Chapt4 Edited Rbm

8/11/2019 Chapt4 Edited Rbm

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Chapter 4

Flow in Pipe


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Objectives

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

2. Calculate the major and minor lossesassociated with pipe flow in piping networksand determine the pumping powerrequirements

3. Understand the different velocity and flowrate measurement techniques and learn theiradvantages and disadvantages


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Introduction

Flows completely bounded by solid surfaces are called

INTERNAL FLOWSwhich include flows through pipes(Round cross section), ducts(NOT Round cross section),nozzles, diffusers, sudden contractions and expansions,valves, and fittings.

The basic principles involved are independent of the cross-sectional shape, although the details of the flow may bedependent on it.

The flow regime (laminar or turbulent) of internal flows isprimarily a function of the Reynolds number.

Laminar flow: Can be solved analytically.

Turbulent flow: Rely Heavily on semi-empirical

theories and experimental data.


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General Characteristics of

Pipe Flow


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Pipe System

A pipe system include the pipes themselves(perhaps of more than one diameter), the various

fittings, the flowrate control devices valves) , and

the pumps or turbines.


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For pipes of constant diameter and

incompressible flow

Vavgstays the same down the pipe, even if the

velocity profile changes

Why? Conservation of Mass

Vavg Vavg

samesame same


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For pipes with variable diameter, mis still the

same due to conservation of mass, but V1 V2

D2

V2

2

1

V1

D1

m m


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Pipe Flow vs. Open Channel Flow

Pipe flow: Flows completely filling the pipe. (a)The pressure gradient along the pipe is main driving

force.

Open channel flow: Flows without completely filling

the pipe. (b)

The gravity alone is the driving force.


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Laminar or Turbulent Flow

The flow of a fluid in a pipe may be Laminar ?

Turbulent ?

Osborne Reynolds, a British scientist and

mathematician, was the first to distinguish the

difference between these classification of flow by

using a simple apparatusas shown.


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Laminar or Turbulent Flow

For small enough flowrate the dye streak will

remain as a well-defined line as it flows along, with onlyslight blurring due to molecular diffusion of the dye into

the surrounding water.

For a somewhat larger intermediate flowratethe dye fluctuates in time and space, and intermittent

bursts of irregular behavior appear along the streak.

For large enough flowrate the dye streak almostimmediately become blurred and spreads across the

entire pipe in a random fashion.


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Indication of Laminar or Turbulent Flow

The term flowrateshould be replaced by Reynolds

number, ,where V is the average velocity inthe pipe.

It is not only the fluid velocitythat determines thecharacter of the flowits density, viscosity, and the pipesize are of equal importance.

For general engineering purpose, the flow in a round pipe

LaminarTransitional

Turbulent

/VDRe

2100Re

4000eR


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Laminar and Turbulent Flows

Critical Reynolds number (Recr) for

flow in a round pipeRe < 2300 laminar2300 Re 4000 transitional

Re > 4000 turbulent

Note that these values are approximate.

For a given application, Recr depends upon

Pipe roughness

Vibrations

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


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Comparison of laminar and turbulent flow

There are some major differences between laminar andturbulent fully developed pipe flows

Laminar

Can solve exactly

Flow is steady Velocity profile is parabolic

Pipe roughness not important


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Turbulent

Cannot 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

Instantaneous

profiles


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Turbulent Flow

Turbulent pipe flow is actually more likely to occur

than laminar flow in practical situations.

Turbulent flow is a very complex process.

Numerous persons have devoted considerable effort

in an attempting to understand the variety of bafflingaspects of turbulence. Although a considerable

amount if knowledge about the topics has been

developed, the field of turbulent flow still remains the

least understood area of fluid mechanics.

Much remains to be learned about the nature of turbulent

flow.


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Due to the macroscopic scale of the randomness inturbulent flow, mixing processes and heat transferprocesses are considerably enhanced in turbulentflow compared to laminar flow.

Such finite-sized random mixing is very effective intransporting energy and mass throughout the flow

field.

[Laminar flow can be thought of as very small butfinite-sized fluid particles flowing smoothly in layer,

one over another. The only randomness and mixingtake place on the molecular scale and result inrelatively small heat, mass, and momentum transferrates.]


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In some situations, turbulent flow characteristics are

advantages. In other situations, laminar flow is

desirable.

Turbulence: mixing of fluids.

Laminar: pressure drop in pipe, aerodynamic drag

on airplane.


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Time Dependence of

Fluid Velocity at a Point


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Entrance Region and Fully Developed Flow

Any fluidflowing in a pipehad to enter thepipe at some location.

The region of flow near where the fluid enters

the pipe is termed the entrance region.


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The fluid typically enters the pipe with a nearly

uniform velocity profile at section (1).

The region of flow near where the fluid enters

the pipe is termed the entrance region.

As the fluid moves through the pipe, viscous

effects cause it to stick to the pipe wall(the no

slip boundary condition).


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A boundary layerin which viscous effects are

important is produced along the pipe wall

such that the initial velocity profile changes

with distance along the pipe,x , until the fluid

reaches the end of the entrance length,

section (2), beyond which the velocityprofile does not vary with x.

The boundary layer has grown in thickness tocompletely fill the pipe.


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eR06.0D

Viscous effects are of considerable importance

within the boundary layer. Outside the

boundary layer, the viscous effects are negligible.

The shape of the velocity profile in the pipe

depends on whether the flow is laminar or

turbulent, as does the length of the entranceregion,

.

