Upload
doanmien
View
262
Download
4
Embed Size (px)
Citation preview
57:020 Mechanics of Fluids and Transport Processes Chapter 6 Professor Fred Stern Fall 2006
1
1
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
57:020 Mechanics of Fluids and Transport Processes Chapter 6 Professor Fred Stern Fall 2006
2
2
Change inδ∀ :
( )u x y z tx
δ δ δ δ δ∂⎛ ⎞∀ = ⎜ ⎟∂⎝ ⎠
the rate at which the volume δ∀ is changing per unit vo-lume due to the gradient ∂u/∂x is
( ) ( )0
1 limt
d u x t udt t xδ
δ δδ δ→
∀ ∂ ∂⎡ ⎤ ∂= =⎢ ⎥∀ ∂⎣ ⎦
If velocity gradients ∂v/∂y and ∂w/∂z are also present, then using a similar analysis it follows that, in the general case,
( )1 d u v wdt x y zδ
δ∀ ∂ ∂ ∂= + + = ∇ ⋅
∀ ∂ ∂ ∂V
This rate of change of the volume per unit volume is called the volumetric dilatation rate. Angular Motion and Deformation For simplicity we will consider motion in the x–y plane, but the results can be readily extended to the more general case.
57:020 Mechanics of Fluids and Transport Processes Chapter 6 Professor Fred Stern Fall 2006
3
3
Angular motion and deformation of a fluid element
The angular velocity of line OA, ωOA, is
0limOA t tδ
δαωδ→
= For small angles
( )tanv x x t v t
x xδ δ
δα δα δδ
∂ ∂ ∂≈ = =∂
so that ( )
0limOA t
v x t vt xδ
δω
δ→
∂ ∂⎡ ⎤ ∂= =⎢ ⎥ ∂⎣ ⎦
Note that if ∂v/∂x is positive, ωOA will be counterclockwise. Similarly, the angular velocity of the line OB is
0limOB t
ut yδ
δβωδ→
∂= =∂
In this instance if ∂u/∂y is positive, ωOB will be clockwise.
57:020 Mechanics of Fluids and Transport Processes Chapter 6 Professor Fred Stern Fall 2006
4
4
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
12z
v ux y
ω ⎛ ⎞∂ ∂= −⎜ ⎟∂ ∂⎝ ⎠
Rotation of the field element about the other two coordinate axes can be obtained in a similar manner:
12x
w vy z
ω ⎛ ⎞∂ ∂= −⎜ ⎟∂ ∂⎝ ⎠
12y
u wz x
ω ∂ ∂⎛ ⎞= −⎜ ⎟∂ ∂⎝ ⎠
The three components, ωx,ωy, and ωz can be combined to give the rotation vector, ω, in the form:
1 12 2x y z curlω ω ω= + + = = ∇ ×ω i j k V V
since
1 12 2 x y z
u v w
∂ ∂ ∂∇ × =∂ ∂ ∂
i j k
V
1 1 12 2 2
w v u w v uy z z x x y
⎛ ⎞ ⎛ ⎞∂ ∂ ∂ ∂ ∂ ∂⎛ ⎞= − + − + −⎜ ⎟ ⎜ ⎟⎜ ⎟∂ ∂ ∂ ∂ ∂ ∂⎝ ⎠⎝ ⎠ ⎝ ⎠i j k
57:020 Mechanics of Fluids and Transport Processes Chapter 6 Professor Fred Stern Fall 2006
5
5
The vorticity, ζ, is defined as a vector that is twice the rota-tion vector; that is,
2ς = = ∇ ×ω V The use of the vorticity to describe the rotational characte-ristics of the fluid simply eliminates the (1/2) factor asso-ciated with the rotation vector. If 0∇ × =V , 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,
δγ δα δβ= + The rate of change of δγ is called the rate of shearing strain or the rate of angular deformation:
lim lim ⁄ ⁄
Similarly,
The rate of angular deformation is related to a correspond-ing shearing stress which causes the fluid element to change in shape.
