INTERNAL INCOMPRESSIBLE
VISCOUS FLOW
Nazaruddin Sinaga
Laboratorium Efisiensi dan Konservasi Energi
Universitas Diponegoro
1
Outline
 Flow Measurements
2
3
4
Osborne Reynolds Experiment
5
6
Laminar and Turbulent Flow
7
Entrance Length
8
Developing Flow
9
10
11
12
13
14
15
16
2
1
5
17
18
19
20
21
22
23
24
25
26
27
Types of Flow
• The physical nature of fluid flow can be categorized into three types, i.e.
laminar, transition and turbulent flow. Reynolds Number (Re) can be
used to characterize these flow.
VD VD
Re 



where
(6.3)
 = density
 = dynamic viscosity
 = kinematic viscosity ( = /)
V = mean velocity
D = pipe diameter
In general, flow in commercial pipes have been found to conform to the
following condition:
Laminar Flow:
Re <2000
Transitional Flow :
2000 < Re <4000
Turbulent Flow :
Re >4000
28
Laminar Flow
 Viscous shears dominate in this type of flow and the
fluid appears to be moving in discreet layers. The
shear stress is governed by Newton’s law of viscosity
du

dy
 In general the shear stress is almost impossible to
measure. But for laminar flow it is possible to
calculate the theoretical value for a given velocity,
fluid and the appropriate geometrical shape.
29
Pressure Loss During A Laminar Flow In A Pipe
- In reality, because fluids are viscous,
energy is lost by flowing fluids due to
friction which must be taken into account.
• - The effect of friction shows itself as a
pressure (or head) loss. In a pipe with a
real fluid flowing, the shear stress at the
wall retard the flow.
• - The shear stress will vary with velocity of
flow and hence with Re. Many experiments
have been done with various fluids
measuring the pressure loss at various
Reynolds numbers.
• - Figure below shows a typical velocity
distribution in a pipe flow. It can be seen
the velocity increases from zero at the wall
to a maximum in the mainstream of the
flow.
30
A typical
velocity
distribution in
a pipe flow
 In laminar flow the paths of individual particles of fluid
do not cross, so the flow may be considered as a series
of concentric cylinders sliding over each other.
 Lets consider a cylinder of fluid with a length L, radius
r, flowing steadily in the center of pipe.
Cylindrical of fluid flowing steadily in a pipe
31
The fluid is in equilibrium, shearing forces equal the pressure forces.
Shearing force = Pressure force
2rL  PA  Pr 2

P r
L 2
(6.5)
Taking the direction of measurement r (measured from the center of
pipe), rather than the use of y (measured from the pipe wall), the
above equation can be written as;
du r
  
dr
Equatting (6.5) with (6.6) will give:
P r
du
 
L 2
dr
du
P r

dr
L 2
(6.6)
 In an integral form this gives an expression for velocity, with the
values of r = 0 (at the pipe center) to r = R (at the pipe wall)
u
P 1 r  R
r  0 rdr
L 2

R 2  r 2  P
ur 
4
(6.7)
L
where P = change in pressure
L = length of pipe
R = pipe radius
r = distance measured from the center of pipe
The maximum velocity is at the center of the pipe, i.e. when r = 0.
R 2 P
u max  
4 L
It can be shown that the mean velocity is half the maximum velocity,
i.e. V=umax/2
Shear stress and velocity distribution in pipe for laminar flow
The discharge may be found using the Hagen-Poiseuille equation,
which is given by the following;
P D 4
Q
L 128
(6.8)
The Hagen-Poiseuille expresses the discharge Q in terms of the
pressure gradient  dP  P  , diameter of pipe, and viscosity of the
 dx L 
fluid.
Pressure drop throughout the length of pipe can then be calculated
by
8LQ 8L
64 V 2 L 64 V 2 L
2
(6.9)
P 
 4 VR  

