FLOWSIN

SIMPLE (NON-COMPLEX)

PIPELINE SYSTEMS

3.1 Pipes connected in SERIESWhenever pipes of different lengths and/ordiameters and/or materials are connected one afteranother referred as pipes connected in SERIES asshown in Fig. 3.1.

Figure 3.1: Three pipes (a, b and c) connected in SERIES.

Two important characteristics are:

..... QQQQ cbatotal

From CONTINUITY equation the DISCHARGE passing

from each pipe should be the SAME,

From ENERGY equation the TOTAL HEAD LOSS of the

system equals to the SUM OF THE INDIVIDUAL HEAD

LOSSES of the respective pipes.

Hence, the head loss due major loss (friction) andminor (local) losses must be evaluated separately foreach pipe size and then added.

...... hhh)h( LcLbLatotalL

Question 3.1 For a flow rate of Q= 40 lt/sec of fully turbulent flow in all

cases, determine the pressure heads and the total energy heads atthe points A, B, C and D. Ignore the effect of minor losses.The pressure head at the point A is .

The elevation of each point from a datum is: zA= 20 m, zB= 25 m,zC= 32.50 m, zD= 37.50 m.

Pipe Length (m) Diameter (cm) f

1 2000 30 0.022

2 1000 20 0.025

3 2000 40 0.021

m 40PA γ

.

3

1 2

A B D C

Solution

m/s 273.1)20.0(

404.0v

22av

π

m/s 566.0)30.0(

404.0

A

Qv

21

1av

π

m/s 318.0)40.0(

404.0v

23av

π

Using the Energy equation;

Total head at point A:

A

2avAA

A zg2

vPH

γm 016.6020

81.9x2

566.040H

2

A

Total head at point B- (just before pipe 2):

g2

v

D

Lf016.60h20

81.9x2

566.040H

21av

1

1121L

2

B

m 621.5762.19

566.0

3.0

2000022.0016.60H

2

B

Pressure head at point B-:

m 32.605 2562.19

566.0621.57z

g2

vH

p2

B

21av

BB γ

Total head at point C- (just before pipe 3):

g2

v

D

Lf621.57hHH

22av

2

2232LBC

m 297.4762.19

273.1

2.0

1000025.0621.57H

2

C

Pressure head at point C-:

m 714.1450.3262.19

273.1297.47

p 2C γ

Total head at point D:

g2

v

D

Lf214.47hHH

23av

3

3343LCD

m 673.4662.19

318.0

4.0

2000021.0214.47H

2

D

Pressure head at point D:

m 168.950.3762.19

318.0673.46

p 2D γ

Question 3.2

a) Water at 20 °C is expected to flow with a rate of 0.030 m3/secfrom reservoir 1 to reservoir 2 through three pipes that areconnected in series as shown. Neglecting the minor losses,determine the difference in water surface elevation betweenthese two reservoirs for the given pipe details.

b) If it is planned to change the existing non-PVC pipes with PVCpipe of the same length determine equivalent pipe diameter forthis new pipe that satisfies the above mentioned details.

Pipe Length (m) Diameter (mm) Material

A 1000 200 PVC

B 1500 180 Commercial Steel

C 2000 220 Asphalted Cast Iron

Solution:a) Connecting reservoir 1 and reservoir 2 by a single representative streamline and writing Bernoulli equation based on this flow line.

L2

22av2

1

21av1 hz

g2

vPz

g2

vP

γγ L21 hzz 21L zzh so

TYPE 1 PROBLEM

But for series connection LCLBLAL hhhh

Hence 21LCLBLA zzhhh

Using the Darcy-Weisbach equation:

g2

v

D

Lfh

2avA

A

AALA

g2

v

D

Lfh

2avB

B

BBLB

g2

v

D

Lfh

2avC

C

CCLC

Therefore the energy equation becomes:

)81.9(2

v

220.0

2000f

)81.9(2

v

180.0

1500f

)81.9(2

v

200.0

1000fzz

2avC

C

2avB

B

2avA

A21

Using continuity equation the velocities are:

m/s 955.0)200.0(

403.0

A

Qv

2A

avA

π

m/s 179.1)180.0(

403.0

A

Qv

2B

avB

π

m/s 789.0)220.0(

403.0

A

Qv

2C

avC

π

Finding roughness ratio, Reynolds number and f values for each pipe:

Pipe Material D (mm) ks/D Re f

A PVC 200 SMOOTH 1.9 x105 0.0151

B Commercial Steel 180 0.00025 2.1x105 0.0173

C Asphalted Cast Iron 220 0.00055 1.7x105 0.0196

2 2 2

1 2 3.849 7.348 9.082avA avB avCz z v v v

2 2 2

1 2 3.849(0.955) 7.348(1.179) 9.082(0.789)z z

1 2 3.182 10.214 5.654z z 1 2 19.050 z z m 3.510 19.370 m

b) So the system will be two pipes in series where the second pipesdiameter is unknown.

