This document summarizes key concepts from Chapter 2 on fluid statics. It discusses static pressure and its characteristics, including that pressure is perpendicular to the surface and the same in all directions at a point. It presents the basic equation of fluid statics and hydrostatics under gravity. It also covers isobaric surfaces, units of pressure measurement, and the relationships between different pressure types such as absolute, gauge, and vacuum pressures.

1. Chapter 2. FLUID STATICS
第二章 流体静力学
双 语 课 程

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2.1 Static Pressure and Characteristics
2.2 Basic Equation of Fluid Statics
2.3 Units and Scales of Pressure Measurement
2.4 Manometers ++
2.5 Forces on Plane Areas #
Home Work
Outline

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2.1 Static Pressure and Characteristics
2.1 静压强及其特性

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2.1.1 Static Pressure
dA
dP
ΔA
ΔP
p
ΔA



lim
0
Definition: the pressure p is the normal force per
unit area in a fluid.
SI unit of pressure is N/m2, called the Pascal. (帕斯卡).
Fluid is at rest, the pressure is called static pressure
(静压强).

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2.1.2 Characteristics of Static Pressure
Two characteristics:
1. The direction of pressure is perpendicular（垂直的） to
the acting surface (作用面) and in the same direction of
inward normal (内法线) of the surface.
Proof : Assuming the fluid is at rest, the direction of static
pressure has an angle α (α≠90°) with tangent of an element
of area, p is decomposed into: tangential pressure pt , normal
pressure pn , as shown in Fig. 2.1.
Figure 2.1 Tangential pressure and normal pressure
(shear stress)

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Discuss: the fluid will deform continuously
when subjected to a shear stress (here is pt ≠ 0). It
means that fluid would flow. That is to say the
assumption of static fluid is a contradiction（矛盾）.
fluid is at rest -> pt =0;
So, If fluid would keep in a rest, the only forces are
the pressure forces(压力) acting on the inward normal
direction of fluid surface.

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2. The pressures at a point in a fluid at rest are
the same in all directions. It has nothing to do with
the direction of the acting surface.
Proof: An element of fluid in arbitrary tetrahedron
[‘tetrə‘hedrən](四面体) ABCD is taken in a fluid at rest, and the
point A is in coincident with the origin of rectangular coordinates,
as shown in Fig. 2.2.

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The total pressure forces on the four face ABC, ACD,
ABD and BCD are:
dydz
p
P x
x
2
1
 dxdz
p
P y
y
2
1
 dxdy
p
P z
z
2
1
 n
n
n dA
p
P 
The total mass forces is:
f
W dxdydz

6
1

in which dAn is the area of face BCD.
The components of force per unit mass are fx, fy, fz in the x, y, z
directions and the components of mass forces in the three axis are :
x
x ρdxdydzf
W
6
1
 y
y ρdxdydzf
W
6
1
 z
z dxdydzf
W 
6
1


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The sum of components of all forces in any axis is equal to zero.
0
x
F 
 0
y
F 
 0
z
F 

In the x direction
0
cos 

 x
n
x W
P
P 
0
6
1
cos
2
1


 x
n
n
x ρdxdydzf
α
dA
p
dydz
p
0
3
1


 dx
f
p
p x
n
x 
Because of
dydz
α
dAn
2
1
cos 
So

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n
x p
p 
The direction of n is arbitrary. Hence in a
static fluid the pressure at a point is the same
in all directions.
Similarly
n
z
y
x p
p
p
p 


The last term on the left of equation is an infinitesimal
[,infini’tesəməl]（无限小,近零值）and can be neglected.
1
3
x
f dx


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2.2 Basic Equation of Fluid Statics
2.2 流体静力学基本方程

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2.2.1 Differential equilibrium equation
In static fluid, a parallelepiped [pærəle‘lepipid]
(平行六面体) element with the sides dx, dy and dz is
selected as shown in Fig. 2.3
Figure 2.3 Forces on a parallelepiped element in the x direction

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Suppose that the static pressure at the center is p. The static
pressures exerted on the centers of six faces can be written in
Taylor series.
In the x direction:

























3
3
3
2
2
2
2
6
1
2
2
1
2
dx
x
p
dx
x
p
dx
x
p
p

























3
3
3
2
2
2
2
6
1
2
2
1
2
dx
x
p
dx
x
p
dx
x
p
p
By neglecting the second and higher order infinitesimals
dx
x
p
p



