ME3560 – Fluid Mechanics
Summer 2016
ME 3560 Fluid Mechanics
Chapter VI. Differential Analysis of
Fluid Flow
Chapter VI. Differential
Analysis of Fluid Flow
1
ME3560 – Fluid Mechanics
Summer 2016
6.1 Fluid Element Kinematics
• In general a fluid particle can undergo translation, linear deformation
rotation and angular deformation.
Chapter VI. Differential
Analysis of Fluid Flow
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ME3560 – Fluid Mechanics
Summer 2016
Translation
• The simplest type of motion that a fluid element can undergo is
translation.
•In a small time interval δt a particle located at point O will move to
point O′.
•If all points in the element have the same velocity (which is only true if
there are no velocity gradients), then the element will simply translate
from one position to another.
•In general, there are velocity gradients
which result in deformation and rotation
of the element as it moves.
Chapter VI. Differential
Analysis of Fluid Flow
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ME3560 – Fluid Mechanics
Summer 2016
Linear Deformation
•Consider the effect of a single velocity gradient, ∂u/∂x, on a small cube
having sides δx, δy, and δz.
•Let the x component of velocity of O and B is u.
•Then at nearby points A and C the x component of the velocity can be
expressed as u + (∂u/∂x) δx.
•This
difference
in
velocity
causes
a
“stretching” of the volume
element by an amount
(∂u/∂x)(δx)(δt) during the
short time interval δt in
which line OA stretches to
OA′ and BC to BC′.
Chapter VI. Differential
Analysis of Fluid Flow
Summer 2016
ME3560 – Fluid Mechanics
•The corresponding change in the original volume, δV= δxδyδz is
δ u
Change in δ V = 
δ
δ x

x (δ yδ z )(δ t )

•The rate at which the volume, δV, is changing per unit volume due to
the gradient ∂u/∂x is
 (∂ u / ∂ x ) δ t  ∂ u
1 d (δ V )
δV
dt
= lim 
δ t →0

δt
 =
 ∂x
1 d (δ V ) ∂ u
=
δ V dt
∂x
Chapter VI. Differential
Analysis of Fluid Flow
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ME3560 – Fluid Mechanics
Summer 2016
•Similar analysis for the y– and z–directions with velocity gradients
∂v/∂y and ∂w/∂z, respectively results in the following expression for the
rate at which the volume, δV, is changing per unit volume:
r
1 d (δ V ) ∂ u ∂ v ∂ w
=
+
+
= ∇ ⋅V
δ V dt
∂x ∂y ∂z
•This rate of change of the volume per unit volume is called the
volumetric dilatation rate.
•The volume of a fluid may change as the element moves from one
location to another in the flow field.
•For an incompressible fluid the volumetric dilatation rate is zero
Chapter VI. Differential
Analysis of Fluid Flow
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ME3560 – Fluid Mechanics
Summer 2016
•Variations in the velocity in the direction of the component itself, as
represented by the derivatives ∂u/∂x, ∂v/∂y, and ∂w/∂z, simply cause a
linear deformation of the element.
•Cross derivatives, such as ∂u/∂y and ∂v/∂x, will cause the element to
rotate and generally to undergo an angular deformation, which changes
the shape of the element.
Chapter VI. Differential
Analysis of Fluid Flow
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ME3560 – Fluid Mechanics
Summer 2016
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
ωOA
δα
= lim
δ t →0 δ t
•For small angles:
(∂ v / ∂ x)δ xδ t ∂ v
tan δ α ≈ δ α =
=
δt
δx
∂x
Chapter VI. Differential
Analysis of Fluid Flow
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Summer 2016
ME3560 – Fluid Mechanics
•Therefore the angular velocity of segment OA is
ωOA
 (∂ v / ∂ x)δ t  ∂ v
= lim 
=

δ t →0
δt

 ∂x
∂v
ωOA =
∂x
• If ∂ v/ ∂ x is positive ωOA is counterclockwise.
Chapter VI. Differential
Analysis of Fluid Flow
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Summer 2016
ME3560 – Fluid Mechanics
•For segment OB the angular velocity is
δβ
ωOB = lim
δ t →0 δ t
•Where
(∂ u / ∂ y )δ yδ t ∂ u
tan δ β ≈ δ β =
=
δt
δy
∂y
(∂ u / ∂ y )δ t
ωOB = lim
δ t →0
δt
ωOB
∂u
=
∂y
• If ∂ u/ ∂ y is positive ωOB is clockwise.
Chapter VI. Differential
Analysis of Fluid Flow
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Summer 2016
ME3560 – Fluid Mechanics
•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:
1  ∂v ∂u 

