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is flow within the boundary walls.
TYPES OF FLOW
Internal flow
Types of internal flow include pipe flow, channel flow, airflow in ducts. Thistype of flow is controlled using valves, fans, pumps.
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is flow outside of a boundary or body.
External flow
Examples of this type of flow include :
flow around immersed bodiesflow over aircraft wings,airflow around buildings/cars
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in general all fluids flow three-dimensionally, with pressures and velocitiesand other flow properties varying in all directions.
In many cases the greatest changes only occur in two directions or evenonly in one.
other direction can be effectively ignored, making analysis much moresimple.
Flow is one dimensional if the flow parameters (such as V, P, d) at a given instant intimeonly vary in the direction of flow and not across the cross-section
An example of one-dimensional flow is the flow in a pipe.
ONE DIMENSION FLOW
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Possibly - but it is onlynecessary if very high accuracy
is required.
A correction factor is thenusually applied.
ONE DIMENSION FLOW
FOR real fluid, flow must be zero at the pipe wall - yet non-zero in the centrethere is a difference of parameters across the cross-section.
Should this be treated as two-dimensional flow?
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LAMINAR AND TURBULENT FLOW.
Fluid flow at low velocities is smooth with the fluid particles movingin straight lines along the direction of flow (LAMINER).
The majority of flows in practice are TURBULENTwith no uniformmotion at the local level but an average velocity in the direction of
flow.
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A typical velocity profile across a pipe
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FLUID FLOW BASIC
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It is a known fact said thatthe fluid continuously and permanently deformed
under shear stress
while solid exhibits a finite deformation which does notchange with time.
It is also said that liquid cannot return to their originalstate after the deformation.
Differences between solids and fluids:
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States of Matter
FluidSolid
Shear Stress t
a fluid, such as water or air, deformscontinuously when acted on by shearing
stresses of any magnitude.- Munson, Young, Okiishi
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The fluids like water or oil, flow on the surface in the form of layerswith the top layer moving at the fastest speed and the bottom layermoving at slowest speed.
Why the speed of the flow of fluid reduces.
Because there is RESISTANCE to the flow of the fluid, which iscause by FRICTION of the adjoining layers.
This property of the fluids is called as viscosity.
FLUID FLOW & VISCOSITY
Viscosity describes a fluid's internal resistance to flowand may be thought of as a measure of fluid friction.
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Viscosity of water
Water at 20 C has a viscosity of 1.0020 cP (centipoise)
1 P = 0.1 Pas,1 cP = 1 mPas = 0.001 Pas.
Dynamic viscosity
Kinematic viscosity
1 St = 1 cm2s1= 104m2s1.1 cSt = 1 mm2s1= 106m2s1.
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Fluid Deformation between Parallel Plates
Side view
Force F causes the top plate to have velocity U.
What other parameters control how much force is
required to get a desired velocity?
Distance between plates (b)
Area of plates (A)
F
b
U
Viscosity!
the velocity of the upper layer of the fluid is faster than the velocity of the lower layers of the fluid.
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Shear Stress
change in velocity with
repect to distance
AFt
2m
N
b
Ut
b
U
dy
dut
b
AUF
AU
Fb
2m
sN
s
1
Tangential stress
Rate of deformation
rate of shear
F
b
U
dy
du
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Shear Stress
b
AUF
bUA
F
AU
Fb
2m
sN
F
b
U
Absolute Viscosity
Shear stess
(dyne/cm2 )
Shear strain rate
(s-1)
Dyne-s/cm2=P (poise)
1000mPa.S1000cPP10Pa.s
Dyne-s/cm2=g-s/cm=P=103 cP
P1010g.s.cm100cm
1000g.s
m
kg.s
m
skg.m.sPa.s
m
sN 11
2
-2
2
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Shear Stress
b
AUF
F
b
U
sm
kg
m
s
s
mkg
..
