FUNDAMENTALS OF
FLUID MECHANICS
Chapter 8 Pipe Flow
1
MAIN TOPICS
 General Characteristics of Pipe Flow
 Fully Developed Laminar Flow
 Fully Developed Turbulent Flow
 Dimensional Analysis of Pipe Flow
 Pipe Flow Examples
 Pipe Flowrate Measurement
2
Introduction
 Flows completely bounded by solid surfaces are called INTERNAL
FLOWS which include flows through pipes (Round cross section),
ducts (NOT Round cross section), nozzles, diffusers, sudden
contractions and expansions, valves, and fittings.
 The basic principles involved are independent of the cross-sectional
shape, although the details of the flow may be dependent on it.
 The flow regime (laminar or turbulent) of internal flows is primarily
a function of the Reynolds number (->inertial force/viscous force).
�Laminar flow: Can be solved analytically.
�Turbulent flow: Rely heavily on semi-empirical theories and
experimental data.
3
General Characteristics of
Pipe Flow
Laminar vs. Turbulent
Entrance Region vs. Fully Developed Flow
4
Pipe System
 A pipe system include the pipes themselves
(perhaps of more than one diameter), the various
fittings, the flowrate control devices valves) , and
the pumps or turbines.
5
Pipe Flow vs. Open Channel Flow
 Pipe flow: Flows completely filling the pipe. (a)
The pressure gradient along the pipe is main driving force.
 Open channel flow: Flows without completely filling the
pipe. (b)
The gravity alone is the driving force.
6
Laminar or Turbulent Flow 1/2
 The flow of a fluid in a pipe may be Laminar ? Or
Turbulent ?
 Osborne Reynolds, a British scientist and mathematician,
was the first to distinguish the difference between these
classification of flow by using a simple apparatus as
shown.
7
Laminar or Turbulent Flow 2/2
�For “small enough flowrate” the dye streak will remain as a
well-defined line as it flows along, with only slight blurring due
to molecular diffusion of the dye into the surrounding water.
�For a somewhat larger “intermediate flowrate” the dye
fluctuates in time and space, and intermittent bursts of irregular
behavior appear along the streak.
�For “large enough flowrate” the dye streak almost
immediately become blurred and spreads across the entire pipe in
a random fashion.
8
Time Dependence of
Fluid Velocity at a Point
9
Indication of
Laminar or Turbulent Flow
 The term flowrate should be replaced by Reynolds
number, Re  VL /  ,where V is the average velocity in the pipe,
and L is the characteristic dimension of a flow. L is usually D
(diameter) in a pipe flow. -> a measure of inertial force to the
viscous force.
 It is not only the fluid velocity that determines the character of the
flow – its density, viscosity, and the pipe size are of equal
importance.
 For general engineering purpose, the flow in a round pipe
�Laminar R e 2100
�Transitional
�Turbulent R e＞4000
10
Entrance Region and
Fully Developed Flow 1/5
 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 (entry) region or developing flow
region..
11
Entrance Region and
Fully Developed Flow 2/5
 The fluid typically enters the pipe with a nearly uniform
velocity profile at section (1).
 As the fluid moves through the pipe, viscous effects cause
it to stick to the pipe wall (the no slip boundary
condition)..
12
Entrance Region and
Fully Developed Flow 3/5
 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, x , until the
fluid reaches the end of the entrance length, section (2),
beyond which the velocity profile does not vary with x..
 The boundary layer has grown in thickness to completely
fill the pipe.
13
Entrance Region and
Fully Developed Flow 4/5
 Viscous effects are of considerable importance within the
boundary layer. Outside the boundary layer, the viscous
effects are negligible.
 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, �� .
For laminar flow
�
D
 0.06Re
For turbulent flow
�
D
 4.4R
1/ 6
e
Dimensionless entrance length
14
Entrance Region and
Fully Developed Flow 5/5
 Once the fluid reaches the end of the entrance region,
section (2), the flow is simpler to describe because the
velocity is a function of only the distance from the pipe
centerline, r, and independent of x..
 The flow between (2) and (3) is termed fully developed..
15
Pressure Distribution along Pipe
In the entrance region of a pipe, the fluid
accelerates or decelerates as it flows. There is
a balance between pressure, viscous, and
inertia (acceleration) force..
p

p
x
�
u
0
x
The magnitude of the
pressure gradient is larger
than that in the fully
developed region.
 constant  0
There is a balance between pressure and
viscous force.
The magnitude of the
pressure gradient is constant.
16
Fully Developed Laminar Flow
There are numerous ways to derive important
results pertaining to fully developed laminar flow:
�From F=ma applied directly to a fluid element.
�From the Navier-Stokes equations of motion
�From dimensional analysis methods
17
From F=ma 1/8
 Considering a fully developed axisymmetric laminar flow
in a long, straight, constant diameter section of a pipe.
 The fluid element is a circular cylinder of fluid of length
l and radius r centered on the axis of a horizontal pipe of
diameter D.
18
From F=ma 2/8
 Because the velocity is not uniform across the pipe, the
initially flat end of the cylinder of fluid at time t become
distorted at time t+t when the fluid element has moved to
its new location along the pipe.
 If the flow is fully developed and steady, the distortion on
each end of the fluid element is the same, and no part of
the fluid experiences any acceleration as it flows.
Steady
V
0
t
u
→
→
Fully developed V  V  u i  0
x
19
From F=ma 3/8
Apply the Newton’s second Law to the cylinder of fluid
Fx  ma x
The force (pressure & friction) balance
p 2 rx