−
ω z = 
2∂x ∂ y
•In a similar analysis the angular velocities in the x and y directions can
be found to be:
1 ∂w ∂v 

ω x = 
−
2∂ y ∂z
Chapter VI. Differential
Analysis of Fluid Flow
1  ∂u ∂ w

ω y = 
−
2∂z ∂x 
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ME3560 – Fluid Mechanics
r
Summer 2016
r 1
r
1
ω = curlV = ∇ × V
2
2
ˆj
iˆ
kˆ
r
ω = ∂/∂ x ∂/∂ y ∂/∂z
ω = ω x iˆ + ω y ˆj + ω z kˆ
u
r
v
z
1  ∂ w ∂ v ˆ 1  ∂ u ∂ w  ˆ 1  ∂ v ∂ u  ˆ
i + 
 j + 
k
−
−
−
ω = 
2  ∂ y ∂ z  2  ∂ z ∂ x r 2  ∂ x ∂ y 
r
•ξ is know as the vorticity: ξ = 2ω = ∇ × V
r
•If ∇×V=0: the angular velocity and vorticity are zero and an
irrotational flow is present.
•If ω = 0 (Irrotational flow) and then it is a POTENTIAL FLOW where:
Chapter VI. Differential
Analysis of Fluid Flow
r
V = ∇Φ
•Φ is a scalar function.
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Summer 2016
ME3560 – Fluid Mechanics
•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.
δγ = δ α + δβ
•δγ is positive if the original right angle is decreasing.
•The rate of change of δγ is called the rate of shearing strain or the rate
of angular deformation:
 (∂ v / ∂ x)δ t + (∂ u / ∂ y )δ t 
δγ
= lim 
γ& = lim

δ t →0 δ t
δ t →0
δt


Chapter VI. Differential
Analysis of Fluid Flow
∂v ∂u
γ& =
+
∂x ∂y
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Summer 2016
ME3560 – Fluid Mechanics
6.2 Conservation of Mass
Mass accumulated
Mass flow rate
=
per unit time within
entering the CV –
the CV
Mass
accumulated
per unit time=
within the
CV
Net mass
flow rate =
crossing the
CS
Chapter VI. Differential
Analysis of Fluid Flow
Mass flow rate
leaving the CV
∂m ∂ρ
=
δ xδ yδ z
∂t
∂t
 ∂ ( ρ u ) ∂ ( ρ u ) ∂ ( ρ w) 
−
δ xδ yδ z
+
+

