aarsh mahavidylaya 3rd mech. chp-8,10 sub-f.m.h.m

7/31/2019 aarsh mahavidylaya 3rd mech. chp-8,10 sub-f.m.h.m.

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Boundary layer concept


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Laminar flow


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Laminar Boundary Layer Flow

The laminar boundary layer is a very smooth

flow, while the turbulent boundary layer contains

swirls or eddies.

The laminar flow creates less skin friction drag

than the turbulent flow, but is less stable.

Boundary layer flow over a wing surface begins

as a smooth laminar flow. As the flow continues

back from the leading edge, the laminar

boundary layer increases in thickness.


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Turbulent Boundary Layer Flow

At some distance back from the leading edge,the smooth laminar flow breaks down and

transitions to a turbulent flow.

From a drag standpoint, it is advisable to have

the transition from laminar to turbulent flow asfar aft on the wing as possible, or have a large

amount of the wing surface within the laminar

portion of the boundary layer.

The low energy laminar flow, however, tends to

break down more suddenly than the turbulent

layer.


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Flow through the pipes in

series


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Pipes in series is defined as the pipes of different lengthsand different diameters connected end to end to form a

pipe line.L1,L2,L3 = length of pipes 1,2 and 3

d1,d2,d3 = diameter of pipes 1,2,3

v1,v2,v3 = velocity of flow through pipes 1,2,3

f1,f2,f3 = coefficient of frictions for pipes 1,2,3

H = difference of water level in the two tanks

The discharge passing through the pipe is same.

Q=A1V1=A2V2=A3V3

The difference in liquid surface levels is equal to the sumof the total head loss in the pipes


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Parallel pipe system

Consider a main pipe which divide into two or more

branches as shown in figure

Again join together downstream to

form a single pipe then the branch pipes are said to be.

increased by connecting pipes in parallel

the rate of flow in the main pipe is equal to the sum of rate

of flow through branch pipes.

hence

Q =Q1+ Q2


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In this arrangement loss of head for each pipe is same

Loss of head for branch pipe1=loss of head for branch

pipe 2


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Moody diagram

Moody Diagram that can be used to estimate

friction coefficients

The Moody friction factor - (or f) - is used in the Darcy-

If the flow is transient - 2300 < Re < 4000- the flowvaries between laminar and turbulent flow and the

friction coefficient is not possible to determine.

The friction factor can usually be interpolated between

the laminar value at Re = 2300and the turbulent value at

Re = 4000


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Moody diagram


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Total energy gradient line


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Total energy gradient line is equal to sum ofpressure head ,velocity head and datum head

EL = H = p / W + v2 / 2 g + h = constant along astreamline

where

(EL ) Energy Line

For a fluid flow without any losses due to friction-

energy line would be at a constant level. In apractical world the energy line decreases alongthe flow due to losses.

A turbine in the flow reduces the energy line and apump or fan in the line increases the energy line


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Hydraulic Grade Line (HGL )

Hydraulic gradient line is the sum of

pressure head and datum headHGL = p / W + h

where

The hydraulic grade line lies one velocity headbelow the energy line.


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Hydro Power Plants

A Hydro Power Harvester.


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A Two-Way Welfare for the Globe


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Specific Speed


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,m

Specific Speed

Hea


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MORE ADAPTED TYPE OF TURBINE AS FUNCTION OF THE SPECIFIC SPEED.

Specific Speed in r.p.m. Turbine type Jump height in m

Until 18 Pelton of an injector 800From 18 to 25 Pelton of an injector 800 to 400

From 26 to 35 Pelton of an injector 400 to 100

From 26 to 35 Pelton of two injectors 800 to 400

From 36 to 50 Pelton of two injectors 400 to 100

From 51 to 72 Pelton of four injectors 400 to 100

From 55 to 70 Very slow Francis 400 to 200

From 70 to 120 Slow Francis 200 to 100

From 120 to 200 Normal Francis 100 to 50

From 200 to 300 Quick Francis 50 to 25

From 300 to 450 Extra-quick Francis 25 to 15

From 400 to 500 Extra-quick helix 15

From 270 to 500 Slow Kaplan 50 to 15

From 500 to 800 Quick Kaplan 15 to 5

From 800 to 1100 Extra-quick Kaplan Less than 5


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Varieties of Hydro Resources Available in Nature

Each Resource is naturally fit in the classification?