6/1

eR4.4D

For laminar flow For turbulent flow

Dimensionless entrance length


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Once the fluid reaches the end of the entrance

region, section (2), the flow is simpler todescribe because the velocity is a function of only

the distance from the pipe centerline, r, and

independent of x.

The flow between (2) and (3) is termedfully

developed.


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The Entrance Region

Consider a round pipe of diameter D. The flow canbe laminar or turbulent. In either case, the profile

develops downstream over several diameters called

theentry lengthLh. Lh/Dis a function of Re.

Lh


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Pressure Distribution along Pipe

In the entrance region of a pipe, the fluid

accelerates or decelerates as it flows. There is a

balance between pressure, viscous, and inertia

(acceleration) force. The magnitude of the pressuregradient is constant.

The magnitude of the pressure gradient is larger than that

in the fully developed region.

0p

x

p


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Shear Stress for Laminar Flow

Laminar flow is modeled as fluid particles that flow

smoothly along in layers, gliding past the slightly sloweror faster ones on either side.

The fluid actually consists of numerous molecules

darting about in an almost random fashion. The motionis not entirely randoma slight bias in one direction.


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The momentum flux in the x direction across

plane A-A give rise to a drag of the lower fluid

on the upper fluid and an equal but oppositeeffect of the upper fluid on the lower fluid.

dy

du

yx

Shear stress is present only if there is a

gradient in u=u(y)


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Shear Stress for Turbulent Flow

The turbulent flow is thought as a series of

random, three-dimensional eddy type motions.These eddies range in size from very small

diameter to fairly large diameter.

This eddy structure greatly promotes mixingwithin the fluid.


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The flow is represented by (time-mean velocity )

plus u and v (time randomly fluctuating velocity

components in the x and y direction).

The shear stress on the plane A-A

u

turbulentarminla'v'udyud

The shear stress is not merely proportional to the

gradient of the time-averaged velocity, .

)y(u


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Structure of Flow in a Pipe

Near the wall(the viscous sublayer), thelaminar shear

stress lamis dominant.

Away fromthe wall (in the outer layer) ,the turbulent

shear stress turbis dominant.

The transition between these two regions occurs in the

overlap layer.


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The relative magnitude of lamcompared toturbis a complex function dependent on the

specific flow involved.

Typically the value of turbis 100 to 1000times greater than lam inthe outer region.


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Analysis of Pipe Flow


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Energy Considerations

Total head loss , hL, is regarded as the sumof major losses, hL major, due to frictional

effects in fully developed flow in constant

area tubes, and minor losses, hL minor

,

resulting from entrance, fitting, area

changes, and so on.

orminmajor LLL hhh


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Major Losses: Friction Factor

L2

2

2221

2

111 hgz2

Vpgz

2

Vp

L1221 h)zz(g

pp

For fully developed flow through a constant area

pipe

For horizontal pipe, z2=z1

L21 h

ppp

The energy equation for steady and

incompressible flow with zero shaft work

C d h d f R l h f L d l


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Consider a pipe with radius of R, length of L and angleof qis flowing a fluid as shown in figure below

Flow is steady and incompressible fluid. Control volume inbetween 1 to 2.


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Continuity equation between point 1 and point 2 :

Q1= Q2or

From Bernoullis equation

When the flow is fully developed

g

Pz

g

Pz

g

P

zg

P

zhf

2

2

1

1

2

22

1

11

A

Qv

A

Qv

fhzgv

gPz

gv

gP 2

2221

2

11

21

21


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Momentum concept in control volumeResultant Force = Rate of Momentum

Change

Divide by gpR2, becomes

Therefore

0)()2(sin)( 2122

vvmLRLRgRP wq

zLbutRg

LL

g

P w

q

q

sin0

2sin

Rg

Lz

g

Ph wf

2

Due to pressure

Due to weightDue to shear

But

dvF )(


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and from analysis dimensional

where f (dimensionless) is call Darcy friction factor

By combination of the above equation

or

The above equation is called Darcy Weisbach equation. It usedfor both flow regime i.e. laminar and turbulence

roughnesspipeis

dvFw

),,,,(

dFf

vw

Re,82

Rg

vLfhf

8

2 2

dg

vLfhf

2

2


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1. Major Losses: Laminar Flow

There are numerous ways to derive importantresults pertaining to laminar flow:

From F=ma applied directly to a fluid element.

From the Navier-Stokes equations of motion

From dimensional analysis methods : Selected


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Based on Dimensional Analysis

Assume that the pressure drop in the horizontalpie, p, is a function of the average velocity of the

fluid in the pipe, V, the length of the pipe, , the pipe

diameter, D, and the viscosity of the fluid, .

),D,,V(Fp Dimensional analysis

DVpD

an unknown function of the

length to diameter ratio of thepipe.


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D

C

V

pD

where C is a constant.

2DVCp

p

4pD)C4/(AVQ

The value of C must be determined by theory or

experiment. For a round pipe, C=32. For duct ofother cross-sectional shapes, the value of C is

different.

2D

V32p

For a round pipe


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f is termed the friction factor, or

sometimes the Darcy friction factor.

For a round pipe

DRe

64

DVD64

V

D/V32

V

p2

21

2

2

21

2

V

Dfp

2

2

V

Dp

f2

2wV

8

Re

64f

For laminar flow

dvvdv

vf w

64)/8(88 22


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Then,

Losses due to friction becomes

or based on HagenPoiseuille equation

gd

vL

gd

vLhf

Re

32

2Re

64 22

4128

dgQLhf


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2. Major Losses: Turbulent Flow

2

V

Dfp

2

Applying dimensional analysis, the result were a correlation

of the form

Experiments show that the nondimensional head loss isdirectly proportional to /D. Hence we can write

D,

D,VD

Vp

2

21

DDV

p

Re,

2

21

DRe,f

g2

V

Dfh

2

Lmajor

Darcy-Weisbach

equation


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Value of friction factor f is dependent to Reynolds's number Reandpipe roughness or relative roughness /d

For laminar flow, f=64/Re, which is independent of the relativeroughness.

