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4.1: Introduction to Forces in Fluids: Surface Force:
Shear/Viscous/Frictional Force
1.
Forces in Fluids
2.
Surface Force and Stress
3.

Surface Force:
Area as A Vector, and Outward Normal to The System

Description of Surface Force:
Stress Vector and Stress Tensor
Newtonian Fluids and Newton’s Viscosity Law

Laminar Flow as A Flow of Layers of Fluids, and
Shear Stress as A “Sliding” Friction Between Adjacent Layers of Fluid
abj
1
Very Brief Summary of Important Points and Equations [1]


Forces in Fluids

Forces in fluids are distributive.

Classified according to geometry of distribution:
 Force 

Line force
 

 Length 

Surface force
 

 Area 

Volume/Body force

  Force 
B  g 

 Volume 
 Force 
=
Pressure
+ Friction
The Importance of A Clear Definition of The SYSTEM of Interest

Convention:
as an outward normal to the system (pointing from system to surroundings)
A = System
B = Surroundings
(eˆ n )
abj
ê n
A = Surroundings
B = System
2
Very Brief Summary of Important Points and Equations [2]
3.
Surface Force/Stress
i j
Index convention:
1.
Component/Index Convention
2.
Sign Convection
Sign Convention
 ij
has a positive numerical value
when the product of the signs of
plane i and force direction j is
positive.
direction
plane
4.
The difference between the
component of stress (tensor)
5.
abj
VS
component of force (vector)
Newton’s Viscosity Law for Newtonian Fluids
1.
Dynamic Viscosity
m
2.
Kinematic Viscosity
n
[VL]
 nt
ut (n)
m
n
ut (n)
 n
n
3
Forces in Fluids
Forces in fluids are distributive
Classified according to the geometry of distribution
Line force
Surface force
Volume (Body) force
Described by
Force

Area
abj
Described by
Stress Tensor 
Force
Volume

Pressure
Friction/Viscous/Shear
(normal)
(tangential)
[neglect the normal component]

g
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Surface Force: Area as A Vector, and Outward Normal to The System
The importance of the system of interest again
1. Area is a vector
2. Its direction is – by convention - OUTWARD NORMAL to the system of interest
A = System
B = Surroundings
A
B
(eˆ n )
ê n
A = Surroundings
B = System
• Whenever you see a surface with a normal vector,
you know – by convention - which side is being considered a system of interest.
abj
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Description of Surface Force: Stress Vector and Stress Tensor
2- and 3-Component Decomposition of The Stress Vector
ê n

Force df
 nn ê n

 df
Stress vector T  
dA
 nt 2  eˆt 2
P
eˆt 2
 
T ( x , t ; eˆ n )   nn eˆ n   nt eˆt
 nt eˆt1
1
eˆt1
 nt  êt
 
T ( x , t ; eˆ n )   nn eˆ n   nt1 eˆt1   nt 2 eˆt 2
ê t
abj
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Decomposition of Three Stress Vectors on Three Mutually Perpendicular Planes at A Point:
Stress Tensor at A Point (Cartesian Coordinates)
Stress Index Convention
     
 
T ( x, t; ˆj )   yx iˆ   yy ˆj   yz kˆ
y
 yy
 yz
 zy
 zz
z
•
 yx
 xy
 zx
 xx
x
 xz
 
    ij
 xx  xy  xz 


  yx  yy  yz 
 zx  zy  zz 


 
 
T ( x, t; iˆ)   xx  iˆ   xy ˆj   xz  kˆ
 
 
T ( x, t; kˆ)   zx  iˆ   zy ˆj   zz  kˆ
Decomposition of three stress vectors on three mutually perpendicular planes at a point
•
results in nine components of a tensor called the stress tensor
 
   ij
abj
  ˆj   xz  kˆ
    ˆj   yz kˆ
  ˆj   zz  kˆ
 
T ( x , t ; iˆ)   xx iˆ   xy
 
T ( x , t ; ˆj )   yx iˆ   yy
 
T ( x , t ; kˆ)   zx iˆ   zy
 xx


  yx


 zx
 xy  xz 
 yy
 zy


 yz 

 zz 


( x, t )
at that point:
i j
Index convention:
direction
plane
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Some Properties (Symmetric Tensor)
y
Symmetry of the stress tensor at a point,
 yy
e.g.,
 yx
 xx  xy  xz 
 =

   yx =  yy  yz ,
 zx  zy=  zz 


2  4
3
e.g.,  2  5 6 
 4 6
8 
.
 xy
 xx
 xx
 xy
 yx

 xy   yx
 ij   ji
x
 yy
From angular momentum consideration with body couple neglected, it follows
Stress tensor at a point is symmetric, i.e.,
 ij   ji
(in matrix notation: 

