)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
2.2.2 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
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
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
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
=
−
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
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.
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.
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.
Mass forces˖
f
x
⋅
ρ
1 6
dxdydz
,
z A
px
dz
pn
py
0 dy dx
pz B x
C y
f
ห้องสมุดไป่ตู้
y
⋅
ρ
1 6
dxdydz
,
fz
⋅
ρ
1 6
dxdydz
(3) Equation of fluid in equilibrium
Consider force components in x direction:
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
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.
condition:
Equilibrium and relative equilibrium Compressible and incompressible flow
2.2 Differential Equations of a Fluid in Equilibrium
fx
−
ρ1
∂p ∂x
=
0
fy
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.
+
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)
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
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