4
R
R
VD 2D Re 2D
Fully Developed Laminar Flow
in a Pipe
• Velocity Distribution
Shear Stress Distribution
35
Fully Developed Laminar Flow
in a Pipe
• Volume Flow Rate
Flow Rate as a Function of Pressure Drop
36
Fully Developed Laminar Flow
in a Pipe
• Average Velocity
Maximum Velocity
37
38
Turbulent Flow
 This is the most commonly occurring flow in engineering
practice in which fluid particles move erratically causing
instantaneous fluctuations in the velocity components.
 These fluctuations cause additional shear stresses. In this
type of flow both viscous and turbulent shear stresses exists.
 Thus, the shear stress in turbulent flow is a combination of
laminar and turbulent shear stresses, and can be written as:
   lam   tur
where
dU
    
dy
 = dynamic viscosity
 = eddy viscosity which is not a fluid property but
depends upon turbulence condition of flow.
39
40
Velocity Profile for Fully Developed
Pipe Flow
41
 The velocity at any point in the cross-section will be proportional
to the one-seventh power of the distance from the wall, which can
be expressed as:
1/ 7
Uy
y
 
 
(6.10)
U CL  R 
where Uy is the velocity at a distance y from the wall, UCL is the
velocity at the centerline of pipe, and R is the radius of pipe. This
equation is known as the Prandtl one-seventh law.
Figure below shows the velocity profile for turbulent flow in a
pipe. The shape of the profile is said to be logarithmic.
Velocity profile for turbulent flow
42
1
0.9
0.8
Uy/UCL
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
y/R
43
For smooth pipe:
U
 yU * 
 5.75 log 10 
  5.5
U*
  
(6.11a)
 For rough pipe:
U
 y
 5.75 log 10    8.5
U*
k
(6.11b)
In the above equations, U represents the velocity at a distance y
from the pipe wall, U* is the shear velocity =  
y is the distance form the pipe wall, k is the surface roughness and
 is the kinematic viscosity of the fluid.
44
Turbulent Velocity Profiles in Fully
Developed Pipe Flow
45
46
47
Head Losses
48
Head Losses
The momentum balance in the flow direction is thus given by
49
50
51
• Head Loss
53
54
Friction Factor
55
56
57
Nikuradse’s Experiments
•
In general, friction factor
f  F (Re,
–
•
e
)
D
Function of Re and roughness
f 
Laminar region
64
f 
Re
–
•
k
Re 
1/ 4
Rough
Blausius
Independent of roughness
Turbulent region
–
Smooth pipe curve
•
–
f 
64
Re
Rough pipe zone
•
f 
All curves coincide @
~Re=2300
All rough pipe curves flatten
out and become
independent of Re
Blausius OK for smooth pipe
0.25

5.74 
 e


log 10 
 3.7 D Re 0.9 

2
Laminar
Transition
Turbulent
Smooth
59
Moody Diagram
60
61
Calculation of Minor Head Loss
• Minor Losses
– Examples: Inlets and Exits; Enlargements and Contractions;
Pipe Bends; Valves and Fittings
62
Pipe Entrance
• Developing flow
– Includes boundary layer
and core,
– viscous effects grow inward
from the wall
• Fully developed flow
– Shape of velocity profile is
Pressure
same at all points along
Entrance
pipe
Fully developed
flow region
Entrance length Le
pressure drop
Le  0.06 Re

D 4.4Re1/6
Laminar flow
Turbulent flow
Region of linear
pressure drop
Le
x
Entrance Loss in a Pipe
• In addition to frictional losses,
there are minor losses due to
– Entrances or exits
– Expansions or contractions
– Bends, elbows, tees, and
other fittings
Abrupt inlet, K ~ 0.5
– Valves
• Losses generally determined by
experiment and then corellated
hL
K

or
with pipe flow characteristics
2
V
• Loss coefficients are generally
2g
given as the ratio of head loss
to velocity head
K – loss coefficent
V2
hL  K
2g
K ~ 0.1 for well-rounded inlet (high Re)
K ~ 1.0 abrupt pipe outlet
K ~ 0.5 abrupt pipe inlet
Elbow Loss in a Pipe
• A piping system may have many minor
losses which are all correlated to
V2/2g
• Sum them up to a total system loss for
pipes of the same diameter
V2
hL  h f   hm 
2g
m
 L

