1. The document discusses principles of pipe flow, including siphon action where a pipeline rises above the hydraulic gradient line. It provides equations to calculate head loss due to friction in pipes. 2. An example problem is presented to calculate residual pressure at the end of a pipe outlet for a pumping system with different pipe fittings, applying equations for head loss calculations. 3. Common pipe flow problems like nodal head, discharge, and diameter problems are introduced and equations are provided to solve each type of problem.

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

2. Pipe Flow Under Siphon Action
A pipeline that rises above its hydraulic gradient line is termed a siphon
the pipeline between the points b and c
crosses a ridge at point e.
If the pipe is long, head loss due to friction is large and the form losses can be
neglected. Thus, the hydraulic gradient line is a straight line that joins the
water surfaces at points A and B.
Pipe Flow Under Siphon Action

3. Pipe Flow Under Siphon Action
The pressure head at any section of the pipe is represented by the vertical
distance between the hydraulic gradient line and the centerline of the pipe.
If the hydraulic gradient line is above the centerline of pipe:
the water pressure is above atmospheric.
if it is below the centerline of the pipe:
the pressure is below atmospheric.
Pipe Flow Under Siphon Action

4. Pipe Flow Under Siphon Action
From Fig. 2.14 points b and c, the water pressure is atmospheric.
whereas between b and c it is less than atmospheric.
At the highest point e, the water pressure is the lowest.
If the pressure head at point e is less than - 2.5 m,
the water starts vaporizing and causes the flow to stop.
Thus, no part of the pipeline should be more than
2.5 m above the hydraulic gradient line.
Pipe Flow Under Siphon Action

5. Example 2.2A.
A pumping system with different pipe fittings is shown in Fig. 2.15. Calculate
residual pressure head at the end of the pipe outlet if the pump is generating an
input head of 50 m at 0.1 m3
/s discharge. The CI pipe diameter 𝐷 is 0.3 m. The
contraction size at point 3 is 0.15 m; pipe size between points 6 and 7 is 0.15 m;
and confusor outlet size 𝑑 = 0.15 m. The rotary valve at point 5 is fully open.
Consider the following pipe lengths between points:
Points 1 and 2 = 100 m, points 2 and 3 = 0.5 m; and points 3 and 4 = 0.5 m
Points 4 and 6 = 400 m, points 6 and 7 = 20 m; and points 7 and 8 = 100 m

6. Solution
Head loss between points 1 and 2.
Pipe length 100 m, flow 0.1 m3
/s, and pipe diameter 0.3 m.
Using Eq. (2.4b), 𝑣 for 20∘C is 1.012 × 10−6 m2/s, similarly using Eq. (2.4c),
Reynolds number 𝐑 = 419,459. Using Table 2.1 for CI pipes, 𝜀 is 0.25 𝐦𝐦. The
friction factor 𝑓 is calculated using Eq. (2.6b) = 0.0197. Using Eq. 2.3b the head
loss ℎ𝑓12 in pipe (1 − 2) is
𝒉𝒇𝟏𝟐 =
𝟖𝒇𝑳𝑸𝟐
𝝅𝟐𝒈𝑫𝟓
=
𝟖 × 𝟎. 𝟎𝟏𝟗𝟕 × 𝟏𝟎𝟎 × 𝟎. 𝟏𝟐
𝟑. 𝟏𝟒𝟏𝟓𝟗𝟐 × 𝟗. 𝟖𝟏 × 𝟎. 𝟑𝟓
= 𝟎. 𝟔𝟕𝟎 𝐦
Example 2.2A continued

7. 2. Head loss between points 2 and 3 (a contraction transition).
For 𝐷 = 0.3, 𝑑 = 0.15, and transition length = 0.5 m, the contraction angle 𝛼𝑐
can be calculated using Eq. (2.13b):
𝛼𝑐 = 2tan−1
𝐷1 − 𝐷2
2𝐿
= 2tan−1
0.3 − 0.15
2 × 0.5
= 0.298 radians .
Using Eq. (2.13a), the form-loss coefficient is
𝑘𝑓 = 0.315𝛼𝑐
1/3
= 0.315 × 0.2981/3 = 0.210
Using Eq. (2.12b), the head loss ℎ𝑚23 = 0.193 m.
Example 2.2A continued

