Brad Peterson, P.E.Brad Peterson, P.E.

New Website:http://njut2009fall.weebly.com

Mr. Peterson’s Email Address: bradpeterson@engineer combradpeterson@engineer.com

Lesson 1, Properties of Fluids, 2009 Sept 04, Rev Sept 18 Lesson 2, Fluid Statics , 2009 Sept 11, Rev Sept 18 Lesson 2, Fluid Statics , 2009 Sept 11, Rev Sept 18 Lesson 3, Hydrostatic Force on Surfaces, 2009 Sept 25 Lesson 4, Buoyancy and Flotation, 2009 Sept 25 Lesson 5, Translation/Rotation of Liquid Masses, 2009 Oct 16

L 6 Di i l A l i /H d li Si ili d b ? Lesson 6, Dimensional Analysis/Hydraulic Similitude, maybe? Lesson 7, Fundamentals of Fluid Flow, 2009 Oct 23, Oct 30 Lesson 8, Flow in Closed Conduits, 2009 Nov 6 Lesson 9 Complex Pipeline Systems 2009 Nov 13 20 Nov Lesson 9, Complex Pipeline Systems, 2009 Nov 13, 20 Nov Lesson 10, Flow in Open Channels Lesson 11, Flow of Compressible Fluids Lesson 12, Measurement of Flow of Fluids

l d b l d Lesson 13, Forces Developed by Moving Fluids Lesson 14, Fluid Machinery

Chapter 8 talked about the flow in a single closed conduit with constant flow.

I real life, we often find:M th d it◦ More than one conduit, or◦ One conduit with changing size.

This chapter talks about some of these: This chapter talks about some of these:◦ Equivalent pipes◦ Pipes in series

ll l◦ Pipes in parallel◦ Branching pipes◦ Pipe networksPipe networks

Pipes are equivalent (same) when:◦ For a given head loss, the same flow rate is

produced in both (or all) pipes Or in other words when: Or, in other words, when:◦ The same Q is produced for the same hL

So, for example, a long pipe would require a larger diameter to be equivalent to a shorter pipe.

Or, a smaller pipe with a much smoother wall might be equivalent to a rougher larger pipemight be equivalent to a rougher, larger pipe.

Pipes that are connected end-to-end so that flow is constant through all pipes.

With pipes in series, flow (Q) is constant throughoutthroughout.

Connected so that flow branches into two or more separate pipes and then comes together again downstream.

Three important principles apply:◦ Total flow entering each joint equals the total flow

leaving the joint◦ Head loss between two joints is the same for eachHead loss between two joints is the same for each

branch connecting these joints◦ Percentage of flow through each branch will be

constant regardless of head loss or flow betweenconstant, regardless of head loss or flow between joints.

One or more pipes that separate into two or more pipes and do not come together again downstream.

In this example, direction of flow will depend on:◦ 1) tank pressures◦ 1) tank pressures

and elevations and ◦ 2) diameters,

l h d k dlengths and kinds of pipes.

Many pipes connected in a complex manner with many entry and exit points.with many entry and exit points.

Analyze using the Hardy Cross Method◦ 1. assume flows for each individual pipe.1. assume flows for each individual pipe.◦ 2. calculate the head loss thru each pipe using

Hazen-Williams.3 find the sum of head losses in each loop◦ 3. find the sum of head losses in each loop.◦ 4. remember, head loss between two joints is the

same for each branch.◦ 5. sum of head losses in each loop must be zero.◦ 6. compute a flow rate correction.◦ 7 adjust the assumed flow rates for all pipes and◦ 7. adjust the assumed flow rates for all pipes and

repeat the process until all corrections are zero.

For a lost head of 5.0m/1000m and C=100 f ll i h 20 ifor all pipes, how many 20cm pipes are equivalent to one 40cm pipe? How may are equivalent to a 60cm pipe?equivalent to a 60cm pipe?

