me 3560 presentation ch vi
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Fluid MechanicsTRANSCRIPT
ME3560 – Fluid Mechanics
Chapter VI. Differential Analysis of Fluid Flow
Summer I 2014
1
ME 3560 Fluid Mechanics
Chapter VI. Differential Analysis of Fluid Flow
ME3560 – Fluid Mechanics
Chapter VI. Differential Analysis of Fluid Flow
Summer I 2014
2
6.1 Fluid Element Kinematics• In general a fluid particle can undergo translation, linear deformationrotation and angular deformation.
ME3560 – Fluid Mechanics
Chapter VI. Differential Analysis of Fluid Flow
Summer I 2014
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Translation• The simplest type of motion that a fluid element can undergo istranslation.•In a small time intervalδt a particle located at pointO will move topointO′.•If all points in the element have the same velocity (which is only true ifthere are no velocity gradients), then the element will simply translatefrom one position to another.
•In general, there are velocity gradientswhich result in deformation and rotationof the element as it moves.
ME3560 – Fluid Mechanics
Chapter VI. Differential Analysis of Fluid Flow
Summer I 2014
•Consider the effect of a single velocity gradient,∂u/∂x, on a small cubehaving sidesδx, δy, andδz.•Let thex component of velocity ofO andB is u.•Then at nearby pointsA andC the x component of the velocity can beexpressed asu + (∂u/∂x) δx.
Linear Deformation
•This difference invelocity causes a“stretching” of the volumeelement by an amount(∂u/∂x)(δx)(δt) during theshort time intervalδt inwhich lineOA stretches toOA′ andBC to BC′.
ME3560 – Fluid Mechanics
Chapter VI. Differential Analysis of Fluid Flow
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( )( )tzyxx
uV δδδδ
δδδ
=inChange
•The corresponding change in the original volume,δV= δxδyδz is
x
u
t
txu
dt
Vd
V t ∂∂=
∂∂=→ δ
δδδ δ
)/(lim
)(10
•The rate at which the volume,δV, is changingper unit volumedue tothe gradient∂u/∂x is
x
u
dt
Vd
V ∂∂=)(1 δ
δ
ME3560 – Fluid Mechanics
Chapter VI. Differential Analysis of Fluid Flow
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•Similar analysis for they– and z–directions with velocity gradients∂v/∂y and∂w/∂z, respectively results in the following expression for therateat which the volume,δV, is changingper unit volume:
•This rate of change of the volume per unit volume is called thevolumetric dilatation rate.
•The volume of a fluid may change as the element moves fromonelocation to another in the flowfield.
•For anincompressible fluidthe volumetric dilatation rate is zero
Vz
w
y
v
x
u
dt
Vd
V
r⋅∇=
∂∂+
∂∂+
∂∂=)(1 δ
δ
ME3560 – Fluid Mechanics
Chapter VI. Differential Analysis of Fluid Flow
Summer I 2014
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•Variations in the velocity in the direction of the component itself, asrepresented by the derivatives∂u/∂x, ∂v/∂y, and∂w/∂z, simply cause alinear deformationof the element.
•Cross derivatives, such as∂u/∂y and ∂v/∂x, will cause the element torotate and generally to undergo anangular deformation, which changesthe shape of the element.
ME3560 – Fluid Mechanics
Chapter VI. Differential Analysis of Fluid Flow
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Angular Motion and Deformation•Consider motion in the x–y plane (results can be extended to 3–D). •The velocity variation that causes rotation and angular deformation is shown Figure (a).
•In δt the line line segments OA and OB will rotate through the angles δα and δβ to the new positions OA′ and OB′, (Figure (b)). •The angular velocity of line OA, ωOA, is
•For small angles:
ttOA δ
αδωδ 0lim
→=
tx
v
x
txxv δδ
δδαδαδ∂∂=∂∂=≈ )/(
tan
ME3560 – Fluid Mechanics
Chapter VI. Differential Analysis of Fluid Flow
Summer I 2014
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•Therefore the angular velocity of segment OA is
• If ∂ v/ ∂ x is positive ωOA is counterclockwise.x
v
x
v
t
txv
OA
tOA
∂∂=
∂∂=
∂∂=→
ω
δδω
δ
)/(lim
0
ME3560 – Fluid Mechanics
Chapter VI. Differential Analysis of Fluid Flow
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•For segment OB the angular velocity is
ttOB δ
βδωδ 0lim
→=
ty
u
y
tyyu δδ
δδβδβδ∂∂=∂∂=≈ )/(
tan•Where
y
u
t
tyu
OB
tOB
∂∂=
∂∂=→
ω
δδω
δ
)/(lim
0
• If ∂ u/ ∂ y is positive ωOB is clockwise.
