maae3300 module 1 part 1of2

8/10/2019 MAAE3300 Module 1 Part 1of2

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Module 1.REVIEW: Elementary Fluid Dynamics and

Finite Volume Analysis

Bernoulli, Momentum and Energy Equations

Carleton University

Department of Mechanical and Aerospace Engineering

Do not reproduce without permission, Prof. Cynthia A. Cruickshank

MAAE 3300

Fluid Mechanics II

(Part 1 of 2)

OUTLINE

Prof. C. A. Cruickshank, Carleton

The Bernoulli Equation (Part 1)

1.1 Newtons Second Law

1.2 Static, Stagnation, Dynamic and Total Pressure

1.3 Examples of Use of the Bernoulli Equation

1.4 Restrictions on the Use of the Bernoulli Equation

Finite Control Volume Analysis (Part 2)1.5 Conservation of Mass The Continuity Equation

1.6 Newtons Second Law The Linear Momentum and

Moment-of-Momentum Equations

1.7 First Law of Thermodynamics The Energy Equation

- Fundamentals of Fluid Dynamics, Munson, Okiishi, Huebsch, Rothm

Chapters 3 and 5

- Fluid Mechanics, Frank White: Chapter 3

Learning Objectives

After completing Module 1 (Part 1 of 2), you should be able to:

- discuss the application of Newtons second law to fluid flows;

- explain the development, uses, and limitations of the Bernoulli

equation;

- use the Bernoulli equation (stand-alone or in combination with

the continuity equation) to solve simple flow problems; and

- apply the concepts of static, stagnation, dynamic and total

pressures.

Prof. C. A. Cruickshank, Carleton University

The Bernoulli Equation

Newtons Second Law of Motion

Newtons second law of motion: the net force acting on a flu

particle must equal its mass times its acceleration.

=

In this module, we consider the motion of inviscidfluids.

We assume that the fluid motion is governed by gravity and

pressure forces only, and examine Newtons second law:

(net gravity force on particle) + (net pressure force on partic

= (particle mass) x (particle accelera



fluid with zero vis


When considering two-dimensional motion, the motion of each

fluid particle is described in terms of its velocity vector, V, which is

defined as the time rate of change of the position of the particle.



The velocity vector

has a magnitude

(the speed, V = |V|)

and a direction.

The particle follows a particular path which is governed by the

velocity of the particle.

n Wiley & Sons, Inc.


If the flow is steady(i.e., no changes with time at a given loca

in the flow field), each particle slides along its path and its ve

vector is tangent to the path (lines tangent to the velocity vec

are called streamlines).



The particle motdescribed in term

distance, =

along the stream

from a convenie

origin and the lo

radius of curvatu

the streamline,

=().

fluid particle

The coordinate normal to the streamline, n, can also be used.

John Wiley & Sons, Inc.


2/7


By definition, accelerationis the time rate of change of the

velocity of the particle, = /.

The acceleration has two components:

1. streamwise acceleration (along the streamline)

= / = / / = /

2. normal acceleration (normal to the streamline)

=2/

Thus, the components of acceleration in the sand ndirections

for steady flow are:



Eq. 1.1

= Along a Streamline

For steady flow, the component of Newtons second law alon

streamline direction, s, is:



Eq

where is the sum of all the forces

acting on the particle in the streamline, the

mass is =, the acceleration in the

sdirection / and the partial

volume =

(is normal

to and )



The free body diagram (FBD)

of a small fluid particle

(size by )is shown.



The important forces

are those of gravity

andpressure.



The gravity force (weight) of the particle can be written as:

=

where =and is the specific weight of the fluid (N/m 3or

Therefore the component of the weight force

in the direction of the streamline is:

sin =

or = sin

= sin



FBD o

fluid p

* If =0 (streamline is horiz.), the weight

of the particle along the streamline would

not contribute to its acceleration in that direction. John Wiley & Sons, Inc.


If the pressureat the center of the particle is given as , then the

average value on the two end faces are + and .