57:020 Mechanics of Fluids and Transport Processes Chapter 6 Professor Fred Stern Fall 2006
6
6
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
∑ ∫ ∂ρ∂−=⋅ρ
CS CVVd
tAV
net outflow of mass = rate of decrease across CS of mass within CV
57:020 Mechanics of Fluids and Transport Processes Chapter 6 Professor Fred Stern Fall 2006
7
7
( ) dydzdxux
u ⎥⎦⎤
⎢⎣⎡ ρ
∂∂+ρ
outlet 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
( )flux outflux influxx ρu ρu dx dydz ρudydz
x∂⎡ ⎤= + −⎢ ⎥∂⎣ ⎦
= dxdydz)u(x
ρ∂∂
V Similarly for the y and z faces
dxdydz)w(z
z
dxdydz)v(y
y
flux
flux
ρ∂∂=
ρ∂∂=
inlet mass flux ρudydz
57:020 Mechanics of Fluids and Transport Processes Chapter 6 Professor Fred Stern Fall 2006
8
8
The total net mass outflux must balance the rate of decrease of mass within the CV which is
dxdydzt∂ρ∂−
Combining the above expressions yields the desired result
0)V(t
0)w(z
)v(y
)u(xt
0dxdydz)w(z
)v(y
)u(xt
=ρ⋅∇+∂ρ∂
=ρ∂∂+ρ
∂∂+ρ
∂∂+
∂ρ∂
=⎥⎦
⎤⎢⎣
⎡ ρ∂∂+ρ
∂∂+ρ
∂∂+
∂ρ∂
ρ∇⋅+⋅∇ρ VV
0VDtD =⋅∇ρ+ρ ∇⋅+
∂∂= VtDt
D
Nonlinear 1st order PDE; ( unless ρ = constant, then linear) Relates V to satisfy kinematic condition of mass conserva-tion Simplifications: 1. Steady flow: 0)V( =ρ⋅∇ 2. ρ = constant: 0V =⋅∇
dV
per unit V differential form of con-tinuity equations
57:020 Mechanics of Fluids and Transport Processes Chapter 6 Professor Fred Stern Fall 2006
9
9
i.e., 0zw
yv
xu =
∂∂+
∂∂+
∂∂ 3D
0yv
xu =
∂∂+
∂∂ 2D
The continuity equation in Cylindrical Polar Coordinates
The velocity at some arbitrary point P can be expressed as
r r z zv v vθ θ= + +V e e e The continuity equation:
( ) ( ) ( )1 1 0r zr v v vt r r r z
θρ ρ ρρθ
∂ ∂ ∂∂ + + + =∂ ∂ ∂ ∂
For steady, compressible flow
( ) ( ) ( )1 1 0r zr v v vr r r z
θρ ρ ρθ
∂ ∂ ∂+ + =
∂ ∂ ∂ For incompressible fluids (for steady or unsteady flow)
( )1 1 0r zrv v vr r r z
θ
θ∂ ∂ ∂+ + =
∂ ∂ ∂
57:020 Mechanics of Fluids and Transport Processes Chapter 6 Professor Fred Stern Fall 2006
10
10
The Stream Function Steady, incompressible, plane, two-dimensional flow represents one of the simplest types of flow of practical im-portance. 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
0yv
xu =
∂∂+
∂∂
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
,u vy xψ ψ∂ ∂= = −
∂ ∂
then the continuity equation is identically satisfied: 2 2
0x y y x x y x y
ψ ψ ψ ψ⎛ ⎞∂ ∂ ∂ ∂ ∂ ∂⎛ ⎞+ − = − =⎜ ⎟ ⎜ ⎟∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂⎝ ⎠⎝ ⎠
Velocity and velocity components along a streamline
57:020 Mechanics of Fluids and Transport Processes Chapter 6 Professor Fred Stern Fall 2006
11
11
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:
0d dx dy vdx udyx yψ ψψ ∂ ∂= + = − + =
∂ ∂
and, therefore, along a line of constant ψ dy vdx u
=
The flow between two streamlines
The 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.
