Hydraulics and Pneumatics
Transmission
FLUID MECHANICS WITH HYDRAULICS
1 Fluid Properties
2 Mechanics of fluids at rest
3 Mechanics of fluids in motion
4 Energy loss of fluids in motion
5 Flow of fluids in clearances and orifices
6 Hydraulically sticking
7 Hydraulically shocking
8 Cavitation
1 Fluid Properties
1.1 Density
Fluid density was defined as mass per unit volume
 
m
3
( kg / m )
V
hydraulic oil ρ=890-910kg/m3
1.1 Compressibility and expansibility
Suppose the volume is the function of pressure and
temperature , or V=f(t,p ) , and t↑→V↑,p↑→V↓
The volumetric increment can be approximated by is
total differential, thus
 V  dV 
V
t
dt 
V  V0  V  V0 
V
p
V
t
V  V 0 (1    t    p )
dp 
t 
V
t
V
p
t 
V
p
p
 p  V 0 [1 
V / t
V0
t  (
V / p
V0
) p ]
1 Fluid Properties
 V  dV 
V
t
dt 
V  V0  V  V0 
V
p
V
t
dp 
t 
V
t
V
p
t 
V
p
 p  V 0 [1 
V  V 0 (1    t    p )
where
 
V / t
  

V0
V / p
V0
p

V / t
t  ( 
V / p
V0
) p ]
V0
expansibility coefficient
compressibility coefficient
Found by experiment:
  ( 8 . 5 ~ 9 . 0 )  10
4
,   ( 5 ~ 7 )  10
 10
In hydraulic transmission problems, p≤32Mpa, the
volume variation caused by pressure variation is
 V / V 0    p  (5 ~ 7)  10
 10
 32  10  1.6 ~ 2.24%
6
Consequently, hydraulic oil may be regarded as uncompressible.
1.3 Viscosity
1.3.1 cohesion force and adhesion force
Y
There are cohesion forces among fluid particles, while
there are adhesion forces among fluid particles and solid
wall.
Adhesion forces are usually greater than cohesion force
except mercury.
1.3.2 dynamic viscosity
Consider two parallel plates, placed a small distance Y
apart, the space between the plates being filled with the
fluid.
U
v
Y
F
The lower surface is assumed to be stationary, while
upper one is moved parallel to it with a velocity U by the
application of a force F, corresponding to some area A of
the moving plates
The particles of the fluid in contact with each plate
will adhere to it. The velocity gradient will be a straight
line. The action is much as if the fluid were made up of a
series of thin sheets.
U
v
Y
dy
F
du
y
u
Experiment has shown that for a large class of fluids
Ff  A
du
dy
If a constant of proportionality μis now introduced, the
shearing stress τbetween any two thin sheet of fluid may
be expressed by
  
du
dy
(1)
1 Fluid Properties
Above equation is called Newtow’s equation of viscosity
and in transposed form it serves to define the
proportional constant
 

( N  s / m ) （泊）
/
2
du / dy
which is called the dynamic viscosity.
The viscosity is the inherent property of fluids, but shown
only in fluid flow.
The viscosity is a measure of its resistance to shear or
angle deformation.
The viscosity accounts for energy losses associated with the
transport of fluids in ducts and pipes.
The friction forces in fluid flow result from the cohesion
and momentum interchange between molecules in the fluid.
1 Fluid Properties
As temperature increases, the viscosity of all liquids
decreases, while the viscosity of all gases increase.
This is because the force of cohesion, which diminishes
with temperature, predominates with liquids; while with
gases the predominating factor in the interchange of
molecules between layers of different velocity.
1.3.3 Kinematic viscosity
In many problems including viscosity there frequently
appear the value of viscosity divided by density. This is
defined as kinematic viscosity
 


2
( m / s ) （斯）
/
E.g. the kinematic viscosity of 30# mechanic oil is 30 厘斯
1 Fluid Properties
1.3.3 Relative viscosity
a. Definition : °E= t1 /t2
t1 –-time the measured liquid (200mL, T°C) passes
through the viscosity-meter orifice(2.8mm);
t2 —time the pure water (200mL, 20°C) passes
through the viscosity-meter orifice(diameter=2.8mm).
b. Transposed relation
 50  (7.31 E 50 
6.31
6
)  10 ( m / s )
2
E 50
The lubrication oil is named according to its kinematic viscosity
E.g. For 30#mechanical oil, its kinematic viscosity is 30 厘斯
E 50  4  10
4
The gasoline is named according to its octane
The diesel oil is named according to its freezing point
2 Mechanics of fluids at rest
Fluid statics is the study of fluids in there is no relative
motion between fluid particles.
The only stress is normal stress, pressure, so it is the
pressure that is of primary interest in fluid statics.
2.1 static pressure property
a. The pressure is defined as the force exerted on a unit
p  dF / dA
area
b. The pressure is the same in all direction
Note geometric relation and
neglect higher order term, thus
px  py  p
pdyds
¦Á
?
px dydz
61?
pdsdy cos   p x dydz  0
1
p y dxdz  pdsdy sin    gdxdydz
2
 0
90 -¦
Á
py dxdy
1
-¦ Ñgdxdydz
2
2 Mechanics of fluids at rest
2.2 Basic differential equation
Assume the pressure is the function of space
 p dx
coordinates, or p=f(x,y,z) .
(p 
)
x 2
Consider a infinitesimal z
element in the figure.
A
dz
Assume that a pressure p
 p dz
z 2
y
(p
 p dy
y 2
x 2
x
)
of sides can be expressed by using chain rule from
calculus with p(x,y,z)
dp 
p
x
dx 
p
y
dy 
p
z
dz
)
 p dx
(p
dx
exists at the center A of this
element, the pressure each
(p
)
2 Mechanics of fluids at rest
Newton’s second law is written z
(p 
 p dx
x
)
(p
2
 p dz
z 2
)
in vector for constant mass system.
ΣF=ma
This results in three component
equations,
(p
(p
(p
 p dx
x 2
 p dy
y 2
 p dz
z 2
) dydz  ( p 
) dydz  ( p 
) dydz  ( p 
 p dx
x 2
 p dy
y 2
 p dz
z 2
A
dz
(p
dx
y
(p
) dydz  (  dxdydz ) a x
 p dy
y 2
 p dx
x 2
)
x
)
) dydz  (  dxdydz ) a y
) dydz  (  gdxdydz )  (  dxdydz ) a z
Where ax,ay,and az are the components of the acceleration of
the element.
2 Mechanics of fluids at rest
Division by the element’s volume dxdydz yields
 p
  a x