p1 r   p1  p r  rx �2 r   0  
p  p1  p2
�
r
2
2
Basic balance in forces needed to drive each fluid particle
along the pipe with constant velocity
Not function of r

p
�

2 rx
r
Not function of r
Independent of r
 ?   rx  Cr
B.C. r=0 rx=0
r=D/2 rx= w < 0
2 wr
 rx 
D
20
From F=ma 4/8
The pressure drop and wall shear stress are related by
2 wr
p 2 rx
4� w


 rx 
p 
D
r
�
D
Valid for both laminar and turbulent flow
Laminar
rx
du

dr
21
From F=ma 5/8
Since
  
du
dr
Laminar


 p 
p 2 rx
 


r
dr
�
r
2� 
 p  2
p
 

r  C1

u

du
rdr


2�
 4� 
du
With the boundary conditions:
u=0 at r=D/2
pD 2
C1  16  �
Velocity distribution
4�
p 
w
D
  2r 2 
pD 2   2r 2 



u (r )  16  � 1      VC 1    
D
D
  2 
   
 w D   r  
1   
u (r )  

4   R  
22
From F=ma 6/8
 The shear stress distribution
rx
du   rp
 2�

dr
 Volume flowrate


2
pD2   2r 
u(r)  16 � 1  
D
  


  2r 2 
  VC 1   
D

   
→
2
R

R
VC
Q   u d A   u (r )2 rdr  ..... 
0
2
A
D 4 p
Poiseuille’s Law
Q 
128  �
Valid for Laminar flow only
23
From F=ma 7/8
 Average velocity
Vaverage
2
Q
Q
pD
 

A R 2
32 �
4
D
p
Q 
128�
 Point of maximum velocity
du
0
dr
u  umax
at r=0
pD2
u(r) 
16�
R 2 p  2Vaverage
 U 
4 �
  2r 2 
  2r 2 
1    VC 1  
  D  
  D  
24
From F=ma 8/8
 Making adjustment to account
for nonhorizontal pipes
p
p  sin
>0 if the flow is uphill
<0 if the flow is downhill
pr 2   p  pr 2 rx �2r   gr 2� sin   0  
p  g� sin    2 rx
r
Specific weight

p  � sin  D 2
 p  � sin  D 4
,  g
Vaverage 
Q 
32 �
128�
25
Example 8.2 Laminar Pipe Flow
 An oil with a viscosity of μ= 0.40 N·s/m2 and density ρ= 900 kg/m3
flows in a pipe of diameter D= 0.20m .
 (a) What pressure drop, p1-p2, is needed to produce a flowrate of Q=2.0×10-5
m3/s if the pipe is horizontal with x1=0 and x2=10 m?
 (b) How steep a hill, θ,must the pipe be on if the oil is to flow through the pipe
at the same rate as in part (a), but with p1=p2?
 (c) For the conditions of part (b), if p1=200 kPa, what is the pressure at section,
x3=5 m, where x is measured along the pipe?
26
Example 8.2 Solution1/2
Re
VD /  2.87  2100
V
Q
A
 0.0637 m / s
The flow is laminar flow
128�Q
p p1  p2 
 ... 20.4kPa
4
D
If the pipe is on the hill of angle θ with Δp=0
128 Q
sin   
 ...    13.34
4
gD
27
Example 8.2 Solution2/2
With p1=p2 the length of the pipe, �, does not appear in the flowrate
equation
Δp=0 for all �
p1  p 2  p3  200 kPa
28
From the Navier-Stokes Equations 1/3
 General motion of an incompressible Newtonian fluid is governed by the
continuity equation and the momentum equation
Mass conservation
Navier-Stokes Equation
in a cylindrical coordinate
Acceleration
29
From the Navier-Stokes Equations 2/3
Simplify the Navier-Stokes equation
axial component: z
The flow is governed by a balance of pressure, weight, and viscous
forces in the flow direction.
For steady, fully developed flow in a pipe, the velocity
contains only an axial component, which is a function of
only the radial coordinate V  u(r) i
30
From the Navier-Stokes Equations 3/3
axial component: x
→
→
V  u(r) i
p
1   u 
 g sin   
 r 
r r  r 
x
Function of, at most, only x
Function of ,at most, only r
p const.  p  p


x
x
�
Integrating
Velocity profile u(r)=
B.C. (1) r = R , u = 0 ;
(2) r = 0 , u < ∞ or u/r=0
31
From Dimensional Analysis 1/3
Assume that the pressure drop in the horizontal pie, Δp, is
a function of the average velocity of the fluid in the pipe,
V, the length of the pipe, �, the pipe diameter, D, and the
viscosity of the fluid, μ.
p  F(V, �, D, )
Dp   � 
  
V
D 
Dimensional analysis
an unknown function of the length to
diameter ratio of the pipe.
k-r = 5 (총 변수) -3 (reference
dimension) =2 dimensionless group
32
From Dimensional Analysis 2/3
Dp  C �
where C is a constant.
V
D
p CV
 2 p 2 ( / 4C)pD4
 2 Q  AV  D
D 
D
4
C
The value of C must be determined by theory or experiment.
For a round pipe, C=32. For duct of other cross-sectional
shapes, the value of C is different
Q
( / 4C)D4
�
p 
1
Flow resistance
32 V
For a round pipe p 
D2
p
V
Analogy: i 
R
33
From Dimensional Analysis 3/3
32 V / D 2
64
p
For a round pipe
 64
 1 2