f

K

m
 D
m


• Where,
hL  Total head loss
h f  Frictional head loss
hm  Minor head loss for fitting m
K m  Minor head loss coefficien t for fitting m
Calculation of Minor Head Loss
66
67
68
69
70
71
72
73
74
Example 1
 Water at 10C is flowing at a rate of 0.03 m3/s through a pipe. The pipe
has 150-mm diameter, 500 m long, and the surface roughness is estimated
at 0.06 mm. Find the head loss and the pressure drop throughout the
length of the pipe.
Solution:
 From Table 1.3 (for water):  = 1000 kg/m3 and  =1.30x10-3 N.s/m2
V = Q/A and
A=R2
A = (0.15/2)2 = 0.01767 m2
V = Q/A =0.03/.0.01767 =1.7 m/s
Re = (1000x1.7x0.15)/(1.30x10-3) = 1.96x105 > 2000  turbulent flow
To find , use Moody Diagram with Re and relative roughness (k/D).
k/D = 0.06x10-3/0.15 = 4x10-4
From Moody diagram,   0.018
The head loss may be computed using the Darcy-Weisbach equation.
500 x 1.7 2
L V2
hf  
 0.018 x
 8.84m.
D 2g
0.15 x 2 x 9.81
The pressure drop along the pipe can be calculated using the relationship:
ΔP=ghf = 1000 x 9.81 x 8.84
ΔP = 8.67 x 104 Pa
75
Example 2
 Determine the energy loss that will occur as 0.06 m3/s water flows
from a 40-mm pipe diameter into a 100-mm pipe diameter through
a sudden expansion.
Solution:
 The head loss through a sudden enlargement is given by;
2
V
hm  K a
2g
Va 
Q
0.06

 3.58 m / s
2
Aa  (0.04 / 2)
Da/Db = 40/100 = 0.4
From Table 6.3: K = 0.70
Thus, the head loss is
3.58 2
h Lm  0.70 x
 0.47m
2 x 9.81
76
Example 3
 Calculate the head added by the pump when
the water system shown below carries a
discharge of 0.27 m3/s. If the efficiency of
the pump is 80%, calculate the power input
required by the pump to maintain the flow.
77
Solution:
Applying Bernoulli equation between section 1 and 2
P1
V12
P2
V2 2
 z1 
 Hp 
 z2 
  H L12
g
2g
g
2g
(1)
P1 = P2 = Patm = 0 (atm) and V1=V2 0
Thus equation (1) reduces to:
H p  z 2  z1   H L12
(2)
HL1-2 = hf + hentrance + hbend + hexit
V2
 H L1 2 
2g
1000


 0. 5  0. 4  1 
 0.015x
0.4


V2
 39.4
2g
From (2):
V2
H p  230  200  39.4
2x9.81
The velocity can be calculated using the
continuity equation:
V
Q
0.27

 2.15 m / s
2
A 0.4 / 2
Thus, the head added by the pump:
p 
Pin 
gQH p
gQH p
p
Hp = 39.3 m
Pin

1000x9.81x 0.27 x39.3
0.8
Pin = 130.117 Watt ≈ 130 kW.
Pipe Flow Analysis
• Pipeline system used in water distribution,
industrial application and in many engineering
systems may range from simple arrangement to
extremely complex one.
• Problems regarding pipelines are usually tackled
by the use of continuity and energy equations.
• The head loss due to friction is usually calculated
using the D-W equation while the minor losses are
computed using equations 6.16, 6.16(a) and
6.16(b) depending on the appropriate conditions.
80
Pipes in Series
 When two or more pipes of
different diameters or
roughness are connected in
such a way that the fluid
follows a single flow path
throughout the system, the
system represents a series
pipeline.
 In a series pipeline the total
energy loss is the sum of the
individual minor losses and
all pipe friction losses.
Pipelines in series
81
 Referring to Figure 6.11, the Bernoulli equation can be
written between points 1 and 2 as follows;
P1
V12 P2
V2 2
 z1 

 z2 
 H L12
g
2g
g
2g
where
(6.18)
P/g = pressure head
z
= elevation head
V2/2g = velocity head
HL1-2 = total energy lost between point 1
and 2
Realizing that P1=P2=Patm, and V1=V2, then equation (6.14)
reduces to
z1-z2 = HL1-2
82
Or we can say that the different of reservoir
water level is equivalent to the total head losses
in the system.
The total head losses are a combination of the all
the friction losses and the sum of the individual
minor losses.
HL1-2 = hfa + hfb + hentrance + hvalve + hexpansion + hexit.
Since the same discharge passes through all the
pipes, the continuity equation can be written as;
Q1 = Q2
83
Pipes in Parallel
• A combination of two
or more pipes
connected between
two points so that the
discharge divides at
the first junction and
rejoins at the next is
known as pipes in
parallel. Here the
head loss between the
two junctions is the
same for all pipes.
Figure 6.12 Pipelines in
parallel
84
 Applying the continuity equation to the system;
Q1 = Qa + Qb = Q2
(6.19)
 The energy equation between point 1 and 2 can be written
as;
P1
V12 P2
V2 2
 z1 