∂y
∂z 
 ∂x
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Summer 2016
ME3560 – Fluid Mechanics
•Thus, the differential form of the mass conservation equation is:
∂ ρ ∂ ( ρ u ) ∂ ( ρ v) ∂ ( ρ w)
+
+
+
=0
∂t
∂x
∂y
∂z
r
∂ρ
+ ∇ ⋅ (ρ V ) = 0
∂t
•If the flow is incompressible:
∂ u ∂v ∂w
+
+
=0
∂x ∂y ∂z
r
∇ ⋅V = 0
Chapter VI. Differential
Analysis of Fluid Flow
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Summer 2016
ME3560 – Fluid Mechanics
Conservation of Mass – Cylindrical Coordinates
•Velocity in Cartesian coordinates:
r
V = u iˆ + v ˆj + w kˆ
•Velocity in Cylindrical coordinates:
r
V = vr eˆr + vθ eˆθ + vz eˆz
•Conservation of mass in cylindrical coordinates
∂ ρ 1 ∂ (r ρ vr ) 1 ∂ ( ρ vθ ) ∂ ( ρ vz )
+
+
+
=0
∂t r
∂r
r ∂θ
∂z
•For incompressible flow:
Chapter VI. Differential
Analysis of Fluid Flow
1 ∂ (r vr ) 1 ∂ vθ ∂ vz
+
+
=0
r ∂r
r ∂θ ∂ z
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Summer 2016
ME3560 – Fluid Mechanics
The Stream Function
•Assuming steady, incompressible, two-dimensional flow, the continuity
∂u ∂v
equation becomes:
∂x
+
∂y
=0
•Introduce a continuous scalar function ψ(x, y),
stream function, such that:
∂ψ
u=
∂y
∂ψ
v=−
∂x
•This definition of u and v in terms of ψ satisfies the continuity equation.
∂ψ
∂ψ
dψ =
dx +
dy
∂x
∂y
dψ = −vdx + udy
dx dy
•If ψ = constant, the equation of stream lines is obtained:
=
u
v
•Thus, ψ = constant lines are STREAMLINES
Chapter VI. Differential
Analysis of Fluid Flow
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ME3560 – Fluid Mechanics
Summer 2016
•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–y plane)
passing between the two streamlines.
•Note that flow never crosses streamlines,
since by definition the velocity is tangent to
the streamline.
Chapter VI. Differential
Analysis of Fluid Flow
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Summer 2016
ME3560 – Fluid Mechanics
•From conservation of mass, the inflow, dq,
crossing the arbitrary surface AC must equal
the net outflow through surfaces AB and BC:
dq = udy − vdx
∂ψ
∂ψ
dq =
dy +
dx
∂y
∂x
dq = dψ
ψ2
dq = ∫ dψ = ψ 2 −ψ 1
ψ1
•The volumetric flow rate is the relative
difference between two stream function values.
Chapter VI. Differential
Analysis of Fluid Flow
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Summer 2016
ME3560 – Fluid Mechanics
Stream Function: Steady, Compressible Flow
∂ψ
ρu =
∂y
∂ψ
ρv=−
∂x
Stream Function–Polar Coordinates ψ =ψ(r, θ)
•In cylindrical coordinates the continuity equation for incompressible,
plane, two-dimensional flow reduces to:
1 ∂ (rvr ) 1 ∂ vθ
+
=0
r ∂r
r ∂θ
•And ψ is defined as:
Chapter VI. Differential
Analysis of Fluid Flow
1 ∂ψ
vr =
r ∂θ
∂ψ
vθ = −
∂r
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ME3560 – Fluid Mechanics
Summer 2016
6.3 Conservation of Linear Momentum
•As determined in the previous chapter, the momentum equation is:
r
r r
r
∂
ρ V dV + ∫ ρ V (V ⋅ nˆ ) dA = ∑ F
∫
∂ t cv
cs
•Next, this equation is applied to an infinitesimal element of fluid.
•The right hand side of the equation can be written as
r
r
r
 ∂V
ρ 
+ (V ⋅ ∇)V 

 ∂t
Chapter VI. Differential
Analysis of Fluid Flow
21
ME3560 – Fluid Mechanics
Summer 2016
6.3 Conservation of Linear Momentum
•Surface forces acting on a differential element of fluid in the x–direction:
 ∂ σ xx ∂ τ yx ∂ τ zx 
δ xδ yδ z
δ Fsx = 
+
+
∂y
∂z 
 ∂x
Chapter VI. Differential
Analysis of Fluid Flow
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ME3560 – Fluid Mechanics
Summer 2016
•A similar analysis for the y– and z–directions yields:
 ∂ τ xy ∂ σ yy ∂ τ zy 
δ xδ yδ z
δ Fsy = 
+
+
∂y
∂z 
 ∂x
 ∂ τ xz ∂ τ yz ∂ σ zz 
δ xδ yδ z
δ Fsz = 
+
+
∂y
∂z 
 ∂x
Chapter VI. Differential
Analysis of Fluid Flow
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ME3560 – Fluid Mechanics
Summer 2016
6.3 Conservation of Linear Momentum
•The momentum applied to a control volume states that the forces acting
on the CV equal the momentum accumulated in the CV minus the net
flux of momentum crossing the CS.
 ∂u
∂ σ xx ∂ τ yx ∂ τ zx
∂u
∂u
∂u 

