CHAPTER 3
VISCOUS FLOW IN PIPES
3.1 Introduction
This chapter will deal almost exclusively with pipes or conduits that are flowing full. A free
surface will not exist in the pipe itself, and the Froude number, which is important to open
channel flows, will accordingly have no significance. The Reynolds number will be the
significant dimensionless parameter, but its importance will also diminish at high Reynolds
numbers as the flow becomes independent of viscous effects. Flow in a circular pipe is
undoubtedly the most common closed fluid flow. It is encountered in the veins and arteries in a
body, in a city’s water system, in a farmer’s irrigation system, in the piping systems transporting
fluids in a factory and in the ink jet of a computer’s printer. For a sufficiently low Reynolds
numbers (Re < 2000 in a pipe) a laminar flow results, and at sufficiently high Reynolds number a
turbulent flow occurs.
For all flows involved in this Chapter, we assume that the pipe is completely filled with the fluid
being transported as is shown in Figure 3.1a Thus, we will not consider a concrete pipe through
which rain water flows without completely filling the pipe, as is shown in Figure 3.1b. Such
flows, called open-channel flow will be treated in next chapters. The difference between the open
channel flow and the pipe flow is in the fundamental mechanism that drives the flow. For open
channel flow, gravity alone is the driving force such that the water flows down a hill like in
streams and rivers. For pipe flow, gravity may be important (the pipe need not be horizontal), but
the main driving force is likely to be a pressure gradient along the pipe. If the pipe is not full, it is
not possible to maintain this pressure difference p1  p 2 .
Figure 3.1 (a) pipe flow and; (b) open channel flow
3.2 Laminar or Turbulent flow
The flow of a fluid in a pipe may be laminar flow or it may be turbulent flow. Osborne Reynolds
(1842-1912), a British scientist and mathematician, was the first to distinguish the difference
between these two classifications of flow by using a simple apparatus as shown in Figure 3.2
-1-
Figure 3.2 (a) Experiment to illustrate type of flow and; (b) typical dye streaks
If water runs through a pipe of diameter D with an average velocity v, the following
characteristics are observed by injecting a dye as shown in Figure 3.2. For “small enough
flowrates” the dye streakline will remain as a well – defined line as it flows along; for a
somewhat larger “intermediate flowrate” the dye streakline fluctuates in time and space. On the
other hand, for “large enough flowrates” the dye streakline almost immediately becomes blurred
and spreads across the entire pipe in a random fashion. These three characteristics are denoted as
laminar, transitional and turbulent flow, respectively.
The curves shown in Figure 3.3 represents the x-component of the velocity as a function of time
at a point A in the flow.
Figure 3.3 Time dependence of fluid velocity at a point
Question 3-1
The 2-cm diameter pipe is used to transport water at 20oC. What is the maximum average
velocity that may exist in the pipe for which laminar flow is guaranteed?
Solution 3-1
The Reynolds Number ranges for which laminar, transitional or turbulent pipe flows are obtained
cannot be precisely given. The actual transition from laminar, transitional or turbulent flow may
take place at various Reynolds numbers, depending on how much the flow is “distributed” by
vibrations of the pipe, roughness of the entrance region, and the like. In general, for engineering
purposes the following values are appropriate.
The flow in a round pipe is laminar if Reynolds number is less than approximately 2100
-2-
The flow in a round pipe is turbulent if Reynolds number is greater than approximately 4000
The flow in a round pipe is in transition if Reynolds number is ranging between 2100 and 4000
The kinematic viscosity of water at 20oC is   10 6 m 2 / sec . Using a Reynolds number of 2000
so that a laminar flow is guaranteed, we can simply find that,
Re 
UD
(1)

2000 
U  (0.02m)
(10 6 m 2 / sec)
U  0.1m / sec
This average velocity is quite small. Such small velocities are not usually encountered in actual
situations. Hence laminar flow is seldom of engineering interest except for specialized topics
such as lubrication. Most internal flows are turbulent flows and thus the study of turbulence
gains much attention.
3.3 Entrance Region and Fully Developed Flow
Any fluid flowing in a pipe had to enter the pipe at some location. The region of flow near where
the fluid enters the pipe is termed the entrance region and is illustrated in the Figure 3.4. As it is
shown in the figure, the fluid typically enters the pipe with a nearly uniform velocity profile. As
the fluid moves through the pipe, viscous effects cause it to stick to the pipe wall. Thus, a
boundary layer in which viscous effects are important is produced along the pipe wall such that
the initial velocity profile changes with distance along the pipe, until the fluid reaches the end of
the entrance length beyond which the velocity profile does not vary with distance x.
Figure 3.4 Entrance region of a laminar flow in a pipe or a wide rectangular channel
Question 3-2
A laminar flow is to occur in an experimental facility with 20oC water flowing through a 4cm
diameter pipe. Calculate the average velocity and the entrance length if the Reynolds number is
8000?
Solution 3-2
The shape of the velocity profile in the pipe depends on whether the flow is laminar or turbulent,
as does the length of the entrance region, LE. As with many other properties of pipe flow, the
dimensionless entrance length, LE/D, correlates quite well with Reynolds number. Typical
entrance lengths are given by
-3-
LE
 0.06 Re for laminar flow
D
(2)
LE
 4.4(Re)1/ 6 for turbulent flow
D
(3)
and
Once the fluid reaches the end of the entrance region, the flow is simpler to describe because the
velocity is a function of only the distance from the pipe centreline, r and independent of x.
UD
Re 