For very large Reynolds numbers, f=(/D), which is independentof the Reynolds numbers.

For flows with very large value of Re, commonly termedcompletely turbulent flow (or wholly turbulent flow or fulldeveloped turbulent), the laminar sublayer is so thin (its thicknessdecrease with increasing Re) that the surface roughness completely

dominates the character of the flow near the wall.

For flows with moderate value of Re, the friction factorf=(Re,/D).

Blausius (1913) was introduced an expression as below

3160


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Prandtl (1935) made a further improvement of Blausiussequation for 4000 < Re < 108iaitu

Nikuradse was introduced the effect of pipe roughness ontofriction factor

Colebrooke (1939) was combined the Prantl and Nikuradseand become

Moody (1944) was simplified the above equation into agraphic but the accuracy is about 15%

pipesmoothandforf 54/1

10Re4000Re

316.0

pipesmoothff

8.0)(Relog21

roughnessrelativedwhered

f /

7.3/log21

f

d

f Re

51.2

7.3

/log2

1

Friction Factor by L F Moody


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Depending on the

specific circumstances

involved.

Friction Factor by L. F. Moody

Roughness for Pipes


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Roughness for Pipes


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Example


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p

Oil with specific gravity of 0.85 and kinematics

viscosity is 6 x 10-4m2/s is flowing in a pipe of 15 cm

diameter at flowrate of 0.020 m3/s. Determine thehead loss of 100 m pipe distance.

Find type of flow regime

Re < 2100: Laminar flow, then

Head Loss, hf

=

E l


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Example

Determine head loss due to friction in a horizontal

pipe of 40 mm diameter and 750 m long when a flowis 67.6 x 10-6m3/s. Given a pipe roughness and

kinematics viscosity is 0.0008 m and 1.14 x 10-6m2/s,

respectively

Re < 2100: Laminar flow, then

Example


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Example

Water at 20oC is flowing at 0.05 m3/s through a

asphalted cast iron pipe of 20 cm diameter. What is a

head loss for 1 kilometer pipe length.

From table, kinematics viscosity at 20oC is

Turbulent flow

Roughness of asphalted cast iron,

From Moody Chart

Mi L


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Minor Losses

Most pipe systems consist of considerably more

than straight pipes. These additionalcomponents (valves, bends, tees, and the like)

add to the overall head loss of the system.

Such losses are termed MINOR LOSS.

The flow pattern through a valve


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The theoretical analysis to predict the

details of flow pattern (through these

additional components) is not, as yet,possible.

The head loss information for essentially allcomponents is given in dimensionless form

and based on experimental data.

The most common method used to

determine these head losses or pressure

drops is to specify the loss coefficient, KL


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2

L2

2

L

L V2

1Kp

V

2

1

p

g2/V

hK ormin

Re),geometry(KL

fDK

g2

V

Df

g2

VKh

Leq

2eq

2

LL ormin

Minor losses are

sometimes given in terms

of an equivalent lengtheq

The actual value of KLis strongly dependent on the

geometry of the component considered. It may alsodependent on the fluid properties. That is


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For many practical applications the Reynoldsnumber is large enough so that the flow throughthe component is dominated by inertial effects,

with viscous effects being of secondaryimportance.

In a flow that is dominated by inertia effectsrather than viscous effects, it is usually found thatpressure drops and head losses correlate directlywith the dynamic pressure.

This is the reason why the friction factor for verylarge Reynolds number, fully developed pipe flowis independent of the Reynolds number.


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This is true for flow through pipe components.

Thus, in most cases of practical interest the loss coefficientsfor components are a function of geometry only,

Therefore minor losses can be calculated in a head loss as

gVKh Lm2

2

)geometry(KL

Mi L C ffi i t Entrance


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Minor Losses Coefficient Entranceflow

Entrance flow conditionand loss coefficient

(a) Reentrant,KL

= 0.8

(b) sharp-edged,KL

= 0.5

(c) slightly rounded,KL

= 0.2

(d) well-rounded,KL

= 0.04

KL

= function of

rounding of the

inlet edge.


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A vena contracta region may result because the fluid

cannot turn a sharp right-angle corner. The flow is saidto separate from the sharp corner.

The maximum velocity at section (2) is greater than

that in the pipe section (3), and the pressure there islower.

If this high speed fluid could slow down efficiently, the

kinetic energy could be converted into pressure.

Such is not the case Although the fluid may be accelerated


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Such is not the case. Although the fluid may be accelerated

very efficiently, it is very difficult to slow down (decelerate)

the fluid efficiently.

(2)(3) The extra kinetic energy of the fluid is partially lostbecause of viscous dissipation, so that the pressure does

not return to the ideal value.

Flow pattern

and pressure

distribution for

a sharp-edgedentrance


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Minor Losses Coefficient Exit flow

Exit flow condition and loss coefficient

(a) Reentrant,KL= 1.0

(b) sharp-edged,KL= 1.0

(c) slightly rounded,KL= 1.0

(d) well-rounded,KL= 1.0

Mi L C ffi i t varied


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Minor Losses Coefficient varieddiameter

Loss coefficient forsudden contraction,

expansion,typical

conical diffuser.