 z2 
 HL
g
2g
g
2g
 The head losses throughout the system are given by;
HL1-2=hLa = hLb
(6.20)
 Equations (6.19) and (6.20) are the governing relationships
for parallel pipe line systems. The system automatically
adjusts the flow in each branch until the total system flow
satisfies these equations.
Pipe Network
 A water distribution system consists of complex
interconnected pipes, service reservoirs and/or pumps,
which deliver water from the treatment plant to the
consumer.
 Water demand is highly variable, whereas supply is
normally constant. Thus, the distribution system must
include storage elements, and must be capable of
flexible operation.
 Pipe network analysis involves the determination of the
pipe flow rates and pressure heads at the outflows
points of the network. The flow rate and pressure
heads must satisfy the continuity and energy
equations.
86
Pipe Network
 The earliest systematic method of network analysis
(Hardy-Cross Method) is known as the head balance or
closed loop method. This method is applicable to
system in which pipes form closed loops. The outflows
from the system are generally assumed to occur at the
nodes junction.
 For a given pipe system with known outflows, the
Hardy-Cross method is an iterative procedure based on
initially iterated flows in the pipes. At each junction
these flows must satisfy the continuity criterion, i.e.
the algebraic sum of the flow rates in the pipe meeting
at a junction, together with any external flows is zero.
87
• Assigning clockwise flows and their associated
head losses are positive, the procedure is as
follows:
 Assume values of Q to satisfy Q = 0.
 Calculate HL from Q using
HL = K1Q2 .
 If HL = 0, then the solution is correct.
88
 If HL  0, then apply a correction factor,
Q, to all Q and repeat from step (2).
 For practical purposes, the calculation is
usually terminated when HL < 0.01 m or Q
< 1 L/s.
 A reasonably efficient value of Q for rapid
convergence is given by;
H

Q  
2 H
(6.21)
L
L
89
Q
Example 4
• A pipe 6-cm in diameter, 1000m long and with  = 0.018 is
connected in parallel between two points M and N with
another pipe 8-cm in diameter, 800-m long and having  =
0.020. A total discharge of 20 L/s enters the parallel pipe
through division at A and rejoins at B. Estimate the
discharge in each of the pipe.
90
Solution:
Continuity: Q = Q1 + Q2
0.02 


(0.06) 2 V12  (0.08) 2 V22
4
4
V1  1.778V2  7.074
Pipes in parallel: hf1 = hf2
L V2
L V2
1 1 1   2 2 2
2gD 1
2gD 2
0.018x1000 2 0.020 x800 2
V1 
V2
0.06
0.08
V1  0.8165V2
(2)
Substitute (2) into (1)
0.8165V2 + 1.778 V2 = 7.074
V2 = 2.73 m/s
91
(1)

Q 2  A 2 V2  (0.08) 2 x 2.73
4
Q2 = 0.0137 m3/s
From (2):
V1 = 0.8165 V2 = 0.8165x2.73 = 2.23 m/s
Q1 = 0.0063 m3/s
Recheck the answer:
Q1+ Q2 = Q
0.0063 + 0.0137 = 0.020
(same as given Q  OK!)
92
Example 6.6
• For the square loop shown, find the discharge in all the
pipes. All pipes are 1 km long and 300 mm in diameter, with
a friction factor of 0.0163. Assume that minor losses can be
neglected.
93
Solution:
 Assume values of Q to satisfy continuity equations all at nodes.
 The head loss is calculated using; HL = K1Q2
 HL = hf + hLm
 But minor losses can be neglected:  hLm = 0
 Thus HL = hf
 Head loss can be calculated using the Darcy-Weisbach equation
L V2
hf  
D 2g
94
L V2
HL  hf  
D 2g
1000
V2
H L  0.0163 x
x
0.3 2 x 9.81
H L  2.77
Q2
 2.77 x
A2
Q2