.
m
sN222
Kinematic Viscosity
densityviscosityabsolute
2s
mkgN 12
3
m
kg
sm
k
sm
g
Stokes
cm
/cms-dyne
s/cmdyne 2
42
2
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ssds
Fluid classification by response to shear stress
Newtonian fluida fluid whose stress versus strain rate curve is
linear and passes through the origin
The constant of proportionality is known as the viscosity.
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FLUID CLASSIFICATION BASED ON VISCOSITY
2) Real fluid:In practice all the fluids are real fluids because all of them have viscosity, smallor high.For the real fluids which are liquids the viscosity reduces as the temperatureincreases and for the gases, the viscosity increases as the temperatureincreases.
1) Ideal FluidThe fluid which is incompressible and has no viscosity is known as the ideal fluid.However, the ideal fluid is only an imaginary fluid, because all the fluids have
viscosity and there is no fluid that doesnt have viscosity (see the fig below)
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3) Newtonian fluid:Is a real fluid that obeys Newtons law of viscosity : the shear stress betweenvarious layers of the fluid is proportional to the rate of shear strain or the
velocity gradient. Fluids, such as water and most gases which have a constantviscosity.
4) Non-Newtonian fluid:Is a real fluids that do not obey the Newtons law of viscosity.In such fluids the shear stress between the various layers of fluid is notproportional to the rate of shear strain or the velocity gradient.
5)Ideal plastic fluid:The fluid in which the shear stress is more than the yield value and shear stressis proportional to the rate of shear strain or velocity gradient is known as idealplastic fluid.
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Bingham plastic fluidsThe viscosity curve does not go through the origin.It behaves as a solid at low stresses but
It flows as a viscous fluid at high stresses.
at a small shear rate (beginning of move), the shear stress can be substantialBut once the fluid is moving, the shear stress is directly proportional to shearrate in exactly the same manner as for Newtonian fluids.
Fluid behave in this manner are :Water suspensions of rock particlesMore familiarly,, a suspension of potatoes in a liquid,flow in the bowl when you stir, flow harder when you stir harder,and assume a mountain peak shape if undisturbed.
Non- Newtonian Fluids:
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Shear thickening:viscosity increaseswith the rate of shear.
Non- Newtonian Fluids:
Pseudoplastic fluids(soo doe plastic)
Shear thinning:
The viscosity of this pseudoplastic fluids decreases with increased shearing.Shear thinning liquids are very commonly, but misleadingly, described asthixotropic.
Molten polymers have this characteristic, which is used to advantage ininjection molding when the material flows through small cross section
gates. Paper pulp suspensions also display pseudoplastic viscosity behavior.
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for these fluids, the shear stress changes with time of shearing.
Thixotropic Fluidsthe viscosity decreases at higher shearing rates and become lessviscousover time when shaken, agitated, or otherwise stressed.
Mayonnaise has this characteristic. its viscosity decreases with higher rates of shearing, but onlyafter a minimum amount of shear is reached. This behavior is related to breaking bonds betweenparticles or molecules or to changes from the at-rest shape to moving shape of long molecules
Rheopectic:
materials which become moreviscous over time when shaken, agitated, or otherwise stressed.
A magnetorheological fluidis a type of "smart fluid" which, when subjected to a magnetic field,greatly increases its apparent viscosity, to the point of becoming aviscoelastic solid.
Shear Time Dependent Fluids
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1) Liquids: As the temperature of the liquid fluid increases itsviscosity decreases.
In the liquids the cohesive forces between the moleculespredominates the molecular momentum transfer between themolecules, mainly because the molecules are closely packed(liquids have lesser volume than gases. The cohesive forces are maximum insolids so the molecules are even more closely packed in them)
When the liquid is heated the cohesive forces between themolecules reduce thus the forces of attraction betweenmolecules reduce, which eventually reduces the viscosity ofthe liquids.