8000 
U  (0.04m)
(10 6 m 2 / sec)
U  0.2m / sec
and since the flow in the pipe is turbulent flow,
LE
 4.4(Re)1 / 6
D
LE
 4.4(8000)1 / 6
0.04
LE  0.04  4.4  (8000)1 / 6  0.787m.
3.4 Pressure change and shear stresses
Fully developed steady flow in a constant diameter pipe is driven by pressure forces. For
horizontal pipe flow, gravity has no effect except for a hydrostatic pressure variation across the
pipe, D , that is usually negligible. It is the pressure difference, p  p1  p 2 between one
section of the horizontal pipe and another which forces the fluid through the pipe. Viscous
effects provide the restraining force that exactly balances the pressure force, thereby allowing the
fluid to flow through the pipe with no acceleration. If viscous effects were absent in such flows,
the pressure would be constant throughout the pipe, except for the hydrostatic variation.
The need for the pressure drop can be viewed from two different standpoints. In terms of a force
balance, the pressure is needed to overcome the viscous forces generated. In terms of an energy
balance, the work done by the pressure force is needed to overcome the viscous dissipation of
energy throughout the fluid.
Figure 3.5: Element of fluid in a pipe
-4-
If gravitational effects are neglected, the pressure is constant across any vertical cross-section of
the pipe, although it varies along the pipe from one section to the next. Thus, if the pressure is
p  p1 at section (1), it is p 2  p1  p at section (2). We can foreseen the fact that the pressure
decreases in the direction of flow so that, p  0 .
The driving force due to pressure (F = Pressure x Area) can then be written as
= Pressure force at 1 - pressure force at 2
( p1 )( A1 )  ( p1  p )( A2 )  ( p1 )( A1 )  ( p1 )( A2 )  ( p )( A2 )
Since the pipe area at section (1) and at section (2) are same, A1=A2=A, the equation reduces to,
 ( p1 )( A)  ( p1 )( A)  (p)( A)
 (p )( A)  (p )(r 2 )
The retarding force is that due to the shear stress by the walls and can be written as
= shear stress × area over which it acts
= τw × area of pipe wall
 ( w )2rl
As the flow is in equilibrium, the driving forces must be equal to the retarding forces.
(p)r 2  ( w )2rl
which can be simplified to give
p 2 w

l
r
(4)
the above equation represents the basic balance in forces needed to drive each fluid particle
along the pipe with constant velocity. It is clear that at r  0 (the centerline of the pipe) there is
no shear stress (   0 ). At r  D 2 (at the pipe wall) the shear stress is maximum, denoted  w ,
the wall shear stress. The pressure drop and wall shear stress are related by
p 
4l w
D
(5)
it is important to note that small shear stress can produce a large pressure differences in the
closed conduits if the pipe is relatively long ( l D  1 ).
The shear stress dependence for turbulent flow is very complex. However, for laminar flow of a
Newtonian fluid, the shear stress is simply proportional to the velocity gradient
  
du
du
  
dy
dy
(6)
the negative sign is included to give   0 with du dr  0 (the velocity decreases from the pipe
centerline to the pipe wall). There are two governing laws for fully developed laminar flow of a
Newtonian fluid within a horizontal pipe. The one is Newton’s second law of motion and the
other is the definition of a Newtonian fluid. By combining these two equations we can obtain,
-5-
 p 
du
 
r
dr
 2l 
(7)
which can be integrated to give the velocity profile as
2
  2r  2 
 pD 2    2r  
 1      Vc 1    
u (r )  
 16l    D  
  D  
(8)
where Vc is the centerline velocity. An alternative expression can be written by using the
relationship between the wall shear stress and the pressure gradient to give
u (r ) 
 wD   r 
1   
4    R 
2



(9)
where R  D 2 is the pipe radius. This velocity profile plotted in Figure 3.5 is parabolic in the
radial coordinate, r, has a maximum velocity, Vc, at the pipe centerline, and a minimum velocity
at the pipe wall. The volume flowrate through the pipe can be obtained by integrating the
velocity profile across the pipe.
r R
R
  r 2 
Q   udA   u (r )2rdr 2Vc  1    rdr
  R  
r 0
0 
or
Q
R 2Vc
(10)
2
By definition, the average velocity is the flowrate divided by the cross-sectional area and can be
attained as the one-half of the maximum velocity.
V
Q
Vc
R 2Vc pD 2
Q
 2 

2 R
2R 2
32 l
D 4 p
128l
(11)
Figure 3.5 Shear stress distributions within the fluid in a pipe and typical velocity profiles.
The above results confirm the following properties of laminar pipe flow. For a horizontal pipe
the flowrate is:
-6-
a. Directly proportional to the pressure drop
b. Inversely proportional to the viscosity
c. Inversely proportional to the pipe length
d. Proportional to the pipe diameter to the fourth power
If we introduce the friction factor f, a quantity of substantial interest in pipe flow, a
dimensionless wall shear, defined by
f 

