12 Fluid Mechanics

8/17/2019 12 Fluid Mechanics

1/66

General Physics : Oscillations, Waves, Sound and Light

Textbook:

- "University Physics, 13th Edition" by Young and Freedman- Ch. 12, 14-16, and 32-36

Material Covered:

- Fluid Mechanics (chapter 12)- Oscillations and Waves, Sound (chapters 14 – 16)- Light and Optics (chapters 32 – 36)

Course website is www.eee.uci.edu Read the detailed syllabus posted there!

Instructor: Ilya Krivorotov (e-mail [email protected], phone: 824-6967)

Office hours:- Wednesday from 4:30 pm to 5:30 pm in Rowland Hall, Room 310E

http://www.eee.uci.edu/mailto:[email protected]:[email protected]://www.eee.uci.edu/


2/66

Textbook purchasing options


3/66


4/66


Grading:

- homework (on-line through MasteringPhysics): 10%

- 2 midterms (closed book, 1 page of handwritten notes, 50 minutes): 40%

- final (closed book, 2 pages of handwritten notes, 1 hour 50 minutes): 50%

- free response questions at the midterm and final exams

- the final course grade will be curved to fit the following distribution:

A/A-: 15%, B+/B/B-: 35%, C+/C/C-: 35%, D+/D/D-/F: 15%


5/66


How to succeed in this course:

- Solving homework problems is the best way to learnand to prepare for exams. Your grade is determined byhow well you can solve problems!

- Attend Discussion Sections to learn how to solveproblems

- Read the relevant sections of the textbook before coming to lectures and discussion sections

- Ask questions, attend my office hours and your TAoffice hours. Utilize the physics tutoring center.


6/66

Fluid Mechanics

States of Matter:

• Solid - Has a definite volume and shape

• Liquid - Has a definite volume but not a definite shape

• Gas - Has neither a definite volume nor shape, occupiesthe volume of the vessel it is confined to.

• Gases and liquids flow and are called fluids

• A branch of physics that deals with macroscopic properties

of liquids and gases is called Fluid Mechanics

vessel

object


7/66

States of Matter

• All of the previous definitions are somewhatartificial:

– Window glass slowly flows although it appears to be a solid – Water and water vapor are indistinguishable at high

pressure and temperature

– Is ice cream a solid or a liquid?

• More generally, the time it takes a particular

substance to change its shape in response to anexternal force determines whether the substanceis treated as a solid, liquid or gas

• A tar drop may be treated as solid on time scales much shorter than 10 years but must be treated asfluid on time scales comparable to 10 years.

Pitch drop experiment set up in1927 at the University of

Queensland in Australia. A tardrop falls every ten years!

http://www.youtube.com/watch?v=w0z8fVrTyzg


8/66

Weather: Tornadoes, Hurricanes Transportation: Aircraft, Ships

Fluid Mechanics and Life

Medicine: Blood flow Beautiful scienceMicrofluidics: Lab on a chip


9/66

Microscopic Structure of Fluids

• From a microscopic point of view, a fluid is a collection of molecules thatare randomly arranged and held together by weak cohesive forces andby forces exerted by the walls of a container.

• Molecules in a solid maintain their positions with respect to each other(thus there is fixed shape). Molecules in a fluid randomly move around(thus fluids flow).

Atoms or molecules

in a typical solidAtoms or moleculesin a typical fluid


10/66

Continuous Model of Fluids

• In fluid mechanics we neglect the atomistic nature of matterand treat fluids as continuous substances

• We can apply the principles we learned in Newtonianmechanics (statics and dynamics) to fluid mechanics

• Similar to Newtonian mechanics, fluid mechanics has twobranches:

• Fluid Statics – Describes fluids at rest

• Fluid Dynamics – Describes fluids in motion


11/66

Forces in Fluids at Rest

• Fluids do not sustain shearing tensile stress

• The only stress that can be exerted on an

object submerged in a static fluid is one thattends to compress the object from all sides

• The force exerted by a static fluid on anobject is always perpendicular to the

surfaces of the object

- force exerted on a solid (e. g. spring)

- force exerted on a fluid (e. g. air)

- an object (e. g. brick)

tension

shear

compression

solid fluid


12/66

Origin of Compressive Forces Exerted by Fluids - I

• What is the origin of the compressive force exerted by a fluid on an

object submerged in it?

