工程流体力学(英文版)第二章.pdf

The n results show that the pressures are independent of direction n because n is arbitrary.
Hence the pressure at a point on a static fluid is the
same in all directions.
dp = 0
fxdx + f ydy + fzdz = 0
Important character of equipressure surface:
f = fxi + fy j + fxk dr = dxi + dyj + dzk
Kinescope Cartoon
f ⋅ dr = 0
mass force of any point on the equipressure surface in equilibrium fluid is perpendicular to the equipressure surface.
−
∂π
∂z
2.2.4 Equipressure Surface
dp = ρ( fx dx + f y dy + fz dz)
Equipressure Surface is a surface that the pressure of every point in
liquid is equal. Common equipressure surfaces are free liquid surface and interface of two unmixed fluids in equilibrium.
p = lim Fn A→ 0 A
Unit: Pa or N/m2
2.2.1 Characteristic
1ǃdirection
Negative √
Normal
Force
Positive--Pulling
Shearing
There is only compressive stress (or pressure ) in a fluid at rest , and the direction of pressure is the same as the direction of inward normal line of acting point . Fluid at rest cannot bear pulling force because of the trends to flow.
−
ρ1
∂p ∂y
=
0
fz
−
ρ1
∂p ∂z
=
0
˄or˅
f − ρ1 gradp = 0
Physical Meaning:
For the fluid in equilibrium, surface force components per mass fluid are equal to mass
force components per mass fluid. Pressure variation rate in axes directions
z
dz
%
•$
&
dx dy
y
x
2.2 Differential Equations of a Fluid in Equilibrium
Consider
force components in y direction˖
f百度文库
(x0
+ ∆x)
=
f
(x0 ) +
f
'(x)∆x +
f
''(x) (∆x)2
˄
∂p ∂x
,
∂p ∂y
,
∂p ∂z
) are equal to mass force components per unit volume in axes directions
˄ρfx, ρfy, ρfz) respectively.
2.2 Differential Equations of a Fluid in Equilibrium
2.2 Differential Equation of Fluid Equilibrium 2.3 Pressure Distribution in the Static Fluid 2.4 Pressure Mearurements 2.5 Fluid in Relative Equilibrium Fluid. 2.6 Fluid Static Force on Plane and Curved Area
condition:
Equilibrium and relative equilibrium Compressible and incompressible flow
2.2 Differential Equations of a Fluid in Equilibrium
fx
−
ρ1
∂p ∂x
=
0
fy
2.2.2 Pressure Difference Equation ( General Differential Equations of a Fluid in
Equilibrium)
Ĩp = p(x,y,z)
fx
−
ρ1
∂p ∂x
=
0
fy
−
ρ1
∂p ∂y
=
0
fz
−
ρ1
∂p ∂z
=
0
ħ
the
total
dz
%
•$
&
dx dy
y
(p−
∂p ∂y
dy 2
)dxdz
−(p
+
∂p ∂y
dy 2
)dxdz
+
f y ρdxdydz
=
0
2.2 Differential Equations of a Fluid in Equilibrium
(p
−
∂p ∂y
dy 2
)dxdz
−(p+
∂p ∂y
dy 2
Mass forces˖
f
x
⋅
ρ
1 6
dxdydz
,
z A
px
dz
pn
py
0 dy dx
pz B x
C y
f
y
⋅
ρ
1 6
dxdydz
,
fz
⋅
ρ
1 6
dxdydz
(3) Equation of fluid in equilibrium
Consider force components in x direction:
Equilibrium (a=0) relative equilibrium ( a = 0)
Characteristic of Fluid at rest
u=0
du=0
τ = µ du = 0
dy
Contents
2.1 Statical pressure intensity and its characteristic
