hydraulic

1. Civil Engineering Department
Hydraulics and Hydrology – CE 352
Prof. Majed Abu-Zreig
Chapter 3
Water Flow in Pipes

2. 3.1 Description of A Pipe Flow
• Water pipes in our homes and the distribution
system
• Pipes carry hydraulic fluid to various components
of vehicles and machines
• Natural systems of “pipes” that carry blood
throughout our body and air into and out of our
lungs.
2

3. • Pipe Flow: refers to a full water flow in a closed
conduits or circular cross section under a certain
pressure gradient.
• The pipe flow at any cross section can be
described by:
 cross section (A),
 elevation (h), measured with respect to a horizontal
reference datum.
 pressure (P), varies from one point to another, for a
given cross section variation is neglected
 The flow velocity (v), v = Q/A.
3

4. Difference between open-channel flow and the pipe flow
Pipe flow Open-channel flow
• The pipe is completely filled • Water flows without
with the fluid being transported. completely filling the pipe.
• The main driving force is likely • Gravity alone is the
to be a pressure gradient along driving force, the water
the pipe. flows down a hill.
4

5. Types of Flow
 Steady and Unsteady flow
The flow parameters such as velocity (v), pressure (P)
and density (r) of a fluid flow are independent of time
in a steady flow. In unsteady flow they are independent.
For a steady flow v t x ,y ,z
o o o
0
For an unsteady flow v t x ,y ,z
o o o
0
If the variations in any fluid’s parameters are small, the
average is constant, then the fluid is considered to be
steady
5

6.  Uniform and non-uniform flow
A flow is uniform if the flow characteristics at any given
instant remain the same at different points in the
direction of flow, otherwise it is termed as non-uniform
flow.
For a uniform flow v s t o
0
For a non-uniform flow v s t o
0
6

7. Examples:
 The flow through a long uniform pipe diameter at a constant rate is
steady uniform flow.
 The flow through a long uniform pipe diameter at a varying rate is
unsteady uniform flow.
 The flow through a diverging pipe diameter at a constant rate is a
steady non-uniform flow.
 The flow through a diverging pipe diameter at a varying rate is an
unsteady non-uniform flow.
7

8.  Laminar and turbulent flow
Laminar flow:
The fluid particles move along smooth well defined path or streamlines
that are parallel, thus particles move in laminas or layers, smoothly
gliding over each other.
Turbulent flow:
The fluid particles do not move in orderly manner and they occupy different
relative positions in successive cross-sections.
There is a small fluctuation in magnitude and direction of the velocity of the
fluid particles
transitional flow
The flow occurs between laminar and turbulent flow
8

9. 3.2 Reynolds Experiment
Reynolds performed a very carefully prepared pipe flow
experiment.
9

10. Increasing
flow
velocity
10

11. Reynolds Experiment
• Reynold found that transition from laminar to turbulent
flow in a pipe depends not only on the velocity, but only
on the pipe diameter and the viscosity of the fluid.
• This relationship between these variables is commonly
known as Reynolds number (NR)
VDr VD
Inertial Forces
NR   
  Viscous Forces
It can be shown that the Reynolds number is a measure of
the ratio of the inertial forces to the viscous forces in the
flow
FI  ma FV   A 11

12. Reynolds number
rVD VD
NR  
 
where V: mean velocity in the pipe [L/T]
D: pipe diameter [L]
r: density of flowing fluid [M/L3]
: dynamic viscosity [M/LT]
: kinematic viscosity [L2/T]
12

13. 13

14. It has been found by many experiments that for flows in
circular pipes, the critical Reynolds number is about 2000
Flow laminar when NR < Critical NR
Flow turbulent when NR > Critical NR
The transition from laminar to turbulent flow does not always
happened at NR = 2000 but varies due to experiments
conditions….….this known as transitional range
14

15. Laminar Vs. Turbulent flows
Laminar flows characterized Turbulent flows characterized
by: by
• low velocities • high velocities
• small length scales • large length scales
• high kinematic viscosities • low kinematic viscosities
• NR < Critical NR • NR > Critical NR
• Viscous forces are • Inertial forces are
dominant. dominant
15