8. 3. Head loss between points 3 and 4 (an expansion transition).
For 𝑑 = 0.15, 𝐷 = 0.3, the expansion ratio 𝑟 = 2, and transition length = 0.5 m.
Using Eq. (2.13d), the expansion angle 𝛼𝑒 = 0.298 radians. Using Eq. (2.13c), the
form-loss coefficient = 0.716. Using Eq. (2.12b), the head loss ℎ𝑚34 = 0.657 m.
4. Headloss between points 4 and 6.
Using Eq. (2.4c), with 𝜈 = 1.012 × 10−6
m2
/s, diameter 0.3, and discharge 0.1 m3
/s,
the Reynolds number = 419,459. With 𝜀 = 0.25 mm using Eq. (2.6b), 𝑓 = 0.0197.
Thus, for pipe length 400 m, using Eq. (2.3b), head loss ℎ𝑓 = 2.681 m.
Example 2.2A continued

9. Example 2.2A continued
5. Head loss at point 5 due to rotary valve (fully open).
For fully open valve 𝛼 = 0. Using Eq. (2.11), form-loss coefficient 𝑘𝑓 = 0 and using
Eq. (2.7b), the form loss ℎ𝑚 = 0.0 m.
6. Head loss at point 6 due to abrupt contraction.
For 𝐷 = 0.3 m and 𝑑 = 0.15 m, using Eq. (2.14b), the form-loss coefficient
𝑘𝑓 = 0.5 1 −
0.15
0.3
2.35
= 0.402.
Using Eq. (2.12b), the form loss ℎ𝑚 = 0.369 m.

10. Example 2.2A continued
7. Head loss in pipe between points 6 and 7.
Pipe length = 20 m, pipe diameter = 0.15 m, and roughness height = 0.25 mm.
Reynolds number = 838,914 and pipe friction factor 𝑓 = 0.0227, head loss ℎ𝑓67 =
4.930 m.
8. Head loss at point 7 (an abrupt expansion).
An abrupt expansion from 0.15 m pipe size to 0.30 m.
Using Eq. (2.14a), 𝑘𝑓 = 1 and using Eq. (2.12b), ℎ𝑚 = 0.918 m.

11. Example 2.2A continued
Head loss in pipe between points 7 and 8.
Pipe length = 100 m, pipe diameter = 0.30 m, and roughness height = 0.25 mm.
Reynolds number = 423,144 and pipe friction factor 𝑓 = 0.0197.
Head loss ℎ𝑓78 = 0.670 m.
Head loss at outlet point 8 (confusor outlet).
Using Eq. (2.17), the form-loss coefficient
𝑘𝑓
= 4.5
𝐷
𝑑
− 3.5 = 4.5 ×
0.30
0.15
− 3.5 = 5.5.
Using Eq. (2.12 b), ℎ𝑚 = 0.560 m.
Total head loss hL = 0.670 + 0.193 + 0.657 + 2.681 + 0.369 + 0 + 4.930 + 0.918 +
0.670 + 0.560 = 11.648 m: Thus, the residual pressure at the end of the pipe outlet =
50 - 11.648 = 38.352 m

12. Example 2.2B.
Design an cxpansion for the pipc diametcrs 1.0 m
and 2.0 m over a distance of 2 m for Fig. 2.9.
Solution
Equation (2.13e) is used for the calculation of
optimal transition profile.
The geometry profile is 𝐷1 = 1.0 m, 𝐷2 = 2.0 m,
and 𝐿 = 2.0 m.
Substituting various values of 𝑥, the corresponding
values of 𝐷 using Eq.
(2.13e) and with linear expansion were computed
and are tabulated in Table 2.3.
𝑥 𝐷 (optimal) 𝐷 (linear)
0.0 1.000 1.000
0.2 1.019 1.100
0.4 1.078 1.200
0.6 1.180 1.300
0.8 1.326 1.400
1.0 1.500 1.500
1.2 1.674 1.600
1.4 1.820 1.700
1.6 1.922 1.800
1.8 1.981 1.900
2.0 2.000 2.000
TABLE 2.3. Pipe Transition
Computations 𝒙 versus 𝑫

13. 2.3. PIPE FLOW PROBLEMS
Nodal Head Problem
Discharge Problem
Diameter Problem
Analysis problem
Design and ansysis problem
Analysis problem

14. 2.3.1.Nodal Head Problem
In the nodal head problem, the known quantities
are 𝐿, 𝐷, ℎ𝐿, 𝑄, 𝜀, 𝑣, and 𝑘𝑓. Using Eqs. (2.2b) and
(2.7b), the nodal head ℎ2 (as shown in Fig. 2.1) is
obtained as
𝒉𝟐 = 𝒉𝟏 + 𝒛𝟏 − 𝒛𝟐 − 𝒌𝒇 +
𝒇𝑳
𝑫
𝟖𝑸𝟐
𝝅𝟐𝒈𝑫𝟒
.
(2:20)

