act on the control surface (such as pressure and viscousforces and reaction forces at points of contact).

Only external forces are considered in the analysis.

The total force acting on a controlvolume is composed of body forcesand surface forces; body force isshown on a differential volumeelement, and surface force is shown

on a differential surface element.


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The most common body force is that of gravity, which exerts a downward forceon every differential element of the control volume.

Surface forces are not as simple toanalyze since they consist of both normal

and tangentialcomponents.Normal stresses are composed ofpressure (which always acts inwardly

normal) and viscous stresses.Shear stresses are composed entirely ofviscous stresses.

The gravitational force acting on a differentialvolume element of fluid is equal to its weight; the

axes have been rotated so that the gravity vectoracts downwardin the negative z-direction.


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A common simplification in the application of Newtons laws of motion is to

subtract the atmospheric pressureand work with gage pressures.

This is because atmospheric pressure acts in all directions, and its effect cancelsout in every direction.

This means we can also ignore the pressure forces at outlet sections where thefluid is discharged to the atmosphere since the discharge pressure in such casesis very near atmospheric pressure at subsonic velocities.

Atmospheric pressure acts in alldirections, and thus it can be ignoredwhen performing force balances sinceits effect cancels out in every direction.

Cross section through a faucetassembly, illustrating the importance ofchoosing a control volume wisely; CV Bis much easier to work with than CV A.


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However, a control volume may have massflows in and out, so how is Newtons 2nd Lawapplied to a control volume ?

Answer: This law can be applied on a net massflowing through the control volume.


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Mass Entering and Leaving CV


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Momentum Flux is the product of mass flow rate andvelocity

It is a vector in the same direction as the velocity


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THE LINEAR MOMENTUM EQUATION

Newtons second law can be stated as the sum of all external

forces acting on a system is equal to the time rate of change of

linear momentum of the system. This statement is valid for acoordinate system that is at rest or moves with a constant velocity,called an inertial coordinate systemor inertial reference frame.


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CHOOSING A CONTROL VOLUMEA control volume can be selected as any arbitraryregion in space through which fluid flows, and itsbounding control surface can be fixed, moving, and

even deforming during flow.

Many flow systems involve stationary hardware firmlyfixed to a stationary surface, and such systems arebest analyzed using fixedcontrol volumes.

When analyzing flow systems that are moving or

deforming, it is usually more convenient to allow thecontrol volume to moveor deform.

In deformingcontrol volume, part of the controlsurface moves relative to other parts.


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Steadyflow

Mass flow rate acrossan inlet or outlet

Momentum flow rate acrossa uniform inlet or outlet:

In a typical engineeringproblem, the control volumemay contain many inlets andoutlets; at each inlet or outletwe define the mass flow rate

and the average velocity.

Special Cases


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Examples of inlets or outlets in which the uniform flowapproximation is reasonable:(a) the well-rounded entrance to a pipe,

(b) the entrance to a wind tunnel test section, and(c) a slice through a free water jet in air.


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Momentum-Flux Correction Factor, The velocity across most inlets and outlets is notuniform.The control surface integral of Eq. 1313 may be converted into algebraic form usinga dimensionless correction factor, called the momentum-flux correction factor.

(13-13)

is always greater than or equal to 1. is close to 1 for turbulent flow and is4/3 for fully developed laminar flow.


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Steady Flow with One Inlet and One Outlet

One inlet andone outlet

Along x-coordinate

A control volume with only oneinlet and one outlet.

The determination by vectoraddition of the reaction force onthe support caused by a changeof direction of water.


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Flow with No External Forces

In the absence of external forces, the rate of change of the

momentum of a control volume is equal to the difference between

the rates of incoming and outgoing momentum flow rates.

The thrust needed to lift the spaceshuttle is generated by the rocketengines as a result of momentum

change of the fuel as it is acceleratedfrom about zero to an exit speed of

about 2000 m/s after combustion.


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Example

Thrust generated by an aircraft engine,assuming that the air pressure all around theengine is atmospheric


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From google book-- Jet propulsion: a simple guide to the aerodynamics and thermodynamic design .

By N. A. Cumpsty

SFx = Fpylon + pressure forcesx= mdot*Vj - mdot*V = mdot*(Vj - V)

zerothrust


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Class Question

13-2 A reducing elbow is used to deflect water flow at a rate of14kg/s in a horizontal pipe upward 30o while accelerating it. Theelbow is 113 cm2 at the inlet and 7 cm2 at the outlet. The

elevation difference between the centers of the outlet and theinlet is 30 cm. The weight of the elbow and the water in it isconsidered to be negligible. Determine (a) the gage pressure atthe centre of the inlet of the elbow and (b) the anchoring forceneeded to hold the elbow in place.


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Practical Applications


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Read and Study

Sections 13.1, 13.4, 13.5 and 13.6

Solve Problems

13-1C, 13-2C, 13-3C, 13-4C, 13-5C,

13-13C, 13-14C, 13-19C


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Next Lecture

More on Angular Momentum Equations

Examples of application of Steady flowMomentum Equations.

Fluid - Lecture 5

Documents

Transcript of Fluid - Lecture 5