16. Example 3.1
40 mm diameter circular pipe carries water at 20oC.
Calculate the largest flow rate (Q) which laminar flow can
be expected.
D  0.04m
  1106 at T  20o C
VD V (0.04)
NR   2000   2000  V  0.05m / sec
 110 6

Q  V . A  0.05  (0.04) 2  6.28 105 m3 / sec
4
16

17. 3.3 Forces in Pipe Flow
• Cross section and elevation of the pipe are varied along
the axial direction of the flow.
17

18. For Incompressible and Steady flows:
Conservation law of mass
r.dVol11'  r.dVol22'  mass flux ( fluid mass)
Mass enters the Mass leaves the
control volume control volume
dVol11' dVol 2 2'
r.  r.
dt dt
dS1 dS 2
r . A1  r . A2  r . A1.V1  r . A2 .V2  r .Q
dt dt
Continuity equation for
Incompressible Steady flow A1.V1  A2 .V2  Q
18

19. Apply Newton’s Second Law:
  
 dV M V 2 M V1

 F  M a  M dt  t
F x  P A1  P2 A2  Fx  Wx
1
Fx is the axial direction force exerted on the control volume
by the wall of the pipe.
but M t  r .Q  mass flow rate
 Fx  r.Q(Vx2  Vx1 ) 
 F r.Q(V

2

V 1)
F y  r .Q(V y2  V y1 )
F
Conservation of
z  r .Q(Vz2  Vz1 ) moment equation
19

20. Example 3.2
dA= 40 mm, dB= 20 mm, PA= 500,000 N/m2, Q=0.01m3/sec.
Determine the reaction force at the hinge.
20

21. 3.4 Energy Head in Pipe Flow
Water flow in pipes may contain energy in three
basic forms:
1- Kinetic energy,
2- potential energy,
3- pressure energy.
21

22. Consider the control volume:
• In time interval dt:
- Water particles at sec.1-1 move to sec. 1`-1` with velocity V1.
- Water particles at sec.2-2 move to sec. 2`-2` with velocity V2.
• To satisfy continuity equation:
A1.V1.dt  A2 .V2 .dt
• The work done by the pressure force
P . A1.ds1  P . A1.V1.dt
1 1
……. on section 1-1
 P2 . A2 .ds2   P2 . A2 .V2 .dt ……. on section 2-2
-ve sign because P2 is in the opposite direction to distance traveled ds2
22

23. • The work done by the gravity force :
Work  W .h  mg.h
m  r .Volume  A1 L  A1V1dt
• The kinetic energy: Work  rg. A1.V1dt.(h1  h2 )
1 1 1
M .V2  M .V1  r. A1.V1.dt (V22  V1 )
2 2 2
2 2 2
The total work done by all forces is equal to the change in
kinetic energy:
1
P .Q.dt  P2 .Q.dt  rg.Q.dt.(h1  h2 )  r.Q.dt (V22  V1 )
2
1
2
Dividing both sides by rgQdt
2 2 Bernoulli Equation
V1 P V P
 1  h1  2  2  h2
2g  2g  Energy per unit weight of water
23
OR: Energy Head

24. Energy head and Head loss in pipe flow
24

25. 2
V2 P2
H2    h2
2g 
Energy = Kinetic + Pressure +
Elevation
head head head head
2
V P
H1  1  1  h1
2g 
Notice that:
• In reality, certain amount of energy loss (hL) occurs when the
water mass flow from one section to another.
• The energy relationship between two sections can be written
as: 2 2
V1 P V2 P2
  h1 
1
  h2  hL
2g  2g  25

26. Example 3.4
The tank is being drained through 12 in pipe. The discharge = 3200 gpm, The
26
Total head loss = 11.5 ft. find the h?

27. Example
In the figure shown:
Where the discharge through the system is 0.05 m3/s, the total losses through
the pipe is 10 v2/2g where v is the velocity of water in 0.15 m diameter pipe,
the water in the final outlet exposed to atmosphere.