 x
 p
  a y

 y
 p
   (a z  g )
 z

(2)
The equation expresses the relation between pressure
variation and acceleration.
2 Mechanics of fluids at rest
2.3 Examples included in fluid statics.
E1: liquid at rest
p0
Solution:
a xa ya z 0
dp
dz
p
g
  g
 z  cons tan t
(3)
The quantity ( p /  g  z ) is referred to as the piezometric head
Another form is written as
p   gh  p 0
(4)
Where p—----the pressure at a point;
ρgh---the pressure caused by liquid column weight;
p0------the pressure caused by external force (either
gases or liquid or solid).
2 Mechanics of fluids at rest
p   gh  p 0
(4)
Explanation:
① The term  gh
is used to convert pressure to a height
of liquid.
② Neglect the pressure caused by the fluid weight when
studying gases.
③ Neglect the pressure caused by the liquid weight and the
pressure caused by atmosphere on hydraulic transmission.
④ The equal-pressure surface is a horizontal plane.
⑤ The free surface is a special equal-pressure surface
(p=pa).
2 Mechanics of fluids at rest
E2:trolley in a linear acceleration
Solution :
pa
The liquid is at rest relative to the trolley,
z
so the reference frame is established on the
x a
trolley.
According to Equation (1)
v
dp 
p
dx 
p
 p
  x    a

p

  g
  z
dz    adx   gdz
x
z
p    ax   gz  c
p    ax   gz  p a
When x=z=0,p=pa, thus
The equal-pressure surface is not a horizontal plane but a
slope
2 Mechanics of fluids at rest
E3: Rotating container
z
pa
Solution :
The the reference frame is established
on the container.
According to Equation (1)
 p
2


x

x

 p
2
x
  y

 y
 p
y
  g

 z
p
p
p
2
2
dp 
dx 
dy 
dz   x  dx   y  dy   gdz
x
y
z
1
2
2
2
p   ( x  y )   gz  c
1
2
2
2
2
p   ( x  y )   gz  p a
When x=y=z=0,p=pa, thus
2
pa
The constant-pressure surface is a paraboloid of revolution
2 Mechanics of fluids at rest
2.4 Absolute pressure, gage pressure and vacuum
If pressure is measured relative to absolute zero, it is
called absolute pressure.
When measured relative to atmosphere as a base, it is
called gage pressure .
If the pressure is below that of the atmosphere, it is
designated as a vacuum .
gage
pressure
fuchsin
vacuum
gage
pressure
atm ospheric
pressure
absolute zero
absolute
pressure
absolute
pressure
2 Mechanics of fluids at rest
2.5 Forces on plane Areas and on curved surface
a. Forces on plane Areas
F  pA
b.Forces on curved surfaces
 F x  pA x

 F y  pA y
 F  pA
z
 z
Where Ax, Ay and Az are project areas in three directions
3 fluid kinematics and dynamics
3.1Description of fluid motion
a. Lagrangian description
In the study of particle mechanics,
attention is focused on individual particles,
motion is observed as a function of time,
the position, velocity, and acceleration of
each particle are listed
x  x ( t ), y  y ( t ), z  z ( t )
dx
dy
dz
ux 
,uy 
, uz 
dt
dt
dt
2
2
2
d x
d y
d z
ax 
,a y 
,az 
2
2
2
dt
dt
dt
1736 ~ 1813
where x, y and z are transient position coordinates。
This description is easily acceptable but difficult as the
number of particles becomes extremely large in a fluid flow.
3 Mechanics of fluids in motion
b. Eulerian description
An alternative to following each fluid
particle separately is to identify points
in space and the observe the velocity of
particles pass each point.
The flow properties, such as velocity,
are functions of both space and time.
u  u( x , y , z , t )
where x, y and z are the position
coordinates of the flow field
a 
 u dx
 x dt

 u dy
 y dt

 u dz
 z dt
Convective acceleration

u
t
1707 ~1783
Local acceleration
3 Mechanics of fluids in motion
3.2 Key concepts
a. Ideal fluid:
A fluid is presumed to have no viscosity
b. Incompressible and compressible fluid
An incompressible fluid is the one whose density remains
relatively constant.
Generally speaking, liquids can be considered as
incompressible fluids while gases as compressible fluids
c. Steady flow:
Where the quantities of interest do not depend on time.
u
t
 0,
p
t
 0,

