Me804306 2 fluid mechanics
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Chapter 3 Fluid Dynamics (Fluid Kinematics). ME804306-2 Fluid Mechanics. Dr. Kamel Mohamed Guedri Mechanical Engineering Department, The College of Engineering and Islamic Architecture, Umm Al- Qura University, Room H1091 Website: https://uqu.edu.sa/kmguedri Email: [email protected]

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ME804306-2 Fluid Mechanics

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Me804306 2 fluid mechanics

Chapter 3

Fluid Dynamics

(Fluid Kinematics)

ME804306-2Fluid Mechanics

Dr. Kamel Mohamed Guedri

Mechanical Engineering Department,The College of Engineering and Islamic Architecture,

Umm Al-Qura University, Room H1091

Website: https://uqu.edu.sa/kmguedriEmail: [email protected]


Outline

OUTLINE

  • Introduction

  • Objectives

  • 3.1 Uniform Flow, Steady Flow

  • 3.1.1 Laminar, Turbulent Flow

  • 3.1.2 Relative Motion

  • 3.1.3 Compressible or Incompressible

  • 3.1.4 One, Two or Three-dimensional Flow

  • 3.1.5 Streamlines

  • 3.1.6 Streamtubes

  • 3.2. Mass and Flow Rate

  • 3.2.1 Mass Flow Rate

  • 3.2.2 Volume Flow Rate

  • 3.3 The Fundamental Equations of Fluid Dynamics

  • 3.3.1 Continuity (Principle of Conservation of Mass)

  • 3.3.2 Work and Energy (Principle of Conservation of Energy)


Outline cont

OUTLINE (Cont)

  • 3.4. Application of Bernoulli Equation

  • 3.4.1 Pitot Tube

  • 3.4.2 Pitot Static Tube

  • 3.4.3 Venturi Meter

  • 3.4.4 Sharp Edge Circular Orifice

  • 3.4.5 Nozzles

  • 3.4.6 Flow over Notches and Weirs

  • 3.5. The Momentum Equation

  • 3.6. Application of the Momentum Equation

  • 3.6.1 The force due the flow around a pipe bend

  • 3.6.2 The force on a pipe nozzle

  • 3.6.3 Impact of a Jet on a plane

  • Summary


Introduction

Introduction

  • Discusses the analysis of fluid in motion: fluid dynamics.

  • When a fluid flows through pipes and channel or around bodies such as aircraft and ships, the shape of the boundaries, the externally applied forces and the fluid properties cause the velocities of the fluid particles to vary from point to point throughout the flow field.

  • The motion of fluids can be predicted using the fundamental laws of physics together with the physical properties of the fluid.

  • The geometry of the motion of fluid particles in space and time is known as the kinematics of the fluid motion.

  • A fluid motion may be specified by either tracing the motion of a particle through the field of flow or examining the motion of all particles as they pass a fixed point in space.

  • This course will use the second method where the emphasis is on the spatial position rather than on the particle, or known as Eulerian Approach.


Objectives

Objectives

  • comprehend the concepts necessary to analyse fluids in motion.

  • identify differences between steady/unsteady, uniform/non-uniform and compressible/incompressible flow.

  • construct streamlines and stream tubes.

  • appreciate the Continuity principle through Conservation of Mass and Control Volumes.

  • derive the Bernoulli (energy) equation.

  • familiarise with the momentum equation for a fluid flow.


3 1 uniform flow steady flow

3.1 Uniform Flow, Steady Flow

uniform flow:flow velocity is the same magnitude and direction at every point in the fluid.

non-uniform:If at a given instant, the velocity is not the same at every point the flow. (In practice, by this definition, every fluid that flows near a solid boundary will be non-uniform - as the fluid at the boundary must take the speed of the boundary, usually zero. However if the size and shape of the of the cross-section of the stream of fluid is constant the flow is considered uniform.)

steady: A steady flow is one in which the conditions (velocity, pressure and cross-section) may differ from point to point but DO NOT change with time.

unsteady:If at any point in the fluid, the conditions change with time, the flow is described as unsteady. (In practice there is always slight variations in velocity and pressure, but if the average values are constant, the flow is considered steady.)


Me804306 2 fluid mechanics

  • 3.1 Uniform Flow, Steady Flow (cont.)

  • Steady uniform flow:

    • Conditions: do not change with position in the stream or with time.

    • Example: the flow of water in a pipe of constant diameter at constant velocity.

  • Steady non-uniform flow:

    • Conditions: change from point to point in the stream but do not change with time.

    • Example: flow in a tapering pipe with constant velocity at the inlet-velocity will change as you move along the length of the pipe toward the exit.

  • Unsteady uniform flow:

    • At a given instant in time the conditions at every point are the same, but will change with time.

