Aliran Dalam Pipa

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Pipe Flow

Pipe Systems and DesignPipeSystemsandDesign

PipeNetworkAnalysis

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Pipe Systems and Design Twomajorconcerns:

Si th i ( f h t & t bl ) Sizethepipe(e.g.fromcharts&tables) Determinetheflowpressurerelationship

Toanalysethesystem,e.g.tofindoutpumppressure Byusingmanualorcomputerbasedmethods

CalculationsforpipelinesorpipenetworksCan be er complicated for branches & loops Canbeverycomplicatedforbranches&loops

Basicparameters:pipediameter,length,frictionfactor,roughness,velocity,pressuredrop

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Pipe Systems and Design

Pipenetworkanalysis Physicalfeaturesareknown Solutionprocesstrytodetermineflow&pressureateverynode

Pipenetworkdesign Variables are unknownVariablesareunknown Trytosolve&selectpipediameters,pumps,valves,etc.

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Basic equationsBasicequations DarcyWeisbachEquation (forfullydevelopedflowsofallNewtonianfluids)

C l b k Whit E ti (f t iti i )

=

=g

VDLfh

gV

DLfp

2or

2

22

ColebrookWhiteEquation (fortransitionregion):

*Theequationisimplicitinf (appearsonbothsides),soiterationsarerequiredtosolveforf.

++=

fDD

f )/Re(3.91log2)/log(214.11

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Basicequations(contd) HazenWilliamsEquation (alternativetoDarcyWeisbachformula;empirical)

g)(1819.61.167852.1

=DC

VLp

C =roughnessfactor(typically,C =150forplasticorcopperpipe,C =140fornewsteelpipe,C

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Table 6.1: Typical pipe roughness (Reference: White, 1999)

Material Roughness, k (mm)Glass smoothBrass new 0 002Brass, new 0.002ConcreteSmoothedRough

0.042.0

IronCast, new

0.260.15

Galvanised, newWrought, new

0.046

SteelCommercial, newRiveted

0.0463

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MinorLossesInadditiontoheadlossduetofriction,thereare always other head losses due to pipearealwaysotherheadlossesduetopipeexpansionsandcontractions,bends,valves,andotherpipefittings.Theselossesareusuallyknownasminorlosses(hLm).

In case of a long pipeline the minor lossesIncaseofalongpipeline,theminorlossesmaybenegligiblecomparedtothefrictionlosses,however,inthecaseofshortpipelines,theircontributionmaybesignificant.

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Losses due to pipe fittings

2V

Kh2

Lm = Type K where hLm= minor loss K = minor loss coefficient V = mean flow velocity

g2Lm Exit (pipe to tank) 1.0Entrance (tank topipe)

0.5

90 elbow 0.945 elbow 0.4

Table 6.2: Typical K values

T-junction 1.8

Gate valve 0.25 - 25

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Loss CoefficientsUse this table to find loss coefficients:

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Example Calculate the head added by the pump when the water

system shown below carries a discharge of 0.27 m3/s. If the efficiency of the pump is 80% calculate the power the efficiency of the pump is 80%, calculate the power input required by the pump to maintain the flow.

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Solution:Applying Bernoulli equation between section 1 and 2

(1)+++=+++ 21L2

22

2p

21

11 H

g2V

zg

PH

g2V

zg

P

P1 = P2 = Patm = 0 (atm) and V1=V2 0Thus equation (1) reduces to: (2)

HL1-2 = hf + hentrance + hbend + hexit+= 21L12p HzzH

From (2):g2

V4.39

14.05.04.0

1000x015.0g2

VH

2

221L

=

+++=

81.9x2V4.39200230H

2p +=

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The velocity can be calculated using the continuity equation:

Thus, the head added by the pump: Hp = 39.3 m( ) s/m15.22/4.0

27.0AQV

2=

==

in

pp P

gQH==

Pin = 130 117 Watt 130 kW.

8.03.39x27.0x81.9x1000gQHP

p

pin =

=

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Pipes in Series When two or more pipes

of different diameters or roughness are connected in such a way that the in such a way that the fluid follows a single flow path throughout the system, the system represents a series pipeline.