• The compressive force comes from the random motion of themolecules of a fluid.

Fluid molecules

Submerged object


13/66

Origin of Compressive Forces Exerted by Fluids - II

• As molecules of a fluid hit the walls of theobject submerged in the fluid (or the walls

of the fluid container), they transfermomentum to the object and thus(second Newton’s law) exert force on theobject.

• Only perpendicular component of themolecule momentum is transferred to thewall of an object, therefore the force exerted by molecules is perpendicular tothe wall

dt

dp

F F – force, p - momentum

Incident molecule

Reflected molecule

force

F

pin

pinx

piny

pr

pr x

pr y


14/66

Pressure

• In order to quantify the force ofcompression exerted by a fluid, we use

pressure

• The pressure P of the fluid at the level towhich an object has been submerged is theratio of the force to the area

• For example, if F is the force acting on thecube top and A is the area of the side ofthe cube, then pressure P exerted by water

on the top of the cube is F/A .

F P A

Cube in water


15/66

Pressure, cont

• Pressure is more convenient than force because force F isproportional to area A (more fluid molecules bouncing off theobject) and thus pressure P = F/A is independent of area!

• Pressure is a scalar (not a vector!) quantity – Because the force by a fluid on an object is always exerted in one

direction – perpendicular to the surface of the object

– Since the direction of the force is pre-determined, there is no need tointroduce vector for pressure

• If the pressure varies over an area, evaluate dF on a surfaceof area dA as dF = P dA

• Unit of pressure is Pascal (Pa) 21 Pa 1 N/m


16/66

Units of Pressure

• SI: 1 Pascal = 1N / 1 m2

• Metrology: 1 bar = 105 Pascal

• 1 atmosphere = 1.013 bar

• 1 Torr = 1/760 atmosphere = 133.3 Pascal

• 1 psi (pounds per square inch) = 6,890 Pa

– 1 atm = 14.7 psi


17/66

High and Low Pressure

• Largest pressure achieved in thelaboratory:

– Diamond anvil cell ~ 200 Giga Pa

– Pressure in the center of the Sun ~3.5*1011 atm

• Degrees of vacuum:

– Atmospheric pressure = 101.3 kilo Pa

– Vacuum Cleaner ~ 80 kilo Pa

– Mechanical Vacuum pump ~ 1 Pa

– Near Earth outer space ~ 100 micro Pa

– Ultra high vacuum chamber ~ 1 nano Pa

– Pressure on the Moon ~ 1 nano Pa

– Interstellar space ~ 1 femto Pa

Ultra highvacuumchamber

Diamondanvil cell


18/66

Measuring Pressure

• The spring is calibrated by aknown force

• The compression length ofthe spring is proportional tothe force x = F/k

• The force due to the fluidpushing on the top of thepiston and compressing thespring

• The force the fluid exerts on

the piston is measured


19/66

Variation of Pressure with Depth

• Fluids exert pressure that varies with depth (or height)

– Air pressure is low at large heights, so cabins of aircraft must be pressurized – Water pressure is very high at large depths, so submarines must be

pressurized

– Pressure increases with depth because the weight of fluid above an object ina fluid pushes on this object

• If a fluid in a container is at rest, all parts of the fluid are in staticequilibrium, which means that the total force on each part of the fluidmust be zero

• We will use the fact that the total force exerted on each part of a fluid iszero to derive the expression for pressure as a function of depth: P(h)


20/66

Pressure and Depth

• Consider water in a beaker

• Examine the darker region - the liquid within an

imaginary cylinder inside the beaker – The cylinder has a cross-sectional area A

– Its cyclinder extends from depth d to d + h belowthe surface

• Water outside of the cylinder exerts compressive

forces on the cylinder: – P 0 is pressure at the top of the cylinder at depth d .

