Modul 2 - Konservasi Energy Bernoully
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Transcript of Modul 2 - Konservasi Energy Bernoully
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Konservasi Energi
(Kontinuitas & Hk. Bernoulli)
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Mass Flux
Tube of flow: bundle of streamlines
A1 A2
1vP
2v
Q
11 1 1 1 1 1 1
1
mass fluxm
m Av t Avt
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Persamaan Kontinuitas
Gambar Aliran masuk dan keluar dari bagian pipa arus
Volume aliran masuk dititik 1 :
A1
. v1
. dt Dimana : A1
= luas penampang dan v1
. dt = tinggi saluran
Volume aliran masuk dititik 2 :
A2. v2. dt Dimana : A2 = luas penampang dan v2. dt = tinggi saluran
Jika kerapatan fluida (densitas) =
Maka massa aliran : . A1. v1 dan . A2. v2Avdtm ..
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Conservation of Mass
IF: no sources and no sinks/drains
1 1 1 2 2 2
1 1 2 2
constant
co fornstant incompressible fluid,
Av A v
Av A v
Example of equation of continuity. Also conservation ofcharge in E&M
Narrower tube = larger speed, fastWider tube = smaller speed, slow
1 = 2
A x V = constan
Hukum kontinuitas
Q = A x V = konstans
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Massa fluida yang mengalir adalah konstan disepanjang saluran, maka :
. A1. v1 = . A2. v2
atau . A . v = konstan Hukum kekekalan massa aliran
Dengan massa aliran dengan massa jenis tetap : , maka :
A1. v1 = A2. v2
A . v = konstanatau Hukum kekekalan volume aliran
Jika : Q = Kapasitas atau jumlah aliran (Debit) (m3/detik)
Q = A . v= konstan Persamaan Kontinuitas
Dimana : A = Luas penampang (m2) dan v= kecepatan aliran (m/detik)
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FLUID DYNAMICSTHE BERNOULLI EQUATION
The laws of Statics that we have learned cannot solveDynamic Problems. There is no way to solve for the flowrate, or Q. Therefore, we need a new dynamic approach
to Fluid Mechanics.
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Example of equation of continuity.
Penampang 1 sebuah saluran yang mengalirkan air dengan kecepatan 3 m/sdan garis tengahnya 2 m, dipenampang 2 garis tengahnya 3 m, berapakahDebit atau kapasitas dan kecepatan dipenampang keluar (2)
1V1
2V2
Penyelesaian :
Persamaan kontinuitas :
Q = A x V = konstans
Q1= A1x V1 Q2= A2x V2&
Q1= Q2=Q
Q=Q2= A2x V2
s
mVdQQ
32
1
2
11 42.93.)2(4
.4
sm
d
x
A
QV /33.1
.
42.94
222
2
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Conservation of Energy
Steady, incompressible, nonviscous, irrotational
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Bernoullis Principle
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Bernoullis Principle
Flow is faster when the pipe is narrower Put your thumb over the end of a garden
hose Energy conservation requires that the
pressure be lower in a gas that is moving
faster Has to do with the work necessary to
compress a gas (PV is energy, more later)
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Bernoullis Principle
When the speed of a fluid increases,internal pressure in the fluid decreases.
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Bernoullis Principle
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Bernoullis Principle
Why the streamlines are compressed isquite complicated and relates to the air
boundary layer, friction and turbulence.
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Bernoullis Principle
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Who Accelerates the Fluid?
Acceleration due to pressure difference.Bernoullis Principle = Conservation of energy
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Bernoullis Equation
1 1 2 2
2 2
1 1 1 2 2 2 2 2 1 1
2 2
1 1 1 2 2 2
2
1 1
2 2
1 12 2
1constant
2
m A x A x
p A x p A x mv mgy mv mgy
p v gy p v gy
p v gy
kinetic E, potential E, external work
Bentuk dan ukuran yang sama
Energi Balance :
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Subskrip 1 dan 2 pada sembarang titik, maka ditulis menjadi :
tan
2
1 2konsvgyp
Persamaan Bernoulli
Jika : = Berat jenisg.