2
1
dx
x
p
p



2
1
the partial (derivative) of p with respect to x

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So the two total pressure forces acting on the left and right
surfaces are respectively:
dydz
dx
x
p
p 








2
1 dydz
dx
x
p
p 








2
1
If the average density of fluid element is ρ, then the
components of mass force along the three coordinate axes are:
ρdxdydz
fx ρdxdydz
fy ρdxdydz
fz
,

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The equilibrium condition of the element of fluid under the
static condition is: the sum of components of the resultant acting
on the element in the three coordinate axes should be zero.
In the x axis,
0
2
1
2
1




















 ρdxdydz
f
dydz
dx
x
p
p
dydz
dx
x
p
p x
Simplifying the above equation
0
1




x
p
fx


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The other two directions yield similar expressions
0
1




y
p
fy

0
1




z
p
fz

In vector form 1
p

 
f
(2.4)
in which  is
gradient(梯度),
















x
x
x
k
j
i
Equation (2.4) is the fluid equilibrium differential equation
( Euler’s equilibrium differential equation).
Physical meaning: in the static fluid, the mass force per unit
mass of fluid and the resultant forces of static pressure at a point
are mutually (相互地) equilibrium.

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total differential (全微分) is
dz
z
p
dy
y
p
dx
x
p
dp









This equation is called the pressure difference formula.
)
( dz
f
dy
f
dx
f
ρ
dp z
y
x 

 (2.5)
Multiplying Eq. (2.4) on both sides by dx, dy, dz respectively and
then adding them together gives
dz
z
p
dy
y
p
dx
x
p
dz
f
dy
f
dx
f
ρ z
y
x










 )
(
)
,
,
( z
y
x
p
p 
thus
Physical meaning: if the increments of coordinates at a spatial
point are dx, dy and dz, the static pressure might have a increment
of dp correspondingly, and the static pressure is also determined
by the mass force.

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2.2.2 Basic equation of hydrostatics under gravity
Gravity G (G=mg) is the only mass force acting on the liquid
fx=0，fy=0，fz= g
Figure 2.4 A vessel containing liquid at rest
0


ρg
dp
dz
rewriting
ρgdz
dp 

From (2.5)
c is the constant of integration (积分常数)
determined by the boundary condition.
c
g
p
z 


integrating
(2.7)

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g
p
z
g
p
z


2
2
1
1 


)
( 2
1
1
2 z
z
g
p
p 

 
For the two points 1 and 2
in the static fluid
For the two points 0 and 1
)
( 1
0
0
1 z
z
g
p
p 

  Figure 2.4 A vessel containing liquid at rest
The pressure at a point in liquid at rest consists of two
parts: the surface pressure, and the pressure caused by
the weight of column of liquid.

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Physical meaning
z —— the position potential energy per unit weight of fluid to
the base level；
p/g —— the pressure potential energy (压强势能) per unit
weight of fluid.
Geometrical meaning
z ——the position height or elevation head (位置水头)
p/g——the pressure head (压力水头) per unit weight of fluid
Sum of the position head (位置水头) and pressure head is
called the hydrostatic head ( 静 水 头 ), also known as the
piezometric head (测压管水头).
The energy per unit weight of fluid can be also
expressed in terms of the length of column of
liquid (液柱), and called the head (水头).

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Three important conclusions：
(2) Under gravity, the increase of static pressure is
directly proportional to the increase of depth measured
vertically downward from the reference - liquid surface.
(3) Each point at the same elevation (h=const.) has the
same static pressure, namely any horizontal plane is an
isobaric surface (等压面) .
(1) The hydrostatic pressure at any point comprises of
two parts: the pressure p0 and the weight of liquid column.

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2.2.3 Isobaric surface（等压面）
p(x, y, z) = C
The interface between liquid and gas, namely the free
surface (自由面) of liquid, is an isobaric surface.
The interface between two kinds of liquid which do not
blend (混合) each other is also the isobaric surface.
In the fluid, isobaric surface is defined by points where
the value of pressure is constant.

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Important property: the isobaric surface and the mass force
are mutually perpendicular (相互垂直).
Proof: Because p is constant, so dp=0, from Eqt.(2.5) the
differential equation of isobaric surface is
Figure 2.5 Inner product of
two vectors
/ 0
x y z
f dx f dy f dz dp 
   
cos 0
x y z
f dx f dy f dz d d 
    
f s f s
To make inner product of two vectors
equal to zero, f and ds must be mutually
perpendicular, or the angle φ= 90.
means that the inner product(内积, 点积)of
the force per unit mass f at point A and the
unit vector of line element ds (dx, dy, dz)
through that point, as shown in Fig. 2.5.