Any modification of the resource is required for better

harvesting of resource?


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The Size of A Power Plant

A Macro Power Plant : > 100 MW.

A Small Power Plant: 5 MW to 25 MW.

A Mini Power Plant: 500 kW to 2 MW.

A Micro Power Plant: 50 kW to 200 kW.

A Pico Power Plant: < 30 kW.


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East Flowing River : The Krishna


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East Flowing River : The Krishna

Catchment Area: 2,58,958 Sq. km

Annual Yield: 57,000 M.cum


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The West Flowing River: The Sharavathi


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MACRO -HYDRO-POWER POTENTIAL IN INDIA

BASINS/RIVERS POWER POTENIAL IN

MW

Basin/Rivers Probable Capacity (MW)

Indus Basin 33,832

India is blessed with immense amount of hydro-electric potential and

ranks 5th in terms of exploitable hydro-potential on global scenario.

,

Central Indian River

system

4,152

Western Flowing

Rivers of southern

India

9,430

Eastern Flowing

Rivers of southern

India

14,511

Brahmaputra Basin 66,065

Total 1,48,701


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LIST OF HYDRO ELECTRIC POWER STATIONS IN THE

COUNTRY WITH STATION CAPACITY ABOVE 25 MW

REGION/NO.OF

STATIONSNO.OF UNITS

CAPACITY

(MW)

NORTHERN 55 187 13678.25

WESTERN 28 101 7392.00

SOUTHERN 66 237 11294.45

EASTERN 15 55 3847.70

EASTERN 9 26 1116.00

ALL INDIA

(TOTAL)173 606 37328.40


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More Details of Macro Hydro Power Plants

STATION NO. OF UNITSX SIZE (MW)

CAPACITY (MW)

Bhakra l hydro electric

power station5*108 540.00

Bhakra r hydro electric

power station5*157 785.00

Ganguwa y ro e ectr c

power station1*29.25+2*24.2 77.65

Kotla hydro electric power

station1*29.25+2*24.2 77.65

Dehar hydro electric power

station6*165 990.00

Pong hydro electric power

station6*66 396.00


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MACRO HYDRO-POWER PROJECTS IN INDIA

S.N o. P roject R iver / State Insta lled

Capacity

1 Bhakra Sutlej & Beas/ H im achal P radesh 1225 M W

2 Dehar Sutlej / H im achal P radesh 990 M W

3 Kalinadi Stage-I Kalinadi 910 M W

4 Sharavathy Sharavathy / K arnataka 891 M W

5 Koyna Koyna / M aharashtra 880 M W

6 Nagarjuna-Sagar Krishna / Andhra P radesh 810 M W

7 Idduki stage-I Idukki, K alavu & Cheru than / K erala 780 M W

8 Srisailam Krishna / Andhra P radesh 770 M W

9 Salal Chenab / Jam m u & K ashm ir 690 M W

10 Ranjit Sagar Ravi / Punjab 600 M W

11 Cham era - I Ravi / H im achal Pradesh 540 M W


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Small Hydro Map of India


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Methodology of HEPP Development

Site Survey: Hydrological & geological Survey.

Estimation of Potential

Regulations & Environmental Concerns

Feasible Supply Tur ne Se ect on

Costing and Payback.


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Hydrological Survey: Flow Duration Curve

To measure the flow-rate vs time at a given site.

Direct Measurement of the flow rate.

The more robust option is to find out the flow-rate bywor ng ou e vo ume o wa er a was en er ng e

river.

This uses the rainfall data from met office.


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Hydrological Cycle


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Catchment Area


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Catchment Area: 2,58,958 Sq. km

Annual Yield: 57,000 M.cum


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Rain Fall Data : Hydrograph


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Flow Duration Curve


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Average Flow duration Curve

Average Flow duration Curve

Mean of 10 30 years

u

mecs

% of time

D

ischarge,

Qm

Q100%

Q95%

Q50%


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Definition of A Turbo MachineDefinition of A Turbo Machine

Turbines are energy developing machines. Turbines convert fluid energy intoTurbines are energy developing machines. Turbines convert fluid energy into

mechanical energy. The mechanical energy developed by the turbines is usedmechanical energy. The mechanical energy developed by the turbines is used

in running an electric generator, which is directly connected, to the shaft of thein running an electric generator, which is directly connected, to the shaft of the

electrical generator.electrical generator.

