Viscous Flow in PipesViscous Flow in Pipes
CEE 331 Fluid Mechanics
April 19, 2023
CEE 331 Fluid Mechanics
April 19, 2023
Types of Engineering ProblemsTypes of Engineering Problems
How big does the pipe have to be to carry a flow of x m3/s?
What will the pressure in the water distribution system be when a fire hydrant is open?
Can we increase the flow in this old pipe by adding a smooth liner?
How big does the pipe have to be to carry a flow of x m3/s?
What will the pressure in the water distribution system be when a fire hydrant is open?
Can we increase the flow in this old pipe by adding a smooth liner?
Example Pipe Flow Problem
D=20 cmL=500 m
valve
100 mFind the discharge, Q.
Describe the process in terms of energy!Describe the process in terms of energy!
cs1cs1
cs2cs2
p Vg
z Hp V
gz H hp t l
11
12
12
222
22 2
p Vg
z Hp V
gz H hp t l
11
12
12
222
22 2
zV
gz hl1
22
22 z
Vg
z hl122
22 V g z z hl2 1 22 a fV g z z hl2 1 22 a f
,Rep
DC f
l Deæ ö=
è ø,Rep
DC f
l Deæ ö=
è ø 2
2C
Vp
p 2
2C
Vp
p Re
VDrm
=ReVDrm
=
Viscous Flow: Dimensional Analysis
Viscous Flow: Dimensional Analysis
Where and
in a bounded region (pipes, rivers): find Cpin a bounded region (pipes, rivers): find Cp
flow around an immersed object : find Cdflow around an immersed object : find Cd
Remember dimensional analysis?
Two important parameters!Re - Laminar or Turbulent /D - Rough or Smooth
Flow geometry internal _______________________________ external _______________________________
Remember dimensional analysis?
Two important parameters!Re - Laminar or Turbulent /D - Rough or Smooth
Flow geometry internal _______________________________ external _______________________________
Turbulent Pipe Flow: OverviewTurbulent Pipe Flow: Overview
Boundary Layer Development Turbulence Velocity Distributions Energy Losses
MajorMinor
Solution Techniques
Boundary Layer Development Turbulence Velocity Distributions Energy Losses
MajorMinor
Solution Techniques
Transition at Re of 2000Transition at Re of 2000
Laminar and Turbulent FlowsLaminar and Turbulent Flows
Reynolds apparatus Reynolds apparatus
ReVDrm
= =ReVDrm
= =dampingdampinginertiainertia
Boundary layer growth: Transition length
Boundary layer growth: Transition length
What does the water near the pipeline wall experience? _________________________Why does the water in the center of the pipeline speed up? _________________________
v v
Drag or shear
Conservation of mass
Non-Uniform Flow
v
Entrance Region LengthEntrance Region Length
1
10
10010 100
1000
1000
0
1000
00
1000
000
1000
0000
1000
0000
0
Re
l e /D
( )1/ 64.4 Reel
D= ( )1/ 6
4.4 ReelD=0.06Reel
D=0.06Reel
D=
laminar turbulent
( )Reel fD= ( )Reel f
D=
Turbulence
A characteristic of the flow. How can we characterize turbulence?
intensity of the velocity fluctuations size of the fluctuations (length scale)
meanvelocitymean
velocityinstantaneous
velocityinstantaneous
velocityvelocity
fluctuationvelocity
fluctuation�
uuu uuu uu
uu
Turbulence: Size of the Fluctuations or Eddies
depth (Re = 500)
diameter (Re = 2000)
Eddies must be smaller than the physical dimension of the flow
Generally the largest eddies are of similar size to the smallest dimension of the flow
Examples of turbulence length scales rivers: ________________ pipes: _________________
Actually a spectrum of eddy sizes
Turbulence: Flow Instability
In turbulent flow (high Reynolds number) the force leading to stability (_________) is small relative to the force leading to instability (_______).
Any disturbance in the flow results in large scale motions superimposed on the mean flow.
Some of the kinetic energy of the flow is transferred to these large scale motions (eddies).
Large scale instabilities gradually lose kinetic energy to smaller scale motions.
The kinetic energy of the smallest eddies is dissipated by viscous resistance and turned into heat. (=___________)head loss
viscosityviscosityinertiainertia
Velocity DistributionsVelocity Distributions
Turbulence causes transfer of momentum from center of pipe to fluid closer to the pipe wall.
Mixing of fluid (transfer of momentum) causes the central region of the pipe to have relatively _______velocity (compared to laminar flow)
Close to the pipe wall eddies are smaller (size proportional to distance to the boundary)
Turbulence causes transfer of momentum from center of pipe to fluid closer to the pipe wall.
