TL2101 Mekanika Fluida I

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TL2101 Mekanika Fluida I

• Benno Rahardyan

Pertemuan 7
2
Mg Topik Sub Topik Tujuan Instruksional (TIK)
1 Pengantar Definisi dan sifat-sifat fluida, berbagai jenis fluida yang berhubungan dengan bidang TL Memahami berbagai kegunaan mekflu dalam bidang TL
Pengaruh tekanan Tekanan dalam fluida, tekanan hidrostatik Mengerti prinsip-2 tekanan statitka
2 Pengenalan jenis aliran fluida Aliran laminar dan turbulen, pengembangan persamaan untuk penentuan jenis aliran bilangan reynolds, freud, dll Mengerti, dapat menghitung dan menggunakan prinsip dasar aliran staedy state
3 Prinsip kekekalan energi dalam aliran Prinsip kontinuitas aliran, komponen energi dalam aliran fluida, penerapan persamaan Bernoulli dalam perpipaan Mengerti, dapat menggunakan dan menghitung sistem prinsi hukum kontinuitas
4 Idem Idem gaya pada bidang terendam Idem
5 Aplikasi kekekalan energi Aplikasi kekekalan energi dalam aplikasi di bidang TL Latihan menggunakan prinsip kekekalan eneri khususnya dalam bidang air minum
Aplikasi kekekalan energi Darcy-Weisbach, headloss, major losses dan minor losses Keseimbangan bidang terapung
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FLUID DYNAMICSTHE BERNOULLI EQUATION
The laws of Statics that we have learned cannot
solve Dynamic Problems. There is no way to solve
for the flow rate, or Q. Therefore, we need a new
dynamic approach to Fluid Mechanics.
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The Bernoulli Equation
By assuming that fluid motion is governed only by
pressure and gravity forces, applying Newtons
second law, F ma, leads us to the Bernoulli
Equation. P/g V2/2g z constant along a
streamline (Ppressure g
specific weight Vvelocity ggravity
zelevation) A streamline is the path of one
particle of water. Therefore, at any two points
along a streamline, the Bernoulli equation can be
applied and, using a set of engineering
assumptions, unknown flows and pressures can
easily be solved for.
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Bernoulli Example Problem Free Jets 2
A small cylindrical tank is filled with water,
and then emptied through a small orifice at the
bottom.
Case 1 What is the flow rate Q exiting through
the hole when the tank is full?
Case 2 What is the flow rate Q exiting through
the hole when the tank is half full?
-Hint- The Continuity Equation is needed
R1
R1
Assumptions Psurf Pout 0 Because its a small
tank, Vsurf ? 0
?H2062.4 lbs/ft3
4
R.5
R.5
2
Q?
Q?
Case 1
Case 2
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Free Jets 2
Case 1
Apply Bernoullis Equation at the Surface and at
the Outlet 0 Vsurf2/2g 4 0 Vout2/2g 0
With two unknowns, we need another equation
The Continuity Equation AsurfVsurfAoutVout
p(1)2 x Vsurf p(.5)2 x Vout ?
Vsurf.25Vout
R1
R1
Substituting back into the Bernoulli Equation
? (.25Vout)2/2g 4 Vout2/2g Solving for Vout
if g 32.2 ft/s2 Vout .257 ft/s Qout AV
.202 ft3/s (cfs)
?H2062.4 lbs/ft3
4
R.5
R.5
2
Q?
Q?
Case 1
Case 2
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Bernoulli Example Problem Free Jets 2
Case 2
Bernoullis Equation at the Surface and at the
Outlet is changed 0 Vsurf2/2g 2 0
Vout2/2g 0 Continuity eqn remains the same.
Substituting back into the Bernoulli Equation
? (.25Vout)2/2g 2 Vout2/2g Solving for Vout
if g 32.2 ft/s2 Vout .182 ft/s Qout AV
.143 cfs Note that velocity is less in Case 2
R1
R1
?H2062.4 lbs/ft3
4
R.5
R.5
2
Q?
Q?
Case 1
Case 2
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Free Jets
The velocity of a jet of water is clearly related
to the depth of water above the hole. The
greater the depth, the higher the velocity.
Similar behavior can be seen as water flows at a
very high velocity from the reservoir behind the
Glen Canyon Dam in Colorado
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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 streamline
This constant is called the total
head (energy), H Because energy is assumed to be
conserved, at any point along the 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 together, will
always equal H Next we will look at this
graphically
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The Energy Line and the Hydraulic Grade Line
Lets first understand this drawing
Measures the Total Head
1 Static Pressure Tap Measures the sum of the
elevation head and the pressure Head. 2 Pilot
Tube Measures the Total Head EL Energy
Line Total Head along a system HGL Hydraulic
Grade line Sum of the elevation and the pressure
heads along a system
Measures the Static Pressure
1
2
1
2
EL
V2/2g
HGL
Q
P/?
Z
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The Energy Line and the Hydraulic Grade Line
Understanding the graphical approach of Energy
Line and the Hydraulic Grade line is key to
understanding what forces are supplying the
energy that water holds.
Point 1 Majority of energy stored in the water
is in the Pressure Head Point 2 Majority of