Maka : tan
2
1 2konsv
gy
p
Atau :
tankonsyg
vp
2
2
Energi Tekanan
Energi Kinetik
Elevasi(Energi Potensial)
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The Bernoulli Equation (unit of L)
At any two points on a streamline:
P1/ + V12/2g + z1= P2/ + V2
2/2g + z2
12
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The Bernoulli Equation (unit of L)
At any two points on a streamline:
P1/ + V12/2g + z1= P2/ + V2
2/2g + z2
12
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A Simple Bernoulli Example
V2Z = air
Determine the difference in pressure between points 1 and 2
Assume a coordinate system fixed to the bike (from this system,the bicycle is stationary, and the world moves past it). Therefore,
the air is moving at the speed of the bicycle. Thus, V2 = Velocity ofthe Biker
Hint: Point 1 is called a stagnation point, because the air particlealong that streamline, when it hits the bikers face, has a zero
velocity (see next slide)
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Stagnation Points
On any body in a flowing fluid, there is a stagnation point. Some fluidflows over and some under the body. The dividing line (the stagnationstreamline) terminates at the stagnation point. The Velocity decreases
as the fluid approaches the stagnation point. The pressure at thestagnation point is the pressure obtained when a flowing fluid is
decelerated to zero speed by a frictionless process
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Apply Bernoulli from 1 to 2
V2Z
Point 1 = Point 2
P1/air+ V12/2g + z1 = P2/air+ V22/2g + z2
Knowing the z1 = z2 and that V1= 0, we can simplifythe equation
P1/air= P2/air+ V22/2g
P1 P2= ( V22
/2g ) air
= air
A Si l B lli E l
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A Simple Bernoulli Example
If Lance Armstrong is traveling at 20 ft/s, what pressure
does he feel on his face if the air= .0765 lbs/ft3?
We can assume P2 = 0 because it is only atmospheric pressure
P1 = ( V22/2g )(air) = P1 = ((20 ft/s)
2/(2(32.2 ft/s2)) x .0765 lbs/ft3
P1 =.475 lbs/ft2
Converting to lbs/in2 (psi)
P1 = .0033 psi (gage pressure)
If the bikers face has a surface area of 60 inches
He feels a force of .0033 x 60 = .198 lbs
B lli A i
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Bernoulli Assumptions
Key Assumption # 1
Velocity = 0
Imagine a swimming pool with a small 1 cm hole on the floor of
the pool. If you apply the Bernoulli equation at the surface, andat the hole, we assume that the volume exiting through the holeis trivial compared to the total volume of the pool, and thereforetheVelocity of a water particle at the surface can be assumed tobe zero
There are three main variables in the Bernoulli EquationPressure Velocity Elevation
To simplify problems, assumptions are often made toeliminate one or more variables
B lli A i
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Bernoulli Assumptions
Key Assumption # 2
Pressure = 0
Whenever the only pressure acting on a point isthe standard atmospheric pressure, then thepressure at that point can be assumed to be zerobecause every point in the system is subject tothat same pressure. Therefore, for any free
surface or free jet, pressure at that point can beassumed to be zero.
B lli A ti
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Bernoulli Assumptions
Key Assumption # 3
The Continuity Equation
In cases where one or both of theprevious assumptions do not apply, then
we might need to use the continuityequation to solve the problem
A1V1=A2V2
Which satisfies that inflow and outfloware equal at any section
B lli E l P bl F J t
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Bernoulli Example Problem: Free Jets
What is the Flow Rate at point 2? What is the velocity at point 3?
1
2
3
H2O
Part 1:Apply Bernoullis eqn between points 1 and 2
P1/H2O + V12/2g + h = P2/H20 + V2
2/2g + 0
simplifies to
h = V22/2g solving for V
V = (2gh)
Q = VA or Q = A2(2gh)0A2
Givens and Assumptions:Because the tank is so large, we assume V1 = 0 (Volout
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1
2
3
H2O
Z = 0A2
Bernoulli Example Problem: Free Jets
Part 2: Find V3?
Apply Bernoullis eq from pt 1 to pt 3
P1/H2O + V12
/2g + h = P3/H20 + V32
/2g HSimplify to h + H = V3
2/2g
Solving for V V3 =( 2g ( h + H ))
Th C ti it E ti
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The Continuity EquationWhy does a hose with a nozzle shoot water further?
Conservation of Mass:In a confined system, all of the mass that enters the system, must also exit
the system at the same time.