    • Example: a pipe of constant diameter connected to a pump pumping at a constant rate which is then switched off.

  • Unsteady non-uniform flow:

    • Every condition of the flow may change from point to point and with time at every point.

    • Example: waves in a channel.


3 1 1 laminar and turbulent flow

  • Typical particles path

3.1.1 Laminar and Turbulent Flow

  • Laminar flow

    • all the particles proceed along smooth parallel paths and all particles on any path will follow it without deviation.

    • Hence all particles have a velocity only in the direction of flow.

  • Figure 3.1a: Laminar flow


Me804306 2 fluid mechanics

  • Particle paths

  • Turbulent Flow

    • The particles move in an irregular manner through the flow field.

    • Each particle has superimposed on its mean velocity fluctuating velocity components both transverse to and in the direction of the net flow.

  • Transition Flow

    • exists between laminar and turbulent flow.

    • In this region, the flow is very unpredictable and often changeable back and forth between laminar and turbulent states.

    • Modern experimentation has demonstrated that this type of flow may comprise short ‘burst’ of turbulence embedded in a laminar flow.

  • Figure 3.1b: Turbulent flow


3 1 2 relative motion

  • Observer

  • Observer

  • Boat stationary

  • Fluid moving past boat pattern stationary relative to boat

  •  does not change with time  STEADY

  • Boat moving

  • Flow pattern moves along channel with boat

  •  changes with time

  • UNSTEADY

3.1.2Relative Motion

  • Figure 3.2: Relative motion


3 1 3 compressible or incompressible

3.1.3 Compressible or Incompressible

  • All fluids are compressible - even water - their density will change as pressure changes.

  • Under steady conditions, and provided that the changes in pressure are small, it is usually possible to simplify analysis of the flow by assuming it is incompressible and has constant density.

  • As you will appreciate, liquids are quite difficult to compress - so under most steady conditions they are treated as incompressible.


3 1 4 one two or three dimensional flow

3.1.4 One, Two or Three-dimensional Flow

  • In general, all fluids flow three-dimensionally, with pressures and velocities and other flow properties varying in all directions.

  • In many cases the greatest changes only occur in two directions or even only in one.

  • In these cases changes in the other direction can be effectively ignored making analysis much more simple.


Me804306 2 fluid mechanics

  • Flow isone dimensionalif the flow parameters (such as velocity, pressure, depth etc.) at a given instant in time only vary in the direction of flow and not across the cross-section. The flow may be unsteady, in this case the parameter vary in time but still not across the cross-section.

  • An example of one-dimensional flow is the flow in a pipe. Note that since flow must be zero at the pipe wall - yet non-zero in the center - there is a difference of parameters across the cross-section.

  • Should this be treated as two-dimensional flow? Possibly - but it is only necessary if very high accuracy is required. A correction factor is then usually applied.


Me804306 2 fluid mechanics

  • Flow is two-dimensionalif it can be assumed that the flow parameters vary in the direction of flow and in one direction at right angles to this direction.

  • Streamlines in two-dimensional flow are curved lines on a plane and are the same on all parallel planes.

  • An example is flow over a weir for which typical streamlines can be seen in the figure below. Over the majority of the length of the weir the flow is the same - only at the two ends does it change slightly. Here correction factors may be applied.


3 1 5 streamlines

3.1.5 Streamlines

  • In analysing fluid flow it is useful to visualise the flow pattern.

  • This can be done by drawing lines joining points of equal velocity - velocity contours. These lines are known as streamlines.

  • Figure 3.5: A streamline


Me804306 2 fluid mechanics

  • Close to a solid boundary streamlines are parallel to that boundary

    • At all points the direction of the streamline is the direction of the fluid velocity: this is how they are defined. Close to the wall the velocity is parallel to the wall so the streamline is also parallel to the wall.

    • It is also important to recognise that the position of streamlines can change with time - this is the case in unsteady flow. In steady flow, the position of streamlines does not change.

  • Some things to know about streamlines

  • Because the fluid is moving in the same direction as the streamlines, fluid can not cross a streamline.

  • Streamlines can not cross each other. If they were to cross this would indicate two different velocities at the same point. This is not physically possible.

  • The above point implies that any particles of fluid starting on one streamline will stay on that same streamline throughout the fluid.


3 1 6 streamtubes

3.1.6 Streamtubes

  • A useful technique in fluid flow analysis is to consider only a part of the total fluid in isolation from the rest.

  • This can be done by imagining a tubular surface formed by streamlines along which the fluid flows.

  • This tubular surface is known as a streamtube.

  • In a two-dimensional flow, we have a streamtube which is flat (in the plane of the paper).

  • Figure 3.7: (a) A streamtube


3 2 1 mass flow rate

  • mass of fluid

  • time taken to collect the fluid

  • mass

  • mass flow rate

  • The "walls" of a streamtube are made of streamlines.