  T)
This reduces the general 9 independent components of the stress tensor to 6 independent
components (good - less variables to deal with).
abj
8
Stress Sign Conventions
y
 yy
 yz
z
 zz
 xx : i  ( x), j  ( x)   xx  ()  ()  ()  0
y
 yx
 zy
z
 xy
 zx  xz
 yx : i  ( y), j  ( x)   yx  ()  ()  ()  0
x
 xy : i  ( x), j  ( x)   xx  ()  ()  ()  0
 xx
x
 xx : i  ( x), j  ( x)   xx  ()  ()  ()  0
y
 yx : i  ( y), j  ( x)   yx  ()  ()  ()  0
x
z
 xy : i  ( x), j  ( x)   xx  ()  ()  ()  0
Sign Conventions: (Note that some books use different index convention, e.g., reverse)
•
The sign of (the numerical value of) the stress components:
sign of the component  i j
•
abj
=
sign of plane i x
sign of direction j
On the negative plane, the sign of the component of stress is opposite to
the sign of the component of force.
i j
Index convention:
direction
plane
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Note the difference:
component of stress (tensor)
VS
component of force (vector)
On the negative plane, the sign of the component of stress is opposite to the
sign of the component of force.
Component of stress:
 xy  0
Fy ( Ax  xy )  0

0 0
 xy  0,
e.g.,
 xy  5 Pa
y
Component of the force due to that stress:
Fy  Ax  xy  0,

Ax  0
x
e.g.,
Fy  10 N
(  )(  )
Area is a vector. It is signed.
abj
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Constitutive Relation:
Newton’s Viscosity Law for
Newtonian Fluids
  
Assuming that the velocity field is a simple shear flow: V  V ( x , t )  (u, v, w)  (u ( y ),0,0)
y
y
u(y)
plane y
 yx ( y; ˆj )  m (du / dy)
Slope du/dy = strain rate
x
y=0
 ĵ
plane y
Top layer of fluid
Bottom layer of fluid
u
Consider the relation between the shear stress yx and the shear deformation rate du/dy.
Newtonian fluids are the fluids in which shear stress is linearly proportional to shear deformation rate.
Newtonian fluids:
 yx ( y ) 
 du 
du( y )
dy
 du 
 yx  m    n  
 dy 
 dy 
 yx   Force
Area
du / dy 
Newton’s viscosity law
1
Time
m = Dynamic/Absolute viscosity, m   Force  Time  Momentum  Mass-Velocity  MV  
Area
n = Kinematic viscosity,
abj
n :
m

Area
Area
n   Velocity  Length  VL
 A 
Mass
M
 
Length  Time  Lt 
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The Importance of A Clear Definition of The SYSTEM of Interest, Again.
A = System
A
B = Surroundings
(eˆ n )
B

f AB
ê n
the external force due to the surrounding B on the system A

df BA
the external force due to the surrounding A on the system B
A = Surroundings
B = System
Are you talking about
• the external force/stress due to the surrounding B on the system A
or
• the external force/stress due to the surrounding A on the system B?
abj
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Although they are action/reaction, they are not the same force.
Some Notes on Newton’s Viscosity Law
y
u(y)
 yx ( y)  m (du / dy)
slope
y
plane y
x
y=0
u
w
1.
 du 
 du 
  n  
 dy 
 dy 
 ĵ
Top layer of fluid
Bottom layer of fluid
 yx ( y )  m 