2

Out of curiousity, let's compare the cross-sectional areas:20

20

2

20A = 100440A 400

40

2

60

A = 400460A = 1200

60A 12004

If area was all that mattered:it would take 20cm pipes to equal 1 40cm pipe and4

it would take 20cm pipes to equal 1-40cm pipe, andit woul

4 d

take 20cm pipes to equal 1-60c12 m pipe

But we must consider head loss. Let’s use two equations:

Q AVand

0.63 0.540.8492andV CR S

2

Q AVd

0.63 0.544

0.8492

dA

V CR S

4dR

5 0.005 /1000100

mS m mm

C

0.63

2 0.54

100

0.8492 100 0.0054

C

dd

4

4Q

0.632 0.540.200.20 0.8492 100 0.005

320

4 0.023 /4

Q m s

0.632 0.54

3

0.400.40 0.8492 100 0.0054 0 143 /Q

3

404 0.143 /

4Q m s

0.632 0.54

360

0.600.60 0.8492 100 0.0054 0.416 /Q m s

60 0.416 /

4Q m s

340 0 143 /Q m s40

320

0.143 / 6.2 20cm pipes equivalent to a 40cm pipe0.023 /

Q m sQ m s

360

320

0.416 / 18.1 20cm pipes equivalent to 60cm pipe0.023 /

Q m sQ m s

A 300mm pipe that is 225m long and a 500m pipe that is 400m long are connected in series. Find the diameter of a 625m long equivalent pipe Assume all pipes areequivalent pipe. Assume all pipes are concrete.

20 63 0 540 8492dQ AV A V CR S 0.63 0.54; ; 0.8492

4Q AV A V CR S

d

; 120 for concrete4dR C

h

31 ; assume 0.1 /h

S Q m sL

0.540.63 hd 0.63

2 10.8492 12040 1

hddL

0.14

for the 300mm dia, 225m long pipe:0.540.63

2 10.300.30 0.8492 1204 225

h 4 2250.1

4

0.5410.1 1.4087

225h

1

2251.678h m

for the 500mm dia, 400m long pipe:0.540.63

2 20.500.50 0.8492 1204 400

h

0 54

4 4000.14

0.5420.1 5.3985

400h

2

000.248h m

total head loss 1.678 0.248 1.926m

Therefore:

0.63 0.54

for a 625m long equivalent pipe:

1 926d 0.63 0.54

2 1.9260.8492 1204 6250 1

dd 0.14

360d mm

A 300mm pipe that is 225m long and a 500m pipe that is 400m long are connected in series. Find the diameter of a 625m long equivalent pipe Assume all pipes areequivalent pipe. Assume all pipes are concrete.

Use Diagram B-3 to solve.

3Assume a flow rate, 0.1 /Q m s

1

From Diagram B-3, for 300mm pipe, 0.0074 /h m m1

2for 500mm pipe, 0.00064 /total head loss =

h m mtotal head loss 0.0074 / 225 0.00064 / 400 1.921f 625 l i 1 921 / 625 0 00

m m m m m m mh

3071for a 625m long pipe, 1.921 / 625 0.00h m m 307

from diagram B-3, 360m

d m

Water flows at a rate of 0.05 m3/s from reservoir A to reservoir B through three concrete pipes connected in series, as shown on the following slide Find the difference inon the following slide. Find the difference in water surface elevations in the reservoirs. Neglect all minor losses.g

Use Diagram B-3

30.05 /Q m sfrom Diagram B-3:for the 400mm pipe 0 00051 /h m m1

1

for the 400mm pipe, 0.00051 /for the 300mm pipe, 0.0020 /

h m mh m m1

1

p p ,for the 200mm pipe, 0.015 /h m m

total head losstotal head loss0.00051 / 2600m m m

0.0020 / 18500 015 / 970

m m m 0.015 / 97019 58

m m mm

19.58m

Use Hazen-Williams Formula2

0.63 0.54; ; 0.84924dQ AV A V CR S

4

; 120 for concrete4dR C

3

4

; 0.05 /hS Q m s ; QL

0.63 0.54d h 0.63 0.5

2 0.8492 12040 05

d hdL

0.05

4

0.540.632 400

for the 400mm dia, 2600m long pipe:

0.400 40 0 8492 120h

2 4000.40 0.8492 120

4 26000.054

300

41.32h m

0.540.632 300

for the 300mm dia, 1850m long pipe:

0.300 30 0 8492 120h

2 3000.30 0.8492 120

4 18500.054

300

43.82h m

for the 200mm dia 970m long pipe:0.540.63

2 200

for the 200mm dia, 970m long pipe:

0.200.20 0.8492 120h

4 9700.054

300 14.44 1.32 3.82 14.44 19.58

h mtotal head m m m m

The flow in pipes AB and EF is 0.850m3/s. All pipes are concrete. Find the flow rate in pipes BCE and BDE.

Assume head loss from B to E 1.001 00

mm

11.00for pipe BCE, 0.00043 /2340

mh m mm

3from Diagram B-3, 0.133 /BCEQ m s

1.00for pipe BDE 0 00031 /mh m m1

3

for pipe BDE, 0.00031 /3200

from Diagram B 3 0 038 /

h m mm

Q m s

from Diagram B-3, 0.038 /BDEQ m s

if head loss from B to E = 1.00m is correct, then

3

sum of the flow rates through BCE and BDE willl 0 850 /3equal 0.850m /s

but, 0.133 0.038 0.171 0.850

head loss of 1.00m is not correct, however,actual flow rates through BCE and BDE willactual flow rates through BCE and BDE willbe at the same proportion. So,

30 133 3BCE

0.133Q = 0.850 0.661 /171

0 038

m s

3BDE

0.038Q = 0.850 0.189 /171

m s

Compute the 

1500120m

C

pflow in each branch.

W Z

30.51 /Q m s

900120m

C

Z

20 63 0 540 8492dQ AV A V CR S 0.63 0.54; ; 0.8492

4dQ AV A V CR S

d h

; 120 for concrete; 4d hR C S

L

0.63 0.542 0 8492 120 d hd 0.8492 120

44

dLQ

Compute the 

1500120m

C

pflow in each branch.

W Z

30.51 /Q m s

900120m

C

Z

assume a head loss of 10m from W to Z

0.63 0.542

then, for the 300mm pipe:

0.3 100.3 0.8492 120 3

300

0.3 0.8492 1204 1500 0.09 /

4and, for the 400mm pipe:

Q m s

0.63 0.542

and, for the 400mm pipe:

0.4 100.4 0.8492 1204 900Q

30 26 /m s400 4

Q 0.26 /m s

1500 , 300120m mm

C

30.09 /Q m s

W Z

30.51 /Q m s

900 , 400120m mm

C

Z

30.26 /Q m s 3 3 30.09 / 0.26 / 0.51 /m s m s m s

3 3 3 30.09 / 0.26 / 0.35 / 0.51 /m s m s m s m s

3300

3

0.09 0.51 0.13 /0.350 26

Q m s

3400

0.26 0.51 0.38 /0.35

Q m s

1500 , 300120m mm

C

30.13 /Q m s

W Z

30.51 /Q m s

900 , 400120m mm

C

Z

30.38 /Q m s 3 3 30.13 / 0.38 / 0.51 /m s m s m s

Next, let’s solve using Hardy Cross Method

Compute the 

1500 , 300120m mm

C

pflow in each branch using 

W Z

30.51 /Q m s

gHardy‐Cross Method.

900 , 400120m mm

C

Z

Many pipes connected in a complex manner with many entry and exit points.

• Analyze using the Hardy Cross Method– 1. assume flows for each individual pipe.– 2. calculate the head loss thru each pipe using

Hazen-WilliamsHazen Williams.– 3. find the sum of head losses in each loop.– 4. remember, head loss between two joints is the

f h b hsame for each branch.– 5. sum of head losses in each loop must be zero.– 6. compute a flow rate correction.6. compute a flow rate correction.– 7. adjust the assumed flow rates for all pipes and

repeat the process until all corrections are zero.

For this problem, we are looking at just one loop:

QQ

QQ

Compute the 

1500 , 300120m mm

C

pflow in each branch using 

W Z

30.51 /Q m s

gHardy‐Cross Method.

900 , 400120m mm

C

Z30.51 /Q m s

• Analyze using the Hardy Cross Method– 1. assume flows for each individual pipe.– 2. calculate the head loss thru each pipe using

Hazen-WilliamsHazen Williams.– 3. find the sum of head losses in each loop.– 4. remember, head loss between two joints is the

f h b hsame for each branch.– 5. sum of head losses in each loop must be zero.– 6. compute a flow rate correction.6. compute a flow rate correction.– 7. adjust the assumed flow rates for all pipes and

repeat the process until all corrections are zero.

For correction, use:

LH

0

3

1.85 ( / )

/

LH Q

flow adjustment m s

, /

flow adjustment m s

LH lost head L S

0 Q assumed initial flow

For Consistency:◦ Clockwise, Q and LH are positive◦ Counterclockwise, Q and LH are negative◦ So in:◦ So, in:

01.85 ( / )

LHLH Q

◦ Sign (+ or -) is important in numerator◦ But denominator is always positiveBut, denominator is always positive

In the table that follows: In the table that follows:

- :S is from the Hazen Williams formula

0.63 0.54

:

0.8492

S is from the Hazen Williams formula

V CR S

:also use

Q VA

Compute the 

1500 , 300120m mm

C

pflow in each branch using 

W Z

30.51 /Q m s

gthe Hardy Cross Method.

900 , 400120m mm

C

Z

See Excel Spreadsheet Or Pdf Version

• Analyze using the Hardy Cross Method– 1. assume flows for each individual pipe.– 2. calculate the head loss thru each pipe using

Hazen-WilliamsHazen Williams.– 3. find the sum of head losses in each loop.– 4. remember, head loss between two joints is the

f h b hsame for each branch.– 5. sum of head losses in each loop must be zero.– 6. compute a flow rate correction.6. compute a flow rate correction.– 7. adjust the assumed flow rates for all pipes and

repeat the process until all corrections are zero.

30 4 /30.4 /m s600300

mmm dia

600300

mmm dia

BA C

400250

mmm dia

400250

mdi

400m

600m 600m

250 mm dia 250 mm dia 250 mm dia

30.4 /m s300 mm dia

600300

mmm diaEF D

• Analyze using the Hardy Cross Method– 1. assume flows for each individual pipe.1. assume flows for each individual pipe.– 2. calculate the head loss thru each pipe using

Hazen-Williams.3 find the sum of head losses in each loop– 3. find the sum of head losses in each loop.

– 4. remember, head loss between two joints is the same for each branch.

– 5. sum of head losses in each loop must be zero.– 6. compute a flow rate correction.

7 adjust the assumed flow rates for all pipes and– 7. adjust the assumed flow rates for all pipes and repeat the process until all corrections are zero.

For this problem, we are looking at only loops:

QQ

QQ

Solve using Excel .pdf Version

0.63 0.54For se Ha en Williams: 0 8492S V CR S

0.63 0.54For , use Hazen-Williams: 0.8492

S V CR SLH total head loss in pipe L S

LH

1.85 /LH Q

30 4 /30.4 /m s600300

mmm dia

600300

mmm dia

BA C3

0 0 200 /Q m s 30 0 100 /Q m s

400250

mmm dia 400

250mmm dia 400

250m

d

0 0.200 /Q m s 0 0.100 /Q m s

600m 600m

250 mm dia 250 mm dia3

0 0.100 /Q m s3

0 0.100 /Q m s3

0 0.200 /Q m s

30.4 /m s

600300

mmm dia 600

300 mmm diaEF D3

0 0.300 /Q m s3

0 0.200 /Q m s

30 4 /30.4 /m s600300

mmm dia

600300

mmm dia

BA C3

0 0 200 /Q sm 3100 /0Q m s

400250

mmm dia 400

250mmm dia 400

250m

d

30

0 0.2410.200 /

/Q sQ s

mm

0

30.15 /.100 /

90Q m s

Q m s

600m 600m

250 mm dia 250 mm dia3

30

0 0.0820.100 /

/Q sQ s

mm

30

30.16 /.200 /

00Q m s

Q m s 3

030.15 /

.100 /9

0Q m sQ m s

30.4 /m s

600300

mmm dia 600

300 mmm diaEF D3

030.16 /

.200 /0

0Q m sQ m s

30

30.24 /.300 /

10Q m s

Q m s