2
1
VD D Re D
2 V
2 V
Dynamic pressure -> Characteristic pressure
2
�
V
p  f

D 2
f is termed the friction factor, or
2

D


V
p

p  2  p* sometimes the Darcy friction
factor -> dimensionless pressure

  * drop for internal flows..
f 
2 
�
�
V
64 8
D
f
2
 w2
Re V
For laminar flow
p  4� w
D
34
Energy Consideration 1/3
 The energy equation for incompressible, steady flow between two
locations
p
 V2
p
 V2
 z1 2  2 2  z2  h L
2g

2g

 V2  V2
1 1
 2 2
2g
2g
 p1
 p

1
 z  2  z  h   p  p   z  z 

2
L
1
2
1
2




1

 
  

1
2�
4� w
 p  g� sin    


r
D
1

1
1
The head loss in a

p  �g� sin    2rrx

2 r
  Dw
pipe is a result of
the viscous shear
stress on the wall.
35
Example 8.3 Laminar Pipe Flow Properties 1/2
 The flowrate, Q, of corn syrup through the horizontal pipe shown in
Figure E8.3 is to be monitored by measuring the pressure difference
between sections (1) and (2). It is proposed that Q=KΔp, where the
calibration constant, K, is a function of temperature, T, because of
the variation of the syrup’s viscosity and density with temperature.
These variations are given in Table E8.3.
 (a) Plot K(T) versus T for 60˚˚F≤T ≤ 160˚˚F.
 (b) Determine the wall shear stress and the pressure drop, Δp=p1-p2, for Q=0.5
ft3/s and T=100˚˚F.
 (c) For the conditions of part (b), determine the nest pressure force.(πD2/4)Δp,
and the nest shear force, πD�τw, on the fluid within the pipe between the
sections (1) and (2).
36
Example 8.3 Laminar Pipe Flow Properties 1/2
37
Example 8.3 Solution1/2
If the flow is laminar (-> should be verified)
Q 
pD4
 Kp
128�
1.60  10 5
K


For T=100˚˚F μ=3.8×10-3 lb·s/ft2, Q=0.5ft3/s
128�Q
2
p 

...

119lb
/
ft
D4
Q
V   ...  10.2ft / s
A
R e VD /  ...  1380  2100
pD
4� w
 w 
p 
 ...  1.24lb / ft2
D
4�
Laminar
38
Example 8.3 Solution2/2
The new pressure force and viscous force on the fluid within the pipe
between sections (1) and (2) is
Fp 
D2
4
Fv 2
D
2
p ... 5.84lb
� w  ... 5.84lb
The values of these two forces are the same. The net
force is zero; there is no acceleration.
39
Fully Developed Turbulent Flow
 Turbulent pipe flow is actually more likely to occur than
laminar flow in practical situations.
 Turbulent flow is a very complex process.
Numerous persons have devoted considerable effort in an
attempting to understand the variety of baffling aspects of
turbulence. Although a considerable amount if knowledge
about the topics has been developed, the field of turbulent
flow still remains the least understood area of fluid
mechanics.
Much remains to be learned about the nature of turbulent flo .
40
Transition from Laminar to Turbulent
Flow in a Pipe 1/2
For any flow geometry, there is one (or more)
dimensionless parameters such as with this parameter
value below a particular value the flow is laminar, whereas
with the parameter value larger than a certain value the
flow is turbulent.
The important parameters involved and their critical
values depend on the specific flow situation involved.
For flow in pipe : Re~4000
For flow along a plate Rex~500000
Turbulence initiated.
Consider a long section of pipe that is
initially filled with a fluid at rest.
41
Energy Considerations 1/7 (1-6: option)
 Considering the steady flow through the piping system, including a
reducing elbow. The basic equation for conservation of energy – the
first law of thermodynamics
→ →
→ →

Q net in  WShaft in CS  nn V  ndA    edV CS e V  ndA
t CV
→ →
→ →

 Q net in  WShaft in   edV CS e V  ndA  CS  nn V  ndA
t CV
Energy equation
-p
→
→
 
p V2
edV  (uˆ    gz)V  ndA  Qnet in  WShaft in

CS
t CV
 2
2
Kinetic energy
V
e  uˆ   gz
2
Internal energy
Note: The shear stress power is negligibly small
on a control surface.
42
Energy Considerations
2/7

When the flow is steady
CV edV  0

t
The integral of
 → →

p V2
 ndA 
 CS û    2  gzV

???
Uniformly distribution

 →
p V2




p V2
p V2
 CS û   2  gzV  ndA    û   2  gzm    û   2  gzm


out 
in 




 
Only one stream
entering and leaving
 → →
 p V2
 ndA
 CS û    2  gzV

  p V2
  p V2


 û

û

gz


gz

   2
 mout    2
 min
out
in


43
Energy Considerations 3/7
If shaft work is involved….
2
2






p
p
g

z
V
V

m uout  uin        out
out  z
in



in


ˆ
ˆ
2
  out   in


One-dimensional energy equation
 Qnet in  Wshaft net in
for steady-in-the-mean flow
p
Enthalpy hˆ  û  

The energy equation is written in terms
of2 enthalpy.