)dxdz
+
f y ρdxdydz
=
0
fy
−
ρ1
∂p ∂y
=
0
Differential Equations of a Fluid in Equilibrium˄Euler Equilibrium Equations˅˖
fx
−
ρ1
∂p ∂x
=
0
fy
−
ρ1
∂p ∂y
=
0
fz
−
ρ1
∂p ∂z
=
0
˄or˅
f − ρ1 gradp = 0
2
+⋅⋅⋅
Surface forces:
left:
pB
=
p−
∂p ∂y
dy 2
(
p
−
∂p ∂y
dy 2
)dxdz
right:
pC
=
p+
∂p ∂y
dy 2
−(
p
+
∂p ∂y
dy 2
)dxdz
z
Mass forces˖
ρ f ydxdydz
There is Fy=0 in y direction because the element is in equilibrium : x
If the density is a constant:
d ( p ρ ) = fxdx + f y dy + fzdz
Define a force potential function:
p
ρ
=
−π
d
p
ρ
=
d
(−π
)
=
−
∂π
∂x
dx
−
∂π
∂y
dy
−
∂π
∂y
dz
fx
=
−
∂π
∂x
fy
=
−
∂π
∂y
fz
=
differential
of
pressure
is:
dp
=
∂p ∂x
dx
+
∂p ∂y
dy
+
∂p ∂z
dz
Multiply every equation in equation group (1) with dx,dy,dz
respectively, then add them:
f x dx
+
f ydy
2ǃmagnitude
The pressure at a point in a fluid at rest is the same in all directions. It has nothing to do with the normal direction of the acting surface.
px = py = pz = pn
z A
px
dz
pn
py
0 dy dx
pz B x
C y
(1) Select a triangular prism element OABC, dx, dy, dz
px, py, pz, pn are pressure intensity acted on the respective surface. n is normal direction of inclined surface ABC.
2.1 Statical pressure intensity and its characteristic
2.2.1 Definition of statical pressure intensity
Normal force acting over per unit area of a static fluid
+
f z dz
=
ρ1
∂p ( ∂x
dx +
∂p ∂y
dy
+
∂p ∂z
dz)
dp = ρ( fxdx + f ydy + fzdz)
2.2 Differential Equations of a Fluid in Equilibrium
2.2.3 Force Potential Function dp = ρ( fxdx + f y dy + fzdz)
p = p (x, y, z)
2.2 Differential Equation of Fluid Equilibrium
1 Differential Equations of a Fluid in Equilibrium ------Euler Equilibrium Equations
2 Pressure Difference Equation 3 Force Potential Function 2 Surface of Equal Pressure
−
pn
⋅
1 2
dydz
+
fx
⋅ρ
1 dxdydz 6
=
0
px
−
pn
+
fx
⋅ρ
1 dx 3
=
0
0 dy dx
pz B x
C y
When the element shrinks to a point o , dx→0ˈthus
Similarly˖
px = pn
px = py = pz = pn
Chapter Two Fluid Statics(⌕ԧ䴭࡯ᄺ˅
What is Fluid Statics
General Rules of fluid at rest, and their engineering application.
fluid at rest
fluid in equilibrium
2.2 Differential Equation of Fluid Equilibrium
2.2.1 Differential Equations of a Fluid in Equilibrium ------Euler Equilibrium Equations
Consider the six surfaces of infinitesimal element in equilibrium fluid. Its sides are dx,dy,dz. Assume the pressure at the center of the element is p(x,y,z)=p.
The surface pressure:
px
⋅
1 2
dydz
,
py
⋅
1 2
dxdz
,
pz
⋅
1 2
dxdy
,
pn ⋅ dA
(2) Force analysis:
Surface forces:
OAC: OAB: OBC: ABC:
px
⋅
1 2
dydz
py
⋅
1 2
dxdz
pz
⋅
1 2
dxdy
pn ⋅ dA
F =0
px
⋅
1 2
dydz
−
pn
⋅ dAcos(n, x) +
fx
⋅ρ
1 dxdydz 6
=
0
px
⋅ 1 dydz 2
−
pn
⋅ dAcos(n, x) +
fx
⋅ρ
1 dxdydz 6
=
0
And:
dAcos(n, x) = 1 dydz 2
z A
dz py
px pn
So:
px
⋅ 1 dydz 2