28. Example
In the figure shown:
Where the discharge through the system is 0.05 m3/s, the total losses through the pipe is 10 v2/2g
where v is the velocity of water in 0.15 m diameter pipe, the water in the final outlet exposed to
atmosphere. Calculate the required height (h =?)
below the tank
Calculate the required
height (h =?)
below the tank
0.05
V Q
  2.83m / s
4 0.15
A  2
0.05
V Q
  6.366m / s
4 0.10 
A  2
p1 V12 p2 V22
  z1    z 2  hL
rg 2 g rg 2 g
0  0  (h  5)  0 
6.3662  20  102.832
2 * 9.81 2 * 9.81
h  21.147 m

29. Without calculation sketch the (E.G.L) and (H.G.L)

30. Basic components of a typical pipe
system
30

31. Calculation of Head (Energy) Losses:
In General:
When a fluid is flowing through a pipe, the fluid experiences some
resistance due to which some of energy (head) of fluid is lost.
Energy Losses
(Head losses)
Major Losses Minor losses
loss of head due to pipe Loss due to the change of
friction and to viscous the velocity of the flowing
dissipation in flowing fluid in the magnitude or in
water direction as it moves
through fitting like Valves,
Tees, Bends and Reducers.
31

32. 3.5 Losses of Head due to Friction
• Energy loss through friction in the length of pipeline is commonly
termed the major loss hf
• This is the loss of head due to pipe friction and to the viscous
dissipation in flowing water.
• Several studies have been found the resistance to flow in a pipe is:
- Independent of pressure under which the water flows
- Linearly proportional to the pipe length, L
- Inversely proportional to some water power of the pipe diameter D
- Proportional to some power of the mean velocity, V
- Related to the roughness of the pipe, if the flow is turbulent

33. The resistance to flow in a pipe is a function of:
• The pipe length, L
• The pipe diameter, D
• The mean velocity, V
• The properties of the fluid ()
• The roughness of the pipe, (the flow is
turbulent).
33

34. Darcy-Weisbach Equation
2 2 Where:
L V 8f LQ
hL  f   f is the friction factor
D 2 g g D5  2 L is pipe length
D is pipe diameter
Q is the flow rate
hL is the loss due to friction
It is conveniently expressed in terms of velocity (kinetic) head in the pipe
The friction factor is function of different terms:
 e  rVD e   VD e 
f  F NR ,   F ,   F
  D , 
 D     D
Renold number Relative roughness

35. Friction Factor: (f)
• For Laminar flow: (NR < 2000) [depends only on
Reynolds’ number and not on the surface roughness]
64
f 
NR
• For turbulent flow in smooth pipes (e/D = 0) with
4000 < NR < 105 is
0.316
f  1/ 4
NR
35

36. Friction Factor f
The thickness of the laminar sublayer  decrease with an increase in NR
laminar flow f independent of relative
NR < 2000 Smooth roughness e/D

e  '  1.7e 64 1 N f 
f   2 log 10  R
 2.51


pipe wall NR f  

f varies with NR and e/D
 e  
transitionally   
 2 log 10    
rough 1 D 2.51 
 e
f  3.7 N R f 
 
pipe wall 0.08e   '  1.7e  
Colebrook formula

turbulent flow
f independent of NR
rough
NR > 4000
e 1  D
   0.08e
'  2 log 10  3.7 
f  e
pipe wall

37. Moody diagram
• A convenient chart was prepared by Lewis F. Moody and commonly
called the Moody diagram of friction factors for pipe flow,
There are 4 zones of pipe flow in the chart:
• A laminar flow zone where f is simple linear function of NR
• A critical zone (shaded) where values are uncertain because
the flow might be neither laminar nor truly turbulent
• A transition zone where f is a function of both NR and relative
roughness
• A zone of fully developed turbulence where the value of f
depends solely on the relative roughness and independent of
the Reynolds Number

38. 38

39. Laminar
Marks Reynolds Number
independence

40. Typical values of the absolute roughness (e) are given in
table 3.1
40

41. Notes: Alternative to Moody Diagram
• Swamee-Jain Equation (1976)
0.25
f  2
  e 5.74 
log10  3.7 D  Re 0.9 
  
Explicit expression 10-6<e/D,10-2; 5000<NR<108
41

42. Problems (head loss)
Three types of problems for uniform flow
in a single pipe:
 Type 1:
Given the kind and size of pipe and the flow rate head loss ?
 Type 2:
Given the kind and size of pipe and the head loss flow rate ?
 Type 3:
Given the kind of pipe, the head loss and flow rate size of pipe ?

