Me 3560 Presentation Ch Vi

ME3560 – Fluid Mechanics

Chapter VI. Differential Analysis of Fluid Flow

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ME 3560 Fluid Mechanics




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6.1 Fluid Element Kinematics• In general a fluid particle can undergo translation, linear deformationrotation and angular deformation.



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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.



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•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′.



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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



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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



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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.



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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



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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



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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.



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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



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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.



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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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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

∂∂+

∂∂+

∂∂− )()()(



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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



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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:



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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



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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.



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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.



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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



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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



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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

∂∂+

∂∂

+∂

∂=



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•A similar analysis for they– andz–directions yields:

zyxzyx

F zyyyxysy δδδ

τστδ

∂∂

+∂

∂+

∂∂

=

zyxzyx

F zzyzxzsz δδδσττδ

∂∂+

∂∂

+∂

∂=



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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



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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 ===−



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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



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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∂=



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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.



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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



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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



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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



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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”



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Uniform Flow)sincos(

)sincos(

ααψαα

xyU

yxU

−=+=Φ

xv

yu

∂∂−=

∂∂= ψψ

yv

xu

∂Φ∂=

∂Φ∂=



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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



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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).



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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.



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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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•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



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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



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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



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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



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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



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• 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(



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• 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.



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•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:

Me 3560 Presentation Ch Vi

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