Viscous Flow in PipesViscous Flow in Pipes

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?

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?

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

Remember dimensional analysis?

Two important parameters! R - Laminar or Turbulent /D - Rough or Smooth

Flow geometry internal _______________________________ external _______________________________

Remember dimensional analysis?

Two important parameters! R - Laminar or Turbulent /D - Rough or Smooth

Flow geometry internal _______________________________ external _______________________________

R,

Df

lD

C p

R,

Df

lD

C p

2

2C

Vp

p 2

2C

Vp

p

VDR

VDR

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

Transition at R of 2000Transition at R of 2000

Laminar and Turbulent FlowsLaminar and Turbulent Flows

Reynolds apparatus Reynolds apparatus

VDR

VD

Rdampingdampinginertiainertia

Boundary layer growth: Transition length

Boundary layer growth: Transition length

Pipe Entrance

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 Flowv

Need equation for entrance length here

Laminar, Incompressible, Steady, Uniform Flow

Laminar, Incompressible, Steady, Uniform Flow

Between Parallel Plates Through circular tubes Hagen-Poiseuille Equation Approach

Because it is laminar flow the shear forces can be easily quantified

Velocity profiles can be determined from a force balance

Don’t need to use dimensional analysis

Between Parallel Plates Through circular tubes Hagen-Poiseuille Equation Approach

Because it is laminar flow the shear forces can be easily quantified

Velocity profiles can be determined from a force balance

Don’t need to use dimensional analysis

Laminar Flow through Circular Tubes

Laminar Flow through Circular Tubes

Different geometry, same equation development (see Streeter, et al. p 268)

Apply equation of motion to cylindrical sleeve (use cylindrical coordinates)

Different geometry, same equation development (see Streeter, et al. p 268)

Apply equation of motion to cylindrical sleeve (use cylindrical coordinates)

Laminar Flow through Circular Tubes: Equations

Laminar Flow through Circular Tubes: Equations

hpdldra

u

4

22

hpdldra

u

4

22

hpdlda

u

4

2

max hpdlda

u

4

2

max

hpdlda

V

8

2

hpdlda

V

8

2

hpdlda

Q

8

4

hpdlda

Q

8

4

Velocity distribution is paraboloid of revolution therefore _____________ _____________

Q = VA =

Max velocity when r = 0

average velocity (V) is 1/2 umax

Vpa2

a is radius of the tubea is radius of the tube

Laminar Flow through Circular Tubes: Diagram

Laminar Flow through Circular Tubes: Diagram

Velocity

Shear

hpdldra

u

4

22

hpdldra

u

4

22

hpdl

dr

dr

du

2

hpdl

dr

dr

du

2

hpdl

dr

dr

du 2

hpdl

dr

dr

du 2

l

hr l

2

l

hr l

2

l

dhl

40

l

dhl

40

True for Laminar or Turbulent flow

Shear at the wallShear at the wall

Laminar flow

The Hagen-Poiseuille EquationThe Hagen-Poiseuille Equation

hpdlda

Q

8

4

hpdlda

Q

8

4

hp

dldD

Q

128

4

hp

dldD

Q

128

4

lhzp

zp 2

2

21

1

1

lhzp

zp 2

2

21

1

1

2

2

21

1

1 zp

zp

hl

2

2

21

1

1 zp

zp

hl

hp

hl

hp

hl

L

hDQ l

128

4

L

hDQ l

128

4

L

hh

pdld l

L

hh

pdld l

L

hDV l

32

2

L

hDV l

32

2

cv pipe flow

Constant cross section

Laminar pipe flow equations

CV equations!CV equations!

h or zh or z

pz

Vg

Hp

zV

gH hp t l

1

11 1

12

2

22 2

22

2 2

Example: Laminar Flow (Team work)

Example: Laminar Flow (Team work)

Calculate the discharge of 20ºCwater through a long vertical section of 0.5 mm ID hypodermic tube. The inlet and outlet pressures are both atmospheric. You may neglect minor losses. What is the total shear force?

What assumption did you make? (Check your assumption!)

Calculate the discharge of 20ºCwater through a long vertical section of 0.5 mm ID hypodermic tube. The inlet and outlet pressures are both atmospheric. You may neglect minor losses. What is the total shear force?

What assumption did you make? (Check your assumption!)

Turbulent Pipe and Channel Flow: Overview

Turbulent Pipe and Channel Flow: Overview

Velocity distributions Energy Losses Steady Incompressible Flow through

Simple Pipes Steady Uniform Flow in Open Channels

Velocity distributions Energy Losses Steady Incompressible Flow through

Simple Pipes Steady Uniform Flow in Open Channels

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

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: _________________ lakes: ____________________

Actually a spectrum of eddy sizes

depth (R = 500)

diameter (R = 2000)

depth to thermocline

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

Turbulent Flow Velocity ProfileTurbulent Flow Velocity Profile

dy

du dy

du

dy

du dy

du

IIul IIul

dy

dulu II

dy

dulu II

dy

dulI

2 dy

dulI

2

Length scale and velocity of “large” eddies

y

Turbulent shear is from momentum transfer

h = eddy viscosity

Dimensional analysis

Turbulent Flow Velocity ProfileTurbulent Flow Velocity Profile

ylI ylI

dy

duy22

dy

duy22

2

22

dy

duy

2

22

dy

duy

dy

duy

dy

duy

dy

dulI

2 dy

dulI

2 dy

du dy

du

Size of the eddies __________ as we move further from the wall.

increases

k = 0.4 (from experiments)