15. 2.3.2. Discharge Problem
For a long pipeline, form losses can be neglected. Thus, in this case the known
quantities are 𝐿, 𝐷, ℎ𝑓, 𝜀, and 𝜈. Swamee and Jain (1976) gave the following
solution for turbulent flow through such a pipeline:
Equation (2.21a) is exact. For
laminar flow, the Hagen-Poiseuille
equation gives the discharge as
𝑸 = −𝟎. 𝟗𝟔𝟓𝑫𝟐 𝒈𝑫 𝒉𝒇 𝑳 𝐥 𝒏
𝜺
𝟑. 𝟕𝑫
+
𝟏. 𝟕𝟖𝒗
𝑫 𝒈𝑫 𝒉𝒇 𝑳
𝑸 =
𝝅𝒈𝑫𝟒
𝒉𝒇
𝟏𝟐𝟖𝒗𝑳
.
(2:21a)
(2:21b)
For laminar
flow

16. Swamee and Swamee (2008) gave the
following equation for pipe discharge that
is valid under laminar, transition,
and turbulent flow conditions:
𝑸 = 𝑫𝟐
𝒈𝑫 𝒉𝒇 𝑳
𝟏𝟐𝟖𝒗
𝝅𝑫 𝒈𝑫 𝒉𝒇 𝑳
𝟒
𝟏 +𝟏. 𝟏𝟓𝟑
𝟒𝟏𝟓𝒗
𝑫 𝒈𝑫 𝒉𝒇 𝑳
𝟖
− 𝐥 𝒏
𝜺
𝟑. 𝟕𝑫
+
𝟏. 𝟕𝟕𝟓𝒗
𝑫 𝒈𝑫 𝒉𝒇 𝑳
−𝟒 −𝟎.𝟐𝟓
Equation (2.21c) is almost exact as the maximum error in the equation is 0.1%.
𝑸 =
(2:21c)

17. 2.3.3. Diameter Problem
In this problem. the known quantities are L.h_f,ε,Q. and v. For a pumping main,
head loss is not known, and one has to select the optimal value of head loss
by minimizing the cost. This has been dealt with in Chapter 6. However, for
turbulent flow in a long gravity main, Swamee and Jain (1976) obtained the
following solution for the pipe diameter:
𝑫 = 𝟎. 𝟔𝟔 𝜺𝟏.𝟐𝟓
𝑳𝑸𝟐
𝒈𝒉𝒇
𝟒.𝟕𝟓
+ 𝒗𝑸𝟗.𝟒
𝑳
𝒈𝒉𝒇
𝟓.𝟐 𝟎.𝟎𝟒
(2:22a)

18. In general, the errors involved in Eq. (2.22a) are less than 1.5%. However, the
maximum error occurring near transition range is about 3%. For laminar flow, the
Hagen Poiseuille equation gives the diameter as
𝑫 =
𝟏𝟐𝟖𝒗𝑸𝑳
𝝅𝒈𝒉𝒇
𝟎.𝟐𝟖
(2:22b)
2.3.3. Diameter Problem

19. Swamee and Swamee (2008) gave the following equation for pipe diameter that
is valid under laminar, transition, and turbulent flow conditions
Equation (2.22c) yields 𝐷 within 2.75%. However, close to transition range,
the error is around 4%.
𝑫 = 𝟎. 𝟔𝟔 𝟐𝟏𝟒. 𝟕𝟓
𝒗𝑳𝑸
𝒈𝒉𝒇
𝟔.𝟐𝟓
+ 𝜺𝟏.𝟐𝟓
𝑳𝑸𝟐
𝒈𝒉𝒇
𝟒.𝟕𝟓
+ 𝒗𝑸𝟗.𝟒
𝑳
𝒈𝒉𝒇
𝟓.𝟐 𝟎.𝟎𝟒
(2:22c)

20. Example 2.3.
As shown in Fig. 2.16, a discharge of 0.1 m3 /s flows through a CI pipe main of 1000 m in
length having a pipe diameter 0.3 m.
A sluice valve of 0.3 m size is placed close to point B.
The elevations of points A and B are 10 m and 5 m, respectively. Assume water
temperature as 20C. Calculate:
(A) Terminal pressure h2 at point B and head loss in the pipe if terminal
pressure h1 at point A is 25 m.
(B) The discharge in the pipe if the head loss is 10 m.
(C) The CI gravity main diameter if the head
loss in the pipe is 10 m and a
discharge of 0.1 m3 /s flows in the pipe.

