Here are the key steps to solve this problem using the Hardy Cross Method: 1. Select a loop (ABDE loop) and make initial guesses for the pipe flows (Q1, Q2, Q3, Q4) 2. Compute the head losses (hf1, hf2, hf3, hf4) in each pipe using the pipe characteristics and guessed flows 3. Compute the algebraic sum of the head losses around the loop. This will not equal zero initially. 4. Use the Hardy Cross formula to calculate flow corrections (ΔQ1, ΔQ2, etc.) needed to balance the head losses. 5. Update the pipe flows by adding the corrections. 6.

1. CVE 341 – Water Resources
Lecture Notes I:
(Combined Chs 7 & 8)
Closed Conduit Flow
Prepared by Ercan Kahya

2. Copyright © 2007 by Nelson, a division of Thomson Canada Limited
A large slurry pipe. (Copyright Terry Vine/CORBIS)
View of a Stunning Example in CE Practices

3. INTRODUCTION
• The flow of water in a conduit may be either open
channel flow or pipe flow.
• Open channel flow must have a free surface whereas
pipe flow has none.
• A free surface subject to atmospheric pressure.
• In pipe flow, there is no direct atmospheric flow but
hydraulic pressure only.

4. INTRODUCTION (Cont`d)
Differences btw open channel flow & pipe flow

5. Steady Uniform Flow
Steady, uniform flow in closed conduits
z: geometric head
P/g: pressure head
V2/2g: velocity head
z
P
g
V
ET 


g

2
2

6. General Energy Consideration
f
T
h
z
P
g
V
z
P
g
V
E 





 2
2
2
2
2
1
1
2
1
1
2
2 g

g

: the velocity coefficient &
set to unity for regular &
symmetrical cross-section
like pipe.

7. Figure 8.2: Layout and hydraulic heads for Example 8.1.
Example 8.1:
Discuss how to attack to this problem…
Given: Q, d, hf, and available pressure at the building
Unknown: z1

8. Forces acting on steady closed conduit flow.
t = g R Sf
t : Boundary shear stress (N/m2)
g : Specific weight of water (N/m3)
R : Hydraulic radius (m)
Sf : Slope of energy grade line
Resistance Application & Friction Losses in Pipes
General Resistance Equation:
from computing the shear stress of a system in dynamic equilibrium

9. Consider the length of pipe to be a control
volume & realize the dynamic equilibrium
P: wetted perimeter
- Dividing this equation by area (A)
R: hydraulic radius (A / P)

10. Note that So: pipe slope
Note that ∆z: change in pipe elevation
→ Piezometric head change

11. 7-11
: the change in the hydraulic grade line & also
equal to the energy loss across the pipe
We refer this loss to as the friction loss (hf) & express as
For circular pipes:
Since ; then
Sf : energy grade line
Finally
The general shear stress relation in all cases of steady uniform

12. Resistance Equations for Steady Uniform Flow
Now a method needed for “ shear stress ↔ velocity ”
From dimensional considerations;
a : constant related to boundary roughness
V : average cross-sectional velocity
Inserting this to the general shear relation,
Solving for V,
Chezy Equation

13. - A general flow function relating flow parameters to
the change in piezometric head in pipes:
- In case of steady uniform flow; the left side equals to friction loss (hf)
Darcy-Weisbach
Equation
This equation can be considered a special case of Chezy formula.

14. Resistance Application & Friction Losses in Pipes
f
RS
C
V  Chezy Equation
From dimensional analysis: Resistance Equation
in terms of the average velocity
In most cases of closed conduit flow, it is customary
to compute energy losses due to resistance by use of
the Darcy-Weishbach equation.
g
V
d
L
f
hf
2
2


15. Development of an analytical relation btw shear stress & velocity
- We earlier had the following relations for the friction loss:
&
For a general case, using d = 4R;
After simplifying;
Note that friction factor is directly
proportional to the boundary shear stress

16. - Let`s define “shear velocity” as
=
Then the above equation becomes
→ important in developing the resistance formula

17. Velocity Distributions in Steady, Uniform Flow
Typical laminar & turbulent velocity distribution for pipes:
To start for velocity profile, let`s recall Newton`s law of viscosity
→ governs flow in the laminar region
Laminar Flow Rn ≤ 2000 Rn: Reynolds number
Turbulence Flow Rn ≥ 4000

18. Laminar Flow
Velocity in terms of radial position:
→ paraboloid distribution
Head loss in a pipe element in terms of average velocity:
Energy loss gradient or friction slope
represents the rate of energy dissipation due to boundary shear stress or friction
- Laminar flow case is governed by the Newtonian viscosity principle.
2
32
d
VL
hf
g