43. Solving Turbulent flow Problems
There are 3 types of problems
• Given: L, D, V solve for hf
– Compute ks/D, Re then f from moody diagram then
find hf and V
• Given: hf, L, D solve for V
3/ 2
D 2 gh
– Compute ks/D the value  L
Then calculate f, V
f
and Q
• Given: Q, L, hf solve for D, Iterative solution
– Iterative process Assume f, calculate V and Re, check f from
moody diagram
– Assume new f, calculate V and Re, then check f

44. Moody Diagram  D3 / 2  2 gh f 
1/ 2
N R f 1/ 2   
   L 


Fully rough pipes
Resistance Coefficient f
Relative roughness e/D
Smooth pipes
Reynolds number

45. Example 1
The water flow in Asphalted cast Iron pipe (e = 0.12mm) has a diameter 20cm
at 20oC. Is 0.05 m3/s. determine the losses due to friction per 1 km
 Type 1:
Given the kind and size of pipe and the flow rate head loss ?
0.05m 3 /s
V  1.59m/s

π/4  0.2 m
2 2

T  20o C  υ  1.0110 6 m 2 /s
e  0.12mm
e 0.12mm
  0.0006 Moody f = 0.018
D 200mm
VD 1.59  0.2
NR    314852  3.15 105
 1.0110 6
L V2  1,000 m  1.59 
2
hf  f  0.018  
D 2g  
 0.20 m  2 9.81 m/s
2
 

45
 11.55 m

46. Example 2
The water flow in commercial steel pipe (e = 0.045mm) has a diameter 0.5m
at 20oC. Q=0.4 m3/s. determine the losses due to friction per 1 km
 Type 1:
Given the kind and size of pipe and the flow rate head loss ?
Q 0.4
V   2.037 m / s
A 0.52 
4
497 10 6 497 10 6
   1.006 10 6
T  42.51.5 20  42.51.5
0.5  2.037
NR   1.012 106
1.006 10 6
e 0.045
  9 10 5
D 0.5 103
Moody f  0.013
 
2
1000 2.037
h f  0.013    5.5 m / km
0.5 2  9.81

47. Example 3
Cast iron pipe (e = 0.26), length = 2 km, diameter = 0.3m. Determine the
max. flow rate Q , If the allowable maximum head loss = 4.6m. T=10oC
 Type 2:
Given the kind and size of pipe and the head loss flow rate ?
2
LV
hF  f
D 2g
2000 V2
4.6  f 
0.3 2  9.81
 1
0.0135
V2  
497 10 6 497 10 6
f    1.3110 6
T  42.51.5 10  42.51.5
0.3  V
NR  6
 2.296 106 V  2

1.3110
e 0.26
  8.67 10 5  0.00009
D 0.3 103

48. Trial 1
f  0.01 V  1.16 m/s
eq1
eq 2 N R  2.668 105
 V2 
0.0135
 1

f
e
 8.67 10  4
D N R  2.296 106V  2

Moody f  0.02

Trial 2
f  0.02 V  0.82 m/s
eq1
eq 2 N R  1.886 105

e
 8.67 10  4
D
Moody f  0.021

V= 0.82 m/s , Q = V*A = 0.058 m3/s

49. Example 3.5
Compute the discharge capacity of a 3-m diameter, wood stave
pipe in its best condition carrying water at 10oC. It is allowed to
have a head loss of 2m/km of pipe length.
 Type 2:
Given the kind and size of pipe and the head loss flow rate ?
Solution 1:
LV2  2ghf 1/ 2   1/ 2
 D 
hf  f V     
D 2g   L    f  
 1000  V
2
0.12
2 f  V 
2
 3  2(9.81) f
Table 3.1 : wood stave pipe: e = 0.18 – 0.9 mm, take e = 0.3 mm
e 0.3
  0.0001
D 3
VD 3V
At T= 10oC,  = 1.31x10-6 m2/sec  N R    2.29  106.V
 1.31 106

50. • Solve by trial and error:
• Iteration 1:
0.12
• Assume f = 0.02  V   V  2.45m / sec
2
0.02
N R  2.29 106.2.45  5.6 106
From moody Diagram: f  0.0122
Iteration 2:
0.12
update f = 0.0122  V2   V  3.14m / sec
0.0122
N R  2.29 106.3.14  7.2 106
From moody Diagram: f  0.0121  0.0122
Iteration f V NR
V2  3.15 m/s
0 0.02 2.45 5.6106   32
1 0.0122 3.14 7.2106 Solution: Q  VA  3.15.
2 0.0121
4
Convergence
 22.27 m3 /s

