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ea e Examples in Fluid Mec hanic s
Lesson plans and solutions
Eann A Patterson Edit ito o
Real eal Lif Life e Example xamples s in in Fluid luid Mechanic Mec hanics s Lesson plans and solutions
First edition January 2011 Second [electronic] edition September 2011
Real Life Examples in Fluid Mechanics Lesson plans and solutions
Copyright © 2011 Eann A Patterson (editor)
This work is licensed under the Creative Commons Common s Attribution-NonCommercial NoDerivs 3.0 Unported License. To view a copy of this license, visit http://creativecommons.org/licenses/by-nc-nd/3.0/ or send a letter to Creative Commons, 444 Castro Street, Suite 900, Mountain View, California, 94041, USA. The Editor has no responsibility for the persistence or accuracy acc uracy of URLs for external or third-party internet websites referred to in this publication, and does not guarantee that any content on such websites is, or will remain, accurate or appropriate.
Bounded printed copies can be purchased on-line at www.engineeringexamples.org
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Real Life Examples in Fluid Mec Mec hanic s
Lesson plans and solutions
Suggest ugg ested ed exempla exemplarrs within within less lesson plans pla ns for Sop Sop homore ho more leve levell cours c ourses es iin n Fl Fluid uid M ec hanics. ha nics. Pr Prep epa a red a s par pa rt o off the NSFNSF-s supp o rted p rojec t (#043 (#04317 1756 56)) entit e ntitled led:: “E “Enha nhanc ncing ing Diversity in the the Undergrad Undergrad uate Mec Me c hanical hanic al Engineeri Engineering ng Po Population pulation throug through h C urr urriculum Cha nge”.
Eann A Patterson, Editor T The he U Uni niver vers sity of Li Liver verpo pool ol
[email protected]
This work was developed during the NSF-supported project (#0431756) entitled: “Enhancing Diversity in the Undergraduate Mechanical Engineering Population through Curriculum Change”. Any opinions, findings, and conclusions or recommendations expressed in this material are those of the authors and do not necessarily reflect the views of the National Science Foundation. The following members of tthe he project contributed to the many discussions which led to this production of this volume: Ilene Busch-Vishniac Patricia B. Campbell Darrell Guillaume Eann Patterson
.
McMaster University ( Project Project leader ) Campbell-Kibler Associates Inc California State University, Los Angeles Michigan State University
Real Life Examples in Fluid Mechanics CONTENTS page no. Introduction
5
INTRODUCTORY CONCEPTS 1.
Fluids and their properties
6
Blowing bubbles, pouring pouring detergent, fl floating oating paperclips, paperclips, fried eggs 2.
Statics
9
Apple bobbing, bath plug, plug, escaping from su submerged bmerged car FLUIDS IN MOTION 3.
Kinematics of fluid motion
14
Cream in coffee, BMW convertible, air-conditioning, bathroom ventilatio ventilation n 4.
Dynamics of fluid motion
18
Floating table tennis tennis ball, hurricanes, hurricanes, hand out of car window, window, vacuum cleaner CONTROL VOLUME ANALYSIS 5.
Momentum
22
Lifting with balloons, balloons, washing cups, umbrellas umbrellas in the wind 6.
Energy
25
Forced air heater, hand-dryer, hand-dryer, cyclist food consu consumption mption MODELING 7.
Similitude and dimensional analysis
28
Bath toys, vacuum cleaner, cleaner, artery flow FLOW 8.
Viscous flow in pipes (internal flow)
32
Vacuum cleaner, water squirter, bicycle pump 9.
Flow over bodies (external flow) Swim suits, heart surgery and pizza delivery
37
10.
Flow in open channels
41
Water slide, curb drain, drainage ditch 11.
Compressible Flow
46
Rampaging bull, bull, flock of sheep, supersonic supersonic flight, factory explosion explosion APPLICATIONS 12.
Turbomachines
51
Toy planes, car water pump, airplane propeller design
3
Real Life Examples in Fluid Mechanics
4
Real Life Examples in Fluid Mechanics INTRODUCTION
These real life examples and supporting materials are designed to enhance the teaching of a sophomore level course in fluid mechanics, increase the accessibility of the principles, and raise the appeal of the subject to students students from diverse backgrounds. The examples have been embedded in skeletal lesson plans using the principle of the 5Es: Engage, Explore, Explain, Elaborate and Evaluate. The 5E outline is not original and was 1 developed by the Biological Sciences Curriculum Study in the 1980s from work by Atkin and Karplus2 in 1962. Today this approach is considered to form part of the constructivist learning theory3 and a number of websites provide easy-to-follow explanations of them. This booklet is intended to be used by instructors and is written in a style that addresses the instructor, however this is not intended to exclude students who should find examples interesting, stimulating and hopefully illuminating, particularly when their instructor is not utilizing them. In the interest of brevity and clarity of presentation, standard derivations, common tables/charts, and definitions are not included since these are readily available in textbooks which this booklet is not intended to replace but rather to supplement and enhance. Similarly, it is is anticipated that these these lesson plans can be used to generate lectures/lessons that supplement those covering the fundamentals of each topic. It is assumed that students have acquired a knowledge and understanding the following topics: first and second law of thermodynamics, Newton’s laws, free-body diagrams, and stresses in pressure vessels. This is the fourth in a series of such notes. The others are entitled ‘Real Life Examples in Mechanics of Solids’ (ISBN: 978-0-615-20394-2), ‘Real Life Examples in Dynamics’ (ISBN: 978-0-9842142-0-4) and ‘Real Life Examples in Thermodynamics’ (ISBN 978-09842142-1-1). They are available on-line at www.engineeringexamples.org.
1
Engleman, Laura (ed.), The BSCS Story: A History of the Biological Sciences Curriculum Study. Colorado Springs: BSCS, 2001. 2 Atkin, J. M. and Karplus, R. (1962). Discovery or invention? Science Teacher 29(5): 45. 3 e.g. Trowbridge, L.W., and Bybee, R.W., Becoming a secondary secondary school science teacher . Merrill Pub. Co. Inc., 1990.
5
Real Life Examples in Fluid Mechanics INTRODUCTORY CONCEPTS
1.
Topic: Fluids and their properties
Engage:
Take into class a pile of drinking straws, some paper cups, a bottle of dish washing detergent and a bottle of water. In front of the students, add detergent to water so that the solution is about one part detergent to ten parts water. Give each student a straw and a paper cup with a small amount of the mixture in the bottom and invite them to blow bubbles. Alternatively if you just want to give a demonstration – kits kits are 4 available on-line (search under soap bubbles) bub bles) as well as advice on solutions etc.
Explore:
Ask students to work in pairs and draw a free-body diagram for a stationary, spherical bubble. If necessary, give them a hint by telling them to treat it like a pressure vessel. Invite a pair to draw their result for the rest of the class. Discuss how the circumferential stress in a solid pressure vessel can be replaced by the surface tension, in in the inner and outer surfaces of the bubble. So that summing the forces in the horizontal direction in the figure we obtain 2r
2
p r
or
p
2 2 r
4
r
pr 2 2r
Explain:
Discuss that athe surface surface of the afluid actshas like te tensioned nsionedsurface membrane. Thea given bubblevolume. prefers to form into sphere because sphere thea smallest area for So in a sphere the molecules of the fluid can be uniformly close to one another. A bubble of pure water would be very unstable whereas the detergent is a surfactant that decreases the surface tension. When a region of the surface of the bubble gets stretched by an external force, the tension in the membrane increases. This causes cau ses detergent to flow away, thus decreasing the concentration of detergent and raising the surface tension of the fluid mixture which prevents the bubble from bursting. bu rsting. Changes in the air temperature will cause expansion or contraction of the bubble and disturb the equilibrium between internal pressure (increases with temperature) and surface tension (decreases with temperature) so that it collapses.
4 http://www.bubbles.org/solutions/ http://www.bubbles.org/solutions/
6
Real Life Examples in Fluid Mechanics Elaborate:
Pour a little pure detergent into a paper cup and, as you do so, point out to the students the slow rate at which it flows compared to water. Explain that this is a result of the high viscosity or internal stickiness of the detergent. detergent. Fluids which have complex molecules that cannot slide easily past one another tend to have a high viscosity. Consider liquid being sucked up a drinking straw (don’t try it with the water-detergent solution!). The molecules in the fluid that are in contact with the inside inside surface of the straw tend to get stuck in the surface roughness of the straw so that they move very slowly relative to those along the center line line of the straw. This creates a velocity gradient in the fluid. When we consider a square element element in the fluid, then tthe he velocity difference between the faces face s parallel pa rallel to the inside surface of the th e straw can be described by v, and after time t the the element will be sheared such that the shear angle, is is given by
vt v or t y y
Element at time, t + t
and in the limit as t → 0 and y → 0 so the rate of shear strain is
vt
Element at time, t
dv
dy
For ‘Newtonian’ fluids, the shear stress is given by
dv dy
y
Wall of the straw
where is is known as the dynamic dyna mic viscosity. Note that if dynamic viscosity is normalized by the density it is referred to as kinematic viscosity, Observe that the fluid mechanics has been discussed using analogies to solids and employing similar language. Another term that can be ttaken aken from mechanics of solids is modulus of elasticity. The bulk modulus of elasticity, elasticity, B of a fluid relates changes in volume, dV to to the change in pressure, dp causing them, i.e. B
dp dV V
or for constant mass, m V B
dp d
and B is a measure of the compressibility of a fluid.
Evaluate:
Invite students to attempt the following examples: Example 1.1
Demonstrate by analysis and by experiment that a paperclip will float in water at room temperature.
7
Real Life Examples in Fluid Mechanics Solution: L
Summing forces in the vertical direction:
L
mg 2 L
substituting for mass 2
steel d Lg 2 L 4
mg
Assuming the density of steel, steel 7850 kg/m3 and surface tension of water, H O 0.073 N/m 2
d
8 steel g
8 0.073 7850 9.81
1.55 10 3 m -3
The paperclips on the editor’s desk have a diameter of about 0.75mm (=0.7510 m) so they should float based on this analysis. If you try to float them from your fingers in a cup of water, then they will almost always sink because you disturb the surface surface tension in process. However, balance the paperclip on a thin piece of tissue and gently lower it to the water surface without touching the surface with your fingers. Then gently push the paper underwater with a pencil without touching the paperclip. The paperclip should be left floating.