Force on the cylinder due to this pressure is -P 0A

– P is pressure at the bottom of the cylinder at depthh+d . Force on the cylinder due to this pressure is P A

– Forces due to pressure on the sides of the cylinder

cancel each other because water does not movelaterally


21/66

Pressure and Depth, cont

• Three forces act on the cylinder of water in the verticaldirection: forces due to water pressure, -P 0A, P A, as well

as the force of gravity• Let us calculate the force of gravity acting on the cylinder

• The liquid has a density r

– Assume the density of water is the same throughout thecylinder

– This means this is an incompressible liquid• The three vertical forces are

– Downward force on the top, P 0A

– Upward force on the bottom, PA

– Gravity force acting downward, M g

• The mass can be found from the density: M V Ah r r


22/66

Pressure and Depth, final

• Since the net force must be zero:

– This chooses upward as positive

• Solving for the pressure gives

P = P 0 + r g h

ˆ ˆ ˆo PA P A M F j j gj = 0


23/66

Example: Force on a Diver’s Eardrum

• What is the force on your eardrum due to the water above

when you are at the bottom of a 5 meter deep pool?• The air inside the middle ear is normally at atmospheric

pressure

• We estimate the area of the eardrum as 1 cm2

gh P P bottom r 0 A P P F bottom 0

=103 kg/m3, g = 9.8 m/s2, h = 5 m F=?


24/66

Pressure and Temperature of Atmosphere versus Elevation

- Atmospheric pressure does not vary linearly with height over large heightvariations because the density of air is not constant (decreases with height).

- Pressure always decreases with increasing height.- Temperature of air is a complex function of height.


25/66

Density

• Density is defined as mass per unit volume: r = M / V

• Gases have much lower densities than liquids

• Liquids have densities similar to solids• r liquid >> r gas because the average inter-molecule

spacing in a gas is much greater than that in a liquid

P = P 0 + r g h


26/66

Density Table


27/66

Atmospheric Pressure

• If a liquid is open to the atmosphere, then pressure at the

surface of the liquid, P 0, is equal to the atmospheric pressure

• P 0 = 1.00 atm = 1.013 x 105 Pa

• The pressure at a given point in a fluid depends on thevalue of pressure at its surface, P 0 , and on depth, h .

• An increase in pressure at the surface must be transmitted

to every other point in the fluid


28/66

Estimate the height of the Earth atmosphere

• Estimate = make approximations

– Assume that the density of air does not depend on height

• The air pressure at the Earth surface, P0 =105 Pa

• The air density is r = 1.29 kg/m3

• h = 105 Pa / (9.8 m/s2 * 1.29 kg/m3) = 7910 m

• Since this is an order-of-magnitude estimate:

• h 105 Pa / (10 m/s2 * 1 kg/m3) = 104 m

Earth surface

A i r c o l u m n a b o v e E a r t h

h

A

P = r g h h =P/( r g )


29/66

Pascal’s Law

• Named after French scientist Blaise Pascal

• A change in the pressu re appl ied to a

f lu id is transm it ted und imin ished to

every point of the f lu id and to the walls

of the con tainer

Blaise Pascal

(1623 – 1662)P 0

h

P 0 + r gh

P 0 +P

P 0 + r gh+ P

F

P = F /A


30/66

Pressure Measurements: Barometer• Invented by Torricelli

• A long closed tube is submerged in adish full of mercury, flipped closed endup and lifted

• The column of mercury is inequilibrium at a certain height

– The closed end is nearly at vacuum

• Measures atmospheric pressure as

• One 1 atm = 0.760 m (of Hg)

0 Hg P gh r Evangelista Torricelli

(1608-1647 )

mercury

P 0 =101.3 kPa, g = 9.8 m/s2

, rHg =13.6 103

kg/m3

76.00

g

P

h Hg r m

http://www.youtube.com/watch?v=-NIdoZELvHk


31/66

Pascal’s Law, Hydraulic Press

• Diagram of a hydraulic press

• A large lifting force, F2, canbe applied by means of asmall force F1

1 2

1 2

1 2

P P

F F

A A

1

212

A

A F F - Force amplification can be very large.

