Chapter 9: Surface Resistanceuser.engineering.uiowa.edu/.../Chap_6_10-24-2012.docx · Web view2012/10/24 · 57:020 Mechanics of Fluids and Transport Processes Chapter 6 36 Professor

57:020 Mechanics of Fluids and Transport Processes Chapter 6Professor Fred Stern Fall 2012

Chapter 6 Differential Analysis of Fluid Flow

Fluid Element Kinematics

Fluid element motion consists of translation, linear defor-mation, rotation, and angular deformation.

Types of motion and deformation for a fluid element.

Linear Motion and Deformation:

Translation of a fluid element

Linear deformation of a fluid element

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Change in :

the rate at which the volume is changing per unit vol-ume due to the gradient ∂u/∂x is

If velocity gradients ∂v/∂y and ∂w/∂z are also present, then using a similar analysis it follows that, in the general case,

This rate of change of the volume per unit volume is called the volumetric dilatation rate.

Angular Motion and DeformationFor simplicity we will consider motion in the x–y plane, but the results can be readily extended to the more general case.

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Angular motion and deformation of a fluid element

The angular velocity of line OA, ωOA, is

For small angles

so that

Note that if ∂v/∂x is positive, ωOA will be counterclockwise.

Similarly, the angular velocity of the line OB is

In this instance if ∂u/∂y is positive, ωOB will be 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. Thus, if counterclockwise rotation is considered to be positive, it follows that

Rotation of the field element about the other two coordinate axes can be obtained in a similar manner:

The three components, ωx,ωy, and ωz can be combined to give the rotation vector, ω, in the form:

since

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The vorticity, ζ, is defined as a vector that is twice the rota-tion vector; that is,

The use of the vorticity to describe the rotational character-istics of the fluid simply eliminates the (1/2) factor associ-ated with the rotation vector. If , the flow is called irrotational.

In addition to the rotation associated with the derivatives ∂u/∂y and ∂v/∂x, these derivatives can cause the fluid ele-ment 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,

The rate of change of δγ is called the rate of shearing strain or the rate of angular deformation:

γ xy= limδt →0

δγδt= lim

δt→ 0

δγδt=[ (∂v /∂ x )δt+ (∂u /∂ y )δt

δt ]=∂v∂ x+ ∂u∂ y

Similarly,γ xz=

∂w∂x+ ∂u∂ z

γ yz=∂w∂ y+ ∂v∂ z

The rate of angular deformation is related to a correspond-ing shearing stress which causes the fluid element to change in shape.

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The Continuity Equation in Differential Form

The governing equations can be expressed in both integral and differential form. Integral form is useful for large-scale control volume analysis, whereas the differential form is useful for relatively small-scale point analysis.

Application of RTT to a fixed elemental control volume yields the differential form of the governing equations. For example for conservation of mass

∑CS

ρV⋅A=−∫CV

∂ ρ∂ t

d V

net outflow of mass = rate of decreaseacross CS of mass within CV

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[ρu+ ∂∂ x ( ρu )dx ]dydzoutlet mass flux


Consider a cubical element oriented so that its sides are to the (x,y,z) axes

Taylor series expansion retaining only first order term

We assume that the element is infinitesimally small such that we can assume that the flow is approximately one di-mensional through each face.

The mass flux terms occur on all six faces, three inlets, and three outlets. Consider the mass flux on the x faces

=∂∂ x( ρu)dxdydz

V

Similarly for the y and z faces

y flux=∂∂ y( ρv )dxdydz

zflux=∂∂ z( ρw )dxdydz

inlet mass fluxudydz

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The total net mass outflux must balance the rate of decrease of mass within the CV which is