2

2

1L

A

A1K

Minor Losses Coefficient - Bend


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Minor Losses Coefficient Bend

Carefully designed guide vanes help direct the flow with less

unwanted swirl and disturbances.

Character of the flow in bend and the associated loss

coefficient.


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Internal Structure of Valves

(a) globe valve

(b) gate valve

(c) swing check valve(d) stop check valve

Loss Coefficients for Pipe Components


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Loss Coefficients for Pipe Components

Losses due to Sudden Enlargement


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Losses due to Sudden Enlargement

A sudden expansion is one of the component for which the

loss coefficient can be obtained by means of simple analysis

Let consider a control volume of ABCDEF as shown in figure

below. Pressure P1 and P2 at point 1 and Point 2 respectively.

Based on continuity equation

1

212211

A

AVVAVA

(a)

Momentum equation in the control volume


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q

Resultant Force in flow direction = Rate of momentum change in flow

direction

where P is pressure acts on annulus that represent by AB and CD

which is at area of A2A1.

We assume the flow is uniform and pressure is constant across the

left-hand side of CV (PAB=PBC=PCD=P1=P) then

Based on Bernoullis equation from 1 to 2.

)()(' 12221211 VVQAPAAPAP

)(

)()(

12221

1222122221

VVVPP

VVAVVVQAPAP

(b)

)(2

1

2

1212

222

1

2

11 horizontalZZhzg

v

g

Pz

g

v

g

Pm

ThenVVPP 22


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Substitute (b) into the above equation becomes

But, from (a)

Then

or

g

VV

g

PP

g

V

g

V

g

P

g

Phm

2

22

22

2121

2121

g

VV

g

VV

g

VVV

g

VV

g

VVVhm

2

)(

22

)(

2

21

2

2

2

121

2

2

2

2

2

112

2

2122

2)/(

gVAAVhm

2

2

1

2

1

2

1

2

2

2 12

12

A

A

g

V

A

A

g

Vhm

)/( 1221 AAVV


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Therefore

or

Losses due to Sudden ContractionThe mechanism by which energy is lost due to sudden contraction is

quite complex. Momentum concept cannot be used since the distribution

of pressure along ABCD is unknown.

2

1

2 1

A

Ak

2

2

11

A

Ak

For this case, however, there is a vena contracta section that


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, ,possible to use sudden enlargement method. Then,

where Cc contraction coefficient between Acand A2.

Generally

where K is losses coefficient and it depend to A2/A1 ratio. Tablebelow is the guideline of some K values.

22

2

2

2

2

2

12

12

c

c

m

C

I

g

V

A

A

g

Vh

gVkhm2

2

2


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A2/A1 0.1 0.3 0.5 0.7 1.0

Cc 0.61 0.632 0.673 0.73 1.0

k 0.41 0.34 0.24 0.14 0

Example


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Water pipe has changes it diameter from 140 mm to 250 mm. If a flow of

water from smaller diameter to bigger diameter experienced of a head loss

is 0.6 m higher than the opposite direction, calculate a velocity in the pipe.V1 V2 V1 V2

V1= flow in small pipe

V2= flow in small pipe

Interpolation methodFrom table

Thereforeand

Flow from big to small pipe : Sudden Contraction

Flow from small to big pipe : Sudden Enlargement


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Example

Water of 1.94 slugs/ft3density and 0.000011 ka2/s kinematics viscosity was

pumped between two tanks at 0.2 ka3/s flowrate through a pipe of 2 inchdiameter and 400 ft length as shown in figure below. Given a relative

roughness of pipe is 0.01. Determine a horsepower required by the pump.

Sharp

entranceGlobe

Valve: Open

90oElbow

Bend of 12 inch

diemater

Gate valve:

Half open

Sharp

Exist

Pump

Energy Equation


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From Moody chart

For minor losses

Sharp entrance

Globe valve: Fully openPipe bend 12 inch diameter

90o Elbow

Gate valve: Half open

Sharp exist

Pump Head

Power of Pump

But

Equivalent Length


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Minor losses are also expressed in terms of equivalent length Le,defined as

where fTis friction factor and D is the diameter of the pipe thatcontains the component.

fTis the friction factor in the pipe to which the valve of fitting isconnected, taken to be in the zone of fully developed turbulence

Leis the length of straight pipe of the same nominal diameter as thevalve that would have the same resistance as the valve

Lecan be one unit of component or total of components in pipesystem

gVk

gdVLfh eTf

22

22

T

ef

dkL

Both approaches are used in practice but the use of loss coefficients is


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Both approaches are used in practice, but the use of loss coefficients is

more common

Therefore, the total head loss in a piping system is determines from (if

f =fT)

Example

A pipe of 0.2 m diameter and 50 m length is consists of 2 unit of 90o

elbows, gate valve, sharp entrance and sharp exist. Calculate the pipeequivalent length and a total head loss whenever a flowrate is 0.2 m3/s

and valve is fully open. Assume a friction factor is 0.005.

dg

VLLfh ef

2

)( 2

90oElbow

Gate valve of fully open

Sharp entrance

Sharp exist


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p

Total losses

2 units of 90o

ElbowsGate valve

Sharp Entrance

Sharp exist

Minor losses

Loss due to friction

Total head losses

Alternative Method

Application of Pipe Flow


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Application of Pipe Flow

g

V

Dfh

majorL 2

2

The energy equation, relating the conditions at

any two points 1 and 2 for a single-path pipesystem

by judicious choice of points 1 and 2 we cananalyze not only the entire pipe system, but also

just a certain section of it that we may beinterested in.

g2

VKh

2

LL ormin

orminmajor LLL22

2221

2

111 hhhzg2

V

g

pz

g2

V

g

p

Major loss Minor loss


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Single pipe whose length may be interrupted byvarious components.

