Download Pressure field and buoyancy. Elementary fluid dynamics. Bernoulli

Pressure in stationary and
moving fluid
Lab-On-Chip: Lecture 2
Fluid Statics
• No shearing stress
•.no relative movement between
adjacent fluid particles, i.e. static or
moving as a single block
Pressure at a point
Question: How pressure depends on the orientation of the plane ?
Newton’s second law:
∑ Fy = p yδ xδ z − psδ xδ s sin θ = ρ
δ xδ yδ z
ay
2
δ xδ yδ z
δ xδ yδ z
=
−
−
=
y
p
x
s
az
F
p
x
cos
δ
δ
δ
δ
θ
γ
ρ
∑ z z
s
2
2
δ y = δ s cos θ δ z = δ s sin θ
δy
p y − ps = pa y
2
pz − ps = ( paz + ρ g )
δz
2
if δ x, δ y, δ z → 0, p y = ps , pz = ps
• Pascal’s law: pressure doesn’t depend on the orientation of
plate (i.e. a scalar number) as long as there are no shearing
stresses
Basic equation for pressure field
Question: What is the pressure distribution in liquid in
absence shearing stress variation from point to point
•
Forces acting on a fluid
element::
– Surface forces (due to
pressure)
– Body forces (due to weight)
Surface forces:
δ Fy = ( p −
∂p δ y
∂p δ y
)δ xδ z − ( p +
)δ xδ z
∂y 2
∂y 2
∂p
δ yδ xδ z
∂y
∂p
δ Fx = − δ yδ xδ z
∂x
∂p
δ Fz = − δ yδ xδ z
∂z
δ Fy = −
Basic equation for pressure field
G
∂p G ∂p G ∂p G
δ F = −( i +
j + k )δ yδ xδ z
Resulting surface force in vector form:
∂x
∂y
∂z
G
G
G
G
∂
∂
∂
δF
If we define a gradient as:
j+ k
∇= i +
= −∇p
∂x
∂y
∂z
δ yδ xδ z
The weight of element is:
Newton’s second law:
G
G
−δ Wk = − ρ g δ yδ xδ z k
G
G
G
δ F − δ Wk = δ ma
G
G
−∇pδ yδ xδ z − ρ g δ yδ xδ z k = − ρ g δ yδ xδ z a
General equation of
motion for a fluid w/o
shearing stresses
G
G
−∇p − ρ gk = − ρ ga
Pressure variation in a fluid at rest
• At rest a=0
G
−∇p − ρ gk = 0
∂p
=0
∂x
• Incompressible fluid
p1 = p2 + ρ gh
∂p
=0
∂y
∂p
= −ρ g
∂z
Fluid statics
Same pressure –
much higher force!
Fluid equilibrium
Transmission of fluid pressure,
e.g. in hydraulic lifts
• Pressure depends on the depth in the solution
not on the lateral coordinate
Compressible fluid
• Example: let’s check pressure variation in the air (in
atmosphere) due to compressibility:
– Much lighter than water, 1.225 kg/m3 against 1000kg/m3 for
water
– Pressure variation for small height variation are negligible
– For large height variation compressibility should be taken
into account:
nRT
= ρ RT
V
dp
gp
= −ρ g = −
dz
RT
p=
⎡ g ( z1 − z2 ) ⎤
−h / H
assuming T = const ⇒ p2 = p1 exp ⎢
⎥ ; p (h) = p0 e
⎣ RT0 ⎦
~8 km
Measurement of pressure
•
Pressures can be designated as absolute or gage (gauge) pressures
patm = γ h + pvapor
very small!
Hydrostatic force on a plane surface
• For fluid in rest, there are no shearing stresses present and
the force must be perpendicular to the surface.
• Air pressure acts on both sides of the wall and will cancel.
Force acting on a side wall in rectangular container:
h
FR = pav A = ρ g bh
2
Example: Pressure force and moment acting on
aquarium walls
• Force acting on the wall
H
H2
FR = ∫ ρ gh dA = ∫ ρ g ( H − y ) ⋅ bdy = ρ g
b
2
A
0
• Generally:
FR = ρ g sin θ ∫ ydA = ρ g sin θ yc A
A
Centroid (first moment of the area)
• Momentum of force acting on the
wall
H
H3
FR yR = ∫ ρ gh ydA = ∫ ρ g ( H − y ) y ⋅ bdy = ρ g
b
6
0
A
yR = H 3
• Generally,
yR =
2
y
∫ dA
A
yc A
Pressure force on a curved surface
Buoyant force: Archimedes principle
• when a body is totally or partially submerged a fluid
force acting on a body is called buoyant force
FB = F2 − F1 − W
F2 − F1 = ρ g (h2 − h1 ) A
FB = ρ g (h2 − h1 ) A − ρ g [ (h2 − h1 ) A − V ]
FB = ρ gV
Buoyant force is acting on the centroid
of displaced volume
Stability of immersed bodies
• Totally immersed body
Stability of immersed bodies
• Floating body
Elementary fluid dynamics:
Bernoulli equation
Bernuolli equation – ”the most used and most abused
equation in fluid mechanics”
Assumptions:
• steady flow: each fluid particle that passes through a given
point will follow a the same path
• inviscid liquid (no viscosity, therefore no thermal
conductivity
F=
F= ma
ma
Net pressure force + Net gravity force
Streamlines
Streamlines: the lines that are tangent to velocity
vector through the flow field
Acceleration along the
streamline:
Centrifugal acceleration:
dv ∂v ∂s ∂v
as =
=
= v
dt ∂s ∂t ∂s
v2
an =
R
Along the streamline
∂v
∑ δFs = mas = ρδV ∂s v
∂p
∑ δFs = δWs + δFps = − ρgδV sin(θ ) − ∂s δV
∂v
∂p
− ρg sin(θ ) −
=ρ v
∂s
∂s
Balancing ball
Pressure variation along the streamline
• Consider inviscid, incompressible, steady flow along the
horizontal streamline A-B in front of a sphere of radius a.
Determine pressure variation along the streamline from point
A to point B. Assume:
⎛ a3 ⎞
V = V0 ⎜1 + 3 ⎟
⎝ x ⎠
Equation of motion:
∂p
∂v
= −ρv
∂s
∂s
3
3
⎛
⎞
a
a
∂v
v = −3v02 ⎜1 + 3 ⎟ 4
∂s
⎝ x ⎠x
∂p 3ρ a 3v02 ⎛ a 3 ⎞
=
⎜1 + 3 ⎟
4
x
∂x
⎝ x ⎠
6
3
⎛
3ρ a 3v02 ⎛ a 3 ⎞
a
a
1
⎛ ⎞ ⎞
2
∆p = ∫
⎜1 + 3 ⎟ dx = − ρ v0 ⎜⎜ 3 + ⎜ ⎟ ⎟⎟
4
x
⎝ x ⎠
−∞
⎝ x 2⎝ x⎠ ⎠
−a
Raindrop shape
The actual shape of a raindrop is a result of
balance between the surface tension and the air
pressure
Bernoulli equation
Integrating
− ρg sin(θ ) −
dz
ds
∂v
∂p
=ρ v
∂s
∂s
dp
, n=const along streamline
ds
dz dp 1 dv 2
−ρ g −
= ρ
ds ds 2 ds
We find
Assuming
incompressible
flow:
dp +
1
ρd (v 2 ) + ρgdz = 0
2
p+
1 2
ρv + ρgz = const
2
Bernoulli equation
Along a streamline
Along a streamline
Example: Bicycle
•
Let’s consider coordinate system fixed to the bike.
Now Bernoulli equation can be applied to
1
2
p2 − p1 = ρv0
2
Pressure variation normal to streamline
v2
∑ δFn = man = ρδV R
∑ δFn = δWn + δFpn = − ρgδV cos(θ ) −
∂p
δV
∂n
dz ∂p
v2
=ρ
− ρg
−
dn ∂n
R
v2
p + ρ ∫ dn + ρgz = const
R
compare
1 2
p + ρv + ρgz = const
2
Across streamlines
Along a streamline
Free vortex
Example: pressure variation normal to streamline
• Let’s consider 2 types of vortices with the velocity
distribution as below:
free vortex
solid body
rotation
dz ∂p
v2
− ρg
−
=ρ
dn ∂n
R
as
∂
∂
∂p ρV 2
=− ,
=
= ρ C12 r
∂n
∂r ∂r
r
1
p = ρ C12 ( r0 2 − r 2 ) + p0
2
∂p ρV 2
C12
=
=ρ 3
r
r
∂r
1
1⎞
2⎛ 1
p = ρ C2 ⎜ 2 − 2 ⎟ + p0
2
⎝ r0 r ⎠
p+
1 2
ρv + ρgz = const
2
Static, Stagnation, Dynamic and TotalPressure
• each term in Bernoulli equation has dimensions of pressure
and can be interpreted as some sort of pressure
1 2
p + ρv + ρgz = const
2
hydrostatic pressure,
dynamic pressure,
static pressure,
point (3)
Velocity can be determined from stagnation pressure:
1 2
ρv
2
Stagnation pressure
p2 = p1 +
On any body in a flowing fluid there is a stagnation
point. Some of the fluid flows "over" and some
"under" the body. The dividing line (the stagnation
streamline) terminates at the stagnation point on
the body.
As indicated by the dye filaments in the water
flowing past a streamlined object, the velocity
decreases as the fluid approaches the stagnation
point. The pressure at the stagnation point (the
stagnation pressure) is that pressure obtained
when a flowing fluid is decelerated to zero speed
by a frictionless process
Pitot-static tube
Steady flow into and out of a tank.
ρ1 A1v1 = ρ 2 A2 v2
Determine the flow rate to keep the height constant
1 2
1
2
p1 + ρv1 + ρgz1 = p2 + ρv2 + ρgz 2
2
2
Q = A1v1 = A2 v2
Venturi channel
Measuring flow rate in pipes
1 2
1
2
p1 + ρv1 = p2 + ρv2
2
2
Q = A1v1 = A2 v2
Probelms
• 2.43 Pipe A contains gasoline (SG=0.7), pipe B contains oil
(SG=0.9). Determine new differential reading of pressure in A
decreased by 25 kPa.
• 2.61 An open tank contains gasoline r=700kg/cm at a depth
of 4m. The gate is 4m high and 2m wide. Water is slowly
added to the empty side of the tank. At what depth h the gate
will open.
Problems
• 3.71 calculate h. Assume water inviscid and
incompressible.