pipe lines lecture ssssssssssssssssssssssssssssssssss cccccc f'wp oww kdw'eow e'fkwe ek ep ks'f' ;fakw 'k ap' ok wpo fkwa kw wapeok w ep owp oe pwoekwepokaew oe we po p aw weof aw oap oawepo kawpokawep k p wepo pokwa po kpofwe ep okwep okfw p weokweapfo kaw epofea po paofk awpe o[e po powpo kawep fowa ak fweofkaw pfokaw fpao ew we p awp ofkwp poawekp[awekofwe pewpo[ fpowef aw f[ powekee [pfodfk kfok' lro powekwepo;l dkpdoko;lfpowe k;lpeo 'lk fpfkoek'd;lfke ;'oeowfkef'kfekpfks;'fkf'aowek'fkepok
2. (2:23)
(3:8)
3.4. ANALYSIS OF DISTRIBUTION MAINS
๐ธ๐ = เท ๐๐โ๐
๐=๐
๐
A pipeline in which there are multiple withdrawals is
called a distribution main.
The discharge flowing in the jth pipe link Qj is given by
3. (2:23)
(3:10)
3.4. ANALYSIS OF DISTRIBUTION MAINS
the nodal head hj is given by
๐๐ = ๐0 + ๐0 โ ๐๐ โ
8
๐ 2๐
เท
๐๐๐ณ๐
๐ซ๐
+ ๐๐๐
๐ธ๐
2
๐ซ๐
4
๐=1
๐
The value of f for pth pipe link is given by
(2:23)
๐๐ = 1.325 ๐ฅ ๐ง
๐บ
3.7๐ซ
+ 4.618 +
๐๐ซ๐
๐ธ๐
0.9 โ2
(3:9)
4. 3.5. PIPE NETWORK GEOMETRY
The water distribution networks have mainly the
following three types of configurations:
1- Branched or tree-like configuration
2- Looped configuration
3- Branched and looped configuration
5.
6.
7. 3.6. ANALYSIS OF BRANCHED NETWORKS
๐ธ๐๐ = เท ๐๐. ๐โ๐.
๐=๐
๐
(3:11)
A branched network, or a tree network, is a distribution
system having no loops.
๐ธ๐ป = เท ๐ธ๐๐
๐=1
๐๐ณ
(3:12)
8. 3.7. ANALYSIS OF LOOPED NETWORKS
A pipe network in which there are one or more closed loops
is called a looped network.
Kirchhoff equations:
1- The algebraic sum of inflow and outflow discharges at
a node is zero; and
2- The algebraic sum of the head loss around a loop is
zero.
9. โloop ๐๐ฒ๐๐ธ๐ ๐ธ๐ = ๐ for all loops ๐ = ๐. ๐. ๐. โฆ . ๐๐ณ
(3:13)
(3:14)
3.7.1. Hardy Cross Method
1. The sum of inflow and outflow at a node should be equal:
โ๐๐ธ๐ = ๐๐ for all nodes ๐ = ๐. ๐. ๐. โฆ . ๐๐ณ
2. The algebraic sum of the head loss in a loop must be equal to zero
โloop ๐๐ฒ๐๐ธ๐ ๐ธ๐ = ๐ for all loops ๐ = ๐. ๐. ๐. โฆ . ๐๐ณ
where
๐ฒ๐ =
8๐๐๐ณ๐
๐ 2๐๐ซ๐
5
(3:15)
10. โloop ๐๐ฒ๐๐ธ๐ ๐ธ๐ = ๐ for all loops ๐
= ๐. ๐. ๐. โฆ . ๐๐ณ
(3:16)
3.7.1. Hardy Cross Method
The modified pipe discharges are determined by applying a correction ฮQk to the
initially assumed pipe flows. Thus,
โloop ๐๐ฒ๐ ๐ธ๐ + ๐ซ๐ธ๐ ๐ธ๐ + ๐ซ๐ธ๐ = ๐
๐๐ธ๐ = โ
โ
loop ๐
๐ฒ๐๐ธ๐ ๐ธ๐
๐โ
loop ๐
๐ฒ๐ ๐ธ๐ (3:17)
๐ธnew = ๐ธiold + ๐ซ๐ธ๐ for all ๐. (3:18)
11. Example 3.3.
A single looped network as shown in Fig. 3.10 has to be analyzed by the
Hardy Cross method for given inflow and outflow discharges.Use Darcyโ
Weisbach head lossโdischarge relationship assuming a constant friction
factor f = 0.02.
12. Solution
Step 1: The pipes, nodes, and loop are numbered
Step 2: Inflow into a node is positive withdrawal negative.
Step 3: Apply continuity equation to obtain pipe discharges.
Assume an arbitrary flow of 0.1 m3/s in pipe 1 (Q1 = 0.1 m3/s),
Q1 + Q4 = q1 or Q4 = q1 ,Q1, hence Q4 = 0:6 - 0:1 =0:5 m3/s:
Also applying continuity equation at node 2:
- Q1 + Q2 = q2 or Q2 = q2 + Q1, hence Q2 = 0 + 0:1 = 0:1 m3/s:
14. Solution
Repeat the process again for the revised pipe discharges as the discharge
correction is quite large in comparison to pipe flows:
Q1 = 0.3 m3/s
Q2 =0.3 m3/s
Q3 = 0.3 m3/s
Q4 = 0.3 m3/s
15. Example 3.4.
The pipe network of two loops as shown in Fig. 3.11 has to be analyzed
by the Hardy Cross method for pipe flows for given pipe lengths L and
pipe diameters D. The nodal inflow at node 1 and nodal outflow at node
3 are shown in the figure. Assume a constant friction factor f = 0.02.
16. Solution
Assume Q1 and calculating the other flow rates:
Q1 = 0.1 m3/s (flow from node 1 to node 2)
Q2 = 0.1 m3/s (flow from node 2 to node 3)
Q3 = 0.4 m3/s (flow from node 4 to node 3)
Q4 = 0.4 m3/s (flow from node 1 to node 4)
Q5 = 0.1 m3/s (flow from node 1 to node 3)
17. Solution
Thus the discharge correction DQ in loop 1 is 0.15 m3/s. The discharges in loop
pipes are corrected as shown in the above table. Applying the same
methodology for calculating DQ for Loop 2:
18.
19.
20.
21.
22. The discharge corrections in the loops are very small after five
iterations, thus the final pipe discharges in the looped pipe network in
Fig. 3.11 are
Q1 = 0.223 m3/s
Q2 = 0.223 m3/s
Q3 = 0.192 m3/s
Q4 = 0.192 m3/s
Q5 = 0.182 m3/s
Solution