−∂ ρ∂ t

dxdydz

Combining the above expressions yields the desired result

[∂ ρ∂ t+∂∂ x( ρu)+∂

∂ y( ρv )+∂

∂ z( ρw )]dxdydz=0

∂ ρ∂ t+∂∂ x( ρu)+∂

∂ y( ρv )+∂

∂ z( ρw )=0

∂ ρ∂ t+∇⋅( ρV )=0

ρ∇⋅V +V⋅∇ ρ

DρDt+ ρ∇⋅V=0 D

Dt= ∂∂ t+V⋅∇

Nonlinear 1st order PDE; ( unless = constant, then linear)Relates V to satisfy kinematic condition of mass conserva-tion

Simplifications:1. Steady flow: ∇⋅( ρV )=0

2. = constant: ∇⋅V=0

dV

per unit Vdifferential form of con-tinuity equations

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i.e.,

∂u∂ x+ ∂ v∂ y+∂w∂ z=0

3D

∂u∂ x+ ∂ v∂ y=0

2D

The continuity equation in Cylindrical Polar Coordinates

The velocity at some arbitrary point P can be expressed as

The continuity equation:

For steady, compressible flow

For incompressible fluids (for steady or unsteady flow)

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The Stream FunctionSteady, incompressible, plane, two-dimensional flow repre-sents one of the simplest types of flow of practical impor-tance. By plane, two-dimensional flow we mean that there are only two velocity components, such as u and v, when the flow is considered to be in the x–y plane. For this flow the continuity equation reduces to

∂u∂ x+ ∂ v∂ y=0

We still have two variables, u and v, to deal with, but they must be related in a special way as indicated. This equation suggests that if we define a function ψ(x, y), called the stream function, which relates the velocities as

then the continuity equation is identically satisfied:

Velocity and velocity components along a streamline

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Another particular advantage of using the stream function is related to the fact that lines along which ψ is constant are streamlines.The change in the value of ψ as we move from one point (x, y) to a nearby point (x + dx, y + dy) along a line of constant ψ is given by the relationship:

and, therefore, along a line of constant ψ

The flow between two streamlinesThe actual numerical value associated with a particular streamline is not of particular significance, but the change in the value of ψ is related to the volume rate of flow. Let dq represent the volume rate of flow (per unit width per-pendicular to the x–y plane) passing between the two streamlines.

Thus, the volume rate of flow, q, between two streamlines such as ψ1 and ψ2, can be determined by integrating to yield:

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In cylindrical coordinates the continuity equation for in-compressible, plane, two-dimensional flow reduces to

and the velocity components, vr and vθ, can be related to the stream function, ψ(r, θ), through the equations

Navier-Stokes Equations

Differential form of momentum equation can be derived by applying control volume form to elemental control volume

The differential equation of linear momentum: elemental fluid volume approach

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∑F= ∂∂ t∫CV

❑

ρV dV⏟

(1)

+∫CS

❑

V ρV ⋅ n dA⏟

(2)

(1) = ∂∂ t

( ρV )dxdydz=( ∂ ρ∂t

V +ρ ∂V∂ t )dxdydz

(2) = [ ∂∂x

( ρuV )⏟

x−face

+ ∂∂ y

(ρvV )⏟

y−face

+ ∂∂ z

(ρw V )⏟

z−face]dxdydz

=[ ρu ∂V∂x+V ∂ ρu

∂ x+ ρv ∂V

∂ y+V ∂ ρv

∂ y+ ρw ∂V

∂ z+V ∂ ρw

∂x ]dxdydz combining and making use of the continuity equation yields

∑F=[V {∂ρ∂ t+∇⋅ (ρV )⏟

¿0}+ρ( ∂V

∂t+V ⋅ ∇V )]dxdydz

∴∑F=ρ DVDt

dxdydz or ∑ f=ρ DVDt

where ∑F=∑Fbody+∑F surface

∑ f=∑ f body+∑ f surface

V ⋅∇=u ∂∂x+v ∂

∂ y+w ∂

∂ z

DDt= ∂

∂ t+V ⋅∇

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Body forces are due to external fields such as gravity or magnetics. Here we only consider a gravitational field; that is,