(12)
(1 / 8) V 2
we see that the change in pressure per unit weight in pipe can be written as
p

 hL  f
L V2
D 2g
(13)
where hL is the head loss with dimension of length. This equation is often referred to as the
Darcy-Weisbach equation, named after Henri P.G. Darcy (1803-1858) and Julius Weisbach
(1806- 1871). Combining above equations:
f 
64
Re
(14)
for laminar flow in a pipe. Substituting this back into above equation we see that
p

 hL  f
32L
D 2
(15)
the head loss is directly proportional to the average velocity (and hence the discharge also) to the
first power, a result that generally is applied to developed, laminar flows in conduits, of shape
other than circular.
Question 3-3
An oil with a viscosity of  =0.4 Ns/m2 and density  = 900 kg/m3 flows in a pipe of diameter
D=0.02m. What pressure drop, p  p1  p 2 is needed to produce a flowrate of 2x10-5 m3/s if the
pipe is horizontal with x1=0 and x2=10m.
Solution 3-3
The question mentions that the Reynolds number is less than 2100.The average velocity in the
pipe can be calculated as:
Q
V
A
V
2  10 5
 0.0637 m/sec.
 (0.01) 2
The Reynolds number is;
Re 
VD

in which  
-7-


(0.0637)  (0.02m)
 2.87
(4.44  10 4 m 2 / sec)
Hence, the flow is laminar. Therefore the pressure difference can be deducted by the following
equation with l  x2  x1  10m .
Re 
p 
p 
128l  Q
D 4
128(0.4)(10)(2  10 5 ) 1024  10 5

 20400 N / m 2  20.4kPa
 (0.02) 4
5.02  10 7
Question 3-4
A small diameter horizontal tube is connected to a supply reservoir as shown in the following
Figure. If 6600 mm3 is captured at the outlet in 10 seconds estimate the viscosity of the water.
Solution 3-4
The tube is very small, so we expect viscous effects to limit the velocity to a small value. Using
Bernoulli’s equation from the surface to the entrance to the tube, and neglecting the velocity
head, we have, letting 0 be a point on the surface:
Ptan k


2
vtan
P
v2
k
 z tan k  tube  tube  ztube

2g
2g
where we have used gage pressure with po  0 and thus ptan k  0 and the velocity head in the
reservoir is accepted as negligible. The equation becomes
P
0  0  2  tube  0  0

2  Ptube
Ptube  2  9800  19600 N / m 2  19.6kPa
at the exit of the tube the pressure is zero; hence,
p 19600

 16.3kPa / m
l
1.2
The average velocity is found to be
V
Q (6600  10 9 ) / 10

 0.840m / s

A
(0.001) 2
4
-8-
Check to make sure the velocity head is negligible: (v2/2g)=0.036m compared with pressure
head. So the assumption of negligible velocity head is valid. The viscosity of the fluid can be
found by:
V
pD 2
32 l

pD 2
32Vl
where

(16300)0.0012
 6.06  10 4 Ns / m 2
32(0.840)
we should check the Reynolds number to determine if our assumption of a laminar flow was
acceptable. It is
VD

Re 
in which  


(1000)(0.84m / sec)  (0.001m)
 138.6
(6.06  10 4 m 2 / sec)
Hence, the flow is laminar. Therefore the pressure difference can be deducted by the following
equation with l  x2  x1  10m .This is obviously a laminar flow since Re<2100, so the
Re 
calculations are valid providing the entrance length is not too long. It is
LE
 0.06 Re
D
LE  D0.06 Re  0.001  0.06  1386  0.083m
this is approximately 8% of the total length, a sufficiently small quantity, hence calculations are
assumed reliable.
3.5 Turbulent Flow in a pipe
The study of turbulent flow in a pipe is of substantial interest in actual flows since most flows
encountered in practical applications are turbulent flows in pipes. In a turbulent flow all three
velocity components are nonzero. If we measure the components as a function of time, the
graphs similar to those shown in Figure 3.6 would result for a flow in a pipe where u, v and w
are in the x-, r- and θ-directions. However, there is seldom interest to the civil engineers in the
details of the randomly fluctuating velocity components; hence, here, only the notation of a timeaveraged quantity will be introduced. The velocity components (u,v,w) are written as
u  u  u'
(16)
vvv
(17)
'
w  w  w'
(18)
where a bar over a quantity denotes a time average and a prime denotes the fluctuating part.
Using the component u as an example, the time average is defined as,
T
u
1
u (t )dt
T 0
(19)
Where T is a time increment large enough to eliminate all time dependence from u . As the flow
in the pipe becomes developed it is observed that the horizontal velocity parameter u would be
-9-
nonzero and v =0, w =0 will be attained.
Figure 3.6 Velocity components in a turbulent pipe flow
All materials are rough when viewed with sufficient magnification, although glass and plastic
assumed to be smooth with e=0 (sometimes mentioned as ks or δ). As noted in the preceding
sections shear stresses are significant only near the wall in the viscous wall layer with thickness
δv. If the thickness δv is sufficiently large, it submerges the wall roughness elements so that they
have negligible effect on the flow; it is as if the wall were smooth. Such a condition is often
referred to as being hydraulically smooth. If the viscous wall layer is relatively thin, the
roughness elements protrude out of this layer and the wall is rough. The relative roughness e/D
and the Reynolds number can be used to determine if a pipe is smooth or rough. (See Figure 3.7)
Figure 3.7: (a) A smooth wall and (b) a rough wall
Question 3-5
Water at 20oC flows in a 10 cm diameter pipe at an average velocity of 1.6 m/sec. If the friction
factor, f of the pipe is 0.018, calculate the wall shear stress and friction (shear) velocity of the
pipe.
Solution 3-5
In the wall region the characteristic velocity and the length are the shear velocity uτ is defined by
u   o / 
(20)
and the wall shear is calculated from
1
8
 o  V 2 f
(21)
Therefore the wall shear stress occurring on the walls of pipe is calculated as:
1
 1000 1.6 2  0.018  5.8Pa
8
- 10 -
and the friction velocity is