ME3560 – Fluid Mechanics
Chapter VI. Differential Analysis of Fluid Flow
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•The rotation, ωz, of the element about the z axis is defined as the average of the angular velocities ωOA and ωOB of the two mutually perpendicular lines OA and OB.1 Thus, if counterclockwise rotation is considered to be positive:
∂∂−
∂∂=
y
u
x
vz 2
1ω
•In a similar analysis the angular velocities in the x and y directions can be found to be:
∂∂−
∂∂=
z
v
y
wx 2
1ω
∂∂−
∂∂=
x
w
z
uy 2
1ω
ME3560 – Fluid Mechanics
Chapter VI. Differential Analysis of Fluid Flow
Summer I 2014
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kji zyxˆˆˆ ωωωω ++=
rVVrrr
×∇==2
1curl
2
1ω
zvu
zyx
kji
∂∂∂∂∂∂= ///
ˆˆˆ
ωr
ky
u
x
vj
x
w
z
ui
z
v
y
w ˆ2
1ˆ2
1ˆ2
1
∂∂−
∂∂+
∂∂−
∂∂+
∂∂−
∂∂=ωr
Vrr
×∇== ωξ 2•ξ is know as the vorticity:
•If ∇×V=0: the angular velocity and vorticity are zero and an irrotational flow is present.
•If ωωωω = 0 (Irrotational flow) and then it is a POTENTIAL FLOW where:
Φ∇=Vr
•Φ is a scalar function.
ME3560 – Fluid Mechanics
Chapter VI. Differential Analysis of Fluid Flow
Summer I 2014
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•In addition to the rotation associated with the derivatives ∂u/∂y and ∂v/∂x, these derivatives can cause the fluid element to undergo an angular deformation, which results in a change in shape of the element. •The change in the original right angle formed by the lines OA and OB is termed the shearing strain, δγ. δβαδδγ +=•δγ is positive if the original right angle is decreasing. •The rate of change of δγ is called the rate of shearing strainor the rate of angular deformation:
∂∂+∂∂==→→ t
tyutxv
t tt δδδ
δγδγ
δδ
)/()/(limlim
00&
y
u
x
v
∂∂+
∂∂=γ&
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Chapter VI. Differential Analysis of Fluid Flow
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6.2 Conservation of MassMass accumulatedper unit time withintheCV
= Mass flow rateentering theCV – Mass flow rate
leaving theCV
Mass accumulated per unit time within the
CV
= zyxtt
m δδδρ∂
∂=∂∂
Net mass flow rate
crossing the CS
= zyxz
w
y
u
x
u δδδρρρ
∂∂+
∂∂+
∂∂− )()()(
ME3560 – Fluid Mechanics
Chapter VI. Differential Analysis of Fluid Flow
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•Thus, the differential formof the mass conservation equation is:
0)()()( =
∂∂+
∂∂+
∂∂+
∂∂
z
w
y
v
x
u
t
ρρρρ 0)( =⋅∇+∂∂
Vt
rρρ
•If the flow is incompressible:
0=⋅∇ Vr
0=∂∂+
∂∂+
∂∂
z
w
y
v
x
u
ME3560 – Fluid Mechanics
Chapter VI. Differential Analysis of Fluid Flow
Summer I 2014
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Conservation of Mass – Cylindrical Coordinates
kwjviuV ˆˆˆ ++=r
•Velocity in Cartesian coordinates:
zzrr evevevV ˆˆˆ ++= θθ
r•Velocity in Cylindrical coordinates:
0)()(1)(1 =
∂∂+
∂∂+
∂∂+
∂∂
z
vv
rr
vr
rtzr ρ
θρρρ θ
•Conservation of mass in cylindrical coordinates
01)(1 =
∂∂+
∂∂+
∂∂
z
vv
rr
vr
rzr
θθ
•For incompressible flow:
ME3560 – Fluid Mechanics
Chapter VI. Differential Analysis of Fluid Flow
Summer I 2014
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The Stream Function
0=∂∂+
∂∂
y
v
x
u•Assuming steady, incompressible, two-dimensional flow, the continuityequation becomes:
•Introduce a continuous scalar functionψ(x, y),streamfunction, such that:
xv
yu
∂∂−=
∂∂= ψψ
•This definition ofu andv in terms ofψ satisfies the continuity equation.