energy stored in the water is in the elevation
head If the tube was symmetrical, then the
velocity would be constant, and the HGL would be
level
EL
V2/2g
V2/2g
HGL
P/?
2
Q
P/?
Z
1
Z
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The Complete Example
Solve for the Pressure Head, Velocity Head, and
Elevation Head at each point, and then plot the
Energy Line and the Hydraulic Grade Line
Assumptions and Hints P1 and P4 0 --- V3 V4
same diameter tube We must work backwards to
solve this problem
1
?H2O 62.4 lbs/ft3
R .5
4
R .25
2
3
4
1
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Point 1 Pressure Head Only atmospheric ? P1/?
0 Velocity Head In a large tank, V1 0 ?
V12/2g 0 Elevation Head Z1 4
1
?H2O 62.4 lbs/ft3
4
R .5
R .25
2
3
4
1
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Point 4 Apply the Bernoulli equation between 1
and 4 0 0 4 0
V42/2(32.2) 1 V4 13.9 ft/s Pressure Head
Only atmospheric ? P4/? 0 Velocity Head
V42/2g 3 Elevation Head Z4 1
1
?H2O 62.4 lbs/ft3
4
R .5
R .25
2
3
4
1
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Point 3 Apply the Bernoulli equation between 3
and 4 (V3V4) P3/62.4
3 1 0 3 1 P3 0 Pressure Head P3/?
0 Velocity Head V32/2g 3 Elevation Head Z3
1
1
?H2O 62.4 lbs/ft3
4
R .5
R .25
2
3
4
1
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Point 2 Apply the Bernoulli equation between 2
and 3 P2/62.4 V22/2(32.2)
1 0 3 1 Apply the Continuity
Equation (?.52)V2 (?.252)x13.9 ? V2 3.475
ft/s P2/62.4 3.4752/2(32.2) 1 4 ? P2
175.5 lbs/ft2
Pressure Head P2/? 2.81 Velocity Head
V22/2g .19 Elevation Head Z2 1
1
?H2O 62.4 lbs/ft3
4
R .5
R .25
2
3
4
1
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Plotting the EL and HGL
Energy Line Sum of the Pressure, Velocity and
Elevation heads Hydraulic Grade Line Sum of the
Pressure and Velocity heads
V2/2g.19
EL
P/? 2.81
V2/2g3
V2/2g3
Z4
HGL
Z1
Z1
Z1
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Pipe Flow and the Energy Equation
For pipe flow, the Bernoulli equation alone is
not sufficient. Friction loss along the pipe,
and momentum loss through diameter changes and
corners take head (energy) out of a system that
theoretically conserves energy. Therefore, to
correctly calculate the flow and pressures in
pipe systems, the Bernoulli Equation must be
modified. P1/? V12/2g z1 P2/? V22/2g z2
Hmaj Hmin Major losses Hmaj Major losses
occur over the entire pipe, as the friction of
the fluid over the pipe walls removes energy from
the system. Each type of pipe as a friction
factor, f, associated with it.
Energy line with no losses
Hmaj
Energy line with major losses
1
2
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Pipe Flow and the Energy Equation
Minor Losses Hmin Momentum losses in Pipe
diameter changes and in pipe bends are called
minor losses. Unlike major losses, minor losses
do not occur over the length of the pipe, but
only at points of momentum loss. Since Minor
losses occur at unique points along a pipe, to
find the total minor loss throughout a pipe, sum
all of the minor losses along the pipe. Each
type of bend, or narrowing has a loss
coefficient, KL to go with it.
Minor Losses
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Major and Minor Losses
Major Losses Hmaj f x (L/D)(V2/2g) f
friction factor L pipe length D pipe
diameter V Velocity g gravity Minor
Losses Hmin KL(V2/2g) Kl sum of loss
coefficients V Velocity g gravity When
solving problems, the loss terms are added to the
system at the second point P1/? V12/2g z1
P2/? V22/2g z2 Hmaj Hmin
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Loss Coefficients
Use this table
to find loss coefficients
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Pipe Flow Example
?oil 8.82 kN/m3 f .035
1
Z1 ?
2
Z2 130 m
60 m
Kout1
7 m
r/D 0
130 m
r/D 2
If oil flows from the upper to lower reservoir at
a velocity of 1.58 m/s in the 15 cm diameter
smooth pipe, what is the elevation of the oil
surface in the upper reservoir? Include major
losses along the pipe, and the minor losses
associated with the entrance, the two bends, and
the outlet.
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Pipe Flow Example
?oil 8.82 kN/m3 f .035
1
Z1 ?
2
Z2 130 m
60 m
Kout1
7 m
r/D 0
130 m
r/D 2
Apply Bernoullis equation between points 1 and
2Assumptions P1 P2 Atmospheric 0 V1
V2 0 (large tank) 0 0 Z1 0 0 130m
Hmaj Hmin Hmaj (fxLxV2)/(Dx2g)(.035 x 197m x
(1.58m/s)2)/(.15 x 2 x 9.8m/s2) Hmaj 5.85m
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Pipe Flow Example
?oil 8.82 kN/m3 f .035
1
Z1 ?
2
Z2 130 m
60 m
Kout1
7 m
r/D 0
130 m
r/D 2
0 0 Z1 0 0 130m 5.85m Hmin Hmin
2KbendV2/2g KentV2/2g KoutV2/2g From Loss
Coefficient table Kbend 0.19 Kent 0.5
Kout 1 Hmin (0.19x2 0.5 1) x
(1.582/2x9.8) Hmin 0.24 m
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Pipe Flow Example
?oil 8.82 kN/m3 f .035
1
Z1 ?
2
Z2 130 m
60 m
Kout1
7 m
r/D 0
130 m
r/D 2
0 0 Z1 0 0 130m Hmaj Hmin 0 0
Z1 0 0 130m 5.85m 0.24m Z1 136.09
meters