5 .8
 0.076 m/sec
1000
Question 3-6
The 4-cm diameter smooth, horizontal pipe as shown in the Figure transports 0.004 m3/sec of
water at 20oC. Calculate (a) the wall shear, (b) the pressure drop over a 10-m length, f of the pipe
is 0.018.
Solution 3-6
The average flow velocity in the pipe can be obtained by using the discharge relation,
Q  AV
Q
0.004

 3.18m / sec
A   0.02 2
As it was same in the previous question, the wall shear stress occurring on the walls of pipe can
be calculated as:
1
 o  1000  3.18 2  0.018  23Pa
8
V
As it was defined before, the pressure drop and wall shear stress are related by
p 
p 
4l o
D
4  10  23
 23000 Pa
0.04
3.6 Losses in Developed Pipe Flow
Perhaps the most calculated quantity in pipe flow is the head loss. If the head loss is known in a
developed flow, the pressure change can be calculated; The head loss that results from the wall
shear in a developed flow is related to the friction factor by the Darcy-Weisbach equation,
namely,
p

 hL  f
L V2
D 2g
(22)
Consequently, if the friction factor is known, we can find the head loss and then the pressure
drop. The friction factor f, depends on the various quantities that affect the flow, written as,
f  f (  ,  , V , D , e)
(23)
where the average wall roughness height e accounts for the influence of the wall roughness
elements. A dimensional analysis provides us with
- 11 -
f  f(
VD e
, )
 D
(24)
where e/D is the relative roughness. Experimental data that relate the friction factor to the
Reynolds number have been obtained for fully developed pipe flow over a wide range of wall
roughnesses. The results of this data are presented in Figure 3.8, which is commonly referred to
as the Moody diagram, named after Lewis F. Moody (1880-1953). There are several features of
the Moody diagram that should be noted.
For a given wall roughness, measured by the relative roughness e/D, there is a sufficiently large
value of Re above which the friction factor is constant, thereby defining the completely turbulent
regime. The average roughness element size e is substantially greater than the viscous wall layer
thickness δv, so that viscous effects are not significant.
For the smaller relative roughness e/D values it is observed that, as Re decreases, the friction
factor increases in the transition zone and eventually becomes the same as that of a smooth pipe.
The roughness elements become submerged in the viscous wall layer so that they produce little
effect on the main flow.
For Reynolds numbers less than 2000, the friction factor of laminar flow is shown. The critical
zone couples the turbulent flow to the laminar flow and may represent an oscillatory flow that
alternately exists between turbulent and laminar flow.
The following empirical equations represent the Moody diagram for Re>4000:
Smooth pipe flow:
1
 0.86 ln Re
f
f  0.8
(25)
Transition zone:
 e
1
2.51
 0.86 ln


3
.
7
D
f
Re f





(26)
Completely turbulent zone:
e
1
 0.86 ln
3.7 D
f
(27)
The transition zone equation that couples the smooth pipe equation to the completely turbulent
regime equation is known as the Colebrook Equation. Smooth pipe flow is the Colebrook
equation with e=0, and completely turbulent zone equation is the Colebrook equation with Re=∞
Three categories of problems can be identified for developed turbulent flow in a pipe of length L.
Catagory
Known
Unknown
1
Q,D,e,ν
hL
2
hL,D,e,ν
Q
3
Q,e,ν,hL
D
A category 1 problem is straightforward and requires no iteration procedure when using the
- 12 -
Moody diagram. Category 2 and 3 problems are more like problems encountered in engineering
design situations and require an iterative trial and-error process when using the Moody diagram.
Each of these types will be illustrated with following examples.
Figure 3.8: Moody diagram (Note that if e/D or ks/D=0.01 and Re=104, the f reads f=0.043
Question 3-7
Water at 20oC is transported for 500m in a 4 cm-diameter wrought iron (commercial steel)
horizontal pipe with a flow rate of 3L/sec. Calculate the pressure drop over the 500-m length of
pipe, using Moody diagram.
Solution 3-7
The average flow velocity in the pipe can be obtained by using the discharge relation,
Q  AV
V
Q
3  10 3

 2.38m / sec
A   0.02 2
The Reynolds number is
2.38  4 100 
 9.52  10 4

1.007  10 6
Obtaining e or ks from the Moody diagram we have, using D=4/100
e 0.046mm

 0.00115
D
40mm
The friction factor is read from the Moody diagram to be
f  0.023
Re 
UD

The head loss is than calculated as
- 13 -
p

 hL  f
L V2
D 2g
500 2.38 2
hL  0.023
 83m
0.04 2  9.81
The pressure drop can also be calculated as
p  hL  9810 83 =814230N/m2 or
p  814.23 kPa
Question 3-8
A pressure drop of 700kPa is measured over a 300-m length of horizontal, 10-cm diameter
commercial pipe that transports oil (S=0.9, ν=10-5 m2/s). Calculate the flow rate using Moody
diagram.
Solution 3-8
The relative roughness is
e 0.046mm

 0.00046
D 100mm
Assuming that the flow is completely turbulent (Re is not needed), the Moody diagram reads to
be
f  0.0165
The head loss is than calculated as
p
L V2
 hL  f

D 2g
hL 
p
 oil

700000
 79.4 m
9800  0.9
note that S 
 oil
 water
The velocity can be calculated from the head loss equation of Darcy Weisbach
hL  f
V 
L V2
D 2g
2 gDhL
fL
0.5
 2  9.81 0.1 79.4 
V 
  5.61m / sec
 0.0165  300 
This provides us with a Reynolds number of
UD 5.61 0.1

