sab2513 hydraulic chapter 7
TRANSCRIPT
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Introduction Water distribution network analysis provides the basis for the
design of new systems and extension of existing systems.
The design criteria are that specified minimum flow rates andpressure head~ must be attained at the outflow points of the
network.
The flow and pressure distributions across a network areaffected by the arrangements and sizes of the pipes and the
distribution of outflows.
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The flow in any network, however complicated, must satisfythe basic relations of continuity and energy as follows:
(a) The algebraic sum of flow rates in the pipes meeting at a
junction is zero
outin QQ
Introduction
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(b) The flow in each pipe must satisfy the pipe-friction laws (Darcy-Weisbach, Hazen-William, etc) for flow in a single pipe.
(c) The algebraic sum of the head losses in the pipe, around any closed loopformed by pipes, is zero
0f
h
n
fkQh
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Hardy Cross Method
2 solutions:(1) Balancing Head ----- Assume Q
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(2) Balancing Flow ----- Assume hf
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(i) At a junction:
(ii) Equation for head loss due to friction, hf:
(k and n values depend on the eqn used (Darcy-
Weisbach, Hazen-William, etc)
outin QQ
n
f kQh
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(a) Darcy-Weisbach Eqn.
(b) Blasius Eqn.
g
v
d
fLhf
2
42
,3
2
5Q
d
fL 2n
,1253
75.1
75.4Q
dLhf ,
125375.4
dLk 75.1n
53d
fLk
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(c) Hazen-William Eqn.
Where;
CHW = Hazen-William roughness coef.
(refer Table)
85.1
87.4
67.10Q
dC
fLh
HW
f
85.1n
87.4
67.10
dC
fLk
HW
Pipe Material CHWPlastic, PYC 150
Asbestos Cement 140
Welded Steel 120
Riveted Steel 110Concrete 130
Asphalted Iron -
Galvanised Iron -
Cast Iron (New) 130
Cast Iron (Old) 100Corrugated Metal -
Table: Hazen-Williams Coefficient,
CHW for different types of pipe
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1. Balancing Head Method
Applicable if the followings are known:
(i) Pipe characteristics such as diameter (d), Length (L)
and friction factor (f)
(ii) Qin, Qout in the system
Where discharges and pressure head are to be
determined????
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Balancing Head Method
Procedures:(i) Label the pipe network into close loops
(ii) Assumed values of Q in each pipe, but make sure:
at any junction: , where
(iii) Depending on the head losses eqn. used, calculate
hf for each pipe, i.e
(Note: sign for hf follows the sign for Q)
0Q outin QQ
n
f kQh
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Balancing Head Method
(iv) For any loop: 0f
h
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Balancing Head Method
Example:
0 fh
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Balancing Head Method
(v) Calculate hf for each pipe in each loop (circuit)
(vi) For each loop/circuit, calculate the value of
(vii) Compute the error, Q:
(Note: n = 2 for Darcy-Weisbach eqn. &
n = 1.85 for Hazen-Williams eqn.)
Q
nhf
Q
nh
hQ
f
f
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Figure below shows layout of pipe network for a
water supply system. Determine the flow rate in
each pipe in the network by using Darcy-
Weisbach approximations.
(Note: assume that friction factor, f = 0.0025 for
all pipes)
Example
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Pipe Network
Pipe data
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Solution
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- Initial assumption for values and directions of Q
in each pipes- (Note: At a junction, Qin = Qout)
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Solution
19
Equations to be used:
(i) Darcy-Weisbach eqn.
(ii)
(iii)
Construct the Table!!!
