pressure in stationary and moving fluid - aalborg...
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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
• 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
sin2
cos2 2
y y s y
z z s z
x y zF p x z p x s a
x y z x y zF p x y p x s a
δ δ δδ δ δ δ θ ρ
δ δ δ δ δ δδ δ δ δ θ γ ρ
= − =
= − − =
∑
∑
Newton’s second law:
cos sin
2
( )2
, , 0, ,
y s y
z s z
y s z s
y s z syp p pa
zp p pa g
if x y z p p p p
δ δ θ δ δ θδ
δρ
δ δ δ
= =
− =
− = +
→ = =
Question: How pressure depends on the orientation of the plane ?
Basic equation for pressure field
• Forces acting on a fluid element::
– Surface forces (due to pressure)
– Body forces (due to weight)
Question: What is the pressure distribution in liquid in absence shearing stress variation from point to point
( ) ( )2 2y
y
x
z
p y p yF p x z p x zy y
pF y x zypF y x zxpF y x zz
δ δδ δ δ δ δ
δ δ δ δ
δ δ δ δ
δ δ δ δ
∂ ∂= − − +
∂ ∂∂
= −∂∂
= −∂∂
= −∂
Surface forces:
Basic equation for pressure field( )p p pF i j k y x z
x y zδ δ δ δ∂ ∂ ∂
= − + +∂ ∂ ∂
Fi j k px y z y x z
δδ δ δ
∂ ∂ ∂∇ = + + = −∇
∂ ∂ ∂
Resulting surface force in vector form:
If we define a gradient as:
The weight of element is: Wk g y x z kδ ρ δ δ δ− = −
Newton’s second law: F Wk ma
p y x z g y x z k g y x z a
δ δ δ
δ δ δ ρ δ δ δ ρ δ δ δ
− =
−∇ − = −
p gk gaρ ρ−∇ − = −General equation of motion for a fluid w/o shearing stresses
Pressure variation in a fluid at rest
• At rest a=0 0p gkρ−∇ − =
0 0p p p gx y z
ρ∂ ∂ ∂= = = −
∂ ∂ ∂
• Incompressible fluid
1 2p p ghρ= +
Fluid statics
• Pressure depends on the depth in the solution not on the lateral coordinate
Fluid equilibrium Transmission of fluid pressure, e.g. in hydraulic lifts
Same pressure –much higher force!
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:
/1 22 1 0
0
( )assuming exp ; ( ) h H
nRTp RTV
dp gpgdz RT
g z zT const p p p h p eRT
ρ
ρ
−
= =
= − = −
⎡ ⎤−= ⇒ = =⎢ ⎥
⎣ ⎦
~8 km
Measurement of pressure• Pressures can be designated as absolute or gage (gauge) pressures
atm vaporp h pγ= +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.
2R avhF p A g bhρ= =Force acting on a side wall in rectangular container:
Example: Pressure force and moment acting on aquarium walls
• Force acting on the wall
( )2
0 2
H
RA
HF gh dA g H y bdy g bρ ρ ρ= = − ⋅ =∫ ∫
Centroid (first moment of the area)
• Generally: sin sinR cA
F g ydA g y Aρ θ ρ θ= =∫
• Momentum of force acting on the wall
( )3
0 63
H
R RA
R
HF y gh ydA g H y y bdy g b
y H
ρ ρ ρ= = − ⋅ =
=
∫ ∫
• Generally,2
AR
c
y dAy
y A=∫
Buoyant force: Archimedes principle• when a body is totally or partially submerged a fluid
force acting on a body is called buoyant force
[ ]
2 1
2 1 2 1
2 1 2 1
( )( ) ( )
B
B
F F F WF F g h h AF g h h A g h h A V
ρρ ρ
= − −− = −
= − − − −
BF gVρ=Buoyant force is acting on the centroidof displaced volume
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= maF= ma
Net pressure force + Net gravity force
Streamlines: the lines that are tangent to velocity vector through the flow field
vsv
ts
sv
dtdvas ∂
∂=
∂∂
∂∂
==
Rvan
2
=
Acceleration along the streamline:
Centrifugal acceleration:
Streamlines
vsvVmaF ss ∂∂
==∑ ρδδ VspVgFWF psss δθδρδδδ∂∂
−−=+=∑ )sin(
vsv
spg
∂∂
=∂∂
−− ρθρ )sin(
Along the streamline
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: 3
0 31 aV Vx
⎛ ⎞= +⎜ ⎟
⎝ ⎠
p vvs s
ρ∂ ∂= −
∂ ∂
Equation of motion:
3 320 3 4
3 2 30
4 3
63 2 3 32004 3 3
3 1
3 1
3 112
a
v a av vs x x
a vp ax x x
a v a a ap dx vx x x x
ρ
ρ ρ−
−∞
⎛ ⎞∂= − +⎜ ⎟∂ ⎝ ⎠
⎛ ⎞∂= +⎜ ⎟∂ ⎝ ⎠
⎛ ⎞⎛ ⎞ ⎛ ⎞∆ = + = − +⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠ ⎝ ⎠∫
Raindrop shape
The actual shape of a raindrop is a result of balance between the surface tension and the air pressure
vsv
spg
∂∂
=∂∂
−− ρθρ )sin(Integrating
dzds
We find 0)(21 2 =++ gdzvddp ρρ
constgzvp =++ ρρ 2
21
Along a streamline
Along a streamline
Bernoulli equation
Bernoulli equation
, n=const along streamlinedpds
212
dz dp dvgds ds ds
ρ ρ− − =
Assuming incompressible flow:
Example: Bicycle
2012 2
1 vpp ρ=−
• Let’s consider coordinate system fixed to the bike.Now Bernoulli equation can be applied to
Pressure variation normal to streamline
RvVmaF nn
2
ρδδ ==∑ VnpVgFWF pnnn δθδρδδδ∂∂
−−=+=∑ )cos(
Rv
np
dndzg
2
ρρ =∂∂
−−
constgzdnRvp =++ ∫ ρρ
2
compare
constgzvp =++ ρρ 2
21
Across streamlines
Along a streamline
Example: pressure variation normal to streamline
Rv
np
dndzg
2
ρρ =∂∂
−−
• Let’s consider 2 types of vortices with the velocity distribution as below:
,n r∂ ∂= −
∂ ∂as
22
1p V C rr r
ρ ρ∂= =
∂
2213
Cp Vr r r
ρ ρ∂= =
∂
solid bodyrotation
free vortex
22 02 2
0
1 1 12
p C pr r
ρ⎛ ⎞
= − +⎜ ⎟⎝ ⎠
( )2 2 21 0 0
12
p C r r pρ= − +
Static, Stagnation, Dynamic and TotalPressure
212 2
1 vpp ρ+=
Stagnation pressure
• each term in Bernoulli equation has dimensions of pressure and can be interpreted as some sort of pressure
constgzvp =++ ρρ 2
21
static pressure,point (3)
dynamic pressure,
hydrostatic pressure,
Velocity can be determined from stagnation pressure:
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
22
2212
11 21
21 gzvpgzvp ρρρρ ++=++
2211 vAvAQ ==
Determine the flow rate to keep the height constant
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.