Re 
 5.6110 4

10 5
using this Reynolds number and e/D=0.00046, the Moody diagram gives the friction factor as
- 14 -
f  0.023
This corrects the original value for f. the velocity is recalculated to be
 2  9.81 0.1 79.4 
V 

0.023  300


0. 5
 4.75m / sec
This provides us with a Reynolds number of
UD 4.75  0.1
Re 

 4.75  10 4
5

10
From the Moody diagram f=0.023 appears to be satisfactory. Thus the flow rate is
Q  AV  4.75    0.05 2  0.037 m3/sec
Question 3-9
Drawn tubing of what diameter should be selected to transport 0.002 m3/sec of 20oC water over a
400-m length so that the head loss does not exceed 30-m? Use Moody diagram to solve the
question
Solution 3-9
In this problem we do not know D. Thus, a trial and error solution is anticipated. The average is
related to D by
Q 0.002 0.00255
V 

A   D2
D2
4
The friction factor and D are related as follows
hL  f
30  f
L V2
D 2g
400 (0.00255 / D 2 ) 2
D
2  9.81
Therefore,
D 5  4.42  10 6 f
The Reynolds number is
Re 
UD


0.00255 D 2550

D
D 2 10 6
Now, let us simply guess a value for f and check with the relations above and the Moody
diagram. The first guess is f=0.03 and the correction is listed in the following table. Note: the
second guess is the value for f found from the calculations of the first guess.
f
0.03
0.02
D(m)
0.0421
0.0388
Re
6.06x104
6.57x104
- 15 -
e/D
0.000036
0.000039
f(from Moody)
0.02
0.02
The value of f=0.02 is acceptable, yielding a diameter of 3.88 cm. Since this diameter would
undoubtedly not be a standard a diameter of
D  4 cm
Would be the tube size selected. This tube would have a head loss less than the limit of 30-m
imposed in the problem statement.
3.6 Losses in Developed noncircular conduits.
Good approximations can be made for the head loss in circuits with noncircular cross sections by
using the hydraulic radius R, defined by:
A
(28)
R
P
Where A is the cross-sectional area and P is the wetted perimeter where the fluid is in contact
with the solid boundary.
Question 3-10
Air at standard conditions (ν=1.5x10-5) is to be transported through a smooth, horizontal, 30 cm
x 20 cm rectangular duct whose length is 500-m. The flow rate is 0.24 m3/sec. Calculate the
pressure drop.
Solution 3-10
For a circular pipe flowing full, the hydraulic radius is;
R
r02 r0

2r0 2
Hence we simply replace the radius with double R and use the Moody diagram with
UD U (4 R)
Re 



And relative roughness as
e
e

D 4R
The head loss therefore becomes,
L V2
L V2
 f
D 2g
4R 2 g
The pressure drop in the duct can be calculated by
hL  f
p
L V2

D 2g
Therefore it is necessary to calculate the head loss in the duct which is directly related with the
friction coefficient of Darcy. On the other hand, since the duct is rectangular it is necessary to
find the hydraulic radius instead of D.
 hL  f
The hydraulic radius is
- 16 -
R
A
0.2  0.3

 0.06 m
P (0.3  0.2)  2
The average velocity is
V 
Q
0.24

 4 m/sec
A 0.3  0.2
This gives a Reynolds number of
4  4  0.06
 6.4  10 4

1.5  10 5
Using the smooth pipe curve of the Moody diagram, there results
f  0.0196
Re 
U (4 R)

Hence,
hL  f
L V2
500
42
 0.0196
 33.3 m
4R 2 g
4  0.06 2  9.81
Therefore, the pressure drop is
p  ghL  1.23  9.8  33.3  402 Pa
The Darcy-Weisbach equation is valid for the laminar or turbulent pipe flow of any
Newtonian fluid. A number of additional equations are also available for evaluating the head loss
in a pipe due to friction. These tend to be more empirical in nature and limited to the turbulent
flow of water. This limitation is rarely restrictive in hydraulic engineering. Their main advantage
is generally their ease of application. Although they are less accurate than the Darcy-Weisbach
equation, many hydraulic engineering applications do not require or justify the higher accuracy
anyway. One such empirical equation is the Hazen-Williams equation:
V  1.32C H R 0.63 S 0.54
(29)
V  0.85C H R 0.63 S 0.54
(30)
or
For use with U.S. customary or SI units, respectively. In either case, CH is a discharge coefficient
with typical values given in the following table
Hazen Williams coefficient for pipes
Pipe description or material
Extremely smooth and straight
Asbestos-cement
New, unlined Steel
New, riveted Steel
Old, riveted Steel
Very smooth Steel
New,unlined Cast iron
5-year-old Cast iron
10-year-old Cast iron
20-year-old Cast iron
40-year-old Cast iron
Smooth wood or masonary
Vitrified clay
Ordinary brick
Old pipe in bad condition
- 17 -
CH
140
140
140-150
110
95
130
130
120
110
100
65-80
120
110
100
60-80
R is the hydraulic radius (equal to D/4 in a full pipe), and S is the slope of the total head or
energy grade line, that is, the ratio of the head loss to the pipe length. There are both nomograph
and tabular solutions of the Hazen-Williams equation that provide rapid solutions to head
loss/discharge problems.
3.7 Minor Losses in Pipe Flow
We now know how to calculate the losses due to a developed flow in a pipe. Pipe systems do,
however, include valves, elbows, enlargements, contractions, inlets, outlets, bends etc. that cause
additional losses. Each of these devices causes a change in the magnitude and/or the direction of
the velocity vectors and hence results in a loss. In general, if the flow is gradually accelerated by
a device, the losses are very small; relatively large losses are associated with sudden
enlargements or contractions.
A minor loss is expressed in terms of a loss coefficient K, defined by
v2
(31)
2g
the values of K have been determined experimentally for the various fittings and geometry
changes of interest in piping systems. One exception is the sudden expansion from Area A1 to
area A2 for which the loss can be calculated as
hm  K

