Physics140
Fluids
Chapter9
Statesofma:er
Solid
Gas
Liquid
Plasma
Physics140,Prof.M.Nikolic
2
Fluids
Afluid–asubstancethatcanflow
(incontrasttoasolid)
ü  Air–gas
ü  Water–liquid
Fluidscomforttotheboundariesofanycontainerinwhichwe
putthem,anddonotmaintainafixedshape.
Physics140,Prof.M.Nikolic
3
Pressure
PressurearisesfromthecollisionsbetweentheparLclesofafluidwith
anotherobject(containerwallsforexample).
Pressure–forceperunitarea:
F
P=
A
SIunits:N/m2=1Pascal[Pa]
NotaLonforpressure:P(again)!
Otherunits:1atmosphere(atm)=1.013x105Pa=760mmHg
Physics140,Prof.M.Nikolic
4
Exercise:Pressure
Whatisthepressureinatmospheresfroma60-kgpersonstandingontheground
whosetwofeetcoveranareaof500cm2.
Whatisgiven:
m=60kg
AreaA=500cm2
F
P=
A
−4
2
10
m
2
Firstweconvertfromunitsofcm2toSIunitsofm2: A = 500 cm ×
=
0
.
05
m
1 cm 2
Theforceappliedistheforceofgravity(weight)ofaperson:F=mg
2
mg 60kg ⋅ 9.8m/s2
P=
=
= 11760 Pa
2
A
0.05 m
P = 11760 Pa ×
Physics140,Prof.M.Nikolic
1 atm
= 0.12 atm
5
1.013 ×10 Pa
5
ConceptualquesLon–Pressure
Q1
Thepressureexertedonthegroundbyamanisgreatestwhen:
A. 
B. 
C. 
D. 
Hestandswithbothfeetflatontheground
Hestandsflatononefoot
Hestandsonthetoesofonefoot
Heliesdownontheground
Nowimaginethe
pressureonherfeet.
F
P=
A
Forceofgravitystaysconstant
ThesmallertheareaèThebiggerthepressure
Physics140,Prof.M.Nikolic
6
Atmosphericpressure
•  Theairaboveushasweightandwearebeingforcedtobearthat
weight.
•  ThepressureoftheEarth’satmospherevarieswithalLtude.
•  Atmosphericpressureatseelevelisabout
1.013x105N/m2=10340kgpereverysquaremeter
Physics140,Prof.M.Nikolic
ThecompensaLonforatmosphericpressurecomes
from,amongotherthings,ourbloodpressure.
7
Pascal’sprinciple
Pascal’sPrincipleisthekeytounderstandingwhyhydrauliclies
and,forthatma:er,hydraulicbrakesworkthewaytheydo.
Physics140,Prof.M.Nikolic
8
Pascal’sprinciple
Achangeinpressureatanypointinconfinedfluidistransmi:ed
everywherethroughoutthefluid.
Or:Thefluidpressuremustbethesameeverywhereinaweightlesssta@cfluid.
Theappliedforceis
transmi:edtothepistonof
cross-secLonalareaA2here.
Physics140,Prof.M.Nikolic
ApplyaforceF1here
toapistonofcrosssecLonalareaA1.
ΔP1 = ΔP2
F1 F2
=
A1 A2
9
Pascal’sprinciple
Volumeofthedisplacedfluidhasto
staythesame
V1 = V2
A1d1 = A2 d 2
F1d1 = F2d2
Theworkdonebybothforces
hastostaythesame.
Whichmeansthatthesamefactorbywhichtheforcegoesupisthefactorby
whichthedistanceoftheinputpistongoesdown.
Physics140,Prof.M.Nikolic
10
Q2
ConceptualquesLon–Lieingacar
Acontainerisfilledwithoilandfi:edonbothendswithpistons.Theareaofthelee
pistonis10mm2;thatoftherightpiston10,000mm2.Whatforcemustbeexerted
ontheleepistontokeepthe10,000-Ncarontherightatthesameheight?
Fcar F
=
Acar A
10,000 N
F
=
10,000 mm2 10 mm2
F = 10 N
A. 
B. 
C. 
D. 
10N
1000N
10000N
106N
Physics140,Prof.M.Nikolic
11
Exercise:Hydrauliclie
Ahydrauliclieislieingacarthatweighs12kN.TheareaofthepistonsupporLngthe
carisA,theareaoftheotherpistonisa,andtheraLoA/a=100.
a)Howfarmustthesmallpistonbepusheddowntoraisethecaradistanceof1cm?