- Since nothing is free, what is the tradeoff

for the force amplification?


32/66

Pascal’s Law, Hydraulic Press cont.

• Since the total volume of liquid isconserved, the volume of liquid pusheddown on the left is equal to the volume ofliquid pushed up on the right:

• Combining the equations:

1 1 2 2 A x A x

1 1 2 2

F x F x

2

1

1

2

x

x

A

A

1

2

1

2

F F

A A We have shown that

We have no gain in work (energy isconserved!). Gain in force is due to thefact that we have to push the liquid

down farther (lever action):21 W W

21 x x


33/66

Pressure Measurements: Manometer

• A device for measuring thepressure of a gas in a vessel

• One end of the U-shaped tube isopen to the atmosphere

• The other end is connected to thepressure to be measured

• Points A and B are at the sameheight, hence pressure at points Aand B is the same

• Pressure at B is P 0+ ρgh

• Therefore P = P 0+ ρgh

liquid

gas


34/66

Absolute vs. Gauge Pressure

• P = P 0 + r g h

• P is the absolutepressure

• The gauge pressure is

P – P 0 – P – P 0 = r g h

– This is what you measure inyour car tires

liquid

gas


35/66

Example: Two Immiscible fluids in a U-tube

Mercury is poured into a U-tube. Theleft arm of the tube has cross-sectionalarea A1 of 10.0 cm

2, and the right armhas a cross-sectional area A2 of 5.00 cm

2. One hundred grams ofwater are then poured into the right arm.

-Determine the length of the watercolumn in the right arm of the U-tube.

- Given that the density of mercury is13.6 g/cm3, what distance h does themercury rise in the left arm?


36/66

(b) Pressures at points A and B in the picture above are equal to each other because thesepoints are at the same level in the same liquid (mercury), and there is the same type of liquidbelow these points.

Lhw

B A

Pressure at point A is: P = P0 + rw g L because this pressure comes from the column ofair (P0) and the column of water (rw g L) above the point A

Pressure at point B is: P = P0 + rm g (h+(L-hw)) because this pressure comes from the

column of air (P0) and the column of mercury (rm g (h+L-hw)) above the point B

Solution:

(a) Using the definition of density:

mw = rw V = rw A2 L

L = mw/(rw A2) == 0.1 kg/(103 kg/m3 *510-4 m2) = 0.2 m


37/66

Pressure values at A and B are equal, thus: P0 + rw g L = P0 + rm g (h+L-hw)

rw g L = rm g (h+L-hw) (rw /rm) L = h+L-hw h = hw-(1-rw /rm) L (1)

Finally, the volume of mercury above the initial mercury level (black dashed line) is the sameas the volume of water below the initial mercury level (so that the total volume of mercury isconserved):

A1 h = A2 (L-hw) hw= L- (A1 /A2)h (2)

Substituting (2) into (1) gives: h = L- (A1 /A2)h -(1-rw /rm) L

Solving this equation for h, we obtain: h +(A1 /A2)h = rw /rm L h = (L rw)/(rm(1+A1 /A2))

0.2 m * 103 kg/m3

13.6103

kg/m3

*(1+ 10/5)

h = = 0.0049 m = 4.9 mm

B t F


38/66

Buoyant Force

• Let us revisit the example of an imaginarycylinder of water inside of a beaker full of

water• The net force from the water surrounding the

cylinder exerted on the cylinder is (P-P 0 )A

• This force is counterbalanced by the gravityforce acting on the water in the cylinder:

M water g• Since the gravity force points down, the force

acting on the cylinder from the surroundingwater must point up (the sum of the twoforces must be zero in equilibrium)

• The buoyant force, FB, is the upward force

exerted by a fluid on any immersed object

A hi d ’ P i i l


39/66

Archimedes’s Principle

• Since the buoyant force exerted on the cylinder of water is

counterbalanced by the gravity force (also called weight)M water g , the magnitude of the buoyant force is

F B = M water g

• This consideration forms the basis of the Archimedes’sPrinciple:

The magnitude of the buoyant force always equals the

weight of the fluid displaced by the object

A hi d ’ P i i l t


40/66

Archimedes’s Principle, cont.

Archimedes’s Principle does not

depend on the material of theobject experiencing the buoyantforce

– The object’s composition is not afactor since the buoyant force is

exerted by the fluid – Every object of a given shape at a

given depth in a fluid experiencesthe same buoyant force = M f luid g ,where M f luid is the fluid mass of

the same volume as the object’svolume

F B

M waterg

F B

M steelg

Water in water:

Steel in water:

F B = M water g

F B = M water g > M steel g <

A hi d ' P i i l T t ll S b d Obj t


41/66

Archimedes's Principle: Totally Submerged Object

• An object is totally submerged in afluid of density r f luid

• The upward buoyant force is:

F B = r f luid g V f luid = r fluid g V object

• The downward gravitaty force is:

F G = mg = robject g V object

• The net force is:

F B - F G = r f luid g V object - robject g V object

F B - F G = ( r f luid - robject ) g V object

A hi d ’ P i i l T t ll S b d Obj t t


42/66

Archimedes’s Principle: Totally Submerged Object, cont

• If the density of the object is less than

the density of the fluid, theunsupported object acceleratesupward

• If the density of the object is morethan the density of the fluid, theunsupported object sinks

• The mot ion of an object in a f lu id isdeterm ined by the densit ies of the

f lu id and the object

Wood in water Steel in water

r

wood rwater

A hi d ’ P i i l C E l


43/66

Archimedes’s Principle, Crown Example

• Archimedes was (supposedly)

asked, “Is the crown made of puregold?”

• It was known that:

– Crown’s weight in air = 7.84 N

– Weight in water (submerged) = 6.84 N

• How did Archimedes solve this

problem?

?

A hi d ’ P i i l C E l


44/66

Archimedes’s Principle, Crown Example

- Gold (Au) was the heaviest substance known at thetime (largest density)

- Archimedes argued:- if rcrown=r Au then the crown must be golden- if rcrown


45/66

Archimedes’s Principle, Crown Example, cont.

• Let T 1 be the weight of the crown in air and T 2 bethe weight of the crown in water

T 1 = F g = M crown g = rcrown V crown g

T 2 = F g -F B = M crown g – M water g

= rcrown V crown g - rwater V crown g

= ( rcrown - rwater )V crown g

• T 2 /T 1= 1- ( rwater / rcrown )= 6.84 (N)/7.84(N)

• rwater = 10 3 (kg/m 3 ); solving for rcrown , we

obtain: rcrown = 7.84 10 3 (kg/m 3 )

Archimedes’s Principle: Floating Object


46/66

Archimedes’s Principle: Floating Object

• Consider an object with

robject


47/66

Archimedes s Principle: Floating Object, cont

The fraction of the volume

of a floating object that isbelow the fluid surface isequal to the ratio of thedensity of the object to thatof the fluid

obj fluid

fluid obj

V

V

r

r

Archimedes’s Principle Iceberg Example


48/66

Archimedes s Principle, Iceberg Example

• An iceberg that appeared to be “small”

sank “Titanic” • What fraction of a typical iceberg is

below water?

