Fluida statik

1. Sasaran
FLUIDA STATIK
HYDROSTATICS PRINCIPLES
Dapat menghitung distribusi tekanan dan gaya pada benda yang
terendam dalam fluida.
Dapat menghitung cara menghitung tekanan menggunakan
differential manometry.
Dapat a menggunakan manometer untuk mengukur tekanan

2. Review FLUIDA STATIK
• TEKANAN
P = F/A
• EFEK GRAVITASI PADA TEKANAN
–P = P0 + ρgd
• GAYA BUOYANT
F = ρ g V
05

3. TEKANAN
Gaya per satuan luas, dimana gaya tegak lurus luasan.
p=
A m2
Nm-2
(Pa)
N
F
Tekanan absolut adalah tekanan relatif terhadap vakum
Tekanan gauge, yaitu tekanan relatif terhadap atmosfir ( p-pa)
patmosfir= 1.013X105
Nm-2
Pa (Pascal)
1 psi = 6895 Pa

4. Seseorang menginjak jari kakimu dengan gaya 500 N pada
luasan 1.0 cm2
. Berapa atmosfir tekanan tersebut .
atm49
Pa10013.1
atm1
N/m1
Pa1
N/m100.5
m101.0
N500
52
26
24-av
=






×






×=
×
==
A
F
P
24
2
2
m100.1
cm100
m1
cm0.1luasan −
×=





=
TEKANAN

5. TEKANAN DALAM BEJANA
Tekanan timbul karena adanya tabrakan
antara partikel fluida dengan dinding
bejana (molecules “bouncing” )
Ada perubahan momentum
(impulse), jika partikel menabrak
dinding, balik arah menjauhi
dinding wadah. Jadi pasti ada
gaya bekerja pada partikel dan
dinding,

6. Pascal’s Principle
Perubahan tekanan pada semua titik dalam
fluida diteruskan keseluruhan fluida.

7. Gaya force F1 bekerja
pada piston A1.
Gaya ditransmisikan ke
piston A2.
F2
1
1
2
2
2
2
1
1
A
A
A
F
A
F
2pointat1pointat
FF
PP






=
=
∆=∆
F2
1 500 N
10 5000 N
100 50,000 N
12 AA
Assume F1 = 500 N
A1A2

8. 0=F
EFEK GRAVITASI PADA TEKANAN FLUIDA
F = gaya dari atas + gaya dari bawah + gaya gravitasi = 0
0=∆∆∆−∆∆−∆∆ zyxgyxPyxP ba ρ
g
z
PP ba
ρ−=
∆
−
Tekanan atas
Pb
Tekanan
bawah Pa
Za
Zb
Densitas=ρ
∆z
g
dz
dP
ρ−=

9. EFEK GRAVITASI PADA TEKANAN FLUIDA
d dgPP ρ+= atm
P
Atm

10. TEKANAN DENGAN KEDALAMAN
g
dz
dP
ρ−=
)( 122 1
zzgPP −−=− ρ
Untuk gas RT
PM
=ρ
g
RT
PM
dz
dP
=
)(ln 12
1
2
zz
RT
gM
P
P
−−=
(gas ideal, isotermal)
Untuk cair

11. Tekanan pada permukaan air danau adalah 105 kPa.
Hitung tekanan pada kedalaman 35.0 m dibawah
permukaan air.
( )( )( )
atm3.4kPa343
m35m/s8.9kg/m1000 23
atm
atm
==
=
=−=∆
+=
dgPPP
dgPP
ρ
ρ
Kerapatan air segar

12. Tekanan di permukaan planet Venus adalah 95 atm.
How far below the surface of the ocean on Earth do
you need to be to experience the same pressure?
( )( )
m950
N/m109.5m/s8.9kg/m1025
N/m109.5atm94
atm1atm95
2623
26
atm
=
×=
×==
+=
+=
d
d
dg
dg
dgPP
ρ
ρ
ρ
Density of sea water

13. PENGUKURAN TEKANAN
A manometer
is a U-shaped
tube that is
partially filled
with liquid.
Both ends of the
tube are open to the
atmosphere.

14. Cylinder
of gas
A container of gas is connected to one end of the U-tube
C
B’
B
A d
gdP
gdPPPP
gdPPP
BCB
CBB
ρ
ρ
ρ
=
=−=−
+==
gauge
atm
'
B'B PP =
atmc PP =
Point A is the original location of the
top of the fluid before the gas
cylinder is connected.

15. Using a u-tube manometer to measure gauge pressure of fluid density 700
kg/m3, and the manometric fluid is mercury, with a relative density of 13.6.
What is the gauge pressure if:
a). h1 = 0.4m and h2 = 0.9m?
b) h1 stayed the same but h2 = -0.1m?
THE U-TUBE MANOMETER.

16. Measuring Pressure
Barometers
A barometer is used to measure
the pressure of the atmosphere.
The simplest type of barometer
consists of a column of fluid.
p1 = 0vacuum
h
p2 = pa
p2 - p1 = ρgh
pa = ρgh
examples
water: h = pa/ρg =105
/(103
*9.8) ~10m
mercury: h = pa/ρg =105
/(13.4*103
*9.8) ~800mm

17. Atmospheric pressure is equivalent to a
column of mercury 76.0 cm tall.
gdP ρ=

18. ARCHIMEDES’ PRINCIPLE
• Buoyant Force (FB) weight of fluid
displaced
–FB = ρfluidVdisplaced g
–Fg = mg = ρobject Vobject g
–object sinks if ρobject > ρfluid
–object floats if ρobject < ρfluid
– object floats if ρfluid = ρobject

19. ARCHIMEDES EXAMPLE
A cube of plastic 4.0 cm
on a side with density
= 0.8 g/cm3
is floating
in the water.
When a 9 gram coin is
placed on the block,
how much sinks below
water surface?
h
koin

20. ARCHIMEDES EXAMPLE
mg
Fb
Mg
Σ F = m a
Fb – Mg – mg = 0
ρ g Vdisp = (M+m) g
Vdisp = (M+m) / ρ
h A = (M+m) / ρ
h = (M + m)/ (ρ A)
= (51.2+9)/(1 x 4 x 4) = 3.76 cm
M = ρplastic Vcube = 0.8x4x4x4
= 51.2 g
h
koin

21. ( )
( )
( )( )
m5.1
m10*m20kg/m1000
kg100.3
0
3
5
=
×
=
=
=
=
==
=
=−=∑
A
m
d
mAd
mV
gmgVgm
wF
wFF
w
b
bw
bww
bwww
B
B
ρ
ρ
ρ
ρ
Wadah segi 4 berdasar datar diisi dengan coal,
massanya 3.0×105
kg. Panjang 20 m dan lebar 10m,
mengambang diair. Berapa kedalaman wadah masuk ke
dalam air.
w
FB

22. 1. Go up, causing the
water to spill out of
the glass.
2. Go down.
3. Stay the same.
CO
RRECT
COBA PIKIRKAN
B = ρW g Vdisplaced
W =ρice gVice → ρW g V
Must be same!
ice-cube
Sebuah balok es mengambang diatas
segelas air, sampai permukaan air rata
pada pinggiran.
Ketika es meleleh maka air di dalam
akan :

23. COBA PIKIRKAN
Which weighs more:
1. A large bathtub filled to the
brim with water.
2. A large bathtub filled to the
brim with water with a
battle-ship floating in it.
3. They will weigh the same.
Tub of water
Tub of water + ship
Overflowed water
Weight of ship = Buoyant force =
Weight of displaced water
16

24. Sepotong logam dilepaskan
dibawah air water. Volume
metal 50.0 cm3
dan SG 5.0.
Hitung percepatan initialnya,
saat v=0 tidak ada gaya drag.
w
FB

25. w
FB
mawFF B =−=∑
VgFB waterρ=
FB adalah berat fluida yang
dipindahkan oleh benda








−=−= 1
ρ
ρ
ρ
ρ
objectobject
water
objectobject
water
V
V
gg
V
Vg
a
g
m
F
m
w
m
F
a BB
−=−=
0.5gravityspecific
water
object
==
ρ
ρ
ρwater = 1000 kg/m3
(at 4 °C).
2
objectobject
water
m/s8.71
0.5
1
1
..
1
1
ρ
ρ
−=





−=





−=








−= g
GS
g
V
V
ga

26. KEDALAMAN OIL RESERVOIR
What the depth
below the surface
of oil reservoir
that produce oil
with relative
density 0.8 and
wellhead pressure
of 120 kN/m2?
ρ water = 1000 kg/m3, and p atmosphere = 101kN/m2.

27. DECANTERS GRAVITY

28. DECANTER
• It is proposed to use a gravity decanter to
separate a light petroleum oil (density 50.0
lbm/ft3) from water (density 62.3 lbm/ft3). Its
desire to maintain a total depth of 30 in. in the
vessel and to have exactly equal depth of oil and
water. What should be the height , expressed in
inch of the water discharge leg above the bottom
of the vessel.

29. TEKANAN DALAM BENDA YANG
BERGERAK
)( 2
2
dt
zd
g
dz
dP
+−= ρ
Gerak lurus

30. Example
An open tank contains water 5m deep.
It is sitting on an elevator.
Calculate the gauge pressure at the bottom of the tank
(a) when the elevator is standing still,
(b) when the elevator is accelerating upward at the rate of
5m/s2, and
(c) when the elevator is accelerating downward at the rate of 5
m/s2.

31. TEKANAN DALAM BENDA RIGID
BERGERAK
Gerak rotasi
Centripetal accelaration =
(angular velocity)2
.radius
rac
2
ω−=
r
dr
dP 2
ρω=
2
2
( cos )c
c
d adP
g
da dt
ρ θ=− +

32. DECANTERS
Centrifugal
ω Aout
BoutrA
rB
ri
R
pA=pB=patm