Whatisgiven:
Fcar=12000N
A/a=100
dcar=1cm
Fcar Fsp
=
A
a
Fcar A
= = 100
Fsp a
Fsp =
Fcar 12, 000 N
=
= 120 N
100
100
Butwealsoknowthattheworkdonebybothforceshastobethesame.
Fsp d sp = Fcar d car
dsp =
Fcar dcar
Fsp
dsp =
12, 000 N ⋅1 cm =
= 100 ⋅1 cm =1 m
120 N
b)Whatistheworkdone?
W = Fsp d sp = Fcar d car
Physics140,Prof.M.Nikolic
W = 12000 N ⋅ 0.01 m = 120 J
12
Density
Density–massperunitvolume
m
ρ=
V
NotaLon:Greekle:errho-ρ
m–massoftheobject
V–itsvolume
SIunits:kg/m3
Example:Whatisthemassoftheairintheclassroomwithdimensions12m
and8mandaheightof4m.Densityofair–1.29kg/m3
Volumeoftherectangularcube: V = h ⋅ l ⋅ w = 12m ⋅ 8m ⋅ 4m = 384m 3
m = ρV = (1.29kg / m 3 )⋅ (12m ⋅ 8m ⋅ 4m) = 495kg
Physics140,Prof.M.Nikolic
13
Gravityandfluidpressure
ü 
ü 
ü 
ü 
AfluidexertsapressureinalldirecLons.
AtanypointatrestthepressureisthesameinalldirecLons
Thepressureincreaseswithdepth
Theforceduetofluidpressurealwaysactsperpendiculartoany
surfaceitisincontactwith
Physics140,Prof.M.Nikolic
14
Gravity’seffectonfluidpressure
Imagineacylinderoffluid
WecandrawanFBDforthefluidcylinder
y
3forcesareexertedonthefluidcylinder
P1A
x
mg
P2A
ApplyNewton’s2ndLaw:
∑F
y
=0
Physics140,Prof.M.Nikolic
1.  Weightofthecylinderèmg
2.  Forceduetopressurefromthe
abovemoleculesèF1=P1A
3.  Forceduetopressurefromthe
moleculesbellowèF2=P2A
P2 = P1 + ρgd
P2 A − P1 A − mg = 0
Trueforanyamountoffluid.
m = ρV = ρAd
Notethatpoint2isdistancedbellow
point1.
15
Gravity’seffectonfluidpressure
Ifthetopofthefluidcolumnisplacedatthesurfaceofthe
fluid,thenP1=Patmifthecontainerisopen.
P = Patm + ρgd
Oh,sowhenIdiveIfeelboth105Papluswaterpressure.
Iftheobjectrisesupintheair,P2willbeaboveP1andthepressurewilldecrease
èP2–P1=-ρgd
Physics140,Prof.M.Nikolic
16
Exercise:Pressureinthelake
Atthesurfaceofafreshwaterlake,thepressureis105kPa.
(a)Whatisthepressureincreaseingoing35.0mbelowthesurface?
Whatisgiven:
P1=105kPa=105x103Pa
d=35m
P2 = P1 + ρgd
Theproblemisaskingfor
thechangeinpressure
ΔP = P2 − P1 = ρgd
Checkthetextbooktable9.1tofindthedensityoflakewater:
→ρ=1000kg/m3
ΔP = 1000 kg/m3 ⋅ 9.8 m/s2 ⋅ 35 m = 343 ×103 Pa
(b)Whatisthepressuredecreaseingoing35.0mabovethesurface?Assumeair
densityof1.2kg/m3?
Inthiscase,thepressureisdecreasingandwecanwrite:
ΔP = P2 − P1 = −ρgd
ΔP = −1.2 kg/m 3 ⋅ 9.8 m/s2 ⋅ 35 m = −411.6 Pa
Physics140,Prof.M.Nikolic
17
Measuringpressure-Manometer
U-shapedtubethatisparLallyfilledwithliquid
Bothendsofthetubeare
opentotheatmosphere
Physics140,Prof.M.Nikolic
Containerofgasisconnected
tooneendofthetube
Thepressuredifferencebetweenthegas
andtheatmospherewillpushthefluid
18
Measuringpressure-Manometer
Wecanseefromthefigure:
1.  PointCisatatmosphericpressure
Pc = Patm
2.  PointsBandB’areatthesamepressure
→sameheight
PB = PB'
ThepressureatpointBisthepressureofthegas.