In a series pipeline the In a series pipeline the total energy loss is the sum of the individual minor losses and all pipe friction losses.

Figure 6.11: Pipelines in series

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Referring to Figure 6.11, the Bernoulli equation can be written between points 1 and 2 as follows;

(6.18)where P/g = pressure head

z = elevation head

21L

22

22

21

11 H

g2V

zg

Pg2

Vz

gP

+++=++

z = elevation headV2/2g = velocity headHL1-2 = total energy lost between point 1 and 2

Realizing that P1=P2=Patm, and V1=V2, then equation (6.14) reduces to

z1-z2 = HL1-2Or we can say that the different of reservoir water level is equivalent to the total head losses in the system.

The total head losses are a combination of the all the friction losses and the sum of the individual minor lossessum of the individual minor losses.

HL1-2 = hfa + hfb + hentrance + hvalve + hexpansion + hexit.

Since the same discharge passes through all the pipes, the continuity equation can be written as;

Q1 = Q2

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Pipes in Parallel

A combination of two or more pipes connected between connected between two points so that the discharge divides at the first junction and rejoins at the next is known as pipes in known as pipes in parallel. Here the head loss between the two junctions is the same for all pipes.

Figure 6.12: Pipelines in parallel

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Applying the continuity equation to the system; Q1 = Q + Qb = Q (6 19) Q1 = Qa + Qb = Q2 (6.19)

The energy equation between point 1 and 2 can be written as;

The head losses throughout the system are given by; HL1-2=hLa = hLb (6.20)

L

22

22

21

11 H

g2V

zg

Pg2

Vz

gP +++=++

L1 2 La Lb ( )

Equations (6.19) and (6.20) are the governing relationships for parallel pipe line systems. The system automatically adjusts the flow in each branch until the total system flow satisfies these equations.

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Pipe Network A water distribution system consists of complex interconnected pipes,

service reservoirs and/or pumps, which deliver water from the treatment plant to the consumer.

Water demand is highly variable whereas supply is normally constant Water demand is highly variable, whereas supply is normally constant. Thus, the distribution system must include storage elements, and must be capable of flexible operation.

Pipe network analysis involves the determination of the pipe flow rates and pressure heads at the outflows points of the network. The flow rate and pressure heads must satisfy the continuity and energy equations.

The earliest systematic method of network analysis (Hardy-Cross Method) is known as the head balance or closed loop method. This method is applicable to system in which pipes form closed loops. The outflows from applicable to system in which pipes form closed loops. The outflows from the system are generally assumed to occur at the nodes junction.

For a given pipe system with known outflows, the Hardy-Cross method is an iterative procedure based on initially iterated flows in the pipes. At each junction these flows must satisfy the continuity criterion, i.e. the algebraic sum of the flow rates in the pipe meeting at a junction, together with any external flows is zero.

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Hardy Cross Methods: Step 1

By careful By careful inspection we may assume the most reasonable distribution of flows in the pipe network p pand make the first guess of the flow pattern.

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Step 1: Sign Convention in Pipe Network

Enter flows at nodes as positive for inflows and negative for outflowsnegative for outflows.

Inflows plus outflows must sum to 0. Enter one pressure in the system and all other

pressures are computed. You do not need to use all the pipes or nodes. Enter a diameter of 0.0 if a pipe does not exist.

If d i d d ll id b If a node is surrounded on all sides by non-existent pipes, the node's flow must be entered as 0.0.

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Step 1: How to Handle Minor Losses

Minor losses such as pipe elbows bends andMinor losses such as pipe elbows, bends, and valves may be included by using the equivalent length of pipe method (Mays, 1999).

Equivalent length (Leq) may be computed using the following calculator which uses the formula

Leq=KD/f.

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Step 1: Summary of Minor Losses Calculation

Enter node flows, elevations, pressure. Select Darcy Weisbach (Moody diagram) or

Hazen Williams friction losses. Include minor losses by equivalent length of pipe.

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Minor Losses If you go by a rigorous method, f is the Darcy-

Weisbach friction factor for the pipe containing theWeisbach friction factor for the pipe containing the fitting, and cannot be known with certainty until afterthe pipe network program is run.

However, since you need to know f ahead of time, a reasonable value to use is f=0.02, which is the default valuevalue.

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Hardy Cross Methods: Step 2 Write head loss condition for each pipe in the form:

H ( h ) K QnHi (or hL) = K Qn

n=2.0 for Darcy Weisbach losses n=1.85 for Hazen Williams losses.

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Friction Losses, H

Hazen Williams equation

Darcy Weisbach equation

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Hardy Cross Methods: Step 3 Compute the algebraic sum of the head losses

around each elementary loop, Hi (or hL,i) = Ki Qin

Consider losses from clockwise flows as positive, counterclockwise negative.

Be careful about the common pipe sections shared by two adjacent loops.by two adjacent loops.

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Hardy Cross Methods: Step 4 Adjust the flow in each loop by a correction Q or

to balance the head in that loop and giveto balance the head in that loop and give K Qn=0

The heart of this method lies in the following determination of Q . For any pipe, we may write:

Q=Q0+ Q Where Q0 is the assumed discharge and Q is the corrected discharge.

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Binomial series gives:Binomial series gives:

If Q is small compared with Q0, we may neglect the terms of the binomial series after the second one:

.....)()( 1000 ++=+== QnQQKQQKKQH nnnn

100+= nn QKnQKQH

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For a loop Hi (or hL i) = Ki Qin =0:For a loop, Hi (or hL, i) Ki Qi 0:

We may solve this equation for Q :

0100 =+== nnL QKnQKQhH

===

0

10

100

10

0

/ Qhnh

KQn

QKQKnQ

KQQ

L

Ln

n

n

n

000 / QhnQnQ L# Sum the numerator algebraically with due account of each sign

# Sum the denominator arithmetically

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Hardy Cross Methods: Step 4 It indicates that clockwise flows may be

id d d i l k i l dconsidered as producing clockwise losses, and counterclockwise flows, counterclockwise losses.

This means that the minus sign is assigned to all counterclockwise conditions in a loop,all counterclockwise conditions in a loop, namely flow Q and lost head hL.

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Friction Losses, H The calculation procedure gives you a choice of

computing friction losses H using the Darcycomputing friction losses H using the Darcy-Weisbach (DW) or the Hazen-Williams (HW) method.

The DW method can be used for any liquid or gas while the HW method can only be used for water at temperatures typical of municipal water supplytemperatures typical of municipal water supply systems.

n=2.0 for Darcy Weisbach losses

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Iteration

After we have given each loop a first correction, te we ave g ve eac oop a st co ect o ,the losses will still not balance, we need to repeat the procedure, arriving at a second correction, and so on, until the corrections become negligible.

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Step 5: Pressure Computation

After computing flowrate Q in each pipe and lossAfter computing flowrate Q in each pipe and loss H in each pipe and using the input node elevations Z and known pressure at one node, pressure P at each node is computed around the network:

Pj = (Zi - Zj - Hpipe) + Pi node j is down-gradient from node i. = fluid density [F/L3].

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Assigning clockwise flows and their associated head losses are positive, the procedure is as follows: Assume values of Q to satisfy Q = 0. Calculate HL from Q using HL = K1Q2 . If HL = 0, then the solution is correct. If HL 0, then apply a correction factor, Q, to all Q and repeat from

step (2). For practical purposes, the calculation is usually terminated when HL <

0.01 m or Q < 1 L/s. A reasonably efficient value of Q for rapid convergence is given by;

(6.21)

=

QH2

HQ

L

L

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Example A pipe 6-cm in diameter, 1000m long and with =

0.018 is connected in parallel between two points M and N with another pipe 8-cm in diameter, 800-m long and having = 0.020. A total discharge of 20 L/s g genters the parallel pipe through division at A and rejoins at B. Estimate the discharge in each of the pipe.