t
 0
d. Path line:
A path line is the locus of points traversed by a given
particle as it travels in the flow field.
Note that a path line is a history of the particle’s locations
(LaGrange description)
3 Mechanics of fluids in motion
e. Streamline:
A streamline is a line possessing following property: the
velocity vector of each particle occupying a point on the
streamline is tangent to the streamline (Eulerian description)
In a steady flow, pathlines and streamlines are all coincident.
3 Mechanics of fluids in motion
f. Stream tube:
A stream tube is a tube whose walls are steam line.
Note that no fluid can cross the walls of a stream tube
since the velocity is tangent to a stream line
People often sketch a stream tube with a infinitesimal
cross section in the interior of flow for demonstration
purposes.
g. Flow cross section
A plane or curved surface at right to the direction of
velocity.
3 Mechanics of fluids in motion
h. Flow rate and mean velocity
The quantity of fluid flowing per unit time across any
section is called the flow rate.
In dealing with incompressible fluids,volume flow rate
is commonly used, whereas mass flow rate is more
convenient with compressible fluids.
The mean value of the velocity in a cross section is
called the mean velocity.
dq  u  dA
q  v A
This indicates that the volume flow rate is equal to the
magnitude of the mean velocity multiplied by the flow
area at right to the direction of velocity.
3 Mechanics of fluids in motion
3.3 Equation of continuity
Assume a incompressible fluid steadily flows in the
infinitesimal stream tube.
The following figure represents a short length of a
2
stream tube
u2
2
1
dA1
dA2
u1 1
The fixed volume between the two fixed sections of the
stream tube is called the control volume.
According to mass conservation law, in the time dt, the
mass flowing in the control volume must be equal to the
mass flowing out the control volume.
 u 1 dA 1 dt   u 2 dA 2 dt
3 Mechanics of fluids in motion
The equation can be simplified, thus
u 1 dA 1  u 2 dA 2
The equation can be integrated along flow section
 q1  q 2

 v 1 A1  v 2 A 2
The equation indicates the mean velocity is inversely
proportional to the flow area.
3 Mechanics of fluids in motion
3.4 Differential equation of steady flow for ideal fluid
dz
ds
Consider steady flow of a ideal fluid.
Use a infinitesimal cylindrical element, with length ds and
cross-section dA, in the
z
(p+ dp)dA
S
s-direction of the stream.
The forces acting on the
element are pressure forces
and the weight.
Summing up the forces in
pdA ¦ È
the s-direction, there results
 F s  p d A  ( p  d p )d A
¦ ÑgdA ds
  g d sd A  co s 
  d p d A   g d sd A  co s 
The acceleration of the s-direction
as 
 u ds
 s dt

u
t
 u
du
ds
x
3 Mechanics of fluids in motion
Apply Newton’s second law , we have
 dpdA   gdsdA  cos    dsdA  u du / ds 
simplify the expression, we have

1 dp
 ds
 g cos  s  u
du
ds
The equation is called Eulerian equation
3.5 Bernoulli equation
a. The Bernoulli equation on following assumptions:
(1) Ideal fluid; (2) Steady flow; (3) An infinitesimal
stream tube; (4) Constant density; (5) Inertial reference
frame.
Consider geometric relation
cos  
Thus
dp
g

1
g
dz
ds
udu  dz  0
3 Mechanics of fluids in motion
The above expression is integrated along the stream line, the result
p1
 z1 
1
2
u1

p2
 z2 
1
2
u2
g
2g
g
2g
where p /  g   P ressu re en erg y p er u n it w eig h t flu id ;
z      P o ten tial en erg y p er u n it w eig h t flu id ;
u / 2 g   K in em atic en erg y p er u n it w eig h t flu id .
2
Bernoulli equation indicates that the total
energy of a fluid flowing from 1 cross section to
2 cross section remains constant though one
energy form can be converted into another.
Bernoulli • Daniel (1700-1782), Swiss
mathematician, who showed that as the
velocity of a fluid increases, the pressure
decreases, a statement known as the Bernoulli
principle.
3 Mechanics of fluids in motion
E1: Manufacture a shower
In order to suck hot water into the tube, the pressure
inside the tube need be lower than atmospheric pressure.
A good idea is to increase
kinematic energy, that is to
热水 温水
冷水
say, to decrease the diameter
of the tube.
E2: The lift force of an airplane
In order to make an airplane lift, the pressure under
the wing need be higher than that on the wing.
A good idea is to make
the wing have different
curve surfaces
3 Mechanics of fluids in motion
b. The Bernoulli equation on following assumptions:
(1) Real fluid; (2) Steady flow; (3) An infinitesimal
stream tube; (4) Constant density; (5) Inertial reference
frame.
The ideal fluid flow or inviscid flow does not cause
energy losses; while a real fluid flow or viscous flow will
cause energy losses.
If energy losses are considered the Bernoulli equation
can be written as following
p1
g
 z1 
1
2g
u1 
2
p2
g
 z2 
1
2g
u 2  h'
2
where h’—energy losses caused by friction forces
3 Mechanics of fluids in motion
c. The Bernoulli equation on following assumptions:
(1) real fluid; (2) steady flow; (3) a real tube;
(4) constant density; (5) inertial reference frame;
(6) cross sections of gradually varied flow
A real tube can be considered as consisting of countless
infinitesimal stream tubes. Consequently, we can
integrate the above equation along the cross-section of a
real tube
(
p1
g
 z1 
1
2g
u 1 ) dQ 
2
(
p2
g
 z2 
1
u 2  h w ) dQ
2
2g
'
Rewrite the integration
(
p1
g
 z 1 ) dQ 
1
 2g
2
u 1 dQ