If we consider the particle to be infinitesimally small (we assume

that the pressure is linear across particle), we obtain:

Thus the net pressure force

on the particle is:

=

( + )

=2

=



FBD of small

fluid particle



Thus, the net force acting in the streamline direction on the par

is given by:

(net gravity force on particle) + (net pressure force on p

(particle mass) x (particle acceleration)

Eq. 1.3 (defi

two previou



Eq. 1.2 (defined

Interpretation: a change in fluid pa

speed is accomplished by the appr

combination of particle weight & p

By combining both equations, we obtain Eq. 1.4:


3/7


Equation 1.4 can be rearranged and integrated as follows:



sin =/

() = 2

restricting to a streamline, and

ay be regarded as only a function of .

is is the Bernoulli Equation!

. 1.5

+1

2

2

+

= 0

Assumptions

1. steady flow

2. incompressible

3. inviscid flow

+

1

2 + = 0

aside:


= Normal to a Streamline

Newtons second law is applied to flows for which the only impo

forces are those due to pressure and gravity viscous effects ar

negligible. The result is the Bernoulli equation (a relationship am

pressure, elevation and velocity variations along the streamline

A similar but less often used equation also exists to described

the variations in these parameters normal to a streamline.



When a fluid particle travels along a curved path, there

is a net force directed toward the center of curvature.

In many instances, the streamlines are nearly straight

( = 0), so centrifugal effects are negligible.

: local

of cu


EXAMPLE (Pressure Variation Along a Streamline)

Consider the flow of air around a bicyclist moving through still air

with velocity V0. Determine the difference in the pressure between

points (1) and (2). Assume the coordinate system is fixed to the bike.




SOLUTION

Recall that the main assumptions in

deriving the Bernoulli equation is

that the fluid is steady,

incompressible and inviscid.

Equation 1.5 (Bernoulli equation) can

be applied along the streamline that

passes through points (1) and (2).



1+

1

+ 1 =2 +

2

+ 2

In a coordinate system fi

the bike, it appears as th

the air is flowing steadily

the bicyclist with speed V

We consider point (1) to be in the free stream so that V1 = V0,

assume 1

= 2

and 2

= 0 (stagnation point).

1 +

1

+ 1 =2 +

2

+ 2

0

2 1 =

1

=


Video: Flow Past a Biker

Full scale wind tunnel experiments are often carried out to obtain

information about the aerodynamic forces on athletes, such as down

hill skiers and bicycle racers. In some instances useful information can

be obtained by using flow visualization techniquessuch as a smoke

wand to show the streamline location.



ideo courtesy of A2 Wind Tunnel, www.A2Wt.com

Wind tunnel experiment is carried out to

analyze the aerodynamics of cycling.

(Source: A2 Wind Tunnel)

n Wiley & Sons, Inc. Prof. C. A. Cruickshank, Carleton


Video: Flow Past a Biker

The Bernoulli equation contains six unknowns - two pressures, two v

ties, and two elevations. To obtain the value of one parameter, the v

of the other five must be known. For the streamlines shown in the v

the elevation at arbitrary points can be measured, but the pressure

velocity are not known. When the smoke wand is placed directly in fthe cover the smoke runs head-on into the cover, producing a stagn

point on the cover at the point of impact. In this case, an additional

parameter is known (velocity is zero) and the stagnation pressure c

calculated if the velocity and pressure in the free-stream are know

A2 Wind Tunnel



4/7

Computational Fluid Dynamics (CFD)

CFD is a branch of fluid mechanics that uses numerical methods and

algorithms to solve and analyze problems that involve fluid flows.



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Physical Interpretation of the Bernoulli Equation

An alternative but equivalent form of the Bernoulli equation is ob

by dividing each term by the specific weight,(recall, =):

The pressure term, /, is called the pressure headand represenheight of a column of fluid that is needed to produce the pressur

The velocity term, 2/, is the velocity head and presents the v

distance needed for the fluid to reach velocity from rest.

The elevation term, , is related to the potential energy of the

particle and is called the elevation head.

The Bernoulli equation states that the sum of the pressure head,

velocity head and elevation head is constant along the streamline



Eq

Static, Stagnation, Dynamic and Total Pressure

Each term of the Bernoulli equation has the dimensions of force

per unit area psi, lb/ft2, N/m2.