ρg x +
+
+
= ρ 
+u
+v
+w
∂x
∂y
∂z
∂x
∂y
∂z
 ∂t
∂ τ xy ∂ σ yy ∂ τ zy
∂v
∂v
∂v
∂v 
ρg y +
+
+
= ρ 
+u
+v
+ w 
∂x
∂y
∂z
∂x
∂y
∂z
 ∂t
∂w
∂ τ xz ∂ τ yz ∂ σ zz
∂w
∂w
∂w

ρg z +
+
+
= ρ 
+u
+v
+w
∂x
∂y
∂z
∂x
∂y
∂z 
 ∂t
Chapter VI. Differential
Analysis of Fluid Flow
24
Summer 2016
ME3560 – Fluid Mechanics
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.
− p = σ xx = σ yy = σ zz
Chapter VI. Differential
Analysis of Fluid Flow
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ME3560 – Fluid Mechanics
Summer 2016
Euler’s Equations of Motion
•For an inviscid flow, τ = 0, and σxx = σyy = σzz= −p, the general
equations of motion become
 ∂u
∂p
∂u
∂u
∂u 

ρg x −
= ρ 
+u
+v
+w
∂x
∂x
∂y
∂z
 ∂t
∂v
∂p
∂v
∂v
∂v 
ρg y −
= ρ 
+u
+v
+ w 
∂y
∂x
∂y
∂z
 ∂t
∂w
∂p
∂w
∂w
∂w

ρg z −
= ρ 
+u
+v
+w
∂z
∂
x
∂
y
∂
z
 ∂t

r
r r
 ∂V
r
ρg − ∇p = ρ 
+ (∇ ⋅ V )V 
 ∂t

Chapter VI. Differential
Analysis of Fluid Flow
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Summer 2016
ME3560 – Fluid Mechanics
Potential Flow
•For an irrotational flow the angular velocity is zero ω = ½∇×V = 0.
•Therefore it is possible to define the velocity in terms of a scalar
r
function Φ(x, y, z, t) as
V = ∇Φ
•Notice that for irrotational flow:
r
∇ × V = ∇ × ∇Φ = 0
•Therefore this definition of the velocity in terms of Φ satisfies the
condition of irrotationallity.
•Φ is called velocity potential.
∂Φ
u=
∂x
Chapter VI. Differential
Analysis of Fluid Flow
∂Φ
v=
∂y
∂Φ
w=
∂z
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Summer 2016
ME3560 – Fluid Mechanics
•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
∂2 Φ ∂2 Φ ∂2 Φ
+
+
=0
2
2
2
∂x
∂y
∂z
∇ Φ=0
2
•The previous equation is the governing equation for potential flow.
•∇⋅∇( )=∇2( ) is called Laplacian operator.
Chapter VI. Differential
Analysis of Fluid Flow
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Summer 2016
ME3560 – Fluid Mechanics
•In cylindrical coordinates the Del operator is:
∂()
∂()
1 ∂()
∇( ) =
eˆr +
eˆθ +
eˆz
∂r
r ∂θ
∂z
•Thus:
•Since
∂Φ
∂Φ
1 ∂Φ
∇Φ =
eˆr +
eˆθ +
eˆz
∂r
r ∂θ
∂z
r
V = vr eˆr + vθ eˆθ + vz eˆz
•It follows
∂Φ
vr =
∂r
1 ∂Φ
vθ =
r ∂r
∂Φ
vz =
∂z
•Thus, the Laplace equation in cylindrical coordinates is:
1 ∂  ∂ Φ  1 ∂2 Φ ∂2 Φ
 r
 + 2
+
=0
2
2
r ∂ r  ∂ r  r ∂θ
∂z
Chapter VI. Differential
Analysis of Fluid Flow
29
Summer 2016
ME3560 – Fluid Mechanics
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 flow
∂Φ
u=
∂x
Chapter VI. Differential
Analysis of Fluid Flow
∂ψ
u=
∂y
∂Φ
v=
∂y
∂ψ
v=−
∂x
∂2 Φ ∂2 Φ
+
=0
2
2
∂x
∂y
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Summer 2016
ME3560 – Fluid Mechanics
•If the flow is irrotational (ωz = ∂ v/ ∂ x – ∂ u/ ∂ y = 0)
∂v ∂u
−
=0
∂x ∂y
∂  ∂ψ  ∂  ∂ψ 
 = 0