PB = PB ' = PC + ρgd
Pgauge = PB − Patm = ρgd
Gaugepressure
Physics140,Prof.M.Nikolic
19
Measuringpressure-Barometer
Mercurybarometer:
•  Thepressureatthetopofthecolumninthetube
iszero.
•  Thepressureatthebo:omofthecolumnis
equaltoatmosphericpressure
Theatmospherepushesonthecontainerofmercury
whichforcesmercuryuptheclosed,invertedtube.
Pgauge = 0 − Patm = ρgd
Thedistancediscalledthebarometricpressure.
Unitsofpressure:1atm=1.013x105Pa=760mmHg.
Physics140,Prof.M.Nikolic
20
ApplicaLon–Asphygmomanometer
Youmustlovethename!
Thesimplestversion:aclosedbag(thecuff)withamercurymanometer
Physics140,Prof.M.Nikolic
Systolicpressure–
whentheheartbeats
Diastolicpressure–
betweentheheartbeats
21
Exercise:UsinganIVfluid
AnIVisconnectedtoapaLent’svein.Thebloodpressureintheveinhasagauge
pressureof12mmofmercury.AtleasthowfarabovetheveinmusttheIVbagbe
placedinorderforfluidtoflowintothevein?AssumethatthedensityoftheIVfluid
isthesameasblood(ρIV=ρblood=1060kg/m3).
dIV
IVfluid
InorderfortheIVfluidtoflowintothevein,thegaugepressureofthe
fluidhastobeatleastequaltothegaugepressureoftheblood
vein
Pgauge, IV ≥ Pgauge, blood
Trustme,youwouldnotlikeit
theotherwayaround.
Thebloodgaugepressureis12mmHgèThesearenotstandardunitsofpressure,weneed
toconverttoPascals:
" 101,300 Pa %
Pgauge,blood = 12 mm Hg × $
' = 1600 Pa
# 760 mm Hg &
TheIVgaugepressure
Pgauge,blood = ρ IV gdIV
Physics140,Prof.M.Nikolic
Pgauge,IV = ρ IV gdIV
dIV =
1600 Pa
= 0.154 m
3
2
1060 kg/m ⋅ 9.8 m/s
22
Buoyantforce
Whenanobjectisimmersedinafluid,thepressureonthelowersurfaceisalways
higherthanthepressureontheuppersurfaceèP2>P1.
AnFBDfortheobject
y
Thetotalforceontheblockduetothe
fluidiscalledthebuoyantforce.
P1A
FB = P2 A − P1 A = (P2 − P1 )A
mg
x
P2A
FB = ρgdA = ρgV
ρ–thedensityofthefluid
V=d*A–thevolumeoftheblock
submergedinfluid
Physics140,Prof.M.Nikolic
23
Archimedes’principleofbuoyancy
Archimedes’principle:Afluidexertsanupwardforceonasubmerged
objectequalinmagnitudetotheweightofthevolumeoffluid
displacedbytheobject
Fb = mfluid g = ρfluid gVfluid
mf–massofthefluidthatisdisplacedby
thebody
Vfluid = Vsubmerged
Fb = ρfluid gVsubmerged
Volumeofthedisplacedfluid=volumeofthesubmergedobject.
Physics140,Prof.M.Nikolic
24
FloaLng–Equilibrium
Whengravita@onalforce=buoyantforce→equilibrium(floa@ng)
FB = mg
ρfluid gVsubmerged = ρobject gVobject
Volumeofthedisplacedfluid=volumeofthesubmergedobject.
Physics140,Prof.M.Nikolic
25
Exercise:Buoyantforce
Aflat-bo:omedbargeloadedwithcoalhasamassof3.0x105kg.Thebargeis20.0
mlongand10.0mwide.Itfloatsinfreshwater(ρ=1000kg/m3).
a)Whatisthebuoyantforceonthebarge?
FB
Whenthebargefloats→equilibrium
(buoyantforce=weightofbarge)
FB = mbarge g = ρbarge gVbarge
mbargeg
FB = 3 ×105 kg ⋅ 9.8 m/s2 = 2.94 ×106 N
b)Whatisthedepthofthebargebelowthewaterline?