Example 1.2
A fried egg of mass 65g sits in a frying pan with a 0.5mm layer of cooking oil between it and the pan surface. surface. If the oil has a viscosity of 0.05 Ns/m2, then how fast will the egg slide across the pan if the pan is tipped to an angle of 20°. Assume that the egg is flat on the underside, is approximately circular with a diameter of 12cm and neglect edge effects. Solution: Equating forces on in the direction of o f sliding using the free-body diagram, mg sin A
so and let
dv dy
dv dy
so
v and y
rearranging
mg sin
dv dy
A
20°
Rnormal
A mg
v mg sin A y v y
mg sin A
0.0005
So your egg will be out of the pan pretty fast!
8
m g s i n
0.065 9.81 sin 20 0.193m/s 0.05 0.12 2 4
Real Life Examples in Fluid Mechanics INTRODUCTORY CONCEPTS
2.
Topic: Statics
Engage:
Take a basin of water and some apples into class. Invite a couple of students to do some apple bobbing, i.e. trying to grab the apples with their teeth. You should probably take some towels into class too!
Explore:
Ask the students who do some apple bobbing and observe how the apples behave. If you can get hold of some small applies then you could pass around an apple in a cup of water and students can experiment for for themselves. Discuss the following with with the students: (i)
when pushed down an apple experiences an upward force restoring it to the surface; and
(ii)
that most apples have a stable orientation.
Explain:
Remind the students that Archimedes principle defines the buoyancy force as equal to the weight of the volume of water displaced, F B gV displaced fluid
Note that, sometimes is used and is known as the the specific weight of the ffluid. luid. So to keep an apple submerged it is necessary to apply a force, S = mg – F B where mg is is the weight of the apple.
When it is submerged, the apple experiences a force acting on its skin due the pressure of the water. This force on the curved surface of the skin must act through the center of curvature of the surface. The pressure is related to the depth in the downward direction, dz and and density of the water, such such that dp g dz
It would be appropriate to explain that absolute pressure is measured relative to absolute zero pressure which could only occur in a perfect vacuum; however it is common engineering practice to to measure pressure relative to atmospheric pressure. A pressure measured in this way is known as gage pressure. pressure. Standard atmospheric pressure pressure is defined as 14.7psi or 101kPa at sea-level.
9
Real Life Examples in Fluid Mechanics Elaborate:
To elaborate on the stability of the apples it is preferable to simplify the situation to a cylinder of uniform material of specific weight, cyl cyl = 9000 N/m3, i.e. slightly less than water, and of length and diameter diameter 55mm. If we consider force equilibrium in the vertical direction for the cylinder floating with its axis vertical, then the weight equals the buoyant force, i.e.
F 0
y
and
mg H O r 2 H
or
cyl r Lg H O r H
2
2
H
2
2
where H is is the submerged depth, L is the length and r the the radius of the cylinder. H
cyl L H O 2
9000 9810
G C 1
m m 5 . 7 2
m m 5 5
55mm
55 10 3 50.5 10 3 m
Now consider the stability of the cylinder. The buoyancy force, W, acts through the centroid of the displaced volume of fluid, C 1, which when the cylinder is rotated by an external force moves to C 2. The weight of cylinder, W, always acts at its centroid, G, so that the two forces W form a couple which tends to return the cylinder to its original position if the original was stable or tends to rotate it further if the original position was unstable. The lines joining the centroid of the cylinder, G, to the centroids of the displaced volume of water, C 1 and C 2, intersect at the metacenter, M. It can be shown that that the distance C M is is given by
M
G G C C 1 1
C 2
W W
1
C 1 M
I O V
where I O is the second moment of area of the water-line section about the axis through its centroid and V is is the submerged volume. Alternatively, given from the geometry that C 1G GM C 1 M
then
GM
I O V
C 1G
When the distance from the centroid of the cylinder to the metacenter, GM is positive, the cylinder is stable. So for the apple idealized as a cylinder, 10
Real Life Examples in Fluid Mechanics
I O
55 10 3
4
d d 4
64
64
4.49 10 7 m4
and, from the diagram 50.5 10 3 C 1G 27.5 10 2 3
so
GM
I O 2
cyl d L H O 2
C 1G
4
2.25 10 3 m 4 .49 10 7
9000 55 10 9810 4
3 3
2.25 10 3 0.00149 m
Thus, the apple is stable in this orientation, with its axis a xis of symmetry or stalk vertical.
Evaluate:
Invite students to attempt the following examples: Example 2.1
Calculate the force needed to push up the 4cm diameterplug in a bath of water 20cm deep, to letthat theoperates water out, if the plug isa mechanical a loose-fit advantage in the plug-hole mechanism the plug provides of 3. and the lever Solution: Pressure on the plug, 2
p H O gz 1000 9.81 0.2 1962 N/m 2
assuming atmospheric pressure on the surface at the surface of the bath water and in the drain under the plug. Force on plug, F pA p
d d 2
4
1962
0.04 2
4
2.46 N
With a mechanical advantage of 3 the force required on mechanism will be 0.82N (=2.46/3). Example 2.2
When a car slides off the road into a river, which is 2.5m deep, it lands on its side at 45 degrees to the vertical such that the passenger side doors are jammed into the soft riverbed. Calculate the force required to open the driver’s door if the car remains watertight and the bottom of the door is just touching the riverbed. Assume the door is a rectangle rectangle of height 1.2m and width 1.1m weighing 33kg with the handle on the opening edge.
m 5 . 2
45°
11
Real Life Examples in Fluid Mechanics Solution: It can be shown that the force acting on an area, F pC A where pC is the pressure at the centroid on the area, A. In this case the centroid is located at L 2cos 45 above the riverbed where L is the height of the car door. Hence, the distance from the water surface to the centroid of the door, h 2.5
1.2 2
cos 45 2.08 m
p H O gz 1000 9.81 2.08 20,400Pa
Thus, the pressure,
2
And, the force acting on the door, F pC A 20363 1.2 1.1 26,900N This force will act at the center of pressure, which will be on the vertical center-line of the door but below the centroid due to the the pressure gradient. Assuming the force to open the door, F open open is applied approximately level with the center of pressure, then taking moments about the hinge, Fw
2
mgw
2
F open w 0
where w is the width of the door, thus F open
F mg
2
2
26879 33 9.81 13601 N 2 2
This is a huge force. It is probably a false assumption that the the car will be watertight; watertight; but nevertheless, if your car is immersed in an accident it will be extremely difficult, if not impossible, to open a door.
Example 2.3
In your car you have a coffee cup of diameter 9cm at the rim and it is full to within 10mm of the rim. When the cup is placed in a cup-holder, how fast can you accelerate without spilling your coffee? Solution: Pressure in a fluid is gz so the force on the left end of the element of fluid in the diagram will be F L gz L A
and for the right end, F R gz R A
z L
L
z R
Direction of travel
In these circumstances it is suggested that you should exit through a window. window. So, your first action on realizing you are going to impact with the water should be to open a window since the electrics will fail quickly on immersion. If you need to break a window, hit it in a corner with a sharp object.
12
Real Life Examples in Fluid Mechanics so the net force in the direction of travel is F gA z L z R
and this must be equal to force due to the acceleration, i.e. F ma thus substituting for the mass of the element gA z L z R ALa or
a
g z L z R L
and using similar triangles a
9.81 10 10 2 9 10
2
2
2.45 m/s2
This equivalent to 0 to 16mph in 3 seconds and is why we put lids on our coffee (or don’t fill your cup too near the top)!
13
Real Life Examples in Fluid Mechanics FLUIDS IN MOTION
3.
Topic: Kinematics of fluid motion
Engage:
Take a flask of coffee, some cream or milk, a glass, and some paper cups into class. Offer a free cup of coffee to students as they come into class and ask them to add the cream slowly and watch what happens. Repeat the the process yourself except use the glass so that the students can see the mixing process in three-dimensions.
Explore: 5
Show a video of How to Cream Coffee (search YouTube using the italicized words). Explain that the cream shows us ‘streaklines’. A streakline is an instantaneous locus of all particles originating from a common point. You might also like like to show the video 6 coffee and cream mix slow motion . Show a video of the flow around a sphere (inH20) in a water tank in which red dye is introduced on the upstream side of the sphere 7. Explain that the fluid dye shows us ‘streaklines’ in the same way as the cream in the coffee. Define a pathline as a history history of particle’s location, and li lines nes that are tangential to the velocity vectors of particles in the flow as streamlines. streamlines. An example of streamlines streamlines would be the lines left in a time-lapsed photograph by the headlights of a moving car. For steady-flow streaklines, pathlines and streamlines are coincident. You could use a 8 video of flow in front of and behind a BMW 3series Convertible 1987 in the wind tunnel to illustrate the difference between steady and unsteady flow. You could talk about the distance between streamlines representing pressure and the relationship with increasing and decreasing velocity. Explain:
Highlight that when watching the coffee and cream or the videos our eyes tend to follow the movement of the cream (or the smoke) in the flow and that this can be equated to the Lagrangian approach to fluid mechanics.
5
www.youtube.com/watch?v=9EsZ7GSUcI4&NR=1 www.youtube.com/watch?v=9EsZ7GSUcI4&NR=1 www.youtube.com/watch?v=M4vB6kAhuN0 www.youtube.com/watch?v=M4vB6kAhuN0 7 www.youtube.com/watch?v=4zHIjyj-vEo www.youtube.com/watch?v=4zHIjyj-vEo 8 www.youtube.com/watch?v=XS3sbYJHkSw&feature=channel www.youtube.com/watch?v=XS3sbYJHkSw&feature=channel 6
14
Real Life Examples in Fluid Mechanics Show the BMW convertible video again and use your laser pointer or the cursor on the screen to highlight a single point. Explain that considering the fluid fluid motion at a single point in this way is an Eulerian approach to fluid mechanics.