  • As we have seen above, fluid cannot flow across a streamline, so fluid cannot cross a streamtube wall.

  • The streamtube can often be viewed as a solid walled pipe.

  • A streamtube is not a pipe - it differs in unsteady flow as the walls will move with time. And it differs because the "wall" is moving with the fluid.

3.2.1 Mass flow rate

3.2 Mass and volume flow rate

  • mass flow rate= m=

  • time =


3 2 2 volume flow rate discharge

  • =

  • m

  • volume of fluid

  • time

  • mass fluid rate

  • density

  • mass of fluid

  • density x time

  • r

  • mass

  • volume

  • (

  • )

  • density =

3.2.2 Volume flow rate - Discharge

  • discharge= Q =

  • =

  • =

  • &


3 2 3 discharge and mean velocity

3.2.3 Discharge and mean velocity

  • If the area of cross section of the pipe at point XisA, and the mean velocity here is um, during a time t, a cylinder of fluid will pass point Xwith a volume A  umt. The volume per unit time (the discharge) will thus be :

  • Q =Q=

  • orum=

  • Letum= V  um = V =

  • Figure 3.8: Discharge in pipe


Me804306 2 fluid mechanics

  • Note how carefully we have called this the mean velocity. This is because the velocity in the pipe is not constant across the cross section.

  • Crossing the centre line of the pipe, the velocity is zero at the walls, increasing to a maximum at the centre then decreasing symmetrically to the other wall.

  • This variation across the section is known as the velocity profile or distribution. A typical one is shown in the figure

  • This idea, that mean velocity multiplied by the area gives the discharge, applies to all situations - not just pipe flow.

  • Figure 3.9: A typical velocity profile across a pipe


Me804306 2 fluid mechanics

  • 8

  • 1.7

  • = 4.7s

  • mass of fluid in bucket

  • time taken to collect the fluid

  • mass

  • mass flow rate

  • 8.0 -2.0

  • 7

  • Example 3.1

  • An empty bucket weighs 2.0 kg. After 7 seconds of collecting water the bucket weighs 8.0 kg, then:

  • mass flow rate= ṁ =

  • == 0.857 kg/s (kg s-1)

  • Example 3.2

  • If we know the mass flow is 1.7 kg/s, how long will it take to fill a container with 8 kg of fluid?

  • time=

  • =


Me804306 2 fluid mechanics

  • 0.857

  • 850

  • =

  • m

  • mass fluid rate

  • density

  • r

  • Example 3.3

  • If the density of the fluid in the above example is 850 kg/m3 what is the volume per unit time (the discharge)?

  • Q=

  • =

  • =0.00108 m3/s (m3s-1)

  • = 1.008  10-3 m3/s

  • but1 litre = 1.0 10-3m3,

  • soQ= 1.008 l/s

  • Example 3.4

  • If the cross-section area, A, is 1.2 x 10-3 m2 and the discharge, Q is 24 l/s, what is the mean velocity, of the fluid?

  • Let mean velocity, um = V

  • um = V = = 2.0 m/s

  • 2.4 x 10-3m3/s

  • 1.2 x 10-3


3 3 the fundamental equations of fluid dynamics

3.3 The Fundamental Equations of Fluid Dynamics

1. The law of conservation of matter

  • stipulates that matter can be neither created nor destroyed, though it may be transformed (e.g. by a chemical process).

  • Since this study of the mechanics of fluids excludes chemical activity from consideration, the law reduces to the principle of conservation of mass.

  • 2. The law of conservation of energy

  • states that energy may be neither created nor destroyed.

  • Energy can be transformed from one guise to another (e.g. potential energy can be transformed into kinetic energy), but none is actually lost.

  • Engineers sometimes loosely refer to ‘energy losses’ due to friction, but in fact the friction transforms some energy into heat, so none is really ‘lost’.


Me804306 2 fluid mechanics

3. The law of conservation of momentum

  • states that a body in motion cannot gain or lose momentum unless some external force is applied.

  • The classical statement of this law is Newton's Second Law of Motion, i.e.

  • force = rate of change of momentum


3 3 1 continuity principle of conservation of mass

  • CONTROL VOLUME

  • Outflow

  • Inflow

  • Control surface

3.3.1Continuity (Principle of Conservation of Mass)

  • Matter cannot be created nor destroyed - (it is simply changed in to a different form of matter).

  • This principle is known as the conservation of massand we use it in the analysis of flowing fluids.