Mixing of fluid (transfer of momentum) causes the central region of the pipe to have relatively _______velocity (compared to laminar flow)
Close to the pipe wall eddies are smaller (size proportional to distance to the boundary)
constant
Shear velocity
Dimensional analysis and measurements
Velocity of large eddies
Log Law for Turbulent, Established Flow, Velocity Profiles
Log Law for Turbulent, Established Flow, Velocity Profiles
*
*
2.5ln 5.0u yuu n= +*
*
2.5ln 5.0u yuu n= +
0* u
0* u
u/umax
rough smoothy
Iuu * Iuu *
0 4lh d
l
gt =0 4
lh d
l
gt =
* 4lgh d
ul
=* 4lgh d
ul
=
Force balance
Turbulence produced by shear!
Pipe Flow: The ProblemPipe Flow: The Problem
We have the control volume energy equation for pipe flow
We need to be able to predict the head loss term.
We will use the results we obtained using dimensional analysis
We have the control volume energy equation for pipe flow
We need to be able to predict the head loss term.
We will use the results we obtained using dimensional analysis
Horizontal pipe
Dimensional Analysis
Darcy-Weisbach equation
Pipe Flow Energy LossesPipe Flow Energy Losses
l
ph z
gæ ö
=- D +ç ÷è øl
ph z
gæ ö
=- D +ç ÷è ø
R,f
Df
L
DC p
R,f
Df
L
DC p
2
2C
Vp
p 2
2C
Vp
p
2
2C
V
ghlp
2
2C
V
ghlp
LD
V
ghl2
2f
LD
V
ghl2
2f
gV
DL
hl 2f
2
g
VDL
hl 2f
2
p Vg
z hp V
gz h hp t l
11
12
12
222
22 2
lh pg =- Dlh pg =- D
Always true (laminar or turbulent)
Friction Factor : Major lossesFriction Factor : Major losses
Laminar flow Turbulent (Smooth, Transition, Rough) Colebrook Formula Moody diagram Swamee-Jain
Laminar flow Turbulent (Smooth, Transition, Rough) Colebrook Formula Moody diagram Swamee-Jain
Hagen-Poiseuille
Darcy-Weisbach
Laminar Flow Friction FactorLaminar Flow Friction Factor
L
hDV l
32
2
L
hDV l
32
2
2
32gD
LVhl
2
32gD
LVhl
gV
DL
hl 2f
2
g
VDL
hl 2f
2
gV
DL
gDLV
2f
32 2
2
gV
DL
gDLV
2f
32 2
2
RVD6464
f
RVD6464
f
Slope of ___ on log-log plot-1
lh Vµlh Vµ
Turbulent Pipe Flow Head LossTurbulent Pipe Flow Head Loss
Proportional
Proportional
Inversely
Increase
independent
___________ to the length of the pipe ___________ to the square of the velocity
(almost) ________ with the diameter (almost) ________ with surface roughness Is a function of density and viscosity Is __________ of pressure
___________ to the length of the pipe ___________ to the square of the velocity
(almost) ________ with the diameter (almost) ________ with surface roughness Is a function of density and viscosity Is __________ of pressure
Smooth, Transition, Rough Turbulent Flow
Hydraulically smooth pipe law (von Karman, 1930)
Rough pipe law (von Karman, 1930)
Transition function for both smooth and rough pipe laws (Colebrook)
51.2
Relog2
1 f
f
51.2
Relog2
1 f
f
D
f
7.3log2
1
D
f
7.3log2
1
g
V
D
Lfh f
2
2
g
V
D
Lfh f
2
2
(used to draw the Moody diagram)
f
D
f Re
51.2
7.3log2
1
f
D
f Re
51.2
7.3log2
1
Moody DiagramMoody Diagram
0.01
0.10
1E+03 1E+04 1E+05 1E+06 1E+07 1E+08R
fric
tion
fact
or
laminar
0.050.04
0.03
0.020.015
0.010.0080.006
0.004
0.002
0.0010.0008
0.0004
0.0002
0.0001
0.00005
smooth
lD
C pf
lD
C pf
D
D
0.02
0.03
0.04
0.050.06
0.08
Swamee-Jain
1976 limitations
/D < 2 x 10-2
Re >3 x 103
less than 3% deviation from results obtained with Moody diagram
easy to program for computer or calculator use
DLQgh
QL
ghf f
FHGIKJ
FHGIKJ
LNMM
OQPP0 66 1 25
24 75
9 4
5 2 0 04
. .
.
.