N

m2

kg

m3m2

s2

kg

m s2

N

m2

= kg

m3m

s2m

kg

m s2

N

m2

Static, Stagnation, Dynamic and Total Pressure



: actual thermodynamic pressure of fluid(called the static pressure,points 1)

: hydrostatic pressure (not actually a pressure but represents thchange in pressure due to potential energy variations)

2: dynamic pressure (the kinetic energy per

unit volume of a fluid particle)

Point 2 is referred to as the stagnation point

(2 = 0, fluid is stationary). The pressure at this

point is called the stagnation pressure,2. The

stagnation pressure is equal to the sum

of the free-stream static pressure andthe free-stream dynamic pressure:

2 =1+

1

2

Bernou

Bernoulli eq. is applied,

2 = 0and z1 =2


Total Pressure and Pitot-Static Tube

The sum of the static pressure, dynamic pressure and hydrostatic

pressure is termed the total pressure.

Knowledge of the values of the static and stagnation pressures

in a fluid implies that the fluid speed can be calculated. This is

the principle on which the Pitot-static tube is based.

Two concentric tubes are attached to two pressure gauges.

The center tube measures the stagnation pressure at its open

tip (2). The outer tube is made with several small holes to

measure static pressure (4 =1 =). If elevation changes

are negligible, then



3 = +

2

gauge

gauge

Fluid upstream

= 2(3 4)/

Eq. 1.8


Total Pressure and Pitot-Static Tube



3 = +

2

Fluid upstream

= 2(3 4)/

Eq. 1.8



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Fluids in the News (Bugged and Plugged Pitot Tubes)

Although a pitot tube is a simple device for measuring aircraft speed,

many airplane accidents have been caused by inaccurate pitot tube

readings. Most of these accidents are the results of having one or more

holes blocked and therefore not indicating the correct pressure (speed).

The most common causes for such a blockage include the pitot tube

cover not being removed, ice build-up, or insects have built their nestwithin the tube and a standard visual check cannot detect it.

One of the most serious accidents caused by a blocked pitot tube

involved a Boeing 757 and occurred shortly after takeoff. Incorrect

airspeed were automatically fed to the computer, causing the autopilot to

change the angle of attack and the engine power. The aircraft stalled and

then plunged killing all aboard.

Investigators concluded that wasps might have nested in the pitot tubes

as the plane had sat grounded for several days.



EXAMPLE (Pitot-Static Tube)

An airplane flies 200 mph at an elevation of 10,000 ft in a stan

atmosphere. Determine:

(a) the pressure at point 1 far ahead of the airplane;

(b) the pressure at the stagnation point on the nose of the air

(point 2); and

(c) the pressure difference indicated by a Pitot-Static probe at

to the fuselage.



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Pitot-static tube


SOLUTION

Part (a): From Table C.1, we find that the static pressure and

density at the 10,000 ft.

1= 10.11 psi, = 0.001756 slug/ft3



Fundamentals of Fluid Mechanics: Munson, Okiishi, Huebsch, Rothmayer


SOLUTION



Part (b): Assuming steady, inviscid and incompressible flow, a

negligible elevation changes and2 = 0(stagnation point at

and coordinates fixed to airplane), the Bernoulli equation be

We know: 1 = 200 mph = 293 ft/s (note: 1 mph = 1.46

1= 10.11 psi = 1456 lb/ft2 (note: 1 psi = 144 lb/

2 =1 +

1

= (1456 lb/ft2 + (0.001756 slugs/ft3)(293 ft/

2= 1531.4 lb/ft2= 10.63 psi

1 +

1

+ 1 =2+

2

+ 20

2 =1 +

2 1 =

1

= 75.4 lb/ft2 = 0.524 psi

Part (c): Pressure difference indicated by Pitot-static tube:

75.4 lb/ft2

EXAMPLE (Q3.27 Fundamentals of Fluid Mechanics 7e, Munson et al.)

A 40-mph wind blowing past a house speeds up as it flows up and

over the roof. Assume the elevation effects are negligible.

(a) Determine the pressure at the point on the roof, in lb/ft2, where

the speed is 60 mph if the pressure in the free stream blowing

towards your house is 14.7 psi. Would this effect tend to push the

roof down against the house or lift the roof?

(b) Determine the pressure, in lb/ft2, on a window facing the wind if

the window is assumed to be the stagnation point.




SOLUTION





6/7



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Fluids in the News (Giraffes Blood Pressure)

A giraffes long neck allows it to graze up to 6 m above the ground. It can

also lower its head to drink at ground level. This elevation change causes a

significant hydrostatic pressure effect ( =) in the circulatory system.