∑ Fbody=d F grav=ρ gdxdydz

and g=−g k for g z

i.e., f body=−ρg k

Surface forces are due to the stresses that act on the sides of the control surfaces

symmetric (ij = ji)ij = - pij + ij 2nd order tensor

normal pressure viscous stress

= -p+xx xy xz

yx -p+yy yz

zx zy -p+zz

As shown before for p alone it is not the stresses them-selves that cause a net force but their gradients.

dFx,surf = [∂∂ x (σ xx )+ ∂∂ y (σ xy )+ ∂∂ z (σ xz )]dxdydz

ij = 1 i = jij = 0 i j

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= [−∂ p∂ x+ ∂∂ x ( τ xx )+ ∂∂ y ( τ xy )+ ∂∂ z ( τ xz )]dxdydz

This can be put in a more compact form by defining vector stress on x-face

τ x=τ xx i+τ xy j+ τxz k

and noting that

dFx,surf = [−∂ p∂ x+∇⋅τ x]dxdydz

fx,surf = −∂ p∂ x+∇⋅τ x

per unit volume

similarly for y and z

fy,surf = −∂ p∂ y+∇⋅τ y τ y=τ yx i+ τ yy j+τ yz k

fz,surf = −∂ p∂ z+∇⋅τ z τ z=τ zx i+τ zy j+ τ zz k

finally if we defineτ ij=τx i+τ y j+τ z k then

f surf=−∇ p+∇⋅τ ij=∇⋅σ ij σ ij=−pδij+τ ij

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Putting together the above results

∑ f=f body+ f surf= ρ DVDt

f body=−ρg kf surface=−∇ p+∇⋅τ ij

inertia body force force surface surface force

due to force due due to viscous gravity to p shear and normal

stresses

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For Newtonian fluid the shear stress is proportional to the rate of strain, which for incompressible flow can be written

τ ij=2μεij=μ ( ∂ui

∂x j+∂u j

∂ x i)

where, μ = coefficient of viscosityε ij = rate of strain tensor

= [∂u∂ x

12 ( ∂ v

∂ x+ ∂u∂ y ) 1

2 (∂w∂x+ ∂u∂ z )

12 ( ∂u

∂ y +∂v∂ x ) ∂v

∂ y12 (∂w

∂ y +∂v∂ z )

12 ( ∂u

∂ z +∂w∂x ) 1

2 (∂ v∂ z +

∂w∂ y ) ∂w

∂ z]

ρa=−ρg k−∇ p+∇ ⋅ (τ ij )

where,∇⋅ (τ ij )=μ ∂

∂ x j (∂ui

∂ x j+∂u j

∂ xi )=μ ( ∂2ui

∂ x j2⏟

∇2V

+ ∂∂x i

∂u j

∂ x j⏟¿ 0)

ρa=−ρg k−∇ p+μ∇2V

ρa=−∇ (p+γz )+μ∇2V Navier-Stokes Equation∇⋅V=0 Continuity Equation

Four equations in four unknowns: V and p

Ex) 1-D flow

τ=μ dudy

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Difficult to solve since 2nd order nonlinear PDE

x: ρ [ ∂u∂ t+u ∂u

∂ x+v ∂u

∂ y+w ∂u

∂z ]=−∂ p∂ x

+μ [ ∂2u∂ x2+

∂2u∂ y2+

∂2 u∂ z2 ]

y: ρ [ ∂v∂ t+u ∂ v

∂ x+v ∂v

∂ y+w ∂v

∂z ]=−∂ p∂ y

+μ[ ∂2v∂ x2 +

∂2 v∂ y2+

∂2 v∂ z2 ]

z: ρ [ ∂w∂ t+u ∂w

∂ x+v ∂w

∂ y+w ∂w

∂ z ]=−∂ p∂ z

−ρg+μ[ ∂2 w∂ x2 +

∂2 w∂ y2 +

∂2w∂ z2 ]

∂u∂ x+ ∂ v∂ y+ ∂w∂ z=0

Navier-Stokes equations can also be written in other coor-dinate systems such as cylindrical, spherical, etc.