V2  V
m ĥ out  ĥin  out
in  g z   z    Q
 Wshaft net / in
in 
out
net / in
2


44
Energy Considerations
4/7
For steady, incompressible flow…One-dimensional energy equation


 p   p  V2  V2
z in   Qnet in
m û out  û   out
in
out
in


in

 
 gz out
2
     


m

2
pout  Vout
pin  Vin2
 û out  ûin qnet in


 gzout 
 gzin

2

2
where qnet
in
 Q net in / m
For steady, incompressible, frictionless flow…
pout

Examples?
2
Vout
Vin2

 zout  pin 
 zin Bernoulli equation
2
2
uˆ out  uˆ in  q net
in
0
Frictionless flow…
45
Energy Considerations
5/7
For steady, incompressible, frictional flow…
û out  û in  q net
in
0
Frictional flow…
p V2
Defining “useful or available energy”…

 2
gz
Defining “loss of useful or available energy”… û out  û in  q net in  loss
2
2
pout  Vout
p
gz
V
 out  in  in  gzin  loss


2

2
46
Energy Considerations 6/7
For steady, incompressible flow with friction and shaft work…


 p   p  V2  V2
mû out  û   out
in
out
in
  z in   Qnet in  Wshaft net in
in





g

z
out


2

     

2
p
V
2
p
Vout
w
q
)
out

 gzout  in  in  gzin  shaft net in  (uˆ out  uˆ in net in

2

2
p
out

g
Shaft head

V2
out
2
 gzout 
p
in


V2
in
2
 gzin  wshaft net in
loss
2
pout V 2
p
V
 out  zout  in  in  zin  hs h L 
 2g

2g
h 
S
wshaft 
g
net in

Wshaft
net in
mg

Wshaft
net in
Q
Head loss h  loss
L
g
47
Energy Considerations 7/7
2
2
pout  Vout
V
 zout  pin  in  zin hs h L

2g
 2g
Total head loss , hL, is regarded as the sum of major losses,
hL major, due to frictional effects in fully developed flow
in constant area tubes, and minor losses, hL minor,
resulting from entrance, fitting, area changes, and so on.
hL  hLmajor  hLminor
48
Major Losses: Friction Factor
 The energy equation for steady and incompressible flow
with zero shaft work
 V2  → →

 2V 2
 p1 1V1


p
2
2
2

  hL


z

2
g  2 g  z 1   




g
2g

A

V  ndA
 2 

 2   1
m V 
 2 


For fully developed flow through a constant area pipe, α1= α2,
V1= V2 , where  is the kinetic energy coefficient and V is the
average velocity. For a uniform velocity, α=1.

p1  p2
 ( z2  z1 )  hL
g
For horizontal pipe, z2=z1

p
h
 p1  p2 
L
g
g
49
Major Losses: Laminar Flow
 In fully developed laminar flow in a horizontal pipe, the
pressure drop
2
128
�Q
128
�V

D
/ 4
� V
p 


32
D 4
D 4
D D
p
 � 64 �


64
1
2
VD D Re D
V
2
� V 2
� V
� V 2 
   64  � V 2



p  f

 h L  32
64
R

D D D
D 2
  eD 2

2
VD

Friction Factor
f  pD / �  / V 2 / 2 
f laminar
64

Re
50







Major Losses: Turbulent Flow 1/3

 In turbulent flow, we cannot evaluate the pressure drop analytically;
we must resort to experimental results and use dimensional analysis
to correlate the experimental data.
 In fully developed turbulent flow the
pressure drop, △p , caused by friction
in a horizontal constant-area pipe is
known to depend on pipe diameter,D,
pipe length, �, pipe roughness,e,
average flow velocity, V, fluid densityρ,
and fluid viscosity,μ.
p  F V, D, �, ,, 
51
Major Losses: Turbulent Flow 2/3
 Applying dimensional analysis, the result were a correlation of the
form p
 VD �  
 
, , 
2
1
  D D 
2 V



 Experiments show that the nondimensional head loss is directly
V
proportional to �/D. Hence we can
write
p  f
2
p
 � Re,  
2
1
D 
D 
2 V
 

f   Re, 
D 

D 2
Darcy-Weisbach equation
h L major
2
�
V
f
D 2g
52
Roughness for Pipes
53
Moody chart
Depending on the specific
circumstances involved.
54
About Moody Chart
 For laminar flow, f=64/Re, which is independent of the
relative roughness..
 For very large Reynolds numbers, f=Φ(ε/D), which is
independent of the Reynolds numbers.
 For flows with very large value of Re, commonly termed
completely turbulent flow (or wholly turbulent flow), the
laminar sublayer is so thin (its thickness decrease with
increasing Re) that the surface roughness completely
dominates the character of the flow near the wall.
 For flows with moderate value of Re, the friction factor
f=Φ(Re,ε/D).
55
Major Losses: Turbulent Flow 3/3
 Colebrook – To avoid having to use a graphical method for
obtaining f for turbulent flows.
Valid for the entire nonlaminar
1
/D
2.51 range of the Moody chart.


 2.0 log 

f
Re f  Colebrook formula -> needs
 3.7
iteration.
 Miler suggests that a single iteration will produce a result within
1 percent if the initial estimate is calculated from
  / D 5.74  2
f 0 0.25 log 