(
p2
g
 z 2 ) dQ 
1
 2g
u 2 dQ 
2
 h w dQ
'
3 Mechanics of fluids in motion
Note that in the cross section of gradually varied flow
p /  g  z  con stan t
Hence
Let
I5 
I1   (
p1
g
 z 1 ) dQ  (
 hw dQ  hw Q I 2 
'
We can obtain
1
p1
g
 2g
p1
g
I3   (
 z1 )Q
u dQ   1
2
1
 z1 
1
2g
1
2g
v
2
1
p2
g
 z 2 ) dQ  (
v Q I4 
2
1

p2
g
1
 2g
 z2 
p2
g
 z 2 )Q
u dQ   2
2
2
2
2g
1
2g
v 2  hw
2
where v1 and v2 ----mean velocities;
α1 andα2----kinetic energy correction factors, α=1~2.
Please note:
2
v2Q
The selected cross sections should ensure that
the stream lines across the cross section are
approximately parallel (gradually varied flow)
3 Mechanics of fluids in motion
Ⅰ
1
2
h1
O
D2
Ⅱ
D1
O
Ⅱ
Ⅰ
h
d. Example Venturi meter
A Venturi meter consists of
one tube with a constricted
throat which produces an
increased velocity accompanied
by a reduction in pressure. The
meter is used for measuring the
flow rate of both compressible
and incompressible. Assuming
D1=200mm, D2=100mm,
the height of the mercury
column h=45mm ， Calculate
the flow rate of water.
O1
O1
图 2-11 文 氏 流 量 计
图 2-11 文 氏 流 量 计
3 Mechanics of fluids in motion
Ⅰ
p1
g
 z1 
1
2g
v 
2
1
p2
g
 z2 
2
2g
1
2
h1
O
p1  p 2


h
O1
图 2-11 文 氏 流 量 计
2
图 2-11 文 氏 流 量 计
2
(v 2  v1 )
2
Ⅱ
O1
v 2  hw
1
O
Ⅰ
Next, calculating parameter z1=z2=0, let  1   2  1 . 0 , h w  0
We can obtain
D2
Ⅱ
D1
Solution
First, selecting two flow cross
section I-I and II-II;
Second, select potential energy base
line O-O;
Then, writing the Bernoulli equation
between cross section I-I and II-II;
2
3 Mechanics of fluids in motion
2
2
(v 2

2
v1
Ⅰ
)
Ⅱ
Use continuity equation v1 A1  v 2 A2
A2


1
2
v1 
2
O
v1 
2
D2
h1
2
D1
v1
2
(v 2  v1 ) 
2
2
v1
4
(
2
2( p1  p 2 )
 ( D 1 / D 2  1)
4
4
Ⅱ
Ⅰ
Inserting this value of v2 in
foregoing expression, we obtain
p1  p 2
1
D1
4
D2
 1)
h
v2 
A1
O
D2


1
D1
p1  p 2
O1
O1
图 2-11 文 氏 流 量 计
图 2-11 文 氏 流 量 计
3 Mechanics of fluids in motion
According to static pressure equation, select equal pressure
Ⅰ
planeO1O1，
D1
p 1   g ( h1  h )  p 2   gh 1   g gh
O
1
2
D2
Ⅱ
O
h1
or p1  p 2  gh (  g   )   gh (  g /   1)
v1 
Ⅰ
2 gh (  g /   1 )
h
thus
( D1 / D 4  1)
4
O1
Finally, the flow rate is
q
2
 D1
4
v1 
Ⅱ
O1
图 2-11 文 氏 流 量 计
图 2-11 文 氏 流 量 计
2
 D1
2 gh (  g /   1 )
4
D1 / D 2  1
4
4
3 Mechanics of fluids in motion
Substituting data for these variables, we obtain the
ideal throat flow rate
q
  0 .2
2
2  9 . 8  0 . 045 ( 13 . 6  1 )
( 0 .2 / 0 .1 )  1
4
4
 2 . 7  10
2
3
m /s
As there is some friction losses between cross section 1-1
and 2-2, the true velocity is slightly less than the value
given by the expression. Hence, we may introduce a
discharge coefficient C, so that the flow rate is given
 D1
2 gh (  g /   1 )
4
D1 / D 2  1
2
qC
4
4
3 Mechanics of fluids in motion
3.6 momentum equation
a. Momentum theorem and d'Alembert principle
The expression of momentum theorem is
F