There are about 80 exact solutions for simple geometries. For practical geometries, the equations are reduced to alge-braic form using finite differences and solved using com-puters.

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Ex) Exact solution for laminar incompressible steady flow in a circular pipe

Use cylindrical coordinates with assumptions

∂∂ z=0 : Fully-developed flow

vθ=0 : Flow is parallel to the wall

Continuity equation:

1r

∂ ( r vr )∂ r

+1r∂ vθ

∂θ+∂vz

∂ z=0

vr=Cr

B.C. vr (r=0 )=0 C=0

i.e., vr=0

Momentum equation:

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ρ( ∂vr

∂ t+vr

∂ vr

∂ r+

vθ

r∂ vr

∂θ−

vθ2

r+vz

∂vr

∂ z ) ¿− ∂ p

∂ r+ρ gr+μ[ 1r ∂

∂r (r ∂ vr

∂ r )− vr

r2 +1r2

∂2 vr

∂θ2 −2r2

∂vθ

∂θ+∂2 vr

∂ z2 ] ρ( ∂vθ

∂ t+vr

∂vθ

∂r+vθ

r∂vθ

∂θ+

vr vθ

r+vz

∂vθ

∂ z ) ¿− ∂ p

∂θ+ρ gθ+μ[ 1r ∂

∂ r (r ∂vθ

∂r )− vθ

r2 +1r2

∂2 vθ

∂θ2 +2r2

∂vr

∂θ+∂2 vθ

∂ z2 ] ρ( ∂v z

∂ t+vr

∂ z∂ r+vθ

r∂v z

∂θ+vz

∂vz

∂ z ) ¿− ∂ p

∂ z+ρ gz+μ [ 1r ∂

∂ r (r ∂ vz

∂ r )+ 1r2

∂2 v z

∂θ2 +∂2 v z

∂ z2 ] or

0=−ρg sin θ−∂ p∂r (1)

0=−ρg cosθ−1r∂ p∂θ (2)

0=−∂ p∂ z

+μ[ 1r ∂∂r (r ∂v z

∂ r )] (3)

where,gr=−g sin θ

gθ=−g cosθ

Equations (1) and (2) can be integrated to givep=−ρg (rsin θ )+ f 1 ( z)=−ρgy+ f 1 ( z )

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pressure p is hydrostatic and ∂ p/∂ z is not a function of r or θ

Equation (3) can be written in the from

1r

∂∂r (r ∂v z

∂r )=1μ

∂ p∂ z

and integrated (using the fact that ∂ p/∂ z = constant) to give

r∂vz

∂r= 1

2μ ( ∂ p∂z )r2+C1

Integrating again we obtain

vz=1

4 μ ( ∂ p∂z )r2+C1 lnr+C2

B.C.vz (r=0 )≠∞ C1=0

vz (r=R )=0 C2=−14 μ ( ∂ p

∂ z )R2

∴ vz=1

4 μ ( ∂ p∂ z )( r2−R2 )

at any cross section the velocity distribution is parabolic

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1) Flow rate Q:

Q=∫0

R

vzdA=2 π∫0

R

vz rdr=−π R4

8μ (∂ p∂ z )

where, dA=(2 πr )dr

If the pressure drops Δ p over a length l: Δ pl=−∂ p

∂ z

Q= π R4 Δ p8 μ l

2) Mean velocity V :

V=QA=( 1

π R2 )( π R4 Δ p8 μ l )=R2Δ p

8μ l

3) Maximum velocity vmax:

vmax=vz (r=0 )=−R2

4 μ ( ∂ p∂ z )= R2 Δ p

4 μ l=2V

v z

vmax=1−( rR )

2

4) Wall shear stress ( τ rz )wall:

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τ rz=μ( ∂vr

∂ z+∂vz

∂r )=μ∂vz

∂r

where

∂v z

∂ r=vmax⏟

¿2V(−2r

R2 )=−4 VrR2

Thus, at the wall (i.e., r=R),

( τ rz )wall=−4 μV

R

and with Q=π R2 V ,

|( τ rz )wall|= 4 μQπ R3

Note: Only valid for laminar flows. In general, the flow re-mains laminar for Reynolds numbers, Re = ρV (2 R)/μ, below 2100. Turbulent flow in tubes is considered in Chapter 8.