Elaborate:
Continue the Eulerian approach and instead of considering a single point, expand it to a box, through which the fluid is flowing. Explain that such a box is known as a control volume in fluid mechanics. Discuss that in in steady flow, the amount amount of fluid entering and leaving the control volume at any instant in time is the same, i.e., the mass of the fluid in the box is constant, which leads to the continuity equation, 1 A1v1 2 A2 v2
constant m
where is is the fluid density, v the fluid velocity and A1 and A2 are cross-section areas at the entry and exit from the control volume, and usually considered to be connected by pathlines, i.e., lines across which no particles flow. In unsteady flow the mass of fluid in the control volume changes with time and hence the continuity equation has to be modified as dmcv dt
m out m in 0
where the first term represents the accumulation of fluid in the control volume and the last two terms represent the outflow and inflow rates across the surfaces of the control volume. Use the continuity equation to calculate the flow rates for the inlet and outlet ducts of the air conditioning in the classroom. classroom. It is recommended recommended that the air in a classroom classroom should 9 be completely changed 12 times per hour . Estimate the volume of the classroom based on floor dimensions of 35ft (10.7m) by 25ft (7.6m) and a ceiling height of 8ft (2.4m), 3
V 10.7 7.6 2.4 195.2 m
So the volumetric flow rate required is 0.65m3/s 195.2 12 60 60 and if we assume constant air density then A1v1
A2v2 0.65m3/s
If the room has five inlet vents of dimensions 3ft (0.914m) by 6in (0.153m) and a 2ft (0.610m) square outlet vent then 0.65 0.914 0.153 5
v1
v2
0.65
0.610 2
0.93 m/s
1.75 m/s
9 http://ateam.lbl.gov/Design-Guide/DGHtm/roomairchangerates.htm
15
Real Life Examples in Fluid Mechanics Evaluate:
Invite students to attempt the following examples: Example 3.1
It is recommended that bathrooms of 100sq.ft. or less should have a ventilation of 1 CFM 10 per sq.ft of area and no less than 50CFM . This is usually usually achieved with a small fan that extracts air. Estimate the volumetric flow rate for your bathroom and the velocity of air flowing under the closed door to replace the air extracted by a fan. Solution: The editor’s bathroom is 8ft by 6ft hence Volumetric flow rate = Floor area 1 = 8 6 48 CFM ≡ 0.023 m3/s If the door is 3ft wide (0.9m) with a half inch (0.0127m) gap underneath it then the velocity of the flow will be given by v
0.023 0.9 0.0127
2 m/s (~ 4.5 mph)
Example 3.2 If a car tire with a slow puncture takes three days (72 hours) to decrease from its recommended pressure of 35psi to 22psi at which the low pressure light activates on the , from the instrument panel, calculate the area of the hole, A, given the mass flow rate, m tire is given by
m
0.66 pA RT
where R is the gas constant, T is is the absolute temperature and p is the tire pressure. You may assume that the change in volume of the tire is negligible. Solution: Using the continuity equation: dmcv dt
so
d V cv dt
m out m in 0 0.66
pA
00
RT
Applying the ideal gas law V cv
pA 0 0.66 dt RT RT d p
assuming A, R (=587 J/kgK), T and V are constant and rearranging
10 www.hvi.org/resourcelibrary/HowMuchVent.html www.hvi.org/resourcelibrary/HowMuchVent.html
16
Real Life Examples in Fluid Mechanics
dt
1.52V dp A RT p
Integrating and applying initial and final pressures as limits t
1.52V A RT
ln
pi p f
Now, the volume of the tire can be calculated by assuming the tire and wheel to be concentric cylinders of outside diameters 0.6m and 0.4m respectively and thickness (= l ) 0.18m (based on measurements taken from the editor’s car), thus V
4
d d l 0.6 0.4 2 tire
2 wheel
4
2 tire
2 wheel
0.18 0.028 m3
So, if the temperature is assumed to be 21 C then A
1.52V
t RT
ln
pi p f
1.52 0 .028 (72 60 60) 587 300
ln
35 22
1.82 10 10 m2
Assuming a round hole, it would have a diameter of 1.5 10-5 (≡ 15m) since 10
d
4 A
4 1.82 10
1.5 10 5 m
17
Real Life Examples in Fluid Mechanics FLUIDS IN MOTION
4.
Topic: Dynamics of fluid motion
Engage:
Take a hair dryer and some table tennis balls into class. Invite a couple of pairs of students to turn on the hair dryer, point it upwards and balance a ball in the air-stream. Ask them to describe what they feel when they try to remove the ball from the air-stream (a force holding the ball in the airstream). There is a good video and explanation of this experiment11.
Explore:
Show a video of flow around a tennis ball with vortex shedding, by searching in Youtube using the words “ Fluid Mechanics - Cool science experiment ” 12, and discuss the presence of areas of laminar and turbulent flow with vortices being b eing formed downstream of the ball. Vortices are low pressures pressures zones formed downstream of blunt objects and the ball tends to move towards the low pressure zone which is why the table tennis ball bounces around in the flow from the air dryer. (p+ p (p+ p ds) dA ss
Explain:
Return their attention to the video and the laminar flow region where the flow visualization with the smoke represents streamlines. Using a Lagrangian approach, consider a particle of fluid on a streamline and apply Newton’s second law in the direction of the streamline, i.e. resolve forces, pdA p p ds dA g dsdA cos a s dsdA s
a v
ds dh R
pdA
( g)dsdA
where a s is the acceleration of the particle along the streamline. streamline. For two points on the streamline and assuming no viscous forces, it can be shown that v12
2
p1
gz 1
v 22
2
p2
gz 2 gH
which is known as Bernoulli’s equation and Bernoulli’s constant, H is is constant along a streamline.
11
www.efluids.com/efluids/gallery_exp/exp_pages/hairdryer.jsp www.efluids.com/efluids/gallery_exp/exp_pages/hairdryer.jsp 12 www.youtube.com/watch?v=7KKFtgx2anY www.youtube.com/watch?v=7KKFtgx2anY
18
Real Life Examples in Fluid Mechanics Note that Bernoulli’s equation can be expressed such that the terms have the units of pressure by normalizing with density, or of length or head by normalizing with z ; total gravitational acceleration, g . Thus, piezometric head can be defined as p g head as v 2 2 g p g z ; and stagnation or total pressure as p v 2 2 .
Elaborate: Highlight that a vortex consists of a system of concentric circular streamlines in which the fluid velocity is inversely proportional to the radius, i.e. vr = constant. When Bernoulli’s constant, H is the same throughout the vortex it is known as a free vortex with rotational flow in which fluid elements rotate in circles but do not deform.
Rotational Rota tional flow without without elemen elementt defo deforma rmation tion Irrotational flow with element deformation
In a free vortex, Bernoulli’s constant requires the pressure to drop as the velocity rises towards the center of a vortex. The pressure drop drop causes:
free surfaces to curve such in the bath around an open plug-hole;
o
water vapor to condense, which generates the vapor streams behind aircraft as
o
vortices shed from the wings; pressure in the center of tornadoes and hurricanes so that a secondary airflow is low established along the ground towards the center and then up into the core.
o
In practice, at the center of most naturally occurring free vortices the viscosity of the fluid causes it to rotate as a solid body creating a compound or Rankine vortex. A category five hurricane (e.g. Katrina in 2005) has wind speeds in excess of 155 mph. When it is over the sea, the outer portion can be modeled as a free vortex so if the maximum wind speed of 160mph (71.5m/s) occurs at 1 mile from the center then the wind speed at 5 miles from the center will be given by vr C so
v5 v1
r 1 r 5
0.2 and
v5
71.5 0.2 14.3 m/s (32mph)
Applying Bernoulli’s equation through the hurricane
19
Real Life Examples in Fluid Mechanics
v12 p1 p 5 2
v 52
2
So the pressure difference between the two locations at the same height ( z 1 z 5 ) is p5
p1
2
v v 1.29 71.5 14.3 3165 Pa 2 1
2 5
2
2
2
This is a large pressure difference and helps to explain the catastrophic damage caused by a category five hurricane, which is the most intense category of storms.
Evaluate:
Invite students to attempt the following examples: Example 4.1
Calculate the force on your hand when you hold it as far out of the car window as you can reach, palm against the wind, when the car is travelling at 70mph. Solution: 2
Measure the area of your hand. For the editor’s hand, tthe he area is approximately 0.018m . Apply Bernoulli’s equation for point some distance in front of your hand and for point just in front of your palm such that both points are on the same streamline,
v12
2
p1 gz 1
v22
2
p2 gz 2
The air velocity at point is 70mph (~31m/s = v1) and zero at your palm (v2 =0). The gage pressure at point will be zero ( p1=0) and there is no height difference between the -3 points if you are on a level road so z 1 z 2 . Thus, with aair ir = 1.29 kg.m p 2
air
v12
2
so p2 air
v12
2
1.29
312 2
620 Pa
Now, F pA 620 0. 018 11.2 N
Example 4.2
If a vacuum cleaner hose is to be able to pick-up a layer of dust 0.5mm thick from a height of 3cm, calculate the maximum average velocity that needs to be achieved at the inlet to the hose. Solution: 3
Let’s assume the dust is dry sand for which density d ensity data is available, i.e. = = 1600 kg/m . If the cross-section area of the hose is A, then the force required to lift the area of sand covered by the hose is F mg Atg and the pressure at the hose, p F / A tg
20
Real Life Examples in Fluid Mechanics where t is is the thickness of the the layer of sand. Applying Bernoulli’s equation for a point on the surface of the sand where the static pressure and velocity are zero and at the hose so v 22 gh1 p 2 2
or
v2
gh2
2 g h1 h2 2tg 2 9 .81 0.03 2 0.0005 9.81 0.76 m/s
21
Real Life Examples in Fluid Mechanics CONTROL VOLUME ANALYSIS
5.
Topic: Momentum
Engage:
Take a hairdryer, stand and clamp, some balloons, and paperclips into class. Set up the hairdryer with the stand and clamp so that it is blowing vertically up on its cold setting. Blow up a balloon and place it in the the air-stream and it will probably be pushed up to the ceiling.
Explore:
Capture the balloon and attach a paperclip to the neck of the balloon and again place it in the airstream. As you add more paperclips the balloon will float lower and lower in the airstream. To save time you could use a series of balloons inflated to tthe he same diameter with various numbers of paperclips attached. There is a nice video of school kids 13
performing this experiment . Explain:
Explain that when the balloon hovers at a stable height, based on Newton’s second law applied to the balloon, the force on the balloon from the jet of air is equal to the weight of the balloon. We can draw a cylindrical control volume around but not including the balloon and equate the change in momentum to the external forces on the control volume, i.e.
F m (v2 – v1) In this case the external force is the weight of the balloon. Sometimes it is easier to consider three scalar components of this vector expression, i.e.
F m v F m v F m v x
2 x
v1
y
2 y
v1
z
2 z
v1
x
y
z
13 www.planet-scicast.com/view_clip.cfm?cit_id=2709 www.planet-scicast.com/view_clip.cfm?cit_id=2709
22
Real Life Examples in Fluid Mechanics Elaborate:
Consider the cylindrical control volume around the balloon. Assume that the velocity of the airstream leaving the hairdryer is a uniform 50m/s and that all of air leaves the control volume at the same speed but at an angle, = = 30° from the horizontal. In practice this angle will vary from about zero close to the bottom of the balloon and approach 90° close to the horizontal diameter of the balloon.