d ( mv )
dt
 lim
t  0
 ( mv )
t
The d'Alembert principle expression of momentum
theorem is
F
 [
d ( mv )
dt
] 0
3 Mechanics of fluids in motion
b. The derivation of momentum equation
Assumptions:
(1) Incompressible fluid; (2) Steady flow; (3) An
infinitesimal stream tube; (4) Constant density; (5) Inertial
reference frame.
Use a infinitesimal stream tube between section 1-1, with
a velocity u1 and cross-section dA1 , and 2-2, with a velocity
u2 and cross-section dA2, as the control volume.
2
2
It may be note that
the control volume is
fixed.
1
2
1
d
1
1
3 Mechanics of fluids in motion
2
2'
2
1
1'
d
2
1
d
1
1
2
2'
1'
Assume that time Δt lapses，the fluid flows from cross
2-12
量方
程 的1’-1’
推导
section1-1 图
and
2-2 to动
cross
section
and 2’-2’.
The variation of the fluid momentum is
t  t
 ( mv )  ( mv ) 1 ' 2 '  ( mv ) 12
t  t
 [( mv ) 1 ' 2
t  t
t
t  t
 ( mv ) 22 ' ]  [( mv ) 11 '  ( mv ) 1 ' 2 ]
t
t  t
 [( mv ) 22 '  ( mv ) 11 ' ]  [( mv ) 1 ' 2
t
Both sides are divided by Δt, then taking limit
t
 ( mv ) 1 ' 2 ]
t
3 Mechanics of fluids in motion
t  t
F
 lim
( mv ) 1 ' 2
 ( mv ) 1 ' 2
t
t  0
t  t
( mv ) 22 '  ( mv ) 11 '
t
 lim
t
t
t  0
Momentum change rate
caused by position
variation
Momentum change
rate caused by time
variation
The expression is written into d'Alembert principle equation
t  t
F
 lim [ 
t  0
( mv ) 1 ' 2  ( mv ) 1 ' 2
t
t  t
t
]  lim [ 
Transient flow force
External resultant force
t  0
( mv ) 22 '  ( mv ) 11 '
t
t
Steady flow force
Inertial force
] =0
3 Mechanics of fluids in motion
The momentum change rate caused by time variation is
equal to zero when flowing steadily.
The momentum change rate caused by position variation is
calculated as following
t  t
lim
( mv ) 22 '  ( mv ) 11 '
t
t
t  0
= lim
＝
lim
( mv ) out  ( mv ) in
t  0
t
 u 2  tdA 2  u 2   u 1  tdA 1  u 1
t
t  0
  u 2 dq 2   u 1 dq 1
The integration of momentum change rate
d ( mv )
dt

  u 2 dq 2    u1 dq 1   2  q 2 v 2   1  q 1 v 1
A2
A1
3 Mechanics of fluids in motion
d ( mv )
dt

  u 2 dq 2    u1 dq 1   2  q 2 v 2   1  q 1 v 1
A2
A1
Where v1, v2 ---mean velocity on cross section 1-1 and 2-2
respectively；
β1, β2 ---momentum correction factors on cross section
1-1 and 2-2 respectively， β=1～4/3.
External forces
The external forces acting on the fluid inside the control volume
can be classified three types: (1)pressure forces on cross sections;
(2) weight force of the fluid inside the control volume;
(3) Restrictive force of the control volume, that is
 F   P W  R
3 Mechanics of fluids in motion
c. Momentum equation of incompressible fluid
Assumptions:
(1) Incompressible fluid; (2) Steady flow; (3) An real tube.
 P  W  R   2  q 2 v 2   1  q1v1
Explanation:
The resultant force acting on the fluid inside the control
volume is equal to that in unit time the momentum fluxing
out the control volume is subtracted by the momentum
fluxing in the control volume.
It may be noted that the equation is vector equation.
Solution steps
①Select a control volume；② Express all external forces in
a figure；③ Select a reference frame；④ Write component
momentum equations；⑤ calculate parameters;⑥Sometimes
the Newton’s third law is applied.
3 Mechanics of fluids in motion
c. Examples
E1: Calculate the force acting on the tube, assume followings
are known: v 1、 v 2、 v 3、 A1、 A2、 A3、 p1、 p 2、 p 3
3
p3 A3
y
v3
3
1
p1 A 1
v1
2
45
Ry
Rx
v2
2
Solution
1
Select the “y” shaped tube as a control volume.
Express all external forces as shown in the Figure
Select the reference frame as shown in the Figure
p2 A 2
x
3 Mechanics of fluids in motion
c. List component equations of momentum
In x direction:
p1 A1  p 2 A2  p 3 A3  cos 45  R x   2  q 2 v 2   3  q 3 v 3  cos 45   1  q1v1
In y direction:
 p 3 A3  sin 45  R y   3  q 3 v 3  sin 45
Consequently
R x   p1 A1  p 2 A2  p 3 A3  cos 45   2  q 2 v 2   3  q 3 v 3  cos 45   1  q1v1
R y   3  q 3 v 3  sin 45  p 3 A3  sin 45
According to Newton’s third law, the forces acting on the
tube are
R X '   R X ， RY '   R y
E2: Stable analysis of directional control valve
θ
θ
(a)
Solution
（ b）
图 2-14
滑阀阀芯所受的稳态液动力
( a ) : R  0   1  q v图
co s  滑
阀
阀
 1芯 所
q v受1 的
co稳s 态
 液(to
动 力th e left)
1 2-14
R '   1  q v1 co s  (to th e rig h t)
( b ) : R   2  q v 2 co s   0   2  q v 2 co s 
R '   2  q v 2 co s 
Both are stable.
(to th e left)
(to th e rig h t)
4 Energy losses of fluids in motion
Energy loss is usually called power loss or head loss.
Head loss is the measure of the reduction in the total head
(sum of elevation head, velocity head and pressure head) of the
fluid as it moves through a fluid system.
p1
g
 z1 
1
2g
v
2
1