FB = ρ water gVsubmerged
Vsubmerged
WecanusethisequaLontofindthevolume
ofthesubmergedpartofthebarge
2.94 ×10 6 N
3
=
=
=
300
m
ρwater g 1000 kg/m 3 ⋅ 9.8 m/s2
FB
Volumeofthesubmergedbarge
(rectangularprismorcube?)
Physics140,Prof.M.Nikolic
Vsubmerged = l × w × h = 20 m ×10 m × h
26
Exercise:Buoyantforce
Aflat-bo:omedbargeloadedwithcoalhasamassof3.0x105kg.Thebargeis20.0m
longand10.0mwide.
b)Whatisthedepthofthebargebelowthewaterline?contd.
3
300 m = 20 m ×10 m × h
300 m 3
h=
= 1.5 m
2
200 m
c)Whatwouldbethebuoyantforceonthebargethatiscompletelysubmergedin
water?Densityofwoodis500kg/m3.
FB = ρ water gVsubmerged
Butnowthewholebargeisunderwater
→Vsubmerged=Vbarge
Volumeofthewholebarge Vbarge =
mb arge
ρb arge
3 ×105 kg
3
=
=
600
m
500 kg/m 3
FB = 1000 kg/m 3 ⋅ 9.8 m/s2 ⋅ 600 m3 = 5.88 ×10 6 N
Physics140,Prof.M.Nikolic
27
Fluiddynamics
Orwhathappenswhenfluidsstarttomove
Ifflowoffluidissmooth
èsteadyorlaminarflow
Aboveacertainspeed,the
flowbecomesturbulentflow
•  Realfluids–feelexternalforcefromthesurface–viscousforce
•  Idealfluids–incompressible,undergoeslaminarflow,noviscosity
Physics140,Prof.M.Nikolic
28
TheconLnuityequaLon–conservaLonofmass
ThemassflowrateisthemassthatpassesagivenpointperunitLme.
Δm ρAΔx
=
= ρAv
Δt
Δt
ΔV
Thevolumeflowrate:
= Av
Δt
Physics140,Prof.M.Nikolic
29
TheconLnuityequaLon–conservaLonofmass
Themassflowratesatanytwopointsmustbeequal,aslongasno
fluidisbeingaddedortakenaway.
Δm1 Δm2
=
Δt
Δt
ρ1 A1v1 = ρ2 A2 v2
Ifthefluidisincompressible→ρ1=ρ2: A1v1 = A2 v2
Physics140,Prof.M.Nikolic
30
ConceptualquesLon–Bloodflow
Q4
A blood platelet drifts along with the flow of blood through an
artery that is partially blocked by deposits. As the platelet moves
from the narrow region to the wider region, its speed
A1v1 = A2 v2
A.  Increases
B.  Remainsthesame
C.  Decreases
CrosssecLonalareaincreasesèspeeddecreases
Exercise:Fluiddynamics
A1v1 = A2 v2
Awaterhosehasadiameterof3.30cm.Waterflowsthroughitataspeedof3.22m/s.
Youuseaspraynozzlewithaholethathasadiameterof0.981cm.Whatisthespeed
ofthewaterthatcomesoutthenozzle?
Whatisgiven:
d1=3.3cm=0.033m
v1=3.22m/s
d2=0.981cm=0.00981m
A1
v2 = v1
A2
A1v1 = A2 v2
2
2
! d1 $
! 0.033 $
−4
2
A
=
π
=
3.14
⋅
#
&
#
& = 8.54 ×10 m
Theareaofthehose: 1
"2%
" 2 %
2
2
!d $
! 0.00981 $
−4
2
Theareaofthenozzle: A2 = π # 2 & = 3.14 ⋅ #
& = 0.755 ×10 m
"2%
" 2 %
8.54 ×10−4 m 2
v2 =
3.22m/s = 36.4 m/s
−4
2
0.755 ×10 m
Asexpected,thespeedoutofthenozzleislarger.