2
 x 0.3 
4

2
H L  554Q 2
H L  K' Q 2
 K '  554
First trial
Pipe
Q (L/s)
HL (m)
HL/Q
AB
60
2.0
0.033
BC
40
0.886
0.0222
CD
0
0
0
AD
-40
-0.886
0.0222
2.00
0.0774

Since HL > 0.01 m, then correction has to be applied.
95
Q  
2
 HL

 12.92 L / s
H
2
x
0
.
0774
2 L
Q
Second trial
Pipe
Q (L/s)
HL (m)
HL/Q
AB
47.08
1.23
0.0261
BC
27.08
0.407
0.015
CD
-12.92
-0.092
0.007
-0.0107
0.07775
Since HL ≈ 0.01 m, then it is OK.
AD
-52.92
-1.555
0.0294
Thus, the discharge in each pipe is as follows (to the nearest integer).

Pipe
Discharge (L/s)
AB
47
BC
27
CD
-13
AD
-53
96
Solution of Pipe Flow Problems
• Single Path
– Find p for a given L, D, and Q
 Use energy equation directly
– Find L for a given p, D, and Q
 Use energy equation directly
Solution of Pipe Flow Problems
• Single Path (Continued)
– Find Q for a given p, L, and D
1. Manually iterate energy equation and friction factor
formula to find V (or Q), or
2. Directly solve, simultaneously, energy equation and
friction factor formula using (for example) Excel
– Find D for a given p, L, and Q
1. Manually iterate energy equation and friction factor
formula to find D, or
2. Directly solve, simultaneously, energy equation and
friction factor formula using (for example) Excel
Solution of Pipe Flow Problems
•
Multiple-Path Systems
Example:
Solution of Pipe Flow Problems
• Multiple-Path Systems
– Solve each branch as for single path
– Two additional rules
1. The net flow out of any node (junction) is zero
2. Each node has a unique pressure head (HGL)
– To complete solution of problem
1. Manually iterate energy equation and friction factor for each
branch to satisfy all constraints, or
2. Directly solve, simultaneously, complete set of equations using
(for example) Excel
Flow Measurement
• Direct Methods
– Examples: Accumulation in a Container; Positive
Displacement Flowmeter
• Restriction Flow Meters for Internal Flows
– Examples: Orifice Plate; Flow Nozzle; Venturi; Laminar
Flow Element
Flow Measurement
• Linear Flow Meters
– Examples: Float Meter (Rotameter); Turbine; Vortex;
Electromagnetic; Magnetic; Ultrasonic
Flow Measurement
• Traversing Methods
– Examples: Pitot (or Pitot Static) Tube; Laser Doppler
Anemometer
The End
Terima kasih
104
Shear Stress in Pipes
•
Steady, uniform flow in a pipe: momentum
flux is zero and pressure distribution across
pipe is hydrostatic, equilibrium exists
between pressure, gravity and shear forces
dp
s ) A  W sin    0 (D )s
ds
dp
dz
0
sA  As   0 (D )s
ds
ds
D
d
p
 0  [  (  z )]
4 ds 
D dh
0  
4 ds
4 L 0
h1  h2  h f 
D
 Fs  0  pA  ( p 
•
•
Since h is constant across the crosssection of the pipe (hydrostatic), and –
dh/ds>0, then the shear stress will be
zero at the center (r = 0) and increase
linearly to a maximum at the wall.
Head loss is due to the shear stress.
•
•
Applicable to either laminar or turbulent
flow
Now we need a relationship for the shear
stress in terms of the Re and pipe
roughness
Darcy-Weisbach Equation
0

V

D
e
ML-1T-2
ML
LT-1
ML-1T-1
L
L
-3
 0  F (  , V ,  , D, e)
 4  F ( 1 ,  2 )
Repeating variables :  , V , D
hf 
4L