p2
g
 z2 
2
2g
v 2  hw
2
Head loss includes friction loss and local loss
As the fluid flows through straight pipes fiction loss occurs.
Losses due to the local disturbance of the flow are called
local losses .
The viscous friction will cause energy loss, the loss is called
the loss caused by viscous force.
The fluid particles do not move in uniform linear motion
but they follow random paths, or exists the loss caused by
inertial force
4 Energy losses of fluids in motion
4.1 Reynolds regime experiment
3
w
2
1
4
5
图 2-15 流 态 试 验 示 意 图
When v is small, a red line appears—laminar flow.
When v is great, the red line disappears—turbulent flow.
4 Energy losses of fluids in motion
'
Laminar flow





v
vc
Turbulent flow
c
vc’—higher critical velocity ; vc----lower critical velocity
Actually, velocity is not the only factor that determines
whether the flow regime is laminar or turbulent.
The flow regime depends on three physical parameters:
Velocity, geometric scale and viscosity.
4.2 Reynolds number
v—mean velocity;
vd
d—tube diameter;
Re 
ν—kinematic viscosity.

Corresponding to two critical velocity，the two critical
Reynolds numbers are
Lower critical Reynolds number: Re c  2320
5
Re
'

8000
~
10
Higher critical Reynolds number: c
4 Energy losses of fluids in motion
To noncircular duct
4 vR v—mean velocity; ν—kinematic
Re 
viscosity; R—hydraulic radius(R=A/x),
 x—wetted perimeter;
4.3 The physical significance of Reynolds number
Re 
vd


 vd

[  ( Ad )] v
du
dy

A
du
dy
[  ( Ad )]

A
dy du
dt dy

du
dy
Ma
A
du

Fi
F
dy
A ratio of the inertial force to the viscous force
The physical nature of laminar flow is that the viscous
force may dominate the inertial force; while the physic
nature of turbulent flow is that the inertial force may
dominate the viscous force.
4 Energy losses of fluids in motion
4.4 The friction loss in laminar flow
a. Velocity profile in laminar flow
Use a small cylindrical element, with length L and
cross-section πr2.
1
2
The forces acting
图 2-16 on
圆 the
管 中element
的 层 流 are pressure
forces and the friction force.
4 Energy losses of fluids in motion
Summing up the forces in the flow direction

F X  r p1   r p 2    2 rl  0
2
2
( p1  p 2 )  2( l / r )
where
    du / dr
Let  p  p1  p 2
2
1
 p   2( l / r )  du / dr
p
du  
r  dr
2l
Integrating and determining the constants of integration
2-16 圆 管 中 的 层 流
from the fact that u=0,when图r=R,
we obtain
u
p
4l
(R  r )
2
2
( parabolic profile )
u m ax 
p
4l
R 
2
p
16  l
d
2
4 Energy losses of fluids in motion
b. Flow rate
q 

udA 
A

p
R
0
2
4l
c. Average velocity q
v
A

( R  r )  2  rdr 
d
2
4
128  l
 p /(
d
4
R
8l
2
)
4
d
p 
d
4
128  l
p
2
32  l
p
d. Kinetic energy correction factor and momentum correction factor
1 2
1 2
 udq 4

  (  u dq ) /( v q )  2
 

2
2
 vq
3
e. Friction loss expression
2
2
2
32  l
64
l v
64
l v
l v
p 
v



 
2
h  
d
2
l v
d 2g
vd
d 2
Re
d 2
(λ-----friction factor)
d 2
4 Energy losses of fluids in motion
4.5 Friction loss in turbulent flow
a. The characteristic of turbulent flow
A distinguishing characteristic of turbulence is its irregularity
and no two particles may have identical even similar motion,
so statistical mean of evaluation must be employed
b. Laminar boundary layer
There can be no turbulence next to a smooth wall.Therefore,
immediately adjacent to a smooth wall there will be a laminar
or viscous sublayer.
The thickness of the laminar boundary layer is
  32.24
d
Re

where d-----diameter of a tube;
Re—Reynolds number;
λ-----friction factor in turbulent flow.
4 Energy losses of fluids in motion
c. Classification of tubes
If the roughness Δ of tubes is considered there will be two type
of tubes.
(a)
(b)
 0 . 25
δ>Δ----Hydraulically
smooth


0
.
316
Re
图 2-17 水 利 光 滑 管 和 水 利 粗 糙 管
δ<Δ时---- Hydraulically rough
When Re is very great

  f (Re, )
 d
  f( )
d
4 Energy losses of fluids in motion
Colebrook equation (Re>4000)
2 . 51 
 