Multiple pipes in different configurationParallelSeries

Network

Pipe flow problems can be categorized by whatparameters are given and what is to be calculated.


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Multiple Path Systems


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Multiple-Path Systems

Series and Parallel Pipe System

321 LLL321 hhhQQQQ

321BA LLLL321 hhhhQQQ



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p y

Multiple Pipe Loop System

3L2L

3L1LB

2

BBA

2

AA

2L1LB

2

BBA

2

AA

321

hh

31hhz

g2

Vpz

g2

Vp

21hhzg2

Vpz

g2

VpQQQ



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Three-Reservoir System

CBhhzg2

Vpz

g2

Vp

BAhhzg2

Vpz

g2

Vp

QQQ

3L2LC

2

CCB

2

BB

2L1LB

2

BBA

2

AA

321

If valve (1) was closed, reservoir Breservoir C

If valve (2) was closed, reservoir Areservoir CIf valve (3) was closed, reservoir Areservoir B

With all valves open

Serial Pipe


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The diameter may be the same or different

Total of head loss is a summation of all losses encounter throughout the

system.

Consider on the figure below.

By applying of Bernoullis equation and B is taken as a datum, then

hL = Entrance loss + Friction loss in pipe 1 +

Valve loss + Sudden enlargement loss +

Friction loss at pipe 2 + Exist loss

Pipe 1

Pipe 2Valve

VVLf)VV(Vk

VLfV50

2

2

2

22

2

21

2

1

2

11

2

1


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Other equation required to solve the related problem is continuityequation

Q1= Q2or A1V1= A2V2

Example

Satu paip tersambung di antara 2 takungan yang mempunyaiperbezaan aras 6 m. Panjang paip 720 m dan menaik kepadaketinggian 3 m di atas takungan atas pada jarak 240 m daripadamasukan sebelum menurun ke takungan bawah. Jika paipberdiameter 1.2 m dan pekali geseran 0.01, apakah kadaralir dantekanan pada titik tertinggi paip tersebut. Abaikan kehilanganmasukan dan keluaran.

g2gd2g2

)(

g2k

gd2g25.0 2

2

22211

1

111


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Parallel Pipe


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Purpose:

1. Stable supplywithout interruption

2. To increase flow without changing the existing pipe system.

Consider a system as shown in figure below

Based on continuity equation

Q1 = QA+ QB+ QC= Q2Based on Bernoullis equation between point 1 and 2

L2

2

22

1

2

11 hzg

v

2

1

g

Pz

g

v

2

1

g

P

where hLis losses between point 1 and 2


91/130

Therefore, the losses generated in this two point is constant andindependent to the direction of flow.

Then

Contoh

Two pipes with a sharp edge of diameter d1= 50 mm and d2= 100mm and both have a 100 m length is connected in parallel to twotanks with a difference of elevation h of h = 10 m as shown in figurebelow. If a friction factor for both pipes f is 0.008 , determine

a) Flowrate in each pipe

b) Diameter D of single pipe of 100 m length to provide the sametotal of flow

LCLBLA hhh


92/130


93/130

Example


94/130

Two pipes connect two reservoirs A and B (figure below), which have a length

difference of 20 m. Pipe 1 has 50 mm diameter and 100 m length. Pipe 2 has 100

mm diameter and 100 m length. Both have entry loss, KL= 0.5 and exist loss, KL=1.0, and friction factor f = 0.032, calculate

a) rate of flow for each pipe

b) the diameter D of a pipe 100 m length that could replace the two pipes and

provide the same flow

Solution


95/130


96/130

Branch Pipe


97/130

Figure below shows a branch pipe system is connecting three waterreservoirs

Based on figure

Water from reservoir A flow to reservoir C through pipe 1 and pipe 3.

Head losses is generated at exist A, junction C and friction in pipe 1 andpipe 3.

Based on Bernoullis equation between reservoir A and reservoir C

2

3V2

3V

3L

3f2

3V2

1V

1L

1f2

1V2

Cv1C

P2

Av1A

P


98/130

PA= PC= Patm= 0VA= VC= 0,

Then

Similar for reservoir B to reservoir C

g2

3

3gd2

333

g2

3k

1gd2

111

g2

15.0

Cz

g

C

2

1

g

C

Az

g

Av

2

1

g

A

g2V

d1Lfk

g2V

dLf5.0ZZ

2

3

3

33

2

1

1

11

CA

g2

V

d

1Lfk

g2

V

d

Lf5.0ZZ

2

3

3

33

2

2

2

22CB

Example


99/130

Water is flowing from reservoir A through diameter pipe d1= 120 mm

and distance L1 = 120 m to a junction D. From D the pipe is branched out

into two which diameter d2

= 75 mm and distance L2

= 60 m to reservoir

B that 16 m below reservoir A and diameter d3= 60 mm and distance L3

= 40 m to reservoir C that 24 m below reservoir A. If a friction factor of

all pipes is 0.01 and minor losses is negligible, calculate flowrate of each

pipe.


100/130


101/130

Example


102/130

A closed tank was filled up by a fluid and connected with a polyethylene pipe of 5

cm diameter as shown in figure below. If,

a) Fluid is water and required to distribute at 60 m3/hr of flowrate, what is a

pressure P1required. Assume water density is 998 kg/m3and neglect the

minor losses

b) Fluid with a specific gravity of 0.68, pressure P1is 700 kPa and flowrate is 27

m3

/jam, what is a viscosity of the fluid. Neglect a minor losses.