Earlier days methodEarlier days method wooden wheelwooden wheel

Overshot WheelOvershot Wheel

Had very good efficiency

Could not handle large quantity of water

Undershot Wheel

Low Efficiency


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General layout of HydroGeneral layout of Hydro--Power PlantPower Plant

a) Reservoir

Reservoirs ensure supply of water through out the year, by storing water

during rainy season and supplying the same during dry season.

b)b) DamDame unct on o t e am s to ncrease t e reservo r capac ty an to ncrease t ee unct on o t e am s to ncrease t e reservo r capac ty an to ncrease t e

working head of the turbine.working head of the turbine.

c) Penstockc) Penstock

A pipe between dam and turbine is known as penstock. It will carry the waterA pipe between dam and turbine is known as penstock. It will carry the water

from dam to turbine. Penstock is commonly made of steel pipes covered withfrom dam to turbine. Penstock is commonly made of steel pipes covered withRCC.RCC.


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d)d) Surge tank/Surge tank/ForebayForebay

When the rate of water flow through the penstock is suddenly decreased, theWhen the rate of water flow through the penstock is suddenly decreased, the

pressure inside the penstock will increase suddenly due to water hammerpressure inside the penstock will increase suddenly due to water hammerand thereby damage the penstock.and thereby damage the penstock.

Surge tank/Surge tank/ForebayForebay is constructed between the dam and turbine. It will actis constructed between the dam and turbine. It will act

as a pressure regulator during variable loads.as a pressure regulator during variable loads.

e)e) TurbineTurbine

Turbines convert the kinetic and potential energy of water into mechanicalTurbines convert the kinetic and potential energy of water into mechanical

energy to produce electric power.energy to produce electric power.

f) Generator and Transformerf) Generator and Transformer

Electric generator converts mechanical energy into electrical energy. A stepElectric generator converts mechanical energy into electrical energy. A step

up transformer will increase the voltage for loss free transmission.up transformer will increase the voltage for loss free transmission.


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General layout of HydroGeneral layout of Hydro--Power PlantPower Plant


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Advantages of hydraulic power plants

Operating cost is very low

Less Maintenance cost and less manpower required

Pollution free

Quick to start and easy to synchronize

Advantages and Disadvantages of HPPAdvantages and Disadvantages of HPP

an e use or rr ga on an oo con ro

Long plant life.

Disadvantages of Hydraulic Power Plants

Initial cost of total plant is comparatively high

Power generation depends on availability of water Cost of transmission is high since most of the plants are in remote areas

Project duration is long.


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1) Gross Head

Difference Between the Head race level and Tail race level

Static (No water flow) / Total Head H1

2) Net or Effective Head

Head of Hydraulic TurbinesHead of Hydraulic Turbines

Head available at the entrance of the turbine: H = H1 - hf

a) Net Head for a Reaction Turbine

H = {(P1/w) + (V12/2g) + Z1} {Z2 + V2

2/2g)}

b) Net Head for Impulse Turbine

H = {(P1/w) + (V12/2g) + Z1} Z2


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1) Hydraulic Efficiency due to hydraulic losses

Power developed by the runner

Net power supplied at the turbine entrance

SI Unit: kW

Efficiencies of Hydraulic TurbinesEfficiencies of Hydraulic Turbines

Metric Unit : Horse Power/Water Horse Power (W.H.P)

2) Mechanical Efficiency Due to mechanical losses ( bearing friction)

Power available at the turbine shaft (P)Power developed by the runner


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3) Volumetric Efficiency due to amt of water slips directly to the tail race

Amount of water striking the runner

Amount of water supplied to the turbine

ContCont

4) Overall Efficiency

Power available at the turbine shaft (P)

Net power supplied at the turbine entrance


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Classification of TurbinesClassification of Turbines

Turbines are classified according to several considerations as indicated below.Turbines are classified according to several considerations as indicated below.

ii)) Based on working principleBased on working principle

a)a) Impulse turbineImpulse turbine

bb Reaction turbineReaction turbine


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Impulse Turbine:The pressure of liquid does not change while flowing through the rotor of the

machine. In Impulse Turbines pressure change occur only in the nozzles of

the machine.

