fluids - ohio state university
TRANSCRIPT
Chapter(10
Fluids
Fluids'in'Motion
In#steady'flow the#velocity#of#the#fluid#particles#at#any#point#is#constant#as#time#passes.
Unsteady'flow exists#whenever#the##velocity#of#the#fluid#particles#at#a#point#changes#as#time#passes.
Turbulent'flow is#an#extreme#kind#of#unsteady#flow#in#which#the##velocity#of#the#fluid#particles#at#a#point#change#erratically#in#both#magnitude#and#direction.
Fluids'in'Motion
Fluid&flow&can&be&compressible or&incompressible.&&Most&liquids&are&nearly&incompressible.
Fluid&flow&can&be&viscous such&that&the&fluid&does¬&flow&readily&due&to&internalfrictional&forces&being&present&(e.g.&honey),
or&nonviscous such&that&the&fluid&flows&readily&due&to&no&internal&frictional&forces&being&present&&(e.g.&water&is&almost&nonviscous).
An&incompressible,&nonviscous fluid&is&called&an&ideal'fluid.
Fluids'in'Motion
Steady'flow'is'also'sometimes'called streamline'or'laminar flow'since'the'neighboring'layers'of'fluid'slide'by'each'other'smoothly,'i.e.'each'particle'of'the'fluid'follows'a'smooth'path'and'the'paths'do'not'cross.
Streamlines are'often'used'to'represent'the'trajectories'of'the'fluid'particles.
Fluids'in'Motion
Making'streamlines'with'dye'ina'flowing'liquid,
and'smoke,'in'a'wind'tunnel.
The$Equation$of$Continuity
The$mass$of$fluid$per$second$that$flows$through$a$tube$is$calledthe$mass$flow$rate.
The$Equation$of$Continuity
2222 vAtm !=""
1111 vAtm !=""
Δm = ρΔV = ρ A vΔtdistance!
Consider)the)steady)(laminar))flow)of)a)fluid)through)a)tube)with)a)varying)cross)sectional)area:
Since mass must be conserved as the fluid flows ⇒ Δm1
Δt=Δm2
Δt
The$Equation$of$Continuity
222111 vAvA !! =
EQUATION)OF)CONTINUITY
The)mass)flow)rate)has)the)same)value)at)every)position)along)a)tube)that)has)a)single)entry)and)a)single)exit)for)fluid)flow.
SI$Unit$of$Mass$Flow$Rate:$$kg/s
The$Equation$of$Continuity
Incompressible$fluid:$$ 2211 vAvA =
Volume$flow$rate$Q:$$ Q ≡ Av = A ΔlΔt
=AΔlΔt
=ΔVΔt
The$Equation$of$Continuity
Example:$$A"Garden"Hose
A"garden"hose"has"an"unobstructed"openingwith"a"cross"sectional"area"of"2.85x10<4m2.""It"fills"a"bucket"with"a"volume"of"8.00x10<3m3
in"30"seconds.
Find"the"speed"of"the"water"that"leaves"the"hosethrough"(a)"the"unobstructed"opening"and"(b)"an"obstructedopening"with"half"as"much"area.
The$Equation$of$Continuity
AvQ =
( ) ( ) sm936.0m102.85
s 30.0m1000.824-
33
=!
!==
"
AQv
(a)
(b) 2211 vAvA =
v2 =A1A2v1 =
A1A12( )v1 = 2( ) 0.936m s( ) =1.87m s
Bernoulli�s*Equation
The$fluid$accelerates$from$higher$to$lower$pressure$regions$from$Newton’s2nd law$due$to$the$unbalancednon9conservative$forces
According$to$the$pressure9depthrelationship,$the$pressure$is$lowerat$higher$levels,$provided$the$areaof$the$pipe$does$not$change.
Consider$the$steady$flow$of$an$incompressible$and$nonviscous$fluid$in$twosituations:
Incorporate*both*of*these*situations*into*a*single*equation!
Bernoulli�s*Equation
Wnc = E1 −E2 = 12mv1
2 +mgy1( )− 12mv2
2 +mgy2( )
Wnc = ΔWnc = ΔF( )s"# $%2→1∑
2→1∑ = ΔP( )As
ΔV!
"
#($
%)2→1∑ = ΔP( )
2→1∑"
#(
$
%)
P2−P1! "# $#
ΔV = P2 −P1( )ΔV
Combined)situation:)
Use)the)Work4Energy)theorem)to)derive)a)relationship)among)P,)v,)and)yat)two)different)points)along)the)tube)! look)at)the)fluid)elements)at)1)and)2.
Bernoulli�s*Equation
P2 −P1( )ΔV = 12mv1
2 +mgy1( )− 12mv2
2 +mgy2( )
( ) ( ) ( )2222
11
212
112 gyvgyvPP !!!! +"+="
BERNOULLI�S*EQUATION
In*steady*flow*of*a*nonviscous,*incompressible*fluid,*the*pressure,*the*fluid*speed,*and*the*elevation*at*two*points*are*related*by:
2222
121
212
11 gyvPgyvP !!!! ++=++
Applications+of Bernoulli�s+Equation
Example:++Efflux&Speed
The$tank$is$open$to$the$atmosphere$atthe$top.$$Find$and$expression$for$the$speed$of$the$liquid$leaving$the$pipe$atthe$bottom.
Applications+of Bernoulli�s+Equation
2222
121
212
11 gyvPgyvP !!!! ++=++
atmPPP == 2102 !v
hyy =! 12
ghv !! =2121
ghv 21 =
For$an$element$offluid$of$mass,$m12mv1
2 =mgy
12 2gh( )
2= gy
⇒ y = h
When%a%moving%fluid%is%contained%in%a%horizontal%pipe,%all%parts%of%ithave%the%same%elevation,%i.e.%y1 = y2,%and%Bernoulli’s%equation%simplifies%to:
P1 + 12 ρv1
2 = P2 + 12 ρv2
2
Thus,%P +%½ρv2 remains%constant,%so%that
If%v increases%! P decreases
If%v decreases%! P increases
Applications+of Bernoulli�s+Equation
Example:)An)enlarged)horizontal)blood)vessel
A1 A2 = 1.7A1
P1P2
v1 = 0.40 m/s v2
! Find%P2 – P1!Blood = 1060 kg/m3
P1 + 12 ρv1
2 = P2 + 12 ρv2
2 ⇒ P2 −P1 = 12 ρ v1
2 − v22( )
Use the continuity equation to find v2
A1v1 = A2v2 ⇒ v2 =A1
A2
v1
∴P2 −P1 = 12 ρ v1
2 −A1
A2
$
%&
'
()
2
v12
*
+,,
-
.//=
12 ρv1
2 1− A1
A2
$
%&
'
()
2*
+,,
-
.//=
12 1060( ) 0.402( ) 1− A1
1.7A1
$
%&
'
()
2*
+,,
-
.//
= 55 Pa = 0.41 mmHg
Applications)of Bernoulli�s)Equation
Applications+of Bernoulli�s+Equation
Conceptual+Example:++Tarpaulins*and*Bernoulli�s*Equation
When%the%truck%is%stationary,%the%tarpaulin%lies%flat,%but%it%bulges%outwardwhen%the%truck%is%speeding%downthe%highway.
Account%for%this%behavior.
Applications+of Bernoulli�s+Equation
Applications+of Bernoulli�s+Equation