D 0
4L
e
V 2 F (Re, )
D
D

e
 1  Re;  2  ;  3  0 2
D
V
0
e
 F (Re, )
D
V 2

L V2

D 2g
e 

8F (Re, D )
e
 0  V 2 F (Re, )
D
hf  f
L V2
D 2g
Darcy-Weisbach Eq.
f  8F (Re,
e
)
D
Friction factor
Laminar Flow in Pipes
•
Laminar flow -- Newton’s law of viscosity is valid:
dV
r dh

dy
2 ds
dV

dr
r dh

2  ds
r dh

dr
2  ds
 
dV
dy
dV
dr
dV
V 
r 2 dh
C
4  ds
r 2 dh
C 0
4  ds
2
r 2 dh   r  
1    
V  0
4  ds   r0  


  r 2 
V  Vmax 1    
  r0  


•
Velocity distribution in a pipe (laminar
flow) is parabolic with maximum at center.
Discharge in Laminar Flow
 dh 2 2
( r0  r )
4  ds
 dh 2 2
Q   VdA  0r0 
( r0  r )( 2rdr )
4  ds
V 
 dh ( r 2  r02 ) 2

4  ds
2
Q
r04 dh
8 ds
D 4 dh

128 ds
V 
Q
A
V 
D 2 dh
32  ds
r0
0
Head Loss in Laminar Flow
V 
D 2 dh
32  ds
32 
dh
 V
ds
D 2
32 
dh  V
ds
D 2
32 
h2  h1  V
( s2  s1 )
D 2
h1  h2  h f
hf 
32 LV
D 2
hf 

32 LV
D 2
32 LV V 2 / 2
D 2 V 2 / 2

L
 64(
)( ) V 2 / 2
V D D

64 L
( ) V 2 / 2
Re D
hf  f
L V 2
D 2
f 
64
Re
EGL & HGL for Losses in a Pipe
•
•
•
Entrances, bends, and other flow transitions
cause the EGL to drop an amount equal to
the head loss produced by the transition.
EGL is steeper at entrance than it is
downstream of there where the slope is
equal the frictional head loss in the pipe.
The HGL also drops sharply downstream of
an entrance
Ex. (10.8)
Given: Kerosene (S=0.94, =0.048 N-s/m2).
Horizontal 5-cm pipe. Q=2x10-3 m3/s.
Find: Pressure drop per 10 m of pipe.
Solution:
V2 p
V2 p
α1 1  1  z1  hL  α2 2  2  z 2
2g
γ
2g
γ
32 μLV
hL 
γD 2
0  0  0.5 
32 μLV
V22
 α2
00
2g
γD 2
α2 2 32 μL
V2 
V  0.5  0
2g
γD 2
2 2
32 * 4 * 10  5 * 10
V2 
V  0.5  0
2g
0.8 * 62.4 * (1 / 32) 2
V22  8.45V  16.1  0
V  1.60 ft / s
0.8 * 1.94 * 1.6 * (0.25 / 12)
Re 
 1293 (laminar)
4 * 10  5
Q  V * A  1.6 *  * (0.25/12) 2 / 4  1.23 * 10  3 cfs
Ex. (10.34)
Given: Glycerin@ 20oC flows commercial
steel pipe.
Find: h
Solution:   12,300 N / m,   0.62 Ns / m2
V2 p
V2 p
α1 1  1  z1  hL  α2 2  2  z 2
2g
γ
2g
γ
p1
p
 z1  hL  2  z 2
γ
γ
p
p
h  1  z1  ( 2  z 2 )  hL
γ
γ
VD VD 0.6 * 0.02
Re 


 23.5 (laminar)


5.1 * 10  4
32 μLV 32(0.62)(1)( 0.6)
h  h L 

 2.42 m
γD 2
12,300 * (0.02) 2
Ex. (10.43)
Given: Figure
Find: Estimate the elevation required in the 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  α2 2  2  z 2
2g
γ
2g
γ
0  0  z1   hL  0  0  z 2
V2 p
V2 p
α1 1  1  z1   hL  αb b  b  zb
2g
γ
2g
γ
V2 p
0  0  z1   hL  1 * b  b  zb
2g
γ
L V 2

 hL   K e  2 K b  K E  f 
D  2g

K e  0.5; K b  0.4 (assumed) ; K E  1.0; f
L
430
 0.025 *
 10.75
D
1
Q
10
V  
 12.73 ft / s
A  / 4 * 12
z1  100  0.5  2 * 0.4  1.0  10.75
12.732
 133 ft
2 * 32.2
pb
V2 
L V 2
 z1  zb  b   K e  K b  f 
γ
2g 
D  2g
300  12.732