. .
f
D
FH IK
LNM
OQP
0 25
3 75 74
0 9
2
.
log.
.Re .
no fno f
Each equation has two terms. Why?Each equation has two terms. Why?
2 1.7840.965 ln
3.7f
f
gDhQ D
L D gDhD
L
e næ öç ÷
=- +ç ÷ç ÷ç ÷è ø
Pipe roughness
pipe materialpipe material pipe roughness pipe roughness ee (mm) (mm)
glass, drawn brass, copperglass, drawn brass, copper 0.00150.0015
commercial steel or wrought ironcommercial steel or wrought iron 0.0450.045
asphalted cast ironasphalted cast iron 0.120.12
galvanized irongalvanized iron 0.150.15
cast ironcast iron 0.260.26
concreteconcrete 0.18-0.60.18-0.6
rivet steelrivet steel 0.9-9.00.9-9.0
corrugated metalcorrugated metal 4545
PVCPVC 0.120.12
d
d Must be
dimensionless! Must be dimensionless!
find head loss given (D, type of pipe, Q)
find flow rate given (head, D, L, type of pipe)
find pipe size given (head, type of pipe,L, Q)
Solution TechniquesSolution Techniques
2 1.7840.965 ln
3.7f
f
gDhQ D
L D gDhD
L
e næ öç ÷
=- +ç ÷ç ÷ç ÷è ø
DLQgh
QL
ghf f
FHGIKJ
FHGIKJ
LNMM
OQPP0 66 1 25
24 75
9 4
5 2 0 04
. .
.
.
. .
h fg
LQDf
82
2
5f
D
FH IK
LNM
OQP
0 25
3 75 74
0 9
2
.
log.
.Re .
Re 4QD
Example: Find a pipe diameterExample: Find a pipe diameter
The marine pipeline for the Lake Source Cooling project will be 3.1 km in length, carry a maximum flow of 2 m3/s, and can withstand a maximum pressure differential between the inside and outside of the pipe of 28 kPa. The pipe roughness is 2 mm. What diameter pipe should be used?
The marine pipeline for the Lake Source Cooling project will be 3.1 km in length, carry a maximum flow of 2 m3/s, and can withstand a maximum pressure differential between the inside and outside of the pipe of 28 kPa. The pipe roughness is 2 mm. What diameter pipe should be used?
Minor LossesMinor Losses
We previously obtained losses through an expansion using conservation of energy, momentum, and mass
Most minor losses can not be obtained analytically, so they must be measured
Minor losses are often expressed as a loss coefficient, K, times the velocity head.
We previously obtained losses through an expansion using conservation of energy, momentum, and mass
Most minor losses can not be obtained analytically, so they must be measured
Minor losses are often expressed as a loss coefficient, K, times the velocity head.
g
VKh
2
2
g
VKh
2
2
( )geometry,RepC f= ( )geometry,RepC f=2
2C
Vp
p 2
2C
Vp
p
2
2C
V
ghlp
2
2C
V
ghlp
g
Vh pl
2C
2
g
Vh pl
2C
2
Venturi
High ReHigh Re
Head Loss due to Gradual Expansion (Diffuser)
1 2
g
VVKh EE
2
221
g
VVKh EE
2
221
diffuser angle ()
00.10.20.30.40.50.60.70.8
0 20 40 60 80
KE
2
1
22
2 12
AA
gV
Kh EE
2
1
22
2 12
AA
gV
Kh EE
( )2
1 2
2E
V Vh
g
-=( )2
1 2
2E
V Vh
g
-= Abrupt expansion
KE < 1
Sudden Contraction
V1 V2
EGL
HGL
vena contracta Losses are reduced with a gradual contraction Equation has same form as expansion equation!
g
V
Ch
c
c
21
1 22
2
g
V
Ch
c
c
21
1 22
2
2A
AC c
c 2A
AC c
c
0.60.650.7
0.750.8
0.850.9
0.951
0 0.2 0.4 0.6 0.8 1
A2/A1
Cc
0.60.650.7
0.750.8
0.850.9
0.951
0 0.2 0.4 0.6 0.8 1
A2/A1
Cc
0.60.650.7
0.750.8
0.850.9
0.951
0 0.2 0.4 0.6 0.8 1
A2/A1
Cc
c 2
g
VKh ee
2
2
g
VKh ee
2
2
0.1eK 0.1eK
5.0eK 5.0eK
04.0eK 04.0eK
Entrance LossesEntrance Losses
Losses can be reduced by accelerating the flow gradually and eliminating the
Losses can be reduced by accelerating the flow gradually and eliminating thevena contracta
Estimate based on contraction equations!