0.9 
Re 
 3.7
56
Example 8.5 Comparison of Laminar or
Turbulent pressure Drop
 Air under standard conditions flows through a 4.0-mm-diameter
drawn tubing with an average velocity of V = 50 m/s. For such
conditions the flow would normally be turbulent. However, if
precautions are taken to eliminate disturbances to the flow (the
entrance to the tube is very smooth, the air is dust free, the tube does
not vibrate, etc.), it may be possible to maintain laminar flow.
 (a) Determine the pressure drop in a 0.1-m section of the tube if the flow is
laminar.
 (b) Repeat the calculations if the flow is turbulent.
57
Example 8.5 Solution1/2
Under standard temperature and pressure conditions
Ρ=1.23kg/m3, μ=1.7910-5Ns/m
The Reynolds number
R e VD /   ...  13,700  Turbulent flow
If the flow were laminar
f=64/Re=…=0.0467
p  f
� 1
D2
V 2  ...  0.179 kPa
58
Example 8.5 Solution2/2
If the flow were turbulent
From Moody chart f=Φ(Re,ε/D) =…0.028
p  f � 1 V 2  ...  1.076 kPa
D2
59
Minor Losses 1/5
Most pipe systems consist of
considerably more than straight
pipes. These additional
components (valves, bends, tees,
and the like) add to the overall
head loss of the system.
Such losses are termed MINOR
LOSS. But, it is not minor at all
and it may be larger than the
major losses.
The flow pattern through a valve
60
Minor Losses 2/5
The theoretical analysis to predict the details of flow
pattern (through these additional components) is not, as
yet, possible.
The head loss information for essentially all components is
given in dimensionless form and based on experimental
data. The most common method used to determine these
head losses or pressure drops is to specify the loss
coefficient, KL..
61
Minor Losses 3/5

1 2
p

V
K L  2 min or  1
 p  K L
2
V / 2g
V2
2
hL


Minor losses are sometimes
given in terms of an equivalent
length �eq
Large K : large pressure
drops for given velocities.
2
� eq V 2
V
 KL
hL
f
min or
2g
D 2g
D
� eq  K L
f
The actual value of KL is strongly dependent on the geometry of
the component considered. It may also dependent on the fluid
properties. That is
K L  (geometry, Re)
62
Minor Losses 4/5
For many practical applications the Reynolds number is
large enough so that the flow through the component is
dominated by inertial effects, with viscous effects being of
secondary importance.
In a flow that is dominated by inertia effects rather than
viscous effects, it is usually found that pressure drops and
head losses correlate directly with the dynamic pressure..
This is the reason why the friction factor for very large
Reynolds number, fully developed pipe flow is
independent of the Reynolds number..
63
Minor Losses 5/5
 This is true for flow through pipe components.
Thus, in most cases of practical interest the loss
coefficients for components are a function of geometry
only,,
K L  (geometry)
64
Minor Losses Coefficient Entrance flow 1/3
Entrance flow condition
and loss coefficient
(a) Reentrant, KL = 0.8
(b) sharp-edged, KL = 0.5
(c) slightly rounded, KL = 0.2
(d) well-rounded KL = 0.04
KL = function of rounding of
the inlet edge.
65
Minor Losses Coefficient Entrance flow 2/3
A vena contract region may result because the fluid cannot
turn a sharp right-angle corner. The flow is said to separate
from the sharp corner.
The maximum velocity velocity at section (2) is greater
than that in the pipe section (3), and the pressure there is
lower.
If this high speed fluid could slow down efficiently, the
kinetic energy could be converted into pressure.
66
Minor Losses Coefficient Entrance flow 3/3
 Such is not the case. Although
the fluid may be accelerated
very efficiently, it is very
difficult to slow down
(decelerate) the fluid efficently.
 (2)(3) The extra kinetic
energy of the fluid is partially
lost because of viscous
dissipation, so that the pressure
does not return to the ideal
value.
Flow pattern and pressure distribution
for a sharp-edged entrance
67
Minor Losses Coefficient Exit flow
Exit flow condition and
loss coefficient
(a) Reentrant, KL = 1.0
(b) sharp-edged, KL = 1.0
(c) slightly rounded, KL = 1.0
(d) well-rounded,, KL = 1.0
68
Minor Losses Coefficient varied diameter
 Loss coefficient for sudden
contraction, expansion,typical
conical diffuser.
2

A
K L  1  1 
 A2 
69







Minor Losses Coefficient Bend

 Character of the flow in bend
and the associated loss
coefficient.
Carefully designed guide vanes
help direct the flow with less
unwanted swirl and disturbances.
70
Internal Structure of Valves
(a) globe valve
(b) gate valve
(c) swing check valve
(d) stop check valve
71
Loss Coefficients for Pipe
Components
72
Example 8.6 Minor Loss 1/2
 Air at standard conditions is to flow through the test section
[between sections (5) and (6)] of the closed-circuit wind tunnel
shown if Figure E8.6 with a velocity of 200 ft/s. The flow is driven
by a fan that essentially increase the static pressure by the amount
p1-p9 that is needed to overcome the head losses experienced by the
fluid as it flows around the circuit. Estimate the value of p1-p9 and
the horsepower supplied to the fluid by the fan.
73
Example 8.6 Minor Loss 2/2
74
Example 8.6 Solution1/3
The maximum velocity within the wind tunnel occurs in the
test section (smallest area). Thus, the maximum Mach number
of the flow is Ma5=V5/c5
V5  200ft / s
c5  (KRT5 )1/ 2  1117ft / s
The energy equation between points (1) and (9)
2
2
p1 V1
p9 V9

z 9 h L19

z



1
2g
 2g

h L19 
p1


p9

The total head loss from (1) to (9).
75
Example 8.6 Solution2/3
The energy across the fan, from (9) to (1)
p1 V12
Hp is the actual head rise supplied
p9 V92 
z



 z 9  h p 
1 by the pump (fan) to the air.