 −
 −
∂x ∂x  ∂y ∂y 
∂2ψ ∂2ψ
+
=0
2
2
∂x
∂y
•As shown next Φ(x, y) and ψ(x, y) are orthogonal functions
∂Φ
∂Φ
dΦ =
dx +
dy = udx + vdy
∂x
∂y
∂ψ
∂ψ
dψ =
dx +
dy = −vdx + udy
∂x
∂y
Chapter VI. Differential
Analysis of Fluid Flow
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ME3560 – Fluid Mechanics
Summer 2016
•For Φ(x, y) = const and ψ(x, y) = const
dy
u
dΦ = udx + vdy = 0
=−
dx
v
dy u
dψ = −vdx + udy = 0
=
dx v
•Thus, for Φ = const and ψ = const the lines are perpendicular generating
a “Flow Net”
Chapter VI. Differential
Analysis of Fluid Flow
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Summer 2016
ME3560 – Fluid Mechanics
Uniform Flow Φ = U ( x cos α + y sin α )
ψ = U ( y cos α − x sin α )
∂Φ
u=
∂x
∂Φ
v=
∂y
Chapter VI. Differential
Analysis of Fluid Flow
∂ψ
u=
∂y
∂ψ
v=−
∂x
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Summer 2016
ME3560 – Fluid Mechanics
Source and Sink (Radial Flow)
m
vr =
2π r
vθ = 0
• m is the source/sink strength
•Its units are m2/s (volume flow rate per
unit width
1 ∂ψ
∂Φ
1 ∂Φ
vr =
vθ =
= 0 vr =
r ∂θ
∂r
r ∂θ
m
Φ=
ln r + c
2π
∂ψ
vθ = −
=0
∂r
m
ψ = θ +c
2π
Chapter VI. Differential
Analysis of Fluid Flow
34
ME3560 – Fluid Mechanics
Summer 2016
Ideal Vortex
• k is the vortex strength
k
vr = 0 vθ =
r
∂ψ
1 ∂Φ
vθ = −
=0
vθ =
∂r
r ∂r
Φ = kθ
ψ = −k ln r
Chapter VI. Differential
Analysis of Fluid Flow
• 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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Summer 2016
ME3560 – Fluid Mechanics
•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
r r
r r
Γ = ∫ V ⋅ ds
V ⋅ ds = udx + vdy
c
∂Φ
∂Φ
dΦ =
dx +
dy = udx + vdy
∂x
∂y
Γ = ∫ dΦ
•For a free vortex
2π
c
Γ = ∫ kdθ = 2π k
0
•Then ψ and Φ are expressed as:
Chapter VI. Differential
Analysis of Fluid Flow
Γ
Φ=
θ
2π
Γ
ψ = − ln r
2π
Γ is often useful when evaluating the forces
developed on bodies immersed in moving fluids.
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Summer 2016
ME3560 – Fluid Mechanics
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:
sin θ
cosθ
ma
ψ = −k
Chapter VI. Differential
Analysis of Fluid Flow
r
Φ=k
r
k=
π
37
ME3560 – Fluid Mechanics
Summer 2016
•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
Chapter VI. Differential
Analysis of Fluid Flow
38
Summer 2016
ME3560 – Fluid Mechanics
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:
∂u
∂w
∂v
σ xx = − p + 2µ
; σ yy = − p + 2µ
; σ zz = − p + 2µ
∂x
∂z
∂y
• Shear stresses:
τ xy = τ yx
 ∂u ∂v 
;
= µ 
+
∂ y ∂x
τ yz
 ∂v ∂w
;
= τ zy = µ 
+
∂z ∂ y 
 ∂ w ∂u 
;
τ xz = τ zx = µ 
+
∂x ∂z
Chapter VI. Differential
Analysis of Fluid Flow
39
ME3560 – Fluid Mechanics
Summer 2016
• Substituting these previous expressions into the momentum equation
(differential form) shown below:
 ∂u
∂ σ xx ∂ τ yx ∂ τ zx
∂u
∂u
∂u 