dq udy vdx dx dy dx yψ ψ ψ∂ ∂= − = + =
∂ ∂
Thus, the volume rate of flow, q, between two streamlines such as ψ1 and ψ2, can be determined by integrating to yield:
57:020 Mechanics of Fluids and Transport Processes Chapter 6 Professor Fred Stern Fall 2006
12
12
2
12 1q d
ψ
ψψ ψ ψ= = −∫
In cylindrical coordinates the continuity equation for in-compressible, plane, two-dimensional flow reduces to
( )1 1 0rrv vr r r
θ
θ∂ ∂+ =
∂ ∂
and the velocity components, vr and vθ, can be related to the stream function, ψ(r, θ), through the equations
1 ,rv vr rθ
ψ ψθ
∂ ∂= = −∂ ∂
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
57:020 Mechanics of Fluids and Transport Processes Chapter 6 Professor Fred Stern Fall 2006
13
13
∑ V
(1) =
(2) =
= combining and making use of the continuity equation yields ∑
∑ or ∑f where ∑ ∑ ∑ ∑f ∑f ∑f
57:020 Mechanics of Fluids and Transport Processes Chapter 6 Professor Fred Stern Fall 2006
14
14
Body forces are due to external fields such as gravity or magnetics. Here we only consider a gravitational field; that is, ∑ ρ== dxdydzgFdF gravbody and kgg −= for g↓ z↑ i.e., kgf body ρ−= Surface forces are due to the stresses that act on the sides of the control surfaces symmetric (σij = σji) σij = - pδij + τij 2nd order tensor normal pressure viscous stress = -p+τxx τxy τxz
τyx -p+τyy τyz
τzx τzy -p+τzz As shown before for p alone it is not the stresses them-selves that cause a net force but their gradients.
dFx,surf = ( ) ( ) ( ) dxdydzzyx xzxyxx ⎥
⎦
⎤⎢⎣
⎡ σ∂∂+σ
∂∂+σ
∂∂
= ( ) ( ) ( ) dxdydzzyxx
pxzxyxx ⎥⎦
⎤⎢⎣
⎡ τ∂∂+τ
∂∂+τ
∂∂+
∂∂−
δij = 1 i = j δij = 0 i ≠ j
57:020 Mechanics of Fluids and Transport Processes Chapter 6 Professor Fred Stern Fall 2006
15
15
This can be put in a more compact form by defining vector stress on x-face kji xzxyxxx τ+τ+τ=τ and noting that
dFx,surf = dxdydzxp
x ⎥⎦⎤
⎢⎣⎡ τ⋅∇+
∂∂−
fx,surf = xxp τ⋅∇+
∂∂− per unit volume
similarly for y and z
fy,surf = yyp τ⋅∇+
∂∂− kji yzyyyxy τ+τ+τ=τ
fz,surf = zzp τ⋅∇+
∂∂− kji zzzyzxz τ+τ+τ=τ
finally if we define
kji zyxij τ+τ+τ=τ then
ijijsurf pf σ⋅∇=τ⋅∇+−∇= ijijij p τ+δ−=σ
57:020 Mechanics of Fluids and Transport Processes Chapter 6 Professor Fred Stern Fall 2006
16
16
Putting together the above results
Dt
VDfff surfbody ρ=+∑ =
kgf body ρ−= ijsurface pf τ⋅∇+−∇=
DV Va V VDt t
∂= = + ⋅∇∂
ˆ
ija gk pρ ρ τ= − − ∇ + ∇ ⋅ inertia body force force surface surface force due to force due due to viscous gravity to p shear and normal
stresses
57:020 Mechanics of Fluids and Transport Processes Chapter 6 Professor Fred Stern Fall 2006
17
17
For Newtonian fluid the shear stress is proportional to the rate of strain, which for incompressible flow can be written 2
where, = coefficient of viscosity
= rate of strain tensor
=
where,
Navier-Stokes Equation 0 Continuity Equation
Ex) 1-D flow
57:020 Mechanics of Fluids and Transport Processes Chapter 6 Professor Fred Stern Fall 2006
18
18
Four equations in four unknowns: V and p Difficult to solve since 2nd order nonlinear PDE
x: y: z:
0zw
yv
xu =
∂∂+
∂∂+
∂∂
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.