  • The principle is applied to fixed volumes, known as control volumes or surfaces

  • Figure 3.10: A control volume


Me804306 2 fluid mechanics

  • For any control volume the principle of conservation of mass says

  • Mass entering = Mass leaving + Increase of mass in the control

  • per unit time per unit time volume per unit time

  • For steady flow:

  • (there is no increase in the mass within the control volume)

  • Mass entering per unit time = Mass leaving per unit time

  • Mass entering per unit time at end 1 = Mass leaving per unit time at end 2

  • Figure 3.11: A streamtube section


Me804306 2 fluid mechanics

  • flow is incompressible, the density of the fluid is constant throughout the fluid continum. Mass flow, m, entering may be calculated by taking the product

    (density of fluid,)  (volume of fluid entering per second Q)

  • Mass flow is therefore represented by the product Q, hence

     Q (entering) =  Q (leaving)

  • But since flow is incompressible, the density is constant, so

    Q (entering) = Q (leaving)(3.5a)

  • This is the ‘continuity equation’ for steady incompressible flow.


Me804306 2 fluid mechanics

  • If the velocity of flow across the entry to the control volume is measured, and that the velocity is constant at V1 m/s. Then, if the cross-sectional area of the streamtube at entry is A1,

    Q (entering) = V1 A1

  • Thus, if the velocity of flow leaving the volume is V2 and the area of the streamtube at exit is A2, then

    Q (leaving) = V2A2

  • Therefore, the continuity equation may also be written as

    V1A1 = V2A2(3.5b)


Me804306 2 fluid mechanics

  • A1V1

  • A2

  • Application of Continuity Equation

  • We can apply the principle of continuity to pipes with cross sections which change along their length.

  • A liquid is flowing from left to right and the pipe is narrowing in the same direction. By the continuity principle, the mass flow ratemust be the same at each section - the mass going into the pipe is equal to the mass going out of the pipe. So we can write:

    1 A1V1= 2 A2V2

  • As we are considering a liquid, usually water, which is not very compressible, the density changes very little so we can say 1 =2 =. This also says that the volume flow rate is constant or that

  • Discharge at section 1 = Discharge at section 2

  • Q1 = Q2

  • A1V1 = A2V2 orV2 =

  • Figure 3.12: Pipe with a contraction


Another example is a diffuser a pipe which expands or diverges as in the figure below

  • ( )

  • 2

  • A1

  • A2

  • pd12/4

  • pd22/4

  • d12

  • d22

  • d12

  • d22

  • V1=

  • V1=

  • V1

  • V1

Another example is a diffuser, a pipe which expands or diverges as in the figure below

  • As the area of the circular pipe is a function of the diameter we can reduce the calculation further,

  • V2 =

  • V2= (3.6)


Me804306 2 fluid mechanics

  • The continuity principle can also be used to determine the velocities in pipes coming from a junction.

  • Total mass flow into the junction = Total mass flow out of the junction

  • 1Q1 = 2Q2 + 3Q3

  • When the flow is incompressible (e.g. water) 1 = 2 = 

  • Q1 = Q2 + Q3

  • A1V1 = A2V2 + A3V3(3.7)


Me804306 2 fluid mechanics

  • ( )

  • A1V1

  • A2

  • 10 x 10-3 x 2.1

  • 3 x 10-3

  • 2

  • 40

  • 30

  • =

  • 3.0 = 5.3m/s

  • Example 3.5

  • If the area in Figure 3.12 A1 = 10  10-3m2 and A1= 10  10-3m2and and the upstream mean velocity, V1 = 2.1 m/s, what is the downstream mean velocity?

  • V2 =

  • = 7.0 m/s

  • Example 3.6

  • If the diameter of a diffuser (Figure 3.13) at section 1 is d1 = 30 mm and at section 2 d2 = 40 mm and the mean velocity at section 2 is V2 = 3.0 m/s. Calculate the velocity entering the diffuser.

  • V2 =


Me804306 2 fluid mechanics

  • Example 3.7

  • For a junction (Figure 3.14), if pipe 1 diameter = 50 mm, mean velocity 2 m/s, pipe 2 diameter 40 mm takes 30% of total discharge and pipe 3 diameter 60 mm. What are the values of discharge and mean velocity in each pipe?

  • Q1 = A1V1 = = 0.00392 m3/s

  • ButQ2 = 0.3Q1 = 0.001178 m3/s

  • AlsoQ1 = Q2 + Q3

  • Q3 = Q1 – 0.3Q1 = 0.7Q1 = 0.00275 m3/s

  • V2 = Q2 / V2 = 0.936 m/s

  • V3 = Q3 / V3 = 0.972 m/s


3 3 2 work and energy principle of conservation of energy

3.3.2Work and Energy(Principle Of Conservation Of Energy)

  • friction: negligible

  • sum of kinetic energy and gravitational potential energy is constant. Recall :

  • Kinetic energy = ½ mV2

  • Gravitational potential energy = mgh

    (m: mass, V: velocity, h: height above the datum).