 2g
2g
hp 
p1


p9

 h L19
The actual power supplied to the air (horsepower, Pa) is obtained
from the fan head by
Pa Qh p  A5V5hp  A5V5h L19
76
Example 8.6 Solution3/3
The total head loss
h L1
9
 h Lcorner 7  h Lcorner 8  h Lcorner 2  h Lcorner 3  h Ldif  h Lnoz  h Lscr
hL
corner
 KL
KLnoz  0.2
V2
 0.2
2g
KLscr
V2
2g
 4.0
hL
dif
 KL dif
V2
2g
 0.6
V2
2g
p1  p9  h L19  (0.765lb / ft 2 )(560ft)  ... 0.298psi
Pa  ... 34300ft  lb / s  62.3hp
77
Noncircular Ducts 1/4
The empirical correlations for pipe flow may be used for
computations involving noncircular ducts, provided their
cross sections are not too exaggerated.
 The correlation for turbulent pipe flow are extended for
use with noncircular geometries by introducing the
hydraulic diameter, defined as
4A
D h 
P
where A is crosssectional area, and P
is wetted perimeter.
78
Noncircular Ducts 2/4
For a circular duct
2
4

R
4A 

Dh 
 D
2R
P
 For a rectangular duct of width b and height h
D 
4A
 4bh  2h
P
2(b  h) 1  ar

h
ar h / b
h
b
The hydraulic diameter concept can be applied in the
approximate range ¼<ar<4. So the correlations for pipe
flow give acceptably accurate results for rectangular ducts.
79
Noncircular Ducts 3/4
The friction factor can be written as f=C/Reh, where the
constant C depends on the particular shape of the duct, and
Reh is the Reynolds number based on the hydraulic
diameter. Note: C = 64 for a circular tube.
 The hydraulic diameter is also used in the definition of the
friction factor, hL  f (� / D h )(V2 / 2g) , and the relative
roughness /Dh..
80
Noncircular Ducts 4/4
 For Laminar flow, the value of C=f·Reh have been
obtained from theory and/or experiment for various shapes.
For turbulent flow in ducts of noncircular cross section,
calculations are carried out by using the Moody chart data
for round pipes with the diameter replaced by the hydraulic
diameter and the Reynolds number based on the hydraulic
diameter.
The Moody chart, developed for round pipes, can also
be used for noncircular ducts.
81
Friction Factor for Laminar Flow in
Noncircular Ducts
Note: C = 64 for a circular tube.
82
Example 8.7 Noncircular Duct
 Air at temperature of 120 F and standard pressure flows from a
furnace through an 8-in.-diameter pipe with an average velocity of
10ft/s. It then passes through a transition section and into a square
duct whose side is of length a. The pipe and duct surfaces are
smooth (ε=0). Determine the duct size, a, if the head loss per foot is
to be the same for the pipe and the duct.
83
Example 8.7 Solution1/3
The head loss per foot for the pipe
f V2

D 2g
�
hL
For given pressure and temperature ν=1.8910-4ft2/s
Re 
VD

 35300
hL
For the square duct
�
D 
h

f Vs2
 0.0512
Dh 2g
4A
P
a
Vs 
Q 3.49
 2
A
a
84
Example 8.7 Solution2/3
hL
�

2
s
f V
Dh 2g
2 2
 0.0512 
f (3.49 / a )  a  1.30f 1/ 5 (1)
a 2(32.2)
The Reynolds number based on the hydraulic diameter
4
2
V
D
1.89
10
(3.49
/
a
)a
s
h

Reh 
 


1.89 10 4
a
(2)
Have three unknown (a, f, and Reh) and three equation –
Eqs. 1, 2, and either in graphical form the Moody chart or
the Colebrook equation
Find a
85
Example 8.7 Solution3/3
Use the Moody chart
Assume the friction factor for the duct is the same as for the pipe.
That is, assume f=0.022.
From Eq. 1 we obtain a=0.606 ft.
From Eq. 2 we have Reh=3.05104
From Moody chart we find f=0.023, which does not quite agree the
assumed value of f.
Try again, using the latest calculated value of f=0.023 as our guess.
…… The final result is f=0.023, Reh=3.05104, and a=0.611ft.
86
Pipe Flow Examples 1/2
The energy equation, relating the conditions at any two
points 1 and 2 for a single-path pipe system, for steady and
incompressible flow with zero shaft work is given by
p

 V2
 p
 V2
 1  1 1  z1   2  2 2  z 2   h L 
2g
2g
  g

 g
h
Lmajor
  h L min or
By judicious choice of points 1 and 2, we can analyze not
only the entire pipe system, but also just a certain section
of it that we may be interested in.
Major loss
hL
major
� V2
f
D 2g
Minor loss
hL
min or
V2
 KL
2g
87
Pipe Flow Examples 2/2
Single pipe whose length may be interrupted by various
components.
 Multiple pipes in different configuration
 Parallel
 Series
 Network
88
Single-Path Systems 1/2
Pipe flow problems can be categorized by what parameters
are given and what is to be calculated..
89
Single-Path Systems 2/2
Type 1: Given pipe (L and D), and flow rate, and Q, find
pressure drop Δp
 Type 1: Given Δp, D, and Q, find L.
 Type 2: Given Δp, L, and D, find Q.
 Type 3: Given Δp, L, and Q, find D.
90
Given L , D, and Q, find Δp
 The energy equation
p