Effect of Temperature on Liquid and Gas Fluids
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Where: - Viscosity of the liquid at t degree Celsius n poiseoViscosity of the fluid at 0
oCelsius in poise, are the constants
For liquids: = o/ (1 + t + t2)
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Effect of Temperature on Liquid and Gas Fluids
Fundamental mechanisms
Liquids - cohesion and momentum transfer
Viscosity decreases as temperature increases. Relatively independent of pressure
(incompressible)
Gases - transfer of molecular momentum
Viscosity __________ as temperature increases.
Viscosity __________ as pressure increases.increases
increases
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In gases there is opposite phenomenon. The viscosity of thegases increases as the temperature of the gas increases. Thereason behind this is again the movement of the molecules andthe forces between them.In the gases the cohesive forces between the molecules is lesser,
while molecular momentum transfer is high. As the temperatureof the gas is increased the molecular momentum transfer rateincreases further which increases the viscosity of the gas.
For gases: = o+ t + t2
Effect temp on Gases:
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Ostwald viscometers measure the viscosity of afluid with a known density.
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GL SS C PILL RY VISCOMETERS
P = Pressure difference across capiller
R = Radius of capiller
L = Length od capiller
V = Volume fluida
= ViscosityLV
t
8
Pr4
ASTM D445
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SIMPLE VISCOMETER :A CALIBRATED HOLE IN THE BOTTOM.
2
1
txV
Dzg o
128
4
)(
xz
xQ
Dzg o
128
4
)(
V
tk
(Poiseuille Eq.)
tk
cP = fluid density X cSt
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0.02m
0.1m D=1mm
=1050kg/m3
2
1
FV
zzg 2
2
2
12 )(
Fzzg )(12
xQ
Dzg o
128
4
)(
PENGUKURAN VISKOSITAS
Fdm
dWVgzP o )2
(2
4
0
128..
DxQ
PF
x V2 diabaikan untuk laminer
z
4( )
128. .
og z D waktu
Vol x
2 1 4
0
128( ) . .g z z Q x
D
H.Newton
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Fd is the frictional force (acting on the interface between the fluid and the particle(in N), is the dynamic viscosity (N s/m2),R is the radius of the spherical object (in m), andvs is the particle's settling velocity (in m/s).
Stokes' law
vs is the particles' settling velocity (m/s)(vertically downwards ifp>f, upwards ifp
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b
AU
F
Co-axial cylinder viscometer
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Example: Measure the viscosity of water
The inner cylinder is 10 cm in diameter and rotates at 10 rpm.The fluid layer is 2 mm thick and 10 cm high. The powerrequired to turn the inner cylinder is 50x10-6watts. What is thedynamic viscosity of the fluid?
Outer
cylinder
Thin layer of water
Inner
cylinder
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Solution Scheme
Restate the goal Identify the given parameters and represent
the parameters using symbols
Outline your solution including the equationsdescribing the physical constraints and anysimplifying assumptions
Solve for the unknown symbolically
Substitute numerical values with units and dothe arithmetic Check your units!
Check the reasonableness of your answer
Solution
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Viscosity Measurement: Solution
hr
Pt32
2
23-
32
6-
s/mN1.16x10
m)(0.1m)(0.05(1.047/s)2
m)(0.002W)10(50
x
t
AUF
UA
t
hrF22
P
t
hr
P
322
Outercylinder
Thin layer of water
Inner
cylinder
r= 5 cm
t= 2 mmh= 10 cmP= 50 x 10-6W10 rpm
r
2rh
Fr
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A rotating disk viscometer has a radius, R = 50 mm, and a clearance,h = 1 mm, as shown in the figure.
a. If the torque (T) required to rotate the disk at n= 900 rpm is 0.537 N
m,
determine the dynamic viscosity of the fluid. You may neglect the viscous forces
acting on the rim of the disk and on the vertical shaft.b. If the uncertainty in each parameter is 1%, determine the uncertainty in the
viscosity.
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PERS. KONTINUITAS
HUKUM KEKEKALAN MASSA
PERSAMAAN NERACA
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Input + Generation Output Consumption = Accumulation.