Differential Analysis of Fluid Flow

We now discuss a couple of exact solutions to the Navier-Stokes equations. Although all known exact solutions (about 80) are for highly simplified geometries and flow conditions, they are very valuable as an aid to our under-

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standing of the character of the NS equations and their so-lutions. Actually the examples to be discussed are for in-ternal flow (Chapter 8) and open channel flow (Chapter 10), but they serve to underscore and display viscous flow. Finally, the derivations to follow utilize differential analy-sis. See the text for derivations using CV analysis.

Couette Flow

boundary conditions

First, consider flow due to the relative motion of two paral-lel plates

Continuity∂u∂ x=0

Momentum0=μ d2u

dy 2

or by CV continuity and momentum equations:ρu1 Δy=ρu2 Δyu1 = u2

∑ F x=∑ uρV⋅d A=ρQ (u2−u1 )=0

u = u(y)v = o∂ p∂ x=∂ p∂ y=0

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=pΔy−( p+ dpdx

Δx)Δy−τΔx+(τ+ dτdy

dy )Δx= 0

dτdy=0

i.e.

ddy (μ du

dy )=0

μ d2udy2=0

from momentum equation

μ dudy=C

u=Cμ

y+D

u(0) = 0 D = 0

u(t) = U C = μ U

t

u=Ut

y

τ=μ dudy=μU

t=

constantGeneralization for inclined flow with a constant pressure gradient

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Continutity∂u∂ x=0

Momentum0=− ∂

∂ x( p+γz )+μ d2u

dy 2

i.e.,μ d2u

dy2=γ dhdx h = p/ +z = constant

plates horizontal dzdx=0

plates vertical dzdx =-1

which can be integrated twice to yield

μ dudy=γ dh

dxy+A

μu=γ dhdx

y2

2+Ay+B

now apply boundary conditions to determine A and Bu(y = 0) = 0 B = 0u(y = t) = U

u = u(y)v = o∂ p∂ y=0

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μU=γ dhdx

t 2

2+At⇒ A=μU

t−γ dh

dxt2

u( y )= γμ

dhdx

y2

2+ 1

μ [ μUt −γ dhdx

t2 ]

=− γ

2μdhdx

(ty− y2)+Ut

y

This equation can be put in non-dimensional form:uU=− γt2

2μUdhdx (1− y

t ) yt+ y

t

define: P = non-dimensional pressure gradient

= − γt2

2μUdhdx

h= pγ+z

Y = y/t=− γz2

2 μU [ 1γ dpdx+ dzdx ]

uU=P⋅Y (1−Y )+Y

parabolic velocity profile

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uU=Py

t−Py2

t2 +yt

q=∫0

t

udy

u=qt=∫0

t

U [ ] dy

t

t uU=∫

0

t

[ Pt y− Pt 2 y2+ y

t ]dy =Pt2−Pt

3+ t

2

uU=P

6+ 1

2⇒u= t2

12μ (−γ dhdx )+U

2

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For laminar flow

u tν<1000

Recrit 1000

The maximum velocity occurs at the value of y for which:dudy=0 d

dy ( uU )=0=P

t−2P

t2 y+1t

⇒ y= t2P

(P+1 )= t2+ t

2P @ umax

∴ umax=u ( ymax )=UP4+U

2+ U

4 P

note: if U = 0:

uumax

=P6/ P

4=2

3

The shape of the velocity profile u(y) depends on P:

1. If P > 0, i.e.,dhdx<0

the pressure decreases in the direction of flow (favorable pressure gradient) and the velocity is positive over the entire width