The mass flow rate at entry, m air Av 1.2
4
0.06 2 50 0.17 kg/s
Assuming the hairdryer has a circular nozzle with with an outlet diameter of 6cm. 6cm. momentum of the fluid crossing AB will be v AB m
The
0.17 50 8.48 N
And in the y-direction for cylindrical surface, given that continuity demands the total mass flow across the surface equals that across AB, v AD sin 0.17 50 sin 30 4.24 N m
So, applying the momentum equation, AB W m v 1 sin 30 4.24 N
This hair dryer could keep a balloon and tail of paperclips weighing 4.24N (=0.42kg) in the air. The airstream is slowed down by entrapment of the surrounding air and sso o the nozzle velocity will not be achieved at all heights above the hairdryer, which is why the balloon sinks down as it is loaded with paperclips.
Evaluate:
Invite students to attempt the following examples: Example 5.1:
Calculate the force exerted by the water from the tap in your kitchen sink on a coffee cup held in the flow when the tap is open to its maximum. Solution: Estimate the water flow by how long it takes to fill a measuring jug. In the editor’s kitchen it took 5 seconds to fill a 1 liter measuring jug, so the mass flow rate is m
H OV 2
T
1000 0.001 0.2 kg/s 5
C
A
B
C
The diameter of the orifice in the tap is 17mm and so the velocity of the flow is
23
Real Life Examples in Fluid Mechanics m
v
A
0.2
0.88 m/s 0.017 2 1000 4
Hence, the momentum of the water leaving the tap at AB is
v AB m
0.2 0 .88 0.176 N
The water is turned through 180° in the cup and so emerges with a momentum of equal v AB 0.352 N. and opposite sign so that the total change in momentum is 2m This is about one third of the weight of the editor’s best porcelain coffee cups.
Example 5.2
Calculate the force required to hold an umbrella against a 30mph wind. Assume the wind is blowing horizontally so you are holding your umbrella head into wind, i.e., the handle is parallel to the wind direction and the umbrella deflects the wind to leave at an angle of 45°. The umbrella is symmetrical symmetrical about the handle with a projected area perpendicular to the wind, which is a circle c ircle of diameter 1.2m. Solution:
F m v x
and
Force, F
A
Consider a control volume such that wind enters at AB with a velocity of 13.4 m/s ( ≡ 30mph) and leaves at 45° to this along BC. We can apply the momentum equation in the direction of the wind such that 2 x
u m b r e l l a
y x
C
v1
B
x
F air Av1 v1 cos 45 v1 x
d n i w
x
x
1.29 1 .2 4
2
13.412 cos 45 1 76.8 N
If you were walking into the wind, then it would be nec necessary essary to construct velocity diagrams for the wind velocity at entry and exit to, what would become, a moving control volume.
24
Real Life Examples in Fluid Mechanics CONTROL VOLUME ANALYSIS
6.
Topic: Energy
Engage:
Take a portable portable forced air heater into into class. The type most most students have in their rooms. Set it up on the desk at the front of class and make a show of warming your hands in front of it. Ask students, working in pairs, to identify the energy transfers being made to the air passing through the fan-heater and then to incorporate them into a control volume diagram.
Explore:
Invite one group to list the energy transfers transfers that they they identified. Discuss how these transfers relate to the first law of thermodynamics, i.e., conservation of energy with electrical energy being converted into heat energy and mechanical energy then the later being transferred as kinetic energy to the air. Invite a second group to sketch their control volume on the board for the class to see. see. Discuss how we can include energy flows as well as mass flows. In this case, mechanical work, W , via the fan and heat transfer, Q, from the heating elements.
W
v1, p1
v2, p2
Q
Explain:
Explain that there is energy contained in the mass flows into and out of the fan-heater, so that these represent energy transfers across the control volume boundary as well as the mechanical or shaft work and the the heat transfer from from the heater. The energy of a mass of fluid has three components: a.
internal energy due to the activity of the molecules;
b.
kinetic energy due to the motion of the fluid; and
c.
potential energy of the fluid as a result of its position (height). (height).
Expand on this by explaining that the first law of thermodynamics can be employed to track or account for these energy transfers and that in this context it is known as the steady flow energy equation. It can be stated in words as ‘ for for any a ny mass system, the net energy supplied (in) to the system equals the increase of energy of the system (stored) plus the energy leaving (out) the system’, i.e. 25
Real Life Examples in Fluid Mechanics
E Q W
where Q is the net heat transfer into the system, W is the net work supplied to the system, and E is the increase in in energy of the system. The Steady Flow Energy Equation (S.F.E.E.) can be derived from this expression by substituting for the three components of the energy (internal, kinetic and potential) in the mass flow to give 2 m h
2
2
h1 2 v2
v 1
1
g z 2 z 1 Q W
Elaborate:
Apply the SFEE to the fan-heater to find the enthalpy (internal energy) change experienced by the air. Assume the fan-heater converts 1000W of electrical energy into heat and work to drive the shaft, i.e. 1000 J/s. Q W
If the change in elevation between the center of the inlet and the center of the outlet is 100mm, then g z 2 z 1 9 .81 0.1 0.981 J
Now if the inlet and outlet ports are the same size, then mass continuity demands that v1=v2 so
1 2
v
v12 0
2 2
If the fan is 110CFM (1100.000472 = 0.052 m3/s), then the mass flow rate will be 0.062 kg/s (=1.20.052) Thus rearranging the SFEE to obtain
h2 h1
Q W
v v g z z 1000 0 0.981 16049 16 kJ/kg
m
1 2
2 2
2 1
2
1
0.062
This is the energy available to warm wa rm the room for you.
Evaluate:
Invite the students to attempt the following examples: Example 6.1
A new design of hand-dryer generates two 400mph sheets of air at approximately room temperature and about 30cm above the air intake. The mass flow rate rate is 68 CFM and its manufacturers claim it will dry your hands in 12 seconds. Calculate the work done by the dryer to dry your hands. Solution: Apply the steady flow energy equation and assume an ideal gas
h2 h1 12 v22 v 12 g z 2 z 1 q w m
26
Real Life Examples in Fluid Mechanics The flow is adiabatic in the dryer (no heat source) so q 0 and the air inlet and outlet are at atmospheric pressure so h1=h2. The mass fl flow ow rate is given by
1.268 0. 000472 0.0385 kg/s m Assume that the air intake is large and so the velocity approximates to zero then: w m
1 2
2 2
v
178.82 2 1 g z z 68 0.000472 2 9.81 0.3 513 J/kg
Over a 12-second period the total mass flow will be 0.462kg (=12 0.0385); and hence, the work done will be 237J (=0.462 513) which is 0.000066 kWh. This compares to the manufacturer’s claim of 0.00468kWh per dry!
Example 6.2
The total force opposing the motion of o f a cyclist is given by 14 R
C d v 2
2
A C r mg
v isand , Acoefficient where the bicycle rider’s speed through the airmof density is the projected frontal area of the rider who have a mass, , and a drag of C dd =0.9. = 0.9. Typical values for the product AC d d are of the order 0.39. Cr is the rolling coefficient and is typically 0.003.
For a rider and bike of combined mass 84kg, calculate the work-rate required to maintain a steady speed of 30mph on a level road; then assuming about 18% of energy obtained from food is converted to mechanical energy, estimate the food intake required for a 90mile bike ride. Solution: Take the cyclist and bicycle as the moving control volume, then the SFEE reduces to 0 Q W
Because there is no change in elevation ( z 1= z 2), no change in velocity ( v1=v2), and no change in enthalpy (h1=h2) across the control volume. The work-rate is the resisting force multiplied by the velocity (13.41 m/s) so,
w
AC d v 3
2
1.2 0.39 13.413 C r mgv 2
0.003 84 9.8113.41 597 J/s
Now from SFEE we have q w so the combustion of calories equals the work-rate once w 597 the metabolic efficiency is accounted for, i.e., q 3319 J/s or about 2880 0.18 Calories per hour. 14
Grappe, F., Candau, R., Belli, A., Rouillon, J.D., Aerodynamic drag in cycling with special reference to the Obree’s position, Ergonomics, 40(12):1299-1311, 1997.
27
Real Life Examples in Fluid Mechanics MODELING
7.
Topic: Similitude and dimensional analysis
Engage:
Take your toy boats from the bath into class. If you don’t have any, then you could either borrow or buy some; or at the end of the previous class invite students to bring in their own bath toys. Show them a video from YouTube of a ship in rough seas (search using ‘ Abeille Flandre’ and show the clip of this name 15.
Explore:
We have all played with boats in the bath. Discuss whether the behavior of toy boats will will provide a good model for predicting the behavior of full-scale ships at sea. We could get out of the bath and conduct the experiment in the controlled environment of the laboratory; but would the behavior in the laboratory be a good prediction of the performance on the high seas? Ask students to construct a list of factors that could differ between the lab and the ocean and then construct a master list on the board. b oard.
Explain:
You should have ended up with a long list of factors that could vary between the lab and the full-scale performance at sea. Explain that it is advantageous to group variables ssuch uch as pressure, density, length, viscosity, velocity into non-dimensional groups and then to conduct experiments to establish the functional relationship between the groups rather than the variables. This considerably reduces the am amount ount of experimentation experimentation required and can also help in ensuring similitude between experiments and the prototype or fullscale version.