PERSAMAAN NERACA
= Deposits Account-penalties- withdrawal+ Interest
ratesavings
imigration + born - emigration -death = City population
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THE CONSERVATION OF MASS
Matter cannot be created or destroyed - (it is simply changed in to adifferent form of matter
Mass entering per unit time = Mass leaving per unit time +Increase of mass in the control volume per unit time
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air
Natural gas
Exhaust Gas
System boundary
No accumulation
gas stove
Mass entering per unit time = Mass leaving per unit time
MASS CONSERVATION LAW ONA STEADY FLOW PROCESS
inflow equals outflow
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222111 vAvAm
CONTINUITY EQUATION(STEADY STATE, one dimension)
Accumulation = 0
Rate of mass entering = Rate of mass leaving
1
2
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Example
Water flows through a 4.0 cm diameter pipe at 5 cm/s.The pipe then narrows downstream and has a diameter of of 2.0 cm.What is the velocity of the water through the smaller pipe?
112
2
2
12
2211
4vvr
rv
vAvA
= 20 cm/s
Flow through a pipe of changing diameter
Example
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If pipe 1 diameter = 50mm, mean velocity 2m/s, pipe 2 diameter40mm takes 30% of total discharge and pipe 3 diameter 60mm. What
are the values of discharge and mean velocity in each pipe?
Exampleto determine the velocities in pipes coming from a junction.
F t
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Faucet :
Explain this phenomenonusing the equation ofcontinuityA1
A2
V1
V2 A1V1=A2V2.
as it falls down, A will bedecreased.
on the other hand V will beincreased because of gravity.therefore, this phenomenon is
appropriate for the equation forcontinuity.
22
A stream of water gets narrower as it falls from a faucet (try it & see)
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steady flow
is the type of flow in which the various parameters at any point donot change with time.
0,,
zyxdt
d
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unsteady or non-steady
is a flow which changes with time.Real flows are generally the latter type
but in completing flow assessmentsit is often more practical
to assume steady flow conditions.
0,,
zyxdt
d
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outinsys mmdt
dM
UNSTEADY FLOWSATU DIMENSI,
Example :
A tank is filled with water flow rate qf and flowing out q. Find height
of the level of the surface .
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dtMdmm oi )(
dt
Ahdqq ooii
)(
dt
Ahdqq ofi
)(
qqdt
Ahdf
)(
)(1)(
qqAdt
hdf
Solution
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BERNOULLI EQUATION
F
dm
dWVgz
P other
)(2
2
What is Bernoulli equation
Bernoulli's principle can be derived from the principle of conservation of energy.
This states that, in a steady flow, the sum of all forms of mechanical energy (kinetic energy and potentialenergy, pressure energy) in a fluid along a streamline is the same (remain constant) at all points
on that streamline.
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It can be expanded to include these simply, by adding the appropriateenergy terms:
Loss
Work done
Heatsupply
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Energy form
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Energy form
Kinetic energy is the energy in moving objects or mass.
Potential energy is any form of stored energy of position
The gravitational potential energy
Objects have mechanical energy if they are in motionand/or if they are at some position relative to a zero
potential
The internal energy the sum of the kinetic and potential energies of theparticles that form the system
the internal energy of the system is still proportional to itstemperature.
ENERGY BALANCE (TERMODYNAMICS)
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Heating cooling jacket
Volume-changing
piston
ENERGY BALANCE (TERMODYNAMICS)
Work
Propeler
Fluida
Internal Energy
Potential Energy
Kinetics EnergyHeat
Volume work
)
2(
2V
gzu
dt
dQ
dtdWother
dt
dWinjection
dt
dmm in
outdm
mdt
Mechanical work
Internal Energy
Potential Energy
Kinetics Energy
Energy is conserved;
Energy can be transferred from the system to its surroundings, or vice versa,but it can't be created or destroyed
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Accumulation = Flow in - Flow out
dQ
sys
Vgzumd )
2(
2
outoutdmV
gzu )2
(2
Heating cooling jacket Volume-changing
piston
otherdW
inindmV
gzu )2
(2
inin dmPv )(
outout dmPv )(
outin dQdQdQ
inoutother dWdWdW
The sign convention for w is
decreases when the system doeswork on its surroundings.