Physics140,Prof.M.Nikolic
32
Bernoulli’sequaLon
ConservaLonofenergyforfluids
ConservaLonofenergyequaLon
Wtot = −ΔU + Wnc = ΔK
1 2
1 2
P1 + ρgy1 + ρv1 = P2 + ρgy 2 + ρv2
2
2
Workperunitvolume
donebythefluid
PotenLalenergy
perunitvolume
Physics140,Prof.M.Nikolic
KineLcenergy
perunitvolume
33
Exercise:Waterflow
Apipeisdesignedtoacceptwaterinwithaspeedof2m/sataheightof1.2mbelow
groundandcarryittoaheightof0.5mabovegroundwithanourlowspeedof0.4m/
satatmosphericpressure.Whatisthechangeinpressure?Densityofwateris1000
kg/m3.
Whatisgiven:
v1=2m/s
y1=-1.2m
v2=0.4m/s
y2=0.5m
1
1
P1 + ρgy1 + ρv12 = P2 + ρgy 2 + ρv22
2
2
1
1
P1 − P2 = ρgy 2 + ρv22 − ρgy1 − ρv12
2
2
P1 − P2 = 1000 kg/m 3 ⋅ 9.8 m/s2 ⋅ 0.5 m − 1000 kg/m 3 ⋅ 9.8 m/s2 ⋅ (−1.2 m)
1
1
2
2
+ 1000 kg/m 3 ⋅ (0.4 m/s) − 1000 kg/m 3 ⋅ (2 m/s)
2
2
P1 − P2 = 18.5 kPa
Physics140,Prof.M.Nikolic
34
Exercise:Nozzleflow
Anozzleisconnectedtoahorizontalhose.Thenozzleshootsoutwatermovingat25
m/s.Whatisthegaugepressureofthewaterinthehose?Neglectviscosityand
assumethattheareaofthenozzleis100smallerthantheareaofthehose.
Whatisgiven:
v1=25m/s
y1=y2=0(horizontalhose)
A2=100A1
1 2
1 2
P1 + ρgy1 + ρv1 = P2 + ρgy 2 + ρv2
2
2
ΔP = P2 − P1 =
WecanusetheconLnuityequaLontofindv2:
1 2 1 2
ρv1 − ρv2
2
2
A1v1 = A2 v2
v2 =
A1
A
v1 = 1 25 m/s = 0.25 m/s
A2
100 A1
1
1
2
2
ΔP = 1000 kg/m 3 ⋅ (25 m/s) − 1000 kg/m 3 ⋅ (0.25 m/s) = 3.12 ×10 5 Pa
2
2
NotethatyoucanalwaysfindtheforcebymulLplyingpressurewitharea.
Physics140,Prof.M.Nikolic
35
Viscousforce–fricLonforceforfluids
Allrealfluidshavesomeviscosity.
Bernoulli’sequaLononlytruefornon
viscousfluids.
Poiseuille’slawdescribestheviscous
flowrate
ΔV π ΔP 4
=
r
Δt 8 Lη
ΔP − change in pressure
L − length of the pipe
r − radius of the pipe
η − viscosity of the fluid
Veryusefulwhendealingwitharterialblockagesèsmallerradiusèbloodneeds
topumpmorepressuretomaintainthesamebloodflow.
36
Physics140,Prof.M.Nikolic
Viscousdrag
Viscousdragisadragforceafluidexertsonanobject
movingthroughthefluidandisgivenwith
Stoke’slaw: FD = 6πηrv
Terminalvelocity–velocityatwhichtheobjectstopsacceleraLng
–viscousdragbecomesequaltothesumofallotherforces
Physics140,Prof.M.Nikolic
37
Review
Pressure
F
P=
A
HydrostaLcpressure
P2 = P1 + ρgd
Pascal’sprinciple
F1 F2
=
A1 A2
Theworkdonebybothforces
hastostaythesame.
F1d1 = F2d2
Density
m
ρ=
V
Physics140,Prof.M.Nikolic
Buoyantforce
FB = ρgdA = ρgV
Fb = mfluid g = ρfluid gVfluid
Vfluid = Vsubmerged
38
Review
TheconLnuityequaLon
ρ1 A1v1 = ρ2 A2 v2
Bernoulli’sequaLon
1 2
1 2
P1 + ρgy1 + ρv1 = P2 + ρgy 2 + ρv2
2
2
Ifthedensitydoesnotchange
A1v1 = A2 v2
Poiseuille’slawdescribestheviscous
flowrate
ΔV π ΔP 4
=
r
Δt 8 Lη
Stoke’slaw: FD = 6πηrv
Physics140,Prof.M.Nikolic
39