Bendalir

60m

P1

Bendalir

10 m

Q

(terbuka keatmosfera)

80m

Solution

60a)


103/130

Pax

xP

gd

flV

g

V

g

P

hZg

V

g

PZ

g

V

g

P

f

PipeSmoothChartMoodyFrom

Vd

smA

QV

f

6

2

1

22

21

2

2

221

2

11

2

1038.2

05.0

1700136.01

81.92

49.8108081.9998

280

20010

22

0136.0

:

424000001.0

05.049.8998Re

/49.8

05.0

4

3600

mkg27

/67999868.0 3b)


104/130

smkg

xx

Vd

PipeSmoothChartMoodyFrom

f

fxg

P

gd

flV

g

VZ

g

P

hZg

V

g

PZ

g

V

g

P

smA

QV

f

./00031.0

05.082.3679416000

Re

416000Re

:

0136.0

05.01701

81.9282.370

22

22

/82.3

05.04

3600

27

2

1

22

21

2

2

221

2

11

2

Example

Air dialirkan daripada reserbor A menerusi satu talian paip sepanjang 4 km dan


105/130

p p p p j g

berdiameter 50 cm kepada reserbor B yang mempunyai perbezaan aras sebanyak

12.5 meter. Jika air perlu dialirkan kepada reserbor C yang mana mempunyai

perbezaan aras sebanyak 15 meter di bawah reserbor A menerusi talian paip

sepanjang 1.5 km yang menyambungkan pada paip utama pada jarak 1 km daripadakeluarannya seperti ditunjukkan dalam rajah di bawah. Jika diabaikan kehilangan

kecil kecuali kehilangan geseran yang ditentukan melalui dan dipertimbangkan nilai

pekali geseran sebagai 0.006, tentukan

a) Kadaralir ke reserbor B jika tidak ada kadaralir ke reserbor C

b) Kadaralir ke reserbor B jika kadaralir ke reserbor C adalah 150 liter per saatc) Diameter paip baru supaya aliran ke dalam ke dua-dua reserbor iaitu B dan C

adalah sama.

C

B

A

12.5 m

15 m

15005.12

30005.0

100040004

DCAB

DBAB

ADAB

mLmZ

mLmD

mLmkmL


106/130

225

5

2

5

2

22

552

5

2

5

2

22

5

2

15.03

5.12

335.1200000

22

15.0

)

/2209.0

4000006.0

5.125.035.123

5.123

35.1200000

22

)

3

006.015

ADDBADAD

DBADDBAD

DB

DBDBDB

AD

ADADAD

fDBfADB

BB

A

AA

ADDB

fABBBB

AAA

DBAD

f

AC

QLQLd

f

ddff

Diberi

d

Qlf

d

Qlf

hhZg

V

g

P

Zg

V

g

P

QQ

BkeAAliranb

smQ

fl

dQ

d

flQ

d

flQ

hZg

V

g

PZ

g

V

g

P

QQQ

BkeAAlirana

d

flQh

fmZ

QQ 15030001000006.0512 22


107/130

sm

QQ

sm

x

xxQ

a

acbbQ

QuadratikPersamaanGuna

QQ

QQQ

QQQ

QQ

ADDB

AD

ADAD

ADADAD

ADADAD

ADAD

/174.0

15.0324.015.0

/324.0

40002

8125.12740004900900

2

4

08125.1279004000

5.67900300010003125.195

0225.03.030001000

5.03

006.05.12

15.0300010005.03

5.12

3

3

2

2

2

22

22

5

2

5

T t i

QQQ

c

DCDBAD

)


108/130

cmmd

d

xx

x

xx

d

Qlf

d

QlfZZ

CkeAAliran

smQQ

Maka

smQ

Q

lld

fQ

d

Qlf

d

QlfZ

g

V

g

PZ

g

V

g

P

BkeAAliran

QQQ

Tetapi

DC

DC

DC

DCDC

AD

ADAD

CA

DCDB

AD

AD

DBADAD

DB

DBDB

AD

ADAD

BBB

AAA

ADDCDB

34.404034.0

3

167.01500006.0

5.03

334.01000006.015

33

/167.02

334.0

/334.0

4

30001000

5.03

006.05.12

2

1

35.12

3322

2

5

2

5

2

5

2

5

2

3

3

5

2

2

5

5

2

5

222

Example


109/130

Rajah menunjukkan suatu sistem paip berdiameter 50 mm

dilengkapkan dengan sebuah pam empar bagi mengepam air dari

sebuah tasik ke sebuah tangki air yang terletak di atas Bukit Kijal.Pam tersebut dapat meningkatkan tenaga per unit berat air yang

melaluinya sebanyak 80 Nm/N. Ketinggian paras air bebas di

dalam tangki dari pam ialah 70m. Dalam sistem paip tersebut,

paip A sehingga paip H berada pada permukaan mendatar yang

sama tetapi paip C sehingga paip D terpaksa membentukseparuh bulatan bagi mengelak sebuah kolam ikan bulat yang

berdiameter 50m. Jika faktor geseran ialah 0.08, kira kadar alir

jisim air bagi sistem paip tersebut. Seterusnya tentukan jenis

aliran di dalam paip penghantaran, berdasarkan nombor Reynold.

Diberi kelikatan air ialah 1.14 x 10-6m2/s. Gunakan persamaan

hf=flv2/(2gd) jika perlu.