INJECTION WORK
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inin
sys
dmPvV
gzumd )()2
(2
INJECTION WORK
inin
sys
dmV
gzuV
gzumd )2
()2
(22
inin
sys
dmV
gzPvuV
gzumd )2
()2
(22
Step 1
Step 2
INJECTION WORK
W=PV
beginningPiston move
Constant pressure
Change of energy in step 1
Piston move back
Change of energy in step 2
Change of energy in step 1 and 2
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inin
sys
dmV
gzPvuV
gzumd )2
()2
(22
otheroutout dWdQdmVgzPvu )2
(
2
inin
sys
dmV
gzhV
gzumd )2
()2
(22
ENTALPY
otheroutout dWdQdmV
gzh )2
(2
ENERGY BALANCE
?
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ADIABATIC THROTTLE TURBIN & COMPRESOR
SIMPLE HEATERSteady state flow,
SIMPLE REACTORSteady state flow,
ENERGY BALANCE IN PROSES SYSTEM?
inin
sys
dmV
gzhV
gzumd )
2
()
2
(22
otheroutout dWdQdm
Vgzh )
2(
2
In many applications of Bernoulli's equation, the change in thegztermalong the streamline is so small compared with the other terms it can be
ignored, and omitted. T
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horizontal
ENERGY BALANCE for THROTTLE PROSES
inin
sys
dmV
gzhV
gzumd )2
()2
(22
otheroutout dWdQdm
Vgzh )
2(
2
outoutinin dmhdmh
P1 P2
ADIABATIC THROTTLE Steady flow
Adiabatic
work=0
V1=V2
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ENERGY BALANCE for TURBIN/COMPRESSOR
inin
sys
dmV
gzhV
gzumd )2
()2
(22
otheroutout dWdQdmV
gzh )2
(2
TURBIN & COMPRESSOR
outinoa hh
dm
dW.
Steady flow
Elevation difference is neglected
Q is neglected
Kinetic energy diff is neglected
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ENERGY BALANCE for Heater
inin
sys
dmVgzhVgzumd )2
()2
(22
otheroutout dWdQdmV
gzh )2
(2
inout hhdm
dQ
SIMPLE HEATERSteady flow
Flowing horizontally.
V is neglected
Work =0
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inin
sys
dmVgzhVgzumd )2
()2
(22
otheroutout dWdQdmV
gzh )2
(2
SIMPLE REACTOR
tanreacproduct hhdm
dQ
outoutinin dmhdmhdQ
ENERGY BALANCE for SIMPLE REACTOR
Steady flow
Flowing horizontally.
V is neglected
Work =0
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Bottle Filling Problem
ininsys dmhmud )(
UNSTEADY STATE SYSTEM
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UNSTEADY STATE SYSTEM
inin
sys
dmVgzhVgzumd )2
()2
(22
otheroutout dWdQdmV
gzh )2
(2
ininsys dmhmud )(
Q=0
Bottle Filling ProblemSteady flow
Flowing horizontally.