It is a field/instantaneous equation, i.e., it is a function of position and time:  yx   yx ( x, t )
•
At the wall, we often refer to the shear stress there as the wall shear stress:
 w   yx ( y  0)  m
2.
 ĵ
plane y
du
dy
y 0
It is a signed equation, the signs of
•
strain rate du/dy , and
•
stress xy (according to the stress sign convention)
should be taken into accounted.
3.
Carefully consider which system and which stress are under consideration:
•
abj
•
Lower/Pink element (unit outward normal is  ĵ ), or
Upper/Grey element (unit outward normal is  ĵ ).
Both have the same value of
13
 yx ( y)  m (du / dy) , but the system and the force are different.
Newton’s Viscosity Law in The Natural Coordinates (n-t )
y
y
y’, n
ên
ên
ên
y
 yx
n
u(y)
x
ên
x
ên
y
x’, t
y’, n
x
 yx   yx
 nt
abj
x
 yx
x’, t
 nt
 yx
y’, n
ên
x’, t
ut(n)
 yx  m
u x ( y )
y
t
 nt  m
u t ( n)
n
Note also that
• ut(n) and u(y) are two different velocity profiles.
•  yx and
 nt are two different stresses acting on two different planes.
u t ( n)
m
n
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Constitutive Relation: Other Types of Fluids
Newtonian and Non-Newtonian fluids’ stress-strain rate relation.
From Fox, R. W., McDonald, A. T., and Pritchard, P. J., 2004, Introduction to Fluid Mechanics, Sixth Edition, Wiley, New York.
Question:
abj
What will happen if shear stress is gradually applied and increased from zero to each of these fluids?
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Dynamic Viscosity (m) VS Temperature
Fox et al. (2010)
abj
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Kinematic Viscosity (n ) VS Temperature
Fox et al. (2010)
abj
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Example 1: Sketch The Velocity Profile and Examine The Shear Force/Stress

Assume that the followings are simple shear flows; that is,
 there is only one dominant velocity component, and
 this velocity component is a function of the traverse/normal coordinate:

V  Vt (n)eˆt

V  Vx ( y)eˆx ,

V  u( y) iˆ
Question:

Sketch the velocity profile along the traverse AB.

State
 the sign ( + / - ) of the component of the stress,
 the sign/direction (+ / - ) of the component of the force due to that stress,
on the surface with the outward normal given.
abj
See next page 
18
y
y
B Still air
B
UW
D
D
Flow
H
C
C
F
A
E
W
E
G
U
LW
A
x
Plate is moving at the speed U.
Flow between two stationary parallel plates
(Channel flow)
UW
D
E
C
Solid wall
element W
abj
D
F
C
G
E
H
LW
19
Laminar Flow as A Flow of Layers of Fluids, and
Shear Stress as A “Sliding” Friction Between Adjacent Layers of Fluid

Laminar flow can be thought of as a stack of layers of fluid sliding over one another.

In between the layers, there is a sliding frictional force.

Vsys / sur floor

Vsys / sur floor
system
system
relative to the surrounding floor.
relative to the surrounding floor.

f

f
Sliding frictional force is opposing the relative velocity of
the system [relative to the adjacent surrounding floor].
A Physical View of Shear Stress as A Sliding Friction in Laminar Flow

The direction of the sliding frictional force is such that
the adjacent surrounding exerts the frictional force opposing the relative velocity/motion of
the system with respect to the adjacent surroundings.
abj
20
Sliding Friction: Drag Backward – Drag Forward
y
u ( y  dy) relatively faster upper layer
Sliding friction
between
the two layers
A) Shear stress on the faster layer:
system
 yx
u( y)

Vsys / sur
u( y)
u ( y  dy)
relatively slower lower layer
B) Shear stress on the slower layer:
 yx
system
u( y)
u ( y  dy)

Vsys / sur
Due to “sliding friction” between the two layers,
faster layer is being dragged backward by the
slower layer is being dragged forward by the
slower layer.
faster layer.
abj
21
Example 1: Sketch The Velocity Profile and Examine The Shear Force/Stress

Assume that the followings are simple shear flows; that is,
 there is only one dominant velocity component, and
 this velocity component is a function of the traverse/normal coordinate:

V  Vt (n)eˆt

V  Vx ( y)eˆx ,

V  u( y) iˆ
Question:

Sketch the velocity profile along the traverse AB.

State
 the sign ( + / - ) of the component of the stress,
 the sign/direction (+ / - ) of the component of the force due to that stress,
on the surface with the outward normal given.
abj
See next page 
22
y
y
B Still air
B
UW
D
D
Flow
H
C
C
F
A
E
W
E
G
U
LW
A
x
Plate is moving at the speed U.
Flow between two stationary parallel plates
(Channel flow)
UW
D
E
C
Solid wall
element W
abj
D
F
C
G
E
H
LW
23
Example
abj
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