γ dhdx=γ d

dx ( pγ +z)=dpdx−γ sinθ

a)dpdx<0

for U = 0, y = t/2

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b)dpdx<γ sinθ

1. If P < 0, i.e., dh /dx>0 the pressure increases in the di-rection of flow (adverse pressure gradient) and the ve-locity over a portion of the width can become negative (backflow) near the stationary wall. In this case the dragging action of the faster layers exerted on the fluid particles near the stationary wall is insufficient to over-come the influence of the adverse pressure gradient.

dpdx−γ sin θ>0

dpdx>γ sinθ

orγ sin θ< dp

dx

2. If P = 0, i.e., dhdx=0

the velocity profile is linear

u=Ut

y

a)dpdx=0

and = 0

b)dpdx=γ sinθ

For U = 0 the form uU=PY (1−Y )+Y

is not appropriateu = UPY(1-Y)+UY

Note: we derived this special case

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= − γt2

2μdhdx

Y (1−Y )+UY

Now let U = 0:u=− γt2

2 μdhdx

Y (1−Y )

3. Shear stress distribution

Non-dimensional velocity distribution

where is the non-dimensional velocity,

is the non-dimensional pressure gradient

is the non-dimensional coordinate.Shear stress

In order to see the effect of pressure gradient on shear stress using the non-dimensional velocity distribution, we define the non-dimensional shear stress:

Then

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where is a positive constant. So the shear stress always varies linearly with across any section.

At the lower wall :

At the upper wall :

For favorable pressure gradient, the lower wall shear stress is always positive:

1. For small favorable pressure gradient :

and

2. For large favorable pressure gradient :

and

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For adverse pressure gradient, the upper wall shear stress is always positive:

1. For small adverse pressure gradient :

and

2. For large adverse pressure gradient :

and

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For , i.e., channel flow, the above non-dimensional form of velocity profile is not appropriate. Let’s use dimen-sional form:

Thus the fluid always flows in the direction of decreasing piezometric pressure or piezometric head because

and . So if is negative, is posi-

tive; if is positive, is negative.

Shear stress:

Since , the sign of shear stress is always oppo-

site to the sign of piezometric pressure gradient , and the magnitude of is always maximum at both walls and zero at centerline of the channel.

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For favorable pressure gradient, ,

For adverse pressure gradient, ,

Flow down an inclined plane

uniform flow velocity and depth do not change in x-direction

Continuitydudx=0

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x-momentum0=− ∂

∂ x( p+γz )+μ d2u

dy 2

y-momentum0=− ∂

∂ y( p+γz )⇒

hydrostatic pressure variation

⇒ dp

dx=0

μ d2udy2=−γ sinθ

dudy=− γ

μsin θy+c

u=− γμ

sinθ y2

2+Cy+D

dudy|y=d=0=− γ

μsin θd+c⇒c =+ γ

μsinθd

u(0) = 0 D = 0

u=− γμ

sin θ y2

2+ γ

μsin θ dy

=

γ2μ

sinθ y (2d− y )

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u(y) =g sin θ

2νy (2d− y )

q=∫0

d

udy= γ2μ

sin θ[dy 2− y3

3 ]0d

=

13

γμd3 sinθ

V avg=qd=1

3γμd2sin θ=gd2

3νsin θ

in terms of the slope So = tan sin

V=gd2 So

3ν

Exp. show Recrit 500, i.e., for Re > 500 the flow will be-come turbulent

∂ p∂ y=−γ cosθ

Recrit=

V dν

500

p=−γ cosθ y+C

discharge per unit width

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p (d )=po=−γ cosθ d+C

i.e., p=γ cosθ (d− y )+ po

* p(d) > po

* if = 0 p = (d y) + po entire weight of fluid imposed

if = /2 p = po

no pressure change through the fluid

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Chapter 9: Surface Resistanceuser.engineering.uiowa.edu/.../Chap_6_10-24-2012.docx · Web view2012/10/24 · 57:020 Mechanics of Fluids and Transport Processes Chapter 6 36 Professor