57:020 Mechanics of Fluids and Transport Processes Chapter 6 Professor Fred Stern Fall 2006
19
19
Ex) Exact solution for laminar incompressible steady flow in a circular pipe
Use cylindrical coordinates with assumptions 0 : Fully-developed flow 0 : Flow is parallel to the wall Continuity equation: 0 B.C. 0 0 ⇒ 0 i.e., 0
57:020 Mechanics of Fluids and Transport Processes Chapter 6 Professor Fred Stern Fall 2006
20
20
Momentum equation:
or 0 sin (1) 0 cos (2) 0 (3) where, sin cos Equations (1) and (2) can be integrated to give sin
⇒ pressure is hydrostatic and ⁄ is not a func-tion of or
57:020 Mechanics of Fluids and Transport Processes Chapter 6 Professor Fred Stern Fall 2006
21
21
Equation (3) can be written in the from 1 1
and integrated (using the fact that ⁄ = constant) to give 12
Integrating again we obtain 14 ln
B.C. 0 ∞ ⇒ 0 0 ⇒ 14
⇒ at any cross section the velocity distribution is parabolic
57:020 Mechanics of Fluids and Transport Processes Chapter 6 Professor Fred Stern Fall 2006
22
22
1) Flow rate : 2 8
where, 2 If the pressure drops Δ over a length ℓ: ℓ Δ8 ℓ
2) Mean velocity : 1 Δ8 ℓ Δ8 ℓ
3) Maximum velocity : 0 4 Δ4 ℓ 2
⇒ 1
57:020Profes
Dif WeStok(aboconstanlutitern10)Finsis. Cou
Firslel p
Con
Mo
or b
0 Mechanics ossor Fred Stern
fferenti
e now dikes equout 80)
nditions, nding ofons. Ac
nal flow, but theally, the See the
uette Flo
st, considplates
ntinuity
omentum
by CV co
f Fluids and Trn Fall 2006
ial Ana
scuss a uations.
are for they are
f the chactually t
w (Chaptey serve e derivate text for
ow
der flow
m
ontinuity
ransport Proce
alysis of
couple o Althouhighly
e very varacter othe examter 8) anto unde
tions to r derivat
bo
w due to t
0xu =
∂∂
dyd0 μ=
y and mo
sses
f Fluid
of exact ugh all simplifi
valuable f the NS
mples to nd open
erscore afollow u
tions usin
oundary
the relati
2
2
yu
omentum
Flow
solutionknown
ied geomas an a
S equatiobe discu
n channeand displutilize dng CV a
conditio
ive moti
m equati
u = u(yv = o
xp
∂∂=
∂∂
ns to theexact
metries aaid to ouons and ussed arel flow lay viscoifferenti
analysis.
ons
ion of tw
ons:
y)
0yp =
∂∂
Chapter 23
2
e Naviersolutionand flowur undertheir so
re for in(Chapte
ous flowal analy
wo paral-
6
23
r-ns w r-o-n-er w. y-
-
57:020Profes
u1ρu1 = ∑F
i.e.
from
dyduμ
u =
u =
=τ
0 Mechanics ossor Fred Stern
uy1 ρ=Δ= u2
∑ ρ= uFx
ypΔ=
0dyd =τ
ddyd⎜⎝
⎛μ
dyud2
2μ
m mome
Cyu =
DyC +μ
=
u(0) =
u(t) = U
ytU=
μ=μdydu
f Fluids and Trn Fall 2006
yu2Δ
=⋅ρ AdV
ddp⎜
⎝⎛ +−
0
0dydu =⎟
⎠
⎞
0=
entum eq
D
0 ⇒ D
U ⇒ C =
=μtU con
ransport Proce
(ρ= uQ 2
yxdxdp Δ⎟
⎠⎞Δ
quation
= 0
= tUμ
nstant
sses
) =− 0u1
xy +Δτ−
0
dyd
⎜⎝
⎛ τ+τ+
xdyy
Δ⎟⎠
⎞τ =
Chapter 24
2
= 0
6
24
57:020Profes
Gengrad
Con
Mo
i.e.,
whi
0 Mechanics ossor Fred Stern
neralizatdient
ntinutity
omentum
, μ
ich can b
μ
f Fluids and Trn Fall 2006
tion for i
y
m
dyud2
2γ=μ
be integr
dd
dydu γ=μ
ransport Proce
inclined
0xu =
∂∂
x0
∂∂−=
dxdhγ