51. Alternative Method for solution of Type 2 problems
1/ 2
D3 / 2  2 gh f 
NR f   
  L 


Type 2. Given the kind and size of pipe and the head loss flow rate ?
Determines relative roughness e/D
Given N R f and e/D we can determine f (Moody diagram)
Use Darcy-Weisbach to determine velocity and flow rate
Because V is unknown we cannot calculate the Reynolds number
However, if we know the friction loss hf, we can use the Darcy-Weisbach equation
to write: LV2  1/ 2   1/ 2
 
2ghf D
hf  f V     
D 2g   L    f  
Re 
VD   1   3 / 2  2ghf 1/ 2
D  
We also know that: Re    1/ 2  
  
 
  f      L  
  
2 1/ 2 
 D  2 gh f
3/

NR f  
1/ 2

 
 unknowns
   L  
Can be calculated based on Quantity plotted along the top of the Moody diagram
available data

52. Moody Diagram  D3 / 2  2 gh f 
1/ 2
N R f 1/ 2   
   L 


Fully rough pipes
Resistance Coefficient f
Relative roughness e/D
Smooth pipes
Reynolds number

53. Example 3.5
Compute the discharge capacity of a 3-m diameter, wood stave pipe in its best
condition carrying water at 10oC. It is allowed to have a head loss of 2m/km
of pipe length.
Type 2: Given the kind and size of pipe and the head loss flow rate ?
Solution 2:
At T= 10oC,  = 1.31x10-6 m2/sec
1/ 2
D  2 gh f
3/ 2
 (3)3 2 2(9.81)(3)
NR f      9.62  105
  L


 1.31 106 1000
Table 3.1 : wood pipe: e = 0.18 – 0.9 mm, take e = 0.3 mm e  0.3  0.0001
D 3
From moody Diagram: f  0.0121
LV 2
 2 gh f 
1/ 2
D
1/ 2
  32
hf  f V  
 L    
 f   3.15m / sec , Q  VA  3.15.
D 2g     4
 22.27 m3 /s

54. f = 0.0121

55. Example (type 2)
1
H = 4 m, L = 200 m, and D = 0.05 m
H
What is the discharge through the
2 galvanized iron pipe?
L
Table : Galvanized iron pipe: e = 0.15 mm e/D = 0.00015/0.05 = 0.003
 = 10-6 m2/s
We can write the energy equation between the water surface in the reservoir and the
free jet at the end of the pipe: 2 2
p1 V1 p V
 h1   2  h2  2  hL
 2g  2g
2
 L V
2
V2
04000  f 
2g  D  2g
2g 4 78.5
V2  
L 1  4000 f
1 f  
D

56. Example (continued)
Assume Initial value for f : fo = 0.026
78.5
Initial estimate for V: V  0.865 m/sec
1  4000  0.026
DV
Calculate the Reynolds number N R   5  104  V  4.3  104

Updated the value of f from the Moody diagram f1 = 0.029
78.5
V  0.819 m/sec
1  4000  0.029
DV2
NR   5  104  V  4.1 104

Iteration f V NR
0 0.026 0.865 4.3104 V 2  0.814 m/s
1 0.029 0.819 4.1104 Solution:   0.05 2
Q VA  0.814 
2 0.0294 0.814 4.07104 4
3 0.0294 Convergence  1.60  103 m3 /s

57. Initial estimate for f
A good initial estimate is to pick the f value that is valid for a fully rough pipe with
the specified relative roughness
fo = 0.026 e/D = 0.003

58. Solution of Type 3 problems-uniform flow in a
single pipe
Given the kind of pipe, the head loss and flow rate size of pipe ?
Determines
equivalent roughness e
Without D we cannot calculate the relative
Problem?
roughness e/D, NR, or N R f
Solution procedure: Iterate on f and D
1. Use the Darcy Weisbach equation and guess an initial value for f
2. Solve for D
3. Calculate e/D
4. Calculate NR
5. Update f
6. Solve for D
7. If new D different from old D go to step 3, otherwise done

