Sasaran

FLUIDA STATIKHYDROSTATICS 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

Review FLUIDA STATIK

• TEKANAN P = F/A

• EFEK GRAVITASI PADA TEKANAN–P = P0 + ρgd

• GAYA BUOYANT F = ρ g V

05

TEKANANGaya per satuan luas, dimana gaya tegak lurus luasan.

p=A m2

Nm-2

(Pa)

NF

Tekanan absolut adalah tekanan relatif terhadap vakum

Tekanan gauge, yaitu tekanan relatif terhadap atmosfir ( p-pa)

patmosfir= 1.013X105 Nm-2Pa (Pascal)

1 psi = 6895 Pa

Seseorang menginjak jari kakimu dengan gaya 500 N pada luasan 1.0 cm2. Berapa atmosfir tekanan tersebut .

atm 49

Pa 10013.1

atm 1

N/m 1

Pa 1N/m 100.5

m 101.0

N 500

5226

24-av

=

×

×=

×==

A

FP

242

2 m 100.1cm 100

m 1cm 0.1 luasan −×=

=

TEKANAN

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,

Pascal’s PrinciplePerubahan tekanan pada semua titik dalam fluida diteruskan keseluruhan fluida.

Gaya force F1 bekerja pada piston A1.

Gaya ditransmisikan ke piston A2.

F2

11

22

2

2

1

1

A

A

A

F

A

F

2point at 1point at

FF

PP

=

=

∆=∆

F2

1 500 N

10 5000 N

100 50,000 N

12 AA

Assume F1 = 500 N

A1A2

0=F

EFEK GRAVITASI PADA TEKANAN FLUIDA

F = gaya dari atas + gaya dari bawah + gaya gravitasi = 0

0=∆∆∆−∆∆−∆∆ zyxgyxPyxP ba ρ

gz

PP ba ρ−=∆−

Tekanan atasPb

Tekanan bawah Pa

Za

Zb

Densitas=ρ

∆z

gdz

dP ρ−=

EFEK GRAVITASI PADA TEKANAN FLUIDA

d dgPP ρ+= atm

P

Atm

TEKANAN DENGAN KEDALAMAN

gdz

dP ρ−=

)( 122 1zzgPP −−=− ρ

Untuk gas RT

PM=ρg

RT

PM

dz

dP =

)(ln 121

2 zzRT

gM

P

P −−=

(gas ideal, isotermal)

Untuk cair

Tekanan pada permukaan air danau adalah 105 kPa. Hitung tekanan pada kedalaman 35.0 m dibawah

permukaan air.

( )( )( )atm 3.4kPa 343

m 35m/s 8.9kg/m 1000

23

atm

atm

===

=−=∆+=

dgPPP

dgPP

ρρ

Kerapatan air segar

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?

( )( )m 950

N/m 109.5m/s 8.9kg/m 1025

N/m 109.5atm 94

atm 1atm 95

2623

26

atm

=×=

×==+=

+=

d

d

dg

dg

dgPP

ρρ

ρ

Density of sea water

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.

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

ρρ

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.

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.

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

Atmospheric pressure is equivalent to a column of mercury 76.0 cm tall.

gdP ρ=

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

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

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

( )

( )

( )( ) m 5.1m 10*m 20kg/m 1000

kg 100.3

0

3

5

=×=

=

==

===

=−=∑

A

md

mAd

mV

gmgVgm

wF

wFF

w

b

bw

bww

bwww

B

B

ρ

ρρ

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

1. Go up, causing the water to spill out of the glass.

2. Go down.

3. Stay the same.

CORRECT

COBA PIKIRKAN

B = ρW g Vdisplaced

W = ρice g Vice → 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 :

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 water16

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

w

FB

mawFF B =−=∑

VgFB waterρ=

FB adalah berat fluida yang dipindahkan oleh benda

−=−= 1

ρ

ρ

ρ

ρ

objectobject

water

objectobject

water

V

Vgg

V

Vga

gm

F

m

w

m

Fa BB −=−=

0.5gravity specificwater

object ==ρρ

ρwater = 1000 kg/m3 (at 4 °C).

2

objectobject

water m/s 8.710.5

11

..

11

ρ

ρ −=

−=

−=

−= g

GSg

V

Vga

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.

DECANTERS GRAVITY

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.

TEKANAN DALAM BENDA YANG BERGERAK

)(2

2

dt

zdg

dz

dP +−= ρ

Gerak lurus

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.

TEKANAN DALAM BENDA RIGID BERGERAK

Gerak rotasiCentripetal accelaration =(angular velocity)2.radius

rac2ω−=

rdr

dP 2ρω=

2

2( cos )c

c

d adPg

da dtρ θ=− +

DECANTERS

Centrifugal

ω Aout

BoutrA

rB

ri

R

pA=pB=patm