  0 . 86 ln 


3
.
7
d

Re  

1
Where λ is the turbulent friction factor.
4 Energy losses of fluids in motion
4.5 Local losses
Loss due to the local disturbance of the flow in conduits
such as changes in cross section, projecting gaskets,elbow,
valves, and similar items are called local losses
Expression of local pressure loss
 p  
v
2
(v--exit velocity)
2
Expression of local head loss
hw  
v
2
2g
Expression of local pressure loss of standardized elements
 pv  (
Q
Qn
) pn
2
4 Energy losses of fluids in motion
Homework: Assuming that a 60°bending pipe is located in
horizontal plane, (fluxing out to atmosphere), the inlet and outlet
diameter are 80 and 40mm respectively, outlet velocity is 8m/s, the
local loss coefficient ξ=0.1. Calculating the x and y component of
the force acting on the pipe ( 1   2   1   2  1 ,   1000 kg / m 3 ).
v2
60°
v1
5 flow of fluids in orifices and clearances
d
D
O
d0
5.1 flow in orifices
a. Classification of orifices
Thin wall orifice:
l / d  0 .5
0 .5  l / d  4
Thick wall orifice:
Long orifice:
l/d  4
b. Flow characteristic in thin wall orifices
Coefficient of contraction
O
Cc 
A0
A
1
2
O
d0
d
D
O
2
1
Selecting cross section 1-1and 2-2. Additionally selecting O-O as
potential energy base line. According to Bernoulli equation
p1
g
 z1 
1
2g
v 
2
1
p2
g
 z2 
2
2g
v 2  hw
2
let  1   2  1 , D  d , v1  v 2 , v1  0 h  h
Inserting these values in Bernoulli equation, we obtain
z1  z 2  0 ,
w
v2 
1
2
1 

( p1  p 2 )  C v
2

p

2

v2
2g
5 flow of fluids in orifices and clearances
q  A0 v 2  C v C c A
2

( p1  p 2 )  C d A
2

 p  C1 A (  p )
c. Flow characteristic in long orifices
According to flow rate expression in a tube
q 
d
2
32  l
 p  C 2 A p
d. Universal expression in a orifice
q  C AT (  p )
m
 m  0 . 5 Thin wall orifice

 m  1 . 0 Long orifice
0.5
5 flow of fluids in orifices and clearances
5.2 flow inside a gap or clearance
Simplifying the foregoing expression
and inserting    du / dy
2
d u
dy
2

+d
p +d p
dy
p
u
y
 (  d  ) bdx  0
h
Geometric characteristic: l  h , b  h
Flow characteristic: laminar
5.3.1 Gap between parallel plates
a. flow by the action of pressure
difference
pbdy  ( p  dp ) bdy   bdx
1 dp
 dx
N ote : u  u ( y ), p  p ( x )
x
dx
N ote : u  u ( y ), p  p ( x )
u has noting to do with p and x; while p is independent
of u and y. Consequently
1 dp
 dx
1 dp
 dx

2
d u
2

p
 C onstant
 dx
The relationship between p and x is
linear, therefore
dy
p
dp
 C onstant 
p1
dy
2
1 dp
dx
x
1 p
 l
1 p
 l
p2
2
d u
l
5 flow of fluids in orifices and clearances
Integrating and determining the constants of integration
from the fact that u=0 when y=0 and u=0 when y=h, we
obtain
p
u
(h  y ) y
2l
The flow rate passing through is
q
bh
3
12  l
p
From the expression we conclude that because the leakage
flow rate through a clearance is proportional to the cube of the
clearance height the clearances among components require
high dimensional accuracy degree and hence the initial cost of
hydraulic elements is very high.
5 flow of fluids in orifices and clearances
a. flow by the action of shearing
2
d u
dy
0
2
Integrating and determining the constants of integration from
the fact that u=0 when y=0 and u=0 when y=h, we obtain
U
u
y
h
The flow rate is
q
1
Ubh
2
The total flow rate is
q
bh
3
12  l
p 
1
2
Ubh
5 flow of fluids in orifices and clearances
h
5.3.2 annular gap
a. Concentric annular gap
p1
p2
r
R
h
O
p
同心环
形缝隙流
动
图 2-22 偏 心 环
Letting b 图
forgoing
expression,
we obtain
 2-21
 d , using
q
 dh
3
12  l
p
h
5 flow of fluids in orifices and clearances
y
b.Eccentric annular gap
y  R  r cos   e cos 
 R  r  e cos   h  e cos 
r
p
 h (1   cos2  )
p1
h
dq 
p
12  l
3
y Rd  
p
12  l
3
3
h (1   cos  ) R d 
Integratingp above expression from 0 to π,
3
 dh
field
2
q 
 p (1  1 .5  )
12心环
l 形缝隙流动
图 2-21 同
R
O
r
O1
e
图 2-22 偏 心 环 形 缝 隙 流 动
From the expression we conclude that when components are
assembled utterly eccentrically the leakage flow rate is 2.5 times
that when utterly concentrically.
We conclude that the clearances among components require
high positional accuracy degree and hence the initial cost of
hydraulic elements is very high.
5 flow of fluids in orifices and clearances
h1
Using a infinitesimal
element, with length dx
in the x-direction of
the stream.
h2
5.3.3 Gap between
nonparallel plates
u
p1
x
p2
dx
l
The gap may be considered as a gap
between parallel plates
图 2-23 不 平 行 平 板 缝 隙 流
due to a infinitesimal length. In this manner, the expression of
gaps between parallel plates can still be used, as long as the
length L is replaced with dx, and the pressure drop replaced
with –dp.
5 flow of fluids in orifices and clearances
dp  
12  q
bh
3
 
b ( h1 
a. Flow rate
q
dx
12  q
h 2  h1
dx
x)
l
2
2
bh 1 h 2
6  l ( h1  h 2 )
b. Pressure distribution
p 
3
p
6q
b ( h 2  h1 )( h1 
h 2  h1
C
x)
2
l
It may be found that p(x) is a nonlinear curve
5 flow of fluids in orifices and clearances
c. Nonlinear factor analysis
p" ( x ) 
36  q
bl ( h1 
h 2  h1
x)
4
l
In order to make analysis convenient, we define
  arctg
h2  h1
l
h 1  h2 ,   0;
h 1  h2 ,   0 .
5 flow of fluids in orifices and clearances
p "( x ) 
36  qtg 
b ( h1  xtg  )
4
It may be seen that:
On one hand,
When h 1  h 2 ,   0 ,
When h 1  h 2 ,   0 ,