59. Example (Type 3)
A pipeline is designed to carry crude oil (S = 0.93,  = 10-5 m2/s) with a discharge of 0.10
m3/s and a head loss per kilometer of 50 m. What diameter of steel pipe is needed?
Available pipe diameters are 20, 22, and 24 cm.
From Table 3.1 : Steel pipe: e = 0.045 mm
Darcy-Weisbach:
2
Q
 
 L V
2
L  A  L Q 4
2 2
1 16 fLQ2
hf  f    hf  f  f   5
 D  2g  D  2g  D D 2 g 2
2 4
D 2g
1/ 5
16  1000  0.102 
1/ 5
16 fLQ  2
D  D f 1/ 5  0.440  f 1/ 5

2 g 2 h f   2  9.81  2  50 


Make an initial guess for f : fo = 0.015 D  0.440 0.0151/ 5  0.190 m
Now we can calculate the relative roughness and the Reynolds number:
e 0.045  103
  0.00024 
D D update f
VD Q D 4Q D 4Q 1 1 f = 0.021
NR      12.7  103   66.8  103
 A  D 2   D D

60. Updated estimate for f
f1 = 0.021
e/D = 0.00024

61. Example Cont’d
D  0.440  f 1/ 5
1 Solution:
N R  12.7  103 
D
D = 0.203
From moody diagram, updated estimated for f :
Use next larger commercial
f1 = 0.021 D = 0.203 m size:
N R  62.5 103 update f
e D = 22 cm
 0.00023
D
Iteration f D NR e/D
0 0.015 0.190 66.8103 0.00024
1 0.021 0.203 62.5103 0.00023
2 0.021 Convergence

62. Example 3.6
Estimate the size of a uniform, horizontal welded-steel pipe installed to carry 14
ft3/sec of water of 70oF (20oC). The allowable pressure loss is 17 ft/mi of
pipe length.
Solution 2:
From Table : Steel pipe: ks = 0.046 mm
LV2 Q 2
 
Darcy-Weisbach: hL  f  
D 2g L  
A L Q 2 42 1 16fLQ 2
hL  f f 
Q VA D 2g D 2g  2D 4 D 5 2g 2
1/ 5
 8 fLQ 2 
 D 2 
 1
/5  g hL 
 8  f  5280 14  2
D  f 1/ 5  4.33  f 1/ 5 a
 9.81  2 17 
Let D = 2.5 ft, then V = Q/A = 2.85 ft/sec
Now by knowing the relative roughness and the Reynolds number:
e 0.003
  0.0012
D 2.5
We get f =0.021
VD 2.85 * 2.5
NR    6.6 *105
 1.08 *10 5

63. A better estimate of D can be obtained by substituting the latter
values into equation a, which gives
D  4.33  f 1/ 5  4.33 * 0.0211/ 5  2.0 ft
A new iteration provide
V = 4.46 ft/sec
NR = 8.3 x 105
e/D = 0.0015
f = 0.022, and
D = 2.0 ft.
More iterations will produce the same results.

64. Major losses formulas
• Several formulas have been developed in the past.
Some of these formulas have faithfully been used in
various hydraulic engineering practices.
1. Darcy-Weisbach formula
2. The Hazen -Williams Formula
3. The Manning Formula
4. The Chezy Formula
5. The Strickler Formula
64

65. Empirical Formulas 1
• Hazen-Williams
D  5cm    V  3.0m / sec
V  1.318CHW Rh .63S 0.54
0
British Units
V  0.85CHW Rh
0.63 0.54
S SI Units
D 2
wetted A D
Rh  hydraulic Radius   4 
Simplified
wetted P D 4
hf
S
L
C HW  Hazen Williams Coefficien t
10.7 L
hf  1.852
Q1.852 SI Units
CHW D 4.87

67. CHW  Hazen Williams Coefficien t

68. CHW  Hazen Williams Coefficien t
68

69. When V  3.0m / sec
0.081
Vo 
CH  C Ho  
V 
Where:
CH = corrected value
CHo = value from table
Vo = velocity at CHo
V = actual velocity
69

70. Empirical Formulas 2
Manning Formula
• This formula has extensively been used
for open channel designs
• It is also quite commonly used for pipe
flows
70