 → Poiseuille Equation
- Darcy friction factor: f = 64 / Rn in laminar flow

19. Turbulent Flow
- More complex relations btw wall shear stress & velocity distribution
- In all flow cases, Prandtl showed “laminar sub-layer” near the boundary
- The thickness of laminar boundary layer decreases as the Re # increases
- Flow is turbulent outside of the boundary layer

20. Turbulent Flow
C
y
y
k o

 ln
1
*


 : instantaneous velocity
* : shear velocity
k : von Karman constant (= 0.4 for water)
y : distance from boundary
yo : hydraulic depth
C : constant
General form for turbulent
velocity distributions
After statistical and dimensional considerations, Von Karman gives
logarithmic dimensionless velocity distribution as

21. Turbulent Flow velocity distribution
1) For smooth-walled conduits
2) For rough-walled conduits
5
.
5
ln
1 *
*


v
yv
k


48
.
8
ln
1
*




 y
k
Function of the laminar sub-layer
properties (wall Reynolds number)
Function of the wall roughness
element

22. Friction Factor in Turbulent flow
In smooth pipes
In rough pipes
74
.
1
)
/
(
2
1

 
o
r
Log
f
8
.
0
)
(Re
2
1

 f
Log
f
14
.
1
)
/
(
2
1

 
d
Log
f
Nikuradse rough pipe equation
Von Karman & Prandtl equation
Nikuradse rough pipe equation
in terms of pipe diameter
Using velocity distribution given previously & shear stress/velocity relation
(Ch7), it is possible to solve for friction factor:

23. Transition region: Characterized by a flow regime in a particular
case follow neither the smooth nor rough pipe
formulations
- Colebrook & White proposed the following semi-empirical function:
)
/
7
.
18
1
log(
2
74
.
1
)
/
log(
2
1
f
R
r
r
f n

 



Friction Factor in Turbulent flow
Note that all analytical expressions are nonlinear; so it is cumbersome to solve !
Lewis Moody developed graphical plots of f as given in preceding expressions
☺

24. Figure 8.4: Friction factors for flow in pipes, the Moody diagram
(From L.F. Moody, “Friction factors for pipe flow,” Trans. ASME, vol.66,1944.)
Moody Diagram

25. How to Read the Moody Diagram
♦ The abscissa has the Reynolds number (Re) as the
ordinate has the resistance coefficient f values.
♦ Each curve corresponds to a constant relative
roughness ks/D (the values of ks/D are given on the right
to find correct relative roughness curve).
♦ Find the given value of Re, then with that value move
up vertically until the given ks/D curve is reached. Finally,
from this point one moves horizontally to the left scale to
read off the value of f.

26. Empirical Resistance Equations
Blasius Equation:
25
.
0
316
.
0
n
R
f 
In case of smooth-walled pipes;
very accurate for Re <100,000
Manning’s Equation: (special case of Chezy)
6
/
1
1
R
n
C 
n : coefficient which is a function
of the boundary roughness and
hydraulic radius
5
.
0
3
/
2
1
S
R
n
V 
For the SI unit system
Velocity: (used in open-channels)

27. Empirical Resistance Equations
5
.
0
3
/
2
485
.
1
S
R
n
V 
Manning’s Equation:
For the BG system
1. Manning’s n is not be a function of turbulence characteristic or Re
number but varies slightly with the flow depth (through the hydraulic
radius)
2. In view of (1), therefore one can say that the Manning equation would
be strictly applicable to rough pipes only, although it has frequently
been employed as a general resistance formula for pipes.
3. It is, however, employed far more frequently in open-channel
situations.

28. Empirical Resistance Equations
54
.
0
63
.
0
f
HW S
R
C
V 
Hazen-Williams Equation:
Important Notes:
1. Widely used in water supply and irrigation works
2. Only valid for water flow under turbulent conditions.
3. It is generally considered to be valid for larger pipe (R>1)
R: hydraulic radius of pipe
Sf : friction slope
CHW : resistance coefficient
(pipe material & roughness conditions)

29. Empirical Resistance Equations
V
RS
C
R
n
g
f f



1
8 6
/
1
•Three well-known 1-D resistance laws are the
Manning, Chezy & Darcy-Weisbach resistance
equations
• The interrelationship between these equations is
as follows:

30. Minor Losses in Pipes
• Energy losses which occur in pipes are due to boundary
friction, changes in pipe diameter or geometry or due to
control devices such as valves and fittings.
• Minor losses also occur at the entrance and exits of pipe
sections.
• Minor losses are normally expresses in units of velocity head
g
V
k
h l
l
2
2

kl : Loss coefficient associated with a
particular type of minor loss and a
function of Re, R/D, bend angle, type
of valve etc.