Flow rate = Q = Area x Velocity
1A1V1(mass inflow rate) = 2A2V2( mass outflow rate)
If the fluid at both points is thesame, then the density drops
out, and you get the continuityequation:
A1V1 =A2V2
ThereforeIf A2 < A1 then V2 >V1
Thus, water exiting a nozzle has
a higher velocity
Q1 = A1V1
A1
V1 ->
Q2 = A2V2
A1V1 = A2V2
A2 V2 ->
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Free Jets
The velocity of a jet of water is clearly related to the depth of waterabove the hole. The greater the depth, the higher the velocity. Similarbehavior can be seen as water flows at a very high velocity from the
reservoir behind a large dam such as Hoover Dam
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example problem
A tank of water has a small nozzle at itsbase as shown. Find the velocity inft/sec and the volumetric flow rate in
ft
3
/sec from the nozzle.
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Fluid Mechanics
Assume the jet is cylindrical and that thepressure is atmospheric as soon as itleaves the nozzle. Apply Bernoullis
equation between a point 1 on the surfaceof the tank and a point 2 at the nozzleexit:
2
2
221
2
11 zg2g
Pzg2g
P
vv
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Fluid Mechanics
The pressure at point 1 and at point 2 isjust that of the atmosphere, so P1 = P2.
At point 1 the height is z1 = H and at point2 the height is z2 = 0. If the size of thetop of the tank is very large compared to
the outlet area, then (V1)2 is significantlyless than (V2)
2 .
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Fluid Mechanics
The Bernoulli Equation becomes
but the terms P1/g and P2/g are equaland cancel because the pressures are thesame, and what is left is
0g2g
P
H0g
P2
221
v
gH22 V
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Fluid Mechanics
And so the exit velocity is
the discharge flow rate is
sec/ft1.32)ft16)(sec/ft2.32(2 22
V
3in ft ft
in sec sec
ft
2
2
2 2 2
41 132.1 2.8
4 412
Q A d
V V
Tank Example
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Tank ExampleSolve for the Pressure Head, Velocity Head, and Elevation Head at each
point, and then plot the Energy Line and the Hydraulic Grade Line
1
23 4
1
4
Assumptions and Hints:
P1 and P4 = 0 --- V3 = V4 same diameter tube
We must work backwards to solve this problem
R = .5
R = .25
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1
23 4
1
4
Point 1:
Pressure Head : Only atmospheric
P1/ = 0Velocity Head : In a large tank,V1 = 0 V12/2g = 0
Elevation Head : Z1= 4
R = .5R = .25
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1
23 4
1
4
H2O= 62.4 lbs/ft3
Point 4:
Apply the Bernoulli equation between 1 and 40 + 0 + 4 = 0 + V4
2/2(32.2) + 1
V4 = 13.9 ft/s
Pressure Head : Only atmospheric P4/ = 0Velocity Head :V4
2/2g = 3
Elevation Head : Z4= 1
R = .5R = .25
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1
23 4
1
4
Point 3:
Apply the Bernoulli equation between 3 and 4 (V3=V4)P3/62.4 + 3 + 1 = 0 + 3 + 1
P3 = 0
Pressure Head : P3/ = 0Velocity Head :V3
2/2g = 3
Elevation Head : Z3= 1
R = .5R = .25
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1
2 3 4
1
4
Point 2:
Apply the Bernoulli equation between 2 and 3P2/62.4 + V2
2/2(32.2) + 1 = 0 + 3 + 1
Apply the Continuity Equation
(P.52)V2 = (P.252)x13.9 V2 = 3.475 ft/s
P2/62.4 + 3.4752/2(32.2) + 1 = 4 P2 = 175.5 lbs/ft
2
R = .5R = .25
Pressure Head :P2/ = 2.81
Velocity Head :V2
2/2g = .19
Elevation Head :Z2= 1
Penerapan Persamaan Bernoulli
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Dengan mengambil dasar tangki sebagai patokan, diperoleh :
2
2
2
12
1
2
1vpghvp a
Atau gh
ppvv a 22
2
1
2
2
Berdasarkan hukum kontinuitas :
1
2
12 v
A
Av
Disebut vena contracta
Jika tangki terbuka dan berhubungan dengan udara luar, maka : p = pa dan p pa = 0
Dan : A1 >> A2, maka : ghv 22 Kecepatan efflux sama dengan kecepatan benda jatuh bebas melaluiTinggi h disebut Dalil Toricelli
Jika pada permukaan tertutup dimana v12 diabaikan dan tekanan p besar maka :
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ppv a
)(22
Penerapan pada gaya dorong sebuah roket
Jika : A = luas lubang, = rapat massa fluida yang keluarv = kecepatan efflux,
maka massa fluida yang mengalir dalam waktu dt :
A v dt
maka momentumnya : A v2 dtkarena kecepatan fluida didalam relatif kecil, maka : A v2Maka gaya reaksi dorongan :
Atau :
ppAAvF o
)(22
)(2 appAF
Gambar 1
Contoh : (gambar 1)
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Sebuah roket dimana disebelah atas air terdapat udara bertekanan p = 2 atm.a). Jika roket ditahan diam, berapakah kecepatan efflux yang keluar dari lubang pada ekor roketb). Berapa tolakan keatas (dorongan) keatas jika luas lubang 0,5 cm2 ?