110/130

22 VPVP

gkitandalamairPermukaan:2Titik

tasikPermukaan:1Titik


111/130

222

f

mf

mf

p

2

pmf2

pmf2

pmf222

111

V1.5805.081.92

V5.71208.0

gd2

flVh

m5.712

54053052530

50501005.78100150404l

m5.7825r

separuhPanjang

08.0f

Diberi

6hh

80hh740

Maka

m80E

m74470Z

EhhZ0

EhhZ00000

EhhZg2

V

g

PZ

g2

V

g

P

bulatan

g

KVh

hKira

m

m

2

2


112/130

GeloraadalahsistemdalamAliran

x

Vd

s

kgAVm

Diketahuis

mV

VV

demikianyangOleh

V

VVVVVh

Jadi

VV

hupInjap

VV

hglobInjap

VV

hKeluaran

VV

hMasukan

VV

hSesiku

g

m

m

m

m

m

m

o

2100Re

140351014.1

05.032.0Re

628.032.005.04

1000

32.0

96.59

6

686.11.580

86.1

01.027.105.003.050.0

01.081.92

19.0:int

27.181.92

)5.12(2:

05.081.92

0.1:

03.081.92

5.0:

50.081.92

75.013:90

6

2

22

2

22222

22

22

22

22

22

Example


113/130

Rajah di bawah menunjukkan sebuah sistem paip yang telah dibina bagi

menghubungkan tangki 1 yang terletak di kaki Bukit Awana dengan

tangki 2 di Kawasan Tinggi Awana. Paip bersaiz 50 mm itu diperbuatdaripada besi tuang (Cast Iron) dan ia digunakan untuk menghantar air

(graviti tentu 9810 N/m3) dan kelikatan kinematik 1 x 10-6m2/s ke

tangki 2. Titik A hingga titik D berada pada permukaan rata yang sama

dengan paip dari titik B dan titik C terpaksa dipasang dengan bentuk

separuh bulatan untuk mengelak kolam ikan bulat. Sesiku B hinggasesiku G adalah daripada jenis 90o. Sebuah injap globe dalam keadaan

terbuka penuh dan sebuah injap pintu juga dalam keadaan terbuka

penuh masing-masing telah dipasang pada kedudukan sebelum pam dan

selepas pam. Ar bebas di dalam tangki 2 ialah 20 m berbanding paras

air bebas di dalam tangki 1. Sebuah pam telah digunakan untuk

mengepam air dalam sistem sehingga kadaalir mencapai 0.05 m3/s. Jika

kecekapan pam adalah 80%, kirakan kuasa kuda yang diperlukan oleh

motor elektrik untuk memacu pam tersebut.


114/130

EhhZg

V

g

PZ

g

V

g

Ppmf

222

222

1

211


115/130

ml

mBCPanjang

md

d

makamIkanKolamKeluasan

f

Jadid

makammIronCastUntuk

DarcyPersamaangunamakageloraadalahaliran

x

xVd

sm

xAQV

hhZE

EhhZ

mfp

pmf

751012612612125

122

65.7

65.7

446

,46

031.0

0052.050

26.0

26.0

,2100Re

1275000101

10505.25Re

5.25

10504

05.0

00000

2

2

6

3

23

2

2

m

flVh f 1541

5.2575031.022


116/130

hpHP

kecekapanBila

hp

gQE

HP

Jadi

mE

demikianyangOlehm

g

V

g

Vg

KVh

mxgd

h

p

p

m

f

17248.0

1379

%80

1379

746

209805.081.91000

746

2098537154120

537

2

19.16

119.075.06105.02

2

1541105081.922

2

2

2

3

Example


117/130

Rajah menunjukkan air dipam dari tangki sedutan A ke tangki tadahan

G melalui suatu siri talian paip yang diperbuat daripada keluli

kemersial. Talian paip ini dilengkapkan dengan sebuah injap pintuterbuka sepenuhnya. Talian paip B ke D berdiameter 140 mm dengan

talian paip seterusnya berdiameter 250 mm. Masukan dan keluaran

adalah tajam. Meter aliran ultrasonik menunjukkan halaju aliran di

dalam paip berdiameter 140 mm ialah 8 m/s. Paras air bebas di dalam

tangki tadahan pula didapati 50 m di atas paras air bebas tangkisedutan. Jika kelikatan air 1.06 x 10-6m2/s, tentukan tenaga per unit

berat yang perlu dibekalkan oleh pam supaya air dapat dialirkan

mengikut perancangan asal. Gunakan hf=flv2/(2gd).


118/130

Example

Rajah di bawah menunjukkan sistem talian paip plastik selari


119/130

Rajah di bawah menunjukkan sistem talian paip plastik selari

mengandungi dua paip yang serba sama yang dipasangkan secara

mendatar. Air memasuki simpang A pada kadaralir 0.4 m3/s dan

meninggalkannya dengan berpecah kepada dua bahagian untukmembekalkan air kepada galas (bearing) yang akan memutarkan aci

turbin. Jika galas (bearing) untuk kedua-dua paip adalah sama dan nilai

pekali kehilangan K adalah 10, diameter paip 250 mm dan kelikatan

kinematik air 1.12 x 10-6m2/s, tentukan:

a) Halaju dalam setiap paip

b) Kejatuhan tekanan yang merentasi cabang (PA- PB)

AVQ

s

mQQQ

makasamaadalahpaipCabanga

T 2.02

4.0

2

)3

21


120/130

m

xxxLe

Jadi

f

makalicinpaippaipKekasaran

x

x

xVD

f

KD

f

KD

f

kD

LeLeLeLe

samaadalahsimpangmerentasiPPtekananKejatuhanb

smA

QVV

AVQ

elbowga lastee

T

elbowga lastee

elbowga lasteeT

BA

203.383

0128.0

25.075.02

0128.0

25.010

0128.0

25.05.12100

0128.0

,

1008.9

1012.1

25.007.4Re

22100

22100

)