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2.3 Units and Scales of Pressure Measurement
2.3 压强单位及度量

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local atmospheric pressure (当地大气压) pa
absolute pressure (绝对压强) pabs gh
p
p 

 a
abs
gauge pressure (表压, 计示压强) = relative pressure
vacuum pressure (真空压强, 真空度) pv , or suction pressure
(吸入压强) , also called negative pressure (负压强)
abs
a
v p
p
p 

relative pressure (相对压强) p abs a
p p p gh

  
It is usually measured in the height of liquid column, such as
millimeters of mercury (mmHg, 毫米水银柱), denoted by hv.
g
p
p
g
p
h a


abs
v
v



Related Pressures

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Local atmospheric pressure
p=pa
Complete vacuum pabs=0
Absolute
pressure
Vacuum
pressure
Gauge
pressure
Figure 2.6 absolute pressure, gauge pressure and vacuum pressure
p
Absolute
pressure
2 p<pa
1 p>pa
O
Relationship Graph

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To avoid any confusion, the convention is
adopted throughout this text that a pressure is
in gauge pressure unless specifically marked
‘abs’, with the exception of a gas, which is
absolute pressure unit.
Attention:

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2.4 Manometers
2.4 测压计 ++

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2.4.1 Piezometer[,paiə’zɔmitə]（压力计）
Structure：
The bottom is connected to
the container with liquid and the
top opening is open to atmosphere.
Measurement principle：
a
e abs
p p p gh

  
pe ——gauge pressure（表压） Figure 2.7 Piezometer
 M
pa
h
It is just used to measure the smaller pressure
( < 9800Pa =1mH2O) , and only for liquid.

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2.4.2 U-tube manometer [mə’nɔmitə]
Structure:
U-shaped glass tube
one end is connected
with the container
other end open to atmosphere
Measurement principle:
p>pa
pM=pMabspa=2gh2  1gh1
p<pa pv=papabs=  1gh1+  2gh2 Figure 2.8 U-tube manometry ( p>pa)
It is used to measure the pressure(< 0.294MPa) high or
below atmospheric pressure(0.1MPa )for liquid or gas.
2
pa
1
M
p
1
h1
2 > 1
h2
Measured fluid
Working medium
Figure 2.9 U-tube manometry (p<pa)
2
pa
1
M
p
1
2
h2
h1

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2.4.3 Differential manometer[mə’nɔmitə](差压计)
used to measure the differences in
pressure for two containers or two
points in a container.
)
(
)
( 1
2
1
1 h
h
g
gh
p
p B
A 



 


Structure：
Measurement principle：
Figure 2.10 Differential
manometer
2
A
A
pA
1
 >A, B
h
h1
B
B
pB
h2
)
( 1
2 h
h
g
gh
gh
p
p A
B
B
A 



 


gh
p
p B
A 


ρA= ρB= ρ1
For two same air, ρA= ρB= 0

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EXAMPLE 2.1
A pressure measurement
apparatus without leakage and
friction between piston and
cylinder wall is shown in Fig.
2.11. The piston diameter is
d=35mm, the relative density of
oil is RDoil=0.92, the relative
density of mercury is RDHg=13.6,
and the height is h=700 mm. If
the piston has a weight of 15N,
calculate the value of height
difference of liquid Δh in the
differential manometer.
1 1
pa
h
RDoil=0.92
RDHg=13.6
d
h
Figure 2.11 Pressure
measurement apparatus
piston
pa

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The pressure on the piston under the weight
Pa)
(
15590
035
.
0
4
15
4
15
2
2






d
p
From the isobaric surface 1-1 the equilibrium equation is
h
g
gh
p 

 Hg
oil 

solving for h
)
m
(
164
.
0
70
.
0
6
.
13
92
.
0
806
.
9
13600
15590
Hg
oil
Hg







 h
g
p
h



Solution:

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A double differential manometer is used to measure the
difference in pressure at two points A and B as shown in Fig.
2.12. If h1=600mm, h2=250mm, h3=200 mm, h4=300mm,
h5=500mm, ρ1=1000kg/m3, ρ2=800kg/m3, ρ3=13598kg/m3.
Determine the pressure difference between point A and point B.
EXAMPLE 2.2
Figure 2.12 Two differential manometers in series
h3
2
1
A
pA
1
3
h1
h2
3
1
4
h4
1
B
pB
4
3
2
2
h5