ρg x +
+
+
= ρ 
+u
+v
+w
∂x
∂y
∂z
∂x
∂y
∂z
 ∂t
∂ τ xy ∂ σ yy ∂ τ zy
∂v
∂v
∂v
∂v 
ρg y +
+
+
= ρ 
+u
+v
+ w 
∂x
∂y
∂z
∂x
∂y
∂z
 ∂t
∂w
∂ τ xz ∂ τ yz ∂ σ zz
∂w
∂w
∂w

ρg z +
+
+
= ρ 
+u
+v
+w
∂x
∂y
∂z
∂x
∂y
∂z 
 ∂t
• The Navier – Stokes Equations are determine as:
r
r
r
DV
2
ρ
= −∇p + ρ g + µ∇ V
Dt
Chapter VI. Differential
Analysis of Fluid Flow
40
ME3560 – Fluid Mechanics
Summer 2016
 ∂2 u ∂2 u ∂2 u 
 ∂u
∂u
∂u
∂u 
∂p
 = −
+v
+w
+ ρg x + µ  2 +
+ 2 
ρ  + u
2
∂x
∂y
∂z
∂x
∂z 
 ∂t
∂x ∂ y
 ∂2 v ∂2 v ∂2 v 
∂v
∂v
∂v
∂v 
∂p
ρ  + u
+v
+ w  = −
+ ρg y + µ  2 +
+ 2 
2
∂x
∂y
∂z
∂y
∂z 
 ∂t
∂x ∂ y
 ∂2w ∂2w ∂2w 
∂w
∂w
∂w
∂w
∂p
 = −
ρ 
+u
+v
+w
+ ρg z + µ  2 + 2 + 2 
∂x
∂y
∂z 
∂z
 ∂t
∂x ∂ y ∂z 
Chapter VI. Differential
Analysis of Fluid Flow
41
ME3560 – Fluid Mechanics
Summer 2016
6.9 Some Simple Solutions for Laminar,
Viscous, Incompressible Fluids
• Steady, Laminar Flow between Fixed Parallel Plates
• 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
Chapter VI. Differential
Analysis of Fluid Flow
42
ME3560 – Fluid Mechanics
Summer 2016
r
r
r
DV
2
ρ
= −∇p + ρ g + µ∇ V
Dt
 ∂2 u ∂2 u ∂2 u 
 ∂u
∂u
∂u
∂u 
∂p
ρ  + u
+v
+ w  = −
+ ρg x + µ  2 +
+ 2 
2
∂x
∂y
∂z
∂x
∂z 
 ∂t
∂x ∂ y
 ∂2 v ∂2 v ∂2 v 
∂v
∂v
∂v
∂v 
∂p
+v
+ w  = −
ρ  + u
+ ρg y + µ  2 +
+ 2 
2
∂x
∂y
∂z
∂y
∂z 
 ∂t
∂x ∂ y
 ∂2w ∂2w ∂2w 
∂w
∂w
∂w
∂w
∂p
 = −
+u
+v
+w
ρ 
+ ρg z + µ  2 + 2 + 2 
∂x
∂y
∂z 
∂z
 ∂t
∂x ∂ y ∂z 
Chapter VI. Differential
Analysis of Fluid Flow
43
Summer 2016
ME3560 – Fluid Mechanics
• 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:
∂u
=0
∂x
∂p
∂2 u
0=−
+µ
∂x
∂ y2
Chapter VI. Differential
Analysis of Fluid Flow
u (0) = 0
u (b) = U
44
ME3560 – Fluid Mechanics
Summer 2016
• 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–Poiseuille or
Poiseuille flow).
•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.
•Under the conditions described above, the continuity equation:
∂ ρ 1 ∂ (r ρ vr ) 1 ∂ ( ρ vθ ) ∂ ( ρ vz )
+
+
+
=0
∂t r
∂r
r ∂θ
∂z
∂ vz
=0
•Becomes:
Chapter VI. Differential
∂z
Analysis of Fluid Flow
45
Summer 2016
ME3560 – Fluid Mechanics
•Whereas the momentum equation in cylindrical coordinates
•Results in:
∂p
r: 0 = − + ρg r
∂r
1∂p
θ :0 = −
+ ρgθ
r ∂θ
Chapter VI. Differential
Analysis of Fluid Flow
z:
 1 ∂  ∂ v z 
∂p
 r

0=−
+ µ
∂z
 r ∂ r  ∂ r 
46