udyvdxddyy
dxx
d +−=∂∂+
∂∂= ψψψψ
•If ψ = constant, the equation of streamlines is obtained:v
dy
u
dx =
•Thus,ψ = constant lines are STREAMLINES
ME3560 – Fluid Mechanics
Chapter VI. Differential Analysis of Fluid Flow
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•The actual numerical value associated with a particular streamline is not of particular significance.•However the change in the value of ψ is related to the volume flowrate.
•Consider two closely spaced streamlines, the lower streamline is ψ, the upper one ψ + dψ.
• Let dq represent the volume rate of flow (per unit width perpendicular to the x–yplane) passing between the two streamlines.
•Note that flow never crosses streamlines, since by definition the velocity is tangent to the streamline.
ME3560 – Fluid Mechanics
Chapter VI. Differential Analysis of Fluid Flow
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•From conservation of mass, the inflow, dq, crossing the arbitrary surface ACmust equal the net outflow through surfaces ABand BC:
vdxudydq −=
dxx
dyy
dq∂∂+
∂∂= ψψ ψddq=
12
2
1
ψψψψ
ψ−== ∫ ddq
•The volumetric flow rate is the relative difference between two stream function values.
ME3560 – Fluid Mechanics
Chapter VI. Differential Analysis of Fluid Flow
Summer I 2014
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Stream Function–Polar Coordinates ψ =ψ(r, θ)•In cylindrical coordinates the continuity equation for incompressible,plane, two-dimensional flowreduces to:
01)(1 =
∂∂+
∂∂
θθv
rr
rv
rr
•And ψ is defined as:
rv
rvr ∂
∂−=∂∂= ψ
θψ
θ1
Stream Function: Steady, Compressible Flow
xv
yu
∂∂−=
∂∂= ψρψρ
ME3560 – Fluid Mechanics
Chapter VI. Differential Analysis of Fluid Flow
Summer I 2014
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6.3 Conservation of Linear Momentum
∑∫∫ =⋅+∂∂
FdAnVVVdVt cscv
rrrr)ˆ(ρρ
•As determined in the previous chapter, the momentum equation is:
•Next, this equation is applied to an infinitesimal element of fluid.
•The right hand side of the equation can be written as
∇⋅+
∂∂
VVt
V rrr
)(ρ
ME3560 – Fluid Mechanics
Chapter VI. Differential Analysis of Fluid Flow
Summer I 2014
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•Surface forces acting on a differential element of fluid in thex–direction:6.3 Conservation of Linear Momentum
zyxzyx
F zxyxxxsx δδδττσδ
∂∂+
∂∂
+∂
∂=
ME3560 – Fluid Mechanics
Chapter VI. Differential Analysis of Fluid Flow
Summer I 2014
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•A similar analysis for they– andz–directions yields:
zyxzyx
F zyyyxysy δδδ
τστδ
∂∂
+∂
∂+
∂∂
=
zyxzyx
F zzyzxzsz δδδσττδ
∂∂+
∂∂
+∂
∂=
ME3560 – Fluid Mechanics
Chapter VI. Differential Analysis of Fluid Flow
Summer I 2014
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6.3 Conservation of Linear Momentum•The momentumapplied to a control volume states that the forces actingon theCV equal the momentumaccumulated in theCV minus the netflux of momentumcrossing theCS.
∂∂+
∂∂+
∂∂+
∂∂=
∂∂+
∂∂
+∂
∂+z
uw
y
uv
x
uu
t
u
zyxg zxyxxx
x ρττσρ
∂∂+
∂∂+
∂∂+
∂∂=
∂∂
+∂
∂+
∂∂
+z
vw
y
vv
x
vu
t
v
zyxg zyyyxy
y ρτστ
ρ
∂∂+
∂∂+
∂∂+
∂∂=
∂∂+
∂∂
+∂
∂+z
ww
y
wv
x
wu
t
w
zyxg zzyzxz
z ρσττρ
ME3560 – Fluid Mechanics
Chapter VI. Differential Analysis of Fluid Flow
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6.4 Inviscid Flow•Shearing stresses develop in a moving fluid because of the viscosity of the fluid.