 V2
 p
 V2
2


1


z

1
1
1
 
 2 2  z2   hL   h L major   h L min or
2g
2g
 g
  g

The flow rate leads to the Reynolds number and hence the
friction factor for the flow.
Tabulated data can be used for minor loss coefficients and
equivalent lengths.
The energy equation can then be used to directly to obtain
the pressure drop.
91
Given Δp, D, and Q, find L
 The energy equation
p

 p
 V2
 V2
2


1


z

1
1
1
 
 2 2  z 2   hL   h L major   h L min or
2g
2g
 g
  g

The flow rate leads to the Reynolds number and hence the
friction factor for the flow.
Tabulated data can be used for minor loss coefficients and
equivalent lengths.
The energy equation can then be rearranged and solved
directly for the pipe length.
92
Given ∆p, L, and D, find Q
These types of problems required either manual iteration
or use of a computer application.
The unknown flow rate or velocity is needed before the
Reynolds number and hence the friction factor can be
found.
First, we make a guess for f* and solve the energy equation
for V in terms of known quantities and the guessed
friction factor f*..
Then we can compute a Reynolds number and hence
obtain a new value for f..
Repeat the iteration process
f*→ V→ Re→ f until convergence (f*=f)
93
Given Δp, L, and Q, find D
These types of problems required either manual iteration
or use of a computer application.
The unknown diameter is needed before the Reynolds
number and relative roughness, and hence the friction
factor can be found.
First, we make a guess for f* and solve the energy equation
for D in terms of known quantities and the guessed
friction factor f*..
Then we can compute a Reynolds number and hence
obtain a new value for f.
Repeat the iteration process
f*→ D→ (Re and ε/D) → f until convergence (f*=f)
94
Example 8.8 Type I Determine Pressure
Drop
 Water at 60 F flows from the basement to the second floor through the
0.75-in. (0.0625-fy)-diameter copper pipe (a drawn tubing) at a rate of
Q = 12.0 gal/min = 0.0267 ft3/s and exits through a faucet of diameter
0.50 in. as shown in Figure E8.8.
Determine the pressure at
point (1) if: (a) all losses
are neglected, (b) the only
losses included are major
losses, or (c) all losses are
included.
95
Example 8.8 Solution1/4
V1 
Q
 ...  8.70ft / s
A1
  2.34 105lb  s / ft2
  1.94slug / ft3
The flow is turbulent
Re  VD /   45000
Nearly uniform velocity profile
The energy equation
p1 1V12
p 2 2 V22

 z1 

 z2  hL
g
g
2g
2g
1   2  1
p1 z2   (V2 V1 )  hL
1
2
2
2
z1  0, z2  20 ft, p2  0( free jet)
V2  Q / A2  ...  19.6 ft / s
V1  Q / A1  8.70 ft / s
Head loss is different for
each of the three cases.
96
Example 8.8 Solution2/4
(a) If all losses are neglected (hL=0)
p  z  1 (V 2  V 2 )  ...  1547lb / ft2  10.7psi
1
2
2
2
1
(b) If the only losses included are the major losses, the head loss is
V12
hL  f
D 2g
 0.000005,
Moody chart
/ D  8 105 , Re  45000
f=0.0215
60ft) V1
2
2
�(
1
p1  z 2  2 (V2  V1 )  f
 ...  3062lb / ft2  21.3psi
D
2
2
97
Example 8.8 Solution3/4
(c) If major and minor losses are included
2
2
V
V
� 1
2
p1 z2  12 (V2 V12 )  f
  K L
2
D 2g
V2
p1 21.3psi  K L
2
2
 
3 (8.70ft / s) Valve (pp. 72) Faucet
[10 4(1.5)  2]
21.3psi  (1.94slugs / ft )
2
Elbow
p1 21.3 ps  9.17psi  30.5psi
i
98
Example 8.9 Type I, Determine Head Loss
 Crude oil at 140˚˚F with γ=53.7 lb/ft3 and μ= 810-5 lb·s/ft2 (about
four times the viscosity of water) is pumped across Alaska through
the Alaska pipeline, a 799-mile-along, 4-ft-diameter steel pipe, at a
maximum rate of Q = 2.4 million barrel/day = 117ft3/s, or
V=Q/A=9.31 ft/s. Determine the horsepower needed for the pumps
that drive this large system.
99
Example 8.9 Solution1/2
The energy equation between points (1) and (2)
p V2
p V2
hP is the head provided to the oil
1
 1 1  z1  hP  2  2 2  z2  hL by the pump.

2g

2g
Assume that z1=z2, p1=p2=V1=V2=0 (<- large, open tank)
Minor losses are negligible because of the large length-todiameter ratio of the relatively straight, uninterrupted pipe.
2
�
V
hL  hP  f
 ...  17700ft
D 2g
Table 8.1
Turbulent (α=1)
f=0.0124 from Moody chart ε/D=(0.00015ft)/(4ft), Re=7.76 x 105
For steel pipe
100
Example 8.9 Solution2/2
The actual power supplied to the fluid.


1hp
 202000hp
Pa  Qh P  ...
550 ft  lb / s 
-> It requires many pump stations to be set up.
101
Example 8.10 Type II, Determine Flowrate
 According to an appliance manufacturer, the 4-in-diameter
galvanized iron vent on a clothes dryer is not to contain more than
20 ft of pipe and four 90 elbows. Under these conditions determine
the air flowrate if the pressure within the dryer is 0.20 inches of
water. Assume a temperature of 100℉ and standard pressure. KL =
0.5 for an entrance and 1.5 for elbows.
102
Example 8.10 Solution1/2
Application of the energy equation between the inside of the dryer,
point (1), and the exit of the vent pipe, point (2) gives
V2
p
p
V
2
2
2
V
V
  KL

 z2  f
1
 1 1  z1 

D 2g
2g
2g

2g
2
2 2
V2
p
Assume that z1=z2, p2=0, V1=0
p1

H2O

1

1 1
p
 z1 
2


V2
2 2
 z2 f
 1ft 
 0.2in  p1  (0.2in.) 
(62.4lb / ft3 )  1.04lb / ft2
12in. 