Elaborate:
An American physicist, Edgar Buckingham (1867-1940), showed that the number of nondimensional groups required to correlate the variables in a certain process is given by n-m where n is the number of variables to be grouped and m is the number of basic dimensions included amongst amongst the variables. So we might expect the force, F , acting on our ship, and toy boat, would be a function of fluid density, ρ; dynamic viscosity, μ; gravity, g ; the speed of the ship, v; and a characteristic dimension of the ship, l . 15 www.youtube.com/watch?v=3408T5A-ApU www.youtube.com/watch?v=3408T5A-ApU
28
Real Life Examples in Fluid Mechanics So,
F f , , g , v, l
The fundamental dimensions of these quantities are F
N ( = kg m s-2) -3 kg m -2 Nsm
MLT-2 -3 ML -1 -1 ML T
g v l
m s-1 ms m
LT-1 LT L
-2
-2
So we have six variables and three basic dimensions (M: mass, L: length, and T: time) and thus we will need three ( 6 3 ) non-dimensional or Pi groups (hence the name Buckingham-Pi approach). This means we can reduce the number of variables from six six to three. To find the first group, 1, we can select the dependent variable, F , and form a nondimensional group by introducing variables with the appropriate dimensions in such a way as to create the non-dimensional group. If a dimension dimension exists on its own as the only dimension of a variable, then it should be considered last. So in this case starting starting with 2 -2 -3 2 -1 [M] we can introduce and and then v to have [T ] but then we have [L ][L ] = [L ] and 2
we need [L] so we must introduce l . So F 1 2 2 v l This is effectively effectively the ratio of the shear force on the the hull to the inertia forces. The process can be repeated choosing to to form the second group
2
vl
Note that the same repeating variables are used to achieve the non-dimensional group. This group is the Reynolds number and describes the ratio of the inertia to shear forces in the fluid. And finally for third group using using g we we can obtain 3 gl v2 This is the Froude number and describes the ratio of inertia to gravitational forces for a fluid with a free surface. So the functional relationship for the behavior of the ship either as a model (toy boat) or full-scale in the ocean is
gl , 2 v l vl v F
2 2
In order to achieve similitude between a model and a prototype Pi groups are made equivalent. So, for instance, if a tug boat in in your bath is 6cm long and an ocean-going tug is 32m long then 3 above would imply that
29
Real Life Examples in Fluid Mechanics
gl bath
2 bath
v
gl ocean 2 ocean
v
or
2 vbath
v
2 ocean
l bath
l ocean
0.6 32
1 533
And the speed of your model needs to be 1/23 (= 1 533 ) of the speed of the ocean-going tug, which is just as well since a typical speed for the ocean version is 11 knots or 5.6 m/s. The model would have to to move at 0.24 m/s m/s (=5.6/23), or about 9 inches/sec, which is quite fast for the bath. Evaluate:
Ask the students to attempt the following examples: Example 7.1
The suction of a vacuum cleaner can be equated with the pressure drop across its fan, p which is in turn related the fan diameter, D; its axial length, l ; the rotational speed, ; the inlet and outlet diameters, d 1 and d 2 and the air density, . Find the functional relationship between these groups. Solution:
p f D, l , , d 1, d 2 , We can express the dimensions of the variables as follows: -2
-1 -2
p
Pa (N m )
ML T
D
m
L
l
m
L
rad s-1
T-1
d 1
m
L
d 2
m
L
kg m
-3
-3
ML
There are seven variables and three basic dimensions so there will be four nondimensional groups. The repeating variables are D, and . The first group can be formed around p and we obtain
1
p 2 D 2
For the second group use the next non-repeating variable and so on to give
2 and
30
p
l D
,
3
d 1 D
and
4
l , d 1 , d 2 2 D 2 D D D
d 2 D
Real Life Examples in Fluid Mechanics Example 7.2
In order to study the interaction of a micro-surgery device and the flow in an artery, a five times scale model of an artery artery is to be constructed. The volume flow rate, Q, in the artery is believed to be a function of frequency of the heart beat, ; artery diameter, D; the fluid density, ; viscosity, ; and the pressure gradient, p/l . Identify the dimensional groups and estimate the volume flow rate required if saline is used as the work fluid instead of blood. Solution:
Q , D, , ,
p l
We can express the dimensions of the variables as follows: Q
m3s-1
L3T-1
s-1
T-1
D
m
L
kg m
-3
-3
ML
-2
p/ l
-1 -1
Nsm -3 Pa/m (N m )
ML T -2 -2 ML T
There are six variables and three basic dimensions so there will be three non-dimensional groups. The repeating variables are f , D and . The first group can be formed around p and we obtain Q
1
D 3
Then taking each of the non-repeating variables in turn
2
2
D
and 3
p l D 2
So, if the working fluid is changed from blood to saline, maintained so s
s b Db2 b s D s2
b
2 10 3 1060 1 b 4 10 3 1200 2 2
2 equivalence must be
0.11 b
assuming the saline is formulated to give approximately the same viscosity as blood, the th model heart beat needs to be about 1/9 of the natural heart rate. Now for flowrate equivalence of the first Pi group is required so Q s
D s3 s 3 b
D b
Qb
23
1
1 0.044
Qb
72Qb
Hence, the volume flowrate would need to be about 70 times the natural value in the body.
31
Real Life Examples in Fluid Mechanics FLOW
8.
Topic: Viscous flow in pipes (internal flow)
Engage:
Take a vacuum cleaner into the class and clean up after the previous class using one on e of the th e tools attached to the pipe. It should attract the attention of the students! Ask them to vote on whether the flow in the pipe is laminar or turbulent. 16 You can use a YouTube video to illustrate ‘laminar flow 17 in pipe’ and a corresponding video : ‘turbulent flow in a pipe’ by searching with the words in italics.
Explore:
Estimate the Reynolds number of the vacuum cleaner pipe: air vD
Re air
The density and viscosity of air at 20°C are 1.2 kg.m-3 and 19.810-6 Pa.s respectively. if the radius of the pipe is approximately 15mm and 150CFM (≡0.0706 m3s-1) is a typical airflow capacity. Thus
0.0706 0.03 0.0152 181,597 19.8 10 6
1.2 Re
Fully turbulent flow occurs with Re>2,300 so those that voted for turbulent flow were correct.
Inviscid core Inviscid core
Boundary layer
r D
l e Fully developed flow
16
www.youtube.com/watch?v=KqqtOb30jWs www.youtube.com/watch?v=KqqtOb30jWs 17 www.youtube.com/watch?v=NplrDarMDF8&NR=1 www.youtube.com/watch?v=NplrDarMDF8&NR=1
32
Real Life Examples in Fluid Mechanics Explain:
Explain how with no tool fitted to the pipe of the vacuum, the air enters with a nearly uniform velocity across across the section. As the air moves along the pipe, viscous effects effects cause it to adhere to the pipe wall creating a boundary layer in which viscous effects dominate. In the central section, beyond the boundary layer viscous effects effects are negligible; however, with distance along the pipe the boundary layer grows and eventually occupies the the whole cross-section. cross-section. The non-dimensional entrance length, l e D correlates well with Reynolds number.
Elaborate:
Explain that for laminar pipe-flow it can be shown that the velocity profile is described by 2 pD 2 2r 1 u r 16 l D
where the first term is the center-line center-line velocity of the fluid. This expression describes a parabolic distribution. By integrating over the cross-section of the pipe p ipe the flowrate, Q, can be obtained as Q
p D 4 128 l
which is known as Poiseuille’s law. The viscous effects at the pipe wall are responsible for energy losses, which are known as head losses, h L. It can be shown that h L
4l W D
stress. The resultant pressure drop is given by where W is the wall shear stress.
p 8 2vl r
where r is is the radius of the pipe. In turbulent flows the velocity approaches a uniform distribution across the pipe with only a small viscous sub-layer sub-layer close to the pipe wall. A number of empirical expressions expressions exist to describe the shear stress and the velocity profil profile. e. A reasonable approximation is obtained using the inviscid Bernoulli equation and assuming a uniform velocity profile. Losses in turbulent flow are assumed to arise from viscous effects in straight pipes and are known as major losses, h Lmajor : and those from head losses in pipe components, known as minor losses, h
L minor
.
33
Real Life Examples in Fluid Mechanics By non-dimensional analysis, it can be shown that the pressure drop along a pipe containing turbulent flow is given by
vD l ' ' 1 v D D 2 p
2
where is is a measure of the roughness of the pipe wall. If we assume the pressure drop is proportional to pipe length, which is supported by experimental evidence, then v 2 l p f 2 D
where f is is the friction factor and f Re, D
And, by substitution in the SFEE, it can be shown that h Lmajor
f
l v 2
D 2 g
For minor losses a loss coefficient, K L, is usually defined such that h L p K L 2 minor v 2 g 12 v 2 Finally for the vacuum tube we can estimate the head losses as follows: for a plastic tube the relative roughness can be taken as zero, and so using a Moody diagram 18 for a 5 Reynolds number of 1.810 we find a friction factor of about 0.0155, hence the pressure drop due to major losses is v 2 l
p f
2 D
0.0706 1.2 0.015 2 0.0155 2
2
1 3092 Pa/m 0.03 19
And for minor losses assuming a 90° flanged elbow e lbow in the pipe for which K =0.3 L
2
p 12 v 2 K L
18
0.0706 0.0152 0.3 =1796Pa.
1.2
2
www.ecourses.ou.edu/cgi-bin/view_anime.cgi?file=d06123.swf&course=fl&chap_sec=06.1 www.ecourses.ou.edu/cgi-bin/view_anime.cgi?file=d06123.swf&course=fl&chap_sec=06.1 19 For example: www.engineeringtoolbox.com/minor-loss-coefficients-pipes-d_626.html www.engineeringtoolbox.com/minor-loss-coefficients-pipes-d_626.html
34
Real Life Examples in Fluid Mechanics Evaluate:
Ask the students to attempt the following examples: Example 8.1
Investigate the mechanism in a spray bottle and identify the size and shape of the fluid pipes though
3mm
40mm
Return Spring Spray
which the liquid is drawn up and then squirted. Estimate whether the flow will be turbulent or laminar and then calculate the pressure drop due to head losses during squirting.
10mm
Non-return valve
Force
Solution: We need to estimate the Reynold’s number, nu mber, Re
H O vD 2
for the pipe to the spray nozzle, D 3mm and assuming we are spraying water so = = 1000 kg/m3 3 3 and = = 110 Pa.s. Let’s say it takes 250 squirts to empty an 8oz (236,584mm ) spray bottle then it dispenses 946mm3 (=236,584/250) at each squirt. You can check this by estimating the stroke volume for the piston; if the piston is of radius 5mm then it would require a stroke of 12mm (=946/(52)) which seems reasonable. If a squirt takes about 1 second, then the velocity will be 134mm/s (=946/( 1.52)). So substituting for the Reynolds number Re
1000 0.134 0.003 110 3
402
This is sufficiently sufficiently low for the flow to remain laminar. The return of the piston is generally slower so the flow will also be laminar. Consequently the pressure drop can be estimated from 8 0.001 0.134 0.04
8 H O vl 2
p
r 2
0.0015 2
19 Pa
This pressure plus the spring stiffness equates to the force needed to squirt the liquid.
Example 8.2
Estimate how hard you would need to pump to induce turbulent flow in the tube connecting a hand-pump to your bicycle tire. For these these circumstances estimate the pressure drop due to major and minor losses. Solution: For the onset of turbulence a Reynolds number of more than 2300 is needed and for fully developed turbulent flow Re>10,000.