•For some common fluids (air and water), the viscosity is small.
•It is reasonable to assume that under some circumstances viscous effects can be neglected (and thus shearing stresses).
•Flow fields in which the shearing stresses are assumed to be negligible are said to be inviscid, nonviscous, or frictionless.
•For fluids in which there are no shearing stresses the normal stress at a point is independent of direction—that is, σxx = σyy = σzz.
zzyyxxp σσσ ===−
ME3560 – Fluid Mechanics
Chapter VI. Differential Analysis of Fluid Flow
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•For an inviscid flow, τ = 0, and σxx = σyy = σzz= −p, the general equations of motion become
Euler’s Equations of Motion
∂∂+
∂∂+
∂∂+
∂∂=
∂∂−
z
uw
y
uv
x
uu
t
u
x
pgx ρρ
∂∂+
∂∂+
∂∂+
∂∂=
∂∂−
z
vw
y
vv
x
vu
t
v
y
pgy ρρ
∂∂+
∂∂+
∂∂+
∂∂=
∂∂−
z
ww
y
wv
x
wu
t
w
z
pgz ρρ
⋅∇+
∂∂=∇− VV
t
Vpg
rrr
r)(ρρ
ME3560 – Fluid Mechanics
Chapter VI. Differential Analysis of Fluid Flow
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Potential Flow•For an irrotational flow the angular velocity is zero ωωωω = ½∇×V = 0.•Therefore it is possible to define the velocity in terms of a scalar function Φ(x, y, z, t) as
•Notice that for irrotational flow:
Φ∇=Vr
0=Φ∇×∇=×∇ Vr
•Therefore this definition of the velocity in terms of Φ satisfies the condition of irrotationallity.
•Φ is called velocity potential.
zw
yv
xu
∂Φ∂=
∂Φ∂=
∂Φ∂=
ME3560 – Fluid Mechanics
Chapter VI. Differential Analysis of Fluid Flow
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•The velocity potential is a consequence of the irrotationality of the flow field.•The stream function is a consequence of conservation of mass.
•The velocity potential can be defined for a general three-dimensional flow.•The stream function is restricted to two-dimensional flows.
•For an Incompressible flow: ∇⋅V = 0
02
2
2
2
2
2
=∂
Φ∂+∂
Φ∂+∂
Φ∂zyx
02 =Φ∇
•The previous equation is the governing equation for potential flow.
•∇⋅∇( )=∇2( ) is called Laplacian operator.
ME3560 – Fluid Mechanics
Chapter VI. Differential Analysis of Fluid Flow
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zr ez
er
er
ˆ)(
ˆ)(1
ˆ)(
)(∂∂+
∂∂+
∂∂=∇ θθ
•In cylindrical coordinates the Del operator is:
•Thus:
zr ez
er
er
ˆˆ1
ˆ∂Φ∂+
∂Φ∂+
∂Φ∂=Φ∇ θθ
zzrr evevevV ˆˆˆ ++= θθ
r•Since
•It follows
zv
rrv
rv zr ∂
Φ∂=∂Φ∂=
∂Φ∂= 1
θ
•Thus, the Laplace equation in cylindrical coordinates is:
011
2
2
2
2
2=
∂Φ∂+
∂Φ∂+
∂Φ∂
∂∂
zrrr
rr θ
ME3560 – Fluid Mechanics
Chapter VI. Differential Analysis of Fluid Flow
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6.5 Some Basic, Plane Potential Flows•Laplace's equation is a linear partial differential equation.
•Since it is linear, various solutions can be added to obtain other solutions—that is, if Φ1(x, y, z) and Φ2 (x, y, z) are two solutions to Laplace's equation, then Φ3=Φ1 +Φ2 is also a solution.
•Thus, basic solutions can be combined to obtain more complicated and interesting solutions.