One such example of impulse turbine is Pelton Wheel.

ContCont

Reaction Turbine:

The pressure of liquid changes while it flows through the rotor of the

machine. The change in fluid velocity and reduction in its pressure causes

a reaction on the turbine blades; this is where from the name Reaction

Turbine may have been derived.

Francis and Kaplan Turbines fall in the category of Reaction Turbines.


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ContCont

ii)ii) Based on working mediaBased on working media

a)a) Hydraulic turbineHydraulic turbine

b)b) Steam turbineSteam turbine

c)c) Gas turbineGas turbine

d)d) Wind TurbineWind Turbine

Head is the elevation difference of reservoir water level and D/S water level.Head is the elevation difference of reservoir water level and D/S water level.

a)a) High head turbineHigh head turbine (Above 250 m)(Above 250 m) Pelton TurbinePelton Turbine

b)b) Medium head turbineMedium head turbine (60(60 250 m)250 m) Francis TurbineFrancis Turbine

c)c) Low head turbineLow head turbine (Below 60 m)(Below 60 m) Kaplan TurbineKaplan Turbine


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iv)iv) Based on specific speedBased on specific speed

Turbines can be classified based on Specific Speed. Specific speed is definedTurbines can be classified based on Specific Speed. Specific speed is defined

as the speed in rpm of a geometrically similar turbine, which is identical inas the speed in rpm of a geometrically similar turbine, which is identical in

shape, dimensions, blade angles and gate openings with the actual turbineshape, dimensions, blade angles and gate openings with the actual turbine

working under unit head and developing unit power. Specific speed is used toworking under unit head and developing unit power. Specific speed is used to

ContCont

..

Specific speedSpecific speed Ns = N P / H5/4

a)a) Low specific speedLow specific speed (8.5(8.5 30)30) -- PeltonPelton TurbineTurbine

b)b) Medium specific speed (50Medium specific speed (50 340)340) -- Francis TurbineFrancis Turbinec)c) High specific speedHigh specific speed (255(255 860)860) -- Kaplan TurbineKaplan Turbine


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v)v) Based on disposition of turbine main shaftBased on disposition of turbine main shaft

a)a) Horizontal shaftHorizontal shaft

b)b) Vertical shaftVertical shaft

vi)vi) Based on flow through the runnerBased on flow through the runner

a)a) Radial flowRadial flow

ContCont

..

2.2. OutwardOutward

b)b) Axial flowAxial flow -- Kaplan TurbineKaplan Turbine

c)c) Mixed flowMixed flow -- Francis TurbineFrancis Turbine

d)d) Tangential flowTangential flow -- PeltonPelton TurbineTurbine


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PeltonPelton Wheel TurbineWheel Turbine

Design of Pelton Wheel Turbine

It has a circular disk with cup shaped blades/buckets,

Water jet emerging from a nozzle is tangential to the circumference of the

wheel.


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Working Principle of Pelton Turbine

Water jets emerging strike the buckets at splitter.

Stream flow along the inner curve of the bucket and leave it in the direction

opposite to that of incoming jet.

The high pressure water can be obtained from any water body situated at

some height or streams of water flowing down the hills.

The change in momentum (direction as well as speed) of water stream

produces an impulse on the blades of the wheel of Pelton Turbine. This

impulse generates the torque and rotation in the shaft of Pelton Turbine.

Horizontal shaft - Not more than 2 jets are used andVertical shaft - Larger no. of jets (upto 6) are used.

Iron/Steel casing to prevent splashing of water and to lead water to the tail

race.


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Chapter 8: Flow in Pipes

Eric G. PatersonDepartment of Mechanical and Nuclear Engineering

The Pennsylvania State University

Spring 2005


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Note to Instructors

These slides were developed1, during the spring semester 2005, as a teaching aid

for the undergraduate Fluid Mechanics course (ME33: Fluid Flow) in the Department of

Mechanical and Nuclear Engineering at Penn State University. This course had two

sections, one taught by myself and one taught by Prof. John Cimbala. While we gave

common homework and exams, we independently developed lecture notes. This was

also the first semester that Fluid Mechanics: Fundamentals and Applicationswas

used at PSU. My section had 93 students and was held in a classroom with a computer,

projector, and blackboard. While slides have been developed for each chapter ofFluid

Chapter 8: Flow in PipesME33 : Fluid Flow 2

,

electronic presentation. In the student evaluations of my course, there were both positive

and negative comments on the use of electronic presentation. Therefore, these slides

should only be integrated into your lectures with careful consideration of your teaching

style and course objectives.