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Solution:
Lines 1-1, 2-2 and 3-3 are all isobaric surface
The basic equation of hydrostatics can be applied to each point
and the pressure at each point is
p1=pA+1gh1
p2=p1 3gh2
p3=p2+ 2gh3
p4=p3 3gh4
pB=p4 1g(h5h4)

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Substituting the former formula into the following formula one
by one
pB=pA+1gh13gh2+2gh33gh41g(h5h4)
pApB
=1g(h5h4)+3gh4+3gh22gh31g h1
= g[1(h5h4 h1)+3(h4+h2)2h3]
=9.806×[1000×(0.50.30.6)+13598×(0.3+0.25)800×0.2]
= 67847(Pa)

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2.5 Forces on Plane Areas
2.5 平面上的力 #

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A plane surface may be inclined, horizontal or vertical. The
inclined surface (斜面) is a common case and the others are its
special case.
y
N
O
yc
yp

F hp
h
hc
b
y
dy
Figure 2.13 Resultant force on an inclined plane in static fluid
M
CA
CP
x
Plane surface
Trace(投影线段)

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h - depth from free surface to an arbitrary point on the inclined plane
y - the corresponding distance in the Oy axis
element strip of the area at depth h, with width b and thickness dy
dA=bdy
The magnitude of force on the element area is
dF=pdA=ρghdA=ρgysindA
A
y
g
ydA
g
F c
A



 sin
sin 
 
Let hc = ycsinθ
F =ρg hc A
2.5.1 The magnitude of the resultant force
area moment (面积矩)
distance from the centroid
（质心） CA to the Ox axis
vertical depth of the centroid. If
it keeps a constant, F will remain
unchanging for any angle .

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2.5.2 Center of pressure
What is center of pressure ?
yp - distance from center of pressure CP to point O in the y direction
x
A
p I
g
dA
y
g
F
y 


 sin
sin 2

 
Ix - inertia moment (惯性矩) of plane surface about the x axis


A
x dA
y
I 2
A
y
I
A
y
g
I
g
y
c
x
c
x
p 





sin
sin
The point where the line of action (作用线) of total hydrostatic
force pierces through in the plane surface submerged in static liquid
is called center of pressure or pressure center.
According to the theorem of resultant moment (合力矩定理),
the moment of total hydrostatic force F about the x axis is

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parallel-axis theorem for moments of inertia (惯性矩平移定理)
cx
c
x I
A
y
I 
 2
A
y
I
y
A
y
I
A
y
y
c
cx
c
c
cx
c
p 



2
Similarly, in the x direction
A
y
I
x
A
y
I
x
c
cxy
c
c
xy
p 


The resultant force passes through the centroid of the horizontal area.
in which Icx is the inertia moment of area about its centroidal axis
(形心轴) which is parallel to the x axis, as shown in Table 2.1.
Usually be neglected.
F=ρghA

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Table 2.1 Geometrical properties of several sections

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EXAMPLE 2.3
The both sides of a rectangular sluice gate (闸门) are partly
submerged in water as shown in Fig. 2.14. If the depth of water on
both sides are h1=2m and h2=4m respectively, determine the
magnitude and point of the net total force per meter width on the
gate.
h1 F1
h2
F2
F
h
Figure 2.14 Rectangular sluice gate
O

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Solution:
The centroid of left side of the rectangular sluice with the depth
h1 below the free surface is
yc=hc=h1/2
The toal force per meter width on the left side of gate is
1
1 c 1
2 2
1
1
2
1 1
1000 9.806 2 19612(N)
2 2
h
F gh A g h
gh
 

  
     
The position of force F1 is 3
1
1 1
1 1
1
1 1
1
2
12
2 3
2
c
p c
c
bh
I h h
y y
h
y A bh
    


45. 45
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The toal force per meter width on the right side of gate is
N)
(
78448
4
9806
2
1
2
1 2
2
2
2 



 gh
F 
The net total force per meter on the sluice gate is
F = F2 F1=7844819612=58836(Ｎ)
Assuming that the distance from the line of action of the net total
force to the bottom is h, the torques should be balanced about the
point O at the bottom of sluice.
2 1
2 2 1 1
2 2
( ) ( )
3 3
h h
Fh F h F h
   
)
m
(
56
.
1
58836
3
2
19612
4
78448
3
1
1
2
2








F
h
F
h
F
h

46. 46
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Home Work
2.1, 2.5, 2.6,
2.8, 2.9
Homework