 133  110.7  1.0  0.5  0.4  0.025

1  2 * 32.2

 1.35 ft
pb  62.4 * ( 1.53)  0.59 psig
Re 
VD


12.73 * 1
1.14 * 10
5
 9 * 105
Ex. (10.68)
Given: Commercial steel pipe to carry 300 cfs of
water at 60oF with a head loss of 1 ft per 1000 ft
of pipe. Assume pipe sizes are available in even
sizes when the diameters are expressed in
inches (i.e., 10 in, 12 in, etc.).
Find: Diameter.
Solution:
  1.22 x105 ft 2 / s; k s  1.5x104 ft
Assume f = 0.015
L V2
hf  f
D 2g
1000 (Q /( / 4) D 2 ) 2
1  0.015 *
D
2g
33,984
1
D5
D  8.06 ft
4
Relative roughness: k s  1.5x10  0.00002
D
8.06
Get better estimate of f
Re 
VD

Q

( / 4) D 2
D

Re 
Q
( / 4) D
300

( / 4)(8.06)1.22 x10
f=0.010
1
22,656
D5
D  7.43 ft  89 in.
Use a 90 in pipe
5
 3.9 x106
Ex. (10.81)
Given: The pressure at a water main is 300 kPa
gage. What size pipe is needed to carry water
from the main at a rate of 0.025 m3/s to a
factory that is 140 m from the main? Assume
galvanized-steel pipe is to be used and that the
pressure required at the factory is 60 kPa gage
at a point 10 m above the main connection.
Find: Size of pipe.
Solution:
hf  f
L V2
L (Q /( / 4) D 2 ) 2
 f
D 2g
D
2g
Assume f = 0.020
1/ 5
 fL Q 2 

D  8
 h f  2g 


1/ 5
 0.02 140 (0.025) 2 

 8
2
 14.45


9
.
81


 0.100 m
Relative roughness:
k s 0.15

 0.0015
D 100
Friction factor:
f  0.022
1/ 5
 fL Q 2 

D  8
 h f  2g 


V2 p
V2 p
α1 1  1  z1   hL  α2 2  2  z 2
2g
γ
2g
γ
300,000
60,000
 hf 
 10
9810
9810
h f  14.45 m
1/ 5
 0.022 
D  0.100

 0.020 
Use 12 cm pipe
 0.102 m
Ex. (10.83)
Given: The 10-cm galvanized-steel pipe is 1000 m
long and discharges water into the atmosphere.
The pipeline has an open globe valve and 4
threaded elbows; h1=3 m and h2 = 15 m.
Find: What is the discharge, and what is the
pressure at A, the midpoint of the line?
Solution:
V12 p1
V22 p2
α1

 z1   hL  α2

 z2
2g
γ
2g
γ
0  0  12  (1  K e  K v  4 K b  f
L V2
)
00
D 2g
D = 10-cm and assume f = 0.025
24 g  (1  0.5  10  4 * 0.9  0.025
1000 2
)V
0.1
24 g
V2 
265.1
V  0.942 m / s
V2 p
V2 p
α A A  A  z A   hL  α2 2  2  z 2
2g
γ
2g
γ
pA
L V2
 15  ( 2 K b  f )
γ
D 2g
pA
500 (0.942) 2
 ( 2 * 0.9  0.025
)
 15  9.6 m
γ
0.1
2g
p A  9810 * ( 9.26)  90.8 kPa
Q  VA  0.942( / 4)( 0.10) 2  0.0074 m 3 / s
Re 
VD


0.942 * 0.1
1.31x10
So f = 0.025
6
 7 x10 4
Near cavitation pressure, not good!
Ex. (10.95)
Given: If the deluge through the system shown is 2
cfs, what horsepower is the pump supplying to
the water? The 4 bends have a radius of 12 in
and the 6-in pipe is smooth.
Find: Horsepower
Solution:
V2 p
V2 p
α1 1  1  z1  h p  α2 2  2  z 2   hL
2g
γ
2g
γ
V2
L
0  0  30  h p  0  60  2 (1  0.5  4 K b  f )
2g
D
Q
2
V  
 10.18 ft / s
A ( / 4)(1 / 2) 2
V22
 1.611 ft
2g
VD 10.18 * (1 / 2)
Re 

 4.17 x105

5

1.22 x10
So f = 0.0135
h p  60  30  1.611(1  0.5  4 * 0.19  0.0135
 107.6 ft
Qh p
p
 24.4 hp
550
1700
)
(1 / 2)