To maintain blood in its head, the giraffe must maintain a high blood

pressure at heart level (2.5 times that of humans).

To prevent rupture of the blood vessels in the high pressure lower leg

regions, giraffes have a tight sheath of thick skin over their lower limbs that

act as an elastic bandage (exactly the same as g-suits for fighter pilots). In

addition, valves in the upper neck prevent backflow into the head when

the giraffe lowers its head to ground level.

EXAMPLE (Q2.17 Fundamentals of Fluid Mechanics 7e, Munson

(a) Determine the change in hydrostatic pressure in a giraffes

as it lowers its head from eating leaves 6 m above the grou

getting a drink of water at ground level as shown. Assume

specific gravity of blood is SG = 1.

(b) Compare the pressure change calculated in part (a) to the

120 mm of mercury pressure in a humans heart. = 133


In this case, the pressure variation in the blood

is equal to:

=(blood) = blood

where blood = blood H2O

Since bloodis equal to 1, blood =H2O

and = blood = H2O = H2O



SOLUTION


= H2O

3

hydrostatic pressure: change in pressure

due to potential energy variations



Free Jets

Consider flow of a liquid from a large reservoir as shown. A jet o

liquid diameter flows from the nozzle with velocity .

In this case, we use the facts that 1 =, 2 = 0, the reservoir i

(1 = 0), open to the atmosphere (1 = 0gauge), and the fluid

as a free jet (2 = 0).



Eq

If the flow is assumed steady, inviscid

incompressible between points (1) a

the Bernoulli equation can be applie0


Video: Flow from a Tank

According to Bernoulli, the velocity of a

fluid flowing through a hole in the side

of an open tank or reservoir is

proportional to the square root of the

depth of fluid above the hole.

The velocity of a jet of water from an

open pop bottle containing four holes is

related to the depth of water above the

hole. The greater the depth, the higher

the velocity.



Eq. 1.9 (defined on

previous slide) Video courtesy of U.S. Bureau of Reclamation


Confined Flows

In some cases, the fluid may be physically constrained within a d

For these situations, it is necessary to use the concept of conse

of mass (the continuity equation) along with the Bernoulli equa

Consider a fluid flowing through a fixed volume that has one inl

one outlet. If the flow is steady, the rate at which the fluid flows

the volume must equal the rate at which is flows out of the volu





7/7

Confined Flows

Thus, the conservation of mass requires:

=

=

m

=

If the density remains constant, then =and the above becomes the continuity equation

for incompressible flow:

where is the volumetric flow rate (

or

)



1.10


EXAMPLE (Q3.44 Fundamentals of Fluid Mechanics 7e, Munson

Water flows steadily through the tanks shown below. Find the wa

depth, , in metres. Assume the reservoirs are large (1 =3 =

open to atmosphere (1 =3 = 0gauge) and that the fluid leave

free jet (2 =4 = 0).



Point (1)

Point (2)

Point (3)

Point (4)


SOLUTION



(1)

(2)

(3)

(4)

Using Bernoulli

Equation, Eq. 1.7

ing Continuity

uation, Eq. 1.10


SOLUTION



(1)

(2)

(3)


Restrictions on the Use of the Bernoulli Equation

1. One of the main assumptions in deriving the Bernoulli equation

is that the fluid is incompressible. Although this is reasonable for

most liquid flows, it can, in some cases, introduce considerable

errors for gases.

2. A second restriction is the assumption that the flow is steady. For

steady flows, on a given streamline, the velocity is only a

function of its location along the streamline, ( = ).

For unsteady flows, the velocity is also a function of time,

= , . Thus, when taking the time derivative of the

velocity to obtain the streamwise acceleration, we obtain:

=

+

The term /does not allow the equation of motion to be

integrated easily without further assumptions.Prof. C. A. Cruickshank, Carleton University


Restrictions on the Use of the Bernoulli Equation

3. Another restriction on the Bernoulli equation is that the flow

inviscid. In the absence of viscous effects, the total energy o

system remains constant. If viscous effects are important, th

system is nonconservative and there are energy losses.

4. The final basic restriction is that there are no mechanical de

(pumps or turbines) in the system between the two points.

These devices represent sources or sinks of energy. The Ber

equation must be altered to include these devices.



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