31. Pipe connections, bends and reducers

32. Sleeve valve. (Courtesy TVA.)

33. Minor Losses in Pipes
• In the case of expansions of cross-sectional
area, the loss function is sometimes written in
terms of the difference between the velocity
heads in the original and expanded section due
to momentum considerations;
 
g
V
V
k
h e
l
2
2
2
1 
 For Gradual Expansion
ke: expansion coefficient

34. Minor Losses in sudden expansions
 
g
V
V
hl
2
2
2
1


For Sudden Expansion
Head loss is caused by a rapid increase in the pressure head

35. Minor Losses in sudden contraction
g
V
k
h l
l
2
2
2

Head loss is caused by a rapid decrease in the pressure head

36. Expansion and contraction coefficients
for threaded fittings
The magnitude of energy loss is a function of the degree and abruptness
of the transition as measured by ratio of diameters & angle θ in Table 8.3.

37. Coefficients of entrance loss for pipes (after Wu et al., 1979).
Coefficients of entrance loss for pipes
Entrance loss coefficient is strongly affected by the nature of the entrance

38. Summary for minor losses
Copyright © Fluid Mechanics & Hydraulics by Ranald V Giles et. al. (Schaum’s outlines series)
Please see Table 4 & 5 of Fluid
Mechanics & Hydraulics by Ranald V
Giles et al. (Schaum’s outlines
series)

39. Bend Loss Coefficients
r : radius of bend
d: diameter of pipe

40. Loss coefficients for
Some typical valves

41. CLASS EXERCISES

42. CLASS EXERCISES

43. Worked EXERCISES

44. An elevated open water tank is used to supply water to a building located 5000 ft
away as shown in the figure. The water is carried through a 12-inch cast iron
pipe (f =0.045) with a flow rate of 2.23 cfs. Fire codes require that a pressure of
30 psi be available on the second floor of the building which is at an elevation of
20 ft above a reference datum.
 What is the required elevation of the water in the tank relative to the reference
datum?
Worked Exercise

45. Worked EXERCISES
H
Reference datum
5000 ft
20 ft
Elevated Tank
D = 12 in & Q = 2.23 ft3/s

46. Three-Reservoir Problem
• Determine the discharge
• Determine the direction of flow

47. Three-Reservoir Problem
If all flows are considered positive
towards the junction then
QA + QB + QC = 0
This implies that one or two of the flows must be outgoing from junction.
The pressure must change through each pipe to provide the same
piezometric head at the junction. In other words, let the HGL at the junction
have the elevation









 D
D
D z
p
h
g
pD: gage pressure

48. Three-Reservoir Problem
Head lost through each pipe,
assuming PA=PB=PC=0 (gage) at
each reservoir surface, must be such
that
D
B
B
B
B h
z
g
V
d
L
f
h 



2
2
D
C
C
C
C h
z
g
V
d
L
f
h 



2
2
D
A
A
A
A h
z
g
V
d
L
f
h 



2
2
1. Guess the value of hD (position of
the intersection node)
2. Assume f for each pipe
3. Solve the equations for VA, VB & VC
and hence for QA, QB & QC
4. Iterate until flow rate balance at the
junction QA+QB+QC=0
If hD too high the QA+QB+QC <0
& reduce hj and vice versa.

49. Example: (Class Exercise)

50. Pipes in Parallel
• Head loss in each pipe must be equal to obtain the same
pressure difference btw A & B (hf1 = hf2)
• Procedure: “trial & error”
• Assume f for each pipe  compute V & Q
• Check whether the continuity is maintained

51. CLASS EXERCISES

52. Pipe Networks
The need  to design the original network
to add additional nodes to an existing network
Two guiding principles (each loop):
1- Continuity must be maintained
2- Head loss btw 2 nodes must be
independent of the route.
0

 i
Q

  fCC
fC h
h
The problem is to determine
flow & pressure at each node

53. Pipe Networks
Consider 2-loop network: Procedure: (inflow + outflow + pipe
characteristics are known)
1- Taking ABCD loop first, Assume Q
in each line
2- Compute head losses in each pipe
& express it in terms of Q
For the loop:
The difference is known:

54. Hardy Cross Method
3- If the first Q assumptions were incorrect, Compute a correction to
the assumed flows that will be added to one side of the loop & subtracted
from the other.
► Suppose we need to subtract a ∆Q from the clockwise side and add
it to the other side for balancing head losses. Then
► After applying Taylor`s series expansion & math manipulations for the above relation:
4- After balancing of flows in the first loop, Move on to the next one

55. Hardy Cross Method --- Example 8.11
Given info: