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Brad Peterson, P.E.Brad Peterson, P.E.
New Website:http://njut2009fall.weebly.com
Mr. Peterson’s Email Address: bradpeterson@engineer [email protected]
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:
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:
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