A 
hm  1  1 
A2 

2
(32)
Question 3-11
If the flow rate through a 10-cm diameter wrought iron pipe is 0.04m3/sec, find the difference in
elevation H of the two reservoirs.
Solution 3-11
The energy equation written for a control volume that contains the two reservoir surfaces can be
given as
v12
P2 v22

 z h h 

 z2
 2g 1 L m  2g
where velocities and pressures at point 1 and point 2 are equivalent and equal to zero.
z1  hL  hm  z 2
P1
Thus letting z1-z2=H, we have
- 18 -
Figure 3.9: Minor Loss coefficients K
- 19 -
H  ( K entrance  K valve  2 K elbow  K exit )
v2
L v2
f
2g
D 2g
The average velocity, Reynolds number and relative roughness are
Q
0.04
 5.09 m/sec
V 
A   0.05 2
This gives a Reynolds number of
U ( D) 5.09  0.01
Re 

 5.09 10 5

10 6
e 0.046mm

 0.00046
D 100mm
Using the Moody diagram we find that
f  0.0173
Using the loss coefficient from the given table, for an entrance, a globe valve, screwed 10-cm
diameter standard elbows, and an exit there results
5.09 2
50 5.09 2
 0.0173
2  9.81
0.1 2  9.81
H  11.2  11.4  22.6m
The minor losses are about equal to the frictional losses as expected, since there are five minor
loss elements in 500 diameters of pipe length.
H  (0.5  5.7  2(0.64)  1)
3.8 Hydraulic and Energy Grade Lines
When the energy equation is written in the form of
v12
P
v2
 z1  hL  hm  2  2  z 2
 2g
 2g
The terms have dimensions of length. This has led to the conventional use of the hydraulic grade
line and the energy grade line. The hydraulic grade line (HGL), the dashed line in Figure 3.10, in
a piping system is formed by the locus of points located a distance p/γ above the center of the
pipe, or (p/γ) + z above a preselected datum; the liquid in a piezometer tube would rise to the
HGL. The energy grade line (EGL), the solid line in Figure 3.10 is formed by the locus of points
a distance v2/2g above the HGL, or the distance (v2/2g) + (p/γ) + z above a preselected datum;
the liquid in a pitot tube would rise to the EGL. The following points are noted relating to the
HGL and the EGL.
P1




As the velocity approaches to zero, the HGL and the EGL approach each other. Thus, in a
reservoir, they are identical and lie on the surface.
The EGL and HGL slope downward in the direction of the flow due to the head loss in
the pipe. The greater the loss per unit length, the greater the slope; As the average
velocity in the pipe increases the loss per unit length increases
A sudden change occurs in the HGL and EGL whenever a loss occurs due to a sudden
geometry change as represented by the valve or the sudden enlargement.
- 20 -


A jump occurs in the HGL and the EGL whenever useful energy is added to the fluid as
occurs with a pump, and a drop occurs if useful energy is extracted from the flow as
occurs with a turbine.
At points where the HGL passes through the centerline of the pipe, the pressure is zero. If
the pipe lies above the HGL, there is a vacuum in the pipe, a condition that is often
avoided. An exception to this condition is the design of a siphon.
Figure 3.10: Hydraulic Grade Line (HGL) and Energy Grade Line (EGL) for a piping system.
Question 3-12
Water at 20oC flows between two reservoirs at the rate of 0.06m3/sec, Sketch the HGL and EGL
What is the minimum diameter DB allowed to avoid the occurrence of cavitation?
Solution 3-12
The EGL and the HGL will be sketched on the figure including sudden changes at the entrance,
contraction, enlargement and the exit. Note the large velocity head (the difference between the
EGL and the HGL) in the smaller pipe because of the high velocity. The velocity, Reynolds
number, and relative roughness in the 20-cm diameter pipe are calculated to be
Q
0.06
V 
 1.91 m/sec
A   0.12
This gives a Reynolds number of
U ( D) 1.91 0.2
Re 

 3.80  10 5

10 6
- 21 -
e 0.26mm

 0.0013
D 200mm
Using the Moody diagram we find that
f  0.022
The velocity, Reynolds number, and relative roughness in the smaller pipe are
0.06
0.0764
Q

V 
2
A
DB
DB2

4
This gives a Reynolds number of
U ( D ) 0.0764  DB 76400
Re 



DB
DB2  10 6
e 0.00026

D
DB
The minimum possible diameter is established by recognizing that the water vapor pressure at
20oC is the minimum allowable pressure. Since the distance between the pipe and the HGL is an
indication of the pressure in the pipe, we can conclude that the minimum pressure will occur at
section 2. Hence the energy equation applied between section 1, the reservoir surface and section
2 gives
P1


v12
v2
v2
L v2
L v2
P v2
 z1  ( K entrance A  K cont B  f A A
 fB B
)  2  B  z2
DA 2 g
2g
2g
2g
DB 2 g
 2g
Where the subscript A refers to the 20-cm diameter pipe. This simplifies to
(0.0764 DB2 ) 2
101000
1.912
30 1.912
 0  20  (0.5
 0.25
 0.022