Eric PatersonPenn State, University Park

August 2005

1 These slides were originally prepared using the LaTeX typesetting system (http://www.tug.org/)and the beamer class (http://latex-beamer.sourceforge.net/), but were translated to PowerPoint forwider dissemination by McGraw-Hill.


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Objectives

1. Have a deeper understanding of laminar andturbulent flow in pipes and the analysis of fully

developed flow

2. Calculate the major and minor losses


and determine the pumping power

requirements

3. Understand the different velocity and flow rate

measurement techniques and learn their

advantages and disadvantages


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Introduction

Average velocity in a pipeRecall - because of the no-slip

condition, the velocity at the walls of

a pipe or duct flow is zero

We are often interested only in Vavg,

which we usually call just V(drop the


su scr p or conven ence

Keep in mind that the no-slip

condition causes shear stress and

friction along the pipe walls

Friction force of wall on fluid


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Introduction

For pipes of constantdiameter and

incompressible flow

Vavgstays the same

down the pipe, even if


the velocity profile

changes

Why? Conservation of

Mass

same

Vavg Vavg

samesame


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Introduction

For pipes with variable diameter, mis still thesame due to conservation of mass, but V1 V2

D1


D2

V2

2

1

V1 m m


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Laminar and Turbulent Flows



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Critical Reynolds number(Recr) for flow in a round pipe

Re < 2300 laminar

2300 Re 4000 transitional

Re > 4000 turbulent

Definition of Reynolds number


approximate.

For a given application, Recrdepends upon

Pipe roughness

VibrationsUpstream fluctuations,disturbances (valves, elbows, etc.that may disturb the flow)


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For non-round pipes, define thehydraulic diameterDh= 4Ac/PAc= cross-section area

P= wetted perimeter


Example: open channel

Ac= 0.15 * 0.4 = 0.06m2

P= 0.15 + 0.15 + 0.5 = 0.8m

Dont count free surface, since it does notcontribute to friction along pipe walls!

Dh= 4Ac/P= 4*0.06/0.8 = 0.3m

What does it mean? This channel flow isequivalent to a round pipe of diameter0.3m (approximately).


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The Entrance Region

Consider a round pipe of diameter D. The flowcan be laminar or turbulent. In either case, the

profile develops downstream over several

diameters called the entry length Lh. Lh/Dis a

function of Re.


Lh


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Fully Developed Pipe Flow

Comparison of laminar and turbulent flowThere are some major differences between laminar and

turbulent fully developed pipe flows

Laminar

Can solve exactly (Chapter 9)


Flow is steady

Velocity profile is parabolic

Pipe roughness not important

It turns out that Vavg = 1/2Umax and u(r)= 2Vavg( 1 - r 2/R2)


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TurbulentCannot solve exactly (too complex)

Flow is unsteady (3D swirling eddies), but it is steady in the mean

Mean velocity profile is fuller (shape more like a top-hat profile,with very sharp slope at the wall)

Pipe roughness is very important


Vavg 85% of Umax (depends on Re a bit)

No analytical solution, but there are some good semi-empiricalexpressions that approximate the velocity profile shape. See text

Logarithmic law (Eq. 8-46)

Power law (Eq. 8-49)

ns an aneous

profiles



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Wall-shear stress

Recall, for simple shear flows u=u(y), we had=du/dy

In fully developed pipe flow, it turns out that

=du/dr


w w

w,turb > w,lamw = shear stress at the wall,

acting on the fluid



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Pressure drop

There is a direct connection between the pressure drop in a pipe andthe shear stress at the wall

Consider a horizontal pipe, fully developed, and incompressible flow

w


Lets apply conservation of mass, momentum, and energy to this CV(good review problem!)