Me804306 2 fluid mechanics

  • mgh = ½ mV2or

  • To apply this to a falling body we have an initial velocity of zero, and it falls through a height of h.

    • Initial kinetic energy = 0

    • Initial potential energy = mgh

    • Final kinetic energy = ½ mV2

    • Final potential energy = 0

  • We know that,

    • kinetic energy + potential energy = constant

  • Initial

  • potential

  • Energy

  • Final

  • Kinetic

  • Energy

  • Final

  • Potential

  • Energy

  • Initial

  • kinetic

  • Energy

  • }+{

  • }={

  • }+{

  • {

  • }


Continuous jet of liquid

continuous jet of liquid

  • Figure 3.15 : The trajectory of a jet of water

  • a continuous jet of water coming from a pipe with velocity V1.

  • One particle of the liquid with mass m travels with the jet and falls from height z1to z2.

  • The velocity also changes from V1 to V2. The jet is traveling in air where the pressure is everywhere atmospheric so there is no force due to pressure acting on the fluid.

  • The only force which is acting is that due to gravity. The sum of the kinetic and potential energies remains constant (as we neglect energy losses due to friction) so :

    mgz1 + mV12 = mgz2 + mV22

  • As m is constant this becomes :

    V12 + gz1 = V22 + gz2


Flow from a reservoir

Flow from a reservoir

  • The level of the water in the reservoir is z1. Considering the energy situation - there is no movement of water so kinetic energy is zero but the gravitational potential energy is mgz1.

  • If a pipe is attached at the bottom water flows along this pipe out of the tank to a level z2. A mass m has flowed from the top of the reservoir to the nozzle and it has gained a velocity V2. The kinetic energy is now ½mV22 and the potential energy mgz2. Summarising :

    • Initial kinetic energy = 0

    • Initial potential energy = mgz1

    • Final kinetic energy = ½ mV22

    • Final potential energy = mgz2

  • So

    mgz1 = ½ mV22 + mgz2

    mg ( z1 - z2 ) = ½ mV22

     V2 = (3.8)

  • Figure 3.16 :

  • Flow from a reservoir


Me804306 2 fluid mechanics

  • Example 3.8

  • A reservoir of water has the surface at 310 m above the outlet nozzle of a pipe with diameter 15mm. What is the

    • velocity;

    • the discharge out of the nozzle; and

    • mass flow rate. (Neglect all friction in the nozzle and the pipe)

  • Solution:

  • a)

  • b) Volume flow rate is equal to the area of the nozzle multiplied by the velocity

  • Q = AV

  • =

  • =

  • = 0.01378 m3/s

  • c) The density of water is 1000 kg/m3 so the mass flow rate is

  • ṁ = density  volume flow rate

  • =  Q

  • = 1000  0.01378

  • = 13.78 kg/s


Bernoulli s equation

Bernoulli's Equation

  • We see that from applying equal pressure or zero velocities we get the two equations from the section above. They are both just special cases of Bernoulli's equation.

  • Bernoulli's equation has some restrictions in its applicability, they are:

    • Flow is steady;

    • Density is constant (which also means the fluid is incompressible);

    • Friction losses are negligible.

    • The equation relates the states at two points along a single streamline, (not conditions on two different streamlines).


Me804306 2 fluid mechanics

  • A fluid of constant density = 960 kg/m3 is flowing steadily through the above tube. The diameters at the sections are d1 = 100mm and d2 = 80mm. The gauge pressure at 1 is P1 = 200 kN/m2and the velocity here is V1 = 5m/s. What is the gauge pressure at section 2.

  • Bernoulli equation is applied along a streamline joining section 1 with section 2.

  • The tube is horizontal, with z1 = z2 so Bernoulli gives us the following equation for pressure at section 2:

  • P2 = P1 + (V12 – V22)

  • Figure 3.19 : A contracting expanding pipe


Me804306 2 fluid mechanics

  • p2 = 200000 -17296.87

  • = 182703 N/m2

  • = 182.7 kN/m2

  • r

  • 2

  • P2 = P1+ (V12 – V22) = 200000 + (52 – 7.81252)

  • 960

  • 2

  • But we do not know the value of V2. We can calculate this from the continuity equation: Discharge into the tube is equal to the discharge out i.e.

  • = 7.8125 m/s

  • So we can now calculate the pressure at section 2


Modifications of bernoulli equation

  • (3.12)

Modifications of Bernoulli Equation

  • In practice, the total energy of a streamline does not remain constant. Energy is ‘lost’ through friction, and external energy may be either :

    • added by means of a pump or

    • extracted by a turbine.