35
Real Life Examples in Fluid Mechanics Re
vD vD
Everything is fixed except the velocity of the air, which is controlled by your pumping, so for a tube of internal diameter 3mm v
Re
2300 19.8 10 6
D
12.65 m/s
1.2 0.003
2 0.003 which requires a volume flow rate, Q vA 12.65 4
8.94 10 5 m3/s
If the internal diameter of the pump is 30mm, then you will need to move the piston at v
Q A piston
8.94 10 5 0.032 4
0.1265 m/s
The pressure drop resulting from major losses can be estimated using v 2 l p f 2 D
For a plasticfactor, tube f it = can be assumed to abepipe smooth, and 0.125m from a Moody chart for Re=2300, the friction = 0.03. Hence for of length v 2 l p f 2 D
1.2 0.12652 0.125 0.03 0.012 Pa 2 0.003
The only significant minor losses are in the valve and a typical loss coefficient is K L = 2, so 1.2 0.1265 2 p v K L 2 1 2
2
2 0.0192 Pa
Thus, in this case, the head loss due to the minor losses is greater than that due to major losses occurring from viscous effects, which is actually more common.
36
Real Life Examples in Fluid Mechanics FLOW
9.
Topic: Flow over bodies (external flow)
Engage:
Wear a swim cap into class – this is bound to engage the students’ attention. Ask them why competition swimmers wear caps and suits [Answer: to reduce drag]. Ask students to estimate whether the flow associated with a swimmer is laminar or turbulent.
Explore:
Estimate the Reynolds number for a competitive swimmer. The world record for for 50m freestyle is about 20s; so v 50 20 2.5 m/s hence for a swimmer of height 1.8m 1 .8m Re
vL vL
1000 2.5 1.8
3 1 10
4.5 106
So the flow is turbulent. You could show some some video footage of freestyle swimmers 20 from Youtube by searching for video “Total Immersion Swimming Freestyle Demo by 21 Shinji Takeuchi” . To illustrate the separation of the boundary layer show again the 22 video of separation around a tennis ball from lesson 4 .
Explain:
Explain that swimmers experience two forms of drag: frictional or skin drag and pressure or form drag. Also that drag due to pressure is almost always greater than drag due to skin friction. Swimmers believe that their swim swim suit and cap decrease skin skin drag by reducing the boundary layer. This is important important because an increase in in turbulence in the boundary layer increases whichspeeds consume reducing the kinetic energythe andeddies speed inof the the boundary swimmer 23layer, . At slow for for aenergy wellwellstreamlined body the the flow will be laminar and frictional resistance will dominate. At higher speeds, the the boundary layer grows and pressure resistance resistance increases. This occurs because the separation of the boundary layer from the body moves closer to the front further increasing the pressure drag but reducing the drag due to skin friction which gives an overall increase in total drag. The pressure resistance force force is given by
20
www.youtube.com/watch?v=rJpFVvho0o4 www.youtube.com/watch?v=rJpFVvho0o4 http://www.youtube.com/watch?v=AInQMmn-0Nw&feature=related)) or search for a video Alternate is http://www.youtube.com/watch?v=AInQMmn-0Nw&feature=related entitled: ‘How to improve your freestyle’ 22 www.youtube.com/watch?v=7KKFtgx2anY www.youtube.com/watch?v=7KKFtgx2anY 23 Zatsiorsky, V.M., Biomechanics in sport: perform performance ance enhancement a and nd injury prevention, prevention, IOC Medical Commission, International Federation of Sports Medicine. 21
37
Real Life Examples in Fluid Mechanics
F C D S M
v 2
2
where S M is is the maximal cross-sectional area interacting with the water and C D is the drag coefficient. Typical values for a swimmer of C D are 0.58 to 1.04 compared to between 0.05 and 0.08 for a dolphin 23. A dolphin has a better streamlined body without the local pressure resistance centers formed by the head, shoulders, buttocks, knees, kne es, heels, etc in a human.
Elaborate:
Explain that a body immersed in and moving relative to a fluid interacts with the fluid and experiences a resultant force in the direction of the upstream flow, known as drag, and a force normal to the the flow, known as lift. lift. A simple explanation of lift lift is given in a 24 short video entitled ‘The Magic of Airfoils’ (search in Youtube using italicized words) you might also want to show a video of a basic student experiment, for example search on YouTube for ‘Wind Tunnel Basic Airfoil Test ’25. It can be shown that lift,
L dF y
p sin dA W cos dA
v
and
pdA
drag, D dF x
p cos dA W sin dA
So that lift and drag coefficients co efficients can be defined as C L
L v 2 A
1 2
and C D
D v 2 A
1 2
wdA y
x
dA
Return to the tennis ball video and highlight that inviscid flow around it are symmetric about the x-axis and so the lift and drag are zero in these conditions. However, if the sphere is spinning about the z -axis, -axis, then the rotation will drag some fluid around so that the flow is no longer symmetric and both lift and drag are created. Ask the students for examples of the application of this phenomenon [answer: curved balls, floaters and sinkers in baseball, hooking or slicing a golf ball, and curving a ball in soccer]. You could show the ‘ Best Best Curve in Soccer History’ search for this title on 26 YouTube . For horizontal flight, the lift generated by spinning of the ball must equal the weight of -1 the ball, so for a soccer ball b all of mass 450g, diameter 22cm traveling at 30ms mg L 12 H 0 v 2 AC L 2
24
www.youtube.com/watch?v=yp7w3p0P2tw www.youtube.com/watch?v=yp7w3p0P2tw www.youtube.com/watch?v=1PXJ3HhV2mE www.youtube.com/watch?v=1PXJ3HhV2mE 26 www.youtube.com/watch?v=-jLD_2ULFxc www.youtube.com/watch?v=-jLD_2ULFxc 25
38
Real Life Examples in Fluid Mechanics
or
C L
mg
1 2
H O v
2
2
4
2
d
1 2
0.45 9.81 1.2 30 2 4 0.22 2
0.215
Empirical data suggests that this can be achieved with d
2v
So
0.9 0.9 2v d
0.9 2 30 0.22
245 rad/s 2343 rpm
It would be relevant to discuss that external ex ternal flows, such as those discussed above beco become me 5 turbulent when the Reynolds Number, Re > 510 (this compares to fully turbulent flow at Re > 4000 and fully laminar flow at Re < 2300 for internal flows).
Evaluate:
Invite the students to attempt the following examples: Example 9.1
Blood cells are damaged by turbulent flow. Estimate the maximum maximum size of a device that could be introduced during surgery on the aortic valve without creating turbulent flow around it. Solution: 27
28
The flow velocity through the aortic valve is typically 6m/s and the density and 29 3 -3 viscosity of arterial blood is 1050kg/m and 410 Pa.s respectively. For laminar flow to be maintained Re < 1000 and Re
so
L
vL vL
1000 4 10 3 Re < 1050 6 v
6.35 10 4 m or 0.6mm
Example 9.2:
When you take a job delivering pizzas you are expected to fit a sign 0.8m 0.3m to the top of your car. If you drive drive at an average of 30mph (≡ 13.41ms-1), calculate the added cost of fuel fuel used per hour. Assume the car is 30% efficient. The drag coefficient for a
27
Mohiaddin, RH, Firmin, DN, Longmore, DB., Age-related changes of human aortic flow wave velocity measured non-invasively by magmetic-resonance imaging, J. Appl. Physiology, 74(1):492-492, 1993. 28 Kenner, T., The measurement of blood density and its meaning, Basic Res Cardiology, 84:111-124, 1989. 29 Dormandy, JA, Influence of blood viscosity on blood flow and the effect of low molecular weight dextran, British Medical J., 4:716-719, 1971.
39
Real Life Examples in Fluid Mechanics long flat plate is 1.9830. If you lived iin n area with some long empty roads so that your average speed increased to 45mph, what would be the cost of the extra fuel? Solution: The definition of the drag coefficient is C
D
D
v A
1 2
2
so, the force (=drag) required to move the sign is D 12 air v 2 AC D
12 1.2 13.412 0.8 0.31.98 51.3 N
Thus, the power required is P Fv
51.3 13 .41 687 W
This needs to be generated by the combustion of fuel in the engine Q
P E
687 0.3 32 109
7 .16 10 8 m3/s
where E (J/kg) (J/kg) is the energy content of gasoline ( 32MJ/liter 31 32,000MJ/m3). And this quantity is equal to 0.068 gallons/hr g allons/hr or 24cents/hr if the price of gas is $3.50 $3.50/gallon /gallon (as it was at the time of writing in Michigan). For an average speed of 45mph (20m/s), D = 114N, P =2280, =2280, Q=2.37510-7 and the cost is 79 cents/hr; so a 50% increase in average speed costs three times as much.
30
See for example: http://www.engineeringtoolbox.com/drag-coefficient-d_627.html 31 http://bioenergy.ornl.gov/papers/misc/energy_conv.html
40
Real Life Examples in Fluid Mechanics FLOW
10.
Topic: Flow in open channels
Engage:
Show a video clip of the ultimate waterslide by searching on YouTube You Tube32 for ‘ Barclay card advert- waterslide 2008 [HQ]’. Ask students, working in pairs, to identify the two essential differences between open channel flow and flow in a pipe.
Explore:
Invite a couple of pairs to present their conclusions. Discuss the obvious one that the presence of the free surface and highlight the perhaps less obvious one that the flow is driven by gravity. It cannot be driven by a pressure difference (e.g. from pump) because of the negligible inertial and viscous effects 33 of the atmosphere above the liquid.
Photograph © BLcarnut/John Clouston 33
The free surface of the fluid in the channel allows waves to exist and the Froude number characterizes their motion. For instance when a stone stone is thrown into a slow flowing canal the resultant waves ripple out more or less equally in all directions, including upstream. This is termed sub-critical flow for which flow velocity, v < wave velocity, c. However, when a stone is thrown into a fast flow river, waves only propagate downstream because v>c and the flow is termed supercritical. The ratio of the flow and wave speed speed is the Froude number Fr
v v gy 2 c
where y is the mean depth and g is is gravitational acceleration. The behavior of subcritical (Fr1) flow can be very different. We have probably all thrown a stone in a river but there is videoclip of ‘ Yosemite National Park water ripples in slow motion’, which can be found by searching on Youtube34 using the title.