9810
2  9.81
2  9.81
0.2 2  9.81
 fB
20 (0.0764 DB2 ) 2
2450 (0.0764 DB2 ) 2
)

0
2  9.81
9810
2  9.81
DB
Which results
98600 
1.25
20
 fB 5
4
DB
DB
Where we assume that Kcont=0.25. This requires a trial and error solution. The following
illustrates the procedure
Let DB=0.1 m
Then (e/D) =0.0026 and Re  7.6 10 5
Therefore f  0.026
?
98600  12500  52000
Let DB=0.09 m
- 22 -
Then (e/D) =0.0029 and Re  8.4  10 5
Therefore f  0.027
?
98600  19000  91000
We can easily observe that 0.1m is too large and 0.09 m is too small. In fact, the value of 0.09 m
is only slightly too small. Consequently, to be safe we must select the next larger pipe size, of
0.1meter diameter. If there were a pipe size of 9.5-cm diameter, that could be selected. Assuming
that such a size is not available we could select DB=10cm
3.9 Pipeline Systems
Pipes in series
One type of pipe system has already been considered in Example 3-11 and 3-12, namely pipes in
series. Inspection of that example would indicate the two basic premises associated with pipes in
series: first, continuity requires that the discharge is the same in all pipes, and second, the total
head loss for the system equals the sum of the individual head losses of the respective pipes.
These principles can be generalized for any number of pipes that are connected in series. It
should be noted that if minor losses are included, they must be referenced to the appropriate pipe
size and velocity head. Similarly, the head loss due to friction must be evaluated separately for
each pipe size. When the Darcy-Weisbach equation is used in the analysis of any type of pipe
system, the resistance coefficients are frequently treated as constants.
Figure 3.11: Three pipes in series system.
Therefore, according to above definition in the pipes connected in series;
The total head loss is the sum of head losses in each pipe:
(hL ) total  hLa  hLb  hLc
The flow rate, in each of the pipes connected is the same.
Qtotal  Qa  Qb  Qc
Pipes in Parallel
A second pipe system is composed of a number of pipes connected in parallel (Fig. 3-12).
Although any number of pipes could be so connected, the development of the concepts will be
based on just three pipes. In passing, note that the pipe sections upstream and downstream of the
parallel pipes could be considered to be in series with the parallel pipe system of our immediate
interest, and once the parallel pipe system is analyzed the remaining system could be treated as a
number of pipes in series.
Returning to the three parallel pipes, continuity can be applied with the result
- 23 -
Qtotal  Qa  Qb  Qc
Since each of the pipes must share common values of piezometric head at the junctions, it
follows that the head loss must be identical in each of the parallel pipes. This gives a second
relationship that may be expressed as
(hL ) total  hLa  hLb  hLc
Figure 3.12: Three pipes in parallel system.
Branching Pipe System
Another type of pipe system may be introducing by considering the three interconnected
reservoirs of Fig. 3-13. Subsequently, we will generalize the analysis to include additional
reservoirs and/or additional junctions. For the present, consider the problem of estimating the
discharge in each of the pipe sections.
Figure 3.13: Three-branch reservoir system.
During the solution of branch systems the necessary numbers of equations are obtained by
utilizing energy and continuity concepts. However, it is difficult to incorporate the velocity head
at the junction, a problem that is circumvented by assuming that it is a small quantity relative to
the other energy terms and may justifiably be ignored. It is usual to make the same assumption
with respect to possible minor losses. A second problem is that the flow direction in pipe DB is
not known at the outset. Our procedure will be to formulate the energy equation so as to
determine the actual flow direction in pipe DB and then proceed on the basis of the outcome of
this first calculation. If the piezometric head (since the velocity head is neglected) at the junction
is assumed equal to the elevation of reservoir surface B, then no flow would occur in pipe DB,
and the continuity equation becomes QAD=QDC. In addition, the following energy equations can
be written (neglecting the velocity heads):
- 24 -
z A  zD 
zB  zD 
zC  z D 
PD

PD

PD

 hL AD
 hLDB
 hLDC
Where the head losses are defined using the Darcy-Weisbach equation.
hLAD  f AD
2
L AD V AD
D AD 2 g
hLDB  f DB
2
LDB VDB
DDB 2 g
hLDC  f DC
2
LDC VDC
DDC 2 g
the continuity equation is
Q AD  QDB  QDC
or
AADV AD  ADBVDB  ADCVDC
By substituting the headloss expressions respectively into equations, the three energy
equations have four unknowns PD/γ, VAD, VDB and VDC. The continuity equation provides the
fourth equation to solve for the four unknowns.
Question 3-13
Water flows at a rate of 0.030-m3/sec from reservoir 1 to reservoir 2 through three pipes
connected in series (f=0.025) as shown in the following figure. Neglecting minor losses,
determine the difference in water surface elevation.
Solution 3-13
Write down the energy equation from reservoir 1 to reservoir 2.
- 25 -
P1


v12
P
v2
 z1   hL  2  B  z 2
2g
 2g
z1   hL  z 2 where
h
L
 z1  z 2
Writing the total head loss in expanded form results in:
 hL  hLA  hLB  hLC
Which results in:
hLA  hLB  hLC  z1  z 2
Using the Darcy-Weisbach equation
hLA  f A
L A V A2
DA 2 g
hLB  f B
LB VB2
DB 2 g
hLC  f C
LC VC2
DC 2 g
Therefore the energy equation is turn out to be
z1  z 2  0.025
VC2
V A2
VB2
1000
1500
2000
 0.025
 0.025
200 / 1000 2(9.81)
180 / 1000 2(9.81)
220 / 1000 2(9.81)
z1  z 2  6.37V A2  10.62VB2  11.58VC2
Use continuity equation to determine the velocities;
0.03  4
Q
VA 

