2013 lect2 basic fluid flow

8/13/2019 2013 Lect2 Basic Fluid Flow

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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


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


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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2013 lect2 basic fluid flow

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