Log Law for Turbulent, Established Flow, Velocity Profiles

Log Law for Turbulent, Established Flow, Velocity Profiles

5.5ln1 *

*

yu

u

u5.5ln

1 *

*

yu

u

u

0* u

0* u

dy

duy

dy

duy

Iuu * Iuu *

Shear velocity

Integration and empirical results

Laminar Turbulent

x

y

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

Pipe Flow Energy LossesPipe Flow Energy Losses

p

hl

p

hl

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

Horizontal pipe

Dimensional Analysis

Darcy-Weisbach equation

p Vg

z hp V

gz h hp t l

11

12

12

222

22 2

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

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

g

VDL

gDLV

2f

32 2

2

RVD6464

f

RVD6464

f

Slope of ___ on log-log plot

Hagen-Poiseuille

Darcy-Weisbach

-1

Turbulent Pipe Flow Head LossTurbulent Pipe Flow Head Loss

___________ 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

Proportional

Proportional

Inversely

Increase

independent

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

Q Dgh

L DD

gh

L

f

f

F

HGGG

I

KJJJ

2 223 7

1785 2

2 3

. log.

./

/

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?

Pipe roughness

pipe materialpipe material pipe roughness pipe roughness (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

Q Dgh

L DD

gh

L

f

f

F

HGGG

I

KJJJ

2 223 7

1785 2

2 3

. log.

./

/

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

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

R,geometryfC p R,geometryfC p 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

High RHigh R

Head Loss due to Gradual Expansion (Diffusor)

g

VVKh EE

2

221

g

VVKh EE

2

221

diffusor 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

Sudden Contraction

losses are reduced with a gradual contraction

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

V1V2

flow separation

Sudden ContractionSudden Contraction

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

hC

Vgc

c

FHG

IKJ

11

2

2

22

Q CA ghorifice orifice 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

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

g

VKh bb

2

2

g

VKh bb

2

2

Kb varies from 0.6 - 0.9

R

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)

g

VKh vv

2

2

g

VKh vv

2

2

Solution TechniquesSolution Techniques

Neglect minor losses Equivalent pipe lengths Iterative Techniques Simultaneous Equations Pipe Network Software

Neglect minor losses Equivalent pipe lengths Iterative Techniques Simultaneous Equations 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

Solution Technique: Head LossSolution Technique: Head Loss

Can be solved directly Can be solved directly

minorfl hhh minorfl hhh

g

VKhminor

2

2

g

VKhminor

2

2

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

42

28

Dg

QKhminor

42

28

Dg

QKhminor

D

Q4Re

D

Q4Re

Solution Technique:Discharge or Pipe Diameter

Solution Technique:Discharge or Pipe Diameter

Iterative technique Set up simultaneous equations in Excel

Iterative technique Set up simultaneous equations in Excel

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.

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

DirectionsDirections

Assume fully turbulent (rough pipe law) find f from Moody (or from von Karman)

Find total head loss Solve for Q using symbols (must include

minor losses) (no iteration required) Obtain values for minor losses from notes

or text

Assume fully turbulent (rough pipe law) find f from Moody (or from von Karman)

Find total head loss Solve for Q using symbols (must include

minor losses) (no iteration required) Obtain values for minor losses from notes

or text

Example (Continued)Example (Continued)

What are the Reynolds number in the two pipes?

Where are we on the Moody Diagram? What value of K would the valve have to

produce to reduce the discharge by 50%? What is the effect of temperature? Why is the effect of temperature so small?

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? Suppose I changed 6” pipe, what is

minimum diameter needed?

Pipe Flow Summary (1)Pipe Flow Summary (1)

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 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 and energy conservation

Turbulent flow losses and velocity distributions require ___________ results

linearly

experimental

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

Columbia Basin Irrigation Project

Columbia Basin Irrigation Project

The Feeder Canal is a concrete lined canal which runs from the outlet of the pumping plant discharge tubes to the north end of Banks Lake (see below). The original canal was completed in 1951 but has since been widened to accommodate the extra water available from the six new pump/generators added to the pumping plant. The canal is 1.8 miles in length, 25 feet deep and 80 feet wide at the base. It has the capacity to carry 16,000 cubic feet of water per second.

The Feeder Canal is a concrete lined canal which runs from the outlet of the pumping plant discharge tubes to the north end of Banks Lake (see below). The original canal was completed in 1951 but has since been widened to accommodate the extra water available from the six new pump/generators added to the pumping plant. The canal is 1.8 miles in length, 25 feet deep and 80 feet wide at the base. It has the capacity to carry 16,000 cubic feet of water per second.

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

GlycerinGlycerin

QUa a d

dlp h

2 12

3

a f

02 12

3

Ua a

ddl

p h a f

Ua

2

6

Um N m

Ns mm s

0 005 12300

6 0 620 083

2 3

2

. /

. /. /

a fc hc h

RVl kg m m s m

Ns m

1254 0 083 0 005

0 620 8

3

2

/ . / .

. /.

c ha fa f

Example: Hypodermic Tubing Flow

Example: Hypodermic Tubing Flow

QD h

Ll

4

128

QN m m

x Ns m

9806 0 0005

128 1 10

3 4

3 2

/ .

/

c hafa fc h

Q x m s 158 10 8 3. /

VQd

4

2V m s 0 0764. /

RVd

R

m s m kg m

x Ns m

0 0764 0 0005 1000

1 10

3

3 2

. / . /

/

a fa fc hc h

R 38

Q L s 158. /ldhl

40

ldhl

40

ldhrl

F lshear 4

2 l

dhrlF l

shear 42

lrFshear 2 lrFshear 2