•For a 2–D incompressible flowx
vy
u∂∂−=
∂∂= ψψ
02
2
2
2
=∂
Φ∂+∂
Φ∂∂
Φ∂=∂
Φ∂=yxy
vx
u
ME3560 – Fluid Mechanics
Chapter VI. Differential Analysis of Fluid Flow
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•If the flow is irrotational (ωz = ∂ v/ ∂ x – ∂ u/ ∂ y = 0)
00 =
∂∂−
∂∂−
∂∂
∂∂=
∂∂−
∂∂
yyxxy
u
x
v ψψ
02
2
2
2
=∂∂+
∂∂
yx
ψψ
•As shown next Φ(x, y) and ψ(x, y) are orthogonal functions
udyvdxdyy
dxx
d
vdyudxdyy
dxx
d
+−=∂∂+
∂∂=
+=∂
Φ∂+∂
Φ∂=Φ
ψψψ
ME3560 – Fluid Mechanics
Chapter VI. Differential Analysis of Fluid Flow
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•For Φ(x, y) = const and ψ(x, y) = const
v
u
dx
dyvdyudxd −==+=Φ 0
v
u
dx
dyudyvdxd ==+−= 0ψ
•Thus, for Φ = const and ψ = const the lines are perpendicular generating a “Flow Net”
ME3560 – Fluid Mechanics
Chapter VI. Differential Analysis of Fluid Flow
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Uniform Flow)sincos(
)sincos(
ααψαα
xyU
yxU
−=+=Φ
xv
yu
∂∂−=
∂∂= ψψ
yv
xu
∂Φ∂=
∂Φ∂=
ME3560 – Fluid Mechanics
Chapter VI. Differential Analysis of Fluid Flow
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Source and Sink (Radial Flow)
02
== θπv
r
mvr
• m is the source/sink strength •Its units are m2/s (volume flow rate per unit width
crm +=Φ ln2π
01 =
∂Φ∂=
∂Φ∂=
θθ rv
rvr
01 =
∂∂−=
∂∂=
rv
rvr
ψθψ
θ
cm += θπ
ψ2
ME3560 – Fluid Mechanics
Chapter VI. Differential Analysis of Fluid Flow
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Ideal Vortex
r
kvvr == θ0
• k is the vortex strength
rkk ln−==Φ ψθrr
v∂Φ∂= 1
θ0=
∂∂−=
rv
ψθ
• The tangential velocity varies inversely with the distance from the origin.•A singularity occurs at r = 0 (where the velocity becomes infinite).
ME3560 – Fluid Mechanics
Chapter VI. Differential Analysis of Fluid Flow
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•The circulation, Γ, is a concept commonly associated with vortex motion •Γ is defined as the line integral of the tangential component of the velocity taken around a closed curve in the flow field
∫ ⋅=Γc
sdVrr
vdyudxsdV +=⋅ rr
vdyudxdyy
dxx
d +=∂
Φ∂+∂
Φ∂=Φ
∫ Φ=Γcd
•For a free vortexkkd πθ
π
22
0
==Γ ∫
rln22 π
ψθπ
Γ−=Γ=Φ•Then ψ and Φ are expressed as:
Γ is often useful when evaluating the forces developed on bodies immersed in moving fluids.
ME3560 – Fluid Mechanics
Chapter VI. Differential Analysis of Fluid Flow
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Doublet•A doublet is formed by combining a source and sink with equal strength.•The source and the sink are approached in a way that m×distance = const•Thus, the distance is decreased while m increases.•Resulting in:
πθθψ ma
kr
kr
k ==Φ−= cossin
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Chapter VI. Differential Analysis of Fluid Flow
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•Study Superposition of Basic Potential Flows:•Source + Uniform Flow �Half body•Source + Sink + Uniform Flow � Rankine Oval•Uniform Flow + Doublet � Flow around a cylinder (no lift).•Uniform Flow + Doublet + Vortex � Flow around a cylinder with lift
ME3560 – Fluid Mechanics
Chapter VI. Differential Analysis of Fluid Flow
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6.8 Viscous Flow• For incompressible Newtonian fluids it is known that the stresses are linearly related to the rates of deformation and can be expressed in Cartesian coordinates as: •Normal stresses:
;2x
upxx ∂
∂+−= µσ ;2y
vpyy ∂
∂+−= µσz
wpzz ∂
∂+−= µσ 2
• Shear stresses:
;
∂∂+
∂∂==
x
v
y
uyxxy µττ ;
∂∂+
∂∂==
y
w
z
vzyyz µττ
;
∂∂+
∂∂==
z
u
x
wzxxz µττ
ME3560 – Fluid Mechanics
Chapter VI. Differential Analysis of Fluid Flow
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∂∂+
∂∂+
∂∂+
∂∂=
∂∂+
∂∂
+∂
∂+z
uw
y
uv
x
uu
t
u
zyxg zxyxxx
x ρττσρ
∂∂+
∂∂+
∂∂+
∂∂=
∂∂
+∂
∂+
∂∂
+z
vw
y
vv
x
vu
t
v
zyxg zyyyxy
y ρτστ
ρ
∂∂+
∂∂+
∂∂+
∂∂=
∂∂+
∂∂
+∂
∂+z
ww
y
wv
x
wu
t
w
zyxg zzyzxz
z ρσττρ
• Substituting these previous expressions into the momentum equation (differential form) shown below:
• The Navier – Stokes Equations are determine as:
VgpDt
VD rrr
2∇++−∇= µρρ
ME3560 – Fluid Mechanics
Chapter VI. Differential Analysis of Fluid Flow
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∂∂+
∂∂+
∂∂++
∂∂−=
∂∂+
∂∂+
∂∂+
∂∂
2
2
2
2
2
2
z
u
y
u
x
ug
x
p
z
uw
y
uv
x
uu
t
ux µρρ
∂∂+
∂∂+
∂∂++
∂∂−=
∂∂+
∂∂+
∂∂+
∂∂
2
2
2
2
2
2
z
v
y
v
x
vg
y
p
z
vw
y
vv
x
vu
t
vy µρρ
∂∂+
∂∂+
∂∂++
∂∂−=
∂∂+
∂∂+
∂∂+
∂∂
2
2
2
2
2
2
z
w
y
w
x
wg
z
p
z
ww
y
wv
x
wu
t
wz µρρ
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6.9 Some Simple Solutions for Laminar, Viscous, Incompressible Fluids
• Flow between the two horizontal, infinite parallel plates. •Fluid particles move in the x direction parallel to the plates v = w = 0.•From continuity equation: ∂u/∂x = 0. •There would be no variation of u in the z direction for infinite plates•For steady flow ∂u/∂t = 0 so that u = u(y). •Under these conditions the Navier–Stokes equations reduce to
• Steady, Laminar Flow between Fixed Parallel Plates
ME3560 – Fluid Mechanics
Chapter VI. Differential Analysis of Fluid Flow
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VgpDt
VD rrr
2∇++−∇= µρρ
∂∂+
∂∂+
∂∂++
∂∂−=
∂∂+
∂∂+
∂∂+
∂∂
2
2
2
2
2
2
z
u
y
u
x
ug
x
p
z
uw
y
uv
x
uu
t
ux µρρ
∂∂+
∂∂+
∂∂++
∂∂−=
∂∂+
∂∂+
∂∂+
∂∂
2
2
2
2
2
2
z
v
y
v
x
vg
y
p
z
vw
y
vv
x
vu
t
vy µρρ
∂∂+
∂∂+
∂∂++
∂∂−=
∂∂+
∂∂+
∂∂+
∂∂
2
2
2
2
2
2
z
w
y
w
x
wg
z
p
z
ww
y
wv
x
wu
t
wz µρρ
ME3560 – Fluid Mechanics
Chapter VI. Differential Analysis of Fluid Flow
Summer I 2014
44
• Couette Flow•Another simple parallel-plate flow can be developed by fixing one plate and letting the other plate move with a constant velocity, U.•The Navier–Stokes equations reduce to:
2
2
0y
u
x
p
∂∂+
∂∂−= µ0=
∂∂
x
uUbu
u
==
)(
0)0(
ME3560 – Fluid Mechanics
Chapter VI. Differential Analysis of Fluid Flow
Summer I 2014
45
• Steady, Laminar Flow in Circular Tubes
•Consider a steady, incompressible, laminar flow through a straight circular tube of constant cross section of radius R (Hagen–Poiseuilleor Poiseuille flow).
0)()(1)(1 =
∂∂+
∂∂+
∂∂+
∂∂
z
vv
rr
vr
rtzr ρ
θρρρ θ
•Becomes: 0=∂∂
z
vz
•Under the conditions described above, the continuity equation:
•Assume that the flow is parallel to the walls so that vr= 0 and vθ= 0.•For steady, axisymmetric flow, the velocity is only a function of r.
ME3560 – Fluid Mechanics
Chapter VI. Differential Analysis of Fluid Flow
Summer I 2014
46
•Whereas the momentum equation in cylindrical coordinates
•Results in:
rgr
p ρ+∂∂−=0
θρθ
gp
r+
∂∂−= 1
0
r:
θ :
∂∂
∂∂+
∂∂−=
r
vr
rrz
p z10 µ
z: