Hydrostatics
e
.
2 .
3
Hydrodynamics
Pressure 4 = -p Navier Stokes :
[Pa N/m2]=
!
-
toot =
Fvolumeforces "Fourface forces
horizontal
altijd gelijk
·
de
Op -
#(x ax)dF F'(x + dx) F 2x)
F(x
+ = +
-g
- -
dF = -
F(x + dx) + F(x)
p(x + dx)S + P(x)S
dric
dF
-
↓
=
x + dx
P(x)]S
[P(x
Sir Gey
si
dF + dx) -
ex
= -
+
When the is the same both sides
pressure on
↑ (x + dx) P(x) and dF 0
#-
iviscastane
=
=
-
F JP(z)
-rw) +zw
Bdz e termen !
Schrijf P in
=
>
-
va n z
7
e
= E
+ +
Spdz
Cart gradient
=
Cross dot
product
thevelocita
and
exex =
1 x 1 x Sin(o2 = water a
because decreases because
xsin()
>
-
- -
& x1
=
er
1
of friction
·
+
=
2x .
COS(0) 1
er
I X
=
- X
ex
x cost Euler Assume and constant
:
yo p
=
x =
dX dz
dV
dy
= C 6
E
· -
Torque i
.
of fluid
* # x rexf Newton on cube at rest
we er = =
>
-
· erx
jp
p
-
M ~xFx 0 +
>
= -
-
=
~
the
7
i and introduce
by
↳infid atrest
xFx divide
Hoal ut de
gradient head p
=
= p
,
,
w
Patm ,
use
piezometric .
Denys000kg/m)
clay
volume
0 a
g(h)
=
=
total volume = -
Pascal :
& usea
/saltwater Patm
Ptop barrel +
pgH For barrel [Path-Pbarrel]S
=
=
(1-0) x pwater + ↓ Xpclay
Pfluid Pin barrel Patm PGH
=
+
PgHS
=
=
eparticles
+ water 1
Poutside barrel Patr
=
:
Integral nemen ove r z
vessel
Viscosi viscosity) ↑ closed :
honey (high viscosity
in
acceleration
advective
Reynolds
,
(20) P(z) number Re
:
-Viscous
- =
#nematic
20
=
2 -
"inertial
force"
2
[v/t] [v2/7= =
(1 +2]
-4
h 20
=
20 + P(z) -
-
2n
-
=
Low Re number >
- laminar flow viscosity v = Torricelli :
-
Stress and :
pressure h Patm temporal advective viscous
forces
Stress T =
=
4
=
y t P9 (unsteady [yv/((p(4)] =
(2/t7]
i
shear strain
v
= =
tan (
A
air Re1 /
- +
v -g-steada =
(
=
shear rate j =
It Re +
v -g- inviscid
=
# therefore
dux
jdy
=
I jaux 93jdy (vx*x Vyy v)
breedte in
= ① de =
papier = + + +
local
O
total advective
jy #(2) pg[ex-eitan derivative derivative
/1117
=
derivative
-
Van Euler naar Bernoulli :
I
& in a 1)
along a streamline c) steady flow
(peconst
wjPgatan
Fz
- friction ( 0) incompressible
-
4)
=
3) no
=
Ex =
T
-z-
PgV
+
-
·
area
Archimedes :
pgz) ( pgz)
P + =
+ p +
No
-
Exercise :
= + h) = h
(*) ydy y ye
+
==
-
or
Fi
pwaterV Ftot
(pcube-pwater) V 2 )
-fly) frictio
Q
= -
=
=
2
use
fly) =
:
Forchimedes Patn]a Pat]
=
-Sugd + +
Grgdta +
Mass conservation :
m(t dt) m(t) drin(t) -
Amout(t)
Rheometer :
#archimedes
+ -
=
The
torque is pugae =+
↑
↑= reix Fe proportional
Shear Stress T
to the
la =
Usub
rF(r) sin) ei =) rF(r)
M(r) =
=
and measured
Notes
device
↑ =
= =
ni
=
h by the
exert an
wal a
a water change jet can o a
↑
=
=
time
of momentum over
Vena contracta :
Rin F
= veugh :S
Q-VSuc Vous
*
=
7
Types of slopes Opdrachten tank -Fer
rotating
.
: v =
rw a=
* river with more
friction -
equilibrium depth
de
-desdo
Mi *
Rotating Cylinder
ww
dar
exa --
Belanger equationisvalidforsteadyou-dimensio
a se
M2 ↳
dexdad
S
Ihethbth
-
My
loss
. -
friction
-,
--
--
experience
&
-
BC
--
* Pitot Lube (Bemoullil
Vort
D
↓ In
&
-
O
Si
App-
In
-
> > de VBz 0
dar
=
constant so point C is
hent -next
how
H =
+ hbend +
VaX 0
-
=
Vcz 0
a
stagnation point he his ha
Bo of fluid in
=
velocity
⑭
dd >
de Vaz =
o Pc =
PB +
pV - => P
·
de
+e
estimated from
z
VDz = 0 tube can be
H-h
dc > d ed
height difference
"
S3
+
longer accelerating
+
, terminal velocity Gur
velocity when
-- is no
-
BC
=
[F = -
and
*
largest errors come from energy losses
thaxit
end
effect corrections (when liquid enters & Whent
exits the
capidary
hEnt + holdw) + hexit
he' Up In
e
.
2 .
3
Hydrodynamics
Pressure 4 = -p Navier Stokes :
[Pa N/m2]=
!
-
toot =
Fvolumeforces "Fourface forces
horizontal
altijd gelijk
·
de
Op -
#(x ax)dF F'(x + dx) F 2x)
F(x
+ = +
-g
- -
dF = -
F(x + dx) + F(x)
p(x + dx)S + P(x)S
dric
dF
-
↓
=
x + dx
P(x)]S
[P(x
Sir Gey
si
dF + dx) -
ex
= -
+
When the is the same both sides
pressure on
↑ (x + dx) P(x) and dF 0
#-
iviscastane
=
=
-
F JP(z)
-rw) +zw
Bdz e termen !
Schrijf P in
=
>
-
va n z
7
e
= E
+ +
Spdz
Cart gradient
=
Cross dot
product
thevelocita
and
exex =
1 x 1 x Sin(o2 = water a
because decreases because
xsin()
>
-
- -
& x1
=
er
1
of friction
·
+
=
2x .
COS(0) 1
er
I X
=
- X
ex
x cost Euler Assume and constant
:
yo p
=
x =
dX dz
dV
dy
= C 6
E
· -
Torque i
.
of fluid
* # x rexf Newton on cube at rest
we er = =
>
-
· erx
jp
p
-
M ~xFx 0 +
>
= -
-
=
~
the
7
i and introduce
by
↳infid atrest
xFx divide
Hoal ut de
gradient head p
=
= p
,
,
w
Patm ,
use
piezometric .
Denys000kg/m)
clay
volume
0 a
g(h)
=
=
total volume = -
Pascal :
& usea
/saltwater Patm
Ptop barrel +
pgH For barrel [Path-Pbarrel]S
=
=
(1-0) x pwater + ↓ Xpclay
Pfluid Pin barrel Patm PGH
=
+
PgHS
=
=
eparticles
+ water 1
Poutside barrel Patr
=
:
Integral nemen ove r z
vessel
Viscosi viscosity) ↑ closed :
honey (high viscosity
in
acceleration
advective
Reynolds
,
(20) P(z) number Re
:
-Viscous
- =
#nematic
20
=
2 -
"inertial
force"
2
[v/t] [v2/7= =
(1 +2]
-4
h 20
=
20 + P(z) -
-
2n
-
=
Low Re number >
- laminar flow viscosity v = Torricelli :
-
Stress and :
pressure h Patm temporal advective viscous
forces
Stress T =
=
4
=
y t P9 (unsteady [yv/((p(4)] =
(2/t7]
i
shear strain
v
= =
tan (
A
air Re1 /
- +
v -g-steada =
(
=
shear rate j =
It Re +
v -g- inviscid
=
# therefore
dux
jdy
=
I jaux 93jdy (vx*x Vyy v)
breedte in
= ① de =
papier = + + +
local
O
total advective
jy #(2) pg[ex-eitan derivative derivative
/1117
=
derivative
-
Van Euler naar Bernoulli :
I
& in a 1)
along a streamline c) steady flow
(peconst
wjPgatan
Fz
- friction ( 0) incompressible
-
4)
=
3) no
=
Ex =
T
-z-
PgV
+
-
·
area
Archimedes :
pgz) ( pgz)
P + =
+ p +
No
-
Exercise :
= + h) = h
(*) ydy y ye
+
==
-
or
Fi
pwaterV Ftot
(pcube-pwater) V 2 )
-fly) frictio
Q
= -
=
=
2
use
fly) =
:
Forchimedes Patn]a Pat]
=
-Sugd + +
Grgdta +
Mass conservation :
m(t dt) m(t) drin(t) -
Amout(t)
Rheometer :
#archimedes
+ -
=
The
torque is pugae =+
↑
↑= reix Fe proportional
Shear Stress T
to the
la =
Usub
rF(r) sin) ei =) rF(r)
M(r) =
=
and measured
Notes
device
↑ =
= =
ni
=
h by the
exert an
wal a
a water change jet can o a
↑
=
=
time
of momentum over
Vena contracta :
Rin F
= veugh :S
Q-VSuc Vous
*
=
7
Types of slopes Opdrachten tank -Fer
rotating
.
: v =
rw a=
* river with more
friction -
equilibrium depth
de
-desdo
Mi *
Rotating Cylinder
ww
dar
exa --
Belanger equationisvalidforsteadyou-dimensio
a se
M2 ↳
dexdad
S
Ihethbth
-
My
loss
. -
friction
-,
--
--
experience
&
-
BC
--
* Pitot Lube (Bemoullil
Vort
D
↓ In
&
-
O
Si
App-
In
-
> > de VBz 0
dar
=
constant so point C is
hent -next
how
H =
+ hbend +
VaX 0
-
=
Vcz 0
a
stagnation point he his ha
Bo of fluid in
=
velocity
⑭
dd >
de Vaz =
o Pc =
PB +
pV - => P
·
de
+e
estimated from
z
VDz = 0 tube can be
H-h
dc > d ed
height difference
"
S3
+
longer accelerating
+
, terminal velocity Gur
velocity when
-- is no
-
BC
=
[F = -
and
*
largest errors come from energy losses
thaxit
end
effect corrections (when liquid enters & Whent
exits the
capidary
hEnt + holdw) + hexit
he' Up In