rated twi
Aydxdh +
sses
flow wi
( zpx
γ+∂
h = p
plates
plates
ice to yie
ith a con
) 2
2
dyudμ+
/γ +z = c
s horizon
s vertica
eld
nstant pre
2u
constant
ntal dxdz =
al dxdz =-1
uv
Chapter 25
2
essure
0=
1
u = u(y)v = o
0yp =
∂∂
6
25
57:020 Mechanics of Fluids and Transport Processes Chapter 6 Professor Fred Stern Fall 2006
26
26
BAy2y
dxdhu
2++γ=μ
now apply boundary conditions to determine A and B u(y = 0) = 0 ⇒ B = 0 u(y = t) = U
2t
dxdh
tUAAt
2t
dxdhU
2γ−μ=⇒+γ=μ
⎥⎦⎤
⎢⎣⎡ γ−μ
μ+
μγ=
2t
dxdh
tU1
2y
dxdh)y(u
2
= ( ) ytUyty
dxdh
22 +−
μγ−
This equation can be put in non-dimensional form:
ty
ty
ty1
dxdh
U2t
Uu 2
+⎟⎠⎞
⎜⎝⎛ −
μγ−=
define: P = non-dimensional pressure gradient
= dxdh
U2t2
μγ− zph +
γ=
Y = y/t ⎥⎦
⎤⎢⎣
⎡ +γμ
γ−=dxdz
dxdp1
U2z2
⇒ Y)Y1(YPUu +−⋅=
parabolic velocity profile
57:020Profes
u
q
=
=
=Uut
Uu =
0 Mechanics ossor Fred Stern
Uu
U
tq
udy
t
0
t
0
∫==
∫=
∫ ⎢⎣⎡ −=
t
0y
tP
21
6P
⇒+=
f Fluids and Trn Fall 2006
tPy
Uu −=
[ ]t
dy
+− 22 y
tP
12tu =⇒
ransport Proce
ty
tPy
2
2+−
⎥⎦⎤+ dy
ty
dd
2t2
⎜⎝⎛ γ−
μ
sses
ty
=2Pt −
2U
dxdh +⎟
⎠⎞
2t
3Pt +−
Chapter 27
2
6
27
57:020 Mechanics of Fluids and Transport Processes Chapter 6 Professor Fred Stern Fall 2006
28
28
For laminar flow 1000tu <ν
Recrit ∼ 1000
The maximum velocity occurs at the value of y for which:
0dydu =
t1y
tP2
tP0
Uu
dyd
2 +−==⎟⎠⎞
⎜⎝⎛
( )P2t
2t1P
P2ty +=+=⇒ @ umax
( )P4
U2U
4UPyuu maxmax ++==∴
note: if U = 0: 32
4P
6P
uu
max==
The shape of the velocity profile u(y) depends on P:
1. If P > 0, i.e., 0dxdh < the pressure decreases in the
direction of flow (favorable pressure gradient) and the velocity is positive over the entire width
θγ−=⎟⎠
⎞⎜⎝
⎛ +γ
γ=γ sindxdpzp
dxd
dxdh
a) 0dxdp <
for U = 0, y = t/2
57:020 Mechanics of Fluids and Transport Processes Chapter 6 Professor Fred Stern Fall 2006
29
29
b) θγ< sindxdp
1. If P < 0, i.e., 0>dxdh 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.
0sindxdp >θγ−
θγ> sindxdp or
dxdpsin <θγ
2. If P = 0, i.e., 0dxdh = the velocity profile is linear
ytUu =
a) 0dxdp = and θ = 0
b) θγ= sindxdp
For U = 0 the form ( ) YY1PYUu +−= is not appropriate
u = UPY(1-Y)+UY
= ( ) UYY1Ydxdh
2t2
+−μ
γ−
Note: we derived this special case
57:020 Mechanics of Fluids and Transport Processes Chapter 6 Professor Fred Stern Fall 2006
30
30
Now let U = 0: ( )Y1Ydxdh
2tu
2−
μγ−=
3. Shear stress distribution Non-dimensional velocity distribution
( )* 1uu P Y Y YU
= = ⋅ − +
where * uuU
≡ is the non-dimensional velocity,
2
2t dhPU dx
γμ
≡ − is the non-dimensional pressure gradient
yYt
≡ is the non-dimensional coordinate. Shear stress
dudy
τ μ=
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:
*
212U
ττρ
=
Then
( )( )
**
2
1 212
Ud u U dutd y t Ut dYU
μτ μρρ
= =
( )2 2 1PY PUtμ
ρ= − + +
( )2 2 1PY PUtμ
ρ= − + +
( )2 1A PY P= − + +
57:020 Mechanics of Fluids and Transport Processes Chapter 6 Professor Fred Stern Fall 2006
31
31
where 2 0AUtμ
ρ≡ > is a positive constant.