Head Loss in Bends
Head loss is a function of the ratio of the bend radius to the pipe diameter (R/D)
Velocity distribution returns to normal several pipe diameters downstream
High pressure
Low pressure
Possible separation from wall
D
R
g
VKh bb
2
2
g
VKh bb
2
2
Kb varies from 0.6 - 0.9
2Vp dn z Cr g+ + =ó
ôõ R
n
Head Loss in Valves
Function of valve type and valve position
The complex flow path through valves can result in high head loss (of course, one of the purposes of a valve is to create head loss when it is not fully open)
see table 8.2 (page 505 in text)
g
VKh vv
2
2
g
VKh vv
2
2
What is the maximum value of Kv?
Solution TechniquesSolution Techniques
Neglect minor losses Equivalent pipe lengths Iterative Techniques
Using Swamee-Jain equations for D and QUsing Swamee-Jain equations for head loss
Pipe Network Software
Neglect minor losses Equivalent pipe lengths Iterative Techniques
Using Swamee-Jain equations for D and QUsing Swamee-Jain equations for head loss
Pipe Network Software
Iterative Techniques for D and Q (given total head loss)
Iterative Techniques for D and Q (given total head loss)
Assume all head loss is major head loss. Calculate D or Q using Swamee-Jain
equations Calculate minor losses Find new major losses by subtracting minor
losses from total head loss
Assume all head loss is major head loss. Calculate D or Q using Swamee-Jain
equations Calculate minor losses Find new major losses by subtracting minor
losses from total head loss
DLQgh
QL
ghf f
FHGIKJ
FHGIKJ
LNMM
OQPP0 66 1 25
24 75
9 4
5 2 0 04
. .
.
.
. .
42
28
Dg
QKhminor
42
28
Dg
QKhminor
f l minorh h h= - åf l minorh h h= - å
2 1.7840.965 ln
3.7f
f
gDhQ D
L D gDhD
L
e næ öç ÷
=- +ç ÷ç ÷ç ÷è ø
Solution Technique: Head LossSolution Technique: Head Loss
Can be solved explicitly Can be solved explicitly
minorfl hhh minorfl hhh
2
2minor
Vh K
g=å
2
2minor
Vh K
g=å
5
2
2
8
D
LQ
gfh f
5
2
2
8
D
LQ
gfh f
2
9.0Re
74.5
7.3log
25.0
D
f
2
9.0Re
74.5
7.3log
25.0
D
f
2
2 4
8minor
Q Kh
g Dp= å
2
2 4
8minor
Q Kh
g Dp= å
D
Q4Re
D
Q4Re
Solution Technique:Discharge or Pipe Diameter
Solution Technique:Discharge or Pipe Diameter
Iterative technique Solve these equations
Iterative technique Solve these equations
minorfl hhh minorfl hhh
42
28
Dg
QKhminor
42
28
Dg
QKhminor
5
2
2
8
D
LQ
gfh f
5
2
2
8
D
LQ
gfh f
2
9.0Re
74.5
7.3log
25.0
D
f
2
9.0Re
74.5
7.3log
25.0
D
f
D
Q4Re
D
Q4Re
Use goal seek or Solver to find discharge that makes the calculated head loss equal the given head loss.
Spreadsheet
Example: Minor and Major Losses
Example: Minor and Major Losses
Find the maximum dependable flow between the reservoirs for a water temperature range of 4ºC to 20ºC.
Find the maximum dependable flow between the reservoirs for a water temperature range of 4ºC to 20ºC.
Water
2500 m of 8” PVC pipe
1500 m of 6” PVC pipeGate valve wide open
Standard elbows
Reentrant pipes at reservoirs
25 m elevation difference in reservoir water levels
Sudden contraction
Directions
Example (Continued)Example (Continued)
What are the Reynolds numbers in the two pipes?
Where are we on the Moody Diagram? What is the effect of temperature? Why is the effect of temperature so small? What value of K would the valve have to
produce to reduce the discharge by 50%?
0.01
0.1
1E+03 1E+04 1E+05 1E+06 1E+07 1E+08Re
fric
tion
fact
or
laminar
0.050.04
0.03
0.020.015
0.010.0080.006
0.004
0.002
0.0010.0008
0.0004
0.0002
0.0001
0.00005
smooth
Example (Continued)Example (Continued)
Were the minor losses negligible? Accuracy of head loss calculations? What happens if the roughness increases by
a factor of 10? If you needed to increase the flow by 30%
what could you do?