)81.9(2

v

D

3500f

)81.9(2

v

200.0

1000fzz

2new av

newnew

2avA

A21

)81.9(2

v

D

3500f

)81.9(2

v

220.0

2000f

)81.9(2

v

180.0

1500f

2new av

newnew

2avC

C

2avB

B

or

2 2

5

0.033500 350010.214 5.254

2(9.81) 12.1

av newnew new

new new

vf f

D D

Type 3 PROBLEM!

5

115.468 0.26 new

new

fD

For initial guess let fnew=0.015So Dnew= 0.191 mCheck ‘f’ : Re=2.01x105 so f=0.0157 = 0.0157 OK

Dnew= 0.191 m

Question 3.3 T:If the water is at 15° C, ignoring the effect of minor

losses, determine:a) the discharge passing from the system pipe,b) the diameter of an equivalent single pvc pipe, if the given pipes

will be replaced with it (total length same!).

zA= 58.25 m

a) Q=0.0333 m3/sb) φ = 0.159 m ≈ 16 cm

3.2 Pipes connected in PARALLEL

Whenever number of pipes of any lengths and/or materials and/or diameters are connected between two common points in parallel are referred pipes connected in PARALLEL, as shown in Fig. 3.2.

Figure 3.2: Three pipes (a, b and c) connected in PARALLEL

Drawing a streamline ψ for the given piping system, forces us to draw separate streamlines of each parallel pipe route; implying different Bernoulli equation set.

So the total head loss between the two common junction points for each pipe, consists of the major head loss due friction and the total minor (local) head losses along that pipe.

..... hhh)h( LcLbLatotalL

Once the Energy and the Continuity equations arewritten for any pipe system, depending on the givendata set different category of problems may begenerated.

To solve these equations by reducing the number ofunknowns, the hydraulically equivalent pipe approachshould be adopted.

This is done by expressing the unknown parameters inone equation set with another unknown parameter ofthe another equation set through applying thecontinuity and/or the energy equation.

This method is referred as Equivalent Pipes Concept.

Question 3.4For the given connection of pipe system as shown; determinedischarges passing in each pipe if the pipe characteristics and thereservoir levels are given. Ignore the effect of the minor lossesand take kinematic viscosity of the fluid =0.89x10-6 m2/s at 250C.

Pipe D (cm) L (m) Material

1 35 4973 Rough riveted steel

2 30 7608 Cast iron

zA=136.5 m

zB=48.3 m

SolutionThe discharges are not known so it is a category 2 typeproblem. Pipe 1 and pipe 2 are parallel so their totallosses are equal to each other. Writing energy equationbetween reservoirs A to B either by pipe 1 or by pipe 2details. Since they are parallel their total head loses shouldbe the same.

1LB

2avBB

A

2avAA hz

g2

vPz

g2

vP

γγ

62.19

v

35.0

4973f3.485.136

21av

1

2 2

1 av1 1 av1LA B 1

1

L v k vh f

D 2g 2g-

S= +

2 2

2 av2 2 av22

2

L v k vf

D 2g 2g

S= +Σ Σ

Assume initially fully turbulent flow for both pipes.

For pipe 1: Using only ks1/D1 = 0.9/35=0.026 → f1 = 0.0496

2

14973136.5 48.3 (0.0496)

0.35 19.62

avv → vav1 = 1.567 m/s

Checking the assumed f1;

ks1/D1 = 0.026

5

1 6

1.567 0.35Re 6.16 10

0.89 10x

→ f1 = 0.0496 (as assumed)

OK.

So2

3

1 1. 1

0.351.567 0.1508 m /s

4av

xQ v A x

For pipe 2: Using only ks2/D2 = 0.026/30=0.0009 → f2 = 0.0199

2

27608136.5 48.3 (0.0199)

0.30 19.62

avv → vav2 = 1.852 m/s

Checking the assumed f2;ks2/D2 = 0.0009

5

2 6

1.852 0.30Re 6.24 10

0.89 10x

→ f2 = 0.0203 (not as assumed) NOT OK.

NEEDS MORE ITERATION

New Trial take f2=0.0203

2

27608136.5 48.3 (0.0203)

0.30 19.62

avv → vav2 = 1.833 m/s

Checking this newly assumed f2;

ks2/D2 = 0.0009

5

2 6

1.833 0.30Re 6.18 10

0.89 10x

→ f2 = 0.0203 (as assumed) OK

So 2

3

2 2. 2

0.301.833 0.1296 m /s

4av

xQ v A x

Total discharge QT carried by to parallel pipes is: QT = 0.1508 + 0.1296 = 0.2804 m3/s.

Question 3.5:If the water is at 15° C, ignoring the effect of minor

losses, determine:a) the discharge passing from the system pipe,b) the diameter of an equivalent single pvc pipe, if the given pipes

will be replaced with it (total length same!). Consider the effectof 10 lt/s also!

a) Q1=0.03378 m3/s Q2=0.02378 m3/sb) φ = 15.5 cm

zA= 58.25 m

Question 3.6:

Ignoring the effect of minor losses, determine thedischarge passing from each pipe if the water is at 15° C.

L1= 2800 mSmoothened surface concrete Square cross-section of each side 15 cmzA= 58.25 m

Q1=0.0338 m3/s ; Q2=0.0246 m3/s , Q3=0.0092 m3/s

Question 3.7 T:The three pipe system shown in below Figure has the following characteristics:

Pipe D (cm) L (m) ks (mm)

A 25 500 0.75

B 30 650 1.20

C 35 1000 0.26

Neglecting minor losses, and take γ=9810 N/m3, υ=10-6 m2/s, find:a) the flow rate of water in each pipe,b) the pressure at point 3,c) if the parallel portion of the system (the pipes A and B) is expected

to be renewed by a single pipe with the similar material pipe (C) of length 575 m, determine the diameter of this new pipe?

Solution

a) Writing the energy equation from reservoir 1 to exit 2.

LminorLmajor2

2

2av21

2

1av1 hhzg2

vPz

g2

vP

γγ

0h80g2

v020000 LMajor

22av

Because the pipes A and B are parallel; the head lossalong pipe 1-A-3 and pipe 1-B-3 are the same so thehead loss for B can also be used in the above energyequation instead of for A. Note that the average velocityat point 2 and at pipe C are the same

LCLBLCLAMajorL hhhhh

The Darcy-Weisbach equation becomes:

This energy equation has 4 unknowns fA, vavA, (or fB vavB) fC and vavC.

By using continuity equation, one can get:

CBA QQQ or avCCavBBavAA vAvAvA

So the number of unknowns increases to five (5) due vavB.

LBLA hh From the parallel pipe conditions it is known that so:

but the number unknowns increases to six (6) due fB.

2 2 2

avC A avA A avAA

A

v L v k v120 f

2g D 2g 2g

S= + +

2 2

C avC C avCC

C

L v k vf

D 2g 2g

S+ +

Σ Σ

2 2

A avA A avAA

A

L v k vf

D 2g 2g

S+

2 2

B avB B avBB

B

L v k vf

D 2g 2g

S= +

Σ Σ

Hence 3 independent equations [Darcy-Weisbach, continuity and parallel pipe head losses] but these equations contains six unknowns (fA, vavA, fB vavB, fC and vavC).

To make the solution possible (3 equations - 3 unknowns)initially the major loss coefficient (fA, fB, fC) values for each pipe are suggested.

Hence for the initial trial only 3 unknows left.

For the initial assumptions, asumme that the flow in eachpipe is fully turbulent where the correction of fA, fB and fC

will be reevaluated at the end of each iteration and theiterations will stop when the previously assumed and thelatest calculated values are reasonably equal to each other.

For fully turbulent flow in:pipe A: ks/D = (0.75/250) = 0.003 → fA = 0.026 ;pipe B: ks/D = (1.20/300) = 0.004 → fB = 0.028;pipe C: ks/D = (0.26/350) = 0.00074 → fC = 0.018

from continuity:

avC

2

avB

2

avA

2

v4

)35.0(v

4

)30.0(v

4

)25.0(πππ

avCavBavA v096.0v071.0v049.0

from so head loss balance due parallel pipes:

81.9x2

v

30.0

650028.0

81.9x2

v

25.0

500026.0

2avB

2avA

2avB

2avA v092.3v650.2

avAavB v926.0v

Substituting into continuity:

avCavAavA v096.0)v926.0(x071.0v049.0

avCavA v096.0v115.0

avCavA v835.0v

Now substituting them into energy equation

)81.9(2

v

35.0

1000018.0

)81.9(2

)v835.0(

25.0

500026.0

)81.9(2

v120

2avC

2avC

2avC

solving for vavC2 2

120 (1 36.256 51.429) 88.6852(9.81) 2(9.81)

avC avCv v

m/s 152.5vavC

m/s 302.4152.5x835.0v835.0v avCavA

m/s 984.3302.4x926.0v926.0v avAavB

So

and

These values are correct if the initially assumed fA, fB and fC

values were ALL correct.