1 2L

P1 P2V



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Pressure drop

Conservation of Mass


Conservation of x-momentum

Terms cancel since 1 = 2and V1 = V2



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Pressure drop

Thus, x-momentum reduces to

Energy equation (in head form)

or


cancel (horizontal pipe)

Velocity terms cancel again because V1 = V2, and 1 = 2 (shape not changing)

hL = irreversible head

loss & it is felt as a pressure

drop in the pipe



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Friction Factor

From momentum CV analysis

From energy CV analysis


qua ng e wo g ves

To predict head loss, we need to be able to calculate w. How?

Laminar flow: solve exactly

Turbulent flow: rely on empirical data (experiments)

In either case, we can benefit from dimensional analysis!



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Friction Factor

w = func( V, , D, ) = average roughness of theinside wall of the pipe

-analysis gives




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Friction Factor

Now go back to equation forhL and substitute fforw


Our problem is now reduced to solving for Darcy friction factorf

Recall

Therefore

Laminar flow: f = 64/Re (exact)

Turbulent flow: Use charts or empirical equations (Moody Chart, a famousplot offvs. Re and /D, See Fig. A-12, p. 898 in text)

But for laminar flow, roughness

does not affect the flow unless itis huge


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Friction Factor

Moody chart was developed for circular pipes, but canbe used for non-circular pipes using hydraulic diameter

Colebrook equation is a curve-fit of the data which isconvenient for computations (e.g., using EES)


Both Moody chart and Colebrook equation are accurateto 15% due to roughness size, experimental error,curve fitting of data, etc.

Implicit equation for f which can be solved

using the root-finding algorithm in EES


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Types of Fluid Flow Problems

In design and analysis of piping systems, 3problem types are encountered

1. Determine p (or hL) given L, D, V (or flow rate)Can be solved directly using Moody chart and Colebrookequation


3. Determine D, given L, p, V (or flow rate)

Types 2 and 3 are common engineeringdesign problems, i.e., selection of pipe

diameters to minimize construction andpumping costs

However, iterative approach required sinceboth Vand Dare in the Reynolds number.


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Types of Fluid Flow Problems

Explicit relations have been developed whicheliminate iteration. They are useful for quick,

direct calculation, but introduce an additional 2%

error



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Minor Losses

Piping systems include fittings, valves, bends, elbows,tees, inlets, exits, enlargements, and contractions.

These components interrupt the smooth flow of fluid and

cause additional losses because of flow separation and

mixing


e n ro uce a re a on or e m nor osses assoc a e

with these components

KL is the loss coefficient.

Is different for each component. Is assumed to be independent of Re.

Typically provided by manufacturer or

generic table (e.g., Table 8-4 in text).


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Minor Losses

Total head loss in a system is comprised ofmajor losses (in the pipe sections) and the minor

losses (in the components)


If the piping system has constant diameter

i pipe sections j components


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Pi i N k d P S l i


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Piping Networks and Pump Selection

Two general types ofnetworks

Pipes in seriesVolume flow rate is

constant


Head loss is thesummation of parts

Pipes in parallelVolume flow rate is thesum of the components

Pressure loss across allbranches is the same

Pi i N t k d P S l ti


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For parallel pipes, perform CV analysis betweenpoints A and B


Since p is the same for all branches, head lossin all branches is the same



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Head loss relationship between branches allows the following ratiosto be developed


Real pipe systems result in a system of non-linear equations. Veryeasy to solve with EES!

Note: the analogy with electrical circuits should be obvious

Flow flow rate (VA) : current (I)Pressure gradient (p) : electrical potential (V)

Head loss (hL): resistance (R), however hL is very nonlinear



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When a piping system involves pumps and/orturbines, pump and turbine head must be included inthe energy equation


The useful head of the pump (hpump,u) or the headextracted by the turbine (hturbine,e), are functions ofvolume flow rate, i.e., they are not constants.

Operating point of system is where the system is inbalance, e.g., where pump head is equal to the headlosses.

P d t


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Pump and systems curves

Supply curve for hpump,u:determine experimentally by

manufacturer. When using EES,

it is easy to build in functional

relationship for hpump,u.

System curve determined from

equations

Operating point is the

intersection of supply and

demand curves

If peak efficiency is far fromoperating point, pump is wrong

for that application.

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aarsh mahavidylaya 3rd mech. chp-8,10 sub-f.m.h.m