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Solution: Continuity: Q = Q1 + Q2

(1)22

221

2 V)08.0(4

V)06.0(4

02.0 += Pipes in parallel: hf1 = hf2

S b tit t (2) i t (1)

44

074.7V778.1V 21 =+V

08.0800x020.0V

06.01000x018.0

gD2VL

gD2VL

22

21

2

222

21

211

1

=

=

Substitute (2) into (1)

0.8165V2 + 1.778 V2 = 7.074 V2 = 2.73 m/s

)2(V8165.0V 21 =

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73.2x)08.0(4

VAQ 2222==

Q2 = 0.0137 m3/s From (2):

V1 = 0.8165 V2 = 0.8165x2.73 = 2.23 m/s

Q1 = 0.0063 m3/sQ1 Recheck the answer:

Q1+ Q2 = Q0.0063 + 0.0137 = 0.020 (same as given Q OK!)

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Example For the square loop shown, find the discharge

in all the pipes. All pipes are 1 km long and 300 mm in diameter, with a friction factor of 0.0163. Assume that minor losses can be neglected.

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Foranalysis,basiccriteriashouldbefollowed: Thetotalflowenteringeachpoint(node)must

equal to the total flow leaving that joint (node)equaltothetotalflowleavingthatjoint(node).

Qin =Qout Theflowinapipemustfollowthepipesfriction

lawforthepipe.hf =kQn

niscoefficientdependonequationused Thealgebrasumoftheheadlossinthepipe

networkmustbezero.hf =0

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LoopMethod Steps:1. AssumeflowsforeachindividualpipeinthenetworkbutQ=0 ateachnode.2. CalculatestheheadlossthrougheachpipewithDarcyor

Hazen WilliamHazenWilliam.3. Namethepipenetworktoseveralloop.4. Findthealgebrasumoftheheadlossesineachloopinthe

pipenetwork.5. Calculate(nhf/Q) foreachloop(ignorepositiveor

negative).6. GetQ valueforcorrection

Q= hf(nhf/Q)

7. CalculatethenewflowrateQnew =Q+Q

8. RepeatthisstepuntilQ=0 .5/19/2008 90

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NodeMethod Thismethodusedwhenpressureheadatcertainnode/point

knownbutflowrateandpressureheadatseveral(different)node/pointshouldbeobtained.Thistypeofproblemgenerallyrequiresatrialanderrorsolution.St Step

1. Assumeanelevationofthehydraulicgradeline(head)atnodeJ.2. CalculatetheQvalueineachpipeusingHJ (assumebefore).Use

DarcyWeisbachorHazenWilliamformula.3. CalculateQin =Qout fromnodeJ.4. CalculateQ/nhfforeachpipethenfind(Q/nhf).

/ /5. Calculatehf= (Q/(nhf/Q)) forcorrection.(Darcy n=2 HazenWilliam n=1.85)

6. CalculateHJnew =HJ hf7. Repeatstep2 5untilhf0

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Solution: Assume values of Q to satisfy continuity equations all

at nodes.

Th h d l i l l t d i H K1Q2 The head loss is calculated using; HL = K1Q2 HL = hf + hLm But minor losses can be neglected: hLm = 0 Thus HL = hf Head loss can be calculated using the Darcy-Weisbach

equationequation

g2V

DLh

2f =

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Qx77.2AQ77.2H

81.9x2Vx

3.01000x0163.0H

g2V

DLhH

22

2

2

2L

2L

2fL

==

=

==

First trial554'K

Q'KH

Q554H

3.0x4

A

2L

2L

2

===

Pipe Q (L/s) HL (m) HL/Q

Since HL > 0.01 m, then correction has to be applied.

AB 60 2.0 0.033

BC 40 0.886 0.0222

CD 0 0 0

AD -40 -0.886 0.0222

2.00 0.07745/19/2008 93

Second trial

s/L92.120774.0x22

QH2

HQ

LL ==

=

Pipe Q (L/s) H (m) H /Q

Since HL 0.01 m, then it is OK.Thus, the discharge in each pipe is as follows (to the nearest integer).

Pipe Q (L/s) HL (m) HL/Q

AB 47.08 1.23 0.0261

BC 27.08 0.407 0.015

CD -12.92 -0.092 0.007

AD -52.92 -1.555 0.0294

-0.0107 0.07775

Pipe Discharge (L/s)

AB 47

BC 27

CD -13

AD -535/19/2008 94

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Thank You

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