Pharos University ME 253 Fluid Mechanics 2
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Transcript Pharos University ME 253 Fluid Mechanics 2
Pharos University
ME 253 Fluid Mechanics 2
Revision for Mid-Term Exam
Dr. A. Shibl
Streamlines
• A Streamline is a curve that is
everywhere tangent to the
instantaneous local velocity
vector.
• Consider an arc length
dr dxi dyj dzk
•
dr must be parallel to the local
velocity vector
V ui vj wk
• Geometric arguments results in
the equation for a streamline
dr dx dy dz
V
u
v
w
Kinematics of Fluid Flow
1 w v
1 u w
1 v u
i j k
2 y z
2 z x
2 x y
2
1 DV
u v w
xx yy zz
V Dt
x y z
1 u v
1 w u
1 v w
xy , zx , yz
2 y x
2 x z
2 z y
Stream Function for
Two-Dimensional
Incompressible Flow
• Two-Dimensional Flow
Stream Function y
Stream Function for
Two-Dimensional
Incompressible Flow
• Cylindrical Coordinates
Stream Function y(r,q)
Is this a possible flow field
Given the y-component
Find the X- Component of the velocity,
Determine the vorticity of flow field described by
Is this flow irrotational?
Momentum Equation
• Newtonian Fluid: Navier–Stokes Equations
Example exact solution
Poiseuille Flow
Example exact solution
Fully Developed Couette Flow
• For the given geometry and BC’s, calculate the velocity and
pressure fields, and estimate the shear force per unit area
acting on the bottom plate
• Step 1: Geometry, dimensions, and properties
Fully Developed Couette Flow
•
Step 2: Assumptions and BC’s
– Assumptions
1.
2.
3.
4.
5.
6.
7.
Plates are infinite in x and z
Flow is steady, /t = 0
Parallel flow, V=0
Incompressible, Newtonian, laminar, constant properties
No pressure gradient
2D, W=0, /z = 0
Gravity acts in the -z direction,
– Boundary conditions
1. Bottom plate (y=0) : u=0, v=0, w=0
2. Top plate (y=h) : u=V, v=0, w=0
Fully Developed Couette Flow
• Step 3: Simplify
3
Note: these numbers refer
to the assumptions on the
previous slide
6
Continuity
This means the flow is “fully developed”
or not changing in the direction of flow
X-momentum
2
Cont.
3
6
5
7
Cont.
6
Fully Developed Couette Flow
• Step 3: Simplify, cont.
Y-momentum
2,3
3
3,6
3
7
3
3
3
Z-momentum
2,6
6
6
6
7
6
6
6
Fully Developed Couette Flow
• Step 4: Integrate
X-momentum
integrate
integrate
Z-momentum
integrate
Fully Developed Couette Flow
• Step 5: Apply BC’s
– y=0, u=0=C1(0) + C2 C2 = 0
– y=h, u=V=C1h C1 = V/h
– This gives
– For pressure, no explicit BC, therefore C3 can remain an
arbitrary constant (recall only P appears in NSE).
• Let p = p0 at z = 0 (C3 renamed p0)
1.
2.
Hydrostatic pressure
Pressure acts independently of flow
Fully Developed Couette Flow
• Step 6: Verify solution by back-substituting into
differential equations
– Given the solution (u,v,w)=(Vy/h, 0, 0)
– Continuity is satisfied
0+0+0=0
– X-momentum is satisfied
Fully Developed Couette Flow
• Finally, calculate shear force on bottom plate
Shear force per unit area acting on the wall
Note that w is equal and opposite to the
shear stress acting on the fluid yx
(Newton’s third law).
Momentum Equation
• Special Case: Euler’s Equation
Inviscid Flow for Steady incompressible
•
For steady incompressible flow, the equation reduces to
v 0
( v ) v p g
where = constant.
•
Integrate from a reference at along any streamline y=C :
p
2
2
v
p v
gz
gz constant
2
2
20
Two-Dimensional Potential Flows
• Therefore, there exists a stream function
such that
y
in the Cartesian coordinate and
y
y
u, v
,
x
y
in the cylindrical coordinate
y
y
ur ,vq
,
r
rq
21
Potential Flow
Two-Dimensional Potential Flows
• The potential function and the stream functiony are conjugate
pair of an analytical function in complex variable analysis.
y
x y
and
y
y
x
• The constant potential line and the constant streamline are
orthogonal, i.e.,
u ,v
to imply that
.
and
y - v,u
y 0
23
Stream and Potential Functions
• If a stream function exists for the velocity field
u = a(x2 -- y2) & v = - 2axy & w = 0
Find it, plot it, and interpret it.
• If a velocity potential exists for this velocity field.
Find it, and plot it.
y
y
u, v
,
x
y
u
x
v
y
Summary
• Elementary Potential Flow Solutions
y
Uniform Stream
U∞y
U∞x
Source/Sink
mq
mln(r)
-Kln(r)
Kq
Vortex
26