36  q tg 
b
4
h
 h  m ean height 
p(x) is concave.
p(x) is convex.
On the other hand,
The curvature of p(x) is inversely proportional to 4-th
power of the gap height.
The more narrow the gap is, the larger The curvature
of p(x) .
6. Hydraulically sticking
6.1 Concept
Moving a cylindrical body in the barrel frequently need to
exert a great force, even not able to move it. This
phenomenon is call hydraulic stick.
6.2 Cause
Shape deviation and position deviation result in
nonparallel clearances.
6.3 Nonlinear pressure force
l
F  b  p ( x)dx
1
2
0
According to the geometric significance of integration, the
magnitude of the force is equal to the area under the cure
6. Hydraulically sticking
6.4 examples
p
p
p
面 积 A1
面积 A2
(a)
p
面 积 A1
p
面 积 A1
面 积A 1
x
x
x
x
x
x
x
x
p
面积 A2
(b)
p
面积
(c)
图 2-24 液 压 卡 紧 的 各 种 情 况 分 析
2
p
面积 A2
(d)
6. Hydraulically sticking
6.4 Solution
a. Raising the accuracy degree of components( dimension
accuracy degree,shape accuracy degree and position accuracy
degree)
b. Machining balancing grooves or anti-stiction grooves
面积
1
面积
2
After machining pressure
图 2 - 2 5 压 力 balancing
平 衡 槽 的 作 用 grooves, pressure
distribution curves are divide into many segments,
lowering unbalance forces to a great degree.
7 hydraulically shocking
The phenomenon encountered when the velocity of a
liquid is abruptly decreased due to valve movement is called
hydraulic shock or water hammer.
It is possible to damage pipes, elements and systems, even
to injure persons.
When studying hydraulic shock the liquid is not
incompressible and the piping is not rigid.
7.1 Physic model and physic process
Consider a single horizontal pipe
of length L and diameter d.
The upstream end of the pipe is
connected to a reservoir and a valve
is situated at downstream end.
图 2-26 液 压 冲 击 的 物 理 模 型
7 hydraulically shocking
Let us assume that the valve is closed instantaneously.
The lamina of liquid next to the valve will be brought to
rest, its kinematic energy is converted into pressure energy.
When the lamina is compressed, the wall of the pipe will be
stretched.
Then , the second lamina will
be brought to rest, its kinematic
energy is converted into pressure
energy.
图 2-26
液 压 冲 击to
的物
理 模 型its
Next , the third lamina will be
brought
rest,
kinematic energy is converted into pressure energy.
……
After the total liquid will be brought to rest, the total
kinematic energy of liquid will be converted into pressure
energy. The pressure will reach maximum p   p
7 hydraulically shocking
(MPa)
Under an excess pressure, some liquid starts to flow back
into the reservoir. The pressure energy is converted into
kinematic energy. The reverse velocity will produce a
pressure drop that will be below the normal pressure.
After the total liquid in the pipe be brought to motion, The
pressure will reach minimum.
If there were not damp, the periodic process would last
12
forever.
10
8
6
4
2
0
0.1
0.2
0.3
0.4
0.5
7 hydraulically shocking
7.2 Calculation of maximum pressure rise
Assume that the valve is closed instantaneously, the velocity
of liquid will abruptly decrease from v to v’. The cross area of
the pipe will become (A+ΔA ), the pressure will rise to (p+Δp),
and the pressure wave from m-m to n-n in time Δt
n
Selecing an element, writing
momentum equation
( m v ) 2  ( m v )1
F 
t
Where
( mv )1   A  lv
m
+Δ
n
Δ
m
( m v ) 2  (     )( A   A )[  l图2-28
 ( 液l压
)]冲
v击
' 计算单元体
 (     )( A   A )  lv '
7 hydraulically shocking
n
Neglect higher order quantity
( m v ) 2   A  lv '
m
+Δ
The resultant force is
F
 p A  ( p   p )( A   A )
n
 pA  ( pA  p  A   pA   p  A )
  p  A   pA   p  A
  pA
Insert these in the momentum equation, we obtain
p  
l
t
m
图 2-28 液 压 冲 击 计 算 单 元 体
Since p  A   pA ,  p  A   pA
F
Δ
( v  v ')   c ( v  v ')
Where c---travel velocity of the shock wave
7 hydraulically shocking
c. Some measures
1 Close valves as slowly as possible
2 Use accumulators to absorb shock
3 Arrange buffering valves and anti-overload valves；
4 Use rubber hoses to connect elements
8. Cavitations
When the local pressure becomes equal to the vapor
pressure of the liquid, small vapor bubbles are generated
and these bubbles collapse when they enter a highpressure region. The collapse is accompanies by very large
local pressures that last for only a small fraction of a
second. These pressure spikes may reach a wall, where
they can, after repeated applications result in significant
damage.
The phenomenon is called cavitations.
In order to prevent cavitations from occurring, the
system pressure, including all local pressures, should
always be ensured to be above the atmospheric pressure.