V is neglected
Work =0
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inin
sys
dmVgzPuVgzumd )()(22
22
otheroutout dWdQdmV
gzP
u )(2
2
1. Electrostatic, magnetic, and surface energy are
negligible2. The content of the system are uniform
3. The inflow and outflow streams are uniform
4. The acceleration of gravity is constant
ENERGY BALANCE
FOR A STEADY INCOMPRESSIBLE FLOW
)(
Puh
inin
sys
dmV
gzhV
gzumd )2
()2
(22
otheroutout dWdQdmV
gzh )2
(2
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inin
sys
dmV
gzP
uV
gzumd )()(22
22
BERNOULLI
dm
dQu
dm
dWVgz
P other)(
2
2
Steady
Fdm
dWVgz
P other
)(2
2
otheroutout dWdQdmV
gzP
u )(2
2
ENERGY BALANCE
dm
dQu
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BERNOULLI EQUATION
Fdm
dWVgz
P other
)(2
2
g
F
gdm
dW
g
Vz
g
P other
)(2
2
HEAD FORM OF BERNOULLI EQUATION
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SATUAN BERNOULLI
g
F
gdm
dW
g
Vz
g
P other
)(2
2
HEAD FORM OF BERNOULLI
EQUATION
zgP
ZERO FLOW =STATIK,
FdmdWVgzP other )(
2
2
What is the Unit ?Energy/massa
lengthWhat is the unit?V=0, W=0, F=0
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BERNOULLI EXAMPLE
SPEED OF FLUIDFLOW OUT OF THE TANK
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A large bucket full of water has two drains.
One is a hole in the side of the bucket at thebottom, and the other is a pipe coming out ofthe bucket near the top, which bent isdownward such that the bottom of this pipeeven with the other hole, like in the picture
below:Though which drain is the water spraying outwith the highest speed?
1. The hole
2. The pipe
3. Same CORRECT
40
Note, the correct height, is where the water reaches theatmosphere, so both are exiting at the same height!
FLOW OUT OF THE TANK
SPEED OF FLUIDFLOW OUT OF THE TANK
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Through which hole will the water come out fastest?
A
B
C
P1+gz1+ v12
= P2+gz2+ v22
Note:
All three holes have same P=1 Atm
1gz1+ v12= 1+ gz2+ v2
2
gz1+ v12= gz2+ v2
2
Smaller y gives larger v. Hole C is fastest
36
v22= 2gz1z2)+ v1
2
FLOW OUT OF THE TANK
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1
we take the datum through the orifice
Between 1 and 2
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Between 1 and 3
If the mouth piece has been removed,
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Example
A closed tank has an orifice 0.025m diameter in oneof its vertical sides. The tank contains oil to a depth of0.61m above the centre of the orifice and thepressure in the air space above the oil is maintainedat 13780 N/m2 above atmospheric. Determine thedischarge from the orifice.(Coefficient of discharge of the orifice is 0.61, relativedensity of oil is 0.9).[0.00195 m3/s]
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Example
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Example
In a vertical pipe carrying water,pressure gauges are inserted at points A and Bwhere the pipe diameters are dA=0.15m and dB=0.075m respectively andpoint B is 2.5m below A.
When the flow rate down the pipe is 0.02 m3/sec, the pressure at B is 14715N/m2 greater than that at A.Assuming the losses in the pipe between A and B can be expressed as
where vis the velocity at A, find the value of k.
If the gauges at A and B are replaced by tubes filled with water andconnected to a U-tube containing mercury of relative density 13.6, give asketch showing how the levels in the two limbs of the U-tube differ andcalculate the value of this difference in metres.[k = 0.319, 0.0794m]
THE SKETCH OF PROBLEM
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THE SKETCH OF PROBLEM
Part i
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By continuity: Q = uAAA= uBAB
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Part ii
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Lift a House
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Calculate the net lift on a 15 m x 15 m house when a 30 m/swind (1.29 kg/m3) blows over the top.
P1+gy1+ v12= P2+gy2+ v2
2
P1P2= (v22v1
2)
= (1.29) (302) N / m2
= 581 N/ m248
F = P A
= 581 N/ m2(15 m)(15 m) = 131,000 N
= 29,000 pounds!
G
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lift
drag
U
q
2
2UACF LL
p > p0positive pressure
DAYA ANGKATSAYAP PESAWAT TERBANG
Sh d P F
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Shear and Pressure Forces:Horizontal and Vertical Components
lift
drag
U
Parallel to the approach velocity
Normal to the approach velocity
q
2
2U
ACF dd
2
2UACF LL
Adefined as projected
area _______ to force!normal
drag
lift
p < p0negative pressure
p > p0positive pressure
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