  • Consider a streamline between two points 1 and 2. If the energy head lost through friction is denoted by Hf and the external energy head added (say by a pump) is or extracted (by a turbine) HE, then Bernoulli's equation may be rewritten as :

  • ± HE = H2 + Hf(3.11)

  • or

  • HE = energy head added/loss due to external source such as pump/turbines

  • This equation is really a restatement of the First Law of Thermodynamics for an incompressible fluid.


The power equation

The Power Equation

  • In the case of work done over a fluid the power input into the flow is :

  • P = gQHE(3.13)

  • where Q = discharge,

  • HE= head added / loss

  • If p = efficiency of the pump, the power input required,

  • Pin =(3.14)


Objectives1

Objectives

3.4 Application of Bernoulli Equation

  • The Bernoulli equation can be applied to a great many situations not just the pipe flow we have been considering up to now.

  • In the following sections we will see some examples of its application to flow measurement from tanks, within pipes as well as in open channels.

  • Acknowledge practical uses of the Bernoulli and momentum equation in the analysis of flow

  • Understand how the momentum equation and principle of conservation of momentum is used to predict forces induced by flowing fluids

  • Apply Bernoulli and Momentum Equations to solve fluid mechanics problems


Me804306 2 fluid mechanics

  • X

  • 0

3.4.1Pitot tube

  • A point in a fluid stream where the velocity is reduced to zero is known as a stagnation point.

  • Any non-rotating obstacle placed in the stream produces a stagnation point next to its upstream surface.

  • The velocity at X is zero: X is a stagnation point.

  • Figure 3.20: Streamlines passing a non-rotating obstacle

  • By Bernoulli's equation the quantity p + ½V2 + gz isconstant along a streamline for the steady frictionless flow of a fluid of constant density.

  • If the velocity V at a particular point is brought to zero the pressure there is increased from p to p + ½V2.

  • For a constant-density fluid the quantity p + ½V2is therefore known as the stagnation pressure of that streamline while ½V2 – that part of the stagnation pressure due to the motion – is termed the dynamic pressure.

  • A manometer connected to the point X would record the stagnation pressure, and if the static pressure p were also known ½V2 could be obtained by subtraction, and hence V calculated.


Me804306 2 fluid mechanics

  • Figure 3.21: Simple Pitot Tube

  • A right-angled glass tube, large enough for capillary effects to be negligible, has one end (A) facing the flow. When equilibrium is attained the fluid at A is stationary and the pressure in the tube exceeds that of the surrounding stream by ½V2. The liquid is forced up the vertical part of the tube to a height :

  • h = p/g = ½V2/g = V2/2g

  • above the surrounding free surface. Measurement of h therefore enables V to be calculated.

  • (3.15)


Me804306 2 fluid mechanics

  • Measurement of the static pressure may be made at the boundary of the flow, as illustrated in (a), provided that the axis of the piezometer is perpendicular to the boundary and the connection is smooth and that the streamlines adjacent to it are not curved

  • A tube projecting into the flow (Tube c) does not give a satisfactory reading because the fluid is accelerating round the end of the tube.

  • Figure 3.22: Piezometers connected to a pipe


Me804306 2 fluid mechanics

  • Two piezometers, one as normal and one as a Pitot tube within the pipe can be used in an arrangement shown below to measure velocity of flow.

  • From the expressions above,

  • Figure 4.23 : A Piezometer and a Pitot tube


3 4 2 pitot static tube

3.4.2 Pitot static tube

  • The tubes recording static pressure and stagnation pressure are frequently combined into one instrument known as a Pitot-static tube

  • The ‘static’ tube surrounds the ‘total head’ tube and two or more small holes are drilled radially through the outer wall into the annular space.

  • The position of these ‘static holes’ is important. This instrument, when connected to a suitable manometer, may be used to measure point velocities in pipes, channels and wind tunnels.

  • Figure 3.24: Pitot static tube


Me804306 2 fluid mechanics

  • Consider the pressures on the level of the centre line of the Pitot tube and using the theory of the manometer,

  • PA = P2 + gX

  • PB = P1 + g(X-h) + maxgh

  • PA = PB

  • P2 + gX = P2 + g(X – h) + mangh

  • We know that , substituting this in to the above gives


3 4 3 venturi meter

3.4.3 Venturi meter

  • The Venturi meter is a device for measuring discharge in a pipe.

  • It consists of a rapidly converging section, which increases the velocity of flow and hence reduces the pressure.

  • It then returns to the original dimensions of the pipe by a gently diverging ‘diffuser’ section. By measuring the pressure differences the discharge can be calculated.

  • This is a particularly accurate method of flow measurement as energy losses are very small.

  • Figure 4.25: A Venturi meter


Me804306 2 fluid mechanics

  • Applying Bernoulli Equation between (1) and (2), and using continuity equation to eliminate V2 will give :

  • (3.16)

  • and Qideal = V1A1

  • To get the actual discharge, taking into consideration of losses due to friction, a coefficient of discharge, Cd, is introduced.