32
www.youtube.com/watch?v=1WlRcXIO5ik www.flickr.com/photos/58017169@N06/5360523177/in/photostream/ 34 www.youtube.com/watch?v=T0tGXxF15-I www.youtube.com/watch?v=T0tGXxF15-I
33
41
Real Life Examples in Fluid Mechanics Explain:
Explain the flow down a water slide. slide. For uniform flow in the slide the gravitational forces would have to equal the frictional and viscous forces associated with the walls of 2 the slide that are dependent on the wall shear stress, W W , which is proportional to v , roughness, and Reynolds number. Reynolds number for open channel flow is defined as Re
vRh
where Rh is the hydraulic radius of the channel, defined as the ratio of the cross-sectional area of the flow to to the wetted perimeter. Generally, open channel flow is laminar when Re < 500 and turbulent when Re > 12,500. At the entrance to the water slide when the velocity is low, the frictional forces will be smaller and the fluid will accelerate under the larger gravitational force until a terminal velocity is reached when the gravitational and frictional forces are equal. uniform flow
varied flow
y1 v1 1
z + g / 1 p
y2 v 2 2
z + g / 2 p
Datum
B
L
F 1
gALsin C
A
gAL
F 2 D
The requirement of continuity in a slide of constant cross-section implies that the depth of water must decrease as the velocity increases. Thus the entry to the slide is characterized by varied flow (changing depth and velocity) prior to the uniform flow region. In the uniform flow region, applying the momentum equation we have F 1 F 2 gAL sin W PL 0 where F 1 and F 2 are the hydrostatic forces acting on the end planes of the section ABCD, and P is is the wetted perimeter. If the velocity is constant, then based on continuity so is the depth and assuming the pressure is given by gz where where z is is depth below the surface, so the mean pressure on AB and CD is gz 2 , and hence F 1 F 2
gyA
2
and for small slopes ( = = sin ) the momentum equation reduces to
42
Real Life Examples in Fluid Mechanics
g W
P W
A
m
where m ( A P ) is known as the mean hydraulic depth. By definition, the friction factor, W
f
v
1 2
so
2
f v 2
g
2 g m C m f
and v
2m
where C is known as the Chezy coefficient (after Antoine Chezy (1718-1798) and is dependent on the friction factor. A civil engineer, Robert Manning (1816-1897) found that the coefficient was also dependent on the hydraulic mean depth so he proposed the empirical relationship 1
C Mm 6
where M is known as the Manning constant. Note that in some US texts the Manning 3 -1 constant is defined as n M 1 and that M has has basic units of [L T ] so it important to take care when using it. Note that M =1.486/n =1.486/n where the units on left side are [m1/3s-1] and on the right, feet and seconds. Now, applying this analysis to a water slide with an approximately rectangular crosssection of width, 0.4m in which we wish to maintain a maximum water depth of 2cm. For plastic pipe, n =0.0135 and 2
Flow cross-section area
= width depth = 0.4 0.002 = 0.0008m
Wetted perimeter
= width + ( 2 depth) = 0.4 + 0.004 = 0.404m m
And, mean hydraulic depth,
A P
0.0008 0.404
0.00198 m
Now combining the Chezy and Manning equations to give
v
So
1
n
2
1
m 2 3
Q Av An 1m 2
3
1
2
AMm
2
3
1
2
And assuming an angle of slope for the slide of about 20 degrees ( ≈0.35 rads), then Q AMm 2
3
1
2
0.0008 148.6 0.00198 0.35 1.1 10 3 or about 66 liters/min. 2
3
1
2
Finally, it is possible to generate discontinuities in open channel flow with or without a change in the channel geometry. A hydraulic jump involves a change in depth from y1 to y2 such that y2 y1
1 2
1 1 8 Fr for y y 2 1
2
1 and Fr 1 1 .
1
35 www.engineeringtoolbox.com/mannings-roughness-d_799.html www.engineeringtoolbox.com/mannings-roughness-d_799.html
43
Real Life Examples in Fluid Mechanics You can show a video of this occurring without a change in channel geometry by searching in YouTube for ‘Water Engineering II S1 - Hydraulic Jump Practical’ 36 or with a geometry change by searching in YouTube for ‘ Hydraulic jump over weir ’37. Evaluate:
Invite the students to attempt the following examples: Example 10.1
During a rain storm water flows along the edge of road. The cross-section of the flow forms a triangle with a base of length 100mm on the surface and base to apex depth of 15mm. If the road is concrete with a gradient of 1 in 25 in the direction of the flow (and traffic), calculate the discharge rate along the curbside gutter.
45° m m 5 1
100mm
Solution: 0.10 0.015 2
7.5 10 4 m3
Cross-section area of flow,
A 12 bh
Wetted perimeter,
P = = (21.2 + 80.2) 10 = 0.101 m
-3
m
So the Mean hydraulic depth,
Manning Constant for finished concrete, 2
3
1
2
4
A P
7.5 10 4 0.101
7.4 10 3 m 38
n = 0.012 so M = 1.486/0.012=124 2
3
1
1
2
4
3
7.5 10 124 0.0074 tan 1 25 7.06 10 m /s Flowrate, Q AMm Or 43 liters/min or 650 gallons/hour.
Example 10.2
A drainage ditch for a local sports field consists of an excavated earth channel with a gravel lining, which gives a Manning constant, M=40. After some time the channel becomes weedy and the Manning constant is reduced to, M=33. If the ditch is semicircular in cross-section with a diameter of 1.2m and on a gradient of 1 in 20, calculate the reduction in flowrate caused by the weeds when it is full. 36
www.youtube.com/watch?v=kM2XAsS4RVo www.youtube.com/watch?v=kM2XAsS4RVo www.youtube.com/watch?v=cRnIsqSTX7Q 38 www.engineeringtoolbox.com/mannings-roughness-d_799.html www.engineeringtoolbox.com/mannings-roughness-d_799.html 37
44
Real Life Examples in Fluid Mechanics Solution: d d 2
Area of flow,
A
Wetted perimeter,
P
d
Mean hydraulic depth,
m
A
8 2 P
1.2 2
d
8
0.565 m2
1.2
0.565 3.08
2
1.2 3.08 m
0.183 m
Difference in Flowrate: dQ dM Am 2
3
1
2
( 40 33 ) 0.565 0.183 tan 1 1 20 0.284 m3/s 2
1
3
2
or 17.5% (= 40 33 40 ) less capacity.
45
Real Life Examples in Fluid Mechanics FLOW
11.
Topic: Compressible flow
Engage:
Show a news-clip entitled ‘ Bull Bull leaps into crowd c rowd in Tafalla Northern Spain 30 hurt ’ by searching for this title on YouTube39. Run the clip once (it is only 29 secs) secs) and then a second time pausing it at about 6 seconds and discuss how the crowd is reacting to the bull. They are moving away; the bull is slowed down by the fence and so people have time to get out of the way. The speed of signaling signaling (about the danger) through the crowd could be considered to be faster than the speed of the bull. Then let the clip40 run to about 10 seconds and pause it again. Discuss the difference difference in crowd behavior. Now the bull has got up some speed and the people in the crowd cannot react faster enough so the bull runs into them. The speed of signaling through the crowd crowd could be considered to be slower than the speed of the bull. If you feel students need to be convinced that crowds can behave in an analogous way to fluid flow, then you might find the following videos useful: a.
a flock of sheep flowing around a stationary car (search on YouTube for ‘ A 41 sheep dog herds sheep near Wanaka, New Zealand ’ ); and
b.
a flock of sheep parting to allow a llow a car through which is analogous to sub-sonic flow with some signaling moving through the flock faster than the motion of 42 the car (search on YouTube for ‘ New New Zealand in 56 secs’ ).
Explore:
Discuss how the signaling mentioned above is achieved in air with an airplane moving through it at subsonic speeds, i.e., that pressure disturbances caused by the plane’s motion are propagated in all directions at the speed of sound and at some distance are attenuated by the viscosity of the air. Ahead of the plane they act as a signal to tthe he air that the plane is coming and the air begins to move out of the way of the plane so that smooth flow is achieved around around the plane. This can occur if the sspeed peed of the plane, v is slower than the speed of signaling, i.e., the speed of sound, c.
39
www.youtube.com/watch?v=2ksQZmxI_Nw www.youtube.com/watch?v=2ksQZmxI_Nw An alternative but longer video can be found at http://www.youtube.com/watch?v=SWTiHCK5ZpM or http://www.youtube.com/watch?v=SWTiHCK5ZpM or by searching on YouTube using ‘ Bull corresponding pause points Bull goes on rampage rampage through crowd ’’.. The corresponding are about 8 seconds and 12 seconds respectively. 41 www.youtube.com/watch?v=sYJaMGcrtWE www.youtube.com/watch?v=sYJaMGcrtWE 42 www.youtube.com/watch?v=WT3dk7HOi1A www.youtube.com/watch?v=WT3dk7HOi1A 40
46
Real Life Examples in Fluid Mechanics You can see this happening using smoke in a wind tunnel, e.g., search on YouTube for a 43 video entitled ‘ How How wings work? Smoke streamlines around an airfoil ’ . Now, ask the students to consider what will happen if v > c? This is analogous to the bull running into the crowd. The air ahead of the plane has no warning and so does not move out of the way. The bull is brought to a stop by the group of compressed people and he moves off to one side; but the plane is not brought to a stop, so what happens? A shock wave forms forms just ahead of the plane. Let us consider shifting our fframe rame of reference to that of the pilot, i.e., a stationary plane with supersonic air moving towards it (v > c). As the air passes through the shock shock wave its velocity is reduced so that v < c and the air has time to part and flow around the plane. This transition across the sshock hock wave not only involves a deceleration of the air but also an increase in its static temperature, pressure and density. For a good illustration of this search on Youtube44 for a video entitled: ‘ F18 Supersonic flyby with sonic boom on CVN69 USS Eisenhower ’. ’.