71. • Manning
1 2 / 3 1/ 2
V  Rh S
n Rh  hydraulic Radius 
wetted A D

wetted P 4
hf
S
Simplified
L
n  Manning Coefficien t
10.3 L nQ 
2
hf  SI Units
D 5.33
71

72. 1 2/ 3 1/ 2
V  Rh S
n
Q2
h f  10.3n 2 L 16 / 3
D
L 2 2
h f  6.35 1.33 n V
D
• n = Manning coefficient of roughness (See Table)
• Rh and S are as defined for Hazen-William
formula.
72

73. 73

74. 74

75. Example
New Cast Iron (CHW = 130, n = 0.011) has length = 6 km and diameter = 30cm.
Q= 0.32 m3/s, T=30o. Calculate the head loss due to friction using:
a) Hazen-William Method
10.7 L
hf  1.852
Q1.852
CHW D 4.87
10.7  6000
hf  0.321.852  333m
1301.852 0.34.87
b) Manning Method
10.3 L nQ 
2
hf 
D 5.33
10.3  6000 0.011 0.32
2
hf  5 .33
 470 m
0.3

76. Minor losses
It is due to the change
of the velocity of the
flowing fluid in the
magnitude or in
direction [turbulence
within bulk flow as it
moves through and
fitting] Flow pattern through a valve
76

77. • The minor losses occurs du to :
• Valves
• Tees
• Bends
• Reducers
• Valves
• And other appurtenances
• It has the common form
V2 Q2
hm  k L  kL
2g 2 gA2
“minor” compared to friction losses in long pipelines but,
can be the dominant cause of head loss in shorter pipelines 77

78. Losses due to contraction
A sudden contraction in a pipe usually causes a marked drop in pressure
in the pipe due to both the increase in velocity and the loss of energy to
turbulence.
Along wall
2
V2
Along centerline hc  kc
2g

79. Value of the coefficient Kc for sudden contraction
V2

80. Different pipe entrance
increasing loss coefficient

81. Loss due to pipe entrance
General formula for head loss at the entrance of a pipe is also
expressed in term of velocity head of the pipe
2
V
hent  K ent
2g
81

82. Head Loss at the entrance of a Pipe
(flow leaving a tank)
Reentrant Sharp
(embeded) edge
KL = 0.8 KL = 0.5
Slightly
rounded
Well
KL = 0.2
rounded
KL = 0.04
V2
hL  K L 82
2g

83. Head Loss Due to a Sudden Contraction
V 22
hL  K L
2g
2
V2
hL 0.5
2g
83

84. Head losses due to pipe contraction may be greatly reduced by
introducing a gradual pipe transition known as a confusor
kc'
2
V2
hc'  kc'
2g

85. Head Loss Due to Gradual Contraction
(reducer or nozzle)
85

86. Losses due to Expansion
A sudden Expansion in a pipe
(V1  V2 ) 2
hE 
2g

87. Note that the drop in the energy line is much
larger than in the case of a contraction
abrupt expansion
gradual expansion
smaller head loss than in the case of an abrupt expansion

88. Head Loss Due to a Sudden Enlargement
V 12
hL  K L
2g
2
 A1 
K L  1  
 A2 
or :
hL 
V1  V2  2
2g
88

89. Head losses due to pipe enlargement may be greatly reduced by
introducing a gradual pipe transition known as a diffusor
V  V2
2 2
hE'  k E' 1
2g

90. Head Loss Due to Gradual Enlargement
(conical diffuser)
hL K L
V
1  V2
2 2

2g
a 100 200 300 400
KL 0.39 0.80 1.00 1.06
90

91. Gibson tests
91

92. Another Typical values for various amount of rounding of
the lip
92

93. Head Loss at the Exit of a Pipe
(flow entering a tank)
KL = 1.0 KL = 1.0
V2
hL 
2g KL = 1.0
KL = 1.0
the entire kinetic energy of the exiting fluid (velocity V1) is
dissipated through viscous effects as the stream of fluid mixes
with the fluid in the tank and eventually comes to rest (V2 = 0).
93

