Pharos University
ME 253 Fluid Mechanics 2
Revision for Final 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
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
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
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
17
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 ,v   
,

r 
 r
18
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
20
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
m
mln(r)
-Kln(r)
K
Vortex
23
Drag
• Drag Coefficient
with
or
DRAG FORCE
• Friction has two effects:
– Skin friction due to shear stress at wall
– Pressure drag due to flow separation
D  D friction  D pressure
Total drag due to
viscous effects
Called Profile Drag
=
Drag due to
skin friction
Less for laminar
More for turbulent
+
Drag due to
separation
More for laminar
Less for turbulent
Drag Coefficient of Blunt and
Streamlined Bodies
• Drag dominated
by viscous drag,
the body is
streamlined
__________.
• Drag dominated
by pressure drag,
the body is
_______.
bluff
Flat plate
Cd 
2Fd
U 2 A
Drag
• Pure Friction Drag: Flat Plate Parallel to the
Flow
• Pure Pressure Drag: Flat Plate Perpendicular to
the Flow
• Friction and Pressure Drag: Flow over a Sphere
and Cylinder
• Streamlining
Drag
• Flow over a Flat Plate Parallel to the Flow: Friction
Drag
Boundary Layer can be 100% laminar,
partly laminar and partly turbulent, or
essentially 100% turbulent; hence
several different drag coefficients are
available
Drag
• Flow over a Flat Plate Perpendicular to the
Flow: Pressure Drag
Drag coefficients are usually obtained empirically
The Boundary-Layer Concept
Boundary Layer Thicknesses
Boundary Layer Thicknesses
• Disturbance Thickness, d
Displacement Thickness, d*
Momentum Thickness, 
Empirical Equations of Laminar B. Layer Parameters
d
5.0

x
Rex
d*
1.721

x
Rex
u
U
 w   | o  
y
x U

0.664

x
Rex
w
0.664
Cf 

1
Rex
U 2
2
1.328
CD 
Re L
DRAG FORCE = CD ( 0.5 Ρ U2 ) A
N
34
Drag on a Flat Plate
• Drag on a flat plate is related to the momentum deficit
D   bU 
2

u
  b
U
0
 u
1  
 U
dD
2 
 b w   bU
dx
x
• Drag and shear stress can be calculated by assuming velocity
profile in the boundary layer