2g
2g
� V2
D 2g
  KL
V2
2g
With γ=0.0709lb/ft3, V2=V, and ν=1.79×10-4 ft2/s.


2


V
20ft

 0.5  4 1.5
1 f
2
4
12ft
0.0709 lb ft3
232.2 ft s

945  (7.5  60 f )V 2
1.04 lb ft2



 (1)
Assume turbulent flow
(α=1)
f is dependent on Re, which is dependent on V, and unknown.
103
Example 8.10 Solution2/2
Re 
VD

 ...  1860V
(2)
For galvanized iron
We have three relationships (Eq. 1, 2, and the ε/D=0.0015 curve of
the Moody chart) from which we can solve for the three unknowns f,
Re and V.
This is done easily by iterative scheme as follows.
Assume f=0.022→V=10.4ft/s (Eq. 1)→Re=19,300 (Eq.2)→f=0.029
Assume f=0.029 →V10.1ft/s→Re=18,800 →f=0.029 (Final value)
Q  AV  ... 0.881ft 3 /s
Turbulent
104
Example 8.11 Type II, Determine Flowrate
 The turbine shown in Figure E8.11 extracts 50 hp from the water
flowing through it. The 1-ft-diameter, 300-ft-long pipe is assumed to
have a friction factor of 0.02. Minor losses are negligible. Determine
the flowrate through the pipe and turbine.
105
Example 8.11 Solution1/2
The energy equation can be applied between the surface of the lake
and the outlet of the pipe as
p
1


V2
1 1
2g
z 
1
p
2


V
2 2
2g
2
 z 2  hL  hT
p
1


V2
1 1
2g
 z 
1
p
2


V
2 2
2g
2
 z 2  hL  hT
Where p1=V1= p2=z2=0, z1=90ft, and V2=V, the fluid velocity in
the pipe
Pa
561
2
...

�
V
2
ft Assume turbulent flow
hL  f
 0.0932V ft hT
(α=1)
V
Q
D 2g
V2
561
2
3
ft
90ft 
ft


0.107V
 90V  561  0

0.0932V
2
2  32.2 ft s
V


There are two real, positive roots: V=6.58 ft/s or V=24.9 ft/s. The third
root is negative (V=-31.4ft/s) and has no physical meaning for this flow.
106
Example 8.11 Solution2/2
Two acceptable flowrates are
 2
Q  D V  ...  5.17ft3 / s
4
 2
Q  D V  ...  19.6ft3 / s
4
At 60oF, 1.21 x 10-5 (ft2/s) Re ≈ 105
107
Example 8.12 Type III Without Minor
Losses, Determine Diameter
 Air at standard temperature and pressure flows through a horizontal,
galvanized iron pipe (ε=0.0005 ft) at a rate of 2.0ft3/s. Determine the
minimum pipe diameter if the pressure drop is to be no more than
0.50 psi per 100 ft of pipe.
108
Example 8.12 Solution1/2
Assume the flow to be incompressible with ρ=0.00238 slugs/ft3 and
μ=3.74×10-7 lb．s/ft2.
If the pipe were too long, the pressure drop from one end to the other,
p1-p2, would not be small relative to the pressure at the beginning, and
compressible flow considerations would be required.
2
2
V2
p2  2V2
�
V

 z2  f

 z1 
 K
D 2g  L 2g

2g

2g
p1
1V1 2
With z1=z2, V1=V2 , The energy equation becomes p1  p2  f
� V2
D g
V2
3
(100ft)
(0.00238slugs / ft )
p1  p2  (0.5)(144)lb / ft2  f
D
2g
Q 2.55
1/5 (1)
V 
D  0.404f
2
A D
109
Example 8.12 Solution2/2
VD
1.62 104
Re 
 ... 
D

 0.0005

D
D
(2)
(3)
We have four equations (Eq. 1, 2, 3, and either the Moody chart or
the Colebrook equation) and four unknowns (f, D, ε/D, and Re)
from which the solution can be obtained by trial-and-error methods.
Repeat the iteration process
f*→ D→ Re and ε/D→ f until convergence
(1)
(2)
(3)
110
Example 8.13 Type III With Minor Losses,
Determine Diameter
 Water at 60 F (ν=1.21×10-5 ft2/s) is to flow from reservoir A to
reservoir B through a pipe of length 1700 ft and roughness 0.0005 ft
at a rate of Q= 26 ft3/s as shown in Figure E8.13. The system
contains a sharp-edged entrance and four flanged 45˚˚ elbow.
Determine the pipe diameter needed.
111
Example 8.13 Solution1/2
The energy equation can be applied between two points on the
surfaces of the reservoirs (p1=V1= p2=z2=V2=0)
p V2
p V2

1

1 1
2g
 z1 
2


2 2
2g
 z2  hL
2
V  �  K 
z1 
L
f

2g  D

Q 33.1
V 
KLent=0.5, KLelbow=0.2, and KLexit=1
2
D
A


V2
 1700  [4(0.2)  0.5  1]

44ft 
f
2 
2(32.2ft / s )  D

f  0.00152D5  0.00135D
(1)
112
Example 8.13 Solution2/2
6
2.74
10
 ... 
Re 
D

 0.0005

(3)
D
D
VD
(2)
We have four equations (Eq. 1, 2, 3, and either the Moody chart or
the Colebrook equation) and four unknowns (f, D, ε/D, and Re)
from which the solution can be obtained by trial-and-error methods.
Repeat the iteration process
D→ f* →Re and ε/D→ f until convergence
(1) (2)
(3)
113