• The iceberg is only partially submerged

and so Vfluid / Vobject = object / fluid

applies• The fraction below the water will be the

ratio of the volumes f :

89.0/1030

/917

3

3

mkg

mkg

V

V f

water

ice

ice

water

r

r

Archimedes


49/66

Archimedes

287-212 BC

The father of integral calculus

and mathematical physics

- Discovered his principle while taking a bath, then

took to the streets naked crying "Eureka!" ("I havefound it!")

- His most famous word: "Give me a lever long enoughand a place to stand, and I will move the world."

- Was killed by a Roman soldier during a siege ofSyracuse. At the moment he was working on ageometry problem drawing on sand and his last wordsto the soldier were: "Don't disturb my circles"

Types of Fluid Flow


50/66

Types of Fluid Flow

• So far we studied fluids at rest, now let us concentrate onmoving (flowing) fluids (fluid dynamics)

• Laminar flow – Steady flow

– Each “particle” of the fluid follows a smooth path (a “particle” consistsof millions of atoms)

– The paths of the different particles never cross each other

– The path taken by the particles is called a streamline

• Turbulent flow – An irregular flow characterized by small whirlpool like regions.

– Turbulent flow occurs when the particles go above some criticalspeed

Types of Fluid Flow Turbulent


51/66

Types of Fluid Flow – Turbulent

• There is a $1,000,000 prize forexplaining turbulent flow

• Turbulent flow of air createsirregular, quasi-chaotic spatialdistribution of density and velocity

of air. This can significantly affectaircraft flight.

• We will only study laminar flow offluids

Image of turbulent flow

Viscosity

http://www.youtube.com/watch?v=XOLl2KeDiOg


52/66

Viscosity

• Characterizes the degree of internal friction in the fluid

• This internal friction, viscous force, is associated with theresistance that two adjacent layers of fluid have due tomoving relative to each other

• Viscosity causes part of the kinetic energy of a fluid to beconverted to internal energy (heat)

• Viscous force is similar to the friction force exerted on anobject sliding on rough surface

• Viscosity is also called “internal friction”

Ideal Fluid Flow


53/66

Ideal Fluid Flow

• We will study a simplified model of a fluid flow. There arefour simplifying assumptions made to the complex flow offluids to make the analysis easier

(1) The f lu id is nonvisco us – internal friction is neglected

(2) The flow is steady – the velocity of each point remains

constant(3) The f lu id is incom pressib le – the density remains

constant

(4) The flow is lam inar – no eddies (vortices) are formed

Streamlines


54/66

Streamlines

• The path the particle takes in

steady flow is a streamline

• The velocity of the particle istangent to the streamline

• A set of streamlines is called atube of flow

• Streamlines do not cross eachother

Equation of Continuity


55/66

Equation of Continuity

• Consider a fluid moving through apipe of nonuniform size (diameter)

• The particles move along streamlinesin steady flow

• The volume of fluid that crosses area

A1 in some time interval is the sameas the volume that crosses area A2 inthat same time interval

• This is because (i) fluid isincompressible and (ii) does not

accumulate in the pipe

Equation of Continuity cont


56/66

Equation of Continuity, cont

• Volume of fluid that flows through area A1 in a

time interval t is:

V 1 =A 1 x 1 =A1 v 1 t

• Volume of fluid that flows through area A1 in a

time interval t is:

V 2 =A 2 x 2 =A2 v 2 t

• Since the fluid is incompressible, V 1 =V 2

• Therefore: A1 v 1 = A2 v 2

– This is called the equation of continuity

– The product of the area and the fluid speed atall points along a pipe is constant for an

incompressible fluid

Equation of Continuity Implications

http://www.youtube.com/watch?v=k8IRxT1sM6s&feature=related


57/66

Equation of Continuity, Implications

• The speed is high where the tube is constricted (small A)

• The speed is low where the tube is wide (large A)

• The product, Av , is called the volume flux or the flow rate

• Av = constant is equivalent to saying the volume thatenters one end of the tube in a given time interval equalsthe volume leaving the other end in the same time