Explain:
Explain that we can analyze the region around a shock wave by enclosing it in a control volume. Applying the momentum equation to the control volume, v1 p2 A m v2 p1 A m
control volume
Continuity of mass flow requires,
T 1
1v1 A 2v2 A
so
p1 p2 v
2 2 2
p1, 2 RT 1, 2
v2
v1
v 2 1 1
1 p1
Assuming an ideal gas both sides of the shock so that 1, 2
T 2
2 p2 shock wave
and substituting p1 p2
p2 v22 RT 2
p1v12 RT 1
v12 v22 p2 1 or p1 1 RT RT 1 2 v
Now defining the Mach number, M a as M a c
p1 p 2
1 k M a 2
v kRT
where k
C p C v
so
2
1 k M 2
a 1
Note that when M a1 and M a 1 p1 and static pressure increases across the shock wave. Assuming adiabatic conditions for the control volume so that the stagnation temperature is constant, i.e. T 0 1 T 0 2 it can be shown that T 2 T 1
T 2
T 0
2 M a 1 k 1 2
T 0 T 1 2 M a 22 k 1
Thus, again, since M a 2 >1 and M a 1 T 1 and temperature increases across the shock wave. Returning to mass continuity for the control volume,
2v2 A 1v1 A And substituting 1, 2
p1, 2
v
and M a c
RT 1, 2
v kRT
we can obtain
2
p M 2 a 2 T p M 1 1 a 1
T 2
Equating this with the previous expression for the temperature ratio we can obtain
2 k 1 M a 12 M a 2 2 2k M a 1 k 1 Elaborate:
Introduce the idea that can produce supersonic flow for testing shapes in a wind tunnel using a convergent-divergent nozzle, known as a Laval nozzle, connected to a large reservoir of gas for which p0, 0, T 0 and v0 are zero. Highlight that the Laval nozzle is the same shape as the exit of a rocket engine. Applying Bernoulli’s equation for any section of the nozzle, k RT 0 T 2 k 1
v2
And normalizing by c v2 c
so
48
2
v M a
Laval nozzle
kRT
2 T 0 1 M a2 k 1 T
k 1 2 1 M a T 2
T 0
Large M a1
Test section
Real Life Examples in Fluid Mechanics T 2
and, recall that
T 1 T t
hence, for conditions at the throat, t
T
T t T 0
T 0
T 1
T 0 T
k 1
T 0 T
Now, for isentropic flow,
2 M a 1 k 1 2
T 0
T 2
2 M a 2 k 1 2
2 k 1 M a2
p p0
2 k 1 M a2 k 1
Applying mass continuity for the nozzle, v
so with M a c
v kRT
we can obtain
k 1
0
t
and hence
k 1 k
1
k 1
vA t vt At At A
T M a t T t
And substituting the values for the density and temperature ratios, At A
k 1 a M 2 k 1 M a2
k
k 1
i.e., for k =1.4 and a Mach number of 3, the ratio of the throat diameter to the section diameter has to be
k 1 At M a 2 2 k 1 M a
k 1
2 k 1
2.4 A 3 2 2 0.4 3
1.4
0.8
A 0.68 A
Evaluate:
Invite students to attempt the following examples: Example 11.1
In a laboratory, if a supersonic wind tunnel has a Laval nozzle with a throat area of 3cm 2 and is supplied from a large reservoir at a pressure of 20kPa and temperature 20°C, calculate the mass of air required to run the tunnel for 30secs. Solution: The mass flow rate through the throat will be given by
t At vt m 2 k 1 and and in the throat M a =1 so vt kRT t also k 1 T 1
T t
2 0 m k 1
1
k 1
At
2kRT 0
2
t
1
k 1
so
k 1
49
Real Life Examples in Fluid Mechanics
and substituting the ideal gas equation, 0
m
At p0 T 0
R k 1 k 2
k 1
2 k 1
RT 0
3 10 2 10 6
p0
4
293
1.4 2 287 2.4
2.4
0.8
1.86 10 4 kg/s
So for 30s we will need 0.0056kg or about 0.0067m3 (=0.00561.2). Example 11.2
An explosion occurs in a processing works and generates a shock wave that propagates outwards radially. At some distance from the works the shock wave has a Mach number of 1.5. Calculate the pressure and velocity just behind the sshock hock wave and comment on its effect. Solution: Assume atmospheric conditions:
T 1=21°C and p1 = 101kPa.
Velocity ahead of the shock wave
v1
2
kRT 2 1.4 287 294
344 m/s
2 k 1 M a 12 2 0.4 1.5 2 0.7 Now, M a 2 2 2.8 1.52 0.4 2k M a 1 k 1 2 1.5 2 0.4 hence, T 2 T 1 294 388 K 2 2 0.7 2 0.4 2 M a 2 k 1 2 M a 1 k 1 2
1 k M a 1
2
and
p 2
p1
1 k M 2
a 2
v
Finally, M a c
v kRT
101105
1 1.4 1.5 2 249 105 Pa 2 1 1.4 0.7
so v2 M a 2 kRT 0 .7 1.4 287 388 276 m/s
So the pressure rises to more than twice atmospheric a tmospheric pressure with winds speeds of 68m/s
v1 v2 344 276 or 152 mph. These values will cause extreme damage!
50
Real Life Examples in Fluid Mechanics APPLICATIONS
12.
Topic: Turbomachines
Engage:
Take a pack or two of ‘ Propeller Propeller Balsa-Wood Model Planes’ into class and share them around the students 45. Invite the students, working in pairs, to put them together, wind up the propeller and let the planes fly around the room.
x Explore:
x
dr
vt =r r
When they have had some fun, ask them to work on a velocity diagram for an element of the blade and determine the velocity of the air relative to the blade. Probably, you will have to get them started by drawing the sketch opposite and the section XX. Invite a pair to draw their velocity diagram for the rest of the class:
v0
vt =r
D 2
vt =r
v0 – airplane airplane velocity
Section XX
The angle of the airflow is given by tan 1
v0 r
and its speed is given by r cos . The angle, is known as the pitch angle and the difference between and and is is the local angle of attack, .
Explain:
Tell the students that it can be b e shown by dimensional analysis that
45
At the time of writing a pack of 12 were available from Amazon.com for $12.99. Try searching Amazon.com using the words in italics above.
51
Real Life Examples in Fluid Mechanics
v0 D 2 ' 2 D 4 D F T
where F T is the propeller thrust, D is the propeller diameter, is is the rotational speed, v0 is T is the plane velocity, and and are the air density and viscosity respectively. The group on the left is known as the thrust coefficient, C T T and the first one on the right as the advance ratio. In most applications the Reynolds number is high and so the thrust coefficient is largely independent of Reynolds number, so C T
v0 D
The advance ratio is related to the relative velocity of the air at the propeller tip by tan 1
2v0 D
A similar analysis can be performed for the power, P , in which case everything is unchanged except the -group on the left becomes the power coefficient,
C p
P 3
5
n D Elaborate:
Discuss how pumps are characterized and selected. selected. An axial flow pump or fan is essentially a propeller and the thrust coefficient can be redefined in terms of the pressure change, p or head difference, H, since
pA HA
F T
and so C T
HA 2
D
4
g H
4 2 D 2
but it is more usual to use the head coefficient, C H
4 g H C T 2 2 . D
A discharge coefficient can be found by dimensional analysis to be C Q
Q D 3
Axial-flow pumps are better suited to applications involving high discharge, Q and low head, H ; while radial-flow pumps, such as centrifugal pumps, are more appropriate for low discharge, discharge, high head applications. A -group known as the specific speed, n s is used when selecting the best pump for an application from a range of geometrically similar pumps.
52
Real Life Examples in Fluid Mechanics 1
nS
C Q2 3
1
C H 4
Q 2
g H
3 4
1 2
3 4
H , which is not non-dimensional, (Note: in the US it is common to use N S NQ since N is is in revolutions per minute, Q in gallons per minute and H in in feet. The specific speed is closely related to the susceptibility to cavitation cavitation on the suction side. Cavitation should not occur, otherwise damage may be incurred in the blades of the pum pump. p. Usually, the specific speed is modified by replacing the pressure head by the difference between the pressure on the suction side of the pump and the vapor pressure of the liquid being pumped, which is called the net positive suction head, NPSH, so 1
nSS
Q 2 3
3
g 4 NPSH 4
where nSS is known as the suction specific speed and in the US is modified to 3
1
N SS NQ 2 NPSH 4 and a critical value of N SS SS is 8500.
Evaluate:
Invite students to attempt the following examples: Example 12.1
The water pump in a car is of the centrifugal type and sucks water from the radiator at about 15psi and 140°F before pumping it into the engine block. If the coolant flow rate required is 100 gallons per minute, calculate the maximum speed the pump can run at to avoid cavitation. Solution: The pressure head, p in psi can be converted to the pressure head, h in feet using h
2.31 p SG
where SG is the specific gravity gravity or relative density of the fluid. Thus, for the radiator h
2.31 p SG
2.31 15 1
34.7 ft
Vapor pressure of water at 140°F is 2.9psi 46 which can also be converted to give 2.31 2.9 1
6.7 ft
So the Net Positive Suction Head, NPSH 1
3
N SS NQ 2 NPSH 4 46 From
34.7 6.7 28 ft
8500
for example http://www.engineeringtoolbox.com/water-vapor-saturation-pressure-air-d_689.html http://www.engineeringtoolbox.com/water-vapor-saturation-pressure-air-d_689.html
53
Real Life Examples in Fluid Mechanics
thus
N
3 4
8500 NPSH Q
1 2
8500 28 100
1 2
3 4
10,300 rpm
Example 12.2
If the tip speed of a propeller a 2m diameter is not too exceed a Mach number of 0.5 in ambient conditions, calculate the maximum maximum speed of rotation. Then if the local angle of attack is to be zero for a forward velocity of 200mph, find the pitch angle as a function of distance along the blade from the tip. Solution: M a
and
v
c
v kRT
vt r so vˆt
equating M a kRT and
so vˆt M a kRT
D
2 M a kRT D
2
D
2
2 0.5 1.4 287 293 171 rad/s or 1640 rpm (=17160/2). 2
89 1 tan 1 r r 171 2r or about 30° at the tip and approaches 80° towards the hub (r = = 0.1m) at the center. For local angle of attack, 0 so tan 1
54
v0
tan 1
Real Life Examples in Fluid Mechanics NOTES
55
Sop homo ho more re Fluid Fluidss C o urse: Sug g e ste ste d e xemp xe mp la rs w ithin le sso sso n p la ns NOT NO TES FOR INS INSTRUC TORS ON EXAMPLE AP APPLIC ATIONS Pre p a re d a s p a rt of Pre o f the NSF NSF-sup -supp p o rted p ro je c t ( #0 #0431756) enti e ntitl tle ed : “Enha “E nhanc nciing Diver Divers sity in the Under nde rg ra d uate ua te M e c hani ha nic c a l Enginee Engineerring Pop Po p ula ula ti tio o n thr thro o ugh C urr urric ulum ulum C hange ha nge” ” Edited by Eann A Patterson, University of Liverpool