94. Head Loss Due to Bends in Pipes
V2
hb  kb
2g
R/D 1 2 4 6 10 16 20
Kb 0.35 0.19 0.17 0.22 0.32 0.38 0.42
94

95. Miter bends
For situations in which space is limited,
95

96. Head Loss Due to Pipe Fittings
(valves, elbows, bends, and tees)
2
V
hv  K v
2g
96

97. 97

98. The loss coefficient for elbows, bends, and tees
98

99. Loss coefficients for pipe components (Table)

100. Minor loss coefficients (Table)

101. Minor loss calculation using equivalent
pipe length
kl D
Le 
f

102. Energy and hydraulic grade lines
Unless local effects are of particular interests the changes in the EGL and HGL are
often shown as abrupt changes (even though the loss occurs over some distance)

103. Example
Given: Figure
Find: Estimate the elevation required in the
b
upper reservoir to produce a water
discharge of 10 cfs in the system. What
is the minimum pressure in the pipeline
and what is the pressure there?
Solution:
V2 p V2 p
α1 1  1  z1   hL  αb b  b  zb
V2 p V2 p 2g γ 2g γ
α1 1  1  z1   hL  α2 2  2  z2
2g γ 2g γ Vb2 pb
0  0  z1   hL  1 *   zb
0  0  z1   hL  0  0  z2 2g γ
 L V 2 pb Vb2  L V 2
 hL   K e  2 K b  K E  f   z1  zb    Ke  Kb  f 
 D  2g γ 2g  D  2g
L 430 2
K e  0.5; K b  0.4 (assumed); K E  1.0; f  0.025 *  10.75  300  12.73
 133  110.7  1.0  0.5  0.4  0.025 
D 1  1  2 * 32.2
Q 10  1.35 ft
V    12.73 ft / s
A  / 4 * 12 pb  62.4 * ( 1.53)  0.59 psig
2
z1  100  0.5  2 * 0.4  1.0  10.75
12.73 VD 12.73 * 1
 133 ft Re    9 * 105
2 * 32.2  1.14 * 10 5

104. Example
In the figure shown two new cast iron pipes in series, D1 =0.6m ,
D2 =0.4m length of the two pipes is 300m, level at A =80m , Q
= 0.5m3/s (T=10oC).there are a sudden contraction between
Pipe 1 and 2, and Sharp entrance at pipe 1.
Fine the water level at B
e = 0.26mm
v = 1.31×10-6
Q = 0.5 m3/s

105. Solution
Z A  ZB  hf
hL  h f 1  h f 2  hent  hc  hexit
2 2 2 2
L1 V1 L2 V2 V1 V2 V22
hL  f1  f2  kent  kc  kexit
D1 2 g D2 2 g 2g 2g 2g
Q 0.5 Q 0.5
V1    1.77 m/ sec , V2    3.98 m/ sec ,
π π
A1 0.62 A2 0.42
4 4
VD VD
Re1  1 1  8.1105 , Re 2  2 2  1.22 106 ,
υ υ
 0.26 
  0.00043,  0.00065,
D1 600 D1
moody f1  0.017
  moody f 2  0.018
 
hent  0.5, hc  0.27, hexit  1

106. 2 2 2 2
L1 V1 L2 V2 V1 V2 V22
hL  f1  f2  kent  kc  kexit
D1 2 g D2 2 g 2g 2g 2g
 300  1.77  300  3.98
2 2
h f  0.017  .  0.018  .
 0.6  2 g  0.4  2 g
 1.77 2   3.982   3.982 
 0.5
 2 g   0.27 2 g    2 g   13.36m
    
     
ZB = 80 – 13.36 = 66.64 m

107. Example
A pipe enlarge suddenly from D1=240mm to D2=480mm. the
H.G.L rises by 10 cm calculate the flow in the pipe

108. Solution p1 V12 p2 V22
  z1    z 2  hL
rg 2 g rg 2 g
V12 V22  p2   p1 
  hL  
 rg  z 2    rg  z1 
  
2g 2g    
V12 V22 V1  V2
 
 2
 0.1
2g 2g 2g
V1 A1  V2 A2
V1 

4

0.24 2  V2 

4 0.482 
V1  4V2
16V22 V22 4V2  V2
 
 
2
 0.1
2g 2g 2g
2
6V2
 0.1
2g
V2  0.57 m / s  Q  V2 A2  0.57   0.482  0.103m 3 / s
4

109. • Note that the above values are average
typical values, actual values will depend
on the make (manufacturer) of the
components.
• See:
– Catalogs
– Hydraulic handbooks !!
109