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

2. Chapter 3: PIPE NETWORK ANALYSIS
Houses are connected through service connections to water distribution
network pipelines for water supply. From these connections, water is
drawn as any of the water taps in a house opens, and the withdrawal
stops as the tap closes. Generally, there are many taps in a house, thus
the withdrawal rate varies in an arbitrary manner.

3. The maximum withdrawal rates occur in morning and evening
hours. The maximum discharge (withdrawal rate) in a pipe is a
function of the number of houses (persons). In the analysis and
design of a pipe ne
The service connections are taken at arbitrary spacing from a pipeline of a
water supply network (Fig. 3.1a).
It is not easy to analyze such a network unless simplifying assumptions are
made.
A conservative assumption is to consider the
withdrawals to be lumped at the two end points of the
pipe link. A more realistic assumption is to consider
the withdrawals to be distributed along the link..

4. Pipe Flow Under Siphon Action
(2:23)
3.2. HEAD LOSS IN A PIPE LINK
(3:1a)
3.2.1. Head Loss in a Lumped Equivalent
Considering q to be the withdrawal rate per unit length of a link, the total
withdrawal rate from the pipe of length L is qL. Lumping the discharges at
the two pipe link ends, the head loss on account of surface resistance is
given by
𝒉𝒇 =
𝟖𝒇𝑳𝑸𝟐
𝝅𝟐𝒈𝑫𝟓
𝟏 −
𝒒𝑳
𝟐𝑸
𝟐
where Q = discharge entering the link.
Equation (2.6b) can be used for calculation of f;
where Reynolds number R is to be taken as
𝑹 =
𝟒(𝑸 − 𝟎. 𝟓𝒒𝑳
𝝅𝝂𝑫 (3:1b)

5. 2.4.1. Pipes in Series
3.2.2. Head Loss in a Distributed Equivalent
The discharge at a distance x from the pipe link entrance end is Q - qx, and
the corresponding head loss in a distance dx is given by
𝒅𝒉𝒇 =
𝟖𝒇(𝑸 − 𝒒𝒙
𝟐
𝒅𝒙
𝝅𝟐𝒈𝑫𝟓
(3:2)
𝒉𝒇 =
𝟖𝒇𝑳𝑸𝟐
𝝅𝟐𝒈𝑫𝟓
𝟏 −
𝒒𝑳
𝑸
+
𝟏
𝟑
𝒒𝑳
𝑸
𝟐
Integrating Eq. (3.2) between the limits x = 0 and L, the following equation is
obtained:
(3:3)
For the calculation of f, R can be obtained by Eq. (3.1b).

6. Example 3.1.
Calculate head loss in a CI pipe of length L = 500 m, discharge Q at
entry node = 0.1m3/s, pipe diameter D = 0.25 m if the withdrawal (Fig.
3.1) is at a rate of 0.0001m3/s per meter length. Assume
(a) lumped idealized withdrawal and
(b) distributed idealized withdrawal patterns.
Solution. Using Table 2.1 and Eq. (2.4b), roughness height 1 of CI pipe =
0.25 mm and kinematic viscosity n of water at 208C = 1.0118 x10^6 m2/s.
(a) Lumped idealized withdrawal (Fig. 3.1b): Applying Eq. (3.2),
𝑹 =
𝟒(𝑸 − 𝟎. 𝟓𝒒𝑳
𝝅𝝂𝑫
=
𝟒(𝟎. 𝟏 − 𝟎. 𝟓 × 𝟎. 𝟎𝟎𝟎𝟏 × 𝟓𝟎𝟎
𝟑. 𝟏𝟒𝟏𝟓 × 𝟏. 𝟎𝟏𝟏𝟖𝟐 × 𝟏𝟎−𝟔 × 𝟎. 𝟐𝟓
= 𝟑𝟕𝟕, 𝟓𝟏𝟑

9. 3.3. ANALYSIS OF WATER
TRANSMISSION LINES
Water transmission lines are long
pipelines having no withdrawals. If water is carried by
gravity, it is called a gravity main (see Fig. 3.2).
In the analysis of a gravity main, it is
required to find the discharge carried
by the pipeline. The available head in
the gravity main is h0 & z0 - zL, and
almost the entire head is lost in
surface resistance.

10. 3.3. ANALYSIS OF WATER
TRANSMISSION LINES
where f is given by Eq. (2.6b). It is difficult to solve Eq. (2.6a) and Eq. (3.4),
however,using Eq. (2.21a) the discharge is obtained as:
𝒉0 + 𝒛0 − 𝒛𝑳 =
8𝒇𝑳𝑸2
𝝅2𝒈𝑫5
(3:4)
𝑸 = −0.965𝑫2
𝒈𝑫 𝒉0 + 𝒛0 − 𝒛𝑳
𝑳
0.5
𝐥 𝐧
𝜺
3.7𝑫
+
1.78𝒗
𝑫
𝑳
𝒈𝑫 𝒉0 + 𝒛0 − 𝒛𝑳
0.5
(3:5)

11. 𝒉𝟎 = 𝑯 + 𝒛𝑳 − 𝒛𝟎 + 𝒌𝒇 +
𝒇𝑳
𝑫
𝟖𝑸𝟐
𝝅𝟐𝒈𝑫𝟒
If water is pumped from an elevation
z0 to zL, the pipeline is called a pumping main
(Fig. 3.3). In the analysis of a pumping main,
one is required to find the pumping head h0
for a given discharge Q.
This can be done by a combination of Eqs. (2.2b), (2.2d), and (2.19b). That is,
(3:6)
where H = the terminal head (i.e., the head at x = L.)
Neglecting the form loss for a long pumping main,
Eq. (3.6) reduces to
𝒉0 = 𝑯 + 𝒛𝑳 − 𝒛0 +
8𝒇𝑳𝑸2
𝝅2𝒈𝑫5 (3:7)

12. Example 3.2.
For a polyvinyl chloride (PVC) gravity main (Fig. 3.2), calculate flow in a
pipe of length 600 m and size 0.3 m. The elevations of reservoir and
outlet are 20 m and 10 m, respectively. The water column in reservoir is
5 m, and a terminal head of 5 m is required at outlet.
Solution. At 20∘C, 𝑣 = 1.012 × 10−6; and from Table 2.1, 𝜀 = 0.05 mm.
With 𝐿 = 600 m, ℎ0 = 5 m, 𝑧𝑜 = 20 m, 𝑧𝐿 = 10 m, and 𝐷 = 0.3 m Eq. (3.5)
gives
𝑸 = −𝟎. 𝟗𝟔𝟓𝑫𝟐
𝒈𝑫 𝒉𝟎 + 𝒛𝟎 − 𝒛𝑳
𝑳
𝟎.𝟓
𝐥 𝒏
𝜺
𝟑. 𝟕𝑫
+
𝟏. 𝟕𝟖𝒗
𝑫
𝑳
𝒈𝑫 𝒉𝟎 + 𝒛𝟎 − 𝒛𝑳
𝟎.𝟓
𝑸 = 𝟎. 𝟐𝟐𝟕 𝒎𝟑 𝒔