Bernoulli’s Equation


58/66

Bernoulli s Equation

• As a fluid moves through a region

where its speed and/or elevationabove the Earth’s surface changes,

the pressure in the fluid varies withthese changes

• The relationship between fluidspeed, pressure and elevation

was first derived by Daniel Bernoulli 1700 - 1782

Bernoulli’s Equation 2


59/66

Bernoulli s Equation, 2

• Consider a segment of a fluid in a pipefrom point 1 to point 2

• In a time interval t the left boundary ofthe segment moves through a distancex 1 while the right boundary of the

segment moves through a distance x 2

• Consider forces exerted on the right and

the left boundaries of the segment asthe segment moves in the pipe:

• Force exerted on the left boundary:F 1 = P 1 A1

• Force exerted on the right boundary:

F 2 = -P 2 A2



60/66


• Work done by force F 1 is W 1 = P 1A1 x 1

• Work done by force F 2 is W 2 = - P 2A2 x 2• Since the fluid is incompressible,

A1 x 1 = A2 x 2 = V

• The net work W done by the two forces

on the segment isW = W 1 +W 2 = (P 1 – P 2 ) V

• Part of this work goes into changing thekinetic energy of the segment and partgoes into changing the gravitational

potential energy of the segment



61/66


– There is no change in the kinetic energy of thegrey portion of the segment (steady-state

streamline flow) – The net effect of the flow in a time interval t is

“displacement” of a the volume of fluid V from

the lower part of the pipe to the upper part

– Initial kinetic energy of the lower portion of

volume V is ½ m v 1 2

– Final kinetic energy of the upper portion of

volume V is ½ m v 2 2

– The net change of kinetic energy of thesegment:

K = ½ m v 2 2 - ½ m v 1 2



62/66


• There is no change in potential energyof the fluid in the grey middle portion of

the section• The change in gravitational potential

energy due to displacement of volumeV from the lower part to the upper partof the pipe is:

U = m g y 2 –

m g y 1

• From the conservation of energy, workdone on the segment of the fluid goesinto changing kinetic energy andpotential energy of the segment:

W = K+ U

Bernoulli’s Equation, 6


63/66


• Substituting the expressions for W , K and U :

(P 1 – P 2 )V =½ m v 2 2 - ½ m v 1

2 + m g y 2 – m g y 1

• Rearranging terms, dividing by V andwriting m/V as density r:

P 1 + ½ r v 1 2 + r g y 1 = P 2 + ½ r v22 + r g y 2

• This is Bernoulli’s equation and is often

expressed as

P + ½ r v 2 + r g y = constant

Bernoulli’s Equation, Final


64/66

Bernoulli s Equation, Final

• Consider a flow in a horizontal pipe. In this case Bernoulli’s equationsimplifies to P + ½ r v 2 = constant . This expression tells us that pressurein a fluid decreases as its speed increases. This means that pressure innarrow sections of pipes is low.

• When the fluid is at rest, Bernoulli’s equation simplifies to P + r g y =constant

• This is equivalent to P 1 –

P 2 = r g h which is consistent with the pressurevariation with depth we found earlier

• The general behavior of pressure with speed is true even for gases(compressible fluids) – As the speed increases, the pressure decreases although quantitatively

Bernoulli’s equation does not apply

Airplane Wing


65/66

Airplane Wing

• Streamline flow around a moving airplanewing

• Li f t is the upward force on the wing fromthe air

• Drag is the resistance

• The lift depends on the speed of theairplane, the area of the wing, its curvature,and the angle between the wing and thehorizontal

Example: Pitot tube


66/66

Example: Pitot tube

A Pitot tube can be used to

determine the velocity of air flowby measuring the differencebetween the total pressure andthe static pressure.

If the fluid in the tube is mercury,density rHg = 13 600 kg/m3, and

h = 5.00 cm, find the speed of airflow.

12 Fluid Mechanics

Documents

Transcript of 12 Fluid Mechanics