  • Qactual = Cd Qideal = (3.17)

  • It can be shown that the discharge can also be expressed in terms of manometer reading :

  • Q actual = (3.18)

  • where man = density of manometer fluid


Me804306 2 fluid mechanics

  • Example 3.9

  • A Venturi meter with an entrance diameter of 0.3 m and a throat diameter of 0.2 m is used to measure the volume of gas flowing through a pipe. The discharge coefficient of the meter is 0.96. Assuming the specific weight of the gas to be constant at 19.62 N/m3, calculate the volume flowing when the pressure difference between the entrance and the throat is measured as 0.06 m on a water U-tube manometer.

  • Solution:

  • What we know from the problem statement :

  • rgg = 19.62 N/m2

  • Cd = 0.96

  • d1 = 0.3m

  • d2 = 0.2m

  • Calculate Q:

  • V1 = Q/0.0707V2 = Q/0.0314


Me804306 2 fluid mechanics

  • For the manometer :

  • --- (1)

  • For the Venturi meter :

  • --- (2)

  • Combining (1) and (2) :


Me804306 2 fluid mechanics

  • (1)

  • h

  • streamline

  • (2)

  • datum

  • 3.4.4 Sharp edge circular orifice

  • Consider a large tank, containing an ideal fluid, having a small sharp-edged circular orifice in one side.

  • If the head, h, causing flow through the orifice of diameter d is constant (h>>d), Bernoulli equation may be applied between two points, (1) on the surface of the fluid in the tank and (2) in the jet of fluid just outside the orifice. Hence :


Me804306 2 fluid mechanics

  • NowP1 = Patmand as the jet in unconfined, P2 = Patm. If the flow is steady, the surface in the tank remains stationary andV1 0 (z2=0, z1=h) and ignoring losses we get :

    or the velocity through the orifice,

    (3.19)

  • This result is known as Toricelli’s equation.

  • Assuming no loses, ideal fluid, V constant across jet at (2), the

  • discharge through the orifice is

  • where A0 is the area of the orifice


Me804306 2 fluid mechanics

  • For the flow of areal fluid,the velocity is less than that givenby eq. 3.19 because offrictional effectsand so theactual velocityV2a, is obtained by introducing a modifying coefficient,Cv, the coefficient of velocity:

  • Velocity, (3.20)

  • or

  • (typically about 0.97)


Me804306 2 fluid mechanics

  • Vena contracta

  • d0

  • P = Patm

  • approx. d0/2

  • As a real fluid cannot turn round a sharp bend, the jet continues to contract for a short distance downstream (about one half of the orifice diameter) and the flow becomes parallel at a point known as thevena contracta(Latin : contracting vein).

  • Figure 3.26: The formation of vena contracta


Me804306 2 fluid mechanics

  • The area of discharge is thus less than the orifice area and a coefficient of contraction, Cc, must be introduced.

  • Hence the actual discharge is :

  • (typically about 0.65)

  • or introducing a coefficient of discharge, Cd, where :

  • (typically 0.63)

  • (3.21)

  • (3.22)

  • Cd = Cc . Cv


3 4 5 nozzles

  • dnozzle

  • contraction within nozzle

3.4.5 Nozzles

  • In a nozzle, the flow contracts gradually to the outlet and hence the area of the jet is equal to the outlet area of the nozzle.

  • i.e.Cc = 1.0

  • therefore Cd = Cv

  • Taking a datum at the nozzle, Torricelli’s equation gives the total energy head in the system as it assumes an ideal fluid and hence no loss of energy, i.e. theoretical head :

  • (3.23)

  • Figure 3.27: Contraction within a nozzle


Me804306 2 fluid mechanics

  • theoretical

  • actual

  • But the actual velocity is :

  • V2a = Cv V2

  • and the actual energy in the jet is :

  • as P2and z2 are zero. Therefore the actual energy head is :

  • (3.24)

  • And the loss of energy head, hf , in the nozzle due to friction is :

  • (3.25)


Me804306 2 fluid mechanics

  • 3.4.6Flow over notches and weirs

  • Notch

  • is an opening in the side of a tank or reservoir, which extends above the surface of the liquid.

  • It is usually a device for measuring discharge.

  • A weir is a notch on a larger scale – usually found in rivers.

  • It may be sharp crested but also may have a substantial width in the direction of flow – it is used as both a flow measuring device and a device to raise water levels.

  • Weir Assumptions

  • assume that the velocity of the fluid approaching the weir is small so that kinetic energy can be neglected.

  • assume that the velocity through any elemental strip depends only on the depth below the free surface.

  • These are acceptable assumptions for tanks with notches or reservoirs with weirs, but for flows where the velocity approaching the weir is substantial the kinetic energy must be taken into account (e.g. a fast moving river).