22. solution
A) The terminal pressure h2 at point B can be calculated using Eq. (2.20). The
friction factor f can be calculated applying Eq. (2.6a) and the roughness height of
CI pipe ¼ 0.25 mm is obtained from Table 2.1. The form-loss coefficient for
sluice valve from Table 2.2 is 0.15. The viscosity of water at 208C can be
calculated using Eq. (2.4b). The coefficient of surface resistance depends on the
Reynolds number R of the flow:
𝑹 =
𝟒𝑸
𝒎𝒗𝑫
= 𝟒𝟏𝟗, 𝟒𝟓𝟗.
Thus, substituting values in Eq. (2.6a), the friction factor
𝒇 =
𝟔𝟒
𝐑
𝟖
+ 𝟗. 𝟓 𝐥 𝒏
𝜺
𝟑. 𝟕𝑫
+
𝟓. 𝟕𝟒
𝐑𝟎.𝟗
−
𝟐𝟓𝟎𝟎
𝐑
𝟔 −𝟏𝟔 𝟎.𝟏𝟐𝟓
= 𝟎. 𝟎𝟏𝟗𝟕

23. solution
Using Eq- (2.20), the terminal head ℎ2 at point 𝐵 is
𝒉𝟐 = 𝒉𝟏 + 𝒛𝟏 − 𝒛𝟐 − 𝒌𝒇 +
𝒇𝑳
𝑫
𝟖𝑸𝟐
𝝅𝟐𝒈𝑫𝟒
𝒉𝟐 = 𝟐𝟓 + 𝟏𝟎 − 𝟓 − 𝟎. 𝟏𝟓 +
𝟎. 𝟎𝟏𝟗𝟕 × 𝟏𝟎𝟎𝟎
𝟎. 𝟑
𝟖 × 𝟎. 𝟏𝟐
𝟑. 𝟏𝟒𝟏𝟓𝟗𝟐 × 𝟗. 𝟖𝟏 × 𝟎. 𝟑𝟓
𝒉𝟐 =
= 𝟑𝟎 − (𝟎. 𝟎𝟏𝟓 + 𝟔. 𝟕𝟎𝟒) = 𝟐𝟑. 𝟐𝟖𝟏 𝐦.

24. = 𝑸 = −𝟎. 𝟗𝟔𝟓𝑫𝟐
𝒈𝑫 𝒉𝒇 𝑳 𝐥 𝒏
𝜺
𝟑. 𝟕𝑫
+
𝟏. 𝟕𝟖𝒗
𝑫 𝒈𝑫𝒉 𝑳
= = −𝟎. 𝟗𝟔𝟓 × 𝟎. 𝟑𝟐 𝟗. 𝟖𝟏 × (𝟏𝟎 𝟏𝟎𝟎𝟎 𝐥 𝒏
𝟎. 𝟐𝟓 × 𝟏𝟎−𝟑
𝟑. 𝟕 × 𝟎. 𝟑
. =
= = 𝟎. 𝟏𝟐𝟑 𝒎𝟑
𝒔
solution
(B) If the total head loss in the pipe is predecided equal to 10 m, the discharge in
Cl pipe of size 0.3 m can be calculated using Eq. ( 2.21a ):
+
𝟏. 𝟕𝟖 × 𝟏. 𝟎𝟏𝟐 × 𝟏𝟎−𝟔
𝟎. 𝟑 )
𝟗. 𝟖𝟏 × 𝟎. 𝟑 × (𝟏𝟎 𝟏𝟎𝟎𝟎

25. solution
(C) Using Eq. (2.22a), the gravity main diameter for preselected head loss of 10 m
and known pipe discharge 0.1 m3
/s is
𝑫 = 𝟎. 𝟔𝟔 𝜺𝟏.𝟐𝟓
𝑳𝑸𝟐
𝒈𝒉𝒇
𝟒.𝟕𝟓
+ 𝒗𝑸𝟗.𝟒
𝑳
𝒈𝒉𝒇
𝟓.𝟐 𝟎.𝟎𝟒
= 𝟎. 𝟔𝟔 𝟎. 𝟎𝟎𝟎𝟐𝟓𝟏.𝟐𝟓
𝟏𝟎𝟎𝟎 × 𝟎. 𝟏𝟐
𝟗. 𝟖𝟏 × 𝟏𝟎
𝟒.𝟕𝟓
+ 𝟏. 𝟎𝟏𝟐 × 𝟏𝟎−𝟔
= × 𝟎. 𝟏𝟗.𝟒
𝟏𝟎𝟎𝟎
𝟗. 𝟖𝟏 × 𝟏𝟎
𝟓.𝟐 𝟎.𝟎𝟒
= =
Also, if head loss is considered = 6.72 m, the pipe diameter is 0.306 m and flow is
0.1 m3
/s.
= 𝟎. 𝟐𝟖𝟒 𝐦.