/07.4

25.0

4

2.0

5

6

221

A B

100 Le(2tee) Le(galas) Le(2elbow)

0420338301280fl 22


121/130

kPa502.162

5649.1681.91000

ghPP

hg

P

g

PZZ

VV

hZg2

V

g

PZ

g2

V

g

P

m5649.16

25.081.92

07.4203.3830128.0

gd2

flVh

fBA

fBA

BA

BA

fB

2BB

A

2AA

22

f

Example

Sebuah pam digunakan untuk mengepam air dari tangki bekalan A ke


122/130

Sebuah pam digunakan untuk mengepam air dari tangki bekalan A ke

tangki agihan B melalui dua batang paip jenis besi tuang terasfalt yang

disambungkan secara bersiri. Aras air di dalam tangki agihan ialah 10 m di

atas aras air di dalam tangki bekalan seperti ditunjukkan dalam rajah. Jikakadaralir air ialah 0.0237 m3/s, kelikatan kinematik ialah 1.004x10-6m2/s

dan kemasukan serta keluaran paip adalah tajam, kira tenaga yang

dibekalkan oleh pam bagi setiap unit berat.

10mD=150mm

L=100m

D=200mm

L=100m

Pam A

B

sm

A

QV /7544.0

2.04

0237.0

21

1


123/130

mZZ

VV

PP

hhhhZg

V

g

PZ

g

V

g

P

f

f

MoodycartaDari

d

d

xx

dV

xx

dV

smA

QV

BA

BA

BA

pmffBBB

AAA

10

0

0

22

0203.0

0199.0

0008.0150

12.0

0006.0

200

12.0

100036.210004.1

15.03411.1Re

105028.110004.1

2.07544.0Re

/3411.1

15.04

0237.0

21

222

1

2

1

5

6

222

5

6

111

22

2


124/130

mh

hxx

K

Jadid

d

JadualmelaluidiperolehitibatibapengecilanbagiK

hg

V

g

VK

g

V

gd

Vlf

gd

Vlf

p

p

p

6590.11

15.0

1000203.02525.01

81.92

3411.1

2.0

1000199.05.0

81.92

7544.010

2525.022.022.035.08.06.0

8.075.0

,75.02.0

15.0

2225.0

2210

22

1

2

2

2

2

2

2

1

2

2

222

1

2

111

Example

Kirakan kejatuhan tekanan P1-P2 yang menerusi talian paip yang


125/130

Kirakan kejatuhan tekanan P1 P2yang menerusi talian paip yang

ditunjukkan dalam rajah. Jika kadaralir minyak yang mempunyai berat

spesifik 8900 N/m2dan kelikatan 1x 10-6m2/s ialah 0.05 m3/s dan diameter

paip ialah 100mm. Jika perbezaan ketinggian antara titik 1 dan 2 adalah 4 mdan panjang keseluruhan paip besi tuang antara titik 1 dan 2 ialah 30 m.

P1

P24m

Globe Valve

(fully Open)

900

elbow

90 0 elbow

2/37.6

1.04

057.0sm

A

QV


126/130

2

21

2

22

1221

22

2

2

22

1

2

11

5

6

/1.314

890029.35

29.35

81.92

37.65.11

1.081.92

37.630025.04

22

5.111075.075.0

2222

025.0

0026.0100

26.0

1037.6100.1

1.037.6Re

mkN

xPP

g

VK

gd

flVZZ

g

PP

Jadi

K

g

VK

gd

flVZ

g

V

g

PZ

g

V

g

P

f

MoodycartaDari

d

xx

Vd

Example

Sebuah reservoir mengalirkan air melalui sebatang paip (paip 1) berdiameter


127/130

Sebua ese vo e ga a a e a u sebata g pa p (pa p ) be a ete200 mm dan 300 m panjang yang dicabangkan kepada dua batang paipberdiameter 150 mm dan 150 m panjang setiap satu (paip 2 dan paip 3). Paip 2

dan paip 3 terbuka penuh pada hujungnya. Paip 3 mempunyai lubang-lubangkeluaran di sepanjang panjangnya. Semua lubang keluaran ini disusunsedemikian rupa supaya separuh daripada air yang masuk ke dalam paip inidialirkan keluar di hujungnya dengan baki air mengalir keluar secara sama ratamelalui semua lubang keluaran tersebut. Kedua-dua hujung paip 2 dan paip 3berketinggian 15 m di bawah permukaan air bebas di reservoir. Abaikan semua

kehilangan kecil. Diberi f = 0.024 untuk semua paip dan ketumpatan air = 1000kg/m3.

a) Dengan menggunakan Persamaan Darcy-Weisbach

buktikan

b) Kira kadaralir untuk setiap paip

gD

fLVhf

2

2

g

V

D

LfHL

212

7 2

3

3

33

Example

In figure below kerosene at 25oC is flowing at 500 L/min from the lower tank to


128/130

In figure below, kerosene at 25 C is flowing at 500 L/min from the lower tank to

the upper tank through 2 inch nominal size type K copper tubing and a valve. Tank

A is very large and tank B is opened to the atmosphere.

a) If the pressure above the fluid in tank A is 103.42 kPa, how much energy loss

occurs in the system

b) Assuming that the energy loss is proportional to the velocity head in the

tubing, compute the pressure in the tank required to cause a flow of 1000

L/min

Solution


129/130


130/130

Chapt4 Edited Rbm

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