Checking f values,

pipe A :

ks/D = (0.75/250) = 0.003

6

6A 10x1.110

25.0x302.4Re

→ fA = 0.026 as before so OK.

pipe B:ks/D = (1.20/300) = 0.004

6

6B 10x2.110

30.0x984.3Re

→ fB = 0.028 as before so OK.

pipe C:ks/D = (0.26/350) = 0.00074

6

6C 10x80.110

35.0x152.5Re

→ fC = 0.018 as before so OK.

Hence the velocity values are correct.

Then the total flow passing from pipe C is:

/sm 496.0152.54

35.0vAQ 3

2

avCCC π

b) The pressure at point 3 can be computed using theenergy equation between* 1 A 3- or* 1 B 3- or * 3+ 2. Using point 3+ and 2:

)2 3(Lmajor2

22av2

3

23av3 hz

g2

vPz

g2

vP

γγ+

Just after the junction vav3 =vav2, then the velocity head terms cancel out:

)2 3(Lmajor

3 h80140P

γ

g2

v

D

Lf60

P 2avC

C

CC

3

γ

)81.9(2

)152.5(

35.0

1000018.060

P 23

γm 58.6960

P3

γm 58.9

P3

γ

3P 9810 9.58 93979.8 Pa+ = ´ =

)(kN/m kPa 980.93P 23

+

Question 3.8:The pipeline system shown in Figure below connects tworeservoirs that have an elevation difference of 20 m. Thispipeline system consists of 250 m of 50-cm diameter pipe (pipeA), that branches into 420-m of 20-cm diameter pipe (pipe B)and 290-m of 40-cm diameter pipe (pipe C) in parallel. Pipes Band C join into a single 30-cm diameter pipe that is 350-m long(pipe D). In this system all the pipes are made up of smooth(PVC). Ignoring minor loss effects and taking γ=9810 N/m3,υ=10-6 m2/s, determine the flow rate in each pipe of thissystem?

Solution

Question 3.9 TThe bathroom plumbing of a building consists of 1.5-cm-diameter copper pipes with threaded connectors as

shown. The gage pressure at the inlet of the system is 200 kPa (at point 1),

a) During the shower while the toilet reservoir is full (no flow in that branch), determine the flow rate at the

shower head. Note that on the shower head, there are 35 holes of each 2.0-mm-diameter. (25 p)

b) Determine the effect of flushing of the toilet on the flow rate through the shower head?

Note that when the toilet is flushed, the float within the reservoir moves and opens the valve so the water

starts to refill the reservoir. (25 p)

In your calculations consider only the given relevant minor loss coefficients ‘KL’ values. Water is @ 17 °C.

K L7= 2

K L2=2

K L6= 8water

KL1=0.9

K L4= 2

Pipe A Pipe B

Pip

e C

K L3=10

K L5= 12

K L8= 14

2

Ψ

𝑝1𝛾

+𝑣𝐴2

2𝑔+ 𝑧1 =

𝑝2𝛾

+𝑣22

2𝑔+ 𝑧2 + 𝑓𝐴

𝐿𝐴𝐷𝐴

𝑣𝐴2

2𝑔+ 𝐾𝐿1

𝑣𝐴2

2𝑔+ 𝑓𝐵

𝐿𝐵𝐷𝐵

𝑣𝐵2

2𝑔+ (𝐾𝐿2+𝐾𝐿3 + 𝐾𝐿4)

𝑣𝐵2

2𝑔+ 𝐾𝐿5

𝑣22

2𝑔

a)

23

ΨΨ

b)

𝑝1𝛾

+𝑣𝐴2

2𝑔+ 𝑧1 =

𝑝2𝛾

+𝑣22

2𝑔+ 𝑧2 + 𝑓𝐴

𝐿𝐴𝐷𝐴

𝑣𝐴2

2𝑔+ 𝐾𝐿1

𝑣𝐴2

2𝑔+ 𝑓𝐵

𝐿𝐵𝐷𝐵

𝑣𝐵2

2𝑔+ (𝐾𝐿2+𝐾𝐿3 + 𝐾𝐿4)

𝑣𝐵2

2𝑔+ 𝐾𝐿5

𝑣22

2𝑔

𝑝1𝛾

+𝑣𝐴2

2𝑔+ 𝑧1 =

𝑝3𝛾

+𝑣32

2𝑔+ 𝑧3 + 𝑓𝐴

𝐿𝐴𝐷𝐴

𝑣𝐴2

2𝑔+ 𝐾𝐿1

𝑣𝐴2

2𝑔+ 𝑓𝐶

𝐿𝐶𝐷𝐶

𝑣𝐶2

2𝑔+ (𝐾𝐿6+𝐾𝐿7)

𝑣𝐶2

2𝑔+ 𝐾𝐿8

𝑣32

2𝑔

Ψ: 1 - 3

Ψ: 1 - 2