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1. BASIC PRINCIPLES
OF PIPE FLOW
Dr/ Ahmed safwat
Amir Ashraf sayed

2. (2:23)
(3:8)
3.4. ANALYSIS OF DISTRIBUTION MAINS
𝑸𝒋 = ෍ 𝒒𝒏−𝒑
𝒑=𝟎
𝒋
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
𝒉𝒋 = 𝒉0 + 𝒛0 − 𝒛𝒊 −
8
𝝅2𝒈
෎
𝒇𝒑𝑳𝒑
𝑫𝒑
+ 𝒌𝒇𝒑
𝑸𝒑
2
𝑫𝒑
4
𝒑=1
𝒋
The value of f for pth pipe link is given by
(2:23)
𝒇𝒑 = 1.325 𝐥 𝐧
𝜺
3.7𝑫
+ 4.618 +
𝒗𝑫𝒑
𝑸𝒑
0.9 −2
(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

7. 3.6. ANALYSIS OF BRANCHED NETWORKS
𝑸𝒊𝒋 = ෍ 𝒒𝒊. 𝒏−𝒑.
𝒑=𝒐
𝒋
(3:11)
A branched network, or a tree network, is a distribution
system having no loops.
𝑸𝑻 = ෍ 𝑸𝒐𝒊
𝒊=1
𝒊𝑳
(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. ∑loop 𝒌𝑲𝒊𝑸𝒊 𝑸𝒊 = 𝟎 for all loops 𝒌 = 𝟏. 𝟐. 𝟑. … . 𝒌𝑳
(3:13)
(3:14)
3.7.1. Hardy Cross Method
1. The sum of inflow and outflow at a node should be equal:
∑𝒊𝑸𝒊 = 𝒒𝒋 for all nodes 𝒋 = 𝟏. 𝟐. 𝟑. … . 𝒋𝑳
2. The algebraic sum of the head loss in a loop must be equal to zero
∑loop 𝒌𝑲𝒊𝑸𝒊 𝑸𝒊 = 𝟎 for all loops 𝒌 = 𝟏. 𝟐. 𝟑. … . 𝒌𝑳
where
𝑲𝒊 =
8𝒇𝒊𝑳𝒊
𝝅2𝒈𝑫𝒊
5
(3:15)

10. ∑loop 𝒌𝑲𝒊𝑸𝒊 𝑸𝒊 = 𝟎 for all loops 𝒌
= 𝟏. 𝟐. 𝟑. … . 𝒌𝑳
(3:16)
3.7.1. Hardy Cross Method
The modified pipe discharges are determined by applying a correction ΔQk to the
initially assumed pipe flows. Thus,
∑loop 𝒌𝑲𝒊 𝑸𝒊 + 𝚫𝑸𝒌 𝑸𝒊 + 𝚫𝑸𝒌 = 𝟎
𝜟𝑸𝒌 = −
∑
loop 𝒌
𝑲𝒊𝑸𝒊 𝑸𝒊
𝟐∑
loop 𝒌
𝑲𝒊 𝑸𝒊 (3:17)
𝑸new = 𝑸iold + 𝚫𝑸𝒌 for all 𝒌. (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 Darcy–
Weisbach head loss–discharge 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:

13. Solution
applying continuity equation at node 3
Q3 = 20.5 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:

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