So the shear stress always varies linearly with Y across any section. At the lower wall ( )0Y = :
( )* 1lw A Pτ = +
At the upper wall ( )1Y = :
( )* 1uw A Pτ = − For favorable pressure gradient, the lower wall shear stress is always positive: 1. For small favorable pressure gradient ( )0 1P< < : * 0lwτ > and * 0uwτ > 2. For large favorable pressure gradient ( )1P > : * 0lwτ > and * 0uwτ < ( )0 1P< < ( )1P > For adverse pressure gradient, the upper wall shear stress is always positive: 1. For small adverse pressure gradient ( )1 0P− < < : * 0lwτ > and * 0uwτ >
τ τ
57:020 Mechanics of Fluids and Transport Processes Chapter 6 Professor Fred Stern Fall 2006
32
32
2. For large adverse pressure gradient ( )1P < − : * 0lwτ < and * 0uwτ >
57:020 Mechanics of Fluids and Transport Processes Chapter 6 Professor Fred Stern Fall 2006
33
33
( )1 0P− < < ( )1P < − For 0U = , i.e., channel flow, the above non-dimensional form of velocity profile is not appropriate. Let’s use dimen-sional form:
( ) ( )2
12 2t dh dhu Y Y y t ydx dx
γ γμ μ
= − − = − −
Thus the fluid always flows in the direction of decreasing piezometric pressure or piezometric head because
0, 02
yγμ
> > and 0t y− > . So if dhdx is negative, u is posi-
tive; if dhdx is positive, u is negative.
Shear stress:
1
2 2du dh t ydy dx
γτ μ ⎛ ⎞= = − −⎜ ⎟⎝ ⎠
Since 1 02
t y⎛ ⎞− >⎜ ⎟⎝ ⎠
, the sign of shear stress τ is always oppo-
site to the sign of piezometric pressure gradient dhdx , and the magnitude of τ is always maximum at both walls and zero at centerline of the channel.
τ τ
57:020Profes
Flow
unif
Con
0 Mechanics ossor Fred Stern
For fav
For ad
w down
form flo
ntinuity
f Fluids and Trn Fall 2006
vorable p
dverse pr
dhdx
<
an incli
ow ⇒ ve ch
dxdu =
ransport Proce
pressure
ressure g
0
ined plan
locity anange in x
0=
τ
sses
e gradien
gradient,
ne
nd depthx-directi
nt, 0dhdx
<
0dhdx
> , τ
h do notion
, 0τ >
0τ <
dd
Chapter 34
3
0dhdx
>
τ
6
34
57:020 Mechanics of Fluids and Transport Processes Chapter 6 Professor Fred Stern Fall 2006
35
35
x-momentum ( ) 2
2
dyudzp
x0 μ+γ+
∂∂−=
y-momentum ( )⇒γ+∂∂−= zpy
0 hydrostatic pressure variation
0dxdp =⇒
θγ−=μ sindy
ud2
2
cysindydu +θ
μγ−=
DCy2ysinu
2++θ
μγ−=
dsinccdsin0dydu
dy
θμγ+=⇒+θ
μγ−==
=
u(0) = 0 ⇒ D = 0
dysin2ysinu
2θ
μγ+θ
μγ−=
= ( )yd2ysin2
−θμγ
57:020 Mechanics of Fluids and Transport Processes Chapter 6 Professor Fred Stern Fall 2006
36
36
u(y) = ( )yd2y2sing −
νθ
d
0
32
d
0 3ydysin
2udyq ⎥
⎦
⎤⎢⎣
⎡−θ
μγ=∫=
= θμγ sind
31 3
θν
=θμγ== sin
3gdsind
31
dqV
22
avg
in terms of the slope So = tan θ ∼ sin θ
ν
=3
SgdV o2
Exp. show Recrit ∼ 500, i.e., for Re > 500 the flow will be-come turbulent
θγ−=∂∂ cosyp
ν= dVRecrit ∼ 500
Cycosp +θγ−= ( ) Cdcospdp o +θγ−==
discharge per unit width
57:020 Mechanics of Fluids and Transport Processes Chapter 6 Professor Fred Stern Fall 2006
37
37
i.e., ( ) opydcosp +−θγ= * p(d) > po * if θ = 0 p = γ(d − y) + po
entire weight of fluid imposed if θ = π/2 p = po no pressure change through the fluid