 0.955 m/sec
AA   (200 / 1000) 2
VB 
0.03  4
Q

 1.180 m/sec
AB   (180 / 1000) 2
VC 
0.03  4
Q

 0.79 m/sec
AC   (220 / 1000) 2
z1  z 2  6.37(0.955) 2  10.62(1.180) 2  11.58(0.790) 2
z1  z 2  5.81  14.79  7.22
z1  z 2  27.82 m
Question 3-14
The three pipe system shown in Figure below has the following characteristics
pipe
A
B
C
D (cm)
25
30
35
L (m)
500
650
1000
f
0.020
0.025
0.030
Find the flowrate of water in each pipe and the pressure at point 3. Neglect minor losses
- 26 -
Solution 3-14
Write down the energy equation from 1 to 2.
P1


v12
P v2
 z1   hL  2  2  z2
2g
 2g
0  0  200   hL  0 
Where V2 is Vc and
h
L
v22
 80
2g
 hL A  hLC and using the Darcy Weisbach equation, we get
120 
vC2
L v2
L v2
 f A A A  fC C C
DA 2 g
2g
DC 2 g
Because the pipes A and B are parallel, the headloss in A is equal to the headloss in B, so the
headloss for B can also be used in the above energy equation instead of for A. This energy
equation has two unknowns VA and VC, so that continuity can be used as a second equation.
QA  QB  QC
AAv A  AB vB  AC vC
Which introduces a third unknown VA. Since hL A  hLB the third equation is
fA
0.020
LA v A2
L v2
 fB B B
DA 2 g
DB 2 g
500 v A2
650 vB2
 0.025
0.25 2(9.81)
0.30 2(9.81)
2.039v A2  2.760vB2
vB  0.859v A
So now, we have three equations and three unknowns. Using the continuity equation, we get

(0.25) 2
(0.30) 2
(0.35) 2
vA  
vB  
vC
4
4
4
0.05v A  0.07vB  0.096vC
Substituting
vB  0.859v A
0.05v A  0.07(0.859)v A  0.096vC
0.11v A  0.096vC
v A  0.872vC
Substituting v A  0.872vC into energy equation
- 27 -
120 
500 (0.872vC ) 2
1000 vC2
vC2
 0.020
 0.03
2(9.81)
0.25 2(9.81)
0.35 2(9.81)
And solve for Vc
120  (1  40  85.71)
vC2
v2
 126.71 C
2(9.81)
2(9.81)
vC  4.31 m/sec
The flow rate is then,
0.352
 4.31  0.414 m3/sec
4
The pressure at 3 can be computed using the energy equation from 1 to 3 or from 3 to 2. Using
Q  AC vC  
v32
P v2
 z3   hL (3  2 )  2  2  z2
 2g
 2g
Since V3=V2, the velocity head terms cancel out:
P3
 140   hL (3 2)  80
P3


P3

P3

 60  f C
 60  0.030
P3

LC vC2
DC 2 g
1000 (4.31) 2
0.35 2(9.81)
 60  81.15
P3

 21.15 m
and
P3  9.81  1000  21.15
P3  207481.5 Pa
P3  207.48 kPa (kN/m2)
Question 3-15
The pipe system shown in Figure below connects two reservoirs that have an elevation
difference of 20m. This pipe system consists of 200m of 50-cm concrete pipe (pipe A), that
branches into 400-m of 20-cm pipe (pipe B) and 400-m of 40-cm pipe (pipe C) in parallel. Pipes
B and C join into a single 50-cm pipe that is 500m long (pipe D). For f=0.030 in all pipes, what
is the flow rate in each pipe of the system?
- 28 -
Solution 3-15
The objective is to compute the velocity in each pipe. We know that VA=VD because they are the
same diameter pipe, hLB  hLC because pipes B and C are in parallel, and
QA  QD  QB  QC
Express hLB  hLC in term of the velocities,
fB
LB vB2
L v2
 fC C C
DB 2 g
DC 2 g
Since f B  f C and LB=LC
vB2
v2
 C
DB DC
vB2
vC2

or
20 100 40 100
vB2 
v
1 2
vC or vB  C or vC  2vB
2
2
Using QA  QB  QC , we get

(50 100) 2
(20 100) 2
(40 100) 2
vA  
vB  
vC
4
4
4
502 v A  202 vB  402 vC
substituting vC  2vB yields
2500v A  400vB  1600 2vB
v A  1.065vB or vB  0.939v A
Next convert the parallel pipes to a single equivalent DA=DD=50-cm diameter pipe, with a length
LE.
f
LB vB2
L v2
 f E A
DB 2 g
DA 2 g
LB 2 LE 2
vB 
vA
DB
DA
LE
400 2
vB 
v A2
20 100
50 100
1000vB2  LE v A2
LE  882 m
Write the energy equation from reservoir surface to reservoir surface
20  f
20  0.030
( LA  LE  LD ) v A2
50 100
2g
(200  882  500) v A2
50 100
2  9.81
v A  2.033 m/sec.
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h
L
 20 m.
QA  
(50 100) 2
(2.033)  0.399 m3/sec.
4
also
v A  vD and QD  QA
vB  0.939(2.033)  1.909 m/sec
QB  
(20 100) 2
(1.909)  0.060 m3/sec.
4
QC  QA  QB
0.399  0.06  0.339 m3/sec.
- 30 -