a. Tekanan relatif p pa = 1 atm 106 dyn cm-2, maka kecepatannya :
ppv a
)(22
3
25
.1
.102
cmgr
cmdynexv
b. Gaya tolak keatas (dorongan) atau gaya reaksi :
dyncmdynxcmxF6262
10.105,02
Pipa Ventury
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2
2
221
2
112
1
2
1gyvpgyvp
Dari persamaan Bernoulli :
y1 = y2
2
22
2
112
1
2
1
vpvp Maka :
Berdasarkan persamaan kontinuitas kecepatan dititik 2 lebih besar dititik 1, maka sebaliknyatekanan dititik 2 (tenggorokan) lebih kecil dari tekanan dititik 1, dengan demikian gaya nettomenuju kekanan memberi percepatan pada fluida ketika memasuki tenggorokan, maka untukmengukur kecepatan dan massa yang mengalir digunakan Ventury meter
Mengukur tekanan didalam fluida yang mengalir
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p adalah tekanan statik fluida yang mengalir didalam saluran tertutup dan jika perbedaan tinggi zatcair h1 yang besarnya adalah sebanding dengan perbedaan tekanan atmosfir pa dan tekanan statikp yaitu :
pa p =mg h1p =mg h1
Dimana : m= rapat massa zat cair
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BEq in Everyday Life
Open a faucet, thestream of water gets
narrower as it falls.
Velocity increases due to
gravity as water flow down,thus, the area must get narrower.
A1
A2
V1
V2
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Q & A on Bernoullis Eq.
A bucket full of water.
One hole andone pipe, bothopen at bottom.
Out of which water flows faster?
Same. It only depends on depth.
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Measuring Pressure
E. Torricelli: Mercury Barometer
hpatm
p=0atm
atm
P ghP
hg
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U-Tube Manometer
1 1 2 2A atmp gh p gh
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The Venturi Meter
Speed changes asdiameter changes.
Can be used tomeasure the speedof the fluid flow.
2 2
1 1 2 2 1 1 2 21 1 ,2 2
p v p v v A v A
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The Pitot Tube
21
2a a b
b a
p v p
p p gh
Th E Li d th H d li G d Li
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The Energy Line and the Hydraulic Grade Line
Looking at the Bernoulli equation again:
P/ + V2/2g + z = constant on a streamlineThis constant is called the total head (energy), H
Because energy is assumed to be conserved, at any point alongthe streamline, the total head is always constant
Each term in the Bernoulli equation is a type of head.
P/ = Pressure Head
V2/2g = Velocity Head
Z = elevation head
These three heads summed equals H = total energy
Next we will look at this graphically
Th E Li d th H d li G d Li
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The Energy Line and the Hydraulic Grade Line
Q
Measures thestatic pressure Pitot measuresthe total head
1
Z
P/
V2
/2g
EL
HGL
2
1: Static Pressure TapMeasures the sum of theelevation head and the
pressure Head.
2: Pitot Tube
Measures the Total Head
EL : Energy Line
Total Head along a systemHGL : Hydraulic Grade line
Sum of the elevation andthe pressure heads along a
system
Th E Li d th H d li G d Li
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The Energy Line and the Hydraulic Grade Line
Q
Z
P/
V2
/2g
EL
HGL
Understanding the graphical approach ofEnergy Line and the Hydraulic Grade line is
key to understanding what forces aresupplying the energy that water holds.
V2/2g
P/
Z
1
2
Point 1:
Majority of energystored in the water is in
the Pressure Head
Point 2:
Majority of energystored in the water is inthe elevation head
If the tube wassymmetrical, then thevelocity would beconstant, and the HGLwould be level
Plotting the EL and HGL
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Plotting the EL and HGLEnergy Line = Sum of the Pressure, Velocity and Elevation heads
Hydraulic Grade Line = Sum of the Pressure and Velocity heads
EL
HGL
Z=1Z=1Z=1
V2/2g=3V2/2g=3
Z=4
P/ =2.81
V2/2g=.19
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