0.01
0.1
1E+03 1E+04 1E+05 1E+06 1E+07 1E+08Re
fric
tion
fact
or
laminar
0.050.04
0.03
0.020.015
0.010.0080.006
0.004
0.002
0.0010.0008
0.0004
0.0002
0.0001
0.00005
smooth
Pipe Flow Summary (1)Pipe Flow Summary (1)
linearly
experimental
Shear increases _________ with distance from the center of the pipe (for both laminar and turbulent flow)
Laminar flow losses and velocity distributions can be derived based on momentum (Navier Stokes) and energy conservation
Turbulent flow losses and velocity distributions require ___________ results
Shear increases _________ with distance from the center of the pipe (for both laminar and turbulent flow)
Laminar flow losses and velocity distributions can be derived based on momentum (Navier Stokes) and energy conservation
Turbulent flow losses and velocity distributions require ___________ results
Pipe Flow Summary (2)Pipe Flow Summary (2)
Energy equation left us with the elusive head loss term
Dimensional analysis gave us the form of the head loss term (pressure coefficient)
Experiments gave us the relationship between the pressure coefficient and the geometric parameters and the Reynolds number (results summarized on Moody diagram)
Energy equation left us with the elusive head loss term
Dimensional analysis gave us the form of the head loss term (pressure coefficient)
Experiments gave us the relationship between the pressure coefficient and the geometric parameters and the Reynolds number (results summarized on Moody diagram)
Pipe Flow Summary (3)Pipe Flow Summary (3)
Dimensionally correct equations fit to the empirical results can be incorporated into computer or calculator solution techniques
Minor losses are obtained from the pressure coefficient based on the fact that the pressure coefficient is _______ at high Reynolds numbers
Solutions for discharge or pipe diameter often require iterative or computer solutions
Dimensionally correct equations fit to the empirical results can be incorporated into computer or calculator solution techniques
Minor losses are obtained from the pressure coefficient based on the fact that the pressure coefficient is _______ at high Reynolds numbers
Solutions for discharge or pipe diameter often require iterative or computer solutions
constant
Pipes are Everywhere!Pipes are Everywhere!
Owner: City of Hammond, INProject: Water Main RelocationPipe Size: 54"
Pipes are Everywhere!Drainage Pipes
Pipes are Everywhere!Drainage Pipes
PipesPipes
Pipes are Everywhere!Water Mains
Pipes are Everywhere!Water Mains
Pressure Coefficient for a Venturi Meter
Pressure Coefficient for a Venturi Meter
1
10
1E+00 1E+01 1E+02 1E+03 1E+04 1E+05 1E+06
Re
Cp
ReVlrm
=ReVlrm
=
2
2C
Vp
p 2
2C
Vp
p
0.01
0.1
1E+03 1E+04 1E+05 1E+06 1E+07 1E+08Re
fric
tion
fact
or
laminar
0.050.04
0.03
0.020.015
0.010.0080.006
0.004
0.002
0.0010.0008
0.0004
0.0002
0.0001
0.00005
smooth
Moody DiagramMoody Diagram
0.01
0.1
1E+03 1E+04 1E+05 1E+06 1E+07 1E+08Re
fric
tion
fact
or
laminar
0.050.04
0.03
0.020.015
0.010.0080.006
0.004
0.002
0.0010.0008
0.0004
0.0002
0.0001
0.00005
smooth
lD
C pf
lD
C pf
D
D
Minor Losses
LSC PipelineLSC Pipeline
cs1cs1
cs2cs2
pz
Vg
pz
Vg
hl1
1 112
22 2
22
2 2
z = 0z = 0
Ignore entrance losses
00
28 kPa is equivalent to 2.85 m of water-2.85 m-2.85 m
D m 154.
DLQgh
QL
ghf f
FHGIKJ
FHGIKJ
LNMM
OQPP0 66 1 25
24 75
9 4
5 2 0 04
. .
.
.
. .
Q m s
m s
L m
m
h mf
2
10
3100
0 002
2 85
3
6 2
/
/
.
.
DirectionsDirections
Assume fully turbulent (rough pipe law) find f from Moody (or from von Karman)
Find total head loss (draw control volume) Solve for Q using symbols (must include
minor losses) (no iteration required)
Assume fully turbulent (rough pipe law) find f from Moody (or from von Karman)
Find total head loss (draw control volume) Solve for Q using symbols (must include
minor losses) (no iteration required)
0.01
0.1
1E+03 1E+04 1E+05 1E+06 1E+07 1E+08Re
fric
tion
fact
or
laminar
0.050.04
0.03
0.020.015
0.010.0080.006
0.004
0.002
0.0010.0008
0.0004
0.0002
0.0001
0.00005
smooth
Pipe roughness