2
53
f
fLh Q
d
Q
hhQ
f
f
2
QQQ
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Solution
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Loop Trial 1
1
pipe d(m) L(m) k=fL/3d5 Q1 hf=kQ12 2hf/Q1 Q1 Q= Q1+Q1
AB 0.5 1000 26.666 0.2 1.0667 10.667 -0.0783 0.1217
AC 0.75 1200 4.2140 -0.3 -0.3793 2.5284 -0.0783 -0.3783
BC 0.5 1500 40.00 0.2 1.600 16.000 0.0581 0.2581
Total, 2.2874 29.1951
0783.01951.292874.2
2
1
1
Q
hhQ
f
f
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Solution
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Loop Trial 2
2
pipe d(m) L(m) k=fL/3d5 Q1 hf=kQ12 2hf/Q1 Q1 Q= Q1+Q1
BC 0.5 1500 40.000 -0.2 -1.6000 16.000 -0.0581 -0.2581
BD 0.25 1000 853.33 0.3 76.800 512.000 -0.1364 0.1636
CD 0.5 700 18.667 -0.3 -1.6800 11.200 -0.1364 -0.4364
Total, 73.52 539.2
1364.02.539
52.732
1
1
Q
hhQ
f
f
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Solution
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Loop Trial 2
1
pipe d(m) L(m) k=fL/3d5 Q2 hf=kQ22 2hf/Q2 Q2 Q= Q2+Q2
AB 0.5 1000 26.666 0.1217 0.3946 6.4880 -0.0810 0.0407
AC 0.75 1200 4.2140 -0.3783 -0.6031 3.1883 -0.0810 -0.4593
BC 0.5 1500 40.00 0.2581 2.6440 20.6440 -0.0283 0.2297
Total, 2.4542 30.3164
0810.03164.304542.2
2
2
2
Q
hhQ
f
f
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Solution
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Loop Trial 2
2
pipe d(m) L(m) k=fL/3d5 Q2 hf=kQ22 2hf/Q2 Q2 Q= Q2+Q2
BC 0.5 1500 40.000 -0.2581 -2.6636 20.6440 0.0284 -0.2297
BD 0.25 1000 853.33 0.1636 22.8534 279.2957 -0.0526 0.1110
CD 0.5 700 18.667 -0.4364 -3.5550 16.2923 -0.0526 -0.4890
Total, 16.6348 316.2320
0526.02320.3166348.16
2
2
2
Q
hhQ
f
f
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Solution
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- Result: (Check the continuity at every junction),
Qin = Qout)
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2. Balancing Flow Method
When pressure head at a point in pipe network is known.
But the flow rates/discharges and pressure head at other
points in the network are to be determined
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Balancing Head Method
(iv) Calculate value for each pipe then determine
the value of
(v) For correction purposes, calculate:
(vi) Recalculate the value of new head at J:
(vii) Repeat the procedures until
f
f
nh
Q
Q
h
fjj hHH
fnh
Q
fnh
Q
0 fh
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Using Darcy-Weisbach approximations, computeand sketch the flow distribution in the pipe network
below:
Example 3
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Solution
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Equations to be used:
(i) hf= difference of level between reservoir and junction J
(ii) Darcy-Weisbach eqn: where
(iii)
nf
k
hQ
1
2,
35
nd
fLk
f
f
nh
Q
Qh
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Solution
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(iv)
- Based on Principle of Flow
10 m < Hj < 50 m
- Assumption: Flow from A to B, C and D30 m < Hj < 50 m
Assume Hj = 35 m
Construct the Table!!!
fjj hHH
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Solution
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Trial 1
Hj = 35 m
pipe d(m) L(m) k=fL/3d5 hf1 Q1=(hf/k)1/2 Q1/(2hf1)
AJ 0.6 500 10.7167 15 1.1813 0.0394
BJ 0.3 1000 685.871 -5 -0.0584 0.0085
CJ 0.3 1500 1028.806 -25 -0.1559 0.0031
DJ 0.45 3000 270.9614 -15 -0.2353 0.0078
Total, 0.7065 0.0589
9884.110589.0
7065.0
1
f
h
9884.46)9884.11(35 J
H
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Solution
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Trial 2
Hj = 46.9884 m
pipe d(m) L(m) k=fL/3d5 hf2 Q2=(hf2/k)1/2 Q2/(2hf2)
AJ 0.6 500 10.7167 3.0116 0.5301 0.0880
BJ 0.3 1000 685.871 -16.9884 -0.1574 0.0046
CJ 0.3 1500 1028.806 -36.9984 -0.1898 0.0026
DJ 0.45 3000 270.9614 -26.9884 -0.3156 0.0058
Total, -0.1325 0.1011
3110.11011.0
)1325.0(
2
f
h
6774.453110.19884.46 JH
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Solution
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- Results (from table): after 4 trials
Check: At junction J:
(OK) Hj = 45.56 m
smQAJ /6436.03
,0Q
3070.01856.01506.06436.0 Q
smQBJ /1506.03
smQCJ /1856.03
smQDJ /3070.0
3
,0Q
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Solution
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The End