Me804306 2 fluid mechanics

  • A General Weir Equation

  • To determine an expression for the theoretical flow through a notch we will consider a horizontal strip of width b and depth h below the free surface, as shown in the figure

  • velocity through the strip

  • V =

  • discharge through the strip,

  • Integrating from the free surface, h = 0, to the weir crest, h = H gives the expression for the total theoretical discharge,

  • Qtheoretical =


Me804306 2 fluid mechanics

  • Rectangular Weir

  • For a rectangular weir the width does not change with depth so there is no relationship between b and depth h. We have the equation, b = constant = B.

  • Substituting this with the general weir equation gives:

  • (3.26)

  • Figure 3.28 : A rectangular weir

  • To calculate the actual discharge we introduce a coefficient of discharge, Cd, which accounts for losses at the edges of the weir and contractions in the area of flow, giving :

  • (3.27)


Me804306 2 fluid mechanics

  • H = 155 mm

  • Cross sectional

  • Area = 0.26 m2

  • = 0.0660 m3/s

Example 3.10

  • Water flows over a sharp-crested weir 600 mm wide. The measured head (relative to the crest) is 155 mm at a point where the cross-sectional area of the stream is 0.26 m2. Calculate the discharge, assuming that Cd = 0.61.

  • As first approximation,


Me804306 2 fluid mechanics

  • Velocity of approach =

  • = 0.254 m/s

  • H + V12/2g = (0.155 + 0.00328) m = 0.1583 m

    Second approximation:

    Further refinement of the value could be obtained by a new calculation of V1 (0.0681 m3/s  0.26 m2), a new calculation of H + V12/2g and so on. One correction is usually sufficient, however, to give a value of Q acceptable to three significant figures.


3 5 the momentum equation

3.5 The momentum equation

  • We have all seen moving fluids exerting forces. The lift force on an aircraft is exerted by the air moving over the wing. A jet of water from a hose exerts a force on whatever it hits. In fluid mechanics the analysis of motion is performed in the same way as in solid mechanics - by use of Newton’s laws of motion. Account is also taken for the special properties of fluids when in motion.

  • The momentum equation is a statement of Newton’s Second Law and relates the sum of the forces acting on an element of fluid to its acceleration or rate of change of momentum. You will probably recognise the equation F = ma which is used in the analysis of solid mechanics to relate applied force to acceleration.

  • In fluid mechanics it is not clear what mass of moving fluid we should use so we use a different form of the equation.


Newton s 2nd law can be written

Newton’s 2nd Law can be written:

  • The Rate of change of momentum of a body is equal to the resultant force acting on the body, and takes place in the direction of the force.

  • To determine the rate of change of momentum for a fluid we will consider a streamtube as we did for the Bernoulli equation, We start by assuming that we have steady flow which is non-uniform flowing in a stream tube.


3 6 application of the momentum equation

3.6 Application of the momentum equation

3.6.1 The force due the flow around a pipe bend

Why do we want to know the forces here? Because the fluid changes direction, a force (very large in the case of water supply pipes,) will act in the bend. If the

bend is not fixed it will move and eventually break at the joints. We need to

know how much force a support (thrust block) must withstand.


Me804306 2 fluid mechanics

3.6.2 The force on a pipe nozzle

Force on the nozzle at the outlet of a pipe. Because the fluid is contracted at the nozzle forces are induced in the nozzle. Anything holding the nozzle (e.g. a fireman) must be strong enough to withstand these forces.


Me804306 2 fluid mechanics

3.6.3 Impact of a Jet on a plane

We will first consider a jet hitting a flat plate (a plane) at an angle of 90 o, as shown in the figure below. We want to find the reaction force of the plate i.e. the force the plate will have to apply to stay in the same position.


Summary

Summary

  • This chapter has outlined and discussed on the fundamental of fluid in motion. Students are aspect to be able to discuss and visualise on the following aspect:

    • Able to classify FOUR (4) types of flow- Steady uniform flow, Steady non-uniform flow, Unsteady uniform flow and Unsteady non-uniform flow

    • The differences between Laminar Flow, Turbulent Flow and also Transition Flow

    • The idea of using the streamline to visualise the flow pattern

    • The calculation of mass flow rate, volume flow rate and the mean velocity of the flow

    • Able to explain and apply the THREE (3) laws-conservation of matter(conservation of mass); conservation of energyandconservation of momentum

  • The important of Bernoulli Equation and the derivation


Summary cont

Summary (cont)

  • Chapter 3 emphasized basically on the application of Bernoulli equation in order to solve problems related to fluid mechanics and the application of momentum equation to solve type of flows problem.

  • Students should concentrate more on the examples given in chapter 3 and try to relate the concept in the real scenario.


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