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